Design of Machinery Solutions Manual Norton 5th Edition (1).pdf

DESIGN OF MACHINERY - 5th Ed

SOLUTION MANUAL 2-1-1

PROBLEM 2-1 Statement:

Find three (or other number as assigned) of the following common devices. Sketch careful kinematic diagrams and find their total degrees of freedom. a. An automobile hood hinge mechanism b. An automobile hatchback lift mechanism c. An electric can opener d. A folding ironing board e. A folding card table f. A folding beach chair g. A baby swing h. A folding baby walker i. A fancy corkscrew as shown in Figure P2-9 j. A windshield wiper mechanism k. A dump-truck dump mechanism l. A trash truck dumpster mechanism m. A pickup tailgate mechanism n. An automobile jack o. A collapsible auto radio antenna

Solution:

See Mathcad file P0201.

Equation 2.1c is used to calculate the mobility (DOF) of each of the models below. a.

An automobile hood hinge mechanism. The hood (3) is linked to the body (1) through two rocker links (2 and 4). Number of links

L  4

Number of full joints

J1  4

Number of half joints

J2  0

M  3  ( L  1 )  2  J1  J2 M1 b.

HOOD 3 2

4

1 BODY

An automobile hatchback lift mechanism. The hatch (2) is pivoted on the body (1) and is linked to the body by the lift arm, which can be modeled as two links (3 and 4) connected through a translating slider joint. HATCH Number of links L  4 Number of full joints

J1  4


J2  0

2 3 1

M  3  ( L  1 )  2  J1  J2

4

M1

1 BODY

c.

An electric can opener has 2 DOF.

d.

A folding ironing board. The board (1) itself has one pivot (full) joint and one pin-in-slot sliding (half) joint. The two legs (2 and 3) hav a common pivot. One leg connects to the pivot joint on the board and the other to the slider joint.

DESIGN OF MACHINERY - 5th Ed


Number of links

L  3


J1  2


J2  1

1 3

2

M  3  ( L  1 )  2  J1  J2 M1 e.

A folding card table has 7 DOF: One for each leg, 2 for location in xy space, and one for angular orientation.

f.

A folding beach chair. The seat (3) and the arms (6) are ternary links. The seat is linked to the front leg(2), the back (5) and a coupling link (4). The arms are linked to the front leg (2), the rear leg (1), and the back (5). Links 1, 2, 4, and 5 are binar links. The analysis below is appropriate when the chair is not fully opened. When fully opened, one or more links are prevented from moving by a stop. Subtract 1 DOF when forced against the stop. Number of links

L  6


J1  7


J2  0

5 6 4

1

M  3  ( L  1 )  2  J1  J2

2 3

M1 g.

A baby swing has 4 DOF: One for the angular orientation of the swing with respect to the frame, and 3 for the location and orientation of the frame with respect to a 2-D frame.

h.

A folding baby walker has 4 DOF: One for the degree to which it is unfolded, and 3 for the location and orientation of the walker with respect to a 2-D frame.

i.

A fancy corkscrew has 2 DOF: The screw can be rotated and the arms rotate to translate the screw.

j.

A windshield wiper mechanism has 1 DOF: The position of the wiper blades is defined by a single input.

k.

A dump-truck dump mechanism has 1 DOF: The angle of the dump body is determined by the length of the hydraulic cylinder that links it to the body of the truck.

l.

A trash truck dumpster mechanism has 2 DOF: These are generally a rotation and a translation.

m. A pickup tailgate mechanism has 1 DOF: n.

An automobile jack has 4 DOF: One is the height of the jack and the other 3 are the position and orientation o the jack with respect to a 2-D frame.

o.

A collapsible auto radio antenna has as many DOF as there are sections, less one.

DESIGN OF MACHINERY - 5th Ed.



How many DOF do you have in your wrist and hand combined?

Solution:


1.

Holding the palm of the hand level and facing toward the floor, the hand can be rotated about an axis through the wrist that is parallel to the floor (and perpendicular to the forearm axis) and one perpendicular to the floor (2 DOF). The wrist can rotate about the forearm axis (1 DOF).

2.

Each finger (and thumb) can rotate up and down and side-to-side about the first joint. Additionally, each finger can rotate about each of the two remaining joints for a total of 4 DOF for each finger (and thumb).

3.

Adding all DOF, the total is Wrist Hand Thumb Fingers 4x4

1 2 4 16

TOTAL

23



How many DOF do the following joints have? a. Your knee b. Your ankle c. Your shoulder d. Your hip e. Your knuckle

Solution:


a.

Your knee. 1 DOF: A rotation about an axis parallel to the ground.

b.

Your ankle. 3 DOF: Three rotations about mutually perpendicular axes.

c.

Your shoulder. 3 DOF: Three rotations about mutually perpendicular axes.

d.

Your hip. 3 DOF: Three rotations about mutually perpendicular axes.

e

Your knuckle. 2 DOF: Two rotations about mutually perpendicular axes.





How many DOF do the following have in their normal environment? a. A submerged submarine b. An earth-orbit satellite c. A surface ship d. A motorcycle (road bike) e. A two-button mouse f. A computer joy stick.

Solution:


a.

A submerged submarine. Using a coordinate frame attached to earth, or an inertial coordinate frame, a submarine has 6 DOF: 3 linear coordinates and 3 angles.

b.

An earth-orbit satellite. If the satellite was just a particle it would have 3 DOF. But, since it probably needs to be oriented with respect to the earth, sun, etc., it has 6 DOF.

c.

A surface ship. There is no difference between a submerged submarine and a surface ship, both have 6 DOF. One might argue that, for an earth-centered frame, the depth of the ship with respect to mean sea level is constant, however that is not strictly true. A ship's position is generally given by two coordinates (longitude and latitude). For a given position, a ship can also have pitch, yaw, and roll angles. Thus, for all practical purposes, a surface ship has 5 DOF.

d.

A motorcycle. At an intersection, the motorcycle's position is given by two coordinates. In addition, it will have some heading angle (turning a corner) and roll angle (if turning). Thus, there are 4 DOF.

e.

A two-button mouse. A two-button mouse has 4 DOF. It can move in the x and y directions and each button has 1 DOF.

f.

A computer joy stick. The joy stick has 2 DOF (x and y) and orientation, for a total of 3 DOF.




Are the joints in Problem 2-3 force closed or form closed?

Solution:


They are force closed by ligaments that hold them together. None are geometrically closed.




Describe the motion of the following items as pure rotation, pure translation, or complex planar motion. a. A windmill b. A bicycle (in the vertical plane, not turning) c. A conventional "double-hung" window d. The keys on a computer keyboard e. The hand of a clock f. A hockey puck on the ice g. A "casement" window

Solution:


a.

A windmill. Pure rotation.

b.

A bicycle (in the vertical plane, not turning). Pure translation for the frame, complex planar motion for the wheels.

c.

A conventional "double-hung" window. Pure translation.

d.

The keys on a computer keyboard. Pure translation.

e.

The hand of a clock. Pure rotation.

f.

A hockey puck on the ice. Complex planar motion.

g.

A "casement" window. Pure rotation.




Calculate the mobility of the linkages assigned from Figure P2-1 part 1 and part 2.

Solution:

See Figure P2-1 and Mathcad file P0207.

1.

Use equation 2.1c (Kutzbach's modification) to calculate the mobility.

a.

Number of links

L  6


J1  7


J2  1

6 3 5 2

M  3  ( L  1 )  2  J1  J2

4 1

M0

(a)

1 3

b.

Number of links

L  3


J1  2


J2  1

1

M  3  ( L  1 )  2  J1  J2 M1

2 1

(b) 4

c.

Number of links

L  4


J1  4


J2  0

1

3

M  3  ( L  1 )  2  J1  J2 2

M1 (c)

1

7

d.

Number of links

L  7


J1  7


J2  1

1

6 5

M  3  ( L  1 )  2  J1  J2 M3

1

2 3

4

(d)



8

5

8 1

5 1

9 10

6

1

1 7

4

4

1

2

2 3

3

1

5 6

2 1

(e)

1

e.

g.

Number of links

L  10

Number of full joints Number of half joints

(f)

Number of links

L  6

J1  13


J1  6

J2  0


J2  2

f.

M  3  ( L  1 )  2  J1  J2

M  3  ( L  1 )  2  J1  J2

M1

M1

Number of links

L  8


J1  9


J2  2

M  3  ( L  1 )  2  J1  J2

4 1

4

7

6

3

7 1

5

8 1

2

2 1

1

1

M1 (g)

2 h.

Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2 M1

1 3 1 4 (h)




Identify the items in Figure P2-1 as mechanisms, structures, or preloaded structures.

Solution:


1.

Use equation 2.1c (Kutzbach's modification) to calculate the mobility and the definitions in Section 2.5 of the text to classify the linkages.

a.

Number of links

L  6


J1  7


J2  1

6 3 5 2

M  3  ( L  1 )  2  J1  J2 M0

4 1

Structure

(a)

1 3

b.

Number of links

L  3


J1  2


J2  1

1

M  3  ( L  1 )  2  J1  J2 M1

Mechanism

2 1

(b) 4

c.

Number of links

L  4


J1  4


J2  0

1

3

M  3  ( L  1 )  2  J1  J2 M1

2

Mechanism (c)

1

7

d.

Number of links

L  7


J1  7


J2  1

1

6 5

M  3  ( L  1 )  2  J1  J2 M3

Mechanism

1

2 3

4

(d)




Use linkage transformation on the linkage of Figure P2-1a to make it a 1-DOF mechanism.

Solution:

See Figure P2-1a and Mathcad file P0209.

1.

The mechanism in Figure P2-1a has mobility: Number of links

L  6


J1  7


J2  1

6 3 5 2

M  3  ( L  1 )  2  J1  J2 M0

4 1 1

2.

Use rule 2, which states: "Any full joint can be replaced by a half joint, but this will increase the DOF by one." One way to do this is to replace one of the pin joints with a pin-in-slot joint such as that shown in Figure 2-3c. Choosing the joint between links 2 and 4, we now have mobility: Number of links

L  6


J1  6


J2  2

6 3 5

M  3  ( L  1 )  2  J1  J2

2 4

M1

1 1




Use linkage transformation on the linkage of Figure P2-1d to make it a 2-DOF mechanism.

Solution:

See Figure P2-1d and Mathcad file P0210.

1.

7

The mechanism in Figure P2-1d has mobility: Number of links

L  7


J1  7


J2  1

M  3  ( L  1 )  2  J1  J2

1

6 5 1

2 3

4

M3 2.

Use rule 3, which states: "Removal of a link will reduce the DOF by one." One way to do this is to remove link 7 such that link 6 pivots on the fixed pin attached to the ground link (1). We now have mobility: Number of links

L  6


J1  6


J2  1

M  3  ( L  1 )  2  J1  J2

1 6 5 1

2 3

M2

4




Use number synthesis to find all the possible link combinations for 2-DOF, up to 9 links, to hexagonal order, using only revolute joints.

Solution:


1.

Use equations 2.4a and 2.6 with DOF = 2 and iterate the solution for valid combinations. Note that the number of links must be odd to have an even DOF (see Eq. 2.4). The smallest possible 2-DOF mechanism is then 5 links since three will give a structure (the delta triplet, see Figure 2-7). L  B  T  Q  P  H

L  3  M  T  2  Q  3  P  4  H L  5  T  2  Q  3  P  4  H

2.

3.

For L  5 0  T  2  Q  3  P  4  H

0=T =Q=P=H

2  T  2  Q  3  P  4  H

H  0

For L  7

Case 1:

Case 2:

4.

B  5

Q  0

Q  1

P  0

T  2  2  Q  3  P  4  H

T2

B  L  T  Q  P  H

B5

T  2  2  Q  3  P  4  H

T0

B  L  T  Q  P  H

B6

T  0

P  0

For L  9 4  T  2  Q  3  P  4  H Case 1:

H  1

Q  0

B  L  T  Q  P  H Case 2a:

H  0

B8

4  T  2  Q  3  P 9  B  T  Q  P

Case 2b:

P  1

1  T  2  Q

Q  0

B  L  T  Q  P  H Case 2c:

P  0

T  1 B7

4  T  2  Q 9  B  T  Q

Case 2c1:

Case 2c2:

Case 2c3:

Q  2 Q  1 Q  0

T  4  2  Q

T0

B  9  T  Q

B7

T  4  2  Q

T2

B  9  T  Q

B6

T  4  2  Q

T4

B  9  T  Q

B5

M  2




Find all of the valid isomers of the eightbar 1-DOF link combinations in Table 2-2 (p. 38) having a. Four binary and four ternary links. b. Five binaries, two ternaries, and one quaternary link. c. Six binaries and two quaternary links. d. Six binaries, one ternary, and one pentagonal link.

Solution:


1.

2.

a.

Table 2-3 lists 16 possible isomers for an eightbar chain. However, Table 2-2 shows that there are five possible link sets, four of which are listed above. Therefore, we expect that the 16 valid isomers are distributed among the five link sets and that there will be fewer than 16 isomers among the four link sets listed above. One method that is helpful in finding isomers is to represent the linkage in terms of molecules as defined in Franke's Condensed Notations for Structural Synthesis. A summary of the rules for obtaining Franke's molecules follows: (1) The links of order greater than 2 are represented by circles. (2) A number is placed within each circle (the "valence" number) to describe the type (ternary, quaternary, etc.) of link. (3) The circles are connected using straight lines. The number of straight lines emanating from a circle must be equal to its valence number. (4) Numbers (0, 1, 2, etc.) are placed on the straight lines to correspond to the number of binary links used in connecting the higher order links. (5) There is one-to-one correspondence between the molecule and the kinematic chain that it represents. Four binary and four ternary links. Draw 4 circles with valence numbers of 3 in each. Then find all unique combinations of straight lines that can be drawn that connect the circles such that there are exactly three lines emanating from each circle and the total of the numbers written on the lines is exactly equal to 4. In this case, there are three valid isomers as depicted by Franke's molecules and kinematic chains below.

8

1 3

3

5

1 0

0

1 3

6

3

3

1

4

2 1

7



8 0 3

3 2

0

5

0

1 3

7

6

3 1

3

4

1 2

8 5

0 3

3

4

2 0

6

0

2 3

3

3 0

7

1 2

The mechanism shown in Figure P2-5b is the same eightbar isomer as that depicted schematically above. b.

Five binaries, two ternaries, and one quaternary link. Draw 2 circles with valence numbers of 3 in each and one with a valence number of 4. Then find all unique combinations of straight lines that can be drawn that connect the circles such that there are exactly three lines emanating from each circle with valence of three and four lines from the circle with valence of four; and the total of the numbers written on the lines is exactly equal to 5. In this case, there are five valid isomers as depicted by Franke's molecules and kinematic chains below.

0

3 2

4

7

0

1

5

3

3

2

6 4

8

1

2



5 1

3 0

3 0

2

6 4

7

3

2 4

8

1

2

5 0

3 1

3

3

7

1

2

4

6

1

2

8

4

1

5 1

3 1

6

3

4

0

1

3 2

7

8 4

1

2

5 1

3 1

6

3

3

1

1

1

8

4

7

2

4

1

c.

Six binaries and two quaternary links. Draw 2 circles with valence numbers of 4 in each. Then find all unique combinations of straight lines that can be drawn that connect the circles such that there are exactly four lines emanating from each circle and the total of the numbers written on the lines is exactly equal to 6. In this case, there are two valid isomers as depicted by Franke's molecules and kinematic chains below.



0 2

7 4

4

4

5

3

6

2

8

2

1

2

1

4

1

7 4

4

d.

3

6

2 2

5

8

2 1

Six binaries, one ternary, and one pentagonal link. There are no valid implementations of 6 binary links with 1 pentagonal link.




Use linkage transformation to create a 1-DOF mechanism with two sliding full joints from a Stephenson's sixbar linkage as shown in Figure 2-14a (p. 47).

Solution:

See Figure 2-14a and Mathcad file P0213.

1.

The mechanism in Figure 2-14a has mobility: Number of links

L  6


J1  7


J2  0

A 4

3

5

B

2

6

M  3  ( L  1 )  2  J1  J2 M1

2.

1

Use rule 1, which states: "Revolute joints in any loop can be replaced by prismatic joints with no change in DOF of the mechanism, provided that at least two revolute joints remain in the loop." One way to do this is to replace pin joints at A and B with translating full slider joints such as that shown in Figure 2-3b. Note that the sliders are attached to links 3 and 5 in such a way that they can not rotate relative to the links. The number of links and 1-DOF joints remains the same. There are no 2-DOF joints in either mechanism.

A 4

3

5 2

1

6 B




Use linkage transformation to create a 1-DOF mechanism with one sliding full joint a a half joint from a Stephenson's sixbar linkage as shown in Figure 2-14b (p. 48).

Solution:


1.

The mechanism in Figure 2-14b has mobility: Number of links

L  6


J1  7


J2  0

3 5 4 2

6

M  3  ( L  1 )  2  J1  J2 1

M1

2.

To get the sliding full joint, use rule 1, which states: "Revolute joints in any loop can be replaced by prismati joints with no change in DOF of the mechanism, provided that at least two revolute joints remain in the loop." One way to do this is to replace pin joint links 3 and 5 with a translating full slider joint such as that shown in Figure 2-3b. Note that the slider is attached to link 3 in such a way that it can not rotate relative to the link. The number of links and 1-DOF joints remains the same.

3

5 4

2

6

1

3.

To get the half joint, use rule 4 on page 42, which states: "The combination of rules 2 and 3 above will keep the original DOF unchanged." One way to do this is to remove link 6 (and its two nodes) and insert a half joint between links 5 and 1. Number of links

L  5


J1  5


3 5 4

J2  1 2

M  3  ( L  1 )  2  J1  J2

1

M1 1




Calculate the Grashof condition of the fourbar mechanisms defined below. Build cardboard models of the linkages and describe the motions of each inversion. Link lengths are in inches (or double given numbers for centimeters). Part 1. a. b. c.

2 2 2

4.5 3.5 4.0

7 7 6

9 9 8

Part 2. d. e. f.

2 2 2

4.5 4.0 3.5

7 7 7

9 9 9

Solution: 1.

See Mathcad file P0215

Use inequality 2.8 to determine the Grashof condition. Condition( a b c d ) 

S  min ( a b c d ) L  max( a b c d ) SL  S  L PQ  a  b  c  d  SL return "Grashof" if SL  PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise

a.

Condition( 2 4.5 7 9 )  "Grashof"

b.

Condition( 2 3.5 7 9 )  "non-Grashof"

c.

Condition( 2 4.0 6 8 )  "Special Grashof" This is a special case Grashof since the sum of the shortest and longest is equal to the sum of the other two link lengths.

d.


e.


f.

Condition( 2 3.5 7 9 )  "non-Grashof"




Which type(s) of electric motor would you specify a. b. c.

Solution:

To drive a load with large inertia. To minimize variation of speed with load variation. To maintain accurate constant speed regardless of load variations.


a.

Motors with high starting torque are suited to drive large inertia loads. Those with this characteristic include series-wound, compound-wound, and shunt-wound DC motors, and capacitor-start AC motors.

b.

Motors with flat torque-speed curves (in the operating range) will minimize variation of speed with load variation. Those with this characteristic include shunt-wound DC motors, and synchronous and capacitor-start AC motors.

b.

Speed-controlled DC motors will maintain accurate constant speed regardless of load variations.




Describe the difference between a cam-follower (half) joint and a pin joint.

Solution:


1.

A pin joint has one rotational DOF. A cam-follower joint has 2 DOF, rotation and translation. The pin joint also captures its lubricant in the annulus between pin and bushing while the cam-follower joint squeezes its lubricant out of the joint.




Examine an automobile hood hinge mechanism of the type described in Section 2.14. Sketch it carefully. Calculate its DOF and Grashof condition. Make a cardboard model. Analyze it with a free-body diagram. Describe how it keeps the hood up.

Solution:

Solution of this problem will depend upon the specific mechanism modeled by the student.




Find an adjustable arm desk lamp of the type shown in Figure P2-2. Sketch it carefully. Measure it and sketch it to scale. Calculate its DOF and Grashof condition. Make a cardboard model. Analyze it with a free-body diagram. Describe how it keeps itself stable. Are there any positions in which it loses stability? Why?

Solution:

Solution of this problem will depend upon the specific mechanism modeled by the student.




The torque-speed curve for a 1/8 hp permanent magnet (PM) DC motor is shown in Figure P2-3. The rated speed for this fractional horsepower motor is 2500 rpm at a rated voltage of 130V. Determine: a) The rated torque in oz-in (ounce-inches, the industry standard for fractional hp motors) b) The no-load speed c) Plot the power-torque curve and determine the maximum power that the motor can deliver.

Given:

Rated speed, N

NR  2500 rpm

R

HR 

Rated power, H

R

1 8

 hp

4000 3500

Speed, rpm

3000 2500 2000 1500 1000 500 0

0

50

100

150

200

250

300

Torque, oz-in

Figure P2-3 Torque-speed Characteristic of a 1/8 HP, 2500 rpm PM DC Motor

Solution: a.


The rated torque is found by dividing the rated power by the rated speed: TR 

Rated torque, TR

HR NR

TR  50 ozf  in

b.

The no-load speed occurs at T = 0. From the graph this is 3000 rpm.

c.

The power is the product of the speed and the torque. From the graph the equation for the torque-speed curve is: 3000 rpm N ( T )    T  3000 rpm 300  ozf  in and the power, therefore, is: H ( T )  10

rpm ozf  in

2

 T  3000 rpm T

Plotting the power as a function of torque over the range T  0  ozf  in 10 ozf  in  300  ozf  in



0.25 0.225 0.2

Power, hp

0.175 0.15 0.125 0.1 0.075 0.05 0.025 0

0

50

100

150

200

250

300

Torque, oz-in

Maximum power occurs when dH/dT = 0. The value of T at maximum power is: ozf  in

Value of T at Hmax

THmax  3000 rpm

Maximum power

Hmax  H  THmax

Hmax  0.223  hp

Speed at max power

NHmax  N  THmax

NHmax  1500 rpm

2  10 rpm

THmax  150  ozf  in

Note that the curve goes through the rated power point of 0.125 hp at the rated torque of 50 oz-in.




Find the mobility of the mechanisms in Figure P2-4.

Solution:


1.


a.

This is a basic fourbar linkage. The input is link 2 and the output is link 4. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1).

Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2

A 2 3 O2

M1

b.

4

C

O4

This is a fourbar linkage. The input is link 2, which in this case is the wheel 2 with a pin at A, and the output is link 4. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1).

Number of links

L  4


J1  4


J2  0

A

2

O2

3

M  3  ( L  1 )  2  J1  J2

4 B

M1

c.

O4

This is a 3-cylinder, rotary, internal combustion engine. The pistons (sliders) 6, 7, and 8 drive the output crank (2) through piston rods (couplers 3, 4, and 5). There are 3 full joints at the crank where rods 3, 4and 5 are pinned to crank 2. The cross-hatched crank-shaft at O2 is supported by the ground link (1) through bearings. Number of links

L  8


J1  10


J2  0

M  3  ( L  1 )  2  J1  J2

6

3

2

4

M1 7

5

8


d.


This is a fourbar linkage. The input is link 2, which in this case is a wheel with a pin at A, and the output is the vertical member on the coupler, link 3. Since the lengths of links 2 and 4 (O2A and O4B) are the same, the coupler link (3) has curvilinear motion and AB remains parallel to O2O4 throughout the cycle. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1).

Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2

B O4

O2

M1

e.

3

A 2

4

This is a fourbar linkage with an output dyad. The input (rocker) is link 2 and the output (rocker) is link 8. Links 5 and 6 are redundant, i.e. the mechanism will have the same motion if they are removed. The input fourbar consists of links 1, 2, 3, and 4. The output dyad consists of links 7 and 8. The cross-hatched pivot pins at O2, O4 and O8 are attached to the ground link (1). In the calculation below, the redundant links and their joints are not counted (subtract 2 links and 4 joints from the totals). A

Number of links

L  6


J1  7


J2  0

O2

4 O4 G

E

3

2 D

5

C

6

7

M  3  ( L  1 )  2  J1  J2 O8

M1 F

H 8

f.

This is a fourbar offset slider-crank linkage. The input is link 2 (crank) and the output is link 4 (slider block). The cross-hatched pivot pin at O2 is attached to the ground link (1).

Number of links

L  4


J1  4


J2  0

4

B

3

M  3  ( L  1 )  2  J1  J2 A

M1 2 O2


g.


This is a fourbar linkage with an alternate output dyad. The input (rocker) is link 2 and the outputs (rockers) are links 4 and 6. The input fourbar consists of links 1, 2, 3, and 4. The alternate output dyad consists of links 5 and 6. The cross-hatched pivot pins at O2, O4 and O6 are attached to the ground link (1). Number of links

L  6


J1  7


J2  0

O6 3

B

A

2

M  3  ( L  1 )  2  J1  J2

4

C

6

O2

M1

5 D O4

h.

This is a ninebar mechanism with three redundant links, which reduces it to a sixbar. Since this mechanism is symmetrical about a vertical centerline, we can split it into two mirrored mechanisms to analyze it. Either links 2, 3 and 5 or links 7, 8 and 9 are redundant. To analyze it, consider 7, 8 and 9 as the redundant links. Analyzing the ninebar, there are two full joints at the pins A, B and C for a total of 12 joints. Number of links

L  9


J1  12


J2  0

6

O2 2

8

7

5

C

B

M  3  ( L  1 )  2  J1  J2

O8

A

9

3

M0 4

D

E

The result is that this mechanism seems to be a structure. By splitting it into mirror halves about the vertical centerline the mobility is found to be 1. Subtract the 3 redundant links and their 5 (6 minus the joint at A) associated joints to determine the mobility of the mechanism. Number of links

L  9  3


J1  12  5


J2  0

6 O2 2

5 B

M  3  ( L  1 )  2  J1  J2

3

M1 D

4

A



PROBLEM 2-22 Statement: Solution: 1.

Find the Grashof condition and Barker classifications of the mechanisms in Figure P2-4a, b, and d. See Figure P2-4 and Mathcad file P0222.

Use inequality 2.8 to determine the Grashof condition and Table 2-4 to determine the Barker classification. Condition( a b c d ) 


a.

This is a basic fourbar linkage. The input is link 2 and the output is link 4. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1). L1  174

L2  116

L3  108

L4  110

A 2 3

Condition L1 L2 L3 L4  "non-Grashof"

O2

4

C

This is a Barker Type 5 RRR1 (non-Grashof, longest link grounded). b.

O4

This is a fourbar linkage. The input is link 2, which in this case is the wheel with a pin at A, and the output is link 4. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1). L1  162

L2  40

L3  96

L4  122

B

A

2

3

O2 4

Condition L1 L2 L3 L4  "Grashof" This is a Barker Type 2 GCRR (Grashof, shortest link is input). d.

This is a fourbar linkage. The input is link 2, which in this case is a wheel with a pin at A, and the output is the vertical member on the coupler, link 3. Since the lengths of links 2 and 4 (O2A and O4B) are the same, the coupler link (3) has curvilinear motion and AB remains parallel to O2O4 throughout the cycle. The cross-hatched pivot pins at O2 and O4 are attached to the ground link (1). L1  150 L2  30 L3  150

L4  30

Condition L1 L2 L3 L4  "Special Grashof" This is a Barker Type 13 S2X (special case Grashof, two equal pairs, parallelogram).

O4

A

3

2 O2

B O4

4




Find the rotability of each loop of the mechanisms in Figure P2-4e, f, and g.

Solution:


1.

Use inequality 2.15 to determine the rotability of each loop in the given mechanisms.

e.

This is a fourbar linkage with an output dyad. The input (rocker) is link 2 and the output (rocker) is link 8. Links 5 and 6 are redundant, i.e. the mechanism will have the same motion if they are removed. The input fourbar consists of links 1, 2, 3, and 4. The output dyad consists of links 7 and 8. The cross-hatched pivot pins at O2, O4 and O8 are attached to the ground link (1). In the calculation below, the redundant links and their joints are not counted (subtract 2 links and 4 joints from the totals).

B

A O2

4 O4 G

E

3

2 D

5

C

6

7

O8

There are two loops in this mechanism. The first loop consists of links 1, 2, 3 (or 5), and 4. The second consists of links 1, 4, 7 (or 6), and 8. By inspection, we see that the sum of the shortest and longest in each loop is equal to the sum of the other two. Thus, both loops are Class III. f.

8

This is a fourbar offset slider-crank linkage. The input is link 2 (crank) and the output is link 4 (slider block). The cross-hatched pivot pin at O2 is attached to the ground link (1).

4

A 2 O2

O6

This is a fourbar linkage with an alternate output dyad. The input (rocker) is link 2 and the outputs (rockers) are links 4 and 6. The input fourbar consists of links 1, 2, 3, and 4. The alternate output dyad consists of links 5 and 6. The cross-hatched pivot pins at O2, O4 and O6 are attached to the ground link (1). r1  87

r2  49

r3  100

r4  153

B

3

We can analyze this linkage if we replace the slider ( 4) with an infinitely long binary link that is pinned at B to link 3 and pinned to ground (1). Then links 1 and 4 for are both infinitely long. Since these two links are equal in length and, if we say they are finite in length but very long, the rotability of the mechanism will be determined by the relative lengths of 2 and 3. Thus, this is a Class I linkage since link 2 is shorter than link 3.

g.

H

F

3

B

2 4

C

A 6

O2 5 D

Using the notation of inequality 2.15, N  4 LN  r4 L1  r2 L2  r1 LN  L1  202

O4

L3  r3 L2  L3  187

Since LN  L1  L2  L3, this is a a class II mechanism.




Find the mobility of the mechanisms in Figure P2-5.

Solution:


1.

Use equation 2.1c (Kutzbach's modification) to calculate the mobility. In the kinematic representations of the linkages below, binary links are depicted as single lines with nodes at their end points whereas higher order links are depicted as 2-D bars.

a.

This is a sixbar linkage with 4 binary (1, 2, 5, and 6) and 2 ternary (3 and 4) links. The inverted U-shaped link at the top of Figure P2-5a is represented here as the binary link 6. Number of links

L  6


J1  7


J2  0

3

5

4

2

M  3  ( L  1 )  2  J1  J2 O2

M1

b.

6

O4

This is an eightbar linkage with 4 binary (1, 4, 7, and 8) and 4 ternary (2, 3, 5, and 6) links. The inverted U-shaped link at the top of Figure P2-5b is represented here as the binary link 8. Number of links

L  8


J1  10


J2  0

M  3  ( L  1 )  2  J1  J2 M1

5

6

2

3

7

8 2 O2

4 O4




Find the mobility of the ice tongs in Figure P2-6. a. When operating them to grab the ice block. b. When clamped to the ice block but before it is picked up (ice grounded). c. When the person is carrying the ice block with the tongs.

Solution:


1.


a.

In this case there are two links and one full joint and 1 DOF. Number of links

L  2


J1  1


J2  0

M  3  ( L  1 )  2  J1  J2 b.

When the block is clamped in the tongs another link and two more full joints are added reducing the DOF to zero (the tongs and ice block form a structure). Number of links

L  2  1


J1  1  2


J2  0

M  3  ( L  1 )  2  J1  J2

c.

M1

M0

When the block is being carried the system has at least 4 DOF: x, y, and z position and orientation about a vertical axis.




Find the mobility of the automotive throttle mechanism shown in Figure P2-7.

Solution:


1.

This is an eightbar linkage with 8 binary links. It is assumed that the joint between the gas pedal (2) and the roller (3) that pivots on link 4 is a full joint, i.e. the roller rolls without slipping. The pivot pins at O2, O4, O6, and O8 are attached to the ground link (1). Use equation 2.1c (Kutzbach's modification) to calculate the mobility. Number of links

L  8


J1  10


J2  0

M  3  ( L  1 )  2  J1  J2

M1

7 6 O6

8

FULL JOINT 5 4 O4 3 2

O2

O8




Sketch a kinematic diagram of the scissors jack shown in Figure P2-8 and determine its mobility. Describe how it works.

Solution:


1.

The scissors jack depicted is a seven link mechanism with eight full and two half joints (see kinematic diagram below). Link 7 is a variable length link. Its length is changed by rotating the screw with the jack handle (not shown). The two blocks at either end of link 7 are an integral part of the link. The block on the left is threaded and acts like a nut. The block on the right is not threaded and acts as a bearing. Both blocks have pins that engage the holes in links 2, 3, 5, and 6. Joints A and B have 2 full joints apiece. For any given length of link 7 the jack is a structure (DOF = 0). When the screw is turned to give the jack a different height the jack has 1 DOF.

4

3

5

7

A

B 2

6 1

Number of links

L  7


J1  8


J2  2

M  3  ( L  1 )  2  J1  J2

M0




Find the mobility of the corkscrew in Figure P2-9.

Solution:


1.

The corkscrew is made from 4 pieces: the body (1), the screw (2), and two arms with teeth (3), one of which is redundant. The second arm is present to balance the forces on the assembly but is not necessary from a kinematic standpoint. So, kinematically, there are 3 links (body, screw, and arm), 2 full joints (sliding joint between the screw and the body, and pin joint where the arm rotates on the body), and 1 half joint where the arm teeth engage the screw "teeth". Using equation 2.1c, the DOF (mobility) is Number of links

L  3


J1  2


J2  1

M  3  ( L  1 )  2  J1  J2

M1




Figure P2-10 shows Watt's sun and planet drive that he used in his steam engine. The beam 2 is driven in oscillation by the piston of the engine. The planet gear is fixed rigidly to link 3 and its center is guided in the fixed track 1. The output rotation is taken from the sun gear 4. Sketch a kinematic diagram of this mechanism and determine its DOF. Can it be classified by the Barker scheme? If so, what Barker class and subclass is it?

Solution:


1.

Sketch a kinematic diagram of the mechanism. The mechanism is shown on the left and a kinematic model of it is sketched on the right. It is a fourbar linkage with 1 DOF (see below).

A

2

2

1

3

3 1

4

B

4

1

2.

C

Use equation 2.1c to determine the DOF (mobility). There are 4 links, 3 full pin joints, 1 half pin-in-slot joint (at B), and 1 half joint (at the interface C between the two gears, shown above by their pitch circles). Links 1 and 3 are ternary. Kutzbach's mobility equation (2.1c) Number of links

L  4


J1  3


J2  2

M  3  ( L  1 )  2  J1  J2 3.

M1

The Barker classification scheme requires that we have 4 link lengths. The motion of link 3 can be modeled by a basic fourbar if the half joint at B is replaced with a full pin joint and a link is added to connect B and the fixed pivot that is coincident with the center of curvature of the slot that guides pin B. L1  2.15

L2  1.25

L3  1.80

L4  0.54

This is a Grashof linkage and the Barker classification is I-4 (type 4) because the shortest link is the output.




Figure P2-11 shows a bicycle hand brake lever assembly. Sketch a kinematic diagram of this device and draw its equivalent linkage. Determine its mobility. Hint: Consider the flexible cable to be a link.

Solution:


1.

The motion of the flexible cable is along a straight line as it leaves the guide provided by the handle bar so it can be modeled as a translating full slider that is supported by the handlebar (link 1). The brake lever is a binary link that pivots on the ground link. Its other node is attached through a full pin joint to a third link, which drives the slider (link 4). Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2 M1

CABLE BRAKE LEVER 3 2

4 1 1




Figure P2-12 shows a bicycle brake caliper assembly. Sketch a kinematic diagram of this device and draw its equivalent linkage. Determine its mobility under two conditions. a. b.

Brake pads not contacting the wheel rim. Brake pads contacting the wheel rim.

Hint: Consider the flexible cable to be replaced by forces in this case. Solution: 1.


The rigging of the cable requires that there be two brake arms. However, kinematically they operate independently and can be analyzed that way. Therefore, we only need to look at one brake arm. When the brake pads are not contacting the wheel rim there is a single lever (link 2) that is pivoted on a full pin joint that is attached to the ground link (1). Thus, there are two links (frame and brake arm) and one full pin joint. Number of links

L  2


J1  1


J2  0

BRAKE ARM

FRAME

2

M  3  ( L  1 )  2  J1  J2 M1

2.

1

When the brake pad contacts the wheel rim we could consider the joint between the pad, which is rigidly attached to the brake arm and is, therefore, a part of link 2, to be a half joint. The brake arm (with pad), wheel (which is constrained from moving laterally by the frame), and the frame constitute a structure. Number of links

L  2


J1  1


J2  1

BRAKE ARM

FRAME

2

M  3  ( L  1 )  2  J1  J2 M0

1 1 HALF JOINT




Find the mobility, the Grashof condition, and the Barker classifications of the mechanism in Figure P2-13.

Solution:


1.

Use equation 2.1c (Kutzbach's modification) to calculate the mobility. When there is no cable in the jaw or before the cable is crimped this is a basic fourbar mechanism with with 4 full pin joints: Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2

M1

When there is a cable in the jaw this is a threebar mechanism with with 3 full pin joints. While the cable is clamped the jaws are stationary with respect to each other so that link 4 is grounded along with link 1, leaving only three operational links.

2.

Number of links

L  3


J1  3


J2  0

M  3  ( L  1 )  2  J1  J2

M0



L1  0.92

L2  0.27

L3  0.50

L4  0.60

Condition L1 L2 L3 L4  "non-Grashof" The Barker classification is II-1 (Type 5) RRR1 (non-Grashof, longest link grounded).




The approximate torque-speed curve and its equation for a 1/4 hp shunt-wound DC motor are shown in Figure P2-14. The rated speed for this fractional horsepower motor is 10000 rpm at a rated voltage of 130V. Determine: a) The rated torque in oz-in (ounce-inches, the industry standard for fractional hp motors) b) The no-load speed c) The operating speed range d) Plot the power-torque curve in the operating range and determine the maximum power that the motor can deliver in the that range.

Given:

Rated speed, N N ( T ) 

NR  10000  rpm

R

0.1 1.7

NR TR NR TR

HR 

Rated power, H

R

1 4

 hp

 T  1.1 NR if T  62.5 ozf  in  T  5.1 NR otherwise

T  0  ozf  in 2.5 ozf  in  75 ozf  in

12000

10000

Speed, rpm

8000

6000

4000

2000

0

0

25

50

75

100

Torque, oz-in

Figure P2-14 Torque-speed Characteristic of a 1/4 HP, 10000 rpm DC Motor Solution: a.


The rated torque is found by dividing the rated power by the rated speed: Rated torque, TR

TR 

HR NR

TR  25 ozf  in

b.

The no-load speed occurs at T = 0. From the graph this is 11000 rpm.

c.

The operating speed range for a shunt-wound DC motor is the speed at which the motor begins to stall up to the no-load speed. For the approximate torque-speed curve given in this problem the minimum speed is defined as the speed at the knee of the curve. Nopmin  N ( 62.5 ozf  in)

Nopmin  8500 rpm



Nopmax  N ( 0  ozf  in)

The power is the product of the speed and the torque. From the graph the equation for the torque-speed curve over the operating range is: N ( T )  

40 rpm ozf  in

 T  11000  rpm

and the power, therefore, is: H ( T )  N ( T )  T Plotting the power as a function of torque over the range T  0  ozf  in 2.5 ozf  in  62.5 ozf  in 0.750 0.700 0.650 0.600 0.550 0.500 Power, hp

d.

Nopmax  11000 rpm

0.450 0.400 0.350 0.300 0.250 0.200 0.150 0.100 0.050 0.000 0.0

12.5

25.0

37.5

50.0

62.5

75.0

Torque, oz-in

Maximum power occurs at the maximum torque in the operating range. The value of T at maximum power is: Value of T at Hmax

THmax  62.5 ozf  in

THmax  62.5 ozf  in

Maximum power

Hmax  H  THmax

Hmax  0.527  hp

Speed at max power

NHmax  N  THmax

NHmax  8500 rpm

Note that the curve goes through the rated power point of 0.25 hp at the rated torque of 25 oz-in.




Figure P2-15 shows a power hacksaw, used to cut metal. Link 5 pivots at O5 and its weight forces the sawblade against the workpiece while the linkage moves the blade (link 4) back and forth within link 5 to cut the part. Sketch its kinematic diagram, determine its mobility and its type (i.e., is it a fourbar, a Watt's sixbar, a Stephenson's sixbar, an eightbar, or what?) Use reverse linkage transformation to determine its pure revolute-jointed equivalent linkage.

Solution:


1.

Sketch a kinematic diagram of the mechanism. The mechanism is shown on the left and a kinematic model of it is sketched on the right. It is a fivebar linkage with 1 DOF (see below). 5

3

5 3

4

4

2

2

2.

1

1

1

Use equation 2.1c to determine the DOF (mobility). There are 5 links, 4 full pin joints, 1 full sliding joint, and 1 half joint (at the interface between the hacksaw blade and the pipe being cut). Kutzbach's mobility equation (2.1c) Number of links

L  5


J1  5


J2  1

M  3  ( L  1 )  2  J1  J2 3.

M1

Use rule 1 to transform the full sliding joint to a full pin joint for no change in DOF. Then use rules 2 and 3 by changing the half joint to a full pin joint and adding a link for no change in DOF. The resulting kinematically equivalent linkage has 6 links, 7 full pin joints, no half joints, and is shown below. Kutzbach's mobility equation (2.1c) Number of links

L  6


J1  7


J2  0

5 4

3 2

M  3  ( L  1 )  2  J1  J2

1 6

M1

1




Figure P2-16 shows a manual press used to compact powdered materials. Sketch its kinematic diagram, determine its mobility and its type (i.e., is it a fourbar, a Watt's sixbar, a Stephenson's sixbar, an eightbar, or what?) Use reverse linkage transformation to determine its pure revolute-jointed equivalent linkage.

Solution:


1.

Sketch a kinematic diagram of the mechanism. The mechanism is shown on the left and a kinematic model of it is sketched on the right. It is a fourbar linkage with 1 DOF (see below).

4 3

4

3 2

2

O2

O2

2.

Use equation 2.1c to determine the DOF (mobility). There are 4 links, 3 full pin joints, 1 full sliding joint, and 0 half joints. This is a fourbar slider-crank. Kutzbach's mobility equation (2.1c) Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2 3.

M1

Use rule 1 to transform the full sliding joint to a full pin joint for no change in DOF. The resulting kinematically equivalent linkage has 4 links, 4 full pin joints, no half joints, and is shown below. 4 O4 3

2 O2




Sketch the equivalent linkage for the cam and follower mechanism in Figure P2-17 in the position shown. Show that it has the same DOF as the original mechanism.

Solution:


1.

The cam follower mechanism is shown on the left and a kinematically equivalent model of it is sketched on the right.

4 1

1 4

3

3

2

2

INSTANTANEOUS CENTER OF CURVATURE OF CAM SURFACE

1

1

2.

Use equation 2.1c to determine the DOF (mobility) of the original mechanism. There are 4 links, 2 full pin joints, 1 full sliding joint, 1 pure rolling joint and 0 half joints. Kutzbach's mobility equation (2.1c) Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2 3.

M1

Use equation 2.1c to determine the DOF (mobility) of the equivalent mechanism. There are 4 links, 3 full pin joints, 1 full sliding joint, and 0 half joints. This is a fourbar slider-crank. Kutzbach's mobility equation (2.1c) Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2

M1


SOLUTIONS MANUAL 2-37-1


Describe the motion of the following rides, commonly found at an amusement park, as pure rotation, pure translation, or complex planar motion. a. A Ferris wheel b. A "bumper" car c. A drag racer ride d. A roller coaster whose foundation is laid out in a straight line e. A boat ride through a maze f. A pendulum ride g. A train ride

Solution:


a.

A Ferris wheel Pure rotation.

b.

A "bumper car" Complex planar motion.

c.

A drag racer ride Pure translation.

d.

A roller coaster whose foundation is laid out in a straight line Complex planar motion.

e.

A boat ride through a maze Complex planar motion.

f.

A pendulum ride Pure rotation.

g.

A train ride Complex planar motion.




Figure P2-1a is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.

Solution:


1.

Label the link numbers and joint letters for Figure P2-1a.

G B

3 C

5

2 A

4

1

1

6

F

E

D 1

a.

Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4 5 6

Link Order Ternary Ternary Binary Ternary Binary Binary

b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G

Joint Order 1 1 1 1 1 2 1

Half/Full Full Full Full Half Full Full Full




Figure P2-1b is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.

Solution:

See Figure P2-1b and Mathcad file P0239.

1.

Label the link numbers and joint letters for Figure P2-1b.

3 C 1 B

2 A 1

a.

Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3

Link Order Binary Binary Binary

b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C

Joint Order 1 1 1

Half/Full Full Half Full




Figure P2-1c is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.

Solution:

See Figure P2-1c and Mathcad file P0240.

1.

Label the link numbers and joint letters for Figure P2-1c. 4 1 C

D

3

2 B

A 1

a.

Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4

Link Order Binary Binary Binary Binary

b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D

Joint Order 1 1 1 1

Half/Full Full Full Full Full




Figure P2-1d is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.

Solution:


1.

Label the link numbers and joint letters for Figure P2-1d. H 1

7 G

6

E 5

F

A 1

D

2 3 B

a.

4 C

Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4 5 6 7

Link Order Binary Binary Ternary Binary Binary Ternary Binary

b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H

Joint Order 1 1 1 1 1 1 1 1

Half/Full Full Full Half Full Full Full Full Full




Find the mobility, Grashof condition and Barker classification of the oil field pump shown in Figure P2-18.

Solution:


1.



4 O4 3

O2

2

This is a basic fourbar linkage. The input is the 14-in-long crank (link 2) and the output is the top beam (link 4). The mobility (DOF) is found using equation 2.1c (Kutzbach's modification): Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2

M1

The link lengths and Grashof condition are L1 

2

( 76  12)  47.5

2

L1  79.701

Condition L1 L2 L3 L4  "Grashof" This is a Barker Type 2 GCRR (Grashof, shortest link is input).

L2  14

L3  80

L4  51.26




Find the mobility, Grashof condition and Barker classification of the aircraft overhead bin shown in Figure P2-19.

Solution:


1.



2.79 O2

6.95

B

2

9.17

9.17 4 3 O4

9.57

A

9.17

This is a basic fourbar linkage. The input is the link 2 and the output is link 4. The mobility (DOF) is found using equation 2.1c (Kutzbach's modification): Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2

The link lengths and Grashof condition are 2

2

L1  7.489

L2  9.17

2

2

L3  12.968

L4  9.57

L1 

2.79  6.95

L3 

9.17  9.17

Condition L1 L2 L3 L4  "non-Grashof" This is a Barker Type 7 RRR3 (non-Grashof, longest link is coupler).

M1




Figure P2-20 shows a "Rube Goldberg" mechanism that turns a light switch on when a room door is opened and off when the door is closed. The pivot at O2 goes through the wall. There are two spring-loaded piston-in cylinder devices in the assembly. An arrangement of ropes and pulleys inside the room transfers the door swing into a rotation of link 2. Door opening rotates link 2 CW, pushing the switch up as shown in the figure, and door closing rotates link 2 CCW, pulling the switch down. Find the mobility of the linkage.

Solution:


1.

2.

Examination of the figure shows 20 links (including the the switch) and 28 full joints. The second piston-in cylinder that actuates the switch is counted as a single binary link of variable length with joints at its ends. The other cylinder consists of two binary links, each link having one pin joint and one slider joint. There are no half joints. Use equation 2.1c to determine the DOF (mobility). Kutzbach's mobility equation (2.1c) Number of links

L  20


J1  28


J2  0

M  3  ( L  1 )  2  J1  J2 3.

M1

An alternative is to ignore the the first piston-in cylinder that acts on the third bellcrank from O2 since it does not affect the the motion of the linkage (it acts only as a damper.) In that case, subtract two links and three full joints, giving L = 18, J1 = 25 and M = 1.




All of the eightbar linkages in Figure 2-11 part 2 have eight possible inversions. Some of these will give motions similar to others. Those that have distinct motions are called distinct inversions. How many distinct inversions does the linkage in row 4, column 1 have?

Solution:

See Figure 2-11, part 2 and Mathcad file P0245.

1.

This isomer has one quaternary, two ternary, and five binary links arranged in a symetrical fashion. Due to this symmetry, grounding link 2 or 7 gives the same inversion, as do grounding 3 or 6 and 4 or 5. This makes 3 of the possible 8 inversions the same leaving 5 distinct inversions. Distinct inversions are obtained by grounding link 1, 2, 3, 4, or 8 (or 1, 5, 6, 7, or 8) for a total of 5 distinct inversions.





Solution:


1.

This isomer has four ternary, and four binary links arranged in a symetrical fashion. Due to this symmetry, grounding link 1 or 5 gives the same inversion, as do grounding 2 or 8, 4 or 6, and 3 or 7. This makes 4 of the possible 8 inversions the same leaving 4 distinct inversions. Distinct inversions are obtained by grounding link 1, 2, 3, or 4 (or 5, 6, 7, or 8) for a total of 4 distinct inversions.





Solution:


1.

This isomer has four ternary, and four binary links arranged in a symetrical fashion. Due to this symmetry, grounding link 2 or 4 gives the same inversion, as does grounding 5 or 7. This makes 2 of the possible 8 inversions the same leaving 6 distinct inversions. Distinct inversions are obtained by grounding link 1, 2, 3, 5, 6 or 8 (or 1, 3, 4, 6, 7, or 8) for a total of 6 distinct inversions.




Find the mobility of the mechanism shown in Figure 3-33.

Solution:

See Figure 3-33 and Mathcad file P0248.

1.

Use equation 2.1c to determine the DOF (mobility). There are 6 links, 7 full pin joints (two at B), and no half-joints.

Kutzbach's mobility equation (2.1c) Number of links

L  6


J1  7


J2  0

M  3  ( L  1 )  2  J1  J2

M1





Solution:


1.

Use equation 2.1c to determine the DOF (mobility). There are 6 links, 7 full pin joints, and no half-joints. Kutzbach's mobility equation (2.1c) Number of links

L  6


J1  7


J2  0

M  3  ( L  1 )  2  J1  J2

M1





Solution:


1.

Use equation 2.1c to determine the DOF (mobility). There are 6 links, 7 full pin joints, and no half-joints. Kutzbach's mobility equation (2.1c) Number of links

L  6


J1  7


J2  0

M  3  ( L  1 )  2  J1  J2

M1





Solution:


1.

Use equation 2.1c to determine the DOF (mobility). There are 8 links, 10 full pin joints (two at O4), and no half-joints. Kutzbach's mobility equation (2.1c) Number of links

L  8


J1  10


J2  0

M  3  ( L  1 )  2  J1  J2

M1





Solution:


1.

Use equation 2.1c to determine the DOF (mobility). There are 6 links, 7 full pin joints (two at O4), and no half-joints. Kutzbach's mobility equation (2.1c) Number of links

L  6


J1  7


J2  0

M  3  ( L  1 )  2  J1  J2

M1




Figure P2-1e is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.

Solution:

See Figure P2-1e and Mathcad file P0253.

1.

Label the link numbers and joint letters for Figure P2-1e. J

K

8

I

8 1

9

1

L 10

M

7

a.

4 D C

A 1

2

2 3

E

H

B 5

F


Link Order 5 nodes Quaternary Binary Binary Binary Binary

Link No. 7 8 9 10

6

G 1

b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H I J K L M

Joint Order 1 1 1 1 1 1 1 1 1 1 1 1 1

Half/Full Full Full Full Full Full Full Full Full Full Full Full Full Full

Joint Classification Grounded rotating joint Moving rotating joint Pure rolling joint Grounded rotating joint Moving rotating joint Moving translating joint Grounded rotating joint Moving rotating joint Moving rotating joint Grounded rotating joint Moving translating joint Moving rotating joint Grounded translating joint

Link Order Binary Ternary Binary Binary




Figure P2-1f is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.

Solution:

See Figure P2-1f and Mathcad file P0254.

1.

Label the link numbers and joint letters for Figure P2-1f.

F 5

E

5 1 6

1

G

H

4

a.

C 3

1

B A 1

2


Link Order Quaternary Binary Ternary Binary Ternary Binary

b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H

Joint Order 1 1 1 1 1 1 1 1

Half/Full Full Full Full Full Full Full Full Full

Joint Classification Grounded rotating joint Moving half joint Grounded translating joint Moving rotating joint Moving rotating joint Grounded rotating joint Moving half joint Grounded translating joint




Figure P2-1g is an example of a mechanism. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.

Solution:

See Figure P2-1g and Mathcad file P0255.

1.

Label the link numbers and joint letters for Figure P2-1g.

I

D 4 E

5

1

4 C 2

1

a.

B

G

A

A

1

F

7

6

3

H

7 1 J

2 1

8 1

K

Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4

Link Order 5 nodes Binary Binary Ternary

Link No. 5 6 7 8

Link Order Binary Binary Ternary Binary

b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H I J K

Joint Order 1 1 1 1 1 1 1 1 1 1 1

Half/Full Full Full Full Full Half Full Full Full Full Half Full

Joint Classification Grounded rotating joint Moving rolling joint Moving rotating joint Grounded rotating joint Moving sliding joint Grounded translating joint Moving rolling joint Moving rotating joint Grounded rotating joint Moving sliding joint Grounded translating joint




For the example linkage shown in Figure 2-4 find the number of links and their respective link orders, the number of joints and their respective orders, and the mobility of the linkage.

Solution:


1.

Label the link numbers and joint letters for Figure 2-4 example.

K

1

9 8

I G

J

6

7 H

D 1

4

E

3 B A 1

C 2

1

5 F 1

2.

Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4 5

3.

Link No. 6 7 8 9

Link Order Ternary Binary Binary Binary

Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H I J K

4.

Link Order 5 nodes Binary Ternary Binary Binary

Joint Order 1 1 1 2 1 1 1 1 1 1 1

Half/Full Full Half Full Full Full Full Full Full Full Full Full

Joint Classification Grounded rotating joint Moving sliding joint Grounded rotating joint Moving rotating joint Moving translating joint Grounded rotating joint Moving rotating joint Grounded rotating joint Moving rotating joint Moving rotating joint Grounded translating joint

Use equation 2.1c to calculate the DOF (mobility). Kutzbach's mobility equation (2.1c) Number of links

L  9

M  3  ( L  1 )  2  J1  J2

Number of full joints M1

J1  11


J2  1




For the linkage shown in Figure 2-5b find the number of joints and their respective orders, and mobility for: a) The condition of a finite load W in the direction shown and a zero F b) The condition of a finite load W and a finite load F both in the directions shown after link 6 is off the stop.

Solution:

See Figure 2-5b and Mathcad file P0257.

1.

Label the link numbers and joint letters for Figure 2-5b. 1

6

O6

W

D 3

A

B

1

F 5

4 O4

2 O2

1

C

1

a)

The condition of a finite load W in the direction shown and a zero F: Using the joint letters, determine each joint's order and whether each is a half or full joint. Link 6 is grounded so joint D is a grounded rotating joint and O6 is not a joint. For this condition there is a total of 6 full joints and no half joints. Joint Letter O2 B C D O4 A

Joint Order 1 1 1 1 1 1

Half/Full Full Full Full Full Full Full

Joint Classification Grounded rotating joint Moving rotating joint Moving rotating joint Grounded rotating joint Grounded rotating joint Moving rotating joint

Use equation 2.1c to calculate the DOF (mobility). Kutzbach's mobility equation (2.1c) Number of links L  5


M  3  ( L  1 )  2  J1  J2

J1  6


J2  0

M0

b) The condition of a finite load W and a finite load F both in the directions shown after link 6 is off the stop. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D O2 O4 O6

Joint Order 1 1 1 1 1 1 1

Half/Full Full Full Full Full Full Full Full

Joint Classification Moving rotating joint Moving rotating joint Moving rotating joint Moving rotating joint Grounded rotating joint Grounded rotating joint Grounded rotating joint

Use equation 2.1c to calculate the DOF (mobility). Kutzbach's mobility equation (2.1c) Number of links L  6 M  3  ( L  1 )  2  J1  J2

Number of full joints M1

J1  7


J2  0




Figure P2-21a shows a "Nuremberg scissors" mechanism. Find its mobility.

Solution:


1.


L  10


J1  13


J2  0

M  3  ( L  1 )  2  J1  J2

M1




Figure P2-21b shows a mechanism. Find its mobility and classify its isomer type.

Solution:


1.


L  6


J1  7


J2  0

M  3  ( L  1 )  2  J1  J2

2.

M1

Using Figure 2-9, we see that the mechanism is a Stephenson's sixbar isomer ( the two ternary links are connected with two binary links and one dyad).




Figure P2-21c shows a circular saw mounted on the coupler of a fourbar linkage. The centerline of the saw blade is at a coupler point that moves in an approximate straight line. Draw its kinematic diagram and determine its mobility.

Solution:

See Figure P2-21c and Mathcad file P0260.

1.

Draw a kinematic diagram of the mechanism. The saw's rotation axis is at point P and the saw is attached to link 3.

A

B 3 4

2 O2

O4

P

1

1 2.


L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2

M1




Figure P2-21d shows a log transporter. Draw a kinematic diagram of the mechanism, specify the number of links and joints, and then determine its mobility: a) For the transporter wheels locked and no log in the "claw" of the mechanism b) For the transporter wheels locked with it lifting a log c) For the transporter moving a log to a destination in a straight line.

Solution:


1.

Draw a kinematic diagram of the mechanism. Link 1 is the frame of the transporter. Joint B is of order 3. Actuators E and F provide two inputs (to get x-y motion) and actuator H provides an additional input for clamping logs. G

D

9 12 H

9

C

I 9

11

8

J 10

4 A

3

2 O2

B

5

7

E O4

1

F

1

6 O6 1

a)

Wheels locked, no log in "claw." Kutzbach's mobility equation (2.1c) Number of links

L  12


J1  15


J2  0

M  3  ( L  1 )  2  J1  J2

M3

b) Wheels locked, log held tighly in the "claw." With a log held tightly between links 9 and 10 a structure will be formed by links 9 through 12 and the log so that there will only be 9 links and 11 joints active. Number of links

L  9


J1  11


J2  0

M  3  ( L  1 )  2  J1  J2 c)

M2

Transporter moving in a straight line with the log holding mechanism inactive. There are two tires, the transporter frame, and the ground, making 4 links and two points of contact with the ground and two axels, making 4 joints.



Number of links

L  4


J1  4


J2  0

M  3  ( L  1 )  2  J1  J2

M1




Figure P2-21d shows a plow mechanism attached to a tractor. Draw its kinematic diagram and find its mobility including the earth as a "link." a) When the tractor is stopped and the turnbuckle is fixed. (Hint: Consider the tractor and the wheel to be one with the earth.) b) When the tractor is stopped and the turnbuckle is being adjusted. (Same hint.) c) When the tractor is moving and the turnbuckle is fixed. (Hint: Add the moving tractor's DOF to those found in part a.)

Solution:

See Figure P2-21e and Mathcad file P0262.

1.

Draw a kinematic diagram of the mechanism with the ground, tractor wheels, and tractor frame as link 1. Joint O4 is of order 2 and joint F is a half joint. The plow and its truss structure attach at joints D and E. Since the turnbuckle is fixed it can be modeled as a single binary link (6).

C

6

5 O4 1

D

B

4

7 7

3 A 2

7

2 O2

E 1

F

7 1

a)

Tractor stopped and turnbuckle fixed. Kutzbach's mobility equation (2.1c) Number of links

L  7


J1  8


J2  1

M  3  ( L  1 )  2  J1  J2

M1

b) When the tractor is stopped and the turnbuckle is being adjusted. Between joints C and D we now have 2 net links (2 links threaded LH and RH on one end and the turnbuckle body) and 1 additional helical full joint. Number of links L  8 Number of full joints

J1  9


J2  1

M  3  ( L  1 )  2  J1  J2 c)

M2

When the tractor is moving and the turnbuckle is fixed. If the tractor moved only in a straight line we would add 1 DOF to the 1 DOF that we got in part a for a total of M = 2. More realistically, the tractor can turn and move up and down hills so that we would add 3 DOF to the 1 DOF of part a to get a total of 4 DOF.




Figure P2-22 shows a Hart's inversor sixbar linkage. a) Is it a Watt or Stephenson linkage? b) Determine its inversion, i.e. is it a type I, II, or III?

Solution:

See Figure P2-22, Figure 2-14, and Mathcad file P0263.

1.

From Figure 2-14 we see that the Watt's sixbar has the two ternary links connected with a common joint while the Stephenson's sixbar has the two ternary links connected by binary links. Thus, Hart's inversor is a Watt's sixbar (links 1 and 2, the ternary links, are connected at a common joint). Further, the Hart's linkage is a Watt's sixbar inversion I since neither of the ternary links is grounded.




Figure P2-23 shows the top view of the partially open doors on one side of an entertainment center cabinet. The wooden doors are hinged to each other and one door is hinged to the cabinet. There is also a ternary, metal link attached to the cabinet and door through pin joints. A spring-loaded piston-in cylinder device attaches to the ternary link and the cabinet through pin joints. Draw a kinematic diagram of the door system and find the mobility of this mechanism.

Solution:


1.

Draw the kinematic diagram of this sixbar mechanism. The spring-loaded piston is just an in-line sliding joint (links 5 and 6, and joint F). The doors are binary links (3 and 4), and the metal ternary link (2) has nodes at A, B, and C. Link 1 is the cabinet.

A Cabinet

1

2

4

E Cabinet

B

D

Door

1

5 Cylinder F 6

Door

2

3

Link

G Cabinet 1 C

2.

Use equation 2.1c (Kutzbach's modification) to calculate the mobility. Number of links

L  6


J1  7


J2  0 M  3  ( L  1 )  2  J1  J2

M1




Figure P2-24a shows the seat and seat-back of a reclining chair with the linkage that connects them to the chair frame. Draw its kinematic diagram and determine its mobility with respect to the frame of the chair.

Solution:


1.

Draw a kinematic diagram of the mechanism. The chair-back attaches to link 2 and the seat with the attached slider slot is link 3. The node at at C is a half-joint as it allows two degrees of freedom.

A 1 2

3

B

C 2.

Determine the mobility of the mechanism. Kutzbach's mobility equation (2.1c) Number of links

L  3


J1  2


J2  1

M  3  ( L  1 )  2  J1  J2

M1




Figure P2-24b shows the mechanism used to extend the foot support on a reclining chair. Draw its kinematic diagram and determine its mobility with respect to the frame of the chair.

Solution:


1.

Draw a kinematic diagram of the mechanism. Link 1 is the frame.

D E 1

J

4 6

3

C 4

5

G

5 6

A 3 1

7

F

H

2

B 2.

Determine the mobility of the mechanism. Kutzbach's mobility equation (2.1c) Number of links

L  8


J1  10


J2  0

M  3  ( L  1 )  2  J1  J2

8

M1

K




Figure P2-24b shows the mechanism used to extend the foot support on a reclining chair. Number the links, starting with 1. (Hint: Don't forget the "ground" link.) Letter the joints alphabetically, starting with A. a. Using the link numbers, describe each link as binary, ternary, etc. b. Using the joint letters, determine each joint's order. c. Using the joint letters, determine whether each is a half or full joint.

Solution:


1.

Label the link numbers and joint letters for Figure P2-24b.

D E 1

J

4 6

3

C 4

5

G

5 6

A 3 1

F

2

8

7

H

B a.

Using the link numbers, describe each link as binary, ternary, etc. Link No. 1 2 3 4 5 6 7 8

Link Order Binary Binary Ternary Ternary Ternary Ternary Binary Binary

b,c. Using the joint letters, determine each joint's order and whether each is a half or full joint. Joint Letter A B C D E F G H J K

Joint Order 1 1 1 1 1 1 1 1 1 1

Half/Full Full Full Full Half Full Full Full Full Full Full

K




Figure P2-24 shows a sixbar linkage. a) Is it a Watt or Stephenson linkage? b) Determine its inversion, i.e. is it a type I, II, or III?

Solution:

See Figure P2-24, Figure 2-14, and Mathcad file P0268.

1.

From Figure 2-14 we see that the Watt's sixbar has the two ternary links connected with a common joint while the Stephenson's sixbar has the two ternary links connected by binary links. Thus, the sixbar linkage shown is a Watt's sixbar (links 3 and 4, the ternary links, are connected at a common joint). Further, the linkage shown is a Watt's sixbar inversion I since neither of the ternary links is grounded.




Define the following examples as path, motion, or function generation cases. a. b. c. d. e.

Solution:

A telescope aiming (star tracking) mechanism A backhoe bucket control mechanism A thermostat adjusting mechanism A computer printing head moving mechanism An XY plotter pen control mechanism


a.

Path generation. A star follows a 2D path in the sky.

b.

Motion generation. To dig a trench, say, the position and orientation of the bucket must be controlled.

c.

Function generation. The output is some desired function of the input over some range of the input.

d.

Path generation. The head must be at some point on a path.

e.

Path generation. The pen follows a straight line from point to point.




Design a fourbar Grashof crank-rocker for 90 deg of output rocker motion with no quick return. (See Example 3-1.) Build a cardboard model and determine the toggle positions and the minimum transmission angle.

Given:

Output angle

Solution:

See Example 3-1 and Mathcad file P0302.

Design choices: 1.

θ  90 deg

Link lengths:

L3  6.000

Link 3

L4  2.500

Link 4

2.

Draw the output link O4B in both extreme positions, B1 and B2, in any convenient location such that the desired angle of motion 4 is subtended. In this solution, link 4 is drawn such that the two extreme positions each make an angle of 45 deg to the vertical. Draw the chord B1B2 and extend it in any convenient direction. In this solution it was extended to the left.

3.

Layout the distance A1B1 along extended line B1B2 equal to the length of link 3. Mark the point A1.

4.

Bisect the line segment B1B2 and layout the length of that radius from point A1 along extended line B1B2. Mark the resulting point O2 and draw a circle of radius O2A1 with center at O2.

5.

Label the other intersection of the circle and extended line B1B2, A2.

6.

Measure the length of the crank (link 2) as O2A1 or O2A2. From the graphical solution, L2  1.76775

7.

Measure the length of the ground link (link 1) as O2O4. From the graphical solution, L1  6.2550 1.7677

6.0000

3.5355

2

A2

A1

3

B2

B1

O2 90.00° 1 4 6.2550

8.

O4

Find the Grashof condition. Condition( a b c d ) 


Condition L1 L2 L3 L4  "Grashof"




Design a fourbar mechanism to give the two positions shown in Figure P3-1 of output rocker motion with no quick-return. (See Example 3-2.) Build a cardboard model and determine the toggle positions and the minimum transmission angle.

Given:

Coordinates of A1, B1, A2, and B2 (with respect to A1):

Solution:

xA1  0.00

xB1  1.721

xA2  2.656

xB2  5.065

yA1  0.00

yB1  1.750

yA2  0.751

yB2  0.281


Design choices:

Link length:

Link 3

L3  5.000

Link 4

L4  2.000

1.

Following the notation used in Example 3-2 and Figure 3-5, change the labels on points A and B in Figure P3-1 to C and D, respectively. Draw the link CD in its two desired positions, C1D1 and C2D2, using the given coordinates.

2.

Draw construction lines from C1 to C2 and D1 to D2.

3.

Bisect line C1C2 and line D1D2 and extend their perpendicular bisectors to intersect at O4.

4.

Using the length of link 4 (design choice) as a radius, draw an arc about O4 to intersect both lines O4C1 and O4C2. Label the intersections B1 and B2.

5.

Draw the chord B1B2 and extend it in any convenient direction. In this solution it was extended to the left.

6.

Layout the distance A1B1 along extended line B1B2 equal to the length of link 3. Mark the point A1.

7.

Bisect the line segment B1B2 and layout the length of that radius from point A1 along extended line B1B2. Mark the resulting point O2 and draw a circle of radius O2A1 with center at O2.

8.

Label the other intersection of the circle and extended line B1B2, A2.

9.

Measure the length of the crank (link 2) as O2A1 or O2A2. From the graphical solution, L2  0.9469

10. Measure the length of the ground link (link 1) as O2O4. From the graphical solution, L1  5.3013

5.3013 5.0000 0.9469 A1 2

O2

O4

1

A2

3

B1

C1

4

R2.000

B2 C2 D1

D2



11. Find the Grashof condition. Condition( a b c d ) 






Design a fourbar mechanism to give the two positions shown in Figure P3-1 of coupler motion. (See Example 3-3.) Build a cardboard model and determine the toggle positions and the minimum transmission angle. Add a driver dyad. (See Example 3-4.)

Given:

Position 1 offsets:

Solution:

See figure below for one possible solution. Input file P0304.mcd from the solutions manual disk to the Mathcad program for this solution, file P03-04.4br to the program FOURBAR to see the fourbar solution linkage, and file P03-04.6br into program SIXBAR to see the complete sixbar with the driver dyad included.

xA1B1  1.721  in

yA1B1  1.750  in

1.

Connect the end points of the two given positions of the line AB with construction lines, i.e., lines from A1 to A2 and B1 to B2.

2.

Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below the bisector of A1A2 was extended downward and the bisector of B1B2 was extended upward.

3.

Select one point on each bisector and label them O4 and O6, respectively. In the solution below the distances O4A and O6B were each selected to be 4.000 in. This resulted in a ground-link-length O4O6 for the fourbar of 6.457 in.

4.

The fourbar stage is now defined as O4ABO6 with link lengths Link 5 (coupler) L5 

2

xA1B1  yA1B1

Link 4 (input)

L4  4.000  in

Ground link 1b

L1b  6.457  in

2

L5  2.454 in Link 6 (output)

L6  4.000  in

5.

Select a point on link 4 (O4A) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it D. (Note that link 4 is now a ternary link with nodes at O4, D, and A.) In the solution below the distance O4D was selected to be 2.000 in.

6.

Draw a construction line through D1D2 and extend it to the left.

7.

Select a point on this line and call it O2. In the solution below the distance CD was selected to be 4.000 in.

8.

Draw a circle about O2 with a radius of one-half the length D1D2 and label the intersections of the circle with the extended line as C1 and C2. In the solution below the radius was measured as 0.6895 in.

9.

The driver fourbar is now defined as O2CDO4 with link lengths Link 2 (crank)

L2  0.6895 in

Link 4a (rocker) L4a  2.000  in

Link 3 (coupler) L3  4.000  in Link 1a (ground) L1a  4.418  in

10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Shortest link

S  L2

S  0.6895 in

Longest link

L  L1a

L  4.4180 in

Other links

P  L3

P  4.0000 in

Q  L4a

Q  2.0000 in


Condition( a b c d ) 



Condition( S L P Q)  "Grashof" O6

6

Ground Link 1b

A1

6 50.231°

5 C1

A2

5

B2

47.893°

2 O2 3

4

C2

B1 D1

4

3 D2

Ground Link 1a

O4

11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4ABO6 is non-Grashoff with toggle positions at 2 = -71.9 deg and +71.9 deg. The minimum transmission angle is 35.5 deg. The fourbar operates between 2 = +21.106 deg and -19.297 deg.




Design a fourbar mechanism to give the three positions of coupler motion with no quick return shown in Figure P3-2. (See also Example 3-5.) Ignore the points O2 and O4 shown. Build a cardboard model and determine the toggle positions and the minimum transmission angle. Add a driver dyad.

Solution:


Design choices: L5  4.250

Length of link 5:

L4b  1.375

Length of link 4b:

1.

Draw link CD in its three design positions C1D1, C2D2, C3D3 in the plane as shown.

2.

Draw construction lines from point C1 to C2 and from point C2 to C3.

3.

Bisect line C1C2 and line C2C3 and extend their perpendicular bisectors until they intersect. Label their intersection O2.

4.

Repeat steps 2 and 3 for lines D1D2 and D2D3. Label the intersection O4.

5.

Connect O2 with C1 and call it link 2. Connect O4 with D1 and call it link 4.

6.

Line C1D1 is link 3. Line O2O4 is link 1 (ground link for the fourbar). The fourbar is now defined as O2CDO4 and has link lengths of Ground link 1a

L1a  0.718

Link 2

L2  2.197

Link 3

L3  2.496

Link 4

L4  3.704 1.230

O6 0.718 1b

2.197 O4 O2

2

C1

6 A

5

1a

4.328 B

2.496

D3

C3

3 4 C2 D1

D2 3.704

7.

Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case.





Condition L1a L2 L3 L4  "Grashof" 8.

9.

Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the solution above the distance O4B was selected to be L4b  1.375 . Draw a construction line through B1B3 and extend it up to the right.

10. Layout the length of link 5 (design choice) along the extended line. Label the other end A. 11. Draw a circle about O6 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L6  1.230. 12. The driver fourbar is now defined as O4BAO6 with link lengths Link 6 (crank)

L6  1.230

Link 5 (coupler) L5  4.250 Link 1b (ground) L1b  4.328 Link 4b (rocker) L4b  1.375 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L4b L5  "Grashof"




Design a fourbar mechanism to give the three positions shown in Figure P3-2 using the fixed pivots O2 and O4 shown. Build a cardboard model and determine the toggle positions and the minimum transmission angle. Add a driver dyad.

Solution:


Design choices: Length of link 5:

L5  5.000

L2b  2.000

Length of link 2b:

1.


2.

Draw the ground link O2O4 in its desired position in the plane with respect to the first coupler position C1D1.

3.

Draw construction arcs from point C2 to O2 and from point D2 to O2 whose radii define the sides of triangle C2O2D2. This defines the relationship of the fixed pivot O2 to the coupler line CD in the second coupler position.

4.


5.

Transfer this relationship back to the first coupler position C1D1 so that the ground plane position O2'O4' bears the same relationship to C1D1 as O2O4 bore to the second coupler position C2D2.

6.

Repeat the process for the third coupler position and transfer the third relative ground link position to the first, or reference, position.

7.

The three inverted positions of the ground link that correspond to the three desired coupler positions are labeled O2O4, O2'O4', and O2"O4" in the first layout below and are renamed E1F1, E2F2, and E3F3, respectively, in the second layout, which is used to find the points G and H.

O2''

C1

D3

O2'

C3 C2

O4'' D1

O4'

8.

O2

Draw construction lines from point E1 to E2 and from point E2 to E3.

D2

O4


9.


Bisect line E1E2 and line E2E3 and extend their perpendicular bisectors until they intersect. Label their intersection G.

10. Repeat steps 2 and 3 for lines F1F2 and F2F3. Label the intersection H. 11. Connect E1 with G and label it link 2. Connect F1 with H and label it link 4. Reinverting, E1 and F1 are the original fixed pivots O2 and O4, respectively. 12. Line GH is link 3. Line O2O4 is link 1a (ground link for the fourbar). The fourbar is now defined as O2GHO4 and has link lengths of Ground link 1a

L1a  4.303

Link 2

L2  8.597

Link 3

L3  1.711

Link 4

L4  7.921

E3

G

3

H

E2 F3

4

2

F2 E1

1a O2

F1 O4

13. Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case. Condition( a b c d ) 


Condition L1a L2 L3 L4  "Grashof" The fourbar that will provide the desired motion is now defined as a Grashof double crank in the crossed configuration. It now remains to add the original points C1 and D1 to the coupler GH and to define the driving dyad.



14. Select a point on link 2 (O2G) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 2 is now a ternary link with nodes at O2, B, and G.) In the solution below, the distance O2B was selected to be L2b  2.000 . 15. Draw a construction line through B1B3 and extend it up to the right. 16. Layout the length of link 5 (design choice) along the extended line. Label the other end A. 17. Draw a circle about O6 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L6  0.412. 18. The driver fourbar is now defined as O2BAO6 with link lengths Link 6 (crank)

L6  0.412

Link 5 (coupler) L5  5.000 Link 1b (ground) L1b  5.369 Link 2b (rocker) L2b  2.000 19. Use the link lengths in step 18 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L2b L5  "Grashof"

G2

G3

H1

G1 3

H2 H3

2

C1

A3 O6

6

D3

C3

A1 5

C2

D1 B3

O2

D2 B1

4

1a

O4




Given: Solution: 1.

Repeat Problem 3-2 with a quick-return time ratio of 1:1.4. (See Example 3.9). Design a fourbar Grashof crank-rocker for 90 degrees of output rocker motion with a quick-return time ratio of 1:1.4. 1 Time ratio Tr  1.4 See figure below for one possible solution. Also see Mathcad file P0307.

Determine the crank rotation angles  and , and the construction angle  from equations 3.1 and 3.2. Tr = Solving for , and 

β 

α

α  β = 360  deg

β 360  deg

β  210 deg

1  Tr

α  360  deg  β

α  150 deg

δ  β  180  deg

δ  30 deg

2.

Start the layout by arbitrarily establishing the point O4 and from it layoff two lines of equal length, 90 deg apart. Label one B1 and the other B2. In the solution below, each line makes an angle of 45 deg with the horizontal and has a length of 2.000 in.

3.

Layoff a line through B1 at an arbitrary angle (but not zero deg). In the solution below, the line is 30 deg to the horizontal.

4.

Layoff a line through B2 that makes an angle  with the line in step 3 (60 deg to the horizontal in this case). The intersection of these two lines establishes the point O2.

5.

From O2 draw an arc that goes through B1. Extend O2B2 to meet this arc. Erect a perpendicular bisector to the extended portion of the line and transfer one half of the line to O2 as the length of the input crank.

3.8637 = b

90.0000°

B2 B1

B2

2.0000 = c

1.0353 = a

LAYOUT

B1

4 3 O4

O4 A1 2 O2

O2

3.0119 = d

A2 LINKAGE DEFINITION 


6.


For this solution, the link lengths are: Ground link (1)

d  3.0119 in

Crank (2)

a  1.0353 in

Coupler (3)

b  3.8637 in

Rocker (4)

c  2.000  in




Design a sixbar drag link quick-return linkage for a time ratio of 1:2, and output rocker motion of 60 degrees. (See Example 3-10.)

Given:

Time ratio

Solution: 1.

Tr 

1 2

See figure below for one possible solution. Also see Mathcad file P0308.

Determine the crank rotation angles  and  from equation 3.1. Tr = Solving for and 

β 

α

α  β = 360  deg

β 360  deg 1  Tr

α  360  deg  β

β  240 deg α  120 deg

2.

Draw a line of centers XX at any convenient location.

3.

Choose a crank pivot location O2 on line XX and draw an axis YY perpendicular to XX through O2.

4.

Draw a circle of convenient radius O2A about center O2. In the solution below, the length of O2A is a  1.000  in.

5.

Lay out angle  with vertex at O2, symmetrical about quadrant one.

6.

Label points A1 and A2 at the intersections of the lines subtending angle  and the circle of radius O2A.

7.

8.

Set the compass to a convenient radius AC long enough to cut XX in two places on either side of O2 when swung from both A1 and A2. Label the intersections C1 and C2. In the solution below, the length of AC is b  1.800  in. The line O2A is the driver crank, link 2, and the line AC is the coupler, link 3.

9.

The distance C1C2 is twice the driven (dragged) crank length. Bisect it to locate the fixed pivot O4.

10. The line O2O4 now defines the ground link. Line O4C is the driven crank, link 4. In the solution below, O4C measures c  2.262  in and O2O4 measures d  0.484  in. 11. Calculate the Grashoff condition. If non-Grashoff, repeat steps 7 through 11 with a shorter radius in step 7. Condition( a b c d ) 


Condition( a b c d )  "Grashof" 12. Invert the method of Example 3-1 to create the output dyad using XX as the chord and O4C1 as the driving crank. The points B1 and B2 will lie on line XX and be spaced apart a distance that is twice the length of O4C (link 4). The pivot point O6 will lie on the perpendicular bisector of B1B2 at a distance from XX which subtends the specified output rocker angle, which is 60 degrees in this problem. In the solution below, the length BC was chosen to be e  5.250  in.



LAYOUT OF SIXBAR DRAG LINK QUICK RETURN WITH TIME RATIO OF 1:2 a = 1.000 b = 1.800 c = 2.262 d = 0.484 e = 5.250 f = 4.524

13. For the design choices made (lengths of links 2, 3 and 5), the length of the output rocker (link 6) was measured as f  4.524  in.




Design a crank-shaper quick-return mechanism for a time ratio of 1:3 (Figure 3-14, p. 112).

Given:

Time ratio

Solution:


TR 

1 3

Design choices:

1.

Length of link 2 (crank)

L2  1.000

Length of link 5 (coupler)

L5  5.000

S  4.000

Length of stroke

Calculate  from equations 3.1. TR 

α β

α  β  360  deg

α 

360  deg 1

α  90.000 deg

1 TR

2.

Draw a vertical line and mark the center of rotation of the crank, O2, on it.

3.

Layout two construction lines from O2, each making an angle /2 to the vertical line through O2.

4.

Using the chosen crank length (see Design Choices), draw a circle with center at O2 and radius equal to the crank length. Label the intersections of the circle and the two lines drawn in step 3 as A1 and A2.

5.

Draw lines through points A1 and A2 that are also tangent to the crank circle (step 2). These two lines will simultaneously intersect the vertical line drawn in step 2. Label the point of intersection as the fixed pivot center O4.

6.

Draw a vertical construction line, parallel and to the right of O2O4, a distance S/2 (one-half of the output stroke length) from the line O2O4.

7.

Extend line O4A1 until it intersects the construction line drawn in step 6. Label the intersection B1.

8.

Draw a horizontal construction line from point B1, either to the left or right. Using point B1 as center, draw an arc of radius equal to the length of link 5 (see Design Choices) to intersect the horizontal construction line. Label the intersection as C1.

9.

Draw the slider blocks at points A1 and C1 and finish by drawing the mechanism in its other extreme position.

STROKE 4.000 C2

2.000

6

C1

B2

B1

5 O2 4

2 A2

3 O4

A1




Find the two cognates of the linkage in Figure 3-17 (p. 116). Draw the Cayley and Roberts diagrams. Check your results with program FOURBAR.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  2

Crank

L2  1

A1P  1.800

δ  34.000 deg

Coupler

L3  3

Rocker

L4  3.5

B1P  1.813

γ  33.727 deg


Draw the original fourbar linkage, which will be cognate #1, and align links 2 and 4 with the coupler. A1

B1

3

2

A1

3

OA

P

B1

4

2 OA P 4 1

OB

OB

2.

Construct lines parallel to all sides of the aligned fourbar linkage to create the Cayley diagram (see Figure 3-24) OA

2

A1

B1

3

OB

4

10

5 A2

B3

P

9 B2

6 8 7

A3

OC A1

3.

4.

Return links 2 and 4 to their fixed pivots OA and OB and establish OC as a fixed pivot by making triangle OAOBOC similar to A1B1P. Separate the three cognates. Point P has the same path motion in each cognate.

2 4

OA

10

P

Calculate the cognate link lengths based on the geometry of the Cayley diagram (Figure 3-24c, p. 114). L5  B1P L6 

L4 L3

 B1P

9

8

6

A2 OC OB

L5  1.813 L6  2.115

B2

A3 7

5.

B1

3

5 B3



P

P

OA

B2

10 A3

9 OC

7

8 A2 OC

6 OB 5

Cognate #2

B3

Cognate #3

L10  A1P

L10  1.800

L7  L9

B1P

L8  L6

A1P

L9 

L2 L3

 A1P

L9  0.600

L7  0.604

A1P

L8  2.100

B1P

From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC  L1AC 

L1 L3 L1 L3

 B1P

L1BC  1.209

 A1P

L1AC  1.200

Calculate the coupler point data for cognates #2 and #3 A3P  L8

A3P  2.100





δ  180  deg  δ  γ

δ  247.727 deg

A2P  L2

A2P  1.000

δ  δ

δ  34.000 deg

SUMMARY OF COGNATE SPECIFICATIONS:

6.

Cognate #1

Cognate #2

Cognate #3

Ground link length

L1  2.000

L1AC  1.200

L1BC  1.209

Crank length

L2  1.000

L10  1.800

L7  0.604

Coupler length

L3  3.000

L9  0.600

L6  2.115

Rocker length

L4  3.500

L8  2.100

L5  1.813

Coupler point

A1P  1.800

A2P  1.000

A3P  2.100

Coupler angle

δ  34.000 deg

δ  34.000 deg

δ  247.727 deg

Verify that the three cognates yield the same coupler curve by entering the original link lengths in program FOURBAR and letting it calculate the cognates.





Note that cognate #2 is a Grashof double rocker and, therefore, cannot trace out the entire coupler curve.




Find the three equivalent geared fivebar linkages for the three fourbar cognates in Figure 3-25a (p. 125). Check your results by comparing the coupler curves with programs FOURBAR and FIVEBAR.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  39.5

Crank

L2  15.5

Coupler

L3  14.0

Rocker

L4  20.0

A1P  26.0


Calculate the length BP and the angle using the law of cosines on the triangle APB. B1P   L3  A1P  2  L3 A1P  cos δ 2

 

2

0.5

B1P  23.270

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

δ  63.000 deg

γ  84.5843 deg

Use the Cayley diagram (see Figure 3-24) to calculate the link lengths of the two cognates. Note that the diagram is made up of three parallelograms and three similar triangles L4

L5  B1P

L5  23.270

L6 

L10  A1P

L10  26.000

L9 

L7  25.763

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  33.243

 A1P

L9  28.786

A1P

L8  37.143

B1P


A3P  20.000

A2P  L2

A2P  15.500

δ  γ

δ  84.584 deg

δ  δ

δ  63.000 deg

From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC 

3.

L1 L3

 B1P

L1BC  65.6548

L1AC 

L1 L3

 A1P

L1AC  73.3571

Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1

Cognate #2

Cognate #3

Ground link length

L1  39.500

L1AC  73.357

L1BC  65.655

Crank length

L2  15.500

L10  26.000

L7  25.763

Coupler length

L3  14.000

L9  28.786

L6  33.243



Rocker length

L4  20.000

L8  37.143

L5  23.270

Coupler point

A1P  26.000

A2P  15.500

A3P  20.000

Coupler angle

δ  63.000 deg

δ  63.000 deg

δ  84.584 deg

OC 8 B2

7

B3 9

P

6 A2 A3 10 3

A1

5

B1 4

2 1

OA

OB

4.

The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAA2PA3OB, OAA1PB3OC, and OBB1PB2OC. They are shown individually below with their associated gears. P

A2 A3 10 5

OA OB

OC



OC

7

B3

OD P

A1 2 OA

OC 8 B2

P

OE

B1 4

OB



SUMMARY OF GEARED FIVEBAR COGNATE SPECIFICATIONS: Cognate #1

Cognate #2

Cognate #3

Ground link length

L1  39.500

L1AC  73.357

L1BC  65.655

Crank length

L10  26.000

L2  15.500

L4  20.000

Coupler length

A2P  15.500

A1P  26.000

L5  23.270

Rocker length

A3P  20.000

L8  37.143

L7  25.763

Crank length

L5  23.270

L7  25.763

L8  37.143

Coupler point

A2P  15.500

A1P  26.000

B1P  23.270

Coupler angle

δ  0.00 deg

δ  0.00 deg

δ  0.00 deg

5.

Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path.

6.

Enter the geared fivebar cognate #1 specifications into program FIVEBAR to get a trace of the coupler path for the geared fivebar (see next page).






Design a sixbar, single-dwell linkage for a dwell of 90 deg of crank motion, with an output rocker motion of 45 deg.

Given:

Crank dwell period: 90 deg. Output rocker motion: 45 deg.

Solution:

See Figures 3-20, 3-21, and Mathcad file P0312.

Design choices: Ground link ratio, L1/L2 = 2.0: GLR  2.0 Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.5: CLR  2.5 Coupler angle, γ  72 deg Crank length, L2  2.000 1.

For the given design choices, determine the remaining link lengths and coupler point specification. Coupler link (3) length

L3  CLR L2

L3  5.000

Rocker link (4) length

L4  CLR L2

L4  5.000

Ground link (1) length

L1  GLR  L2

L1  4.000

Angle PAB

δ 

Length AP on coupler 2.

180  deg  γ 2

AP  2  L3 cos δ

δ  54.000 deg AP  5.878

Enter the above data into program FOURBAR, plot the coupler curve, and determine the coordinates of the coupler curve in the selected range of crank motion, which in this case will be from 135 to 225 deg..



FOURBAR for Windows Angle Step Deg 135 140 145 150 155 160 165 170 175 180 185 190 195 200 205 210 215 220 225 3.

File P03-12.DAT

Coupler Pt X

Coupler Pt Y

Coupler Pt Mag

-1.961 -2.178 -2.393 -2.603 -2.809 -3.008 -3.201 -3.386 -3.563 -3.731 -3.890 -4.038 -4.176 -4.302 -4.417 -4.520 -4.610 -4.688 -4.753

7.267 7.128 6.977 6.813 6.638 6.453 6.257 6.052 5.839 5.617 5.389 5.155 4.915 4.671 4.424 4.175 3.924 3.673 3.424

7.527 7.453 7.375 7.293 7.208 7.119 7.028 6.935 6.840 6.744 6.646 6.548 6.450 6.351 6.252 6.153 6.054 5.956 5.858

Coupler Pt Ang

105.099 106.992 108.930 110.911 112.933 114.994 117.093 119.228 121.396 123.595 125.822 128.075 130.351 132.646 134.955 137.274 139.598 141.921 144.235

Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Use the points at crank angles of 135, 180, and 225 deg to define the pseudo-arc. Find the center of the pseudo-arc erecting perpendicular bisectors to the chords defined by the selected coupler curve points. The center will lie at the intersection of the perpendicular bisectors, label this point D. The radius of this circle is the length of link 5.

y 135 P

PSEUDO-ARC

180 B

225 3 D A

4

2 x O2

O4


4.


The position of the end of link 5 at point D will remain nearly stationary while the crank moves from 135 to 225 deg. As the crank motion causes the coupler point to move around the coupler curve there will be another extreme position of the end of link 5 that was originally at D. Since a symmetrical linkage was chosen, the other extreme position will be located along a line through the axis of symmetry (see Figure 3-20) a distance equal to the length of link 5 measured from the point where the axis of symmetry intersects the coupler curve near the 0 deg coupler point. Establish this point and label it E. FOURBAR for Windows Angle Step Deg 300 310 320 330 340 350 0 10 20 30 40 50 60

File P03-12.DAT

Coupler Pt X

Coupler Pt Y

Coupler Pt Mag

-4.271 -4.054 -3.811 -3.526 -3.159 -2.651 -1.968 -1.181 -0.441 0.126 0.478 0.631 0.617

0.869 0.926 1.165 1.628 2.343 3.286 4.336 5.310 6.085 6.654 7.068 7.373 7.598

4.359 4.158 3.985 3.883 3.933 4.222 4.762 5.440 6.101 6.656 7.085 7.400 7.623

Coupler Pt Ang

168.495 167.133 162.998 155.215 143.437 128.892 114.414 102.534 94.142 88.914 86.129 85.111 85.354

y 135 P

PSEUDO-ARC

180 B

5

4

225 3

AXIS OF SYMMETRY

D A

E

2

x O2 5.

O4

The line segment DE represents the maximum displacement that a link of the length equal to link 5, attached at P, will reach along the axis of symmetry. Construct a perpendicular bisector of the line segment DE and extend it to the right (or left, which ever is convenient). Locate fixed pivot O6 on the bisector of DE such that the lines O6D and O6E subtend the desired output angle, in this case 45 deg. Draw link 6 from D through O6 and extend it to any convenient length. This is the output link that will dwell during the specified motion of the crank. See next page for the completed layout and further linkage specifications.



y 135 P

PSEUDO-ARC

180

45.000° B

5

4

225

O6 BISECTOR

3 D A

E

2

x O4

O2

SUMMARY OF LINKAGE SPECIFICATIONS Original fourbar: Ground link

L1  4.000

Crank

L2  2.000

Coupler

L3  5.000

Rocker

L4  5.000

Coupler point

AP  5.878

δ  54.000 deg

Added dyad: Coupler

L5  6.363

Output

L6  2.855

Pivot O6

x  3.833

y  3.375




Design a sixbar double-dwell linkage for a dwell of 90 deg of crank motion, with an output of rocker motion of 60 deg, followed by a second dwell of about 60 deg of crank motion.

Given:

Initial crank dwell period: 90 deg Final crank dwell period: 60 deg (approx.) Output rocker motion between dwells: 60 deg

Solution:


Design choices:

1.

Ground link length

L1  5.000

Crank length

L2  2.000

Coupler link length

L3  5.000

Rocker length

L2  5.500

Coupler point data:

AP  8.750

δ  50 deg

In the absence of a linkage atlas it is difficult to find a coupler curve that meets the specifications. One approach is to start with a symmetrical linkage, using the data in Figure 3-21. Then, using program FOURBAR and by trial-and-error, adjust the link lengths and coupler point data until a satisfactory coupler curve is found. The link lengths and coupler point data given above were found this way. The resulting coupler curve is shown below and a printout of the coupler curve coordinates taken from FOURBAR is also printed below.



FOURBAR for Windows Angle Step Deg 0.000 10.000 20.000 30.000 40.000 50.000 60.000 70.000 80.000 90.000 100.000 110.000 120.000 130.000 140.000 150.000 160.000 170.000 180.000 190.000 200.000 210.000 220.000 230.000 240.000 250.000 260.000 270.000 280.000 290.000 300.000 310.000 320.000 330.000 340.000 350.000 360.000

File P03-13.DAT

Cpler Pt Cpler Pt Cpler Pt Cpler Pt X Y Mag Ang

9.353 9.846 10.167 10.286 10.226 10.031 9.746 9.406 9.039 8.665 8.301 7.958 7.647 7.376 7.151 6.977 6.853 6.778 6.748 6.755 6.792 6.847 6.912 6.976 7.031 7.073 7.099 7.112 7.120 7.137 7.184 7.288 7.481 7.792 8.233 8.779 9.353

4.742 4.159 3.491 2.840 2.274 1.815 1.457 1.180 0.963 0.787 0.637 0.507 0.391 0.291 0.209 0.151 0.126 0.140 0.201 0.316 0.488 0.719 1.008 1.351 1.741 2.170 2.626 3.098 3.570 4.030 4.458 4.834 5.131 5.312 5.332 5.147 4.742

10.487 10.688 10.750 10.671 10.476 10.194 9.854 9.480 9.090 8.701 8.325 7.974 7.657 7.382 7.154 6.978 6.854 6.779 6.751 6.763 6.809 6.885 6.985 7.105 7.243 7.398 7.569 7.757 7.965 8.196 8.455 8.746 9.072 9.430 9.809 10.177 10.487

26.886 22.900 18.951 15.437 12.537 10.257 8.503 7.152 6.081 5.187 4.391 3.644 2.928 2.256 1.671 1.242 1.051 1.182 1.708 2.678 4.110 5.996 8.300 10.963 13.911 17.057 20.302 23.536 26.632 29.448 31.819 33.555 34.446 34.286 32.931 30.384 26.886

2.

Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Fit tangent lines to the nearly straight portions of the curve. Label their intersection O6. The coordinates of O6 are (6.729, 0.046).

3.

Design link 6 to lie along these straight tangents, pivoted at O6. Provide a slot in link 6 to accommodate slider block 5, which pivots on the coupler point P. (See next page).

4.

The beginning and ending crank angles for the dwell portions of the motion are indicated on the layout and in the table above by boldface entries.



y 6 B 60.000°

260 5 4

P

3

90 O2 2

A

170 x O4

150 O6




Figure P3-3 shows a treadle-operated grinding wheel driven by a fourbar linkage. Make a cardboard model of the linkage to any convenient scale. Determine its minimum transmission angles. Comment on its operation. Will it work? If so, explain how it does.

Given:

Link lengths:

Link 2

L2  0.60 m

Link 3

L3  0.75 m

Link 4

L4  0.13 m

Link 1

L1  0.90 m

Grashof condition function: Condition( a b c d ) 


Solution: 1.


Determine the Grashof condition of the mechanism from inequality 2.8 and its Barker classification from Table 2-4. Grashof condition: Barker classification:

Condition L1 L2 L3 L4  "Grashof" Class I-4, Grashof rocker-rocker-crank, GRRC, since the shortest link is the output link.

2.

As a Grashof rocker-crank, the minimum transmission angle will be 0 deg, twice per revolution of the output (link 4) crank.

3.

Despite having transmission angles of 0 deg twice per revolution, the mechanism will work. That is, one will be able to drive the grinding wheel from the treadle (link 2). The reason is that the grinding wheel will act as a flywheel and will carry the linkage through the periods when the transmission angle is low. Typically, the operator will start the motion by rotating the wheel by hand.




Figure P3-4 shows a non-Grashof fourbar linkage that is driven from link O2A. All dimensions are in centimeters (cm). (a) (b) (c) (d)

Given:

Solution: 1.

Find the transmission angle at the position shown. Find the toggle positions in terms of angle AO2O4. Find the maximum and minimum transmission angles over its range of motion. Draw the coupler curve of point P over its range of motion.

Link lengths: Link 1 (ground)

L1  95 mm

Link 2 (driver)

L2  50 mm

Link 3 (coupler)

L3  44 mm

Link 4 (driven)

L4  50 mm


To find the transmission angle at the position shown, draw the linkage to scale in the position shown and measure the transmission angle ABO4. P

y 77.097°

B

3 A 2

4

O4

50.000° 1

x

O2

The measured transmission angle at the position shown is 77.097 deg. 2.

The toggle positions will be symmetric with respect to the O2O4 axis and will occur when links 3 and 4 are colinear. Use the law of cosines to calculate the angle of link 2 when links 3 and 4 are in toggle.

 L3  L42  L12  L22  2 L1 L2 cosθ where 2 is the angle AO2O4. Solving for 2,

 L12  L22   L3  L4 2 θ  acos   2  L1 L2   The other toggle position occurs at θ  73.558 deg

θ  73.558 deg


3.


Use the program FOURBAR to find the maximum and minimum transmission angles.

FOURBAR for Windows

File P03-15

Design #

1

Angle Step Deg

Theta2 Mag degrees

Theta3 Mag degrees

Theta4 Mag degrees

Trans Ang Mag degrees

-73.557 -58.846 -44.134 -29.423 -14.711 0.000 14.711 29.423 44.134 58.846 73.557

-73.557 -58.846 -44.134 -29.423 -14.711 0.000 14.711 29.423 44.134 58.846 73.557

30.861 64.075 77.168 83.147 80.604 68.350 50.145 32.106 16.173 0.566 -30.486

-149.490 -176.312 170.696 157.514 142.103 125.123 111.644 106.473 109.701 120.179 149.159

0.352 60.387 86.472 74.367 61.499 56.773 61.499 74.367 86.472 60.387 0.355

A partial output from FOURBAR is shown above. From it, we see that the maximum transmission angle is approximately 86.5 deg and the minimum is zero deg. 4.

Use program FOURBAR to draw the coupler curve with respect to a coordinate frame through O2O4.




Draw the Roberts diagram for the linkage in Figure P3-4 and find its two cognates. Are they Grashof or non-Grashof?

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  9.5

Crank

L2  5

Coupler

L3  4.4

Rocker

L4  5

A1P  8.90



 

2

0.5

B1P  7.401

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

δ  56.000 deg

γ  94.4701 deg


L5  B1P

L5  7.401

L6 

L10  A1P

L10  8.900

L9 

L7  8.410

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  8.410

 A1P

L9  10.114

A1P

L8  10.114

B1P


A3P  5.000

A2P  L2

A2P  5.000

δ  γ

δ  94.470 deg

δ  δ

δ  56.000 deg


3.

L1 L3

 B1P

L1BC  15.9793

L1AC 

L1 L3

 A1P

L1AC  19.2159


Cognate #2

Cognate #3

Ground link length

L1  9.500

L1AC  19.216

L1BC  15.979

Crank length

L2  5.000

L10  8.900

L7  8.410

Coupler length

L3  4.400

L9  10.114

L6  8.410



Rocker length

L4  5.000

L8  10.114

L5  7.401

Coupler point

A1P  8.900

A2P  5.000

A3P  5.000

Coupler angle

δ  56.000 deg

δ  56.000 deg

δ  94.470 deg

B2

OC

8

7

B3

P 9 6 A3

A2

5

3

B1

10

4

A1 2

OB 1

OA

6.

Determine the Grashof condition of each of the two additional cognates. Condition( a b c d ) 


Cognate #2:

Condition L10 L1AC L8 L9  "non-Grashof"

Cognate #3:

Condition L5 L1BC L6 L7  "non-Grashof"




Design a Watt-I sixbar to give parallel motion that follows the coupler path of point P of the linkage in Figure P3-4.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  9.5

Crank

L2  5

Coupler

L3  4.4

Rocker

L4  5

A1P  8.90



 

2

0.5

B1P  7.401

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

δ  56.000 deg

γ  94.4701 deg


L5  B1P

L5  7.401

L6 

L10  A1P

L10  8.900

L9 

L7  8.410

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  8.410

 A1P

L9  10.114

A1P

L8  10.114

B1P


A3P  5.000

A2P  L2

A2P  5.000

δ  γ

δ  94.470 deg

δ  δ

δ  56.000 deg


3.

L1 L3

 B1P

L1BC  15.9793

L1AC 

L1 L3

 A1P

L1AC  19.2159


Cognate #2

Cognate #3

Ground link length

L1  9.500

L1AC  19.216

L1BC  15.979

Crank length

L2  5.000

L10  8.900

L7  8.410

Coupler length

L3  4.400

L9  10.114

L6  8.410


B2


Rocker length

L4  5.000

L8  10.114

L5  7.401

Coupler point

A1P  8.900

A2P  5.000

A3P  5.000

Coupler angle

δ  56.000 deg

δ  56.000 deg

δ  94.470 deg

OC

8

7

B3

P 9 6 A3

A2

P

5

3 10

B1 4

A1 2

OB 1 3

OA

4

A1

4.

5.

All three of these cognates are non-Grashof and will, therefore, have limited motion. However, following Example 3-11, discard cognate #2 and retain cognates #1 and #3. Draw line qq parallel to line OAOC and through point OB. Without allowing links 5, 6, and 7 to rotate, slide them as an assembly along lines OAOC and qq until the free end of link 7 is at OA. The free end of link 5 will then be at point O'B and point P on link 6 will be at P'. Add a new link of length OAOC between P and P'. This is the new output link 8 and all points on it describe the original coupler curve. Join links 2 and 7, making one ternary link. Remove link 5 and reduce link 6 to a binary link. The result is a Watt-I sixbar with links numbered 1, 2, 3, 4, 6, and 8 (see next page). Link 8 is in curvilinear translation and follows the coupler path of the original point P.

B1 q

2

OB 1

OA 7

B3 P'

6 A3 5

q

O'B



P

B1 3

4

A1 8 2

OB

1

OA

B3 P'

6




Design a Watt-I sixbar to give parallel motion that follows the coupler path of point P of the linkage in Figure P3-4 and add a driver dyad to drive it over its possible range of motion with no quick return. (The result will be an 8-bar linkage).

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  9.5

Crank

L2  5

Coupler

L3  4.4

Rocker

L4  5

A1P  8.90



 

2

0.5

B1P  7.401

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

δ  56.000 deg

γ  94.4701 deg


L5  B1P

L5  7.401

L6 

L10  A1P

L10  8.900

L9 

L7  8.410

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  8.410

 A1P

L9  10.114

A1P

L8  10.114

B1P


A3P  5.000

A2P  L2

A2P  5.000

δ  γ

δ  94.470 deg

δ  δ

δ  56.000 deg

From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC  3.

L1 L3

 B1P

L1BC  15.9793

L1AC 

L1 L3

 A1P

L1AC  19.2159


Cognate #2

Cognate #3

Ground link length

L1  9.500

L1AC  19.216

L1BC  15.979

Crank length

L2  5.000

L10  8.900

L7  8.410

Coupler length

L3  4.400

L9  10.114

L6  8.410


B2


Rocker length

L4  5.000

L8  10.114

L5  7.401

Coupler point

A1P  8.900

A2P  5.000

A3P  5.000

Coupler angle

δ  56.000 deg

δ  56.000 deg

δ  94.470 deg

OC

8

7

B3

P 9 6 A3

A2

5

3 10

P

B1 4

A1 2

OB 1

OA

3

4

A1

4.

5.

6.

All three of these cognates are non-Grashof and will, therefore, have limited motion. However, following Example 3-11, discard cognate #2 and retain cognates #1 and #3. Draw line qq parallel to line OAOC and through point OB. Without allowing links 5, 6, and 7 to rotate, slide them as an assembly along lines OAOC and qq until the free end of link 7 is at OA. The free end of link 5 will then be at point O'B and point P on link 6 will be at P'. Add a new link of length OAOC between P P' and P'. This is the new output link 8 and all points on it describe the original coupler curve. Join links 2 and 7, making one ternary link. Remove link 5 and reduce link 6 to a binary link. The result is a Watt-I sixbar with links numbered 1, 2, 3, 4, 6, and 8 (see next page). Link 8 is in curvilinear translation and follows the coupler path of the original point P.

B1 q

2

OB 1

OA 7

B3

6 A3 5

Add a driver dyad following Example 3-4.

q

O'B



P

B1 3

4

A1 8 2

OB

1

OA

B3 P'

6

P

B1 3

4

A1 8 2

OB

1

OA

P'

6

B3

OC




Design a pin-jointed linkage that will guide the forks of the fork lift truck in Figure P3-5 up and down in an approximate straight line over the range of motion shown. Arrange the fixed pivots so they are close to some part of the existing frame or body of the truck.

Given:

Length of straight line motion of the forks: Δx  1800 mm

Solution:


Design choices: Use a Hoeken-type straight line mechanism optimized for straightness. Maximum allowable error in straightness of line: ΔCy  0.096  % 1.

Using Table 3-1 and the required length of straight-line motion, determine the link lengths. Link ratios from Table 3-1 for ΔCy  0.096 %: L1overL2  2.200 L3overL2  2.800

ΔxoverL2  4.181

Link lengths:

2.

L2 

Coupler

L3  L3overL2  L2

L3  1205.5 mm

Ground link


L1  947.1 mm

Rocker

L4  L3

L4  1205.5 mm

Coupler point

BP  L3

BP  1205.5 mm

L2  430.5 mm

ΔxoverL2

Calculate the distance from point P to pivot O4 (Cy). Cy 

3.

Δx

Crank

2 L32   L1  L22

Cy  1978.5 mm

Draw the fork lift truck to scale with the mechanism defined in step 1 superimposed on it..

1978.5mm

O4

P

620.0mm

4 947.1mm B

900.0mm

487.1mm 3

O2 2 A




Figure P3-6 shows a "V-link" off-loading mechanism for a paper roll conveyor. Design a pinjointed linkage to replace the air cylinder driver that will rotate the rocker arm and V-link through the 90 deg motion shown. Keep the fixed pivots as close to the existing frame as possible. Your fourbar linkage should be Grashof and be in toggle at each extreme position of the rocker arm.

Given:

Dimensions scaled from Figure P3-6: Rocker arm (link 4) distance between pin centers:

Solution:

L4  320  mm


Design choices: 1. Use the same rocker arm that was used with the air cylinder driver. 2. Place the pivot O2 80 mm to the right of the right leg and on a horizontal line with the center of the pin on the rocker arm. 3. Design for two-position, 90 deg of output rocker motion with no quick return, similar to Example 3-2. 1.

Draw the rocker arm (link 4) O4B in both extreme positions, B1 and B2, in any convenient location such that the desired angle of motion 4 is subtended. In this solution, link 4 is drawn such that the two extreme positions each make an angle of 45 deg to the vertical.

2.

Draw the chord B1B2 and extend it in any convenient direction. In this solution it was extended horizontally to the left.

3.

Mark the center O2 on the extended line such that it is 80 mm to the right of the right leg. This will allow sufficient space for a supporting pillow block bearing.

4.

Bisect the line segment B1B2 and draw a circle of that radius about O2.

5.

Label the two intersections of the circle and extended line B1B2, A1 and A2.

6.

Measure the length of the coupler (link 3) as A1B1 or A2B2. From the graphical solution, L3  1045 mm

7.

Measure the length of the crank (link 2) as O2A1 or O2A2. From the graphical solution, L2  226.274  mm

8.

Measure the length of the ground link (link 1) as O2O4. From the graphical solution, L1  1069.217 mm 1045.000 80.000 1069.217 320.000

1 4 3

1045.000

9.

Find the Grashof condition.

2

226.274









Figure P3-7 shows a walking-beam transport mechanism that uses a fourbar coupler curve, replicated with a parallelogram linkage for parallel motion. Note the duplicate crank and coupler shown ghosted in the right half of the mechanism - they are redundant and have been removed from the duplicate fourbar linkage. Using the same fourbar driving stage (links 1, 2, 3, 4 with coupler point P), design a Watt-I sixbar linkage that will drive link 8 in the same parallel motion using two fewer links.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  2.22

Crank

L2  1

Coupler

L3  2.06

Rocker

L4  2.33

A1P  3.06



 

2

0.5

B1P  1.674

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

δ  31.000 deg

γ  109.6560 deg


L5  B1P

L5  1.674

L6 

L10  A1P

L10  3.060

L9 

L7  0.812

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  1.893

 A1P

L9  1.485

A1P

L8  3.461

B1P


A3P  3.461





δ  180  deg  δ  γ 

A2P  L2

A2P  1.000

δ  δ

δ  31.000 deg

δ  39.344 deg From the Roberts diagram, calculate the ground link lengths for cognates #2 and #3 L1BC 

3.

L1 L3

 B1P

L1BC  1.8035

L1AC 

L1 L3

 A1P

L1AC  3.2977

Using the calculated link lengths, draw the Roberts diagram (see next page). SUMMARY OF COGNATE SPECIFICATIONS: Cognate #1 Ground link length

L1  2.220

Cognate #2

Cognate #3

L1AC  3.298

L1BC  1.804



Crank length

L2  1.000

L10  3.060

L7  0.812

Coupler length

L3  2.060

L9  1.485

L6  1.893

Rocker length

L4  2.330

L8  3.461

L5  1.674

Coupler point

A1P  3.060

A2P  1.000

A3P  3.461

Coupler angle

δ  31.000 deg

δ  31.000 deg

δ  39.344 deg

OC 7 6

B3

A3

8

5

OB

9

4

1

B2

A2

10

P

OA 2 A1 3 B1

4.

Determine the Grashof condition of each of the two additional cognates. Condition( a b c d ) 


5.

Cognate #2:

Condition L8 L9 L10 L1AC  "Grashof"

Cognate #3:

Condition L5 L6 L7 L1BC  "Grashof"

Both of these cognates are Grashof but cognate #3 is a crank rocker. Following Example 3-11, discard cognate #2 and retain cognates #1 and #3. Draw line qq parallel to line OAOC and through point OB. Without allowing links 5, 6, and 7 to rotate, slide them as an assembly along lines OAOC and qq until the free end of link 7 is at OA. The free end of link 5 will then be at point O'B and point P on link 6 will be at P'. Add a new link of length OAOC between P and P'. This is the new output link 8 and all points on it describe the original coupler curve.



q OB 1 4 P

OA

2

7

B3

6 A3 A1

3 B1 8

5 O'B q P'

6.

Join links 2 and 7, making one ternary link. Remove link 5 and reduce link 6 to a binary link. The result is a Watt-I sixbar with links numbered 1, 2, 3, 4, 6, and 8 (see next page). Link 8 is in curvilinear translation and follows the coupler path of the original point P. The walking-beam (link 8 in Figure P3-7) is rigidly attached to link 8 below.

OB 1 4 P

OA

2

A3 A1 3 B1

6

8

P'




Find the maximum and minimum transmission angles of the fourbar driving stage (links L1, L2, L3, L4) in Figure P3-7 (to graphical accuracy).

Given:

Link lengths:

Link 2

L2  1.00

Link 3

L3  2.06

Link 4

L4  2.33

Link 1

L1  2.22



Solution: 1.



2.

Condition L1 L2 L3 L4  "Grashof" Class I-2, Grashof crank-rocker-rocker, GCRR, since the shortest link is the input link.

It can be shown (see Section 4.10) that the minimum transmission angle for a fourbar GCRR linkage occurs when links 2 and 1 (ground link) are colinear. Draw the linkage in these two positions and measure the transmission angles. O4

O4

A O2

31.510°

O2 A 85.843°

B

3.

B

As measured from the layout, the minimum transmission angle is 31.5 deg. The maximum is 90 deg.




Figure P3-8 shows a fourbar linkage used in a power loom to drive a comb-like reed against the thread, "beating it up" into the cloth. Determine its Grashof condition and its minimum and maximum transmission angles to graphical accuracy.

Given:

Link lengths:

Link 2

L2  2.00 in

Link 3

L3  8.375  in

Link 4

L4  7.187  in

Link 1

L1  9.625  in



Solution: 1.



2.

Condition L1 L2 L3 L4  "Grashof" Class I-2, Grashof crank-rocker-rocker, GCRR, since the shortest link is the input link.

It can be shown (see Section 4.10) that the minimum transmission angle for a fourbar GCRR linkage occurs when links 2 and 1 (ground link) are colinear. Draw the linkage in these two positions and measure the transmission angles.

83.634°

58.078°

3.

As measured from the layout, the minimum transmission angle is 58.1 deg. The maximum is 90.0 deg.




Draw the Roberts diagram and find the cognates for the linkage in Figure P3-9.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  2.22

Crank

L2  1.0

Coupler

L3  2.06

Rocker

L4  2.33

A1P  3.06



 

2

0.5

B1P  1.674

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

δ  31.00  deg

γ  109.6560 deg


L5  B1P

L5  1.674

L6 

L10  A1P

L10  3.060

L9 

L7  0.812

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  1.893

 A1P

L9  1.485

A1P

L8  3.461

B1P


A3P  3.461

δ  180  deg  δ  γ

δ  39.344 deg

A2P  L2

A2P  1.000

δ  δ

δ  31.000 deg


L1 L3

 B1P

L1BC  1.8035

L1AC 

L1 L3

 A1P

L1AC  3.2977


Cognate #2

Cognate #3

Ground link length

L1  2.220

L1AC  3.298

L1BC  1.804

Crank length

L2  1.000

L10  3.060

L7  0.812

Coupler length

L3  2.060

L9  1.485

L6  1.893

Rocker length

L4  2.330

L8  3.461

L5  1.674

Coupler point

A1P  3.060

A2P  1.000

A3P  3.461

Coupler angle

δ  31.000 deg

δ  31.000 deg

δ  39.344 deg



B1 P B2 3

2

9

4

A1

A2

10

8 OB

5

1

OA

1BC 1AC

B3 6

A3

7

OC




Find the equivalent geared fivebar mechanism cognate of the linkage in Figure P3-9.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  2.22

Crank

L2  1.0

Coupler

L3  2.06

Rocker

L4  2.33

A1P  3.06



 

2

0.5

B1P  1.674

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

δ  31.00  deg

γ  109.6560 deg


L5  B1P

L5  1.674

L6 

L10  A1P

L10  3.060

L9 

L7  0.812

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  1.893

 A1P

L9  1.485

A1P

L8  3.461

B1P


A3P  3.461

δ  180  deg  δ  γ

δ  39.344 deg

A2P  L2

A2P  1.000

δ  δ

δ  31.000 deg


3.

L1 L3

 B1P

L1BC  1.8035

L1AC 

L1 L3

 A1P

L1AC  3.2977


Cognate #2

Cognate #3

Ground link length

L1  2.220

L1AC  3.298

L1BC  1.804

Crank length

L2  1.000

L10  3.060

L7  0.812

Coupler length

L3  2.060

L9  1.485

L6  1.893

Rocker length

L4  2.330

L8  3.461

L5  1.674



Coupler point

A1P  3.060

A2P  1.000

A3P  3.461

Coupler angle

δ  31.000 deg

δ  31.000 deg

δ  39.344 deg

B1 P B2 3

2

9

4

A1

A2

10

8 OB

5

1

OA

1BC 1AC

B3 6

A3

4.

7

OC

The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAA2PB3OB, OAA1PA3OC, and OBB1PB2OC. The three geared fivebar cognates are summarized in the table below. SUMMARY OF GEARED FIVEBAR COGNATE SPECIFICATIONS: Cognate #1

Cognate #2

Cognate #3

Ground link length

L1  2.220

L1AC  3.298

L1BC  1.804

Crank length

L10  3.060

L2  1.000

L4  2.330

Coupler length

A2P  1.000

A1P  3.060

L5  1.674

Rocker length

L4  2.330

L8  3.461

L7  0.812

Crank length

L5  1.674

L7  0.812

L8  3.461

Coupler point

A2P  1.000

A1P  3.060

B1P  1.674

Coupler angle

δ  0.00 deg

δ  0.00 deg

δ  0.00 deg

5.

Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path (see next page)

6.







Use the linkage in Figure P3-9 to design an eightbar double-dwell mechanism that has a rocker output through 45 deg.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  2.22

Crank

L2  1.0

Coupler

L3  2.06

Rocker

L4  2.33

A1P  3.06

δ  31.00  deg


Enter the given data into program FOURBAR and print out the resulting coupler point coordinates (see table below). FOURBAR for Windows

File P03-26.DAT

Angle Step Deg


0.000 10.000 20.000 30.000 40.000 50.000 60.000 70.000 80.000 90.000 100.000 110.000 120.000 130.000 140.000 150.000 160.000 170.000 180.000 190.000 200.000 210.000 220.000 230.000 240.000 250.000 260.000 270.000 280.000 290.000 300.000 310.000 320.000 330.000 340.000 350.000 360.000

2.731 3.077 3.350 3.515 3.576 3.554 3.473 3.350 3.203 3.040 2.872 2.706 2.548 2.403 2.274 2.164 2.075 2.005 1.953 1.917 1.892 1.875 1.862 1.848 1.832 1.810 1.784 1.754 1.723 1.698 1.687 1.702 1.761 1.883 2.088 2.380 2.731

2.523 2.407 2.228 2.032 1.855 1.708 1.592 1.499 1.420 1.348 1.278 1.207 1.135 1.062 0.990 0.925 0.869 0.826 0.802 0.798 0.817 0.860 0.925 1.011 1.115 1.235 1.367 1.508 1.654 1.804 1.955 2.105 2.251 2.386 2.494 2.550 2.523

3.718 3.906 4.023 4.060 4.028 3.943 3.820 3.671 3.503 3.326 3.144 2.963 2.789 2.627 2.480 2.354 2.249 2.168 2.111 2.076 2.061 2.063 2.079 2.107 2.145 2.192 2.248 2.313 2.388 2.477 2.582 2.707 2.858 3.040 3.253 3.488 3.718

42.731 38.029 33.626 30.035 27.412 25.672 24.635 24.107 23.915 23.915 23.988 24.039 24.001 23.834 23.533 23.134 22.719 22.404 22.326 22.614 23.365 24.632 26.417 28.678 31.340 34.306 37.463 40.683 43.826 46.730 49.207 51.038 51.965 51.715 50.064 46.967 42.731



2.

Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Fit tangent lines to the nearly straight portions of the curve. Label their intersection O6.

3.

Design link 6 to lie along these straight tangents, pivoted at O6. Provide a guide on link 6 to accommodate slider block 5, which pivots on the coupler point P. O8

8 45.000° F

E D

7

B 70.140°

C 6

3 P 5 A O6 2

4

O4

O2

4.

Extend link 6 a convenient distance to point C. Draw an arc through point C with center at O6. Label the intersection of the arc with the other tangent line as point D. Attach link 7 to the pivot at C. The length of link 7 is CE, a design choice. Extend line CDE from point E a distance equal to CD. Label the end point F. Layout two intersecting lines through E and F such that they subtend an angle of 45 deg. Label their intersection O8. The link joining O8 and point E is link 8. The link lengths and locations of O6 and O8 are: Link 6

L6  2.330

Fixed pivot O6:

Link 7

x  1.892 y  0.762

L7  3.000 Fixed pivot O8:

Link 8 x  1.379 y  6.690

L8  3.498




Use the linkage in Figure P3-9 to design an eightbar double-dwell mechanism that has a slider output stroke of 5 crank units.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  2.22

Crank

L2  1.0

Coupler

L3  2.06

Rocker

L4  2.33

A1P  3.06

δ  31.00  deg


Enter the given data into program FOURBAR and print out the resulting coupler point coordinates (see table below). FOURBAR for Windows

File P03-26.DAT

Angle Step Deg


0.000 10.000 20.000 30.000 40.000 50.000 60.000 70.000 80.000 90.000 100.000 110.000 120.000 130.000 140.000 150.000 160.000 170.000 180.000 190.000 200.000 210.000 220.000 230.000 240.000 250.000 260.000 270.000 280.000 290.000 300.000 310.000 320.000 330.000 340.000 350.000 360.000

2.731 3.077 3.350 3.515 3.576 3.554 3.473 3.350 3.203 3.040 2.872 2.706 2.548 2.403 2.274 2.164 2.075 2.005 1.953 1.917 1.892 1.875 1.862 1.848 1.832 1.810 1.784 1.754 1.723 1.698 1.687 1.702 1.761 1.883 2.088 2.380 2.731

2.523 2.407 2.228 2.032 1.855 1.708 1.592 1.499 1.420 1.348 1.278 1.207 1.135 1.062 0.990 0.925 0.869 0.826 0.802 0.798 0.817 0.860 0.925 1.011 1.115 1.235 1.367 1.508 1.654 1.804 1.955 2.105 2.251 2.386 2.494 2.550 2.523

3.718 3.906 4.023 4.060 4.028 3.943 3.820 3.671 3.503 3.326 3.144 2.963 2.789 2.627 2.480 2.354 2.249 2.168 2.111 2.076 2.061 2.063 2.079 2.107 2.145 2.192 2.248 2.313 2.388 2.477 2.582 2.707 2.858 3.040 3.253 3.488 3.718

42.731 38.029 33.626 30.035 27.412 25.672 24.635 24.107 23.915 23.915 23.988 24.039 24.001 23.834 23.533 23.134 22.719 22.404 22.326 22.614 23.365 24.632 26.417 28.678 31.340 34.306 37.463 40.683 43.826 46.730 49.207 51.038 51.965 51.715 50.064 46.967 42.731



2.

Layout this linkage to scale, including the coupler curve whose coordinates are in the table above. Fit tangent lines to the nearly straight portions of the curve. Label their intersection O6.

3.

Design link 6 to lie along these straight tangents, pivoted at O6. Provide a guide on link 6 to accommodate slider block 5, which pivots on the coupler point P. F

E

8

D 7

C

B 6 70.140° 3 P 5 A O6 2

4

O4

O2

4.

Extend link 6 and the other tangent line until points C and E are 5 units apart. Attach link 7 to the pivot at C. The length of link 7 is CD, a design choice. Extend line CDE from point D a distance equal to CE. Label the end point F. As link 6 travels from C to E, slider block 8 will travel from D to F, a distance of 5 units. The link lengths and location of O6: Link 6

L6  4.351

Fixed pivot O6:

Link 7

x  1.892 y  0.762

L7  2.000




Use two of the cognates in Figure 3-26 (p. 126) to design a Watt-I sixbar parallel motion mechanism that carries a link through the same coupler curve at all points. Comment on its similarities to the original Roberts diagram.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  45

Crank

L2  56

Coupler

L3  22.5

Rocker

L4  56

A1P  11.25 δ  0.000  deg



 

2

0.5

B1P  11.250

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

γ  0.0000 deg


L5  B1P

L5  11.250

L6 

L10  A1P

L10  11.250

L9 

L7  28.000

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  28.000

 A1P

L9  28.000

A1P

L8  28.000

B1P


A3P  56.000

A2P  L2

A2P  56.000

δ  δ

δ  0.000 deg

δ  δ

δ  0.000 deg


3.

L1 L3

 B1P

L1BC  22.5000

L1AC 

L1 L3

 A1P

L1AC  22.5000


Cognate #2

Cognate #3

Ground link length

L1  45.000

L1AC  22.500

L1BC  22.500

Crank length

L2  56.000

L10  11.250

L7  28.000

Coupler length

L3  22.500

L9  28.000

L6  28.000



Rocker length

L4  56.000

L8  28.000

L5  11.250

Coupler point

A1P  11.250

A2P  56.000

A3P  56.000

Coupler angle

δ  0.000 deg

δ  0.000 deg

δ  0.000 deg

P

3

B1

A1

B2

B3

6 9

8

2

A2

7

5

10 OA

4.

4

A3

OB

OC

Both of these cognates are identical. Following Example 3-11, discard cognate #2 and retain cognates #1 and #3. Without allowing links 5, 6, and 7 to rotate, slide them as an assembly along line OAOC until the free end of link 7 is at OA. The free end of link 5 will then be at point O'B and point P on link 6 will be at P'. Add a new link of length OAOC between P and P'. This is the new output link 8 and all points on it describe the original coupler curve. P'

B1

8

3

P

A1

B3

6

7 2

4

5 OA

O'B

A3 OB


5.


Join links 2 and 7, making one ternary link. Remove link 5 and reduce link 6 to a binary link. The result is a Watt-I sixbar with links numbered 1, 2, 3, 4, 6, and 8. Link 8 is in curvilinear translation and follows the coupler path of the original point P. Link 8 is a binary link with nodes at P and P'. It does not attach to link 4 at B1.

P'

8

B1

3

P

A1

6

B3

2

OA

4

OB




Find the cognates of the Watt straight-line mechanism in Figure 3-29a (p. 131).

Given:

Link lengths:

Solution:

Coupler point data:

Ground link

L1  4

Crank

L2  2

A1P  0.500

δ  0.00 deg

Coupler

L3  1

Rocker

L4  2

B1P  0.500

γ  0.00 deg


1.

Input the link dimensions and coupler point data into program FOURBAR.

2.

Use the Cognate pull-down menu to get the link lengths for cognates #2 and #3 (see next page). Note that, for this mechanism, cognates #2 and #3 are identical. All three mechanisms are non-Grashof with limited crank angles.






Find the cognates of the Roberts straight-line mechanism in Figure 3-29b.

Given:

Link lengths:

Solution:

Coupler point data:

Ground link

L1  2

Crank

L2  1

A1P  1.000

δ  60.0 deg

Coupler

L3  1

Rocker

L4  1

B1P  1.000

γ  60.0 deg


1.

Input the link dimensions and coupler point data into program FOURBAR.

2.

Note that, for this mechanism, cognates #2 and #3 are identical with cognate #1 because of the symmetry of the linkage (draw the Cayley diagram to see this). All three mechanisms are non-Grashof with limited crank angles.




Design a Hoeken straight-line linkage to give minimum error in velocity over 22% of the cycle for a 15-cm-long straight line motion. Specify all linkage parameters.

Given:

Length of straight line motion: Δx  150  mm Percentage of cycle over which straight line motion takes place: 22%

Solution:


1.

Using Table 3-1 and the required length of straight-line motion, determine the link lengths. Link ratios from Table 3-1 for 22% cycle: L1overL2  1.975

L3overL2  2.463


Link lengths:

2.

L2 

Coupler


L3  200.24 mm

Ground link


L1  160.57 mm

Rocker

L4  L3

L4  200.24 mm

Coupler point

AP  2  L3

AP  400.49 mm

L2  81.30 mm

ΔxoverL2

Calculate the distance from point P to pivot O4 (Cy) when crank angle is 180 deg. Cy 

3.

Δx

Crank

2 L32   L1  L22

Cy  319.20 mm

Enter the link lengths into program FOURBAR to verify the design (see next page for coupler point curve). Using the PRINT facility, determine the x,y coordinates of the coupler curve and the x,y components of the coupler point velocity in the straight line region. A table of these values is printed below. Notice the small deviations over the range of crank angles from the y-coordinate and the x-velocity at a crank angle of 180 deg.

FOURBAR for Windows

File P03-31.DOC

Angle Step Deg

Cpler Pt X mm

Cpler Pt Y mm

Veloc CP X mm/sec

Veloc CP Y mm/sec

140 150 160 170 180 190 200 210 220

235.60 216.84 198.06 179.31 160.58 141.85 123.09 104.31 85.55

319.95 319.72 319.46 319.27 319.20 319.27 319.47 319.72 319.95

-1,072.61 -1,076.20 -1,075.51 -1,073.75 -1,072.93 -1,073.75 -1,075.52 -1,076.22 -1,072.63

-14.74 -13.54 -7.99 0.02 8.03 13.58 14.78 10.76

-10.73






Design a Hoeken straight-line linkage to give minimum error in straightness over 39% of the cycle for a 20-cm-long straight line motion. Specify all linkage parameters.

Given:

Length of straight line motion: Δx  200  mm Percentage of cycle over which straight line motion takes place: 39%

Solution:


1.

Using Table 3-1 and the required length of straight-line motion, determine the link lengths. Link ratios from Table 3-1 for 39% cycle: L1overL2  2.500

L3overL2  3.250


Δx ΔxoverL2

L2  55.20 mm

Link lengths:

2.

Crank

L2 

Coupler


L3  179.41 mm

Ground link


L1  138.01 mm

Rocker

L4  L3

L4  179.41 mm

Coupler point

AP  2  L3

AP  358.82 mm

Calculate the distance from point P to pivot O4 (Cy) when crank angle is 180 deg. Cy 

3.

2 L32   L1  L22

Cy  302.36 mm

Enter the link lengths into program FOURBAR to verify the design (see next page for coupler point curve). Using the PRINT facility, determine the x,y coordinates of the coupler curve and the x,y components of the coupler point velocity in the straight line region. A table of these values is printed below. Notice the small deviations over the range of crank angles from the y-coordinate and the x-velocity from a crank angle of 180 deg. FOURBAR for Windows

File P03-32.DAT

Angle Step Deg

Coupler Pt X mm

Coupler Pt Y mm

Veloc CP X mm/sec

110 120 130 140 150 160 170 180 190 200 210 220 230 240 250

237.992 225.289 211.710 197.521 182.927 168.076 153.076 138.010 122.944 107.944 93.093 78.499 64.311 50.731 38.028

302.408 302.361 302.378 302.398 302.399 302.385 302.368 302.360 302.368 302.385 302.399 302.398 302.378 302.361 302.408

-696.591 -755.847 -797.695 -826.217 -844.774 -856.043 -861.994 -863.841 -861.994 -856.043 -844.774 -826.217 -797.695 -755.847 -696.591

Veloc CP Y mm/sec -6.416 -0.019 1.426 0.664 -0.483 -1.052 -0.800 0.000 0.800 1.052 0.483 -0.664 -1.426 0.019 6.416






Design a linkage that will give a symmetrical "kidney bean" shaped coupler curve as shown in Figure 3-16 (p. 114 and 115). Use the data in Figure 3-21 (p. 120) to determine the required link ratios and generate the coupler curve with program FOURBAR.

Solution:


Design choices: Ground link ratio, L1/L2 = 2.0: GLR  2.0 Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.5: CLR  2.5 Coupler angle, γ  72 deg Crank length, L2  2.000 1.


L3  CLR L2

L3  5.000


L4  CLR L2

L4  5.000


L1  GLR  L2

L1  4.000

Angle PAB

δ 


180  deg  γ 2

AP  2  L3 cos δ

δ  54.000 deg AP  5.878

Enter the above data into program FOURBAR and plot the coupler curve.




Design a linkage that will give a symmetrical "double straight" shaped coupler curve as shown in Figure 3-16. Use the data in Figure 3-21 to determine the required link ratios and generate the coupler curve with program FOURBAR.

Solution:


Design choices: Ground link ratio, L1/L2 = 2.5: GLR  2.5 Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.5: CLR  2.5 Coupler angle, γ  252  deg Crank length, L2  2.000 1.


L3  CLR L2

L3  5.000


L4  CLR L2

L4  5.000


L1  GLR  L2

L1  5.000

Angle PAB

δ 


180  deg  γ 2

AP  2  L3 cos δ

δ  36.000 deg AP  8.090





Design a linkage that will give a symmetrical "scimitar" shaped coupler curve as shown in Figure 3-16. Use the data in Figure 3-21 to determine the required link ratios and generate the coupler curve with program FOURBAR. Show that there are (or are not) true cusps on the curve.

Solution:


Design choices: Ground link ratio, L1/L2 = 2.0: GLR  2.0 Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.5: CLR  2.5 Coupler angle, γ  144  deg Crank length, L2  2.000 1.


L3  CLR L2

L3  5.000


L4  CLR L2

L4  5.000


L1  GLR  L2

L1  4.000

Angle PAB

δ 


180  deg  γ 2

AP  2  L3 cos δ

δ  18.000 deg AP  9.511



3.


The points at the ends of the "scimitar" will be true cusps if the velocity of the coupler point is zero at these points. Using FOURBAR's plotting utility, plot the magnitude and angle of the coupler point velocity vector. As seen below for the range of crank angle from 50 to 70 degrees, the magnitude of the velocity does not quite reach zero. Therefore, these are not true cusps.




Find the Grashof condition, inversion, any limit positions, and the extreme values of the transmission angle (to graphical accuracy) of the linkage in Figure P3-10.

Given:

Link lengths:

Link 2

L2  0.785

Link 3

L3  0.356

Link 4

L4  0.950

Link 1

L1  0.544


S  min ( a b c d ) L  max( a b c d ) SL  S  L PQ  a  b  c  d  SL return "Grashof" if SL  PQ return "Special Grashof" if SL = PQ

Solution: 1.

return "non-Grashof" otherwise See Figure P3-10 and Mathcad file P0336.

Determine the Grashof condition of the mechanism from inequality 2.8 and its Barker classification from Table 2-4. Condition L1 L2 L3 L4  "Grashof"

Grashof condition: Barker classification:

2.

Class I-3, Grashof rocker-crank-rocker, GRCR, since the shortest link is the coupler link.

A GRCR linkage will have two toggle positions. Draw the linkage in these two positions and measure the input link angles. A

B

158.286° O4

O2

O4

O2 B A

158.286°

3.

As measured from the layout, the input link angles at the toggle positions are: +158.3 and -158.3 deg.

4.

Since the coupler link in a GRCR linkage can make a full rotation with respect to the input and output rockers, the minimum transmission angle is 0 deg and the maximum is 90 deg.





Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  0.544 Crank

L2  0.785

Coupler

L3  0.356 Rocker

L4  0.950

δ  0.00 deg



 

2

0.5

B1P  0.734

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

A1P  1.09

γ  180.0000 deg


L5  B1P

L5  0.734

L6 

L10  A1P

L10  1.090

L9 

L7  1.619

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  1.959

 A1P

L9  2.404

A1P

L8  2.909

B1P


A3P  0.950

A2P  L2

A2P  0.785

δ  180  deg  δ

δ  180.000 deg

δ  δ

δ  0.000 deg


3.

L1 L3

 B1P

L1BC  1.1216

L1AC 

L1 L3

 A1P

L1AC  1.6656


Cognate #2

Cognate #3

Ground link length

L1  0.544

L1AC  1.666

L1BC  1.122

Crank length

L2  0.785

L10  1.090

L7  1.619

Coupler length

L3  0.356

L9  2.404

L6  1.959



Rocker length

L4  0.950

L8  2.909

L5  0.734

Coupler point

A1P  1.090

A2P  0.785

A3P  0.950

Coupler angle

δ  0.000 deg

δ  0.000 deg

δ  180.000 deg

B2

9

8

P B1 A1

3 4

A2

2

5

10 OA

1

A3 OC

OB 6

7

B3




Find the three geared fivebar cognates of the linkage in Figure P3-10.

Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  0.544 Crank

L2  0.785

Coupler

L3  0.356 Rocker

L4  0.950

δ  0.00 deg



 

2

0.5

B1P  0.734

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

A1P  1.09

γ  180.0000 deg


L5  B1P

L5  0.734

L6 

L10  A1P

L10  1.090

L9 

L7  1.619

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  1.959

 A1P

L9  2.404

A1P

L8  2.909

B1P


A3P  0.950

A2P  L2

A2P  0.785

δ  180  deg  δ

δ  180.000 deg

δ  δ

δ  0.000 deg


L1 L3

 B1P

L1BC  1.1216

L1AC 

L1 L3

 A1P

L1AC  1.6656


Cognate #2

Cognate #3

Ground link length

L1  0.544

L1AC  1.666

L1BC  1.122

Crank length

L2  0.785

L10  1.090

L7  1.619

Coupler length

L3  0.356

L9  2.404

L6  1.959

Rocker length

L4  0.950

L8  2.909

L5  0.734

Coupler point

A1P  1.090

A2P  0.785

A3P  0.950



δ  0.000 deg

Coupler angle

δ  0.000 deg

δ  180.000 deg

B2

9

8

P B1 A1

3 4

A2

2

5

10 OA

1

A3 OC

OB 6

7

B3 4.

The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAA2PA3OB, OAA1PB3OC, and OBB1PB2OC. They are specified in the summary table below. SUMMARY OF GEARED FIVEBAR COGNATE SPECIFICATIONS: Cognate #1

Cognate #2

Cognate #3

Ground link length

L1  0.544

L1AC  1.666

L1BC  1.122

Crank length

L10  1.090

L2  0.785

L4  0.950

Coupler length

A2P  0.785

A1P  1.090

L5  0.734

Rocker length

A3P  0.950

L8  2.909

L7  1.619

Crank length

L5  0.734

L7  1.619

L8  2.909

Coupler point

A2P  0.785

A1P  1.090

B1P  0.734

Coupler angle

δ  0.00 deg

δ  0.00 deg

δ  0.00 deg

5.

Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path (see next page).

6.







Find the Grashof condition, any limit positions, and the extreme values of the transmission angle (to graphical accuracy) of the linkage in Figure P3-11.

Given:

Link lengths:

Link 2

L2  0.86

Link 3

L3  1.85

Link 4

L4  0.86

Link 1

L1  2.22



Solution: 1.


Determine the Grashof condition of the mechanism from inequality 2.8 and its Barker classification from Table 2-4. Condition L1 L2 L3 L4  "non-Grashof"


2.

Class II-1, non-Grashof triple rocker, RRR1, since the longest link is the ground link.

An RRR1 linkage will have two toggle positions. Draw the linkage in these two positions and measure the input link angles. 116.037° A B

O4

O2

O4

O2 B A 116.037°

88.2° B

A 67.3° O2

O4

3.

As measured from the layout, the input link angles at the toggle positions are: +116 and -116 deg.

4.

Since the coupler link in an RRR1 linkage cannot make a full rotation with respect to the input and output rockers, the minimum transmission angle is 0 deg and the maximum is 88 deg.





Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  2.22

Crank

L2  0.86

Coupler

L3  1.85

Rocker

L4  0.86

δ  0.00 deg



 

2

0.5

B1P  0.520

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

A1P  1.33

γ  0.0000 deg


L5  B1P

L5  0.520

L6 

L10  A1P

L10  1.330

L9 

L7  0.242

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  0.242

 A1P

L9  0.618

A1P

L8  0.618

B1P


A3P  0.618

A2P  L7

A2P  0.242

δ  180  deg

δ  180.000 deg

δ  180  deg

δ  180.000 deg


L1 L3

 B1P

L1BC  0.6240

L1AC 

L1 L3

 A1P

L1AC  1.5960


Cognate #2

Cognate #3

Ground link length

L1  2.220

L1AC  1.596

L1BC  0.624

Crank length

L2  0.860

L10  1.330

L7  0.242

Coupler length

L3  1.850

L9  0.618

L6  0.242

Rocker length

L4  0.860

L8  0.618

L5  0.520

Coupler point

A1P  1.330

A2P  0.242

A3P  0.618

Coupler angle

δ  0.000 deg

δ  180.000 deg

δ  180.000 deg



B2

B1

9 A2

3

10

P

4

8

OC

OA

A3

7 6

2 A1

OB 5

B3





Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  2.22

Crank

L2  0.86

Coupler

L3  1.85

Rocker

L4  0.86

δ  0.00 deg



 

2

0.5

B1P  0.520

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

A1P  1.33

γ  0.0000 deg


L5  B1P

L5  0.520

L6 

L10  A1P

L10  1.330

L9 

L7  0.242

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  0.242

 A1P

L9  0.618

A1P

L8  0.618

B1P


A3P  0.618

A2P  L7

A2P  0.242

δ  180  deg

δ  180.000 deg

δ  180  deg

δ  180.000 deg


3.

L1 L3

 B1P

L1BC  0.6240

L1AC 

L1 L3

 A1P

L1AC  1.5960


Cognate #2

Cognate #3

Ground link length

L1  2.220

L1AC  1.596

L1BC  0.624

Crank length

L2  0.860

L10  1.330

L7  0.242

Coupler length

L3  1.850

L9  0.618

L6  0.242

Rocker length

L4  0.860

L8  0.618

L5  0.520



Coupler point

A1P  1.330

A2P  0.242

A3P  0.618

Coupler angle

δ  0.000 deg

δ  180.000 deg

δ  180.000 deg

B2

B1

9 A2

3

10

P

4

8

OC

OA

A3

7 6

2

5 B3

A1 4.

OB

The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAB2PB3OB, OAA1PA3OC, and OBB1PA2OC. The three geared fivebar cognates are summarized in the table below. SUMMARY OF GEARED FIVEBAR COGNATE SPECIFICATIONS: Cognate #1

Cognate #2

Cognate #3

Ground link length

L1  2.220

L1AC  1.596

L1BC  0.624

Crank length

L10  1.330

L2  0.860

L4  0.860

Coupler length

L2  0.860

A1P  1.330

L5  0.520

Rocker length

L4  0.860

L8  0.618

L7  0.242

Crank length

L5  0.520

L7  0.242

L8  0.618

Coupler point

L2  0.860

A1P  1.330

B1P  0.520

Coupler angle

δ  0.00 deg

δ  0.00 deg

δ  0.00 deg

5.

Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path (see next page)

6.







Find the Grashof condition, any limit positions, and the extreme values of the transmission angle (to graphical accuracy) of the linkage in Figure P3-12.

Given:

Link lengths:

Link 2

L2  0.72

Link 3

L3  0.68

Link 4

L4  0.85

Link 1

L1  1.82



Solution: 1.


Determine the Grashof condition of the mechanism from inequality 2.8 and its Barker classification from Table 2-4. Condition L1 L2 L3 L4  "non-Grashof"


2.

Class II-1, non-Grashof triple rocker, RRR1, since the longest link is the ground link.

An RRR1 linkage will have two toggle positions. Draw the linkage in these two positions and measure the input link angles.

A

55.4° B

O4

O2

O4

O2

B A

55.4°

B 88.8° O4 O2

A

3.

As measured from the layout, the input link angles at the toggle positions are: +55.4 and -55.4 deg.

4.

Since the coupler link in an RRR1 linkage it cannot make a full rotation with respect to the input and output rockers, the minimum transmission angle is 0 deg and the maximum is 88.8 deg.





Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  1.82

Crank

L2  0.72

Coupler

L3  0.68

Rocker

L4  0.85

δ  54.0 deg



 

2

0.5

B1P  0.792

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

A1P  0.97

γ  82.0315 deg


L5  B1P

L5  0.792

L6 

L10  A1P

L10  0.970

L9 

L7  0.839

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  0.990

 A1P

L9  1.027

A1P

L8  1.212

B1P


A3P  0.850

A2P  L2

A2P  0.720

δ  γ

δ  82.032 deg

δ  δ

δ  54.000 deg


L1 L3

 B1P

L1BC  2.1208

L1AC 

L1 L3

 A1P

L1AC  2.5962


Cognate #2

Cognate #3

Ground link length

L1  1.820

L1AC  2.596

L1BC  2.121

Crank length

L2  0.720

L10  0.970

L7  0.839

Coupler length

L3  0.680

L9  1.027

L6  0.990

Rocker length

L4  0.850

L8  1.212

L5  0.792

Coupler point

A1P  0.970

A2P  0.720

A3P  0.850



δ  54.000 deg

Coupler angle

δ  54.000 deg

OC

8 B2

7

9

B3

P 6 A2 1AC 10 3

OA

B1

5 4

A1 2

1BC

A3

1 OB

δ  82.032 deg





Given:

Link lengths:

Solution: 1.

Coupler point data:

Ground link

L1  1.82

Crank

L2  0.72

Coupler

L3  0.68

Rocker

L4  0.85

δ  54.0 deg



 

2

0.5

B1P  0.792

 L32  B1P 2  A1P 2  γ  acos   2  L3 B1P   2.

A1P  0.97

γ  82.0315 deg


L5  B1P

L5  0.792

L6 

L10  A1P

L10  0.970

L9 

L7  0.839

L8  L6

L7  L9

B1P A1P

L3 L2 L3

 B1P

L6  0.990

 A1P

L9  1.027

A1P

L8  1.212

B1P


A3P  0.850

A2P  L2

A2P  0.720

δ  γ

δ  82.032 deg

δ  δ

δ  54.000 deg


3.

L1 L3

 B1P

L1BC  2.1208

L1AC 

L1 L3

 A1P

L1AC  2.5962


Cognate #2

Cognate #3

Ground link length

L1  1.820

L1AC  2.596

L1BC  2.121

Crank length

L2  0.720

L10  0.970

L7  0.839

Coupler length

L3  0.680

L9  1.027

L6  0.990

Rocker length

L4  0.850

L8  1.212

L5  0.792

Coupler point

A1P  0.970

A2P  0.720

A3P  0.850



δ  54.000 deg

Coupler angle

δ  54.000 deg

δ  82.032 deg

OC

8 B2

7

9

B3

P 6 A2 1AC 10 3

B1

5 4

A1 2

1BC

A3

1 OB

OA

4.

The three geared fivebar cognates can be seen in the Roberts diagram. They are: OAA2PA3OB, OAA1PB3OC, and OBB1PB2OC. SUMMARY OF GEARED FIVEBAR COGNATE SPECIFICATIONS: Cognate #1

Cognate #2

Cognate #3

Ground link length

L1  1.820

L1AC  2.596

L1BC  2.121

Crank length

L10  0.970

L2  0.720

L4  0.850

Coupler length

A2P  0.720

A1P  0.970

L5  0.792

Rocker length

A3P  0.850

L8  1.212

L7  0.839

Crank length

L5  0.792

L7  0.839

L8  1.212

Coupler point

A2P  0.720

A1P  0.970

B1P  0.792

Coupler angle

δ  0.00 deg

δ  0.00 deg

δ  0.00 deg

5.

Enter the cognate #1 specifications into program FOURBAR to get a trace of the coupler path (see next page).

6.







Prove that the relationships between the angular velocities of various links in the Roberts diagram as shown in Figure 3-25 (p. 125) are true.

Given:

OAA1PA2, OCB2PB3, and OBB1PA3 are parallelograms for any position of link 2..

Proof: 1.

OAA1 and A2P are opposite sides of a parallelogram and are, therefore, always parallel.

2.

Any change in the angle of OAA1 (link 2) will result in an identical change in the angle of A2P.

3.

Angular velocity is the change in angle per unit time.

4.

Since OAA1 and A2P have identical changes in angle, their angular velocities are identical.

5.

A2P is a line on link 9 and all lines on a rigid body have the same angular velocity. Therefore, link 9 has the same angular velocity as link 2.

6.

OCB3 (link 7) and B2P are opposite sides of a parallelogram and are, therefore, always parallel.

7.

B2P is a line on link 9 and all lines on a rigid body have the same angular velocity. Therefore, link 7 has the same angular velocity as links 9 and 2.

8.

The same argument holds for links 3, 5, and 10; and links 4, 6, and 8.




Design a fourbar linkage to move the object in Figure P3-13 from position 1 to 2 using points A and B for attachment. Add a driver dyad to limit its motion to the range of positions shown, making it a sixbar. All fixed pivots should be on the base.

Given:

Length of coupler link: L3  52.000

Solution:


Design choices: Length of link 2

L2  130

Length of link 2b

L2b  40

L4  110

Length of link 4

1.

Connect the end points of the two given positions of the line AB with construction limes, i.e., lines from A1 to A2 and B1 to B2.

2.

Bisect these lines and extend their perpendicular bisectors into the base.

3.

Select one point on each bisector and label them O2 and O4, respectively. In the solution below the distances O2A was selected to be L2  130.000 and O4B to be L4  110.000 . This resulted in a ground-link-length O2O4 for the fourbar of 27.080.

4.

The fourbar stage is now defined as O2ABO4 with link lengths Ground link 1a

L1a  27.080

Link 3 (coupler)

L3  52.000

Link 2 (input)

L2  130.000

Link 4 (output)

L4  110.000

5.

Select a point on link 2 (O2A) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 2 is now a ternary link with nodes at O2, C, and A.) In the solution below the distance O2C was selected to be L2b  40.000 .

6.

Draw a construction line through C1C2 and extend it to the left.

7.

Select a point on this line and call it O6. In the solution below O6 was placed 20 units from the left edge of the base.

8.

Draw a circle about O6 with a radius of one-half the length C1C2 and label the intersections of the circle with the extended line as D1 and D2. In the solution below the radius was measured as 23.003 units.

A1

B1

3

A2

6

D1

O6

2 D2

C1 5

9.


L6  23.003

Link 5 (coupler) L5  106.866 Link 1b (ground) L1b  111.764 Link 2b (rocker) L2b  40.000

23.003 106.866 111.764

B2

4 C2

40.000 O2

O4 27.080



10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition( a b c d ) 


Condition L1b L2b L5 L6  "Grashof" min  L1b L2b L5 L6  23.003





Given:


Solution:



L2  130

Length of link 4b

L4b  40

L4  225

Length of link 4

1.


2.


3.


4.


L1a  111.758

Link 2 (input)

L2  130.000

Link 3 (coupler)

L3  52.000

Link 4 (output)

L4  225.000

5.

Select a point on link 4 (O4B) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the solution below the distance O4C was selected to be L4b  40.000 .

6.

Draw a construction line through C2C3 and extend it downward.

7.

Select a point on this line and call it O6. In the solution below O6 was placed 20 units from the bottom of the base.

8.

9.


A

2

B 111.758

C3

O4 83.977

Link 5 (coupler) L5  83.977

1b

6 O6

92.425

A

5 D2

10.480

Link 4b (rocker) L4b  40.000

O2

1a

L6  10.480

Link 1b (ground) L1b  92.425

4

C2


3

D3

B









Design a fourbar linkage to move the object in Figure P3-13 through the three positions shown using points A and B for attachment. Add a driver dyad to limit its motion to the range of positions shown, making it a sixbar. All fixed pivots should be on the base.

Given:


Solution:


Design choices: Length of link 4b

L4b  50

1.

Draw link AB in its three design positions A1B1, A2B2, A3B3 in the plane as shown.

2.

Draw construction lines from point A1 to A2 and from point A2 to A3.

3.

Bisect line A1A2 and line A2A3 and extend their perpendicular bisectors until they intersect. Label their intersection O2.

4.

Repeat steps 2 and 3 for lines B1B2 and B2B3. Label the intersection O4.

5.

Connect O2 with A1 and call it link 2. Connect O4 with B1 and call it link 4.

6.

Line A1B1 is link 3. Line O2O4 is link 1 (ground link for the fourbar). The fourbar is now defined as O2ABO4 and has link lengths of Ground link 1a

L1a  20.736

Link 2

L2  127.287

Link 3

L3  52.000

Link 4

L4  120.254

B1

A1 3

A2 4

D3

B2

2

O6

D1

6

5 A3 C3 O2 C1

C2 O4

7.

B3







Select a point on link 4 (O4B) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the solution above the distance O4C was selected to be L4b  50.000 .

9.


10. Select a point on this line and call it O6. In the solution above O6 was placed 20 units from the left edge of the base. 11. Draw a circle about O6 with a radius of one-half the length C1C3 and label the intersections of the circle with the extended line as D1 and D3. In the solution below the radius was measured as L6  45.719. 12. The driver fourbar is now defined as O4CDO6 with link lengths Link 6 (crank)

L6  45.719

Link 5 (coupler) L5  126.875 Link 1b (ground) L1b  128.545 Link 4b (rocker) L4b  50.000 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L4b L5  "Grashof" min  L6 L1b L4b L5  45.719





Given:


Solution:



L2  125

Length of link 2b

L4b  50

Length of link 4

L4  140

1.


2.


3.


4.


L1a  97.195

Link 3 (coupler)

L3  86.000

Link 2 (input)

L2  125.000

Link 4 (output)

L4  140.000

5.


6.


7.


8.

9.


A1

3 2 B1 B2

6 D1

O6

L6  25.808


4

D2

O2 1a

97.195

5


A2

1b

C1

C2

25.808 130.479

137.327

O4



10. Use the link lengths in step 9 to find the Grashof condition of the driving fourbar (it must be Grashof and the shortest link must be link 6). Condition( a b c d ) 







Given:


Solution:



L2  130

Length of link 2b

L2b  50

L4  130

Length of link 4

1.


2.


3.


4.

The fourbar stage is now defined as O2ABO4 with link lengths Ground link 1a Link 3 (coupler)

5.

6. 7.

8.

9.

L1a  67.395 L3  86.000

Link 2 (input)

L2  130.000

Link 4 (output)

L4  130.000

Select a point on link 2 (O2A) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 4 is now a ternary link with nodes at O2, C, and A.) In the solution below the distance O2C was selected to be L2b  50.000 and the link was extended away from A to give a better position for the driving dyad. Draw a construction line through C2C3 and extend it downward. Select a point on this line and call it O6. In the solution below O6 was placed 35 units from the bottom of the base. A2

Draw a circle about O6 with a radius of one-half the length C1C2 and label the intersections of the circle with the extended line as D2 and D3. In the solution below the radius was measured as 24.647 units. The driver fourbar is now defined as O2CDO6 with link lengths Link 6 (crank)

3 C3

155°

107.974

O2

C2

1a

4

67.395

1b 5


Link 2b (rocker) L2b  50.000

B2 A3

L6  24.647

Link 1b (ground) L1b  107.974

2

D3

O4 6

98.822

B3

O6

24.647 D2










Given:


Solution:


Design choices: Length of link 4b

L4b  50

1.


2.


3.


4.


5.


6.


L1a  61.667

Link 2

L2  142.357

Link 3

L3  86.000

Link 4

L4  124.668

A1 A2

3

2 B1

D3

B2 O6

O2 6

D1

4

1b

A3

1a

5 C3

O4

7.

B3 C2

C1







Select a point on link 4 (O4B) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 4 is now a ternary link with nodes at O4, C, and B.) In the solution above the distance O4C was selected to be L4b  50.000 .

9.



L6  45.178

Link 5 (coupler) L5  140.583 Link 1b (ground) L1b  142.205 Link 4b (rocker) L4b  50.000 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L4b L5  "Grashof" 14. Unfortunately, although the solution presented appears to meet the design specification, a simple cardboard model will quickly demonstrate that it has a branch defect. That is, in the first position shown, the linkage is in the "open" configuration, but in the 2nd and 3rd positions it is in the "crossed" configuration. The linkage cannot get from one circuit to the other without removing a pin and reassembling after moving the linkage. The remedy is to attach the points A and B to the coupler, but not at the joints between links 2 and 3 and links 3 and 4.





Given:


Solution:



L2  100

Length of link 4b

L4b  40

L4  160

Length of link 4

1.


2.


3.


4.


L1a  81.463

Link 3 (coupler)

L3  52.000

Link 2 (input)

L2  100.000

Link 4 (output)

L4  160.000

5.


6.


7.


8.


A2

B1

A1 3

B2

2 14.351

9.


L6  14.351


1a

O2 C2

5

O6

1b

C1

6 132.962 O4

Link 1b (ground) L1b  138.105 Link 4b (rocker) L4b  40.000

138.105

81.463

4

D2

D1










Given:


Solution:



L2  150

Length of link 4b

L4b  50

L4  200

Length of link 4

1.


2.


3.

Select one point on each bisector and label them O2 and O4, respectively. In the solution below the distances O2A was selected to be L2  150.000 and O4B to be L4  200.000 . This resulted in a ground-link-length O2O4 for the fourbar of L1a  80.864.

4.

The fourbar stage is now defined as O2ABO4 with link lengths Ground link 1a Link 3 (coupler)

L1a  80.864 L3  52.000

Link 2 (input)

L2  150.000

Link 4 (output)

L4  200.000

5.


6.

Draw a construction line through C2C3 and extend it downward.

7.

Select a point on this line and call it O6. In the solution below O6 was placed 25 units from the bottom of the base.

8.

9.

Draw a circle about O6 with a radius of one-half the length C1C2 and label the intersections of the circle with the extended line as D2 and D3. In the solution below the radius was measured as L6  12.763. The driver fourbar is now defined as O4CDO6 with link lengths

A2 3

C2

B2

A3

4 2

O4 C3 1a

Link 6 (crank)

L6  12.763

B3


5

1b

O2 112.498

80.864

D2

122.445 O6

D3

12.763










Given:


Solution:


Design choices: L2b  40

Length of link 2b 1.


2.


3.


4.


5.


6.


L1a  53.439

Link 2

L2  134.341

Link 3

L3  52.000

Link 4

L4  90.203

A1

A2

B1 3

B2

6 D1

4

O6 2

D3 5

C1

C2

A3

O4

1b C3 O2

7.

1a B3






Condition L1a L2 L3 L4  "non-Grashof" Although this fourbar is non-Grashof, there are no toggle points within the required range of motion. 8.

9.

Select a point on link 2 (O2A) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it C. (Note that link 2 is now a ternary link with nodes at O2, C, and A.) In the solution above the distance O2C was selected to be L2b  40.000 . Draw a construction line through C1C3 and extend it to the left.


L6  29.760

Link 5 (coupler) L5  119.665 Link 1b (ground) L1b  122.613 Link 2b (rocker) L2b  40.000 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L2b L5  "Grashof"




Design a fourbar mechanism to move the link shown in Figure P3-16 from position 1 to position 2. Ignore the third position and the fixed pivots O2 and O4 shown. Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.

Given:

Position 1 offsets:

Solution:


xC1D1  3.744  in

yC1D1  2.497  in

1.

Connect the end points of the two given positions of the line CD with construction lines, i.e., lines from C1 to C2 and D1 to D2.

2.

Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below the bisector of C1C2 was extended downward and the bisector of D1D2 was extended upward.

3.

Select one point on each bisector and label them O4 and O6, respectively. In the solution below the distances O4D and O6C were each selected to be 7.500 in. This resulted in a ground-link-length O4O6 for the fourbar of 15.366 in.

4.

The fourbar stage is now defined as O4CDO6 with link lengths Link 5 (coupler) L5 

2

xC1D1  yC1D1

Link 4 (input)

L4  7.500  in

Ground link 1b

L1b  15.366 in

2


L6  7.500  in

5.

Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, B, and D.) In the solution below the distance O4B was selected to be 4.000 in.

6.

Draw a construction line through B1B2 and extend it to the right.

7.

Select a point on this line and call it O2. In the solution below the distance AB was selected to be 6.000 in.

8.

Draw a circle about O2 with a radius of one-half the length B1B2 and label the intersections of the circle with the extended line as A1 and A2. In the solution below the radius was measured as 1.370 in.

9.

The driver fourbar is now defined as O2ABO4 with link lengths Link 2 (crank)

L2  1.370  in







Condition L1a L2 L3 L4a  "Grashof" min  L1a L2 L3 L4a  1.370 in

O4 4.000 7.500

B2

B1

A1 2 O2

4

4 D1 5 5

A2

3

C2

D2 15.366

C1 6 6 7.500 O6

11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4DCO6 is non-Grashoff with toggle positions at 4 = -49.9 deg and +49.9 deg. The fourbar operates between 4 = +28.104 deg and -11.968 deg.





Given:

Position 2 offsets:

Solution:


xC2D2  4.355  in

yC2D2  1.134  in

1.


2.


3.


4.

The fourbar stage is now defined as O4DCO6 with link lengths Link 5 (coupler) L5 

2

xC2D2  yC2D2

Link 4 (input)

L4  6.000  in

Ground link 1b

L1b  14.200 in

2


L6  6.000  in

5.


6.


7.


8.


9.


L2  1.271  in








6.000 O4

4.000

7.099

4 4

B1 D2 C2

B2 3

5

A1

5 C3 6.000

6

D3

2

O2

A2

6

O6

14.200





Design a fourbar mechanism to give the three positions shown in Figure P3-16. Ignore the points O2 and O4 shown. Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.

Solution:



Length of link 3:

Length of link 4b:

L4b  4.500

1.


2.


3.


4.


5.


6.

Line C1D1 is link 5. Line O6O4 is link 1a (ground link for the fourbar). The fourbar is now defined as O6CDO4 and has link lengths of Ground link 1a

L1a  2.616

Link 6

L6  6.080

Link 5

L5  4.500

Link 4

L4  6.901

D2

D1 5 5 C1

C3

C2

D3

B2

B1 6

2.765

5

4

3

4 4

6

B3

6 O4

6.080

A1 O2

2

6.901

O6 10.611

2.616

7.

Check the Grashof condition. Note that any Grashof condition is potentially acceptable in this case. Condition( a b c d ) 


A3




9.

Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, D, and B.) In the solution above the distance O4B was selected to be L4b  4.500 . Draw a construction line through B1B3 and extend it up to the right.

10. Layout the length of link 3 (design choice) along the extended line. Label the other end A. 11. Draw a circle about O2 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L2  2.765. 12. The driver fourbar is now defined as O4BAO2 with link lengths Link 2 (crank)

L2  2.765

Link 3 (coupler) L3  10.000 Link 1b (ground) L1b  10.611 Link 4b (rocker) L4b  4.500 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Condition L2 L3 L1b L4b  "Grashof" min  L2 L3 L1b L4b  2.765




Design a fourbar mechanism to give the three positions shown in Figure P3-16 using the fixed pivots O2 and O4 shown. (See Example 3-7.) Build a cardboard model and add a driver dyad to limit its motion to the range of positions designed, making it a sixbar.

Solution:


Design choices:

L5  5.000

Length of link 5:

L2b  2.500

Length of link 2b:

1.


2.


3.


4.


5.


6.


7.

The three inverted positions of the ground link that correspond to the three desired coupler positions are labeled O2O4, O2'O4', and O2"O4" in the first layout below and are renamed E1F1, E2F2, and E3F3, respectively, in the second layout, which is used to find the points G and H. D2 D1

C2 C3

C1

O2

O4''

O2''

O2'

O4

O4'

D3



8.


9.



L1a  3.000

Link 2

L2  8.597

Link 3

L3  1.711

Link 4

L4  7.921

G 3

H

2

4

F1

E 1 O2 1a

F3

E3

O4

F2

E2



Condition L1a L2 L3 L4  "Grashof" The fourbar that will provide the desired motion is now defined as a Grashof double crank in the crossed configuration. It now remains to add the original points C1 and D1 to the coupler GH and to define the driving dyad.



14. Select a point on link 2 (O2G) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 2 is now a ternary link with nodes at O2, B, and G.) In the solution below, the distance O2B was selected to be L2b  2.500 . 15. Draw a construction line through B1B3 and extend it up to the left. 16. Layout the length of link 5 (design choice) along the extended line. Label the other end A. 17. Draw a circle about O6 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L6  1.541. 18. The driver fourbar is now defined as O2BAO6 with link lengths Link 6 (crank)

L6  1.541

Link 5 (coupler) L5  5.000 Link 1b (ground) L1b  5.374 Link 2b (rocker) L2b  2.500 19. Use the link lengths in step 18 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L5 L1b L2b  "Grashof"

D2

D1

C2

3

D3

C3 G3

G2 C1

3

H1

3

H2

2

G1 2

B3

2

4

H3

4 4

5 O6

B1

A3 O2

A1

6

1a

O4





Given:

Position 1 offsets:

Solution:


xC1D1  1.896  in

yC1D1  1.212  in

1.


2.


3.

Select one point on each bisector and label them O4 and O6, respectively. In the solution below the distances O6C and O4D were each selected to be 6.500 in. This resulted in a ground-link-length O4O6 for the fourbar of 14.722 in.

4.


2

xC1D1  yC1D1

Link 4 (input)

L4  6.500  in

Ground link 1b

L1b  14.722 in

2


L6  6.500  in

5.


6.


7.


8.


9.


L2  0.645  in



10. Use the link lengths in step 9 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Condition( a b c d )  S  min ( a b c d ) L  max( a b c d ) SL  S  L PQ  a  b  c  d  SL return "Grashof" if SL  PQ return "Special Grashof" if SL = PQ return "non-Grashof" otherwise




6.500 4.500 B2

4 4

D2

B1

5

6.500

C2 5 C1

O6

7.472

3

6 6

O4

D1 14.722

A2 O2

2 A1

0.645

11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4CDO6 is non-Grashoff with toggle positions at 4 = -17.1 deg and +17.1 deg. The fourbar operates between 4 = +5.216 deg and -11.273 deg.





Given:

Position 2 offsets:

Solution:


xC2D2  0.834  in

yC2D2  2.090  in

1.


2.


3.


4.


2

xC2D2  yC2D2

Link 4 (input)

L4  5.000  in

Ground link 1b

L1b  12.933 in

2


L6  5.000  in

5.


6.


7.


8.


9.


L2  0.741  in







Condition L1a L2 L3 L4  "Grashof"

O4 5.500

4.000 4 B3

7.173 4 B2

D3 D2

5 C3

5

3

A3

O2 2

A2

C2 6 6

12.933

O6






Solution:



Length of link 3:

L4b  2.500

Length of link 4b:

1.


2.


3.


4.


5.


6.


L1a  1.835

Link 6

L6  2.967

Link 5

L5  2.250

Link 4

L4  3.323

D3 B3 3.323

D2

B2

4 5

4

5

C3 1.835

4 C2

O4 6 O6

6

D1

B1 3 5 C1

6

1.403 A3 O2 2

2.967

A1

6.347

7.







Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, D, and B.) In the solution above the distance O4B was selected to be L4b  2.500 .

9.

Draw a construction line through B1B3 and extend it up to the right.

10. Layout the length of link 3 (design choice) along the extended line. Label the other end A. 11. Draw a circle about O2 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L2  1.403. 12. The driver fourbar is now defined as O2ABO4 with link lengths Link 2 (crank)

L2  1.403

Link 3 (coupler) L3  6.000 Link 1b (ground) L1b  6.347 Link 4b (rocker) L4b  2.500 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Condition L1b L2 L3 L4b  "Grashof" min  L1b L2 L3 L4b  1.403





Solution:


Design choices:

L5  4.000

Length of link 5:

Length of link 2b:

L2b  0.791

1.


2.


3.


4.


5.


6.


7.


D3 D2

C3

D1 C2

C1 O2

O4 O2'' O2'

O4' O4''



8.


9.



L1a  3.000

Link 2

L2  0.791

Link 3

L3  1.222

Link 4

L4  1.950

E1 O2

F1 O4

1a 2

G

4

3 E3 E2

H

F2 F3



Condition L1a L2 L3 L4  "non-Grashof" The fourbar that will provide the desired motion is now defined as a non-Grashof double rocker in the crossed configuration. It now remains to add the original points C1 and D1 to the coupler GH and to define the driving dyad, which in this case will drive link 4 rather than link 2. 14. Select a point on link 2 (O2G) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 2 is now a ternary link with nodes at O2, B, and G.) In the solution below, the distance O2B was selected to be L2b  0.791 . Thus, in this case B and G coincide. 15. Draw a construction line through B1B3 and extend it up to the left. 16. Layout the length of link 5 (design choice) along the extended line. Label the other end A. 17. Draw a circle about O6 with a radius of one-half the length B1B3 and label the intersections of the circle



with the extended line as A1 and A3. In the solution below the radius was measured as L6  0.727.

18. The driver fourbar is now defined as O2BAO6 with link lengths Link 6 (crank)

L6  0.727

Link 5 (coupler) L5  4.000 Link 1b (ground) L1b  4.012 Link 2b (rocker) L2b  0.791 19. Use the link lengths in step 18 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L5 L1b L2b  "Grashof"

D3 D2 A3 O6

D1

6 C3

A1

C2 3

5

3

1b

3

C1

4 O2

O4

4

2 G

H

4





Given:

Position 1 offsets:

Solution:


xC1D1  1.591  in

yC1D1  1.591  in

1.


2.


3.

Select one point on each bisector and label them O4 and O6, respectively. In the solution below the distances O4C and O6D were each selected to be 5.000 in. This resulted in a ground-link-length O4O6 for the fourbar of 10.457 in.

4.


2

xC1D1  yC1D1

Link 4 (input)

L4  5.000  in

Ground link 1b

L1b  10.457 in

2


L6  5.000  in

5.

Select a point on link 4 (O4C) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, B, and C.) In the solution below the distance O4B was selected to be 3.750 in.

6.


7.


8.


9.


L2  0.882  in





Condition L1a L2 L3 L4a  "Grashof"



O6

6 6

10.457

C2

5

5

A2 O2 2

C1

B2

B1

A1

3

4

4 5.000

D1

D2

3.750

7.020 O4

11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4CDO6 is non-Grashoff with toggle positions at 4 = -38.5 deg and +38.5 deg. The fourbar operates between 4 = +15.206 deg and -12.009 deg.





Given:

Position 2 offsets:

Solution:


xC2D2  2.053  in

yC2D2  0.920  in

1.


2.


3.


4.


2

xC2D2  yC2D2

Link 4 (input)

L4  5.000  in

Ground link 1b

L1b  8.773  in

2


L6  5.000  in

5.


6.


7.


8.


9.


L2  0.892  in







Condition L1a L2 L3 L4a  "Grashof"

7.019

O4

5.000

4

3.750 4 B3

B2 D2

D3 5 8.773

C3

A3 O2 2

A2

3

5 C2

6 6

O6

11. Using the program FOURBAR and the link lengths given above, it was found that the fourbar O4DCO6 is non-Grashoff with toggle positions at 4 = -55.7 deg and +55.7 deg. The fourbar operates between 4 = -7.688 deg and -35.202 deg.





Solution:



Length of link 3:

Length of link 4b:

L4b  5.000

1.


2.


3.


4.


5.


6.


L1a  8.869

Link 6

L6  1.831

Link 5

L5  2.250

Link 4

L4  6.953

7.646

O4 4

O2 A1

6.953 8.869

B3

4 B1

2 1.593

D3 D2

D 1 C2 5 C1

3

A3

5 C3

6 O6

6 1.831

7.






Condition L6 L1a L4 L5  "non-Grashof" 8.

9.

Select a point on link 4 (O4D) at a suitable distance from O4 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 4 is now a ternary link with nodes at O4, D, and B.) In the solution above the distance O4B was selected to be L4b  5.000 . Draw a construction line through B1B3 and extend it up to the right.

10. Layout the length of link 3 (design choice) along the extended line. Label the other end A. 11. Draw a circle about O2 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L2  1.593. 12. The driver fourbar is now defined as O2ABO4 with link lengths Link 2 (crank)

L2  1.593

Link 3 (coupler) L3  6.000 Link 1b (ground) L1b  7.646 Link 4b (rocker) L4b  5.000 13. Use the link lengths in step 12 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 2). Condition L1b L2 L3 L4b  "Grashof" min  L1b L2 L3 L4b  1.593





Solution:


Design choices:

Length of link 5:

L5  4.000

Length of link 2b:

L2b  2.000

1.


2.


3.


4.


5.


6.


7.


D1 D2

D3

C1 C2

O2''

O4'

C3 O2

O4 O2'

O4''

8.


9.


10. Repeat steps 2 and 3 for lines F1F2 and F2F3. Label the intersection H.



11. Connect E1 with G and label it link 2. Connect F1 with H and label it link 4. Reinverting, E1 and F1 are the original fixed pivots O2 and O4, respectively. 12. Line GH is link 3. Line O2O4 is link 1a (ground link for the fourbar). The fourbar is now defined as O2GHO4 and has link lengths of Ground link 1a

L1a  4.000

Link 2

L2  2.000

Link 3

L3  6.002

Link 4

L4  7.002

H

3

E3

4

F2

G 2 O2

O4 E 2 F1

E1

F3



Condition L1a L2 L3 L4  "Grashof" The fourbar that will provide the desired motion is now defined as a non-Grashof crank rocker in the open configuration. It now remains to add the original points C1 and D1 to the coupler GH and to define the driving dyad, which in this case will drive link 4 rather than link 2.



14. Select a point on link 2 (O2G) at a suitable distance from O2 as the pivot point to which the driver dyad will be connected and label it B. (Note that link 2 is now a ternary link with nodes at O2, B, and G.) In the solution below, the distance O2B was selected to be L2b  2.000 . Thus, in this case B and G coincide. 15. Draw a construction line through B1B3 and extend it up to the left. 16. Layout the length of link 5 (design choice) along the extended line. Label the other end A. 17. Draw a circle about O6 with a radius of one-half the length B1B3 and label the intersections of the circle with the extended line as A1 and A3. In the solution below the radius was measured as L6  1.399. 18. The driver fourbar is now defined as O2BAO6 with link lengths L6  1.399

Link 6 (crank)

Link 5 (coupler) L5  4.000 Link 1b (ground) L1b  4.257 Link 2b (rocker) L2b  2.000 19. Use the link lengths in step 18 to find the Grashoff condition of the driving fourbar (it must be Grashoff and the shortest link must be link 6). Condition L6 L1b L2b L5  "Grashof" H1

H2

3 D1

3 D2

H3

D3 C1 C2 4 4 G2

2

3

C3 2

G1

O2

1a 2

1b

G3 5

A1 6 O6

A3

4 O4




Design a fourbar Grashof crank-rocker for 120 degrees of output rocker motion with a quick-return time ratio of 1:1.2. (See Example 3-9.)

Given:

Time ratio

Solution: 1.

Tr 

1 1.2



β 

α

α  β = 360  deg

β 360  deg

β  196  deg

1  Tr

α  360  deg  β

α  164  deg

δ  β  180  deg

δ  16 deg

2.


3.

Layoff a line through B1 at an arbitrary angle (but not zero deg). In the solution below the line is 60 deg to the horizontal.

0. 95 3

=

c

90.00°

B2 B2

B1

B1 4 O4

d

O4 3

4.4 91 =

b

3.8 33 =

0° .0 16

LAYOUT

A2

A1 2

O2

a

LINKAGE DEFINITION 0.2 55 =

O2



4.


5.


6.


d  3.833  in

Coupler (3)

b  4.491  in

Crank (2)

a  0.255  in

Rocker (4)

c  0.953  in





Given:

Time ratio

Solution: 1.

Tr 

1 1.5



β 

α

α  β = 360  deg

β 360  deg

β  216 deg

1  Tr

α  360  deg  β

α  144 deg

δ  β  180  deg

δ  36 deg

2.


3.

Layoff a line through B1 at an arbitrary angle (but not zero deg). In the solution below the line is 20 deg to the horizontal.

4.


5.


B1

B2

B2

B1

3.0524 = b O4

3

O2

4 1.2694 = a

2

A1

LAYOUT O4

O2

2.0000 = c

A2 LINKAGE DEFINITION 2.5364 = d


6.



d  2.5364 in

Coupler (3)

b  3.0524 in

Crank (2)

a  1.2694 in

Rocker (4)

c  2.000  in





Given:

Time ratio

Solution: 1.

Tr 

1 1.33



β 

α

α  β = 360  deg

β 360  deg

β  205  deg

1  Tr

α  360  deg  β

α  155  deg

δ  β  180  deg

δ  25 deg

2.


3.

Layoff a line through B1 at an arbitrary angle (but not zero deg). In the solution below the line is 150 deg to the horizontal. 2. 00 0= c

90.00°

B2

B2

B1

B1 4

O4 25.0 0°

O4 3

6.2 32 =

b

4.7 63 =

d

LAYOUT

A2

A1 2

LINKAGE DEFINITION

a

O2

0.4 35 =

O2



4.


5.


6.


d  4.763  in

Coupler (3)

b  6.232  in

Crank (2)

a  0.435  in

Rocker (4)

c  2.000  in




Design a sixbar drag link quick-return linkage for a time ratio of 1:4 and output rocker motion of 50 degrees. (See Example 3-10.)

Given:

Time ratio

Solution: 1.

Tr 

1 4


Determine the crank rotation angles  and  from equation 3.1. Tr = Solving for and 

β 

α

α  β = 360  deg

β 360  deg 1  Tr

α  360  deg  β

β  288 deg α  72 deg

2.

Draw a line of centers XX at any convenient location.

3.

Choose a crank pivot location O2 on line XX and draw an axis YY perpendicular to XX through O2.

4.

Draw a circle of convenient radius O2A about center O2. In the solution below, the length of O2A is a  1.000  in.

5.

Lay out angle  with vertex at O2, symmetrical about quadrant one.

6.

Label points A1 and A2 at the intersections of the lines subtending angle  and the circle of radius O2A.

7.

8.

Set the compass to a convenient radius AC long enough to cut XX in two places on either side of O2 when swung from both A1 and A2. Label the intersections C1 and C2. In the solution below, the length of AC is b  2.000  in. The line O2A is the driver crank, link 2, and the line AC is the coupler, link 3.

9.

The distance C1C2 is twice the driven (dragged) crank length. Bisect it to locate the fixed pivot O4.

10. The line O2O4 now defines the ground link. Line O4C is the driven crank, link 4. In the solution below, O4C measures c  2.282  in and O2O4 measures d  0.699  in. 11. Calculate the Grashoff condition. If non-Grashoff, repeat steps 7 through 11 with a shorter radius in step 7. Condition( a b c d ) 


Condition( a b c d )  "Grashof" 12. Invert the method of Example 3-1 to create the output dyad using XX as the chord and O4C1 as the driving crank. The points B1 and B2 will lie on line XX and be spaced apart a distance that is twice the length of O4C (link 4). The pivot point O6 will lie on the perpendicular bisector of B1B2 at a distance from XX which subtends the specified output rocker angle, which is 50 degrees in this problem. In the solution below, the length BC was chosen to be e  5.250  in.



9.000°

72.000°

LAYOUT OF SIXBAR DRAG LINK QUICK RETURN WITH TIME RATIO OF 1:4 a = 1.000 b = 2.000 c = 2.282 d = 0.699 e = 5.250 f = 5.400

13. For the design choices made (lengths of links 2, 3 and 5), the length of the output rocker (link 6) was measured as f  5.400  in.




Design a crank-shaper quick-return mechanism for a time ratio of 1:2.5 (Figure 3-14, p. 112).

Given:

Time ratio

Solution:


TR 

1 2.5

Design choices:

1.

Length of link 2 (crank)

L2  1.000

Length of link 5 (coupler)

L5  5.000

S  4.000

Length of stroke

Calculate  from equations 3.1. TR 

α β

α  β  360  deg

α 

360  deg 1

α  102.86 deg

1 TR

2.

Draw a vertical line and mark the center of rotation of the crank, O2, on it.

3.

Layout two construction lines from O2, each making an angle /2 to the vertical line through O2.

4.

Using the chosen crank length (see Design Choices), draw a circle with center at O2 and radius equal to the crank length. Label the intersections of the circle and the two lines drawn in step 3 as A1 and A2.

5.

Draw lines through points A1 and A2 that are also tangent to the crank circle (step 2). These two lines will simultaneously intersect the vertical line drawn in step 2. Label the point of intersection as the fixed pivot center O4.

6.

Draw a vertical construction line, parallel and to the right of O2O4, a distance S/2 (one-half of the output stroke length) from the line O2O4.

7.

Extend line O4A1 until it intersects the construction line drawn in step 6. Label the intersection B1.

8.

Draw a horizontal construction line from point B1, either to the left or right. Using point B1 as center, draw an arc of radius equal to the length of link 5 (see Design Choices) to intersect the horizontal construction line. Label the intersection as C1.

9.

Draw the slider blocks at points A1 and C1 and finish by drawing the mechanism in its other extreme position.

STROKE 4.000 2.000 C2

6

C1

B2

B1

5

O2 4

2 A2

3 O4

A1




Design a sixbar, single-dwell linkage for a dwell of 70 deg of crank motion, with an output rocker motion of 30 deg using a symmetrical fourbar linkage with the following parameter values: ground link ratio = 2.0, common link ratio = 2.0, and coupler angle  = 40 deg. (See Example 3-13.)

Given:

Crank dwell period: 70 deg. Output rocker motion: 30 deg. Ground link ratio, L1/L2 = 2.0: GLR  2.0 Common link ratio, L3/L2 = L4/L2 = BP/L2 = 2.0: CLR  2.0 Coupler angle, γ  40 deg

Design choice: Crank length, L2  2.000 Solution: 1.

See Figures 3-20 and 3-21 and Mathcad file P0372.

For the given design choice, determine the remaining link lengths and coupler point specification. Coupler link (3) length

L3  CLR L2

L3  4.000


L4  CLR L2

L4  4.000


L1  GLR  L2

L1  4.000

Angle PAB

δ 


180  deg  γ 2

AP  2  L3 cos δ

δ  70.000 deg AP  2.736

Enter the above data into program FOURBAR, plot the coupler curve, and determine the coordinates of the coupler curve in the selected range of crank motion, which in this case will be from 145 to 215 deg.



FOURBAR for Windows

3.

File

P03-72

Angle Coupler Pt Step X Deg in

Coupler Pt Y in

Coupler Pt Mag in

Coupler Pt Ang in

145 150 155 160 165 170 175 180 185 190 195 200 205 210 215

3.818 3.661 3.494 3.319 3.135 2.945 2.749 2.547 2.342 2.133 1.923 1.711 1.499 1.289 1.080

4.422 4.360 4.295 4.226 4.156 4.083 4.009 3.935 3.859 3.783 3.707 3.631 3.555 3.479 3.403

120.297 122.895 125.549 128.259 131.025 133.846 136.723 139.655 142.639 145.674 148.757 151.886 155.055 158.261 161.498

-2.231 -2.368 -2.497 -2.617 -2.728 -2.829 -2.919 -2.999 -3.067 -3.124 -3.169 -3.202 -3.223 -3.232 -3.227


y

145 P

B

180

3 4 D

215 A 2

x

PSEUDO-ARC O2

4.

O4

The position of the end of link 5 at point D will remain nearly stationary while the crank moves from 145 to 215 deg. As the crank motion causes the coupler point to move around the coupler curve there will be another extreme position of the end of link 5 that was originally at D. Since a symmetrical linkage was chosen, the other extreme position will be located along a line through the axis of symmetry (see Figure 3-20) a distance equal to the length of link 5 measured from the point where the axis of symmetry intersects the coupler curve near the 0 deg coupler point. Establish this point and label it E.



FOURBAR for Windows

File

P03-72

Angle Step Deg

Coupler Pt X in

Coupler Pt Y in

Coupler Pt Mag n

Coupler Pt Ang in

340.000 345.000 350.000 355.000 0.000 5.000 10.000 15.000

-0.718 -0.615 -0.506 -0.386 -0.255 -0.117 0.022 0.155

0.175 0.481 0.818 1.178 1.549 1.917 2.269 2.598

0.739 0.781 0.962 1.240 1.570 1.920 2.269 2.603

166.325 142.001 121.717 108.135 99.365 93.499 89.434 86.581

y 145 P

B 5

180

3 5

AXIS OF SYMMETRY

4 D

215

355 A

E

2

x

PSEUDO-ARC

O4

O2

5.

The line segment DE represents the maximum displacement that a link of the length equal to link 5, attached at P, will reach along the axis of symmetry. Construct a perpendicular bisector of the line segment DE and extend it to the right (or left, which ever is convenient). Locate fixed pivot O6 on the bisector of DE such that the lines O6D and O6E subtend the desired output angle, in this case 30 deg. Draw link 6 from D through O6 and extend it to any convenient length. This is the output link that will dwell during the specified motion of the crank. SUMMARY OF LINKAGE SPECIFICATIONS Original fourbar: O6

y 6

145 P

B 5

180

30.000° 3 5

L2  2.000

Coupler

L3  4.000

Rocker

L4  4.000

Coupler point

AP  2.736 δ  70.000 deg

E 2

x O2

Crank

Added dyad:

355 A

L1  4.000

4 D

215

Ground link

O4

Coupler

L5  3.840

Output

L6  5.595

Pivot O6

x  3.841 y  5.809





Given:





L3  CLR L2

L3  5.000


L4  CLR L2

L4  5.000


L1  GLR  L2

L1  4.000

Angle PAB

δ 


180  deg  γ 2

AP  2  L3 cos δ

δ  60.000 deg AP  5.000




FOURBAR for Windows

3.

File

P03-73


Coupler Pt Y in

Coupler Pt Mag in

Coupler Pt Ang in

130 140 150 160 170 180 190 200 210 220 230

6.449 6.171 5.840 5.464 5.047 4.598 4.123 3.631 3.130 2.629 2.138

6.812 6.695 6.559 6.408 6.244 6.071 5.892 5.709 5.523 5.336 5.146

108.774 112.833 117.078 121.493 126.060 130.765 135.588 140.504 145.482 150.482 155.454

-2.192 -2.598 -2.986 -3.347 -3.675 -3.964 -4.209 -4.405 -4.551 -4.643 -4.681


y 130 P

B

180

D

230

4

3

PSEUDO-ARC

A 2

x O2

4.

O4




FOURBAR for Windows

File

P03-73


Coupler Pt Y in

Coupler Pt Mag in

Coupler Pt Ang in

340 350 0 10 20

1.429 2.316 3.316 4.265 5.047

3.013 3.237 3.746 4.414 5.078

151.688 134.332 117.727 104.920 96.371

-2.652 -2.262 -1.743 -1.137 -0.564

y 130 P 20 180

B

10 5

AXIS OF SYMMETRY

0

D

350

230

4

3 340 A

PSEUDO-ARC

E

2

x O4

O2

5.

The line segment DE represents the maximum displacement that a link of the length equal to link 5, attached at P, will reach along the axis of symmetry. Construct a perpendicular bisector of the line segment DE and extend it to the right (or left, which ever is convenient). Locate fixed pivot O6 on the bisector of DE such that the lines O6D and O6E subtend the desired output angle, in this case 30 deg. Draw link 6 from D through O6 and extend it to any convenient length. This is the output link that will dwell during the specified motion of the crank. SUMMARY OF LINKAGE y SPECIFICATIONS 130

Original fourbar: P

Ground link

L1  4.000

Crank

L2  2.000

Coupler

L3  5.000

Rocker

L4  5.000

Coupler point

AP  5.000

20 180

50.000°

B

10 5 0

O6 6

230

δ  60.000 deg

D

350

Added dyad:

4

3 340 PSEUDO-ARC

A

E

2

x O2

O4

Coupler

L5  5.395

Output

L6  2.998

Pivot O6

x  3.166 y  3.656





Given:





L3  CLR L2

L3  3.500


L4  CLR L2

L4  3.500


L1  GLR  L2

L1  4.000

Angle PAB

δ 


180  deg  γ 2

AP  2  L3 cos δ

δ  55.000 deg AP  4.015




FOURBAR for Windows

3.

File

P03-74


Coupler Pt Y in

Coupler Pt Mag in

Coupler Pt Ang in

140 150 160 170 180 190 200 210 220

5.208 4.940 4.645 4.332 4.005 3.668 3.322 2.969 2.613

5.252 5.032 4.804 4.578 4.359 4.152 3.958 3.779 3.612

97.395 100.971 104.781 108.860 113.242 117.942 122.946 128.210 133.663

-0.676 -0.958 -1.226 -1.480 -1.720 -1.945 -2.153 -2.337 -2.493


y

140 P 180

220

B

PSEUDO-ARC

4

3 A

O4

2 O2

4.

x D




FOURBAR for Windows

File

P03-74


Coupler Pt Y in

Coupler Pt Mag in

Coupler Pt Ang in

340 350 0 10 20

1.658 2.360 3.147 3.886 4.490

2.158 2.562 3.185 3.887 4.530

129.810 112.856 98.919 88.916 82.372

-1.382 -0.995 -0.494 0.074 0.601

y

140 P 20 180

10

0

220

AXIS OF SYMMETRY

B

350 PSEUDO-ARC

A

4

340 3

O4

2 O2

x

D E

5.

The line segment DE represents the maximum displacement that a link of the length equal to link 5, attached at P, will reach along the axis of symmetry. Construct a perpendicular bisector of the line segment DE and extend it to the right (or left, which ever is convenient). Locate fixed pivot O6 on the bisector of DE such that the lines O6D and O6E subtend the desired output angle, in this case 30 deg. Draw link 6 from D through O6 and extend it to any convenient length. This is the output link that will dwell during the specified motion of the crank.



y

SUMMARY OF LINKAGE SPECIFICATIONS Original fourbar:

140 P 20 180

10

0

220

Ground link

L1  4.000

Crank

L2  2.000

Coupler

L3  3.500

Rocker

L4  3.500

Coupler point

AP  4.015

B

δ  55.000 deg

350 PSEUDO-ARC

A

45.000°

340 3

4

O4

2 O2

O6

x

Added dyad:

D E

Coupler

L5  7.676

Output

L6  1.979

Pivot O6

x  6.217 y  0.653




Using the method of Example 3-11, show that the sixbar Chebychev straight-line linkage of Figure P2-5 is a combination of the fourbar Chebychev straight-line linkage of Figure 3-29d and its Hoeken's cognate of Figure 3-29e. See also Figure 3-26 for additional information useful to this solution. Graphically construct the Chebychev sixbar parallel motion linkage of Figure P2-5a from its two fourbar linkage constituents and build a physical or computer model of the result.

Solution:

See Figures P2-5, 3-29d, 3-29e, and 3-26 and Mathcad file P0375.

1.

Following Example 3-11and Figure 3-26 for the Chebyschev linkage of Figure 3-29d, the fixed pivot OC is found by laying out the triangle OAOBOC, which is similar to A1B1P. In this case, A1B1P is a striaght line with P halfway between A1 and B1 and therefore OAOBOC is also a straightline with OC halfway between OA and OB. As shown below and in Figure 3-26, cognate #1 is made up of links numbered 1, 2, 3, and 4. Cognate #2 is links numbered 1, 5, 6, and 7. Cognate #3 is links numbered 1, 8, 9, and 10.

3

P

3 3

B1

2

4

9

OC

B2

4

6

5 1

1

6

Links Removed

10 OA

2.

2

6

7

8

A3

4

B2

P2

A1

P1

6

9 B3

B1

A1

OB

A2

5 1

A2 OA

OC

OB

Discard cognate #3 and shift link 5 from the fixed pivot OB to OC and shift link 7 from OC to OB. Note that due to the symmetry of the figure above, L5 = 0.5 L3, L6 = L2, L7 = 0.5 L2 and OCOB = 0.5 OAOB. Thus, cognate #2 is, in fact, the Hoeken straight-line linkage. The original Chebyschev linkage with the Hoeken linkage superimposed is shown above right with the link 5 rotated to 180 deg. Links 2 and 6 will now have the same velocity as will 7 and 4. Thus, link 5 can be removed and link 6 can be reduced to a binary link supported and constrained by link 4. The resulting sixbar is the linkage shown in Figure P2-5.




Design a driver dyad to drive link 2 of the Evans straigh-line linkage in Figure 3-29f from 150 deg to 210 deg. Make a model of the resulting sixbar linkage and trace the couple curve.

Given:

Output angle

Solution:

See Figjre 3-29f, Example 3-1, and Mathcad file P0376.

Design choices:

θ  60 deg

Link lengths:

L2  2.000

Link 2

Link 5

L5  3.000

1.

Draw the input link O2A in both extreme positions, A1 and A2, at the specified angles such that the desired angle of motion 2 is subtended.

2.

Draw the chord A1A2 and extend it in any convenient direction. In this solution it was extended downward.

3.

Layout the distance A1C1 along extended line A1A2 equal to the length of link 5. Mark the point C1.

4.

Bisect the line segment A1A2 and layout the length of that radius from point C1 along extended line A1A2. Mark the resulting point O6 and draw a circle of radius O6C1 with center at O6.

5.

Label the other intersection of the circle and extended line A1A2, C2.

6.

7.

A1

Measure the length of the crank (link 6) as O6C1 or O6C2. From the graphical solution, L6  1.000 Measure the length of the ground link (link 1) as O2O6. From the graphical solution, L1  3.073

P2 3 B1 , B 2

2 O2

4 A2

P1

1

5 C1

O4

6 O6

3.073" C2

8.


2.932"



0.922" L1 = 2.4 L2 = 2 L3 = 3.2 L4 = 2.078 L5 = 3.00 L6 = 1.00 AP = 5.38




Design a driver dyad to drive link 2 of the Evans straigh-line linkage in Figure 3-29g from -40 deg to 40 deg. Make a model of the resulting sixbar linkage and trace the couple curve.

Given:

Output angle

Solution:

See Figjre 3-29G, Example 3-1, and Mathcad file P0377.

Design choices:

θ  80 deg

Link lengths:

L2  2.000

Link 2

Link 5

L5  3.000

1.

Draw the input link O2A in both extreme positions, A1 and A2, at the specified angles such that the desired angle of motion 2 is subtended.

2.

Draw the line A1C1 and extend it in any convenient direction. In this solution it was extended at a 30-deg angle from A1O2 (see note below) .

3.

Layout the distance A1C1 along extended line A1C1 equal to the length of link 5. Mark the point C1.

4.

Bisect the line segment A1A2 and layout the length of that radius from point C1 along extended line A1C1. Mark the resulting point O6 and draw a circle of radius O6C1 with center at O6.

5.

6.

7.

C2

O6 P2

Extend a line from A2 through O6. Label the other intersection of the circle and extended line A2O6, C2. Measure the length of the crank (link 6) as O6C1 or O6C2. From the graphical solution, L6  1.735

6

3.165"

C1

Measure the length of the ground link (link 1) as O2O6. From the graphical solution, L1  3.165


O2 B2

B1

3 2

4 A1

1

O4

P1



A2 5

Note: If the angle between link 2 and link 5 is zero the resulting driving fourbar will be a special Grashof. For angles greater than zero but less than 33.68 degrees it is a Grashof crank-rocker. For angles greater than 33.68 it is a non-Grashof double rocker. 8.

L1 = 4.61 L2 = 2 L3 = 2.4 L4 = 2.334 L5 = 3.00 L6 = 1.735 AP = 3.00




Figure 6 on page ix of the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD) shows a 50-point coupler that was used to generate the curves in the atlas. Using the definition of the vector R given in Figure 3-17b of the text, determine the 10 possible pairs of values of  and R for the first row of points above the horizontal axis if the gridpoint spacing is one half the length of the unit crank.

Given:

Grid module g  0.5

Solution:

See Figure 6 H&N Atlas, Figure 3-17b, and Mathcad file P0378.

1.

The moving pivot point is located on the 3rd grid line from the bottom and the third grid line from the left when the crank angle is  radians. Let the number of horizontal grid spaces from the left end of the coupler to the coupler point be n  2 1  7 and the number of vertical grid spaces from the coupler to the coupler point be m  2 1  2

2.

For the first row of points above the horizontal axis shown in Figure 6, n  2 1  7 and m  1.

3.

The angle, , between the coupler and the line from the coupler/crank pivot to the coupler point is

π π ϕ( m n )  if  n  0 atan2( n m) if  m = 0 0 if  m  0     



4.



2

2



The distance, R, from the pivot to the coupler point along the same line is 2

R( m n )  g  m  n

2

ϕ( m n ) n 

5.



deg



R( m n ) 

-2.000

153.435

1.118

-1.000

135.000

0.707

0.000

90.000

0.500

1.000

45.000

0.707

2.000

26.565

1.118

3.000

18.435

1.581

4.000

14.036

2.062

5.000

11.310

2.550

6.000

9.462

3.041

7.000

8.130

3.536

The coupler point distance, R, like the link lengths A, B, and C is a ratio of the given length to the the length of the driving crank.




The set of coupler curves in the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD, page 16 of the PDF file) has A = B = C = 1.5. Model this linkage with program FOURBAR using the coupler point fartherest to the left in the row shown on page 1 and plot the resulting coupler curve.

Given:

A  1.5

Solution:

See Figure on page 1 H&N Atlas, Figure 3-17b, and Mathcad file P0379.

B  1.5

C  1.5

1.


2.

For the second column of points to the left of the coupler pivot and the second row of points above the horizontal axis n  2 and m  2. The grid spacing is g  0.5

3.





4.



2

6.

2

2



R( m n )  1.414

Determine the values needed for input to FOURBAR. Link 2 (Crank)

a  1

Link 3 (Coupler)

b  A  a

b  1.500

Link 4 (Rocker)

c  B a

c  1.500

Link 1 (Ground)

d  C a

d  1.500

Distance to coupler point

R( m n )  1.414

Angle from link 3 to coupler point

ϕ( m n )  135.000 deg

Calculate the coordinates of O4. Let the angle between links 2 and 3 be  , then

 A 2  ( 1  C) 2  B2   2  A  ( 1  C) 

7.

2

The distance from the pivot to the coupler point, R, along the same line is R( m n )  g  m  n

5.



α  acos

α  33.557 deg

xO4  C cos α

xO4  1.250

yO4  C sin α

yO4  0.829

Enter this data into FOURBAR and then plot the coupler curve. (See next page)

ϕ( m n )  135.000 deg






The set of coupler curves on page 17 in the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD, page 32 of the PDF file) has A = 1.5, B = C = 3.0. Model this linkage with program FOURBAR using the coupler point fartherest to the right in the row shown and plot the resulting coupler curve.

Given:

A  1.5

Solution:


B  3.0

C  3.0

1.


2.

For the fifth column of points to the right of the coupler pivot and the first row of points above the horizontal axis n  5 and m  1. The grid spacing is g  0.5

3.





4.



2

6.

2

2



R( m n )  2.550


a  1

Link 3 (Coupler)

b  A  a

b  1.500

Link 4 (Rocker)

c  B a

c  3.000

Link 1 (Ground)

d  C a

d  3.000


R( m n )  2.550


ϕ( m n )  11.310 deg


 A 2  ( 1  C) 2  B2   2  A  ( 1  C) 

7.

2


5.



α  acos

α  39.571 deg


xO4  2.313

yO4  C sin α

yO4  1.911


ϕ( m n )  11.310 deg






The set of coupler curves on page 21 in the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD, page 36 of the PDF file) has A = 1.5, B = C = 3.5. Model this linkage with program FOURBAR using the coupler point fartherest to the right in the row shown and plot the resulting coupler curve.

Given:

A  1.5

Solution:


B  3.5

C  3.5

1.


2.

For the fourth column of points to the right of the coupler pivot and the second row of points above the horizontal axis n  4 and m  2. The grid spacing is g  0.5

3.





4.



2

6.

2

2



R( m n )  2.236


a  1

Link 3 (Coupler)

b  A  a

b  1.500

Link 4 (Rocker)

c  B a

c  3.500

Link 1 (Ground)

d  C a

d  3.500


R( m n )  2.236


ϕ( m n )  26.565 deg


 A 2  ( 1  C) 2  B2   2  A  ( 1  C) 

7.

2


5.



α  acos

α  40.601 deg


xO4  2.657

yO4  C sin α

yO4  2.278


ϕ( m n )  26.565 deg






The set of coupler curves on page 34 in the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD, page 49 of the PDF file) has A = 2.0, B = 1.5, C = 2.0. Model this linkage with program FOURBAR using the coupler point fartherest to the right in the row shown and plot the resulting coupler curve.

Given:

A  2.0

Solution:


B  1.5

C  2.0

1.


2.

For the sixth column of points to the right of the coupler pivot and the first row of points below the horizontal axis n  6 and m  1. The grid spacing is g  0.5

3.





4.



2

6.

2

2



R( m n )  3.041


a  1

Link 3 (Coupler)

b  A  a

b  2.000

Link 4 (Rocker)

c  B a

c  1.500

Link 1 (Ground)

d  C a

d  2.000


R( m n )  3.041


ϕ( m n )  9.462 deg


 A 2  ( 1  C) 2  B2   2  A  ( 1  C) 

7.

2


5.



α  acos

α  26.384 deg


xO4  1.792

yO4  C sin α

yO4  0.889


ϕ( m n )  9.462 deg






The set of coupler curves on page 115 in the Hrones and Nelson atlas of fourbar coupler curves (on the book DVD, page 130 of the PDF file) has A = 2.5, B = 1.5, C = 2.5. Model this linkage with program FOURBAR using the coupler point fartherest to the right in the row shown and plot the resulting coupler curve.

Given:

A  2.5

Solution:


B  1.5

C  2.5

1.


2.

For the second column of points to the right of the coupler pivot and the second row of points below the horizontal axis n  2 and m  2. The grid spacing is g  0.5

3.





4.



2

6.

2

2



R( m n )  1.414


a  1

Link 3 (Coupler)

b  A  a

b  2.500

Link 4 (Rocker)

c  B a

c  1.500

Link 1 (Ground)

d  C a

d  2.500


R( m n )  1.414


ϕ( m n )  45.000 deg


 A 2  ( 1  C) 2  B2   2  A  ( 1  C) 

7.

2


5.



α  acos

α  21.787 deg


xO4  2.321

yO4  C sin α

yO4  0.928


ϕ( m n )  45.000 deg






Design a fourbar mechanism to move the link shown in Figure P3-19 from position 1 to position 2. Ignore the third position and the fixed pivots O2 and O4 shown. Build a cardboard model that demonstrates the required movement.

Given:

Position 1 offsets:

Solution:

See figure below and Mathcad file P0384 for one possible solution.

xC1D1  17.186 in

yC1D1  0.604  in

1.


2.

Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below the bisector of C1C2 was extended upward and the bisector of D1D2 was also extended upward.

3.

Select one point on each bisector and label them O2 and O4, respectively. In the solution below the distances O2C and O4D were selected to be 15.000 in. and 8.625 in, respectively. This resulted in a ground-link-length O2O4 for the fourbar of 9.351 in.

4.

The fourbar is now defined as O2CDO4 with link lengths Link 3 (coupler) L3 

2

xC1D1  yC1D1

Link 2 (input)

L2  14.000 in

Ground link 1

L1  9.351  in

2

L3  17.197 in Link 4 (output)

L4  7.000  in

9.35 1

15 .00 0

O2

O4

17.197

8.6 25

D2

C1

D1

C2




Design a fourbar mechanism to move the link shown in Figure P3-19 from position 2 to position 3. Ignore the first position and the fixed pivots O2 and O4 shown. Build a cardboard model that demonstrates the required movement.

Given:

Position 2 offsets:

Solution:

See figure below and Mathcad file P0385 for one possible solution.

xC2D2  15.524 in

yC2D2  7.397  in

1.


2.

Bisect these lines and extend their perpendicular bisectors in any convenient direction. In the solution below the bisector of C2C3 was extended upward and the bisector of D2D3 was also extended upward.

3.

Select one point on each bisector and label them O2 and O4, respectively. In the solution below the distances O2C and O4D were selected to be 15.000 in and 8.625 in, respectively. This resulted in a ground-link-length O2O4 for the fourbar of 9.470 in.

4.


2

xC2D2  yC2D2

Link 2 (input)

L2  15.000 in

Ground link 1b

L1b  9.470  in

2

L3  17.196 in Link 4 (output)

L6  8.625  in

8.625

D3

9.47 0

O2 O4

15.000

D2 96 17.1

C3

C2







Design a fourbar mechanism to give the three positions shown in Figure P3-19. Ignore the points O2 and O4 shown. Build a cardboard model that has stops to limit its motion to the range of positions designed.

Solution:


1.


2.


3.


4.


5.


6.

Line C1D1 is link 3. Line O2O4 is link 1 (ground link for the fourbar). The fourbar is now defined as O2CDO4 an has link lengths of Ground link 1

L1  9.187

Link 2

L2  14.973

Link 3

L3  17.197

Link 4

L4  8.815 D3

8.815

9.18 7 O2

14 .97 3

O4

2 4

D2

17.197 C1

D1

3

C3 C2




Design a fourbar mechanism to give the three positions shown in Figure P3-17 using the fixed pivots O2 and O4 shown. (See Example 3-7.) Build a cardboard model that has stops to limit its motion to the range of positions designed.

Solution:


1.


2.


3.


4.


5.


6.


7.

The three inverted positions of the ground link that correspond to the three desired coupler positions are labeled O2O4, O2'O4', and O2"O4" in the first layout below and are renamed E1F1, E2F2, and E3F3, respectively, in the second layout, which is used to find the points G and H. D3

O'2 O2

O"2

O4

O'4

D2

C1

O" 4

D1 C3

C2

First layout for steps 1 through 7



E2 O'2 E1 O2 E3 O" 2

F1 O4

O'4 F2 2 4 F3 O"4 3

G

H

Second layout for steps 8 through 12 8.


9.



L1a  9.216

Link 2

L2  16.385

Link 3

L3  18.017

Link 4

L4  8.786





Condition L1a L2 L3 L4  "non-Grashof" The fourbar that will provide the desired motion is now defined as a non-Grashof double rocker in the open configuration. It now remains to add the original points C1 and D1 to the coupler GH.

9.21 6

O2

16 .38 5

O4

4

C1

3

D1 H

G

18.017

8.786

2


SOLUTION MANUAL 4-7a-1

PROBLEM 4-7a Statement:

Given:

The link lengths and value of 2 for some fourbar linkages are defined in Table P4-1. The linkage configuration and terminology are shown in Figure P4-1. For row a, find all possible solutions (both open and crossed) for angles 3 and 4 using the vector loop method. Determine the Grashof condition. Link 2 Link 1 d  6  in a  2  in b  7  in

Link 3 Solution: 1.

c  9  in

Link 4

See Mathcad file P0407a.

Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a. K1 

d

K2 

a

K1  3.0000

2

d

K3 

c

K2  0.6667

 

 

 

2 a c

B  1.0000

 

C  K1   K2  1   cos θ  K3

C  3.5566

Use equation 4.10b to find values of 4 for the open and crossed circuits. Open:





2

θ  2  atan2 2  A B 

B  4 A  C



θ  242.714  deg

θ  θ  360  deg



θ  602.714  deg



2

Crossed: θ  2  atan2 2  A B 

B  4 A  C



θ  216.340  deg

Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. K4 

2

d

K5 

b

2

2

c d a b

 

2 a b

2

K4  0.8571 K5  0.2857

 

D  cos θ  K1  K4 cos θ  K5

D  1.6774

 

E  2  sin θ

E  1.0000

 

F  K1   K4  1   cos θ  K5 4.

2

A  0.7113

B  2  sin θ

3.

2

a b c d

K3  2.0000

A  cos θ  K1  K2 cos θ  K3

2.

θ  30 deg

F  2.5906

Use equation 4.13 to find values of 3 for the open and crossed circuits. Open:





θ  2  atan2 2  D E 

2

E  4  D F



θ  θ  360  deg





Crossed: θ  2  atan2 2  D E 

θ  271.163  deg θ  631.163  deg

2

E  4  D F



θ  244.789  deg

2


5.

Check the Grashof condition. Condition( a b c d ) 


Condition( a b c d )  "Grashof"





A position vector is defined as having length equal to your height in inches (or centimeters). The tangent of its angle is defined as your weight in lbs (or kg) divided by your age in years. Calculate the data for this vector and: a. Draw the position vector to scale on Cartesian axes. b. Write an expression for the position vector using unit vector notation. c. Write an expression for the position vector using complex number notation, in both polar and Cartesian forms.

Assumptions: Height  70, weight  160, age  20 Solution:

The magnitude of the vector is R  Height. The angle that the vector makes with the x-axis is θ  atan

weight 

  age 

a.

θ  82.875 deg

θ  1.446 rad

Draw the position vector to scale on Cartesian axes. y 100 80

R

60 70.000

1.


40

82.875°

20 0

b.

x 20

40

60

80

100

Write an expression for the position vector using unit vector notation.

 cos θ    sin θ 

R  R 

R

 8.682     69.459 

R = 8.682 i + 69.459 j

c. Write an expression for the position vector using complex number notation, in both polar and Cartesian forms. j  1.446

Polar form:

R  68 e

Cartesian form:

R  8.682  j  69.459




A particle is traveling along an arc of 6.5 inch radius. The arc center is at the origin of a coordinate system. When the particle is at position A, its position vector makes a 45-deg angle with the X axis. At position B, its vector makes a 75-deg angle with the X axis. Draw this system to some convenient scale and: a. b. c.

d.

Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. Write a vector equation for the position difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. Check the result of part c with a graphical method.

Given:

Circle radius and vector magnitude, R  6.5 in; vector angles: θA  45 deg

Solution:


θB  75 deg

1.

Establish an X-Y coordinate frame and draw a circle with center at the origin and radius R.

2.

Draw lines from the origin that make angles of 45 and 75 deg with respect to the X axis. Label the intersections of the lines with the circles as A and B, respectively. Make the line segment OA a vector by putting an arrowhead at A, pointing away from the origin. Label the vector RA. Repeat for the line segment OB, labeling it RB.

Y 8 B 6 A 4

RB RA

2

0 a.

b.

X 2

4

6

8

Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. j  θA

Polar form:

RA  R e

Cartesian form:

RA  R  cos θA  j  sin θA 

j

RA  6.5 e

π 4

RA  ( 4.596  4.596j) in

Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms.


c.


j

j  θB

Polar form:

RB  R e

RB  6.5 e

Cartesian form:

RB  R  cos θB  j  sin θB 

180

RB  ( 1.682  6.279j) in

Write a vector equation for the position difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. RBA  RB  RA

d.

75 π

RBA  ( 2.914  1.682j) in

Check the result of part c with a graphical method.

Y 8

3.365 B

6

RBA

1.682

A 4

RB

2.914

2 RA 0

X 2

4

6

8

On the layout above the X and Y components of RBA are equal to the real and imaginary components calculated, confirming that the calculation is correct.




Two particles are traveling along an arc of 6.5 inch radius. The arc center is at the origin of a coordinate system. When one particle is at position A, its position vector makes a 45-deg angle with the X axis. Simultaneously, the other particle is at position B, where its vector makes a 75deg angle with the X axis. Draw this system to some convenient scale and: a. b. c.

d.

Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. Write an exp ession for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. Write a vector equation for the relative position of the particle at B with respect to the particle at A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. Check the result of part c with a graphical method.

Given:

Circle radius and vector magnitude, R  6.5 in; vector angles: θA  45 deg

Solution:


θB  75 deg

1.

Establish an X-Y coordinate frame and draw a circle with center at the origin and radius R.

2.

Draw lines from the origin that make angles of 45 and 75 deg with respect to the X axis. Label the intersections of the lines with the circles as A and B, respectively. Make the line segment OA a vector by putting an arrowhead at A, pointing away from the origin. Label the vector RA. Repeat for the line segment OB, labeling it RB.

Y 8 B 6 A 4

RB RA

2

0 a.

b.

X 2

4

6

8


Polar form:

RA  R e

Cartesian form:

RA  R  cos θA  j  sin θA 

j

RA  6.5 e

π 4

RA  ( 4.596  4.596j) in

Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms.


c.


j

j  θB

Polar form:

RB  R e

RB  6.5 e

Cartesian form:

RB  R  cos θB  j  sin θB 

180

RB  ( 1.682  6.279j) in

Write a vector equation for the relative position of the particle at B with respect to the particle at A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. RBA  RB  RA

d.

75 π

RBA  ( 2.914  1.682j) in


Y 8

3.365 B

6

RBA

1.682

A 4

RB

2.914

2 RA 0

X 2

4

6

8

On the layout above the X and Y components of RBA are equal to the real and imaginary components calculated, confirming that the calculation is correct.




A particle is traveling along the line y = -2x + 10. When the particle is at position A, its position vector makes a 45-deg angle with the X axis. At position B, its vector makes a 75-deg angle with the X axis. Draw this system to some convenient scale and: a. b. c.

d.

Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. Write a vector equation for the position difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. Check the result of part c with a graphical method.

Given:

Vector angles: θA  45 deg

Solution:


θB  75 deg

1.

Establish an X-Y coordinate frame and draw the line y = -2x + 10.

2.

Draw lines from the origin that make angles of 45 and 75 deg with respect to the X axis. Label the intersections of the lines with the line in step 1 as A and B, respectively. Make the line segment OA a vector by putting an arrowhead at A, pointing away from the origin. Label the vector RA. Repeat for the line segment OB, labeling it RB.

Y 10 y = -2x + 10 8 B 6

4

RB

2

0

3.

A

RA X 2

4

6

8

Calculate the coordinates of points A and B. xA tan  θA = 2  xA  10

xA 

10

2  tan  θA

yA  xA tan  θA xB tan  θB = 2  xB  10

xB 

10

2  tan  θB

xA  3.333 yA  3.333 xB  1.745



yB  xB tan  θB 4.

a.

b.

yB  6.511

Calculate the distances of points A and B from the origin. 2

2

RA  4.714

2

2

RB  6.741

RA 

xA  yA

RB 

xB  yB


j

Polar form:

RA  RA e

Cartesian form:

RA  RA  cos θA  j  sin θA 

RA  4.714  e

π 4

RA  3.333  3.333j

Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. j

j  θB

Polar form:

RB  RB e

RB  6.741  e

Cartesian form:

RB  RB  cos θB  j  sin θB 

75 π 180

RB  1.745  6.511j

Y c.

10

Write a vector equation for the position difference between points B and A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically.

y = -2x + 10 8 B

RBA  RB  RA RBA  1.589  3.178j d.

Check the result of part c with a graphical method. On the layout above the X and Y components of RBA are equal to the real and imaginary components calculated, confirming that the calculation is correct.

6 3.178 4

3.553

RBA RB A

2

0

RA X 2

4

6 1.589

8




Two particles are traveling along the line y = -2x2 - 2x +10. When one particle is at position A, its position vector makes a 45-deg angle with the X axis. Simultaneously, the other particle is at position B, where its vector makes a 75-deg angle with the X axis. Draw this system to some convenient scale and: a. b. c.

d.

Write an expression for the particle's position vector in position A using complex number notation, in both polar and Cartesian forms. Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. Write a vector equation for the relative position of the particle at B with respect to the particle at A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. Check the result of part c with a graphical method.

Given:

Vector angles: θA  45 deg

Solution:


θB  75 deg

1.

Establish an X-Y coordinate frame and draw the line y = -2x2 - 2x +10.

2.

Draw lines from the origin that make angles of 45 and 75 deg with respect to the X axis. Label the intersections of the lines with the line drawn in step 1 as A and B, respectively. Make the line segment OA a vector by putting an arrowhead at A, pointing away from the origin. Label the vector RA. Repeat for the line segment OB, labeling it RB.

Y 10 y = -2x^2 - 2x + 10 8

6 B 4 RB 2

A RA

0

3.

X 2

4

6

8

Calculate the coordinates of points A and B. xA tan  θA = 2  xA  2  xA  10 2

2  tan  θA  tan  θA    xA    1    1    2  2  2   

1

 20 

xA  1.608



yA  xA tan  θA

yA  1.608

xB tan  θB = 2  xB  2  xB  10 2

2  tan  θB  tan  θB    xB    1    1    2  2  2   

1

yB  xB tan  θB 4.

a.

b.

c.

xB  1.223

yB  4.564

Calculate the distances of points A and B from the origin. 2

2

RA  2.275

2

2

RB  4.725

RA 

xA  yA

RB 

xB  yB


Polar form:

RA  R e

Cartesian form:

RA  RA  cos θA  j  sin θA 

j

RA  2.275  e

π 4

RA  1.608  1.608j

Write an expression for the particle's position vector in position B using complex number notation, in both polar and Cartesian forms. j  θB

Polar form:

RB  R e

Cartesian form:

RB  RB  cos θB  j  sin θB 

j

RB  4.725  e

75 π 180

RB  1.223  4.564j

Write a vector equation for the relative position of the particle at B with respect to the particle at A. Substitute the complex number notation for the vectors in this equation and solve for the position difference numerically. RBA  RB  RA

d.

 20 

RBA  0.386  2.955j


On the layout on the next page the X and Y components of RBA are equal to the real and imaginary components calculated, confirming that the calculation is correct.



Y 10 y = -2x^2 - 2x + 10 8

6 B 4 RB

2.955

RBA

2.980

2 A RA 0

X 4

2 0.386

6

8




The link lengths and value of 2 for some fourbar linkages are defined in Table P4-1. The linkage configuration and terminology are shown in Figure P4-1. For row a, draw the linkage to scale and graphically find all possible solutions (both open and crossed) for angles 3 and 4. Determine the Grashoff condition.

Given:

Link 1

d  6  in

Link 2

a  2  in

Link 3

b  7  in

Link 4

c  9  in

Solution:

θ2  30 deg

See figure below for one possible solution. Also see Mathcad file P0406a.

1.

Lay out an xy-axis system. Its origin will be the link 2 pivot, O2.

2.

Draw link 2 to some convenient scale at its given angle.

3.

Draw a circle with center at the free end of link 2 and a radius equal to the given length of link 3.

4.

Locate pivot O4 on the x-axis at a distance from the origin equal to the given length of link 1.

5.

Draw a circle with center at O4 and a radius equal to the given length of link 4.

6.

The two intersections of the circles (if any) are the two solutions to the position analysis problem, crossed and open. If the circles don't intersect, there is no solution.

7.

Draw links 3 and 4 in their two possible positions (shown as solid for open and dashed for crossed in the figure) and measure their angles 3 and 4 with respect to the x-axis. From the solution below, OPEN

θ θ

CROSSED

θ θ

8.

31 41 32 42

 88.84  deg  117.29 deg  360  deg  115.21 deg

θ

 360  deg  143.66 deg

θ

42

 244.790 deg  216.340 deg

y


32

B

S  min ( a b c d )

OPEN

L  max( a b c d ) SL  S  L

3

PQ  a  b  c  d  SL

4

return "Grashof" if SL  PQ 88.837°

return "Special Grashof" if SL = PQ

117.286°

A

return "non-Grashof" otherwise

2 O2


115.211°

O4 143.660°

CROSSED B'

x




Given:

The link lengths and value of 2 for some fourbar linkages are defined in Table P4-1. The linkage configuration and terminology are shown in Figure P4-1. For row a, find all possible solutions (both open and crossed) for angles 3 and 4 using the vector loop method. Determine the Grashof condition. Link 2 Link 1 d  6  in a  2  in b  7  in

Link 3

c  9  in

Link 4

θ  30 deg

Two argument inverse tangent atan2( x y ) 

return 0.5 π if x = 0  y  0 return 1.5 π if x = 0  y  0 return atan 

y 

  if x  0  x 

atan 

y 

   π otherwise  x 

Solution: 1.



d

K2 

a

K1  3.0000

2

d

K3 

c

K2  0.6667

 

 

2 a c

A  0.7113

 

B  2  sin θ

B  1.0000

 

C  K1   K2  1   cos θ  K3

C  3.5566

Use equation 4.10b to find values of 4 for the open and crossed circuits.

Open:





2

θ  2  atan2 2  A B 

B  4 A  C



θ  477.286 deg

θ  θ  360  deg



θ  117.286 deg



2

Crossed: θ  2  atan2 2  A B  3.

2

K3  2.0000

A  cos θ  K1  K2 cos θ  K3

2.

2

a b c d

B  4 A  C



θ  216.340 deg


2

d

K5 

b

2

2

c d a b

 

2 a b

 

D  cos θ  K1  K4 cos θ  K5

 

E  2  sin θ

2

K4  0.8571 K5  0.2857 D  1.6774 E  1.0000

 

F  K1   K4  1   cos θ  K5

F  2.5906

2


4.


Use equation 4.13 to find values of 3 for the open and crossed circuits. Open:





θ  2  atan2 2  D E 

2

E  4  D F



θ  θ  360  deg





Crossed: θ  2  atan2 2  D E  5.

θ  88.837 deg 2

E  4  D F






θ  448.837 deg

θ  244.789 deg




Expand equation 4.7b and prove that it reduces to equation 4.7c (p. 157).

Solution:


1.

Write equation 4.7b and expand the two terms that are squared. 2



 

 2  a cosθ  c cosθ  d2

b  a  sin θ  c sin θ

2.

(4.7b)

a sinθ  c sinθ2  a2 sinθ2  2 a c sinθ sinθ  c2 sinθ2

(a)

a cosθ  c cosθ  d2  a2 cosθ2  2 a c cosθ cosθ  2 a d cosθ  2 2 2  2  c d  cos θ  c  cos θ  d

(b)

Add the two expanded terms, equations a and b, noting the identity sin 2x + cos2x = 1. 2

2

2

2

 

 

    

 

 

b  a  c  d  2  a  d  cos θ  2  c d  cos θ  2  a  c sin θ  sin θ  cos θ  cos θ This is equation 4.7c.




The link lengths, value of 2, and offset for some fourbar slider-crank linkages are defined in Table P4-2. The linkage configuration and terminology are shown in Figure P4-2. For row a, draw the linkage to scale and graphically find all possible solutions (both open and crossed) for angles 3 and slider position d.

Given:

Link 2

a  1.4 in

Link 3

Offset

c  1  in

θ  45 deg

Solution:

b  4  in


1.


2.


3.


4.

Draw a horizontal line through y = c (the offset).

5.

The two intersections of the circle with the horizontal line (if any) are the two solutions to the position analysis problem, crossed and open. If the circle and line don't intersect, there is no solution.

6.

Draw link 3 and the slider block in their two possible positions (shown as solid for open and dashed for crossed in the figure) and measure the angle 3 and length d for each circuit. From the solution below, θ31  360  deg  179.856  deg

θ31  180.144 deg

θ32  0.144  deg

d  3.010  in

d  4.990  in 1

2

Y d2 = 3.010

d1 = 4.990

3(CROSSED)

B'

A 2

0.144° O2

45.000°

3 (OPEN)

B 179.856° 1.000 X




Given:

Solution:

The link lengths, value of 2, and offset for some fourbar slider-crank linkages are defined in Table P4-2. The linkage configuration and terminology are shown in Figure P4-2. For row a, using the vector loop method, find all possible solutions (both open and crossed) for angles 3 and slider position d. Link 2 Offset

a  1.4 in c  1  in

Link 3 b  4  in θ  45 deg

See Figure P4-2 and Mathcad file P0410a. Y d2 = 3.010

d1 = 4.990

3(CROSSED)

B'

A 2

0.144°

45.000°

3 (OPEN)

B 179.856° 1.000 X

O2

1.

Determine 3 and d using equations 4.16 and 4.17. Crossed:

 a sin θ  c   b  

θ  0.144 deg

 

d 2  3.010 in

θ  asin

 

d 2  a  cos θ  b  cos θ Open:

 a  sin θ  c  π b  

θ  180.144 deg

 

d 1  4.990 in

θ  asin 

 

d 1  a  cos θ  b  cos θ




The link lengths and the value of 2 and  for some inverted fourbar slider-crank linkages are defined in Table P4-3. The linkage configuration and terminology are shown in Figure P4-3. For row a, draw the linkage to scale and graphically find both open and closed solutions for 3 and 4 and vector RB.

Given:

Link 1

d  6  in

Link 2

Link 4

c  4  in

γ  90 deg

Solution:

a  2  in θ  30 deg


1.


2.


3a. If  = 90 deg, locate O4 on the x-axis at a distance equal the length of link 1 (d) from the origin. Draw a circle with center at O4 and radius equal to the length of link 4 (c). From point A, draw two lines that are tangent to the circle. The points of tangency define the location of the points B for the open and crossed circuits. 3b. When  is not 90 deg there are two approaches to a graphical solution for link 3 and the location of point B: 1) establish the position of link 4 and the angle  by trial and error, or 2) calculate the distance from point A to point B (the instantaneous length of link 3). Using the second approach, from triangle O2AO4

y

B



b

 c

A a 2 d

x 04

02

2

2

 

2

AO4 = a  d  2  a  d  cos θ

and, from triangle AO4B (for the open circuit)

AO4 = b  c  2  b  c cos π  γ 2

2

2

where a, b, c, and d are the lengths of links 2, 3, 4, and 1, respectively. Eliminating AO4 and solving for the unknown distance b for the open branch, b 1 

1 2



 2  c cos π  γ 

2 c cos π  γ 2  4  c2  a2  d2  2 a d cos θ

b 1  1.7932 in 2

2

2

For the closed branch: AO4 = b  c  2  b  c cos( γ) b 2 

1 2



 2  c cos γ 

and

 2 c cosγ 2  4  c2  a2  d2  2 a d cos θ



b 2  1.7932 in Draw a circle with center at point A and radius b 1. Draw a circle with center at O4 and radius equal to the length of link 4 (c). The intersections of these two circles is the solution for the open and crossed locations of the point B. 4.

Draw the complete linkage for the open and crossed circuits, including the slider. The results from the graphical solution below are: θ  127.333  deg

OPEN

CROSSED

θ  100.959  deg

θ  142.666  deg

θ  169.040  deg

RB1  3.719 at 40.708 deg

RB2  2.208 at -20.146 deg

B y 90.0°

b

127.333° c

A

142.666°

a 30.000°

d

x 04

02 B'

169.040° 79.041°




Given:

The link lengths and the value of 2 and  for some inverted fourbar slider-crank linkages are defined in Table P4-3. The linkage configuration and terminology are shown in Figure P4-3. For row a, using the vector loop method, find both open and closed solutions for 3 and 4 and vector RB. Link 1 Link 2 d  6  in a  2  in Link 4

Solution: 1.

c  4  in

γ  90 deg

θ  30 deg


Determine the values of the constants needed for finding 4 from equations 4.25 and 4.26.

 



 



P  a  sin θ  sin γ  a  cos θ  d  cos γ

 

2.

3.

4.

5.



 

P  1.000 in



Q  a  sin θ  cos γ  a  cos θ  d  sin γ

Q  4.268 in

R  c sin γ

R  4.000 in

T  2  P

T  2.000 in

S  R  Q

S  0.268 in

U  Q  R

U  8.268 in

Use equation 4.26c to find values of 4 for the open and crossed circuits.



T  4 S U



T  4 S U

OPEN

θ  2  atan2 2  S T 

CROSSED

θ  2  atan2 2  S T 

2



θ  142.667 deg

2



θ  169.041 deg

Use equation 4.22 to find values of 3 for the open and crossed circuits. OPEN

θ  θ  γ

θ  232.667 deg

CROSSED

θ  θ  γ

θ  79.041 deg

Determine the magnitude of the instantaneous "length" of link 3 from equation 4.24a. OPEN

b 1 

CROSSED

b 2 

    sin θ  γ

a  sin θ  c sin θ

b 1  1.793 in

    sin θ  γ

a  sin θ  c sin θ

b 2  1.793 in

Find the position vector RB from the definition given in the text. OPEN

  

   b1 cosθ  j  sinθ

RB1  a  cos θ  j  sin θ RB1  RB1

RB1  3.719 in

θ  arg RB1

θ  40.707 deg


CROSSED


  

   b2 cosθ  j  sinθ

RB2  a  cos θ  j  sin θ RB2  RB2

RB2  3.091 in

θ  arg RB2

θ  63.254 deg

DESIGN OF MACHINERY

SOLUTION MANUAL 4-12c-1

PROBLEM 4-12c Statement:

Given:

The link lengths and the value of 2 and for some inverted fourbar slider-crank linkages are defined in Table P4-3. The linkage configuration and terminology are shown in Figure P4-3. For row c, using the vector loop method, find both open and closed solutions for 3 and 4 and vector RB . Link 1

d  3 in

Link 2

Link 4

c  6 in

 45 deg

a 10 in  45 deg

Two argument inverse tangent atan2 (x  y)   return 0.5  if x = 0 y 0 return 1.5  if x = 0 y 0

 y if x 0 return atan  x     y  atan  x    otherwise    Solution: 1.

See Mathcad file P0412c.

Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a.



  

P a  sin  sin   a cos  d  cos 



2.

P 7.879 in

  

Q a  sin  cos   a cos  d  sin 

Q 2.121 in

R c sin 

R 4.243 in

T  2 P

T 15.757 in

S R Q

S 2.121 in

U Q R

U 6.364 in

Use equation 4.22c to find values of 4 for the open and crossed circuits. OPEN



2



46.400 deg



2



163.739 deg

  2 atan2 2  S T  T 4 S U 360 deg

CROSSED   2 atan2 2  S T  T 4 S U 360 deg 3.

4.

Use equation 4.18 to find values of 3 for the open and crossed circuits. OPEN

 

91.400 deg

CROSSED

 

118.739 deg

Determine the magnitude of the instantaneous "length" of link 3 from equation 4.20a. OPEN

b1  

CROSSED

b2  

   sin   

a sin  c sin 

   sin 

a sin  c  sin 

b1 2.727 in

b2 11.212 in

DESIGN OF MACHINERY

5.


Find the position vector RB from the definition given on page 162 of the text. OPEN

CROSSED

 

 

  

 

RB1  acos  j  sin  b 1cos  j  sin  RB1   RB1

RB1 8.356in

 arg  RB1

31.331 deg

 

 

  

 

RB2  acos  j  sin  b 2cos  j  sin  RB2   RB2

RB2 12.764 in

 arg  RB2

12.488 deg




Find the transmission angles of the linkage in row a of Table P4-1.

Given:

Link 1

d  6  in

Link 2

a  2  in

Link 3

b  7  in

Link 4

c  9  in

Solution: 1.



d a

2

d

K2 

K1  3.0000

K3 

c

K2  0.6667

 

 

   

2 a c

C  3.5566

Use equation 4.10b to find 4 for the open circuit.





2

θ  2  atan2 2  A B 

B  4 A  C



θ  242.714 deg


2

d

K5 

b

2

2

c d a b

 

2 a b

2

K4  0.8571 K5  0.2857

 

D  cos θ  K1  K4 cos θ  K5

D  1.6774

 

E  2  sin θ

E  1.0000

 

F  K1   K4  1   cos θ  K5

F  2.5906

Use equation 4.13 to find 3 for the open circuit.





θ  2  atan2 2  D E 

2

E  4  D F

θ  θ  360  deg 5.

2

B  1.0000

C  K1   K2  1   cos θ  K3

4.

2

A  0.7113

B  2  sin θ

3.

2

a b c d

K3  2.0000

A  cos θ  K1  K2 cos θ  K3

2.

θ  30 deg



θ  271.163 deg θ  631.163 deg

Use equations 4.32 to find the transmission angle.

θtrans θ θ 

t  θ  θ

θtrans θ θ  208.449 deg

return t if t  0.5 π π  t otherwise 6.

It can be shown that the triangle ABO4 in Figure 4-17 is symmetric with respect to the line AO4 for the crossed branch and, therefore, the transmission angle for the crossed branch is identical to that for the open branch.




Find the minimum and maximum values of the transmission angle for all the Grashof crankrocker linkages in Table P4-1.

Given:

Table P4-1 data:

i  1 2  14 Row  i

"a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" Solution: 1.

d 

a 

b 

c 

6 7 3 8 8 5 6 20 4 20 4 9 9 9

2 9 10 5 5 8 8 10 5 10 6 7 7 7

7 3 6 7 8 8 8 10 2 5 10 10 11 11

9 8 8 6 6 9 9 10 5 10 7 7 8 6

i

i

i

i

See Table P4-1 and Mathcad file P0414.

Determine which of the linkages in Table P4-1 are Grashof. Condition( a b c d ) 


Row a

Condition( 6 2 7 9 )  "Grashof"

Row b


Row c


Row d

Condition( 8 5 7 6 )  "Special Grashof"

Row e


Row f


Row g


Row h

Condition( 20 10 10 10)  "non-Grashof"

Row i


Row j


Row k

Condition( 4 6 10 7 )  "non-Grashof"

Row l




Row m


Row n


2.

Determine which of the Grashof linkages are crank-rockers. To be a Grashof crank-rocker, the linkage must be Grashof and the shortest link is either 2 or 4. This is true of rows a, d, and e.

3.

Use equations 4.32 and 4.33 to calculate the maximum and minimum transmission angles.

Row a

i  1

 b 2  c 2  d  a 2   i  i  i i  μ  acos   2 b  c i i    

μ  if  μ 

π 2

 

π  μ μ

 b 2  c 2  d  a 2   i  i  i i  μ  acos   2 b  c i i  

Row d

i  4

 

π 2

 

π  μ μ


i  5

μ  25.209 deg

 b 2  c 2  d  a 2   i  i  i i  μ  acos   2 b  c i i   μ  if  μ 

Row e

μ  58.412 deg

μ  0.000 deg

μ  25.209 deg

 b 2  c 2  d  a 2   i  i  i i  μ  acos   2 b  c i i    

μ  if  μ 

π 2

 

π  μ μ


μ  44.049 deg

μ  18.573 deg




Find the input angles corresponding to the toggle positions of the non-Grashof linkages in Table P4-1.

Given:

Table P4-1 data:

i  1 2  14 Row  i

"a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" Solution: 1.

d 

a 

b 

c 

6 7 3 8 8 5 6 20 4 20 4 9 9 9

2 9 10 5 5 8 8 10 5 10 6 7 7 7

7 3 6 7 8 8 8 10 2 5 10 10 11 11

9 8 8 6 6 9 9 10 5 10 7 7 8 6

i

i

i

i

See Table P4-1 and Mathcad file P0415.

Determine which of the linkages in Table P4-1 are Grashof. Condition( a b c d ) 


Row a


Row b


Row c


Row d

Condition( 8 5 7 6 )  "Special Grashof"

Row e


Row f


Row g


Row h


Row i


Row j


Row k


Row l



2.


Row m


Row n


There are six non-Grashof rows in the Table: Rows h, and j through n. For each row there are two possible arguments to the arccos function given in equation (4.37). They are: i  8

Row  "h"

j  1

i

ai  di  bi  ci  2

arg

j 1

2

2

2

b c 

2 a  d

i i

arg

j 2

i i

ai  di  bi  ci  2

2

2

2

2 a  d

Row  "j"

i i

ai  di  bi  ci  2

j 1

2

2

2

b c 

2 a  d

i i

j 2

2

2

2

2 a  d

Row  "k"

2

2

2

b c 

2 a  d

i i

j 2

2

2

2

b c 

2 a  d

i i

i  12

Row  "l"

i i

ai  di  bi  ci  2

j 1

2

2

2

b c 

2 a  d

i i

j 2

2

2

2

2 a  d

Row  "m"

2

2

2

b c 

2 a  d

i i

j 2

2

2

2 a  d

i i

i i

a d

i i

ai  di  bi  ci  2

arg

a d

i i

ai  di  bi  ci  2

j 1

i i

j  5

i

arg

a d b c



i i

i  13

i i i i

ai  di  bi  ci  2

arg

i i

a d

j  4

i

arg

i i

a d

i i

ai  di  bi  ci  2

arg

a d

i i

ai  di  bi  ci  2

j 1

i i

j  3

i

arg

a d b c



i i

i  11

i i i i

ai  di  bi  ci  2

arg

i i

a d

j  2

i

arg

b c 

i i

i  10

i i

a d

2

b c 

i i

a d

i i


i  14


Row  "n"

j  6

i

ai  di  bi  ci  2

arg

j 1

2

2

2

b c 

2 a  d

i i

arg

j 2

2

2

2 a  d

2

b c 

i i

 1.250  1.188  0.896 arg    0.960  0.960   0.833 3.

a d

i i

ai  di  bi  ci  2

i i

i i

a d

i i

 0.688   4.938  1.262   1.833  1.262  0.250

Choose the argument values that lie between plus and minus 1,



1 2



θ2h  75.5 deg



2 2



θ2j  46.6 deg



3 1



θ2k  26.4 deg



4 1



θ2l  16.2 deg

θ2h  acos arg θ2j  acos arg

θ2k  acos arg θ2l  acos arg



5 1



θ2m  16.2 deg



6 1



θ2n  33.6 deg

θ2m  acos arg θ2n  acos arg




The link lengths, gear ratio, phase angle, and the value of 2 for some geared fivebar linkages are defined in Table P4-4. The linkage configuration and terminology are shown in Figure P4-4. For row a, draw the linkage to scale and graphically find all possible solutions for angles 3 and 4.

Given:

Link 1

d  4  in

Link 2

a  1  in

Link 3

b  7  in

Link 4

c  9  in

Link 5

f  6  in

Gear ratio

λ  2.0

Phase angle

ϕ  30 deg

Input angle

θ  60 deg

Solution: 1.


Determine whether or not an idler is required. idler 

"required" if λ  0 "not-required" otherwise

idler  "required" 2.

Choose radii for gears 2 and 5 by making a design choice for their center distance (which must be increased if an idler is required). Let the standard center distance when no idler is required be C  0.5 c then C = r2  r5

and

λ =

r2 r5

Solving for r2 and r5, r5 

C λ 1

r2  r5 λ

r5  1.500 in r2  3.000 in

If an idler is required, increase the center distance. C  if  idler = "required" C  r5 C

C  6.000 in

Note that the amount by which C is increased if an idler is required is a design choice that is made based on the size of the gears and the space available. 3.

Using equation 4.27c, determine the angular position of link 5 corresponding to the position of link 2. θ  λ θ  ϕ

θ  150 deg

4.


5.


6.


7.

Locate pivot O4 on the x-axis at a distance from the origin equal to the given length of link 1.

8.

Draw link 5 to some convenient scale at its calculated angle.

9.


10. The two intersections of the circles (if any) are the two solutions to the position analysis problem, crossed and open. If the circles don't intersect, there is no solution. 11. Draw links 3 and 4 in their two possible positions (shown as solid for open and dashed for crossed in the figure) and measure their angles 3 and 4 with respect to the x-axis. From the solution below,



θ  173.64 deg

OPEN

θ  360  deg  177.715  deg θ  182.285 deg

CROSSED

θ  360  deg  115.407  deg

θ  360  deg  124.050  deg

θ  244.593 deg

θ  235.950 deg

12. Draw gears 2 and 5 schematically at their calculated radii. If an idler is required, draw it tangent to gears 2 and 5. Its diameter is a design choice that will be made on strength and space requirements. It does not affect the gear ratio.

y C

4

B

177.7152°

173.6421° 3

5 2

B`

124.0501° x

O2

3

150.0000°

115.4074°

4

O5

DESIGN OF MACHINERY - 5th ed.



Given:

Solution: 1.

The link lengths, gear ratio, phase angle, and the value of 2 for some geared fivebar linkages are defined in Table P4-4. The linkage configuration and terminology are shown in Figure P4-4. For row a, using the vector loop method, find all possible solutions for angles 3 and 4. Link 1

d  4  in

Link 2

a  1  in

Link 3

b  7  in

Link 4

c  9  in

Link 5

f  6  in

Gear ratio

λ  2.0

Phase angle

ϕ  30 deg

Input angle

θ  60 deg


Determine the values of the constants needed for finding 3 and 4 from equations 4.27h and 4.27i.







 







 

A  2  c d  cos λ θ  ϕ  a  cos θ  f



2

A  36.6462 in 2

B  2  c d  sin λ θ  ϕ  a  sin θ

2

2

2

2



2

B  20.412 in

  

C  a  b  c  d  f  2  a  f  cos θ   2  d  a  cos θ  f  cos λ θ  ϕ    2  a  d  sin θ  sin λ θ  ϕ



     





2

C  37.4308 in

2

D  C  A

D  0.78461 in

E  2  B

E  40.823 in

F  A  C

F  74.077 in

2 2





  a cosθ  f 

G  28.503 in





  a sinθ

H  15.876 in

2

G  2  b   d  cos λ θ  ϕ

2

H  2  b   d  sin λ θ  ϕ

2

2.

2

2

2



2

  

K  a  b  c  d  f  2  a  f  cos θ   2  d  a  cos θ  f  cos λ θ  ϕ    2  a  d  sin θ  sin λ θ  ϕ

K  26.569 in

L  K  G

L  1.933 in

M  2  H

M  31.751 in

N  G  K

N  55.072 in



     





2

2 2

2

Use equations 4.28h and 4.28i to find values of 3 and 4 for the open and crossed circuits. OPEN





M  4  L N





E  4  D F





M  4  L N





E  4  D F

θ  2  atan2 2  L M  θ  2  atan2 2  D E 

CROSSED

θ  2  atan2 2  L M  θ  2  atan2 2  D E 

2

2

2

2



 



θ  173.642 deg θ  177.715 deg θ  115.407 deg θ  124.050 deg




The angle between the X and x axes is 25 deg. Find the angular displacement of link 4 when link 2 rotates clockwise from the position shown (+37 deg) to horizontal (0 deg). How does the transmission angle vary and what is its minimum between those two positions? Find the toggle positions of this linkage in terms of the angle of link 2.

Given:

Link lengths: Crank

L2  116

Coupler

L3  108

Rocker

L4  110

Ground link

L1  174

Crank angle for position shown (relative to O2O4):

θ  62 deg

Y y

A

2

Crank rotation angle from position shown to horizontal:

37°

Δθ  37 deg 3 X O2

25°

B

O4

4

x

Solution: 1.

See Figure P4-5a and Mathcad file P0418a.

Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the crossed circuit). K1 

L1

K2 

L2

K1  1.5000

 

2

L1

K3 

L3

K2  1.6111

 

2

2

L2  L3  L4  L1

2

2  L2 L4

K3  1.7307

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ  2.

Determine 4 for the position shown and after the crank has moved to the horizontal position.

 

θ  θ θ



θ  θ θ  Δθ 3.

θ  183.5 deg



θ  212.8 deg

Subtract the two values of 4 to find the angular displacement of link 3 when link 2 rotates clockwise from the position shown to the horizontal.   θ  θ

4.

 2  4 A θ Cθ 

B θ

  29.2 deg


L1 L3

2

K5 

2

2

L4  L1  L2  L3 2  L2 L3

2

K4  1.6111

K5  1.7280


 


 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

5.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

Use equation 4.13 to find values of 3 for the crossed circuit.



 



 

 

θ θ  2   atan2 2  D θ E θ  6.

Determine 3 for the position shown and after the crank has moved to the horizontal position.

 

θ  θ θ

θ  275.1 deg



θ  θ θ  Δθ 7.

 2  4 Dθ F θ 

E θ



θ  256.1 deg

Use equations 4.28 to find the transmission angles. μ  π  θ  θ

μ  88.4 deg

μ  θ  θ

μ  43.4 deg

The transmission angle is smaller when the crank is in the horizontal position. 8.

Check the Grashof condition of the linkage. Condition( a b c d ) 


Condition L1 L2 L3 L4  "non-Grashof" 9.

Using equations 4.37, determine the crank angles (relative to the XY axes) at which links 3 and 4 are in toggle. 2

arg1 

2

2

2  L2 L1 2

arg2 

2

L2  L1  L3  L4

2

2

2

L2  L1  L3  L4 2  L2 L1





L3 L4 L2 L1 L3 L4 L2 L1

θ2toggle  acos arg2 The other toggle angle is the negative of this.

arg1  1.083

arg2  0.094

θ2toggle  95.4 deg


SOLUTION MANUAL 4-18b-1

PROBLEM 4-18b Statement:

Find and plot the angular position of links 3 and 4 and the transmission angle as a function of the angle of link 2 as it rotates through one revolution.

Given:

Link lengths: Wheel (crank)

L2  40

a  L2

Coupler

L3  96

b  L3

Rocker

L4  122

c  L4

Ground link

L1  162

d  L1

Two argument inverse tangent atan2( x y ) 

Y

return 0.5 π if x = 0  y  0 return atan 

y 

B

A

2

return 1.5 π if x = 0  y  0

y 3 X

  if x  0  x 

O2 4

y atan     π otherwise  x  Solution: 1.

See Figure P4-5b and Mathcad file P0418b. O4


x


Condition( a b c d )  "Grashof" 2.

Define one cycle of the input crank: θ  0  deg 1  deg  360  deg

3.

Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). K1 

d

K2 

a

K1  4.0500

 

2

d

K3 

c

K2  1.3279

 

2

2

a b c d

2

2 a c

K3  3.4336

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ  4.

 2  4 A θ Cθ 

B θ

If the calculated value of 4 is greater than 2, subtract 2 from it. If it is negative, make it positive.

 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ


5.



2

d

K5 

b

 

2

2

c d a b

2

K4  1.6875

2 a b

 

K5  2.8875

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

6.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ


 





 

 

θ θ  2   atan2 2  D θ E θ  7.

 2  4 Dθ F θ 

E θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ

8.

Plot 3 and 4 as functions of the crank angle 2 (measured from the ground link). Angular Displacement of Coupler & Rocker

Coupler or Rocker angle, deg

200

 

150

θ θ  deg

  100

θ θ  deg

50

0

0

45

90

135

180

225

270

θ deg Crank angle, deg

9.

Use equations 4.32 to find the transmission angle.

 

 

 

Tran θ  θ θ  θ θ

 

 

 

Trans θ  if  Tran θ 

π 2

 

 

 π  Tran θ Tran θ



315

360



10. Plot the transmission angle.

Transmission Angle 90

Transmission Angle, deg

80

 

Trans θ

70

deg 60

50

40

0

45

90

135

180 θ deg

Wheel angle, deg

225

270

315

360




Find and plot the position of any one piston as a function of the angle of crank 2 as it rotates through one revolution. Once one piston's motion is defined, find the motions of the other two pistons and their phase relationship to the first piston. Y

Given: L2  19

a  L2

Piston-rod length L3  70

b  L3

Crank length

6

3

2

5

8 X

c  0

Offset

4

Solution:

See Figure P4-5c and Mathcad file P0418c. 7

1.

Let pistons 1, 2, and 3 be links 7, 6, and 8, respectively.

2.

Solve first for piston 6. Establish 2 as a range variable: θ  0  deg 2  deg  360  deg

3.

Determine 3 and d using equations 4.16 and 4.17.

 

 a sin θ  b 

θ θ  asin 

4.

c

 

  

d 1 θ  a  cos θ  b  cos θ θ

For each piston (slider) the crank angle is measured counter-clock-wise from the centerline of the piston, which goes through the O2 in all cases. Thus, when the crank angle for piston 1 is 0 deg, it is 120 deg for piston 2 and 240 deg for piston 3. Thus, the crank angles for pistons 2 and 3 are

 

 

θ θ  θ  120  deg 5.

 

π 

θ θ  θ  240  deg

Determine 3 and d for pistons 2 and 3.

 

 a sin θ θ   b 

θ θ  asin 

 

c

π 

    b cosθθ

d 2 θ  a  cos θ θ

 

 a sin θ θ   c  π b  

θ θ  asin 

 

    b cosθθ

d 3 θ  a  cos θ θ

6.

Plot the piston displacements as a function of crank angle (referenced to line AC (see next page).



Piston Displacement (1, 2, and 3) 90

Piston displacement, mm

80

  d2 θ 70 d3 θ d1 θ

60

50

0

60

120

180

240

300

θ deg Piston 1 crank angle, deg.

The solid line is piston 1, the dotted line is piston 2, and the dashed line is piston 3.

360


SOLUTION MANUAL 4-18d-1

PROBLEM 4-18d Statement:

Find the total angular displacement of link 3 and the total stroke of the box as link 2 makes a complete revolution.

Given:

Ground link

L1  150

Input crank

L2  30

Coupler link

L3  150

Output crank

L4  30

Solution:

See Figure P4-5d and Mathcad file P0418d.

Y 3 2

B

A

O2

O4

X A 4

1.

This is a special-case Grashof mechanism in the parallelogram form (see Figure 2-17 in the text). As such, the coupler link 3 executes curvilinear motion and is always parallel to the ground link 1. Thus, the total angular motion of link 3 as crank 2 makes one complete revolution is zero degrees.

2.

The stroke of the box will be equal to twice the length of the crank link in one complete revolution of the crank stroke  2  L2

stroke  60


SOLUTION MANUAL 4-18e-1

PROBLEM 4-18e Statement:

Determine the ratio of angular displacement between links 8 and 2 as a function of angular displacement of input crank 2. Plot the transmission angle at point B for one revolution of crank 2. Comment on the behavior of this linkage. Can it make a full revolution as shown?

Given:

Link lengths:

Crank (O2A)

a 1  20

Coupler (L3)

b 1  160

Crank (O4B)

c1  20

B

A

Ground link (O2O4) d 1  160

O2

3

4 O4 G

E

2 D

5

C

6

7

Ground link (O4O8) d 2  120

Solution:

Crank (O4G)

a 2  30

Coupler (L6)

b 2  120

Crank (O8F)

c2  30

O8 H

F 8

See Figure P4-5e and Mathcad file P0418e.

1.

This is an eightbar, 1-DOF linkage with two redundant links (3 and 6 or 5 and 7) making it, effectively, a sixbar. It is composed of a fourbar (1, 2, 3, and 4) with an output dyad (7 and 8). The input fourbar is a special-case Grashof in the parallelogram configuration. Thus, the output angle is equal to the input angle and the couplers execute curvilinear motion with links 3 and 5 always parallel to the horizontal. The output dyad also behaves like a special-case Grashof with parallelogram configuration so that the angular motion of link 8 is equal to that of link 4. Therefore, the ratio of angular displacement between links 8 and 2 is unity. The mechanism is not capable of making a full revolution. The couplers 3 and 5 (also 6 and 7) cannot pass by each other near 2 = 0 and 180 deg because of interference with the pins that connect them to their cranks.

2.

Define the approximate range of motion of the input crank: θ  0  deg 2  deg  180  deg

3.

Define 3 and 4.

 

θ  0.deg

Use equations 4.32 to find and plot the transmission angle.

 

 

 

tran θ  θ  θ θ

 

 



 

 

 

Tran θ  if tran θ  π tran θ  π tran θ

 

Trans θ  if  Tran θ 

π 2

 

 

 π  Tran θ Tran θ



Transmission Angle at B Transmission Angle, deg

4.

θ θ  θ

90 80 70 60 Trans θ 50 40 deg 30 20 10 0

 

0

45

90 θ deg Crank angle, deg

135

180


SOLUTION MANUAL 4-18f-1

PROBLEM 4-18f Statement:

Find and plot the displacement of piston 4 and the angular displacement of link 3 as a function of the angular displacement of crank 2.

Given:

Link lengths: Crank length, L2

Solution:

a  63

Piston-rod length, L3

b  130

Offset

c  52

4

B Y, x

3

See Figure P4-5f and Mathcad file P0418f.

1.

Establish 2 as a range variable: θ  0  deg 1  deg  360  deg

2.

Determine 3 and d in global XY coord using equations 4.16 and 4.17.

A

 a sin θ  90 deg  c  θ θ  asin  π b  

 





2 y

X

O2

  

d θ  a  cos θ  90 deg  b  cos θ θ

Plot the piston displacement (directly below) and rod angle (next page) as functions of crank angle in the global XY coordinate frame. Piston Displacement 200

150 Piston displacement, mm

3.

 

d θ 100

50

0

0

60

120

180 θ deg Crank angle, deg.

240

300

360



Piston-Rod Angular Displacement 260

Angular displacement, deg

240

 

220

θ θ  deg

200

180

160

0

60

120


240

300

360


SOLUTION MANUAL 4-18g-1

PROBLEM 4-18g Statement:

Find and plot the angular displacement of link 6 versus the angle of input link 2 as it is rotated from the position shown (+30 deg) to a vertical position (+90 deg). Find the toggle positions of this linkage in terms of the angle of link 2.

Given:

Link lengths:

Y

Input (L2)

a  49

Rocker (L4)

c  153

B

Coupler (L3)

b  100

Ground link (L1)

d  87

3 30° A

2 4

Angle from x axis to X axis:

α  121  deg

Starting angle:

θ  30 deg

Crank rotation angle from position shown to vertical:

Solution:

O6

X

6

C O2 5

D

y

O4

Δθ  60 deg

x

121°

See Figure P4-5g and Mathcad file P0418g.

1.

Define one cycle of the input crank in global coord: θ  θ θ  1  deg  θ  Δθ

2.


d

K2 

a

K1  1.7755

 

2

d

K3 

c

K2  0.5686







2

2

a b c d

2

2 a c

K3  1.5592



A θ  cos θ  α  K1  K2 cos θ  α  K3

 





 

 





 

 

θ θ  2   atan2 2  A θ B θ  3.





C θ  K1   K2  1   cos θ  α  K3

B θ  2  sin θ  α

 2  4 A θ Cθ   α

B θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ

4.

Plot 4 as a function of the crank angle 2 (measured from the X-axis) as it rotates from the position shown to the vertical position.



Angular Displacement of Rocker Link 4 120

Rocker angle, deg

110

 

θ θ 

100

deg

90

80 30

40

50

60

70

80

90


4.



Condition( a b c d )  "non-Grashof" 5.

Using equations 4.37, determine the crank angles (relative to the x-axis) at which links 3 and 4 are in toggle. 2

arg1 

2

2



2 a d 2

arg2 

2

a d b c

2

2

a d b c 2 a d

θ2toggle  acos arg1

2



b c a d b c a d

arg1  0.840

arg2  6.338


The other toggle angle is the negative of this. Thus, in the global XY frame the toggle positions are:

θ2XYtoggle  θ2toggle  α

θ2XYtoggle  88.130 deg

θ2XYtoggle  θ2toggle  α

θ2XYtoggle  153.870 deg


SOLUTION MANUAL 4-18h-1

PROBLEM 4-18h Statement:

Find link 4's maximum displacement vertically downward from the position shown. What will the angle of input link 2 be at that position?

Given:

Link lengths:

c2

a  19.8 mm

Crank length, L2 or L8

Solution:

Coupler length, L3 or L5

b1  19.4 mm

Offset of 1, 2, 3, 4

c1  4.5 mm

Distance from O2 to O8

L1  45.8 mm

Coupler length, L5 or L7

b2  13.3 mm

Offset of 1, 2, 5, 6

c2  22.9 mm

Angle of link 2 as shown

θ  47 deg

6

O2

O8

A

2

8

7

5

C

B

9

3

4

D

E

c1

See Figure P4-5h and Mathcad file P0418h.

1.

Links 1, 2, 3, 4, 5, and 6 make up two offset slider-cranks with a common crank, link 2. Links 7, 8, and 9 are kinematically redundant and contribute only to equalizing the forces in the mirror image links. Slider-crank 1, 2, 3, 4 is in the open circuit, and slider-crank 1, 2, 5, 6 is in the crossed circuit.

2.

Calculate the displacement of link 4 with respect to link 2 angle for the position shown in Figure P4-5h using equations 4.17 and 4.16b.

 a sin θ  c1  π b1  

θ  149.038 deg

 

d 10  30.14 mm

θ  asin 





d 10  a  cos θ  b1 cos θ 3.

Link 4 will reach its maximum downward displacement when links 8 and 9 and links 2 and 3 are in the toggle position. However, it is possible that they may not be able to reach this position because links 5 and 7 may be too short to allow links 2 and 8 to rotate far enough to reach toggle with 3 and 9, respectively.

4.

Using equation 4.16a, determine the angle that the crank will make with the x axis (see layout below) when links 5 and 7 are horizontal (5 = -90 deg). This will be the least value of the angle 2.

 a sin θ  c2  θ  asin  b2  

22.9 = c2 O2

y

θ  90 deg

A

a b2

 

29.0°

sin θ  1.000

 

a  sin θ  c2 b2

B

47°

A'

D' b1

 1.000

c2  b2  θ  asin   a 

D 5.90 D'

θ  29.00 deg 4.5= c1


5.

SOLUTION MANUAL 4-18h-2

Use equations 4.17 and 4.16b to determine the displacement of D' with respect to O2.

 a  sin θ  c1  π b1  

θ  164.759 deg

 

d 1  36.03 mm

θ  asin 

 

d 1  a  cos θ  b1 cos θ 6.

The maximum displacement of link 4 from the position shown in Figure P4-5h is the difference between the displacement found in step 5 and that found in step 2.

Δdmax  d 1  d 10

Δdmax  5.90 mm




For one revolution of the driving link 2 of the walking-beam indexing and pick-and-place mechanism in Figure P4-6, find the horizontal stroke of link 3 for the portion of their motion where their tips are above the platen. Express the stroke as a percentage of the crank length O2B. What portion of a revolution of link 2 does this stroke correspond to? Also find the total angular displacement of link 6 over one revolution of link 2.

Given:

Measured lengths: Input crank length (O2A)

a  40

Coupler length (L3)

b  108

Output crank length (L4)

c  40

95

Q

3

A

64

2

4

O4

p  119.81

Coupler data (finger at Q) Distance from O2 to the platen surface

1.

D

C

Ground link length (O2O4) d  108

Solution:

73

E

6

δ  37.54  deg e  64

B

O2

7

O6 O5 185


Links 1, 2, 3 and 4 are a special-case Grashof linkage in the parallelogram form. The tip of the finger at point Q (left end of the coupler) is used as the coupler point. The distance from the tip to the platen is . Top platen surface Q p

 

b D

A a

c d

x

O2

O4

y

2.

Define the crank angle as a range variable and define 3 ,which is constant because the coupler has curvilinear motion.. θ  0  deg 1  deg  360  deg θ  0  deg

3.

82

5

Use equations 4.27 to define the y-component of the vector RP. RP  RA  RPA

  

 

RA  a  cos θ  j  sin θ

 







RPA  p  cos θ  δ  j  sin θ  δ

 

 





RPy θ  a  sin θ  p  sin θ  δ


4.


Define the distance of point Q above the platen (note the direction of the positive y axis in the figure above).

 

 

ε θ  e  RPy θ 5.

Plot  as a function of crank angle 2. Height of Q Above Platen 60

14

168

Height Above Platen

40

 

20

ε θ

0

 20

 40

0

60

120

180

240

300

360

θ deg

6.

From the graph we see that the coupler point Q is above the platen when the crank angle is greater than 168 deg and less than 14 deg. To find the horizontal stroke during that range of 2, calculate the x-components of any point on the coupler, say point A, for those two crank angles and subtract them. Ax1  a  cos( 14 deg)

Ax1  38.812

Ax2  a  cos( 168  deg)

Ax2  39.126

Horizontal stroke when above the platen normalized by dividing by the crank length Stroke  7.

Ax1  Ax2

Stroke  1.95

a

times the crank length

Links 1, 4, 5, and 6 constitute a Grashoff crank-rocker-rocker. The extreme positions of the output rocker (link 6) occur when links 4 and 5 are in extended and overlapping toggle positions (see Figure 3-1b in the text for example, but in this case the mechanism is in the crossed circuit). C2

29.609°

C1

6

O6

B1

O5 5

B2



Given link lengths: LO5B  13

L7  193

LO6C  92

LO5O6  128

In the first position (links 5 and 7 extended), the angle between link 6 and the ground link is:

 L 2  L 2   L  L  2  O6C O5O6 O5B 7  α  acos     L 2 L O6C O5O6  

α  138.312 deg

In the second position (links 5 and 7 overlapping), the angle between link 6 and the ground link is:

 LO6C2  LO5O62   L7  LO5B 2 α  acos   2  LO6C LO5O6  

α  108.702 deg

The total angular displacement of link 6 is the difference between these two angles.

Δ12  α  α

Δ12  29.609 deg




Figure P4-7 shows a power hacksaw, used to cut metal. Link 5 pivots at O5 and its weight forces the saw blade against the workpiece while the linkage moves the blade (link 4) back and forth on link 5 to cut the part. It is an offset slider-crank mechanism. The dimensions are shown in the figure. For one revolution of the driving link 2 of the hacksaw mechanism on the cutting stroke, find and plot the horizontal stroke of the saw blade as a function of the angle of link 2.

Given:

Link Lengths:

3

Crank length, L2

a  75 mm

Coupler length, L3

b  170  mm

Offset

c  45 mm

B

A 4

5

2 O2 O5

1

Assumptions: The arm that guides the slider (hacksaw blade carrier) remains horizontal throughout the stroke. Solution:


1.

This is a slider-crank mechanism in the crossed circuit. The offset is the vertical distance from the horizontal centerline through O2 to point B.

2.

Establish 2 as a range variable: θ  0  deg 2  deg  360  deg

3.

Determine 3 and d using equations 4.16a and 4.17.

 

 a  sin θ  c   b  

θ θ  asin

 

 

  

d θ  a  cos θ  b  cos θ θ

Plot the blade (point B) displacement as a function of crank angle. Hacksaw Blade Stroke  50

 100 Blade displacement, mm

4.

 

d θ

 150

mm

 200

 250

0

60

120


240

300

360




Given:

For the linkage in Figure P4-8, find its limit (toggle) positions in terms of the angle of link O2A referenced to the line of centers O2O4 when driven from link O2A. Then calculate and plot the xy coordinates of coupler point P between those limits, referenced to the line of centers O2O4. P Link lengths: Input (O2A)

a  5.00 in

Coupler (AB)

b  4.40 in

Rocker (O4B)

c  5.00 in

Ground link

d  9.50 in

y

Y

Coupler point data:

B

p  8.90 in δ  56 deg

3 A

4

Coordinate transformation angle: x

2

α  14 deg

O4 1

14.000° X

O2

See Figure P4-8 and Mathcad file P0421. Solution: 1. Define the coordinate systems. The local frame has origin at O2 with the positive x axis going through O4. Let the global frame also have its origin at O2 with the positive X axis to the right. 2.




Using equations 4.37, determine the crank angles (relative to the line AD) at which links 3 and 4 are in toggle. 2

arg1 

2

2



2 a d 2

arg2 

2

a d b c 2

2

a d b c 2 a d

2





arg1  1.209 arg2  0.283

θ2toggle  73.6 deg

The other toggle angle is the negative of this. 4.

Define one cycle of the input crank between limit positions: θ  θ2toggle θ2toggle  1  deg  θ2toggle

5.

Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12.


K1 


d a

K1  1.9000

 

2

d

K4 

K5 

b

K4  2.1591

 

2

2

c d a b

2

2 a b

K5  2.4911

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

6.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

Use equation 4.13 to find values of 3 for the open circuit.

 





 

 2  4 Dθ F θ 

 

θ θ  2   atan2 2  D θ E θ  7.

E θ

Use equations 4.31 to define the x- and y-components of the vector RP. RP  RA  RPA

  

 


 








 

 

  



 

RPx θ  a  cos θ  p  cos θ θ  δ 8.

 



Transform the coupler point coordinates in the local frame to the global frame using coordinate transformation equations.

 

 

 

 

 

 

XP θ  RPx θ  cos α  RPy θ  sin α YP θ  RPx θ  sin α  RPy θ  cos α Plot the coordinates of the coupler point in the global system.

COUPLER CURVE 1.2

1

0.8

Y

9.

  

RPy θ  a  sin θ  p  sin θ θ  δ

0.6

0.4

0.2

0  0.4

 0.2

0

0.2 X

0.4

0.6




For the walking beam mechanism of Figure P4-9, calculate and plot the x and y components of the position of the coupler point P for one complete revolution of the crank O2A. Hint: Calculate them first with respect to the ground link O2O4 and then transform them into the global XY coordinate system (i.e., horizontal and vertical in the figure).

Given:

Link lengths:

Coupler point data:

Ground link

d  2.22

Crank

a  1

Coupler

b  2.06

Rocker

c  2.33

1.

δ  31.000 deg

α  26.5 deg

Coordinate transformation angle: Solution:

p  3.06


Define the coordinate systems. The local frame has origin at O2 with the positive x axis going through O4. Let the global frame also have its origin at O2 with the positive X axis to the right. Y

x

y O4 4

1

26.500° X

O2

P

2 A 3 B

2.

Define one revolution of the input crank: θ  0  deg 2  deg  360  deg

3.


d

K4 

a

K1  2.2200

 

2

d

K5 

b

K4  1.0777

 

2

2 a b

K5  1.1512

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

4.


 





 

 

θ θ  2   atan2 2  D θ E θ  5.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

 2  4 Dθ F θ 

E θ


2

c d a b

2



  

 


 








 

 

  



 


 

  

Transform the coupler point coordinates in the local frame to the global frame using coordinate transformation equations.

 

 

 

 

 

 

XP θ  RPx θ  cos α  RPy θ  sin α YP θ  RPx θ  sin α  RPy θ  cos α Plot the coordinates of the coupler point in the global system.

COUPLER CURVE 0.5

0

Y

7.




 0.5

1

 1.5

2

2.5

3

3.5 X

4

4.5




For the linkage in Figure P4-10, calculate and plot the angular displacement of links 3 and 4 and the path coordinates of point P with respect to the angle of the input crank O2A for one revolution.

Given:

Link lengths:

B

y 3

Ground link

d  2.22

Crank

a  1.0

Coupler

b  2.06

Rocker

c  2.33

b

Coupler point data: p  3.06

P p

A

δ  31.00  deg

2

a

4

2

4

c d

Solution:

x

1

O2

O4


1.


2.


d a

K1  2.2200

 

d

K2 

2

K3 

c

K2  0.9528

 

2

2

a b c d

2

2 a c

K3  1.5265

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 

 



 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  3.

B θ

If the calculated value of 4 is greater than 2, subtract 2 from it.

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ 4.


2

d

K5 

b

 

2

2

c d a b

2

K4  1.0777

2 a b

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

5.


 





 

 

θ θ  2   atan2 2  D θ E θ  6.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

 2  4 Dθ F θ 

E θ


 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

K5  1.1512


7.


Plot 3 and 4 as functions of the crank angle 2 (measured from the ground link). Angular Displacement of Coupler  240

Coupler angle, deg

 260

   280

θ θ  deg

 300  320  340

0

60

120

180

240

300

360

300

360


Angular Displacement of Rocker  200

Rocker angle, deg

 220

 

θ θ 

 240

deg  260

 280

0

60

120

180

240


8.


  

 


 








 

 

  




 

 

  




Plot the coordinates of the coupler point in the local xy coordinate system.



COUPLER POINT CURVE 3

2.5

Y

2

1.5

1

0.5 1.5

2

2.5

3 X

3.5

4




For the linkage in Figure P4-11, calculate and plot the angular displacement of links 3 and 4 with respect to the angle of the input crank O2A for one revolution.

Given:

Link lengths: Link 2

a  2.00 in

Link 3

b  8.375  in

Link 4

c  7.187  in

Link 1

d  9.625  in

A 3 2

B

2

O2 4

1

O4

Solution:


1.


2.


d a

K1  4.8125

 

2

d

K2 

K3 

c

K2  1.3392

 

2

2

a b c d

2

2 a c

K3  2.7186

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 

 



 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  3.

B θ


 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ 4.


2

d

K5 

b

 

2

2

c d a b

2

K4  1.1493

2 a b

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

5.


 





 

 

θ θ  2   atan2 2  D θ E θ  6.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

 2  4 Dθ F θ 

E θ


 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

K5  3.4367


Plot 3 and 4 as functions of the crank angle 2 (measured from the ground link).

Angular Displacement of Coupler

Coupler angle, deg

 290

 300

 310

 320

 330

0

60

120

180

240

300

360

300

360

Crank angle, deg

Angular Displacement of Rocker  220

 230 Rocker angle, deg

7.


 240

 250

 260

0

60

120

180 Crank angle, deg

240




For the linkage in Figure P4-12, find its limit (toggle) positions in terms of the angle of link O2A referenced to the line of centers O2O4 when driven from link O2A. Then calculate and plot the angular displacement of links 3 and 4 and the path coordinates of point P with respect to the angle of the input crank O2A over its possible range of motion referenced to the line of centers O2O4.

Given:

Link lengths: Input (O2A)

a  0.785

Coupler (AB)

b  0.356

Rocker (O4B)

c  0.950

Ground link

d  0.544

A

1.

B

158.286° c

a O2

Coupler point data: p  1.09 Solution:

b

O4

d

δ  0  deg





double rocker

Using the geometry defined in Figure 3-1a in the text, determine the input crank angles (relative to the line O2O4) at which links 2 and 3, and 3 and 4 are in toggle.

 d2  ( a  b ) 2  c2  θ  acos  2 d ( a  b) 

θ  55.937 deg

 a2  d 2  ( b  c) 2  2 a d  

θ  acos 3.

θ  158.286 deg

Define one cycle of the input crank between limit positions: θ  θ θ  1  deg  θ

4.


d

K2 

a

K1  0.6930

 

2

d

K3 

c

K2  0.5726

 

K3  1.1317

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 

 



 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ 

 2  4 A θ Cθ 

B θ

2

2

a b c d 2 a c

2


5.



 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 6.

Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. 2

d

K4 

K5 

b

 

2

2

c d a b

2

K4  1.5281

2 a b

 

K5  0.2440

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

7.




 



 

 

θ θ  2   atan2 2  D θ E θ  8.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

 2  4 Dθ F θ 

E θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 9.



360 300 240 180 120 60 0 50

60

70

80

90

100

110

120

130

Crank angle, deg Coupler Rocker

10. Use equations 4.31 to define the x- and y-components of the vector RP. RP  RA  RPA

  

 


140

150

160


 









 

 

  



RPx θ  a  cos θ  p  cos θ θ  δ

 

 

11. Plot the coordinates of the coupler point in the local xy coordinate system.

COUPLER POINT PATH

Coupler Point Coordinate - y

2

1.5

1

0.5

0

0

0.5

  




1

Coupler Point Coordinate - x

1.5

2





Given:

Link lengths: a  0.86

Input (O2A) Coupler (AB)

b  1.85

Rocker (O4B)

c  0.86

Ground link

d  2.22

1.

B c

a O2

Coupler point data: p  1.33 Solution:

116.037° b

A

O4

d

δ  0  deg






arg1 

2

2



2 a d 2

arg2 

2

a d b c

2

2

a d b c 2 a d

2



b c

arg1  1.228

a d b c

arg2  0.439

a d




Define one cycle of the input crank between limit positions: θ  θ2toggle θ2toggle  1  deg  θ2toggle

4.


d

K2 

a

K1  2.5814

 

d c

K2  2.5814

 

2

K3 

K3  2.0181

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

2

2

a b c d

C θ  K1   K2  1   cos θ  K3

2 a c

2


 






 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 6.


d

K4 

K5 

b

 

2

2

c d a b

2

K4  1.2000

2 a b

 

K5  2.6244

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

7.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ


 





 

 

θ θ  2   atan2 2  D θ E θ  8.

 2  4 Dθ F θ 

E θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 9.



360 300 240 180 120 60 0  120

 80

 40

0

40



  

 


80

120


 









 

 

  



 


 

  


COUPLER POINT PATH 1

Y

0.5

0

 0.5 0.5

1




1.5 X

2

2.5




For the linkage in Figure P4-13, find its limit (toggle) positions in terms of the angle of link O4B referenced to the line of centers O4O2 when driven from link O4B. Then calculate and plot the angular displacement of links 2 and 3 and the path coordinates of point P with respect to the angle of the input crank O4B over its possible range of motion referenced to the line of centers O4O2.

Given:

Link lengths:

116.037° b 3 4

a  0.86

Input (O4B)

b  1.85

Coupler (AB) Rocker (O2A)

c  0.86

Ground link

d  2.22

x


c A

O2

B a O4

d

δ  0  deg y

Solution: 1.





Using equations 4.33, determine the crank angles (relative to the line O4O2) at which links 2 and 3 are in toggle. 2

arg1 

2

2



2 a d 2

arg2 

2

a d b c

2

2

a d b c 2 a d

2





arg1  1.228

arg2  0.439



Define one cycle of the input crank between limit positions: θ  θ4toggle θ4toggle  1  deg  θ4toggle

4.


d a

K2 

d c

2

K3 

K1  2.5814

K2  2.5814

    B θ  2  sin θ

  C θ  K1   K2  1   cos θ  K3

A θ  cos θ  K1  K2 cos θ  K3

2

2

a b c d

K3  2.0181

2 a c

2


 






 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 6.


d

K4 

 

b

K5 

2

2

c d a b

2

K4  1.2000

2 a b

 

K5  2.6244

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

7.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ


 





 

 

θ θ  2   atan2 2  D θ E θ  8.

 2  4 Dθ F θ 

E θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 9.



360 300 240 180 120 60 0  120

 80

 40

0

40


10. Use equations 4.31 to define the x- and y-components of the vector RP. RP  RB  RPB

  

 

RB  a  cos θ  j  sin θ

80

120


 








RPB  p  cos θ  δ  j  sin θ  δ

 

 

  



 

RPx θ  a  cos θ  p  cos θ θ  δ

 

  

11. Plot the coordinates of the coupler point in the local xy coordinate system. COUPLER POINT PATH 1

Y

0.5

0

 0.5

1

0

0.25

0.5



RPy θ  a  sin θ  p  sin θ θ  δ

0.75 X

1

1.25

1.5




For the rocker-crank linkage in Figure P4-14, find the maximum angular displacement possible for the treadle link (to which force F is applied). Determine the toggle positions. How does this work? Explain why the grinding wheel is able to fully rotate despite the presence of toggle positions when driven from the treadle. How would you get it started if it was in a toggle position?

Given:

Link lengths:

x

Input (O2A)

a  600  mm

Coupler (AB)

b  750  mm

Rocker (O4B)

c  130  mm

Ground link

d  900  mm

B

B''

c

O4

B'

b

d

43.331° A'' 25.182° a

A

O2

A'

y

Solution: 1.


Use Figure 3-1(b) in the text to calculate the angles that link O2A makes with the ground link in the toggle positions.

 a2  d 2  ( b  c) 2  θ  acos 2 a d  

θ  43.331 deg

 a2  d 2  ( b  c) 2  2 a d  

θ  68.513 deg

θ  acos 2.

Subtract these two angles to get the maximum angular displacement of the treadle.   θ  θ

3.

  25.182 deg

Despite having transmission angles of 0 deg twice per revolution, the mechanism will work. That is, one will be able to drive the grinding wheel from the treadle (link 2). The reason is that the grinding wheel will act as a flywheel and will carry the linkage through the periods when the transmission angle is low. Typically, the operator will start the motion by rotating the wheel by hand if it is in or near a toggle position.





Given:


a  0.72

Coupler (AB)

b  0.68

Rocker (O4B)

c  0.85

Ground link

d  1.82

P A B


1.

O4

O2

δ  54 deg Solution:

55.355°






arg1 

2

2



2 a d 2

arg2 

2

a d b c 2

2

a d b c 2 a d

2



b c

arg1  1.451

a d b c

arg2  0.568

a d




Define one cycle of the input crank between limit positions: θ  θ2toggle θ2toggle  0.5 deg  θ2toggle

4.


d a

K1  2.5278

 

K2 

 

d c

K2  2.1412

 

2

K3 

K3  3.3422

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

2

2

a b c d

C θ  K1   K2  1   cos θ  K3

2 a c

2


 






 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 6.


2

d

K5 

b

 

2

2

c d a b

2

K4  2.6765

2 a b

 

K5  3.6465

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

7.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

Use equation 4.13 to find values of θ3 for the open circuit.

 





 

 

θ θ  2   atan2 2  D θ E θ  8.

 2  4 Dθ F θ 

E θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 9.

Plot 3 and 4 as functions of the crank angle 2 (measured from the ground link).

Angular Displacement of Coupler & Rocker


360 300 240 180 120 60 0  60

 45

 30

 15

0

15

30



45

60



  

 


 








 

 

  



 


 

  


COUPLER POINT PATH 1.4

1.2

Y

1

0.8

0.6

0.4

0.2 0.2

0.4

0.6




0.8 X

1

1.2

1.4




For the linkage in Figure P4-15, find its limit (toggle) positions in terms of the angle of link O4B referenced to the line of centers O4O2 when driven from link O4B. Then calculate and plot the angular displacement of links 2 and 3 and the path coordinates of point P with respect to the angle of the input crank O4B over its possible range of motion referenced to the line of centers O4O2.

Given:

Link lengths: Input (O4B)

a  0.85

Coupler (AB)

b  0.68

Rocker (O2A)

c  0.72

Ground link

d  1.82

P 47.885° c

Coupler point data: p  0.792 δ  82.032 deg Solution: 1.

B A

a

b

O4

O2





Using equations 4.37, determine the crank angles (relative to the line O4O2) at which links 2 and 3 are in toggle. 2

arg1 

2

2



2 a d 2

arg2 

2

a d b c 2

2

a d b c 2 a d


2



b c

arg1  1.304

a d b c

arg2  0.671

a d



Define one cycle of the input crank between limit positions: θ  θ4toggle θ4toggle  0.5 deg  θ4toggle

4.


d

K2 

a

K1  2.1412

 

d c

K2  2.5278

 

2

K3 

K3  3.3422

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

B θ  2  sin θ

 

 

2

2

a b c d

C θ  K1   K2  1   cos θ  K3

2 a c

2


 






 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  5.

B θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 6.


2

d

K5 

b

 

2

2

c d a b

2

K4  2.6765

2 a b

 

K5  3.4420

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

7.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ


 





 

 

θ θ  2   atan2 2  D θ E θ  8.

 2  4 Dθ F θ 

E θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 9.



360 300 240 180 120 60 0  60

 45

 30

 15

0

15

30


10. Use equations 4.27 to define the x- and y-components of the vector RP. RP  RB  RPB

45

60



  

 

RB  a  cos θ  j  sin θ

 







RPB  p  cos θ  δ  j  sin θ  δ

 

 

  



 

RPx θ  a  cos θ  p  cos θ θ  δ

 

  


COUPLER POINT PATH 1.5

Y

1

0.5

0

0

0.2

0.4

0.6 X



RPy θ  a  sin θ  p  sin θ θ  δ

0.8

1




Write a computer program (or use an equation solver such as Mathcad, Matlab, or TKSolver) to find the roots of y = 9x2 + 50x - 40. Hint: Plot the function to determine good guess values.

Solution:


1.

Plot the function. 2

x  10 9.5  10

f ( x)  9  x  50 x  40

200

100

f ( x)

0

 100

 200  10

8

6

4

2

0

2

x

2.

From the graph, make guesses of x1  6 , x2  1

3.

Define the program using the pseudo code in the text. nroot( f df x) 

y  f ( x) y  TOL

return x if

while y  TOL xx

y df ( x)

y  f ( x) x

where,

3

TOL  1.000  10

4.

Define the derivative of the given function. df ( x)  18 x  50

5.

Use the program to find the roots. r1  nroot f df x1

r1  6.265

r2  nroot f df x2

r2  0.709

4

6

8

10




Write a computer program (or use an equation solver such as Mathcad, Matlab, or TKSolver) to find the roots of y = -x3 - 4x2 + 80x - 40. Hint: Plot the function to determine good guess values.

Solution:


1.

Plot the function. 3

x  15 14.5  10

2

f ( x)  x  4  x  80 x  40

200

100

f ( x)

0

 100

 200  20

 15

 10

5

0

x

2.

From the graph, make guesses of x1  11 , x2  0 , x3  7

3.

Define the program using the pseudo code in the text. nroot( f df x) 

y  f ( x) y  TOL

return x if


y df ( x)

y  f ( x) x

where,

3

TOL  1.000  10

2

4.

Define the derivative of the given function. df ( x)  3  x  8  x  80

5.

Use the program to find the roots. r1  nroot f df x1

r1  11.355


r2  0.515


r3  6.840

5

10




Figure 4-18 (p. 193) plots the cubic function from equation 4.34. Write a computer program (or use an equation solver such as Mathcad, Matlab, or TKSolver) to investigate the behavior of the Newton-Raphson algorithm as the initial guess value is varied from x = 1.8 to 2.5 in steps of 0.1. Determine the guess value at which the convergence switches roots. Explain this root-switching phenomenon based on your observations from this exercise.

Solution:


1.

Define the range of the guess value, the function, and the derivative of the function. xguess  1.8 1.9  2.5 3

2

2

f ( x)  x  2  x  50 x  60 2.

df ( x)  3  x  4  x  50

Define the root-finding program using the pseudo code in the text. nroot( f df x) 

y  f ( x) y  TOL

return x if


y df ( x)

y  f ( x) x 3.

Find the roots that correspond to the guess values. r( xguess)  nroot( f df xguess) 1

1

f ( xguess) df ( xguess)

1

1

1

1.800

1

33.080

1

-2.362

1

-1.177

2

1.900

2

31.570

2

-2.564

2

-1.177

3

2.000

3

30.000

3

-2.800

3

-1.177

2.100 df ( xguess)  4

28.370

nextx( xguess)  4

-3.079

r( xguess)  4

-1.177

xguess  4

4.

nextx( xguess)  xguess 

5

2.200

5

26.680

5

-3.410

5

-1.177

6

2.300

6

24.930

6

-3.807

6

-1.177

7

2.400

7

23.120

7

-4.289

7

6.740

8

2.500

8

21.250

8

-4.882

8

-7.562

Find the roots of the derivative (values of x where the slope is zero). ddf ( x)  6  x  4

5.

xz1  nroot( df ddf 5 )

xz1  4.803

xz2  nroot( df ddf 4 )

xz2  3.470

For guess values up to 2.3, the root found is that whose slope is nearly the same as the slope of the function at the guess value. At 2.4, the value of x that is calculated next results in a slope that throws the next x-value to the right of the extreme function value at x = 3.470. Subsequent estimates of x then follow down the slope to x = 6.740. At a guess value of 2.5, the value of x that is calculated next is to the left of the extreme function value at x = -4.803. Subsequent estimates of x follow up the slope to x = -7.562.




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular position of link 4 and the position of slider 6 in Figure 3-33 as a function of the angle of input link 2.

Given:

Link lengths: Input crank (L2)

a  2.170

Fourbar coupler (L3)

b  2.067

Output crank (L4)

c  2.310

Sllider coupler (L5)

e  5.40

Fourbar ground link (L1) Solution:

d  1.000


1.

This sixbar drag-link mechanism can be analyzed as a fourbar Grashof double crank in series with a slidercrank mechanism using the output of the fourbar, link 4, as the input to the slider-crank.

2.


3.

Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit) in the global XY system. K1 

d

K2 

a

K1  0.4608

 

2

d

K3 

c

K2  0.4329

 

2

2

a b c d

2

2 a c

K3  0.6755

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 

 





 

 

θ θ  2   atan2 2  A θ B θ  4.

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ

 2  4 A θ Cθ   102 deg

B θ

If the calculated value of 4 is greater than 2, subtract 2 from it and if it is negative, make it positive.

 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 5.

Determine the slider-crank motion using equations 4.16 and 4.17 with 4 as the input angle.

 

 c sin θ θ   π e  

 

    e cosθθ

θ θ  asin

f θ  c cos θ θ 6.

Plot the angular position of link 4 and the position of link 6 as functions of the angle of input link 2. See next page.



Angular Position of Link 4 360 315 270

 

θ  θ

225 180

deg 135 90 45 0

0

45

90

135

180

225

270

315

360

315

360

θ deg

Position of Slider 6 With Respect to O4 8 7.167 6.333

 

f θ

5.5 4.667 3.833 3

0

45

90

135

180 θ deg

225

270




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angles at points B and C of the linkage in Figure 3-33 as a function of the angle of input link 2.

Given:

Link lengths: Input crank (L2)

a  2.170

Fourbar coupler (L3)

b  2.067

Output crank (L4)

c  2.310

Sllider coupler (L5)

e  5.40

d  1.000

Fourbar ground link (L1) Solution:


1.

This sixbar drag-link mechanism can be analyzed as a fourbar Grashof double crank in series with a slidercrank mechanism using the output of the fourbar, link 4, as the input to the slider-crank.

2.


3.


d a

K1  0.4608

 

2

d

K2 

K3 

c

K2  0.4329

 

2

2

a b c d

2

2 a c

K3  0.6755

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 

 



 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  4.

B θ


 

  

 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 5.


 

2

d b

K5 

2

2

c d a b

2

K4  0.4838

2 a b

 

K5  0.5178

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

 

F θ  K1   K4  1   cos θ  K5 6.

Use equation 4.13 to find 3 for the open circuit.

 





 

 

θ θ  2   atan2 2  D θ E θ  7.

 2  4 Dθ F θ 

E θ


 

  

 

 

θ θ  if θ θ  2  π θ θ  2  π θ θ


 


  

 

 

θ θ  if θ θ  0 θ θ  2  π θ θ 8.

Calculate (using equations 4.32) and plot the transmission angle at B.

θtransB1 θ  θ θ  θ θ

 

θtransB θ  if  θtransB1 θ 

π 2

 

 

π  θtransB1 θ θtransB1 θ



Transmission Angle at B 40 35 30

θtransB θ

25 20

deg 15 10 5 0

0

45

90

135

180

225

270

315

360

θ deg

9.

Determine the slider-crank motion using equations 4.16 and 4.17 with 4 as the input angle.

 c sin θ θ   π e  

 

θ θ  asin

10. Calculate (using equations 4.32) and plot the transmission angle at C.

θtransC1 θ  θ θ

 

θtransC θ  if  θtransC1 θ 

π 2

 

 

π  θtransC1 θ θtransC1 θ



Transmission Angle at C 30 25

θtransC θ

20 15

deg 10 5 0

0

45

90

135

180 θ deg

225

270

315

360




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of the coupler point of the approximate straight-line linkage shown in Figure 3-29f (p. 142). Use program Fourbar to check your result.

Given:


a  1.000

Coupler (AB)

b  1.600

Rocker (O4B)

c  1.039

Ground link

d  1.200

p  2.690

Coupler point data:

α  60 deg

Coordinate rotation angle: Solution: 1.

δ  0  deg

See Figure 3-29f and Mathcad file P0436.

Check the Grashof condition of the linkage and determine its Baker classification. Condition( a b c d ) 


Condition( a b c d )  "non-Grashof" Since b (link 3) is the longest link and the linkage is non-Grashof, this is a Class 3 triple rocker. Using Figure 3-1a as a guide, determine the limiting values of 2 at the toggle positions. For links 2 and 3 colinear:

 ( b  a) 2  d2  c2 π  2 d ( b  a) 

θ  acos

θ  240 deg

For links 3 and 4 colinear:

 a2  d 2  ( b  c) 2  θ  acos 2 d a   2.

θ  27.683 deg

θ  θ

Define one cycle of the input crank (driving through the links 2-3 toggle position): θ  θ θ  1  deg  360  deg  θ

3.


d

K4 

a

K1  1.2000

 

2

d

K5 

b

K4  0.7500

 

2

2 a b

K5  1.2251

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ 4.

 

 

F θ  K1   K4  1   cos θ  K5


2

c d a b

2




 




 

 

θ θ  2   atan2 2  D θ E θ  5.

 2  4 Dθ F θ 

E θ


     RPA  p   cos θ  δ  j  sin θ  δ  RA  a  cos θ  j  sin θ

 

 

  



 


  



Plot the coordinates of the coupler point in the global X,Y coordinate system using equations 4.0b to rotate the local coordinates to a global frame.

 

 

 

 

 

 

PX θ  RPx θ  cos( α)  RPy θ  sin( α) PY θ  RPx θ  sin( α)  RPy θ  cos( α) COUPLER POINT PATH 2

1

Coupler Point Coordinate - Y

6.

 


0

1

2

3

4

0

1

2 Coupler Point Coordinate - X

3

4




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular position of link 6 in Figure 3-34 as a function of the angle of input link 2.

Given:

Link lengths:

Solution: 1.

Input crank (L2)

g  1.556

First coupler (L3)

f  4.248

First rocker (L4)

c  2.125

Third coupler (CD)

b  2.158

Output rocker (L6)

a  1.542

Second ground link (O4O6) d  1.000

Angle CDB

δ  36 deg

Distance (BD)

O2O4 ground link offsets:

h X  3.259

h Y  2.905


Calculate the length of the O2O4 ground link and the angle that it makes with the global XY system. h 

2.

p  3.274

2

hX  hY

2

 hY    hX 

γ  atan

h  4.366

γ  41.713 deg

Calculate the distance BC on link 5. This is the length of vector R51. Also, calculate the angle between vectors R51 and R52 e 

b  p  2  b  p  cos δ 2

2

e  1.986

Second coupler (BC)

 b 2  e2  p 2    2 b e 

α  acos

α  104.305 deg

β  π  α

β  75.695 deg

3.

This is a Stephenson's sixbar linkage similar to the one shown in Figure 4-13. Since the output link 6 is known to rotate 180 deg and return for a full revolution of link 2 we can use links 6, 5, and 4 as a first-stage fourbar with known input (link 6) and then solve for vector loop equations to get the corresponding motion of link 2.

4.

Define the rotation of the output crank: θ  90 deg 91 deg  270  deg

5.

Use equations 4.8a and 4.10 to calculate 4 in the local xy coordinate system as a function of 6 (for the crossed circuit). K1 

d

K2 

a

K1  0.6485

 

2

d

K3 

c

K2  0.4706

 

2

2

a b c d

2

2 a c

K3  0.4938

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ  6.

 2  4 A θ Cθ 

B θ

Use equations 4.12 and 4.13 to calculate 5 in the local xy coordinate system as a function of 6 (for the crossed circuit). K4 

d b

K4  0.463

2

K5 

2

2

c d a b

K5  0.529

2 a b

2


 


 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5



 



 

 

θ θ  2   atan2 2  D θ E θ  7.

 2  4 Dθ F θ 

E θ

Transform the angles for 4 and 52 into the global XY system and define 51 in the global system.

 

 

θ θ  θ θ  90 deg

 

 

 

 

θ θ  θ θ  90 deg θ θ  θ θ  β 8.

Define a vector loop for the remaining links and solve the resulting vector equation by separating it into real and imaginary parts using the method of section 4.5 and the identities of equations 4.9.

Y X

O2 2 R2

R1

6

A

O6

y

D R 5 R52 3 C R4

3

O4 4 x

R51 B R1 + R4 + R51 + R3 - R2 = 0. In this equation the unknowns are 3 and 2. Following the method of Section 4.5, substitute the complex number notation for each position vector and separate the resulting equations into real and imaginary parts:

 

 

 

 

f  cos θ = g  cos θ  G1 f  sin θ = g  sin θ  G2 where

 

    e cosθθ

G1 θ  h  cos γ  c cos θ θ

 



    e sinθθ

G2 θ  h  sin γ  c sin θ θ 9.

Solve these equations in the manner of equations 4.11 and 4.12 using the identities of equations 4.9 gives:

 2  G2θ2  f 2

2

 

g  G1 θ

 

 

G3 θ 

2 g

 

A' θ  G1 θ  G3 θ

 

 

B' θ  2  G2 θ

 

 

 

C' θ  G1 θ  G3 θ



 θ  B'  B'2  4 A'  C' = 2  A' 2

tan 



 



 

 

θ θ  2   atan2 2  A' θ B' θ 

 2  4 A' θ C'θ 

B' θ

10. Plot 6 vs 2 in global XY coordinates: Rotation of Link 6 vs Link 2 320 300 280 260

 

θ  θ

240 220

deg 200 180 160 140 120

0

20

40

60

80 θ deg

100  90

120

140

160

180




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angles at points B, C, and D of the linkage in Figure 3-34 as a function of the angle of input link 2.

Given:

Link lengths:

Solution: 1.

Input crank (L2)

g  1.556

First coupler (L3)

f  4.248

First rocker (L4)

c  2.125

Third coupler (CD)

b  2.158

Output rocker (L6)

a  1.542

Second ground link (O4O6) d  1.000

Angle CDB

δ  36 deg

Distance (BD)

O2O4 ground link offsets:

h X  3.259

h Y  2.905


Calculate the length of the O2O4 ground link and the angle that it makes with the global XY system. h 

2.

p  3.274

2

hX  hY

2

 hY    hX 

γ  atan

h  4.366

γ  41.713 deg

Calculate the distance BC on link 5. This is the length of vector R51. Also, calculate the angle between vectors R51 and R52 e 

b  p  2  b  p  cos δ 2

2

e  1.986

Second coupler (BC)

 b 2  e2  p 2    2 b e 

α  acos

α  104.305 deg

β  π  α

β  75.695 deg

3.

This is a Stephenson's sixbar linkage similar to the one shown in Figure 4-13. Since the output link 6 is known to rotate 180 deg and return for a full revolution of link 2 we can use links 6, 5, and 4 as a first-stage fourbar with known input (link 6) and then solve for vector loop equations to get the corresponding motion of link 2.

4.

Define the rotation of the output crank: θ  90 deg 91 deg  270  deg

5.

Use equations 4.8a and 4.10 to calculate 4 in the local xy coordinate system as a function of 6 (for the crossed circuit). K1 

d

K2 

a

K1  0.6485

 

2

d

K3 

c

K2  0.4706

 

2

2

a b c d

2

2 a c

K3  0.4938

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ  6.

 2  4 A θ Cθ 

B θ

Use equations 4.12 and 4.13 to calculate 5 in the local xy coordinate system as a function of 6 (for the crossed circuit). K4 

d b

K4  0.463

2

K5 

2

2

c d a b

K5  0.529

2 a b

2


 


 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5



 



 

 2  4 Dθ F θ 

 

θ θ  2   atan2 2  D θ E θ  7.

E θ

Transform the angles for 4 and 52 into the global XY system and define 51 in the global system.

 

 

θ θ  θ θ  90 deg

 

 

 

 

θ θ  θ θ  90 deg θ θ  θ θ  β 8.

Define a vector loop for the remaining links and solve the resulting vector equation by separating it into real and imaginary parts using the method of section 4.5 and the identities of equations 4.9.

Y X

O2 2 R2

R1

6

A

O6

y

D R 5 R52 3 C R4

3

O4 4 x

R51 B R1 + R4 + R51 + R3 - R2 = 0. In this equation the unknowns are 3 and 2. Following the method of Section 4.5, substitute the complex number notation for each position vector and separate the resulting equations into real and imaginary parts:

 

 

 

 

g  cos θ = f  cos θ  G1 g  sin θ = f  sin θ  G2 where

G1 θ  h  cos γ  c cos θ θ

 

    e cosθθ

G2 θ  h  sin γ  c sin θ θ

 

9.



    e sinθθ

Solve these equations for 2 in the manner of equations 4.11 and 4.12 using the identities of equations 4.9 gives:

 2  G2θ2  f 2

2

 

g  G1 θ

 

 

G3 θ 

2 g

 

A' θ  G1 θ  G3 θ

 

 

B' θ  2  G2 θ

 

 

 

C' θ  G1 θ  G3 θ



 θ  B'  B'2  4 A'  C' = 2  A' 2

tan 



 



 

 

θ θ  2   atan2 2  A' θ B' θ 

 2  4 A' θ C'θ 

B' θ

10. Solve these equations for 3 in the manner of equations 4.11 and 4.12 using the identities of equations 4.9 gives:

 2  G2θ2  f 2

2

 

g  G1 θ

 

 

G4 θ 

2 f

 

 

D' θ  G1 θ  G4 θ

 

E' θ  2  G2 θ

 

 

 

F' θ  G1 θ  G4 θ

 θ  E'  E'2  4 D' F' tan   = 2  D' 2

 





 

 

θ θ  2   atan2 2  D' θ E' θ 

 2  4 D'θ F'θ 

E' θ

11. Calculate (using equations 4.32) and plot the transmission angle at B.

θtransB1 θ  θ θ  θ θ

 

θtransB θ  if  θtransB1 θ 

π 2

 

 

π  θtransB1 θ θtransB1 θ



Transmission Angle at B 90 80 70 60

θtransB θ 50 deg

40 30 20 10 0 125

150

175

200

225

 

θ θ  deg


θtransC1 θ  θ θ  θ θ

250

275

300

325



 

θtransC θ  if  θtransC1 θ 

π 2

 

 

π  θtransC1 θ θtransC1 θ



Transmission Angle at C 60 55 50

θtransC θ

45

deg 40 35 30 125

150

175

200

225

 

250

275

300

325

θ θ  deg

13. Calculate (using equations 4.32) and plot the transmission angle at D.

θtransD1 θ  θ  θ θ

 

θtransD θ  if  θtransD1 θ 

π 2

 

 

π  θtransD1 θ θtransD1 θ



Transmission Angle at D 60 50 40

θtransD θ

30

deg 20 10 0 125

150

175

200

225

 

θ  θ deg

250

275

300

325




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of the coupler point of the approximate straight-line linkage shown in Figure 3-29g (p. 142). Use program Fourbar to check your result.

Given:


a  1.000

Coupler (AB)

b  1.200

Rocker (O4B)

c  1.167

Ground link

d  2.305

p  1.5

Coupler point data:

α  30 deg


δ  180  deg

See Figure 3-29g and Mathcad file P0439.



Condition( a b c d )  "non-Grashof" Since d (link 1) is the longest link and the linkage is non-Grashof, this is a Class 1 triple rocker. Using Figure 3-1a as a guide, determine the limiting values of 2 at the toggle positions. For links 3 and 4 colinear:

 a2  d 2  ( b  c) 2  2 d a  

θ  acos 1.

θ  81.136 deg

θ  θ

Define one cycle of the input crank: θ  θ θ  0.5 deg  θ

2.


d

K4 

a

K1  2.3050

 

2

d

K5 

b

K4  1.9208

 

2

2 a b

K5  2.6630

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

3.


 





 

 

θ θ  2   atan2 2  D θ E θ  4.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

 2  4 Dθ F θ 

E θ

Use equations 4.31 to define the x- and y-components of the vector RP.

2

c d a b

2



RP  RA  RPA


 

 

  



 


  



Plot the coordinates of the coupler point in the global X,Y coordinate system using equations 4.0b to rotate the local coordinates to a global frame.

 

 

 

 

 

 

PX θ  RPx θ  cos( α)  RPy θ  sin( α) PY θ  RPx θ  sin( α)  RPy θ  cos( α) COUPLER POINT PATH 2

1


5.

 


0

1

2 2

 1.5

1 Coupler Point Coordinate - X

 0.5

0





Given:

Link lengths:

Solution:

Input crank (L2)

a  1.00

First coupler (L3)

b  3.80

Common rocker (O4B)

c  1.29

Second coupler (L5)

b'  1.29

First ground link (O2O4)

d  3.86

Common rocker (O4C)

a'  1.43

Output rocker (L6)

c'  0.77

Second ground link (O4O6) d'  0.78

Angle BO4C

α  157  deg


1.

This sixbar drag-link mechanism can be analyzed as two fourbar linkages in series that use the output of the first fourbar, link 4, as the input to the second fourbar.

2.


3.


d

K2 

a

K1  3.8600

 

2

d

K3 

c

K2  2.9922

 

2

2

a b c d

2

2 a c

K3  1.2107

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  4.

B θ

Use equations 4.8a and 4.10 to calculate 6 as a function of 2 (for the open circuit).

 

K' 1 

d'

K' 2 

a'

K' 1  0.5455

 

 

θ θ  θ θ  α

Input angle to second fourbar:

2

d'

K' 3 

c'

K' 2  1.0130

2

2

2  a' c'

K' 3  0.7184

    K'1  K'2 cosθθ  K'3

A' θ  cos θ θ

 

  

 

 





 

 

θ θ  2   atan2 2  A' θ B' θ 

5.

    K'3

C' θ  K' 1   K' 2  1   cos θ θ

B' θ  2  sin θ θ

 2  4 A' θ C'θ 

B' θ

Plot the angular position of link 6 as a function of the angle of input link 2.

2

a'  b'  c'  d'



Angular Position of Link 6 100 75 50 25

 

θ  θ

0  25

deg  50  75  100  125  150

0

45

90

135

180 θ deg

225

270

315

360




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angles at points B, C, and D of the linkage in Figure 3-35 as a function of the angle of input link 2.

Given:

Link lengths:

Solution:

Input crank (L2)

a  1.00

First coupler (L3)

b  3.80

Common rocker (O4B)

c  1.29

Second coupler (L5)

b'  1.29


d  3.86

Common rocker (O4C)

a'  1.43

Output rocker (L6)

c'  0.77


Angle BO4C

α  157  deg


1.

This sixbar drag-link mechanism can be analyzed as two fourbar linkages in series that use the output of the first fourbar, link 4, as the input to the second fourbar.

2.


3.


d

K2 

a

K1  3.8600

 

2

d

K3 

c

K2  2.9922

 

2

2 a c

K3  1.2107

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ  4.

 2  4 A θ Cθ 

B θ

Use equations 4.12 and 4.13 to calculate 3 as a function of 2 (for the crossed circuit). K4 

2

d

K5 

b

K4  1.016

 

2

2

c d a b

2

2 a b

K5  3.773

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5

 





 

 

θ θ  2   atan2 2  D θ E θ  5.

 2  4 Dθ F θ 

E θ


θtransB1 θ  θ θ  θ θ

 


π 2

 

2

a b c d

 




2



Transmission Angle at B 90 82.5 75 67.5


60

deg 52.5 45 37.5 30

0

45

90

135

180

225

270

315

360

θ deg

6.


 

K' 1 

d'

K' 2 

a'

K' 1  0.5455

 

 

θ θ  θ θ  α


2

d'

K' 3 

c'

K' 2  1.0130

2

2

2

a'  b'  c'  d' 2  a' c'

K' 3  0.7184

    K'1  K'2 cosθθ  K'3

A' θ  cos θ θ

 

  

 



 



 

 2  4 A' θ C'θ 

 

θ θ  2   atan2 2  A' θ B' θ  7.

    K'3

C' θ  K' 1   K' 2  1   cos θ θ

B' θ  2  sin θ θ

B' θ

Use equations 4.12 and 4.13 to calculate 5 as a function of 2 (for the crossed circuit). K' 4 

2

d'

K' 5 

b'

K' 4  0.605

 

2

2

2

c'  d'  a'  b' 2  a' b'

K' 5  1.010

    K'1  K'4 cosθθ  K'5

D' θ  cos θ θ

 

 

    K'5

F' θ  K' 1   K' 4  1   cos θ θ

 





 

 

θ θ  2   atan2 2  D' θ E' θ  8.

 2  4 D'θ F'θ 

E' θ

Calculate (using equations 4.32) and plot the transmission angle at C.

θtransC1 θ  θ θ  θ θ

 


π 2

 

  

E' θ  2  sin θ θ

 






Transmission Angle at C 30

24


18

deg 12

6

0

0

45

90

135

180

225

270

315

360

θ deg

9.

Calculate (using equations 4.32) and plot the transmission angle at D.

θtransD1 θ  θ θ  θ θ

 

θtransD θ  if  θtransD1 θ 

π 2

 

 

π  θtransD1 θ θtransD1 θ



Transmission Angle at D 120 100 80

θtransD θ

60

deg 40 20 0

0

45

90

135

180 θ deg

225

270

315

360




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of the coupler point of the approximate straight-line linkage shown in Figure 3-29h (p. 142). Use program Fourbar to check your result.

Given:


a  1.000

Coupler (AB)

b  1.000

Rocker (O4B)

c  1.000

Ground link

d  2.000

p  2.0

Coupler point data: Solution: 1.

δ  0  deg

See Figure 3-29h and Mathcad file P0442.




 a2  d 2  ( b  c) 2  2 d a  

θ  acos 1.

θ  75.522 deg

θ  θ

Define one cycle of the input crank: θ  θ θ  0.25 deg  θ

2.

Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. d

K1 

K4 

a

K1  2.0000

 

2

d

K5 

b

K4  2.0000

 

2

2 a b

K5  2.5000

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

3.


 





 

 

θ θ  2   atan2 2  D θ E θ  4.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ

 2  4 Dθ F θ 

E θ


  

 


2

c d a b

2


 









 

 

  



 


 

  

COUPLER POINT PATH 2


1.5

1

0.5

0

1

1.5




2 Coupler Point Coordinate - X

2.5

3





Given:

Link lengths:

Solution:

Input crank (L2)

a  0.450

First coupler (L3)

b  0.990

Common rocker (O4B)

c  0.590


d  1.000

Common rocker (O4C)

a'  0.590

Second coupler (CD)

b'  0.325

Output rocker (L6)

c'  0.325


Link 7 (L7)

e  0.938

Link 8 (L8)

f  0.572

Link 5 extension (DE)

p  0.823

Angle DCE

δ  7.0 deg

Angle BO4C

α  128.6  deg


1.

This eightbar can be analyzed as a fourbar (links 1, 2, 3, and 4) with its output (link 4) as the input to another fourbar (links 1, 4, 5, and 6). Since links 1 and 4 are common to both, we have an eightbar linkage with links 7 & 8 included. Start by analyzing the input fourbar.

2.


3.

Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). d

K1 

K2 

a

K1  2.2222

 

2

d

K3 

c

K2  1.6949

 

2

2

a b c d

2

2 a c

K3  1.0744

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ  4.

 2  4 A θ Cθ 

B θ


 

K' 1 

d'

K' 2 

a'

K' 1  0.7102

 

 

θ θ  θ θ  α


2

d'

K' 3 

c'

K' 2  1.2892

2

2

2  a' c'

K' 3  1.3655

    K'1  K'2 cosθθ  K'3

A' θ  cos θ θ

 

  

 

 





 

 

θ θ  2   atan2 2  A' θ B' θ  5.

    K'3

C' θ  K' 1   K' 2  1   cos θ θ

B' θ  2  sin θ θ

 2  4 A' θ C'θ 

B' θ

Use equations 4.11b, 4.12 and 4.13 to calculate 5 as a function of 2.

2

a'  b'  c'  d'


K' 4 

SOLUTION MANUAL 4-43-2 2

d'

K' 5 

b'

K' 4  1.289

 

2

2

2

c'  d'  a'  b' 2  a' b'

K' 5  1.365

    K'1  K'4 cosθθ  K'5

D' θ  cos θ θ

 

  

E' θ  2  sin θ θ

 

    K'5

F' θ  K' 1   K' 4  1   cos θ θ

 





 

 

θ θ  2   atan2 2  D' θ E' θ  6.

 2  4 D'θ F'θ 

E' θ

Define a vector loop for links 1, 5, 6, 7, and 8 as shown below and write the vector loop equation.

Y C

4 R12

O4

5 O6

X D

R6 6

8

R8 RDE

F

R7 7

E

R12 + R6 +RDE + R7 - R8 = 0. Solving for R7 gives R7 = R8 - R12 - R6 - RDE. In this equation the only unknowns are 7 and 8. Following the method of Section 4.5, substitute the complex number notation for each position vector and separate the resulting equation into real and imaginary parts:

 

 

 

 

e cos θ = f  cos θ  D1 e sin θ = f  sin θ  D2 where

 

    p  cosθθ  δ

D1 θ  d'  c'  cos θ θ

 

    p  sinθθ  δ

D2 θ  c'  sin θ θ 7.

Solve these equations in the manner of equations 4.11 and 4.12 using the identities of equations 4.9 gives:


 

D3 θ 

 

f

2


 2  D2θ2  e2

 D1 θ

2 f

 

 

 

A' θ  D1 θ  D3 θ

 

 

B' θ  2  D2 θ

 

 

C' θ  D1 θ  D3 θ

 θ  B'  B'2  4 A'  C' = 2  A' 2

tan 

 





 

 

θ θ  2   atan2 2  A' θ B' θ  8.

 2  4 A' θ C'θ 

B' θ

Plot the angular position of link 8 as a function of the angle of input link 2. If 81 is greater than 360 deg, subtra 2 from it.

 

 

θ θ  if  θ 

π

    θ θ  0  θ θ  2 π θ θ 

4

Angular Position of Link 8 360 315 270 225

  180

θ  θ

135

deg 90 45 0  45  90

0

45

90

135

180

225

270

θ deg

The graph shows that link 8 rotates 360 deg between 2 = 19 deg and 2 = 209 deg. θ( 19 deg)  42 deg θ( 209  deg)  θ( 19 deg)  360.0 deg

θ( 209  deg)  2  π  42 deg

315

360




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angles at points B, C, D, E, and F of the linkage in Figure 3-36 as a function of the angle of input link 2.

Given:

Link lengths:

Solution:

Input crank (L2)

a  0.450

First coupler (L3)

b  0.990

Common rocker (O4B)

c  0.590


d  1.000

Common rocker (O4C)

a'  0.590

Second coupler (CD)

b'  0.325

Output rocker (L6)

c'  0.325


Link 7 (L7)

e  0.938

Link 8 (L8)

f  0.572

Link 5 extension (DE)

p  0.823

Angle DCE

δ  7.0 deg

Angle BO4C

α  128.6  deg


1.

This eightbar can be analyzed as a fourbar (links 1, 2, 3, and 4) with its output (link 4) as the input to another fourbar (links 1, 4, 5, and 6). Since links 1 and 4 are common to both, we have an eightbar linkage with links 7 & 8 included. Start by analyzing the input fourbar.

2.


3.

Use equations 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit). d

K1 

K2 

a

K1  2.2222

 

2

d

K3 

c

K2  1.6949

 

2

2

a b c d 2 a c

K3  1.0744

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ  4.

 2  4 A θ Cθ 

B θ

Use equations 4.12 and 4.13 to calculate 3 as a function of 2 (for the open circuit). K4 

2

d

K5 

b

K4  1.010

 

2

2

c d a b

2

2 a b

K5  2.059

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

F θ  K1   K4  1   cos θ  K5

 





 

 

θ θ  2   atan2 2  D θ E θ  5.

 2  4 Dθ F θ 

E θ

Use equations 4.8a and 4.10 to calculate 6 as a function of 2 (for the open circuit). Input angle to second fourbar:

 

E θ  2  sin θ

 

 

θ θ  θ θ  α

2


K' 1 


d'

K' 2 

a'

K' 1  0.7102

 

2

d'

K' 3 

c'

K' 2  1.2892

2

2

2

a'  b'  c'  d' 2  a' c'

K' 3  1.3655

    K'1  K'2 cosθθ  K'3

A' θ  cos θ θ

 

  

 



 



 

 

θ θ  2   atan2 2  A' θ B' θ  6.

    K'3

C' θ  K' 1   K' 2  1   cos θ θ

B' θ  2  sin θ θ

 2  4 A' θ C'θ 

B' θ

Use equations 4.11b, 4.12 and 4.13 to calculate 5 as a function of 2. K' 4 

2

d' b'

K' 4  1.289

 

2

2

2

c'  d'  a'  b'

K' 5 

2  a' b'

K' 5  1.365

    K'1  K'4 cosθθ  K'5

D' θ  cos θ θ

 

  

E' θ  2  sin θ θ

 

    K'5

F' θ  K' 1   K' 4  1   cos θ θ

 





 

  

 



 

 

θ θ  2   atan2 2  D' θ E' θ 

 

 2  4 D'θ F'θ 

E' θ

 

θ θ  if θ θ  2  π θ θ  2  π θ θ

   



 

 

θ θ  if  105  deg  θ  322  deg  θ θ  0 θ θ  2  π θ θ  7.

Define a vector loop for links 1, 5, 6, 7, and 8 as shown on the next page and write the vector loop equation. R12 + R6 +RDE + R7 - R8 = 0. Solving for R7 gives R7 = R8 - R12 - R6 - RDE. In this equation the only unknowns are 7 and 8. Following the method of Section 4.5, substitute the complex number notation for each position vector and separate the resulting equation into real and imaginary parts:

 

 

 

 

e cos θ = f  cos θ  D1 e sin θ = f  sin θ  D2 where

 

    p  cosθθ  δ

D1 θ  d'  c'  cos θ θ

 

    p  sinθθ  δ

D2 θ  c'  sin θ θ



Y C

4 R12

O4

5 X

O6

D

R6 6

R8

8

RDE

R7

F

E

7 8.

Solve these equations in the manner of equations 4.10 using the identities of equations 4.9 gives:

 

D3 θ 

f

2

 2  D2θ2  e2

 D1 θ

2 f

 

 

 

A' θ  D1 θ  D3 θ

 

 

B' θ  2  D2 θ

 

 

 

C' θ  D1 θ  D3 θ

 θ  B'  B'2  4 A'  C' = 2  A' 2

tan 



 



 

 

θ θ  2   atan2 2  A' θ B' θ 

 

 

θ θ  if  θ  9.

 2  4 A' θ C'θ 

B' θ

π

    θ θ  0  θ θ  2 π θ θ 4 

Similarly, solve these equations in the manner of equations 4.11, 4.12 and 4.13 using the identities of equations 4.9 gives:

 

D4 θ 

 

f

2

 2  D2θ2  e2

 D1 θ

2 e

 

 

D' θ  D1 θ  D4 θ

 θ  E'  E'2  4 D' F' = 2  D' 2

tan 

 

 

E' θ  2  D2 θ

 

 

 

F' θ  D1 θ  D4 θ




 




 

 2  4 D'θ F'θ 

 

θ θ  2   atan2 2  D' θ E' θ 

E' θ

10. Calculate (using equations 4.32) and plot the transmission angle at B.

θtransB1 θ  θ θ  θ θ

 


π 2

 

 





θtransB θ 60 deg

50 40 30 20

0

45

90

135

180

225

270

315

360

θ deg


θtransC1 θ  θ θ  θ θ θtransC2 θ  if 

π

2

 

 

 

 θtransC1 θ  2  π π  θtransC1 θ θtransC1 θ




100

θtransC θ deg 50

0

0

45

90

135

180 θ deg

225

270

315

360



12. Calculate (using equations 4.32) and plot the transmission angle at D.

θtransD1 θ  θ θ  θ θ  π θtransD2 θ  if 

π

2

 

 

 

 θtransD1 θ  π π  θtransD1 θ θtransD1 θ



Transmission Angle at D 90 80 70 60

θtransD θ 50 deg

40 30 20 10 0

0

45

90

135

180

225

270

315

360

315

360

θ deg

13. Calculate (using equations 4.32) and plot the transmission angle at E.

θtransE1 θ  θ θ  θ θ

 

θtransE θ  if  θtransE1 θ 

π 2

 

 

π  θtransE1 θ θtransE1 θ



Transmission Angle at E 90 80 70

θtransE θ

60

deg 50 40 30

0

45

90

135

180 θ deg

225

270




Model the linkage shown in Figure 3-37a in Fourbar. Export the coupler curve coordinates to Excel and calculate the error function versus a true circle.

Given:


a  0.136

Coupler (AB)

b  1.000

Rocker (O4B)

c  1.000

Ground link

d  1.414


p  2.000

δ  0  deg


Model the linkage in Fourbar.

2.

Write coupler point coordinates to a data file.

3.

Import the data file into Excell and add columns for the true circle coordinates and radius. Analyze the coupler point radius to determine its mean, maximum deviation from mean, and average absolute deviation from its mean (See next two pages, note that the last four columns were added to the imported data).


FOURBAR P0445 Tom Cook


Design #

Selected Linkage Parameters a = 0.136 b = 1.000 Angle Step Deg

2

8/9/2006

c = 1.000

d = 1.414

Coupler Pt Coupler Pt Coupler Pt Coupler Pt True Circ True Circ True Circ Coupler Pt X Y Mag Ang X Y R R in in in in in in in in 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100 105 110 115 120 125 130 135 140 145 150 155 160 165 170 175 180 185 190 195 200 205 210

1.414 1.4283 1.4423 1.4561 1.4693 1.4819 1.4937 1.5046 1.5145 1.5233 1.5309 1.5373 1.5425 1.5465 1.5493 1.5508 1.5512 1.5505 1.5488 1.546 1.5424 1.5379 1.5326 1.5266 1.52 1.5128 1.5052 1.4971 1.4887 1.4799 1.471 1.4618 1.4524 1.4429 1.4333 1.4237 1.414 1.4043 1.3947 1.3851 1.3756 1.3662 1.357

1.5384 1.5379 1.5363 1.5336 1.5299 1.5251 1.5195 1.5129 1.5055 1.4974 1.4885 1.479 1.469 1.4585 1.4475 1.4363 1.4248 1.4131 1.4014 1.3897 1.378 1.3665 1.3552 1.3442 1.3336 1.3235 1.314 1.3051 1.2969 1.2895 1.2829 1.2772 1.2725 1.2688 1.2661 1.2645 1.2639 1.2645 1.2661 1.2688 1.2725 1.2772 1.2829

2.0895 2.0988 2.1072 2.1147 2.1212 2.1265 2.1307 2.1337 2.1355 2.136 2.1353 2.1333 2.1301 2.1258 2.1203 2.1138 2.1063 2.0979 2.0887 2.0788 2.0683 2.0572 2.0458 2.0341 2.0221 2.0101 1.998 1.9861 1.9744 1.9629 1.9518 1.9411 1.931 1.9214 1.9124 1.9041 1.8965 1.8897 1.8836 1.8784 1.8739 1.8703 1.8674

47.413 47.1164 46.8058 46.4846 46.1562 45.8237 45.4901 45.1581 44.8303 44.5088 44.1957 43.8925 43.6007 43.3213 43.0555 42.8038 42.567 42.3456 42.1401 41.9507 41.7781 41.6225 41.4846 41.3647 41.2636 41.1819 41.1204 41.0799 41.0612 41.0654 41.0931 41.1453 41.2227 41.3257 41.455 41.6105 41.7924 42.0001 42.2329 42.49 42.7699 43.0708 43.3907

1.4140 1.4021 1.3904 1.3788 1.3675 1.3565 1.3460 1.3360 1.3266 1.3178 1.3098 1.3026 1.2962 1.2907 1.2862 1.2826 1.2801 1.2785 1.2780 1.2785 1.2801 1.2826 1.2862 1.2907 1.2962 1.3026 1.3098 1.3178 1.3266 1.3360 1.3460 1.3565 1.3675 1.3788 1.3904 1.4021 1.4140 1.4259 1.4376 1.4492 1.4605 1.4715 1.4820

1.5500 1.5495 1.5479 1.5454 1.5418 1.5373 1.5318 1.5254 1.5182 1.5102 1.5014 1.4920 1.4820 1.4715 1.4605 1.4492 1.4376 1.4259 1.4140 1.4021 1.3904 1.3788 1.3675 1.3565 1.3460 1.3360 1.3266 1.3178 1.3098 1.3026 1.2962 1.2907 1.2862 1.2826 1.2801 1.2785 1.2780 1.2785 1.2801 1.2826 1.2862 1.2907 1.2962

0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360

0.1244 0.1247 0.1255 0.1268 0.1284 0.1302 0.1322 0.1341 0.1359 0.1375 0.1386 0.1394 0.1398 0.1398 0.1394 0.1386 0.1376 0.1365 0.1354 0.1342 0.1334 0.1327 0.1324 0.1325 0.1330 0.1340 0.1353 0.1370 0.1389 0.1409 0.1430 0.1449 0.1466 0.1480 0.1492 0.1498 0.1501 0.1498 0.1492 0.1480 0.1466 0.1449 0.1430


215 220 225 230 235 240 245 250 255 260 265 270 275 280 285 290 295 300 305 310 315 320 325 330 335 340 345 350 355 360

1.3481 1.3393 1.3309 1.3228 1.3152 1.308 1.3014 1.2954 1.2901 1.2856 1.282 1.2792 1.2775 1.2768 1.2772 1.2787 1.2815 1.2855 1.2907 1.2971 1.3047 1.3135 1.3234 1.3343 1.3461 1.3587 1.3719 1.3857 1.3997 1.414

1.2895 1.2969 1.3051 1.314 1.3235 1.3336 1.3442 1.3552 1.3665 1.378 1.3897 1.4014 1.4131 1.4248 1.4363 1.4475 1.4585 1.469 1.479 1.4885 1.4974 1.5055 1.5129 1.5195 1.5251 1.5299 1.5336 1.5363 1.5379 1.5384


1.8655 1.8643 1.864 1.8645 1.8659 1.868 1.871 1.8747 1.8793 1.8846 1.8907 1.8975 1.905 1.9132 1.922 1.9314 1.9415 1.952 1.963 1.9744 1.9861 1.998 2.0101 2.0222 2.0342 2.0461 2.0577 2.0688 2.0795 2.0895

43.7272 44.0777 44.439 44.8081 45.1815 45.5556 45.9269 46.2915 46.6458 46.986 47.3085 47.61 47.8871 48.1368 48.3563 48.5433 48.6956 48.8117 48.8904 48.9308 48.9328 48.8966 48.823 48.713 48.5685 48.3915 48.1846 47.9504 47.6921 47.413

1.4920 1.5014 1.5102 1.5182 1.5254 1.5318 1.5373 1.5418 1.5454 1.5479 1.5495 1.5500 1.5495 1.5479 1.5454 1.5418 1.5373 1.5318 1.5254 1.5182 1.5102 1.5014 1.4920 1.4820 1.4715 1.4605 1.4492 1.4376 1.4259 1.4140

1.3026 1.3098 1.3178 1.3266 1.3360 1.3460 1.3565 1.3675 1.3788 1.3904 1.4021 1.4140 1.4259 1.4376 1.4492 1.4605 1.4715 1.4820 1.4920 1.5014 1.5102 1.5182 1.5254 1.5318 1.5373 1.5418 1.5454 1.5479 1.5495 1.5500

The mean value of the coupler point radius is

0.1366

The maximum deviation from the mean is

0.0135

The average absolute deviation from the mean is

0.005127

0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360 0.1360

0.1409 0.1389 0.1370 0.1353 0.1340 0.1330 0.1325 0.1324 0.1327 0.1334 0.1342 0.1354 0.1365 0.1376 0.1386 0.1394 0.1398 0.1398 0.1394 0.1386 0.1375 0.1359 0.1341 0.1322 0.1302 0.1284 0.1268 0.1255 0.1247 0.1244




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of point P in Figure 3-37a as a function of the angle of input link 2. Also plot the variation (error) in the path of point P versus that of point A.

Given:


a  0.136

Coupler (AB)

b  1.000

Rocker (O4B)

c  1.000

Ground link

d  1.414

p  2.000


δ  0  deg



S  min ( a b c d ) L  max( a b c d ) SL  S  L PQ  a  b  c  d  SL return "Grashof" if SL  PQ return "Special Grashof" if SL = PQ

return "non-Grashof" otherwise crank rocker Condition( a b c d )  "Grashof" 1.


2.

Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12. d

K1 

K4 

a

K1  10.3971

 

2

d

K5 

b

K4  1.4140

 

2

2

c d a b

2

2 a b

K5  7.4187

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

3.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ




 



 

 

θ θ  2   atan2 2  D θ E θ  4.

 2  4 Dθ F θ 

E θ


  

 


 








 

 

  




 

 

  




Plot the coordinates of the coupler point in the local xy coordinate system.



COUPLER POINT PATH

Coupler Point Coordinate - y

 1.2

1.414

 1.3

 1.4

 1.414

 1.5

 1.6 1.2

1.3

1.4

1.5

1.6

Coupler Point Coordinate - x

6.

Replot, transforming the coupler path to 0,0 and plot the path of point A.

 

 

 

XA θ  a  cos θ

 

YA θ  a  sin θ

PATHS OF POINTS A &P 0.2

0.1

0

 0.1

 0.2  0.2

 0.1 Point P Point A

0

0.1

0.2




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angle at point B of the linkage in Figure 3-37a as a function of the angle of input link 2.

Given:


a  0.136

Coupler (AB)

b  1.000

Rocker (O4B)

c  1.000

Ground link

d  1.414

p  2.000


δ  0  deg





crank rocker


2.


d

K2 

a

K1  10.3971

 

2

d

K3 

c

K2  1.4140

 

2

2 a c

K3  7.4187

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 

θ θ  2   atan2 2  A θ B θ  3.

 2  4 A θ Cθ 

B θ


2

d

K5 

b

K4  1.414

 

2

2

c d a b

2

2 a b

K5  7.419

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 





 

 

 

E θ  2  sin θ

F θ  K1   K4  1   cos θ  K5 θ θ  2   atan2 2  D θ E θ 

 2  4 Dθ F θ 

E θ

2

a b c d

2


4.



θtransB1 θ  θ θ  θ θ

 


π 2

 

 




Transmission Angle at B 90

85

θtransB θ deg 80

75

0

45

90

135

180 θ deg

225

270

315

360

DESIGN OF MACHINERY - 5thEd.



Figure 3-29f shows Evan's approximate straight-line linkage #1. Determine the range of motion of link 2 for which the point P varies no more than 0.0025 from the straight line X = 1.690 (assuming that O2 is the origin of a global coordinate frame whose positive X axis is rotated 60 deg from O2O4).

Given:


a  1.000

Coupler (AB)

b  1.600

Rocker (O4B)

c  1.039

Ground link

d  1.200

p  2.690

Coupler point data:

α  60 deg


δ  0  deg

See Figure 3-29f and Mathcad file P0448.



Condition( a b c d )  "non-Grashof" Since b (link 3) is the longest link and the linkage is non-Grashof, this is a Class 3 triple rocker. Using Figure 3-1a as a guide, determine the limiting values of 2 at the toggle positions. For links 2 and 3 colinear:

 ( b  a) 2  d2  c2 π  2 d ( b  a) 

θ  acos

θ  240 deg

For links 3 and 4 colinear:

 a2  d 2  ( b  c) 2  2 d a  

θ  acos 1.

θ  27.683 deg

θ  θ

Define one cycle of the input crank (driving through the links 2-3 toggle position): θ  θ θ  1  deg  360  deg  θ

2.


d

K4 

a

K1  1.2000

 

2

d

K5 

b

K4  0.7500

 

2

2 a b

K5  1.2251

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ 3.

 

 

F θ  K1   K4  1   cos θ  K5


2

c d a b

2

DESIGN OF MACHINERY - 5thEd.

 






 

 

θ θ  2   atan2 2  D θ E θ  4.

 2  4 Dθ F θ 

E θ

Use equations 4.31 to define the x-components of the vector RP. RP  RA  RPA


 

 

  



 


 

  

Plot the X coordinate of the coupler point in the global X,Y coordinate system using equations 4.0b to rotate the local coordinates to a global frame.

 

 

 

PX θ  RPx θ  cos( α)  RPy θ  sin( α) X COORDINATE 1.7 1.698 Coupler Point Coordinate - X

1.696 1.694 1.692 1.69 1.688 1.686 1.684 1.682 1.68 180

210

240

270

300

Input Angle - Theta2

6.




Using the graph for guess values, solve by trial and error to find 2 for X = 1.6900 +/-0.0025. PX ( 201.525  deg)  1.69250

θ2min  201.525  deg

PX ( 273.450  deg)  1.69250

θ2max  273.450  deg




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of point P in Figure 3-37b as a function of the angle of input link 2.

Given:

Link lengths:

Solution: 1.

Input crank (L2)

a  0.50

First coupler (AB)

b  1.00

Rocker 4 (O4B)

c  1.00

Rocker 5 (L5)

c'  1.00

Ground link (O2O4)

d  0.75

Second coupler 6 (CD)

b'  1.00

Coupler point (DP)

p  1.00

Distance to OP (O2OP)

d'  1.50


Links 4, 5, BC, and CD form a parallelogram whose opposite sides remain parallel throughout the motion of the fourbar 1, 2, AB, 4. Define a position vector whose tail is at point D and whose tip is at point P and another whose tail is at O4 and whose tip is at point D. Then, since R5 = RAB and RDP = -R4, the position vector from O2 to P is P = R1 + RAB - R4. Separating this vector equation into real and imaginary parts gives the equations for the X and Y coordinates of the coupler point P.

 

 

 

XP = d  b  cos θ  c cos θ

 

YP = b  sin θ  c sin θ

2.


3.


d a

K1  1.5000

 

2

d

K2 

K3 

c

K2  0.7500

 

2

2

a b c d

2

2 a c

K3  0.8125

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 

 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ



 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  4.

B θ


 

2

d

K5 

b

2

2

c d a b

2

2 a b

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

K4  0.7500

K5  0.8125

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5 5.


 





 

 

θ θ  2   atan2 2  D θ E θ  6.

 2  4 Dθ F θ 

E θ

Define a local xy coordinate system with origin at OP and with the positive x axis to the right. The coordinates of P are transformed to xP = XP - d', yP = YP.

 

    c cosθθ  d'

xP θ  d  b  cos θ θ

 

    c sinθθ

yP θ  b  sin θ θ


7.


Plot the path of P as a function of the angle of link 2.

Path of Coupler Point P About OP 0.6 0.5 0.4 0.3 0.2 0.1

 

yP θ

0

 0.1  0.2  0.3  0.4  0.5  0.6  0.6  0.5

 0.4  0.3

 0.2  0.1

0

 

xP θ

0.1

0.2

0.3

0.4

0.5

0.6




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the transmission angles at points B, C, and D of the linkage in Figure 3-37b as a function of the angle of input link 2.

Given:

Link lengths:

Solution:

Input crank (L2)

a  0.50

First coupler (AB)

b  1.00

Rocker 4 (O4B)

c  1.00

Rocker 5 (L5)

c'  1.00

Ground link (O2O4)

d  0.75

Second coupler 6 (CD)

b'  1.00

Coupler point (DP)

p  1.00

Distance to OP (O2OP)

d'  1.50


1.

Links 4, 5, BC, and CD form a parallelogram whose opposite sides remain parallel throughout the motion of the fourbar 1, 2, AB, 4. Therefore, the transmission angles at points B and D will be the same and the transmission angle at point C will be the complement of the angle at B.

2.


3.


d a

K1  1.5000

 

2

d

K2 

K3 

c

K2  0.7500

 

2

2

a b c d

2

2 a c

K3  0.8125

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 



 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ

 

 2  4 A θ Cθ 

 

θ θ  2   atan2 2  A θ B θ  4.

B θ


 

2

d

K5 

b

2

2

c d a b

2

K4  0.7500

2 a b

 

 

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

E θ  2  sin θ

 

F θ  K1   K4  1   cos θ  K5 5.


 





 

 

θ θ  2   atan2 2  D θ E θ  6.

 2  4 Dθ F θ 

E θ

Calculate (using equations 4.32) and plot the transmission angles at B and D.

θtransB1 θ  θ θ  θ θ

 


π 2

K5  0.8125

 

 






Transmission Angles at B and D 80

60


40

deg

20

0

0

45

90

135

180

225

270

315

360

θ deg

6.

Calculate and plot the transmission angle at C.

θtransC1 θ  180  deg  θtransB θ

 


π 2

 

 





60


40

deg

20

0

0

45

90

135

180 θ deg

225

270

315

360




Figure 3-29g shows Evan's approximate straight-line linkage #2. Determine the range of motion of link 2 for which the point P varies no more than 0.005 from the straight line X = -0.500 (assuming that O2 is the origin of a global coordinate frame whose positive X axis is rotated 30 deg from O2O4).

Given:


a  1.000

Coupler (AB)

b  1.200

Rocker (O4B)

c  1.167

Ground link

d  2.305

p  1.50

Coupler point data:

α  30 deg


δ  180  deg

See Figure 3-29g and Mathcad file P0451.




 a2  d 2  ( b  c) 2  2 d a  

θ  acos 1.

θ  81.136 deg

θ  θ

Define one cycle of the input crank (driving through the links 2-3 toggle position): θ  θ θ  0.5 deg  θ

2.


d

K4 

a

K1  2.3050

 

2

d

K5 

b

K4  1.9208

 

2

2 a b

K5  2.6630

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

3.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ


 





 

 

θ θ  2   atan2 2  D θ E θ 

 2  4 Dθ F θ 

E θ

2

c d a b

2


4.


Use equations 4.31 to define the x and y-components of the vector RP. RP  RA  RPA


 

 

  



 


 

  

Plot the X coordinate of the coupler point in the global X,Y coordinate system using equations 4.0b to rotate the local coordinates to a global frame.

 

 

 

PX θ  RPx θ  cos( α)  RPy θ  sin( α) X COORDINATE  0.48

Coupler Point Coordinate - X

 0.485  0.49  0.495

 

PX θ

 0.5  0.505  0.51  0.515  0.52

0

15

30

45

60

θ deg Input Angle - Theta2

6.




Using the graph for guess values, solve by trial and error to find 2 for X = -0.500 +/-0.005 PX ( 11.59  deg)  0.49500

θ2min  11.59  deg

PX ( 57.80  deg)  0.50500

θ2max  57.80  deg

75




For the linkage in Figure P4-16, what are the angles that link 2 makes with the positive X-axis when links 2 and 3 are in toggle positions?

Given:

Link lengths:

Solution: 1.

Input (O2A)

a  14

Rocker (O4B)

c  51.26

b  80

Coupler (AB)

O4 offset in XY coordinates:

O4X  47.5

Ground link:

d 

2

O4X  O4Y

O4Y  76  12 2

O4Y  64.000

d  79.701





Define the coordinate frame transformation angle:

 O4Y    O4X 

δ  π  atan 3.

crank rocker

δ  126.582 deg

Calculate the angle of link 2 in the XY system when links 2 and 3 are in the overlapped toggle position.

 ( b  a) 2  d2  c2 δ  2 ( b  a)  d 

θ21XY  acos 4.

θ21XY  86.765 deg

Calculate the angle of link 2 in the XY system when links 2 and 3 are in the extended toggle position.

 ( b  a) 2  d2  c2 δ  2 ( b  a)  d 

θ22XY  acos

θ22XY  93.542 deg




The coordinates of the point P1 on link 4 in Figure P4-16 are (114.68, 33.19) with respect to the xy coordinate system when link 2 is in the position shown. When link 2 is in another position the coordinates of P2 with respect to the xy system are (100.41, 43.78). Calculate the coordinates of P1 and P2 in the XY system for the two positions of link 2. What is the salient feature of the coordinates of P1 and P2 in the XY system?

Given:

Vertical and horizontal offsets from O2 to O4. O2O4X  47.5 in

O2O4Y  64 in

Coordinates of P1 and P2 in the local system

Solution: 1.

P1x  114.68 in

P1y  33.19  in

P2x  100.41 in

P2y  43.78  in


Calculate the angle from the global X axis to the local x axis.

 O2O4Y    O2O4X 

δ  180  deg  atan 2.

3.

δ  126.582 deg

Use equations 4.0b to transform the given coordinates from the local to the global system. P1X  P1x cos( δ)  P1y sin( δ)

P1X  95.00 in

P1Y  P1x sin( δ)  P1y cos( δ)

P1Y  72.31 in

P2X  P2x cos( δ)  P2y sin( δ)

P2X  95.00 in

P2Y  P2x sin( δ)  P2y cos( δ)

P2Y  54.54 in

In the global XY system the X-coordinates are the same for each point, which indicates that the head on the end of the rocker beam 4 is designed such that its tangent is always parallel to the Y-axis.




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the angular position of link 4 with respect to the XY coordinate frame and the transmission angle at point B of the linkage in Figure P4-16 as a function of the angle of input link 2 with respect to the XY frame.

Given:

Link lengths:

Solution: 1.

Input (O2A)

a  14

Coupler (AB)

b  80

Rocker (O4B)

c  51.26

Ground link

d  79.70





Define the coordinate frame transformation angle: δ  90 deg  atan

47.5 

  64 

3.

crank rocker

δ  126.582 deg

Define one cycle of the input crank with respect to the XY frame:

θ2 θ2XY   θ2XY  δ

θ2XY  0  deg 1  deg  360  deg 4.


d

K2 

a

K1  5.6929

2

d

K3 

c

K2  1.5548

2

2

a b c d

2

2 a c

K3  1.9339

A  θ2XY   cos θ2 θ2XY    K1  K2 cos θ2 θ2XY    K3 B θ2XY   2  sin θ2 θ2XY  



C θ2XY   K1   K2  1   cos θ2 θ2XY    K3



θ θ2XY   2   atan2 2  A  θ2XY  B θ2XY  

2

θ θ2XY   θ θ2XY   δ  2  π 5.


d b

K4  0.996

2

K5 

2

2

c d a b

K5  4.607

2 a b



B θ2XY   4  A  θ2XY   C θ2XY  

2



D θ2XY   cos θ2 θ2XY    K1  K4 cos θ2 θ2XY    K5 E θ2XY   2  sin θ2 θ2XY   F  θ2XY   K1   K4  1   cos θ2 θ2XY    K5





θ θ2XY   2   atan2 2  D θ2XY  E θ2XY   6.



E θ2XY   4  D θ2XY   F  θ2XY    2

Plot the angular position of link 4 as a function of the input angle of link 2 with respect to the XY frame.

Angular Position of Link 4 40 35 30 θ θ2XY 25 20 deg 15 10 5





0

0

45

90

135

180

225

270

315

360

θ2XY deg

7.


θtransB1 θ2XY   θ θ2XY   θ θ2XY 

 

θtransB θ2XY   if  θtransB1 θ2XY  

π 2

 

π  θtransB1 θ2XY  θtransB1 θ2XY  


θtransB θ2XY 

75 70

deg 65 60 55 50

0

45

90

135

180

θ2XY deg

225

270

315

360




For the linkage in Figure P4-17, calculate the maximum CW rotation of link 2 from the position shown, which is -20.60 deg with respect to the local xy system. What angles do link 3 and link 4 rotate through for that excursion of link 2?

Given:


a  9.17

Coupler (AB)

b  12.97

Rocker (O4B)

c  9.57

Ground link

d  7.49

Initial position of link 2: θ  26.00  deg  2  π (with respect to xy system) Solution: 1.




Condition( d b a c)  "non-Grashof" 2.

Using equations 4.37, determine the crank angles (relative to the line O2O4) at which links 3 and 4 are in toggle 2

arg1 

2

2



2 a d 2

arg2 

2

a d b c

2

2

a d b c

2



2 a d

b c

arg1  0.936

a d b c

arg2  2.678

a d

θ  acos arg1

θ  20.55 deg

The other toggle angle is the negative of this. θ  θ  2  π 3.

θ  339.45 deg

Calculate the CW rotation of link 2 from the initial position to the toggle position. Δ  θ  θ

4.

Δ  313.45 deg


d

K4 

a

K1  0.8168

 

2

d

K5 

b

K4  0.5775

 

2

2 a b

K5  0.9115

 

D θ  cos θ  K1  K4 cos θ  K5

 

 

 

6.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ


 





 

 

θ θ  2   atan2 2  D θ E θ 

 2  4 Dθ F θ 

E θ

2

c d a b

2



 

Initial angular position of link 3:

θ θ  2  π  647.755 deg

Final angular position of link 3:

θ θ  0.001  deg  250.764 deg







   

  θ θ  0.001  deg  θ θ  2  π 7.



  898.518 deg


d

K2 

a

K1  0.8168

 

2

d

K3 

c

K2  0.7827

 

2

2 a c

K3  0.3621

 

A θ  cos θ  K1  K2 cos θ  K3

 

 

 



 



 

C θ  K1   K2  1   cos θ  K3

B θ  2  sin θ

 

 

θ θ  2   atan2 2  A θ B θ 

 2  4 A θ Cθ 

B θ

 

Initial angular position of link 4:

θ θ  2  π  659.462 deg

Final angular position of link 4:

θ θ  0.001  deg  250.615 deg





   

  θ θ  0.001  deg  θ θ  2  π





2

a b c d

  910.077 deg

2




Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of point P in Figure P4-17 with respect to the XY coordinate system as a function of the angle of input link 2 with respect to the XY coordinate system.

Given:


a  9.174

Coupler (AB)

b  12.971

Rocker (O4B)

c  9.573

Ground link

d  7.487

p  15.00


δ  0  deg





Using equations 4.37, determine the crank angles (relative to the line O2O4) at which links 3 and 4 are in toggle 2

arg1 

2

2



2 a d 2

arg2 

2

a d b c

2

2

a d b c

2



2 a d

b c

arg1  0.937

a d b c

arg2  2.679

a d




Define the coordinate transformation angle. Transformation angle:

4.

α  atan

6.95 

  2.79 

α  68.128 deg

Define one cycle of the input crank between limit positions: θ  θ2toggle θ2toggle  1  deg  2  π  θ2toggle

5.


d

K4 

a

K1  0.8161

 

2

d

K5 

b

K4  0.5772

 

 

D θ  cos θ  K1  K4 cos θ  K5

2

2

c d a b 2 a b

K5  0.9110

2


 


 

 

6.

 

F θ  K1   K4  1   cos θ  K5

E θ  2  sin θ


 





 

 

θ θ  2   atan2 2  D θ E θ  7.

 2  4 Dθ F θ 

E θ



 

 

  



 


 

  

Transform these local xy coordinates to the global XY coordinate system using equations 4.0b.

 

 

 

 

 

 

RPX θ  RPx θ  cos α  RPy θ  sin α RPY θ  RPx θ  sin α  RPy θ  cos α Plot the coordinates of the coupler point in the global XY coordinate system. COUPLER POINT PATH 5

0


9.

5

 10

 15

 20  10

5

0




5

10

Coupler Point Coordinate - X

15

20




For the linkage in Figure P4-17, calculate the coordinates of the point P in the XY coordinate syste if its coordinates in the xy system are (2.71, 10.54).

Given:

Vertical and horizontal offsets from O2 to O4. O2O4X  2.790  in

O2O4Y  6.948  in

Coordinates of P in the local system Px  12.816 in Solution: 1.


Calculate the angle from the global X axis to the local x axis.

 O2O4Y    O2O4X 

δ  atan 2.

Py  10.234 in

δ  68.122 deg

Use equations 4.0b to transform the given coordinates from the local to the global system. PX  Px cos( δ)  Py sin( δ)

PX  14.273 in

PY  Px sin( δ)  Py cos( δ)

PY  8.079 in




The elliptical trammel in Figure P4-18 must be driven by rotating link 3 in a full circle. Derive analytical expressions for the positions of points A, B, and a point C on link 3 midway between A and B as a function of 3 and the length AB of link 3. Use a vector loop equation. (Hint: Place the global origin off the mechanism, preferably below and to the left and use a total of 5 vectors.) Code your solution in an equation solver such as Mathcad, Matlab, or TKSolver to calculate and plot the path of point C for one revolution of link 3.

Solution:


1.

Establish the global XY system such that the coordinates of the intersection of the slot centerlines is at (d X,d Y ). Then, define position vectors R1X, R1Y , R2, R3, and R4 as shown below. Y

4 B

3 R3



3

C

R2 2

R4

A 1

R1Y

X R1X

2.

Write the vector loop equation: R1Y + R2 + R3 - R1X - R4 = 0 then substitute the complex number notation for each position vector. The equation then becomes:

 π  π j   j  θ 3 2 j ( 0) j ( 0) 2 dY  e    a e  c e  dX  e  b e   = 0 j

3.

Substituting the Euler identity into this equation gives: d Y  j  a  c  cos θ3  j  sin θ3   d X  b  j = 0

4.

Separate this equation into its real (x component) and imaginary (y component) parts, setting each equal to zero. a  c cos θ3  d X = 0

5.

Solve for the two unknowns a and b in terms of the constants d X and d Y and the independent variable 3. Where (a,d Y ) and (d X,b) are the coordinates of points A and B, respectively, and c is the length of link 3. With no loss of generality, let d X = d Y = d. Then, a = d  c cos θ3

6.

b = d  c sin θ3

The coordinates of the point C are: CX = d  0.5 cos θ3

7.

d Y  c sin θ3  b = 0

CY = d  0.5 c sin θ3

Using a local coordinate system whose origin is located at the intersection of the centerlines of the two slots and transforming the above functions to the local xy system:


8.


a x = c cos θ3

ay = 0

bx = 0

b y = c sin θ3

To plot the path of point C as a function of 3, let c  1 and define a range function for 3

θ3  0  deg 1  deg  360  deg Cx θ3  0.5 c cos θ3

Cy θ3  0.5 sin θ3 Path of Point C

1

0.5

 

Cy θ3

0

 0.5

1 1

 0.5

0

 

Cx θ3

0.5

1




Calculate and plot the angular position of link 6 in Figure P4-19 as a function of the angle of input link 2.

Given:

Link lengths:

Solution: 1.

Input crank (L2)

a  1.75

First coupler (AB)

b  1.00

First rocker (O4B)

c  1.75

Ground link (O2O4)

d  1.00

Second input (BC)

e  1.00

Second coupler (L5)

f  1.75

Output rocker (L6)

g  1.00

Third coupler (BE)

h  1.75

Ternary link (O4C)

i  2.60


Because the linkage is symmetrical and composed of two parallelograms the analysis can be done with simple trigonometry.

39.582°

C B

5

D

3

3

A

2

4

6 E

O4

2.

1

O2

Calculate the fixed angle that line BC makes with the extension of line O4B using the law of cosines.

 c2  e2  i 2    2  c e 

δ  π  acos

δ  39.582 deg

3.


4.

Because links 1, 2, 3, and 4 are a parallelogram, link 4 will have the same angle as link 2 and AB will always be parallel to O2O4. And because links BC, 5, 6, and BE are also a parallelogram, link 6 will have the same angle as link BC. Thus,

 

θ θ  θ  δ 5.

Plot the angular position of link 6 as a function of the angle of input link 2 (see next page).



Angular Position of Link 6 400 360 320 280 240

 

θ  θ

200

deg 160 120 80 40 0

0

45

90

135

180 θ deg

225

270

315

360



PROBLEM 4-60a The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row a, draw the linkage to scale and graphically find all possible solutions (both open and crossed) for angles 2 and θ3.

Statement:

Given:

Link 2

a  1.4 in

Link 3

b  4  in

Offset

c  1  in

Slider position

d  2.5 in

See Figure P4-2, Table P4-5, and Mathcad file P0460a.

Solution: 1.


2.

Draw a circle centered at the origin with radius equal to a at some convenient scale.

3.

Draw construction lines to define the point (d,c).

4.

From the point (d,c) draw an arc with radius equal to b.

5.

The two intersections of the circle and arc are the two solutions to the position analysis problem, crossed and open. If the circle and line don't intersect, there is no solution.

6.

Draw links 2 and 3 in their two possible positions (shown as solid for branch 1 and dashed for branch 2 in the figure) and measure the angles θ2 and 3 for each branch. From the solution below, Branch 1:

θ21  176.041  deg

θ31  ( 180  13.052)  deg

θ31  193.052 deg

Branch 2:

θ22  132.439  deg

θ32  ( 180  30.551)  deg

θ32  210.551 deg

13.052° 30.551°

176.041° 000 b = 4.

c = 1.000"

a = 1.400

132.439°

d = 2.500"




Given:

Solution: 1.

The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row a, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2

a  1.4 in

Link 3

Offset

c  1  in

Slider position

d  2.5 in

See Figure P4-2, Table P4-5, and Mathcad file P0461a.

Determine both values of 2 using equations 4.20 and 4.21. 2

2

2

K1  a  b  c  d

2

2

K1  6.790  in

2

K2  2  a  c

K2  2.800  in

K3  2  a  d

K3  7.000  in

A  K1  K3

A  0.210  in

B  2  K2

B  5.600  in

C  K1  K3

C  13.790 in

2

2 2 2

  2  atan2 2  A B 

θ21  2  atan2 2  A B  θ22 2.

b  4  in

 2 B  4  A  C 2

B  4 A  C

θ21  176.041  deg θ22  132.439  deg

Determine both values of 3 using equation 4.16a or 4.17.

β ( a b c d α) 

θ  asin



a  sin( α)  c 

 

b

d 1  a  cos( α)  b  cos( θ ) return θ if d 1 = d asin 



a  sin( α)  c  b

  π otherwise 

θ31  β  a b c d θ21

θ31  193.052 deg

θ32  β  a b c d θ22

θ32  210.551 deg



PROBLEM 4-61b Statement:

Given:

Solution: 1.

The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row b, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2

a  2  in

Link 3

Offset

c  3  in

Slider position

d  5  in

See Figure P4-2, Table P4-5, and Mathcad file P0461b.


2

2

K1  a  b  c  d

2

2

K1  2.000  in

2

K2  2  a  c

K2  12.000 in

K3  2  a  d

K3  20.000 in

A  K1  K3

A  22.000 in

B  2  K2

B  24.000 in

C  K1  K3

C  18.000 in

2

2

2 2

  2  atan2 2  A B 

θ21  2  atan2 2  A B  θ22 2.

b  6  in

 2 B  4  A  C 2

B  4 A  C

θ21  54.117 deg θ22  116.045  deg


β ( a b c d α) 

θ  asin



a  sin( α)  c 

 

b




a  sin( α)  c  b


θ31  β  a b c d θ21

θ31  129.640 deg

θ32  β  a b c d θ22

θ32  168.433 deg




Given:

Solution: 1.

The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row c, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2

a  3  in

Link 3

Offset

c  2  in

Slider position

d  8  in

See Figure P4-2, Table P4-5, and Mathcad file P0461c.


2

2

K1  a  b  c  d

2

2

K1  13.000 in

2

K2  2  a  c

K2  12.000 in

K3  2  a  d

K3  48.000 in

A  K1  K3

A  61.000 in

B  2  K2

B  24.000 in

C  K1  K3

C  35.000 in

2

2 2 2

  2  atan2 2  A B 

θ21  2  atan2 2  A B  θ22 2.

b  8  in

 2 B  4  A  C 2

B  4 A  C

θ21  88.803 deg θ22  60.731 deg


β ( a b c d α) 

θ  asin



a  sin( α)  c 

 

b




a  sin( α)  c  b


θ31  β  a b c d θ21

θ31  172.824  deg

θ32  β  a b c d θ22

θ32  215.249  deg


SOLUTION MANUAL 4-61d-1

PROBLEM 4-61d Statement:

Given:

Solution: 1.

The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row d, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2

a  3.5 in

Link 3

Offset

c  1  in

Slider position

d  8  in

See Figure P4-2, Table P4-5, and Mathcad file P0461d.


2

2

K1  a  b  c  d

2

2

K1  22.750 in 2

K2  2  a  c

K2  7.000  in

K3  2  a  d

K3  56.000 in

A  K1  K3

A  78.750 in

B  2  K2

B  14.000 in

C  K1  K3

C  33.250 in

2 2

2

2

  2  atan2 2  A B 

θ21  2  atan2 2  A B  θ22 2.

b  10 in

 2 B  4  A  C 2

B  4 A  C

θ21  286.648  deg θ22  300.898  deg


β ( a b c d α) 

θ  asin



a  sin( α)  c 

 

b




a  sin( α)  c  b


θ31  β  a b c d θ21

θ31  25.806 deg

θ32  β  a b c d θ22

θ32  11.556 deg


SOLUTION MANUAL 4-61e-1

PROBLEM 4-61e Statement:

Given:

Solution: 1.

The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row e, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2

a  5  in

Link 3

Offset

c  5  in

Slider position

d  15 in

See Figure P4-2, Table P4-5, and Mathcad file P0461e.


2

2

K1  a  b  c  d

2

2

K1  125.000  in 2

K2  2  a  c

K2  50.000 in

K3  2  a  d

K3  150.000  in

A  K1  K3

A  25.000 in

B  2  K2

B  100.000  in

C  K1  K3

C  275.000  in

2

2 2 2

  2  atan2 2  A B 

θ21  2  atan2 2  A B  θ22 2.

b  20 in

 2 B  4  A  C 2

B  4 A  C

θ21  123.804  deg θ22  160.674  deg


β ( a b c d α) 

θ  asin



a  sin( α)  c 

 

b




a  sin( α)  c  b


θ31  β  a b c d θ21

θ31  152.759  deg

θ32  β  a b c d θ22

θ32  170.371  deg



PROBLEM 4-61f Statement:

Given:

Solution: 1.

The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row f, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2

a  3  in

Link 3

Offset

c  0  in

Slider position

d  12 in

See Figure P4-2, Table P4-5, and Mathcad file P0461f.


2

2

K1  a  b  c  d

2

2

K1  16.000 in 2

K2  2  a  c

K2  0.000  in

K3  2  a  d

K3  72.000 in

A  K1  K3

A  88.000 in

B  2  K2

B  0.000  in

C  K1  K3

C  56.000 in

2 2

2 2

  2  atan2 2  A B 

θ21  2  atan2 2  A B  θ22 2.

b  13 in

 2 B  4  A  C 2

B  4 A  C

θ21  282.840  deg θ22  282.840  deg


β ( a b c d α) 

θ  asin



a  sin( α)  c 

 

b




a  sin( α)  c  b


θ31  β  a b c d θ21

θ31  13.003 deg

θ32  β  a b c d θ22

θ32  13.003 deg



PROBLEM 4-61g Statement:

Given:

Solution: 1.

The link lengths, offset, and value of d for some fourbar slider-crank linkages are defined in Table P4-5. The linkage configuration and terminology are shown in Figure P4-2. For row g, using the vector loop method, find all possible solutions (both open and crossed) for angles 2 and θ3. Link 2

a  7  in

Link 3

Offset

c  10 in

Slider position

d  25 in

See Figure P4-2, Table P4-5, and Mathcad file P0461g.


2

2

K1  a  b  c  d

2

2

K1  149.000  in

2

K2  2  a  c

K2  140.000  in

K3  2  a  d

K3  350.000  in

A  K1  K3

A  499.000  in

B  2  K2

B  280.000  in

C  K1  K3

C  201.000  in

2

2 2 2

  2  atan2 2  A B 

θ21  2  atan2 2  A B  θ22 2.

b  25 in

 2 B  4  A  C 2

B  4 A  C

θ21  88.519 deg θ22  44.916 deg


β ( a b c d α) 

θ  asin



a  sin( α)  c 

 

b




a  sin( α)  c  b


θ31  β  a b c d θ21

θ31  186.898  deg

θ32  β  a b c d θ22

θ32  216.705  deg




Design a fourbar mechanism to give the two positions shown in Figure P3-1 of output rocker motion with no quick-return. (See Problem 3-3).

Given:

Coordinates of the points C1, D1, C2, and D2 with respect to C1: C1x  0.0

D1x  1.721

C2x  2.656

D2x  5.065

C1y  0.0

D1y  1.750

C2y  0.751

D2y  0.281

Assumptions: Use the pivot point between links 3 and 4 (C1 and C2 )as the precision points P1 and P2. Define position vectors in the global frame whose origin is at C1. Solution:

See solution to Problem 3-3 and Mathcad file P0501.

1.

Note that this is a two-position function generation (FG) problem because the output is specified as an angular displacement of the rocker, link 4. See Section 5.13 which details the 3-position FG solution. See also Section 5.3 in which the equations for the two-position motion generation problem are derived. These are really the same problem and have the same solution. The method of Section 5.3 will be used here.

2.

Two solution methods are derived in Section 5.3 and are presented in equations 5.7 and 5.8 for the left dyad and equations 5.11 and 5.12 for the right dyad. The first method (equations 5.7 and 5.11) looks like the better one to use in this case since it allows us to choose the link's angular positions and excursions and solve for the lengths of links 2 and 3 (w and z). Unfortunately, this method fails in this problem because of the requirement for a non-quick-return Grashof linkage, which requires the angular displacement of link 3 in going from position 1 to position 2 to be zero (2 = 0) causing a divide-by-zero error in equations 5.7d.

3.

Method 2 (equations 5.8 and 5.12) requires the choosing of two angles and a length for each dyad.

4.

To obtain the same solution as was done graphically in Problem 3-4, we need to know the location of the fixed pivot O4 with respect to the given CD. While we could take the results from Problem 3-3 and use them here to establish the location of O4, that won't be done. Instead, we will use the point C as the joint between links 3 and 4 as well as the precision point P.

5.

Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1. R1 

 C1x     C1y 

R2 

 C2x     C2y 

 P21x     R2  R1  P21y 

P21x  2.656 P21y  0.751

p 21  6.

2

2

P21x  P21y

p 21  2.760

From the trigonometric relationships given in Figure 5-1, determine 2. From the requirement for a non-quick-return, α  0. δ  atan2 P21x P21y

7.

δ  15.789 deg

From a graphical solution (see figure below), determine the values necessary for input to equations 5.8. z  5.000

β  180  deg

ϕ  δ



3.483 5.0000

1.380

2.027 5.621

O4

1

A1 O2

2

4 2.095

0.989

3

56.519°

A2 C1

D2 C2 2.922

D1

8.

Solve for the WZ dyad using equations 5.8. Z1x  z cos ϕ

Z1x  4.811

 

A  2.000

D  sin α

 

B  0.000

E  p 21  cos δ

 

C  0.000

F  p 21  sin δ

A  cos β  1 B  sin β

C  cos α  1

W1x 

W1y  w 

Z1y  z sin ϕ

 

D  0.000

 

E  2.656

 

F  0.751

A   C Z1x  D Z1y  E  B  C Z1y  D Z1x  F 

W1x  1.328

2  A A   C Z1y  D Z1x  F   B  C Z1x  D Z1y  E

W1y  0.375

2  A 2

Z1y  1.360

2

W1x  W1y

w  1.380

θ  atan2 W1x W1y

θ  164.211  deg

This is the expected value of w (one half of p 21) based on the design choices made in the graphical solution and the assumptions made in this problem. 9.

From the graphical solution (see figure above), determine the values necessary for input to equations 5.12. s  0

γ  56.519 deg

ψ  0  deg

10. Solve for the US dyad using equations 5.12. S 1x  s cos ψ

S 1x  0.000

S 1y  s sin ψ

 

A  0.448

D  sin α

 

B  0.834

E  p 21  cos δ

A  cos γ  1 B  sin γ

 

S 1y  0.000 D  0.000

 

E  2.656



 

C  cos α  1

C  0.000

 

F  p 21  sin δ

A   C S 1x  D S 1y  E  B  C S 1y  D S 1x  F 

U1x 

2  A A   C S 1y  D S 1x  F   B  C S 1x  D S 1y  E

U1y 

2

U1x  2.027

U1y  2.095

2  A

u 

F  0.751

2

U1x  U1y

u  2.915

σ  atan2 U1x U1y

σ  134.048  deg

This is the expected value of u based on the design choices made in the graphical solution and the assumptions made in this problem. 11. Solve for links 3 and 1 using the vector definitions of V and G. Link 3:

V1x  z cos ϕ  s cos ψ

V1x  4.811

V1y  z sin ϕ  s sin ψ

V1y  1.360

θ  atan2 V1x V1y

θ  15.789 deg

v  Link 1:

2

2

V1x  V1y

v  5.000

 

G1x  w cos θ  v cos θ  u  cos σ

 

G1x  5.510

G1y  w sin θ  v sin θ  u  sin σ

G1y  1.110

θ  atan2 G1x G1y

θ  11.391 deg

g 

2

2

G1x  G1y

g  5.621

12. Determine the initial and final values of the input crank with respect to the vector G.

θ2i  θ  θ

θ2i  152.820  deg

θ2f  θ2i  β

θ2f  332.820  deg

13. Define the coupler point with respect to point A and the vector V. rp  z

δp  ϕ  θ

rp  5.000

δp  0.000  deg

which is correct for the assumption that the precision point is at C. 14. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  0  deg R1 

2

ρ  0.000  deg 2

C1x  C1y

R1  0.000



 

O2x  3.483

 

O2y  0.985

 

O4x  2.027

 

O4y  2.095

O2x  R1 cos ρ  z cos ϕ  w cos θ O2y  R1 sin ρ  z sin ϕ  w sin θ O4x  R1 cos ρ  s cos ψ  u  cos σ O4y  R1 sin ρ  s sin ψ  u  sin σ

15. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis the line O2O4.

θrot  atan2 O4x  O2x  O4y  O2y

θrot  11.391 deg

16. Determine the Grashof condition. Condition( a b c d ) 


Condition( g u v w)  "Grashof" 17. DESIGN SUMMARY Link 2:

w  1.380

θ  164.211  deg

Link 3:

v  5.000

θ  15.789 deg

Link 4:

u  2.915

σ  134.048  deg

Link 1:

g  5.621

θ  11.391 deg

Coupler:

rp  5.000

δp  0.000  deg

Crank angles:

θ2i  152.820  deg θ2f  332.820  deg




Design a fourbar mechanism to give the two positions shown in Figure P3-1 of coupler motion.

Given:

Coordinates of the points A1, B1, A2, and B2 with respect to A1: A1x  0.0

B1x  1.721

A2x  2.656

B2x  5.065

A1y  0.0

B1y  1.750

A2y  0.751

B2y  0.281

Assumptions: Use the points A1 and A2 as the precision points P1 and P2. Define position vectors in the global frame whose origin is at A1. Solution:

See solution to Problem 3-4 and Mathcad file P0502.

1.

Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. See Section 5.3 in which the equations for the two-position motion generation problem are derived.

2.

Two solution methods are derived in Section 5.3 and are presented in equations 5.7 and 5.8 for the left dyad and equations 5.11 and 5.12 for the right dyad.

3.

Method 1 (equations 5.7 and 5.11) requires the choosing of three angles for each dyad. Method 2 (equations 5.8 and 5.12) requires the choosing of two angles and a length for each dyad. Method 1 is used in this solution.

4.

In order to obtain the same solution as was done graphically in Problem 3-4, the necessary assumed values were taken from that solution as shown below.

5.


 A1x     A1y 

R2 

 A2x     A2y 

 P21x     R2  R1  P21y 

P21x  2.656 P21y  0.751

p 21  6.

7.

2

2

P21x  P21y

p 21  2.760

From the trigonometric relationships given in Figure 5-1, determine 2 and 2. α  atan2 A2x  B2x  A2y  B2y   atan2 A1x  B1x  A1y  B1y

α  303.481  deg

δ  atan2 P21x P21y

δ  15.789 deg

From the graphical solution (see figure below), determine the values necessary for input to equations 5.7. θ  94.394 deg

β  40.366 deg

ϕ  45.479 deg



O4 93.449° jY 54.330° 2.760 u

A1 X 45.479°

0.281

P 21

15.789°

B2

0.751

v 1.750

A2 134.521° 2.656

2.409

B1 w

40.366°

94.394° 75.124° O2

8.

Solve for the WZ dyad using equations 5.7.

  



 

  



 

  



 

  



 

A  cos θ  cos β  1  sin θ  sin β

B  cos ϕ  cos α  1  sin ϕ  sin α

 

C  p 21  cos δ

D  sin θ  cos β  1  cos θ  sin β

E  sin ϕ  cos α  1  cos ϕ  sin α w 

C  E  B F

 

F  p 21  sin δ z 

A  E  B D

w  4.000

A  F  C D A  E  B D

z  0.000

These are the expected values of w and z based on the design choices made in the graphical solution and the assumptions made in this problem. 9.

From the graphical solution (see figure above), determine the values necessary for input to equations 5.11. σ  93.449 deg

γ  54.330 deg

ψ  134.521  deg

10. Solve for the US dyad using equations 5.11.

  



 

  



 

A'  cos σ  cos γ  1  sin σ  sin γ

B'  cos ψ  cos α  1  sin ψ  sin α

 

C  p 21  cos δ



  



 

  



 

D'  sin σ  cos γ  1  cos σ  sin γ

E'  sin ψ  cos α  1  cos ψ  sin α u 

C E'  B' F

 

F  p 21  sin δ s 

A'  E'  B' D'

u  4.000

A'  F  C D' A'  E'  B' D'

s  2.455

These are the expected values of u and s based on the design choices made in the graphical solution and the assumptions made in this problem. 11. Solve for the links 3 and 1 using the vector definitions of V and G. Link 3:


V1x  1.721


V1y  1.750

θ  atan2 V1x V1y

θ  45.479 deg

v 

Link 1:

2

2

V1x  V1y

v  2.455

 


 

G1x  1.655


G1y  6.231

θ  atan2 G1x G1y

θ  75.123 deg

g 

2

2

G1x  G1y

g  6.447

12. Determine the initial and final values of the input crank with respect to the vector G.

θ2i  θ  θ

θ2i  19.271 deg

θ2f  θ2i  β

θ2f  21.095 deg

13. Define the coupler point with respect to point A and the vector V. rp  z

δp  ϕ  θ

rp  0.000

δp  0.000  deg

which is correct for the assumption that the precision point is at A. 14. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  0  deg R1 

ρ  0.000  deg

2

A1x  A1y

2

R1  0.000

 

O2x  0.306

 

O2y  3.988

O2x  R1 cos ρ  z cos ϕ  w cos θ O2y  R1 sin ρ  z sin ϕ  w sin θ



 

O4x  1.962

 

O4y  2.243

O4x  R1 cos ρ  s cos ψ  u  cos σ O4y  R1 sin ρ  s sin ψ  u  sin σ

15. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.


θrot  75.123 deg



Condition( g u v w)  "non-Grashof" 17. DESIGN SUMMARY Link 2:

w  4.000

θ  94.394 deg

Link 3:

v  2.455

θ  45.479 deg

Link 4:

u  4.000

σ  93.449 deg

Link 1:

g  6.447

θ  75.123 deg

Coupler:

rp  0.000

δp  0.000  deg

Crank angles:

θ2i  19.271 deg θ2f  21.095 deg




Design a fourbar mechanism to give the three positions of coupler motion with no quick return shown in Figure P3-2. (See Problem 3-5). Ignore the fixed pivot points in the Figure.

Given:

Coordinates of points A and B with respect to point A1: A1x  0.0

A1y  0.0

B1x  0.741

B1y  2.383

A2x  2.019

A2y  1.905

B2x  4.428

B2y  2.557

A3x  3.933

A3y  1.035

B3x  6.304

B3y  0.256

Assumptions: Let points A1, A2, and A3 be the precision points P1, P2, and P3, respectively. Solution: 1.

2.

3.

See Figure P3-2 Mathcad file P0503.

Determine the magnitudes and orientation of the position difference vectors. 2

2

p 21  2.776

δ  atan2 A2x A2y 

δ  43.336 deg

2

2

p 31  4.067

δ  atan2 A3x A3y 

δ  14.744 deg

p 21 

A2x  A2y

p 31 

A3x  A3y

Determine the angle changes of the coupler between precision points.

θP1  atan2 A1x  B1x  A1y  B1y

θP1  107.273  deg

θP2  atan2 A2x  B2x  A2y  B2y

θP2  164.856  deg

θP3  atan2 A3x  B3x  A3y  B3y

θP3  161.812  deg

α  θP2  θP1

α  57.582 deg

α  θP3  θP1

α  269.085  deg

The free choices for this linkage are (from the graphical solution to Problem 3-5): β  78.375 deg

4.

β  135.560  deg

γ  59.771 deg

γ  107.023  deg

Evaluate terms in the WZ coefficient matrix and constant vector from equations 5.26 and form the matrix and vector:

  D  sin α G  sin β L  p 31  cos δ

A  cos β  1

 A F AA   B G 

B C D 

 G H K  A D C   F K H 

 

B  sin β

  H  cos α  1 M  p 21  sin δ E  p 21  cos δ

 E  L CC    M  N   

  F  cos β  1 K  sin α N  p 31  sin δ C  cos α  1

 W1x     W1y   AA  1 CC  Z1x     Z1y 



The components of the W and Z vectors are: W1x  2.178

W1y  0.286

Z1x  0.000

Z1y  0.000


θ  172.523  deg

ϕ  atan2 Z1x Z1y

ϕ  174.344  deg

 W1x2  W1y2 , w  2.197  

The length of link 2 is: w 

 Z1x2  Z1y2 , z  0.000  

The length of vector Z is: z  5.

Evaluate terms in the US coefficient matrix and constant vector from equations 5.31 and form the matrix and vector:

 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

F'  cos γ  1

 

M  p 21  sin δ

B' C D 

 E  L CC    M  N   



G' H K 

  F' K H 

A'

D C

N  p 31  sin δ

 U1x  U   1y   AA  1 CC  S1x     S1y 

The components of the U and S vectors are: U1x  1.995

U1y  3.121

S 1x  0.741

S 1y  2.383


σ  122.591  deg

ψ  atan2 S 1x S 1y

ψ  107.267  deg

The length of link 4 is: u 

 U 2  U 2 , u  3.704 1y   1x

The length of vector S is: s  6.

 S 1x2  S 1y2 , s  2.495  

Solve for links 3 and 1 using the vector definitions of V and G. Link 3:

V1x  Z1x  S 1x

V1x  0.741

V1y  Z1y  S 1y

V1y  2.383

θ  atan2 V1x V1y

θ  72.734 deg

v  Link 1:

2

2

V1x  V1y

G1x  W1x  V1x  U1x

v  2.495 G1x  0.558


G1y  W1y  V1y  U1y

G1y  0.452

θ  atan2 G1x G1y

θ  39.009 deg

g  7.

8.

9.


2

2

G1x  G1y

g  0.718

Determine the initial and final values of the input crank with respect to the vector G.

θ2i  θ  θ

θ2i  211.532  deg

θ2f  θ2i  β

θ2f  75.972 deg

Define the coupler point with respect to point A and the vector V. rp  z

δp  ϕ  θ

rp  0.000

δp  247.078  deg

Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. O2x  z cos ϕ  w cos θ

O2x  2.178

O2y  z sin ϕ  w sin θ

O2y  0.286

O4x  s cos ψ  u  cos σ

O4x  2.736

O4y  s sin ψ  u  sin σ

O4y  0.738



θrot  39.009 deg




w  2.197

θ  172.523  deg

Link 3:

v  2.495

θ  72.734 deg

Link 4:

u  3.704

σ  122.591  deg

Link 1:

g  0.718

θ  39.009 deg

Coupler:

rp  0.000

δp  247.078  deg



Crank angles:

θ2i  211.532  deg θ2f  75.972 deg 13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.

Y

0.718 2.197 O4 O2

2

A1

1a

B3

2.496

A3

3 4 A2 B1

B2 3.704

This is the same result as that found in Problem 3-5.

X




Design a fourbar mechanism to give the three positions shown in Figure P3-2 (see Problem 3-6). Use analytical synthesis and design it for the fixed pivots shown.

Given:

Link end points (with respect to A1): A1x  0.0

B1x  0.741

A2x  2.019

B2x  4.428

A3x  3.933

B3x  6.304

A1y  0.0

B1y  2.383

A2y  1.905

B2y  2.557

A3y  1.035

B3y  0.256

Fixed pivot points (with respect to A1): O2x  0.995 Solution: 1.

2.

3.

O2y  5.086

O4x  5.298

O4y  5.086


Determine the angle changes between precision points from the body angles given.

θP1  atan2 A1x  B1x A1y  B1y

θP1  107.273  deg

θP2  atan2 A2x  B2x A2y  B2y

θP2  164.856  deg

θP3  atan2 A3x  B3x A3y  B3y

θP3  161.812  deg

α  θP2  θP1

α  57.582 deg

α  θP3  θP1

α  269.085  deg

Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. P21x  A2x

P21x  2.019

P31x  A3x

P31x  3.933

P21y  A2y

P21y  1.905

P31y  A3y

P31y  1.035

R1x  O2x

R1x  0.995

R1y  O2y

R1y  5.086

R2x  R1x  P21x

R2x  1.024

R2y  R1y  P21y

R2y  3.181

R3x  R1x  P31x

R3x  2.938

R3y  R1y  P31y

R3y  4.051

2

2

R1  5.182

2

2

R2  3.342

2

2

R3  5.004

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

Using Figure 5-6, determine the angles that R1, R2, and R3 make with the x axis. ζ  atan2 R1x R1y

ζ  101.069  deg

ζ  atan2 R2x R2y

ζ  72.156 deg



ζ  atan2 R3x R3y 4.

ζ  54.048 deg

Solve for 2 and 3 using equations 5.34





















 

C3  8.007

 

C4  5.127

 

C5  5.851

 

C6  1.294

C1  R3 cos α  ζ  R2 cos α  ζ C2  R3 sin α  ζ  R2 sin α  ζ C3  R1 cos α  ζ  R3 cos ζ





C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

2

C2  3.679

A1  C3  C4

A1  90.406

A2  C3 C6  C4 C5

A2  19.633

A3  C4 C6  C3 C5

A3  53.487

A4  C2 C3  C1 C4

A4  22.524

A5  C4 C5  C3 C6

A5  19.633

A6  C1 C3  C2 C4

A6  29.689

K1  A2  A4  A3  A6

K1  1.146  10

K2  A3  A4  A5  A6

K2  1.788  10

2

K3 

3 3

2

2

2

A1  A2  A3  A4  A6

2

2

3

K3  1.769  10

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  90.915 deg

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  23.770 deg

The first value is the same as 3, so use the second value

β  β

 A5  sin β  A3  cos β  A6   A1  

β  16.790 deg

 A3  sin β  A2  cos β  A4   A1  

β  16.790 deg

β  acos

β  asin

Since both values are the same, 5.

C1  1.352

Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2.

β  β



R1x  O4x

6.

7.

R1x  5.298

R1y  O4y

R2x  R1x  P21x

R2x  3.279

R2y  R1y  P21y

R2y  3.181

R3x  R1x  P31x

R3x  1.365

R3y  R1y  P31y

R3y  4.051

2

2

R1  7.344

2

2

R2  4.568

2

2

R3  4.275

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  5.086


ζ  136.170  deg

ζ  atan2 R2x R2y

ζ  135.869  deg

ζ  atan2 R3x R3y

ζ  108.621  deg

Solve for 2 and 3 using equations 5.34





















 

C3  3.636

 

C4  9.430

 

C5  3.855

 

C6  4.927






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

2

C1  1.023 C2  4.349

A1  C3  C4

A1  102.135

A2  C3 C6  C4 C5

A2  18.434

A3  C4 C6  C3 C5

A3  60.472

A4  C2 C3  C1 C4

A4  25.460

A5  C4 C5  C3 C6

A5  18.434

A6  C1 C3  C2 C4

A6  37.287

K1  A2  A4  A3  A6

K1  1.785  10

K2  A3  A4  A5  A6

K2  2.227  10

3 3


2

K3 


2

2

2

A1  A2  A3  A4  A6

2 3

K3  2.198  10

2

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  90.915 deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  11.643 deg


γ  

 A5  sin γ  A3  cos γ  A6   A1  

  11.069 deg

 A3  sin γ  A2  cos γ  A4   A1  

  11.069 deg

  acos

  asin

γ  

Since both angles are the same, 8.

Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. p 21 

2

2

P21x  P21y

p 21  2.776

δ  atan2 P21x P21y p 31 

2

δ  43.336 deg

2

P31x  P31y

p 31  4.067

δ  atan2 P31x P31y 9.

δ  14.744 deg

Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector: A  cos β  1 B  sin β C  cos α  1

 

 

 

 

E  p 21  cos δ

 

H  cos α  1

D  sin α G  sin β

 

L  p 31  cos δ

 A F AA   B G 

B C D 

 G H K  A D C   F K H 

10. The components of the W and Z vectors are:

 

 

 

M  p 21  sin δ

 E  L CC    M  N   

 

F  cos β  1

 

K  sin α

 

N  p 31  sin δ

 W1x   W1y   AA  1 CC  Z1x   Z1y   



W1x  3.594

W1y  7.810 2

w 

11. The length of link 2 is:

Z1x  4.589

2

W1x  W1y

Z1y  2.724

w  8.597

12. Evaluate terms in the US coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:

 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

K  sin α

 

L  p 31  cos δ

 A'  F' AA    B'  G' 

 

F'  cos γ  1

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

E   L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x     U1y   AA  1 CC  S1x   S1y   

13. The components of the W and Z vectors are: U1x  2.400


U1y  7.549 2

u 

U1x  U1y

S1x  2.898

2

S1y  2.463

u  7.921

15. Solving for links 3 and 1 from equations 5.2a and 5.2b. V1x  Z1x  S1x

V1x  1.691

V1y  Z1y  S1y

V1y  0.261

The length of link 3 is:

v 

2

2

V1x  V1y

v  1.711

G1x  W1x  V1x  U1x

G1x  4.303

G1y  W1y  V1y  U1y

G1y  5.329  10


g 

2

G1x  G1y

2

 14

g  4.303

16. Check the location of the fixed pivots with respect to the global frame using the calculated vectors W1, Z1, U1, and S1. O2x  Z1x  W1x

O2x  0.995

O2y  Z1y  W1y

O2y  5.086

O4x  S1x  U1x

O4x  5.298

O4y  S1y  U1y

O4y  5.086

These check with Figure P3-2.



17. Determine the location of the coupler point with respect to point A and line AB. 2

2

z  5.336

2

2

s  3.803

Distance from A to P

z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y

rP  z

ψ  atan2( S1x S1y)

ψ  139.639  deg

ϕ  atan2( Z1x Z1y )

ϕ  149.309  deg

θ  atan2 z cos ϕ  s cos ψ z sin ϕ  s sin ψ  θ  171.233  deg

δp  ϕ  θ

δp  21.924 deg

18. DESIGN SUMMARY Link 1:

g  4.303

Link 2:

w  8.597

Link 3:

v  1.711

Link 4:

u  7.921

Coupler point:

rP  5.336

δp  21.924 deg

19. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct.




See Project P3-8. Define three positions of the boat and analytically synthesize a linkage to move through them.

Assumptions: Launch ramp angle is 15 deg to the horizontal. Solution: 1.

See Project P3-8 and Mathcad file P0505.

This is an open-ended design problem that has many valid solutions. First define the problem more completely than is stated by deciding on three positions for the boat to move through. The figure below shows one such set of positions (dimensions are in mm). Y 3539 1453 1 deg. P2

15 deg.

725 X

P1 1179

1261

1431

0 deg. P3

O4

O2

WATER LEVEL 331 2694 RAMP

2.

From the figure, the design choices are: P21x  1453

P21y  725

P31x  3539

P31y  1261

O2x  331

O2y  1179

O4x  2694

O4y  1431

Body angles:

θP1  15 deg

θP2  1  deg

θP3  0  deg

3.

The methods of Section 5.8 are used to get a solution for this problem. The solution is sensitive to small changes in the design choices so a trial-and-error approach is warranted.

4.


5.

α  θP2  θP1

α  14.000 deg

α  θP3  θP1

α  15.000 deg

Using Figure 5-6, determine the magnitudes of R1, R2, and R3 and their x and y components. R1x  O2x

R1x  331.000

R1y  O2y 3

R2x  R1x  P21x

R2x  1.122  10

R2y  R1y  P21y

R2y  1.904  10

R3x  R1x  P31x

R3x  3.208  10

R3y  R1y  P31y

R3y  82.000

3 3

R1y  1.179  10

3


6.

7.


2

2

R1  1.225  10

3

2

2

R2  2.210  10

3

2

2

R3  3.209  10

3

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y


ζ  74.318 deg

ζ  atan2 R2x R2y

ζ  120.510  deg

ζ  atan2 R3x R3y

ζ  178.536  deg






















 

C3  3.833  10

 

C4  1.135  10

 

C5  1.728  10

 

C6  840.097

C1  R3 cos α  ζ  R2 cos α  ζ C2  R3 sin α  ζ  R2 sin α  ζ



C4  R1 sin α  ζ  R3 sin ζ







C6  R1 sin α  ζ  R2 sin ζ 2

C2  1.433  10

3

3

3

C5  R1 cos α  ζ  R2 cos ζ



3

3

C3  R1 cos α  ζ  R3 cos ζ



C1  2.542  10

2

7

A1  C3  C4

A1  1.598  10

A2  C3 C6  C4 C5

A2  5.182  10

A3  C4 C6  C3 C5

A3  5.671  10

6

A4  C2 C3  C1 C4

A4  2.607  10

6

A5  C4 C5  C3 C6

A5  5.182  10

6

A6  C1 C3  C2 C4

A6  1.137  10

7

K1  A2  A4  A3  A6

K1  5.096  10

K2  A3  A4  A5  A6

K2  7.370  10

2

K3 

6

13 13

2

2

2

A1  A2  A3  A4  A6

2

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

13

K3  3.015  10

β  125.675  deg



 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  15.000 deg

The second value is the same as 3, so use the first value

β  β

 A5  sin β  A3  cos β  A6   A1  

β  39.836 deg

 A3  sin β  A2  cos β  A4   A1  

β  39.836 deg



β  β


Repeat steps 2, 3, and 4 for the right-hand dyad to find 1 and 2. R1x  O4x

9.

R1x  2.694  10

3

R1y  O4y

R2x  R1x  P21x

R2x  1.241  10

3

R2y  R1y  P21y

R2y  2.156  10

3

R3x  R1x  P31x

R3x  845.000

R3y  R1y  P31y

R3y  170.000

2

2

R1  3.050  10

3

2

2

R2  2.488  10

3

2

2

R3  861.931

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  1.431  10


ζ  27.976 deg

ζ  atan2 R2x R2y

ζ  60.075 deg

ζ  atan2 R3x R3y

ζ  168.625  deg

10. Solve for 2 and 3 using equations 5.34





















 

C3  3.818  10

 

C4  514.981

 

C5  1.719  10

 

C6  1.419  10






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ

C1  2.536  10

3

C2  1.392  10

3

3

3 3

3

DESIGN OF MACHINERY - 5th Ed. 2


2

7

A1  C3  C4

A1  1.484  10

A2  C3 C6  C4 C5

A2  6.303  10

A3  C4 C6  C3 C5

A3  5.832  10

6

A4  C2 C3  C1 C4

A4  4.008  10

6

A5  C4 C5  C3 C6

A5  6.303  10

6

A6  C1 C3  C2 C4

A6  1.040  10

7

K1  A2  A4  A3  A6

K1  3.537  10

K2  A3  A4  A5  A6

K2  8.891  10

2

K3 

6

13 13

2

2

2

A1  A2  A3  A4  A6

2

2

13

K3  1.115  10

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  151.615  deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  15.000 deg


γ  γ

 A5  sin γ  A3  cos γ  A6   A1  

  56.167 deg

 A3  sin γ  A2  cos γ  A4   A1  

  56.167 deg

  acos

  asin

γ  

Since both angles are the same,

11. Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. p 21 

2

2

P21x  P21y

δ  atan2 P21x P21y

p 31 

2

2

P31x  P31y

δ  atan2 P31x P31y

3

p 21  1.624  10

δ  153.482  deg

3

p 31  3.757  10

δ  160.388  deg



12. Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:

 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 

M  p 21  sin δ

B C D 

 A F AA   B G 

 

C  cos α  1

N  p 31  sin δ

 E  L CC    M  N   

 G H K  A D C   F K H 

 W1x   W1y   AA  1 CC  Z1x   Z1y   

13. The components of the W and Z vectors are: W1x  1.331  10

3

w 


W1y  1.653  10 2

3

Z1x  1.000  10

2

W1x  W1y

3

Z1y  474.187

w  2122.473


 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

F'  cos γ  1

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   

16. The components of the W and Z vectors are: 3

U1x  1.690  10


u 

U1y  951.273 2

U1x  U1y

2

S1x  1.004  10

3

S1y  479.727

u  1939.291


V1x  2.004  10

V1y  Z1y  S1y

V1y  953.914


v 

2

2

V1x  V1y

v  2219.601

3



G1x  W1x  V1x  U1x

G1x  2.363  10

G1y  W1y  V1y  U1y

G1y  252.000


2

g 

G1x  G1y

2

3

g  2376.399


O2x  331.000

O2y  Z1y  W1y

O2y  1179.000

O4x  S1x  U1x

O4x  2694.000

O4y  S1y  U1y

O4y  1431.000

These check with the design choices shown in the figure above. 20. Determine the location of the coupler point with respect to point A and line AB. 2

2

z  1106.834

2

2

s  1112.770


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y

rP  z


ψ  25.538 deg

ϕ  atan2( Z1x Z1y )

ϕ  154.633  deg


δp  ϕ  θ

δp  0.086  deg


g  2376.4

Link 2:

w  2122.5

Link 3:

v  2219.6

Link 4:

u  1939.3

Coupler point:

rP  1106.8

δp  0.086  deg

22. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct. The solution is drawn below to show the locations of the moving pivots for the three positions chosen (see next page). 23. The design needs to be checked for the presence of toggle positions within its desired range of motion. This design has none, but is close to toggle at position 1. This could be used as a locking feature. 24. The transmission angles need to be checked also. This design has poor transmission angles, especially in and near positions 1 and 3. Unfortunately, this is where a large overturning moment is created by the mass of the boat. A large mechanical advantage input device will need to be used here, such as a hydraulic cylinder or geared drive. Note that the drive mechanism must also resist being overdriven (back driven) by the load as the boat descends from its high point onto the trailer.



A2 A1

B2

B1 A3 B3 WATER LEVEL

RAMP

O4

O2




See Project P3-20. Define three positions of the dumpster and analytically synthesize a linkage to move through them. The fixed pivots must be located on the existing truck.

Solution:


1.

This is an open-ended design problem that has many valid solutions. First define the problem more completely than it is stated by deciding on three positions for the dumpster box to move through. The figure below shows one such set of positions (dimensions are in mm). 1999 59.1 deg. 590

30.3 deg.

P3 P2 1817 0 deg.

1202

226 311

P1 O2 O4

2036 2094

2.

From the figure, the design choices are: P21x  590

P21y  1202

P31x  1999

P31y  1817

O2x  2094

O2y  226

O4x  2036

O4y  311

Body angles:

θP1  0  deg

θP2  30.3 deg

θP3  59.1 deg

3.


4.


5.

α  θP2  θP1

α  30.300 deg

α  θP3  θP1

α  59.100 deg


R1x  2.094  10

3

R1y  O2y

R2x  R1x  P21x

R2x  1.504  10

3

R2y  R1y  P21y

R2y  1.428  10

3

R3x  R1x  P31x

R3x  95.000

R1y  226.000



R3y  R1y  P31y

6.

7.

R3y  2.043  10

3

2

2

R1  2.106  10

3

2

2

R2  2.074  10

3

2

2

R3  2.045  10

3

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y


ζ  6.160  deg

ζ  atan2 R2x R2y

ζ  43.515 deg

ζ  atan2 R3x R3y

ζ  87.338 deg






















 

C3  786.433

 

C4  130.152

 

C5  189.927

 

C6  176.392






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

2

C1  495.777 C2  212.019

5

A1  C3  C4

A1  6.354  10

A2  C3 C6  C4 C5

A2  1.140  10

A3  C4 C6  C3 C5

A3  1.723  10

5

A4  C2 C3  C1 C4

A4  2.313  10

5

A5  C4 C5  C3 C6

A5  1.140  10

5

A6  C1 C3  C2 C4

A6  3.623  10

5

K1  A2  A4  A3  A6

K1  3.607  10

K2  A3  A4  A5  A6

K2  8.115  10

2

K3 

5

10 10

2

2

2

A1  A2  A3  A4  A6

2

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

10

K3  8.816  10

β  72.976 deg



 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  59.100 deg


β  β

 A5  sin β  A3  cos β  A6   A1  

β  34.802 deg

 A3  sin β  A2  cos β  A4   A1  

β  34.802 deg



β  β



9.

R1x  2.036  10

3

R1y  O4y

R2x  R1x  P21x

R2x  1.446  10

3

R2y  R1y  P21y

R2y  1.513  10

3

R3x  R1x  P31x

R3x  37.000

R3y  R1y  P31y

R3y  2.128  10

3

2

2

R1  2.060  10

3

2

2

R2  2.093  10

3

2

2

R3  2.128  10

3

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  311.000


ζ  8.685  deg

ζ  atan2 R2x R2y

ζ  46.297 deg

ζ  atan2 R3x R3y

ζ  89.004 deg






















 

C3  741.712

 

C4  221.269

 

C5  154.965

 

C6  217.266






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ

C1  486.018 C2  161.777



2

5

A1  C3  C4

A1  5.991  10

A2  C3 C6  C4 C5

A2  1.269  10

A3  C4 C6  C3 C5

A3  1.630  10

5

A4  C2 C3  C1 C4

A4  2.275  10

5

A5  C4 C5  C3 C6

A5  1.269  10

5

A6  C1 C3  C2 C4

A6  3.247  10

5

K1  A2  A4  A3  A6

K1  2.406  10

K2  A3  A4  A5  A6

K2  7.828  10

2

K3 

5

10 10

2

2

2

A1  A2  A3  A4  A6

2

2

10

K3  7.953  10

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  86.724 deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  59.100 deg


γ  γ

 A5  sin γ  A3  cos γ  A6   A1  

  39.743 deg

 A3  sin γ  A2  cos γ  A4   A1  

  39.743 deg

  acos

  asin

γ  



2

2

P21x  P21y

δ  atan2 P21x P21y p 31 

2

2

P31x  P31y

δ  atan2 P31x P31y

p 21  1338.994 δ  116.144  deg p 31  2701.387 δ  137.731  deg




 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 

M  p 21  sin δ

B C D 

 A F AA   B G 

 

C  cos α  1

 E  L CC    M  N   

 G H K  A D C   F K H 

N  p 31  sin δ

 W1x   W1y   AA  1 CC  Z1x   Z1y   

13. The components of the W and Z vectors are: W1x  3.194  10

3

W1y  829.763

Z1x  1.100  10

3

Z1y  603.763

14. The length of link 2 is: w 

2

2

W1x  W1y

w  3299.543


 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

F'  cos γ  1

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   

16. The components of the W and Z vectors are: 3

U1x  1.621  10

U1y  13.492

S1x  415.016

S1y  297.508

17. The length of link 4 is: u 

2

U1x  U1y

2

u  1621.040


V1x  1.515  10

V1y  Z1y  S1y

V1y  901.272

3




v 

2

V1x  V1y

v  1762.404

G1x  W1x  V1x  U1x

G1x  58.000

G1y  W1y  V1y  U1y

G1y  85.000


2

g 

G1x  G1y

2

g  102.903


O2x  2094.000

O2y  Z1y  W1y

O2y  226.000

O4x  S1x  U1x

O4x  2036.000

O4y  S1y  U1y

O4y  311.000


2

z  1254.370

2

2

s  510.637


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y

rP  z


ψ  35.635 deg

ϕ  atan2( Z1x Z1y )

ϕ  151.228  deg


δp  ϕ  θ

δp  1.984  deg


g  102.9

Link 2:

w  3299.5

Link 3:

v  1762.4

Link 4:

u  1621.0

Coupler point:

rP  1254.4

δp  1.984  deg

22. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct. The solution is drawn below to show the locations of the moving pivots for the three positions chosen (see next page). 23. The design needs to be checked for the presence of toggle positions within its desired range of motion. This design has none, but is close to toggle at position 3.



24. The transmission angles need to be checked also. A large mechanical advantage input device will need to be used here, such as a hydraulic cylinder. Note that the drive mechanism must also resist being overdriven (back driven) by the load as the dumpster descends from its high point onto the truck.

A3

A2

B3

B2

O2

B1 O4

A1




See Project P3-7. Define three positions of the computer monitor and analytically synthesize a linkage to move through them. The fixed pivots must be located on the floor or wall.

Solution:


1.

This is an open-ended design problem that has many valid solutions. First define the problem more completely than it is stated by deciding on three positions for the computer monitor to move through. The figure below shows one such set of positions (dimensions are in inches). Y 33.816

O2 97 deg. 11.580 P1

O4

X

90 deg. 7.812

1.272

14.472

P2

WALL 85 deg.

P3

2.148

5.736

2.

From the figure, the design choices are: P21x  2.148

P21y  7.812

P31x  5.736

P31y  14.472

O2x  33.816

O2y  11.580

O4x  33.816

O4y  1.272

Body angles:

θP1  97 deg

θP2  90 deg

θP3  85 deg

3.


4.


5.

α  θP2  θP1

α  7.000  deg

α  θP3  θP1

α  12.000 deg


R1x  33.816

R1y  O2y

R2x  R1x  P21x

R2x  31.668

R2y  R1y  P21y

R2y  19.392

R3x  R1x  P31x

R3x  28.080

R3y  R1y  P31y

R3y  26.052

R1y  11.580


6.

7.


2

2

R1  35.744

2

2

R2  37.134

2

2

R3  38.304

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y


ζ  161.097  deg

ζ  atan2 R2x R2y

ζ  148.519  deg

ζ  atan2 R3x R3y

ζ  137.146  deg






















 

C3  7.405

 

C4  21.756

 

C5  3.307

 

C6  12.019






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

2

C1  3.962 C2  10.052

A1  C3  C4

A1  528.143

A2  C3 C6  C4 C5

A2  17.049

A3  C4 C6  C3 C5

A3  285.981

A4  C2 C3  C1 C4

A4  11.771

A5  C4 C5  C3 C6

A5  17.049

A6  C1 C3  C2 C4

A6  248.020

K1  A2  A4  A3  A6

K1  7.073  10

K2  A3  A4  A5  A6

K2  7.595  10

2

K3 

4 3

2

2

2

A1  A2  A3  A4  A6

2

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

4

K3  6.760  10

β  24.258 deg



 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  12.000 deg


β  β

 A5  sin β  A3  cos β  A6   A1  

β  12.435 deg

 A3  sin β  A2  cos β  A4   A1  

β  12.435 deg



β  β



9.

R1x  33.816

R1y  O4y

R2x  R1x  P21x

R2x  31.668

R2y  R1y  P21y

R2y  6.540

R3x  R1x  P31x

R3x  28.080

R3y  R1y  P31y

R3y  13.200

2

2

R1  33.840

2

2

R2  32.336

2

2

R3  31.028

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  1.272


ζ  177.846  deg

ζ  atan2 R2x R2y

ζ  168.331  deg

ζ  atan2 R3x R3y

ζ  154.822  deg






















 

C3  4.733

 

C4  21.475

 

C5  1.741

 

C6  11.924






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ

C1  2.856 C2  9.867



2

A1  C3  C4

A1  483.571

A2  C3 C6  C4 C5

A2  19.043

A3  C4 C6  C3 C5

A3  264.299

A4  C2 C3  C1 C4

A4  14.646

A5  C4 C5  C3 C6

A5  19.043

A6  C1 C3  C2 C4

A6  225.402

K1  A2  A4  A3  A6

K1  5.929  10

K2  A3  A4  A5  A6

K2  8.163  10

2

K3 

4 3

2

2

2

A1  A2  A3  A4  A6

2

2

4

K3  5.630  10

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  27.678 deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  12.000 deg


γ  γ

 A5  sin γ  A3  cos γ  A6   A1  

  14.435 deg

 A3  sin γ  A2  cos γ  A4   A1  

  14.435 deg

  acos

  asin

γ  



2

2

P21x  P21y

δ  atan2 P21x P21y

p 31 

2

2

P31x  P31y

δ  atan2 P31x P31y

p 21  8.102 δ  74.626 deg

p 31  15.567 δ  68.379 deg




 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 

M  p 21  sin δ

B C D 

 A F AA   B G 

 

C  cos α  1

N  p 31  sin δ

 E  L CC    M  N   

 G H K  A D C   F K H 

 W1x   W1y   AA  1 CC  Z1x   Z1y   

13. The components of the W and Z vectors are: W1x  36.030

W1y  8.098

Z1x  2.214

Z1y  3.482

14. The length of link 2 is: w 

2

2

W1x  W1y

w  36.929


 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

F'  cos γ  1

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   


U1y  2.592

S1x  1.522

17. The length of link 4 is: u 

2

U1x  U1y

2

u  32.398


V1x  3.736

V1y  Z1y  S1y

V1y  7.346

S1y  3.864




v 

2

V1x  V1y

v  8.241

G1x  W1x  V1x  U1x

G1x  0.000

G1y  W1y  V1y  U1y

G1y  12.852


2

g 

G1x  G1y

2

g  12.852


O2x  33.816

O2y  Z1y  W1y

O2y  11.580

O4x  S1x  U1x

O4x  33.816

O4y  S1y  U1y

O4y  1.272


2

z  4.126

2

2

s  4.153


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y


ψ  111.494  deg

ϕ  atan2( Z1x Z1y )

ϕ  57.551 deg

rP  z

θ  atan2 z cos ϕ  s cos ψ z sin ϕ  s sin ψ  θ  63.046 deg

δp  ϕ  θ

δp  5.495  deg


g  12.852

Link 2:

w  36.929

Link 3:

v  8.241

Link 4:

u  32.398

Coupler point:

rP  4.126

δp  5.495  deg

22. VERIFICATION: The calculated values of g (length of the ground link) and of the coordinates of O2 and O4 give the same values as those on the problem statement, verifying that the calculated values for the other links and the coupler point are correct. The solution is drawn below to show the locations of the moving pivots for the three positions chosen (see next page). 23. The design needs to be checked for the presence of toggle positions within its desired range of motion. This design has none.



24. The transmission angles need to be checked also. A means to support the weight of the monitor must be provided. The figure below shows a spring placed between links to provide the balancing moment. Further design and analysis needs to be done to optimize the spring placement in order to compensate for its change in force with deflection and the change in moment arm as the linkage moves.

SPRING O2

A1

A2 O4

B1

WALL A3 B2

B3




Design a linkage to carry the body in Figure P5-1 through the two positions P1 and P2 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown. Use the free choices given below.

Given:

Coordinates of the points P1 and P2 with respect to P1: P1x  0.0

P1y  0.0

P2x  1.236

P2y  2.138

Angles made by the body in positions 1 and 2:

θP1  210  deg

θP2  147.5  deg

Free choices for the WZ dyad : z  1.075

β  27.0 deg

ϕ  204.4  deg

γ  40.0 deg

ψ  74.0 deg

Free choices for the US dyad : s  1.240 Solution:


1.

Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. Because of the data given in the hint, the second method of Section 5.3 will be used here.

2.


 P1x     P1y 

R2 

 P2x     P2y 

 P21x     R2  R1  P21y 

P21x  1.236 P21y  2.138

p 21  3.

4.

2

2

P21x  P21y

p 21  2.470

From the trigonometric relationships given in Figure 5-1, determine 2 and 2. α  θP2  θP1

α  62.500 deg

δ  atan2 P21x P21y

δ  120.033  deg



 

A  0.109

D  sin α

 

B  0.454

E  p 21  cos δ

 

C  0.538

F  p 21  sin δ

A  cos β  1 B  sin β

C  cos α  1

W1x 

Z1x  0.979

 

A   C Z1x  D Z1y  E  B  C Z1y  D Z1x  F  2  A

Z1y  0.444 D  0.887

 

E  1.236

 

F  2.138

W1x  1.462


W1y  w 


A   C Z1y  D Z1x  F   B  C Z1x  D Z1y  E

W1y  3.367

2  A 2

2

W1x  W1y

w  3.670

θ  atan2 W1x W1y 5.

θ  113.472  deg

Solve for the US dyad using equations 5.12. S 1x  s cos ψ

S 1x  0.342

 

A  0.234

D  sin α

 

B  0.643

E  p 21  cos δ

 

C  0.538

F  p 21  sin δ

A  cos γ  1 B  sin γ

C  cos α  1

U1x 

U1y  u 

S 1y  s sin ψ

 

D  0.887

 

E  1.236

 

F  2.138

A   C S 1x  D S 1y  E  B  C S 1y  D S 1x  F  2  A A   C S 1y  D S 1x  F   B  C S 1x  D S 1y  E

2

U1x  U1y

u  5.461

σ  atan2 U1x U1y 6.

σ  125.619  deg



V1x  1.321


V1y  1.636

θ  atan2 V1x V1y

θ  128.914  deg

v  Link 1:

2

2

V1x  V1y

v  2.103

 


 

G1x  0.398


G1y  0.564

θ  atan2 G1x G1y

θ  54.796 deg

g  7.

U1x  3.180

U1y  4.439

2  A 2

S 1y  1.192

2

2

G1x  G1y

g  0.690


θ2i  θ  θ

θ2i  58.677 deg



θ2f  θ2i  β 8.

9.

θ2f  85.677 deg


δp  ϕ  θ

rp  1.075

δp  333.314  deg

Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  0  deg R1 

ρ  0.000  deg

2

2

P1x  P1y

R1  0.000

 

O2x  2.441

 

O2y  3.811

 

O4x  2.838

 

O4y  3.247



θrot  atan2 O4x  O2x  O4y  O2y

θrot  54.796 deg




w  3.670

θ  113.472  deg

Link 3:

v  2.103

θ  128.914  deg

Link 4:

u  5.461

σ  125.619  deg

Link 1:

g  0.690

θ  54.796 deg

Coupler:

rp  1.075

δp  333.314  deg

Crank angles:

θ2i  58.677 deg θ2f  85.677 deg



13. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design.

O2

1.236

G1 Y

O4 U2

P2 S2

B2

V2 2.138

W2

U1

Z2 A2

Z1 62.5°

W1

A1 X

P1 V1

S1 B1




Design a linkage to carry the body in Figure P5-1 through the two positions P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown. Hint: First try a rough graphical solution to create realistic values for free choices.

Given:

Coordinates of the points P2 and P3 with respect to P1: P2x  1.236

P2y  2.138

P3x  2.500

P3y  2.931


θP2  147.5  deg Solution:

θP3  110.2  deg


1.

Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. The second method of Section 5.3 will be used here.

2.

Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1.

 P2x     P2y 

R1 

R2 

 P3x     P3y 

 P21x     R2  R1  P21y 

P21x  1.264 P21y  0.793

p 21  3.

2

2

P21x  P21y

p 21  1.492


α  37.300 deg

δ  atan2 P21x P21y

δ  147.897  deg O4

4.

From a graphical solution (see figure at right), determine the values necessary for input to equations 5.8.

Y

1.250

P3 43.806°

z  0.0 B3

β  43.806 deg

P2 32.500°

57.012° B2

O2

ϕ  32.500 deg

P1

5.


Z1x  0.000


 

A  0.278

D  sin α

 

B  0.692

E  p 21  cos δ

 

C  0.205

F  p 21  sin δ

A  cos β  1 B  sin β

C  cos α  1

 

Z1y  0.000 D  0.606

 

E  1.264

 

F  0.793

X


W1x 

W1y  w 



W1x  1.618


W1y  1.175

2  A 2

2

W1x  W1y

w  2.000

θ  atan2 W1x W1y 5.

θ  35.994 deg

From the graphical solution (see figure above), determine the values necessary for input to equations 5.12. s  1.250

6.

γ  57.012 deg

ψ  147.5  deg


S 1x  1.054

 

A  0.456

D  sin α

 

B  0.839

E  p 21  cos δ

 

C  0.205

F  p 21  sin δ

A  cos γ  1 B  sin γ

C  cos α  1

U1x 

U1y  u 

S 1y  s sin ψ

 

A   C S 1x  D S 1y  E  B  C S 1y  D S 1x  F  2  A A   C S 1y  D S 1x  F   B  C S 1x  D S 1y  E 2  A 2

2

U1x  U1y

 

E  1.264

 

F  0.793

U1x  0.675

U1y  1.883

σ  70.278 deg



V1x  1.054


V1y  0.672

θ  atan2 V1x V1y

θ  32.500 deg

v  Link 1:

2

2

V1x  V1y

v  1.250

 


 

G1x  1.997


G1y  2.386

θ  atan2 G1x G1y

θ  50.070 deg

g  8.

D  0.606

u  2.000

σ  atan2 U1x U1y 7.

S 1y  0.672

2

2

G1x  G1y

g  3.112



9.


θ2i  θ  θ

θ2i  14.077 deg

θ2f  θ2i  β

θ2f  29.729 deg


δp  ϕ  θ

rp  0.000

δp  0.000  deg

which is correct for the assumption that the precision point is at C. 10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  atan2 P2x P2y R1 

2

ρ  120.033  deg

2

P2x  P2y

R1  2.470

 

O2x  2.854

 

O2y  0.963

 

O4x  0.857

 

O4y  3.349



θrot  atan2 O4x  O2x  O4y  O2y 12. Determine the Grashof condition. Condition( a b c d ) 

θrot  50.070 deg



w  2.000

θ  35.994 deg

Link 3:

v  1.250

θ  32.500 deg

Link 4:

u  2.000

σ  70.278 deg

Link 1:

g  3.112

θ  50.070 deg

Coupler:

rp  0.000

δp  0.000  deg

Crank angles:

θ2i  14.077 deg θ2f  29.729 deg




Design a linkage to carry the body in Figure P5-1 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown. Use the free choices given below.

Given:


P1y  0.0

P2x  1.236

P3x  2.500

P3y  2.931

P2y  2.138

Angles made by the body in positions 1, 2 and 3:

θP1  210  deg

θP2  147.5  deg

θP3  110.2  deg

Free choices for the WZ dyad : β  30.0 deg

β  60.0 deg

Free choices for the US dyad : γ  10.0 deg Solution: 1.

2.

3.

γ  25.0 deg



2

p 21  2.470

δ  atan2 P2x P2y

δ  120.033  deg

2

2

p 31  3.852

δ  atan2 P3x P3y

δ  130.463  deg

p 21 

P2x  P2y

p 31 

P3x  P3y

Determine the angle changes of the coupler between precision points. α  θP2  θP1

α  62.500 deg

α  θP3  θP1

α  99.800 deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

 

K  sin α

 

L  p 31  cos δ

 

M  p 21  sin δ

B C D 

 E  L CC    M  N   



G H K 

  F K H  A

 

F  cos β  1

 

G  sin β

 A F AA   B G 

 

C  cos α  1

D C

N  p 31  sin δ

 W1x  W   1y   AA  1 CC  Z1x     Z1y 

The components of the W and Z vectors are: W1x  2.920

W1y  1.720

Z1x  0.756

Z1y  0.442




ϕ  149.697  deg

 W1x2  W1y2 , w  3.389  


 Z1x2  Z1y2 , z  0.876  



θ  30.493 deg


 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

F'  cos γ  1

 

M  p 21  sin δ

B' C D 

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

N  p 31  sin δ

 U1x  U   1y   AA  1 CC  S1x     S1y 


U1y  2.693

S 1x  0.792

S 1y  2.418


σ  110.545  deg

ψ  atan2 S 1x S 1y

ψ  108.125  deg


 U1x2  U1y2 , u  2.875  


 S 1x2  S 1y2 , s  2.544  


V1x  Z1x  S 1x

V1x  0.036

V1y  Z1y  S 1y

V1y  1.976

θ  atan2 V1x V1y

θ  88.968 deg

v  Link 1:

2

2

V1x  V1y

v  1.977

G1x  W1x  V1x  U1x

G1x  3.965

G1y  W1y  V1y  U1y

G1y  1.003

θ  atan2 G1x G1y

θ  14.202 deg


g  6.

7.

8.

9.

2


2

G1x  G1y

g  4.090


θ2i  θ  θ

θ2i  16.291 deg

θ2f  θ2i  β

θ2f  76.291 deg


δp  ϕ  θ

rp  0.876

δp  238.665  deg


O2x  2.164

O2y  z sin ϕ  w sin θ

O2y  1.278

O4x  s cos ψ  u  cos σ

O4x  1.801

O4y  s sin ψ  u  sin σ

O4y  0.274

Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.


θrot  14.202 deg




w  3.389

θ  30.493 deg

Link 3:

v  1.977

θ  88.968 deg

Link 4:

u  2.875

σ  110.545  deg

Link 1:

g  4.090

θ  14.202 deg

Coupler:

rp  0.876

δp  238.665  deg

Crank angles:

θ2i  16.291 deg θ2f  76.291 deg




Y

P3 S3

Z3 A3

B1

S2 P2

B3

V3

V2 S1

A2

W3

60.0°

B2

10.0° 25.0° V1

Z1

30.0°

A1

U2 U1 X

W2

P1 W1 G1

O2

O4




Given:

Solution: 1.

2.

Design a linkage to carry the body in Figure P5-1 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis and design it for the fixed pivots shown. P21x  1.236 O2x  2.164

P21y  2.138 O2y  1.260

P31x  2.500 O4x  2.190

P31y  2.931 O4y  1.260

Body angles:

θP1  210  deg

θP2  147.5  deg

θP3  110.2  deg


Determine the angle changes between precision points from the body angles given. α  θP2  θP1

α  62.500 deg

α  θP3  θP1

α  99.800 deg


3.

4.

R1x  2.164

R1y  O2y

R2x  R1x  P21x

R2x  0.928

R2y  R1y  P21y

R2y  3.398

R3x  R1x  P31x

R3x  0.336

R3y  R1y  P31y

R3y  4.191

2

2

R1  2.504

2

2

R2  3.522

2

2

R3  4.204

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  1.260


ζ  30.210 deg

ζ  atan2 R2x R2y

ζ  74.725 deg

ζ  atan2 R3x R3y

ζ  94.584 deg






















 

C3  1.209

 

C4  6.538

 

C5  1.189






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ

C1  0.372 C2  3.726







 

C6  R1 sin α  ζ  R2 sin ζ 2

C6  4.736

2

A1  C3  C4

A1  44.206

A2  C3 C6  C4 C5

A2  2.046

A3  C4 C6  C3 C5

A3  32.399

A4  C2 C3  C1 C4

A4  6.937

A5  C4 C5  C3 C6

A5  2.046

A6  C1 C3  C2 C4

A6  23.911

K1  A2  A4  A3  A6

K1  760.497

K2  A3  A4  A5  A6

K2  273.669

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

K3  140.232

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  60.217 deg

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  99.800 deg


β  β

 A5  sin β  A3  cos β  A6   A1  

β  30.143 deg

 A3  sin β  A2  cos β  A4   A1  

β  30.143 deg



β  β



R1x  2.190

R1y  O4y

R2x  R1x  P21x

R2x  3.426

R2y  R1y  P21y

R2y  3.398

R3x  R1x  P31x

R3x  4.690

R3y  R1y  P31y

R3y  4.191

R1 

2

2

R1x  R1y

R1  2.527

R1y  1.260


6.

7.


2

2

R2  4.825

2

2

R3  6.290

R2 

R2x  R2y

R3 

R3x  R3y


ζ  150.086  deg

ζ  atan2 R2x R2y

ζ  135.235  deg

ζ  atan2 R3x R3y

ζ  138.216  deg






















 

C3  6.304

 

C4  2.247

 

C5  3.532

 

C6  0.874






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

2

C1  2.380 C2  3.298

A1  C3  C4

A1  44.796

A2  C3 C6  C4 C5

A2  2.431

A3  C4 C6  C3 C5

A3  24.233

A4  C2 C3  C1 C4

A4  15.441

A5  C4 C5  C3 C6

A5  2.431

A6  C1 C3  C2 C4

A6  22.414

K1  A2  A4  A3  A6

K1  505.612

K2  A3  A4  A5  A6

K2  428.679

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

2

K3  336.363

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  19.215 deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  99.800 deg




γ  γ

 A5  sin γ  A3  cos γ  A6   A1  

  6.628  deg

 A3  sin γ  A2  cos γ  A4   A1  

  6.628  deg

  acos

  asin

Since 2 is not in the first quadrant , 8.

γ  

Use the method of Section 5.7 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. 2

p 21 

2

P21x  P21y

p 21  2.470

δ  atan2 P21x P21y 2

p 31 

δ  120.033  deg

2

P31x  P31y

p 31  3.852

δ  atan2 P31x P31y 9.

δ  130.463  deg

Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector:

 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

A  F AA   B G 

 

C  cos α  1

 

M  p 21  sin δ

B C D 

E   L CC    M  N   

 G H K  A D C   F K H 

N  p 31  sin δ

 W1x     W1y   AA  1 CC  Z1x   Z1y   

10. The components of the W and Z vectors are: W1x  2.915 11. The length of link 2 is:

w 

W1y  1.702 2

Z1x  0.751

2

W1x  W1y

Z1y  0.442

w  3.376


 

A'  cos γ  1

 

D  sin α

 

B'  sin γ

 

E  p 21  cos δ

 

C  cos α  1

 

F'  cos γ  1



 

 

G'  sin γ

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

H  cos α  1

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   



U1y  3.634 2

u 

U1x  U1y

S1x  0.819

2

S1y  2.374

u  3.884


V1x  0.068

V1y  Z1y  S1y

V1y  1.932 v 


2

2

V1x  V1y

v  1.933

G1x  W1x  V1x  U1x

G1x  4.354

G1y  W1y  V1y  U1y

G1y  2.220  10

g 


2

G1x  G1y

2

 15

g  4.354


O2x  2.164

O2y  Z1y  W1y

O2y  1.260

O4x  S1x  U1x

O4x  2.190

O4y  S1y  U1y

O4y  1.260

These check with Figure P5-1. 17. Determine the location of the coupler point with respect to point A and line AB. 2

2

z  0.871

2

2

s  2.511


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y

rP  z


ψ  109.037  deg

ϕ  atan2( Z1x Z1y )

ϕ  149.555  deg

θ  atan2 z cos ϕ  s cos ψ z sin ϕ  s sin ψ 



θ  87.994 deg

δp  ϕ  θ

δp  237.549  deg


g  4.354

Link 2:

w  3.376

Link 3:

v  1.933

Link 4:

u  3.884

Coupler point:

rP  0.871

δp  237.549  deg





Design a linkage to carry the body in Figure P5-2 through the two positions P1 and P2 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown. Use the free choices given below.

Given:


P1y  0.0

P2x  1.903

P2y  1.347


θP1  101.0  deg

θP2  62.0 deg


β  30.0 deg

ϕ  150.0  deg

γ  40.0 deg

ψ  50.0 deg



1.

Note that this is a two-position motion generation (MG) problem because the output is specified as a complex motion of the coupler, link 3. Because of the data given in the hint, the second method of Section 5.3 will be used here.

2.


 P1x     P1y 

R2 

 P2x     P2y 

 P21x     R2  R1  P21y 

P21x  1.903 P21y  1.347

p 21  3.

4.

2

2

P21x  P21y

p 21  2.331


α  39.000 deg

δ  atan2 P21x P21y

δ  35.292 deg



A  0.134

D  sin α

 

B  0.500

E  p 21  cos δ

 

E  1.903

 

C  0.223

F  p 21  sin δ

 

F  1.347

B  sin β

C  cos α  1

 

Z1y  1.000

 

A  cos β  1

W1x 

Z1x  1.732

A   C Z1x  D Z1y  E  B  C Z1y  D Z1x  F  2  A

D  0.629

W1x  0.452


W1y  w 


A   C Z1y  D Z1x  F   B  C Z1x  D Z1y  E

W1y  1.896

2  A 2

2

W1x  W1y

w  1.949

θ  atan2 W1x W1y 5.

θ  76.607 deg


S 1x  1.928 A  0.234

D  sin α

 

B  0.643

E  p 21  cos δ

 

E  1.903

 

C  0.223

F  p 21  sin δ

 

F  1.347

B  sin γ

C  cos α  1

U1y  u 

 

D  0.629


2

2

U1x  U1y

u  6.284 σ  81.540 deg



V1x  3.660


V1y  3.298

θ  atan2 V1x V1y

θ  137.980  deg

v  Link 1:

2

2

V1x  V1y

v  4.927

 


 

G1x  4.133


G1y  7.617

θ  atan2 G1x G1y

θ  118.485  deg

g  7.

U1x  0.924

U1y  6.216

2  A

σ  atan2 U1x U1y 6.

S 1y  2.298

 

A  cos γ  1

U1x 

S 1y  s sin ψ

2

2

G1x  G1y

g  8.667


θ2i  θ  θ

θ2i  195.092  deg



θ2f  θ2i  β 8.

9.

θ2f  165.092  deg


δp  ϕ  θ

rp  2.000

δp  12.020 deg


ρ  0.000  deg

2

2

P1x  P1y

R1  0.000

 

O2x  1.281

 

O2y  0.896

 

O4x  2.853

 

O4y  8.514




θrot  118.485  deg




w  1.949

θ  76.607 deg

Link 3:

v  4.927

θ  137.980  deg

Link 4:

u  6.284

σ  81.540 deg

Link 1:

g  8.667

θ  118.485  deg

Coupler:

rp  2.000

δp  12.020 deg

Crank angles:

θ2i  195.092  deg θ2f  165.092  deg




O4

Y

1.903

U2 G1

U1

B2

S2

39.0° B1

V2 P2

V1 O2 S1

Z2 1.347

W1

W2

P1

X Z1 A1

A2




Design a linkage to carry the body in Figure P5-2 through the two positions P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown. Hint: First try a rough graphical solution to create realistic values for free choices.

Given:


P2y  1.347

P3x  1.389

P3y  1.830


θP2  62.0 deg Solution:

θP3  39.0 deg


1.


2.

Define the position vectors R1 and R2 and the vector P21 using Figure 5-1 and equation 5.1.

 P2x     P2y 

R1 

R2 

 P3x     P3y 

 P21x     R2  R1  P21y 

P21x  0.514 P21y  0.483

p 21  3.

4.

2

2

P21x  P21y


α  23.000 deg

δ  atan2 P21x P21y

δ  136.781  deg

From a graphical solution (see figure next page), determine the values necessary for input to equations 5.8. z  3

5.

p 21  0.705

β  45.0 deg

ϕ  100  deg


Z1x  0.521


 

A  0.293

D  sin α

 

B  0.707

E  p 21  cos δ

A  cos β  1 B  sin β

 

W1x 

W1y  w 

D  0.391

 

C  0.079 F  p 21  sin δ   A   C Z1x  D Z1y  E  B  C Z1y  D Z1x  F 

C  cos α  1

2  A A   C Z1y  D Z1x  F   B  C Z1x  D Z1y  E 2  A 2

2

W1x  W1y


Z1y  2.954

E  0.514 F  0.483 W1x  1.476

W1y  1.807 w  2.333 θ  50.759 deg



Y

B3 2.000

O4

20.000°

P3

B2

P2

20.0°

3.000 X

P1

100.0°

23.0° A3 45.000° A2

O2

5.

From the graphical solution (see figure above), determine the values necessary for input to equations 5.12. s  2.000

6.

γ  20.0 deg

ψ  20.0 deg


S 1x  1.879

 

A  0.060

D  sin α

 

B  0.342

E  p 21  cos δ

 

C  0.079

F  p 21  sin δ

A  cos γ  1 B  sin γ

C  cos α  1

U1x 

U1y 

u 

S 1y  s sin ψ

 

A   C S 1x  D S 1y  E  B  C S 1y  D S 1x  F  2  A A   C S 1y  D S 1x  F   B  C S 1x  D S 1y  E 2  A 2

2

U1x  U1y

σ  atan2 U1x U1y 7.

S 1y  0.684 D  0.391

 

E  0.514

 

F  0.483

U1x  3.346

U1y  0.306

u  3.360 σ  5.216  deg



V1x  2.400



V1y  z sin ϕ  s sin ψ θ  atan2 V1x V1y v  Link 1:

2

θ  123.413  deg

2

V1x  V1y

v  4.359

 


 

9.

G1x  4.271


G1y  5.751

θ  atan2 G1x G1y

θ  126.601  deg

g  8.

V1y  3.638

2

2

G1x  G1y

g  7.163


θ2i  θ  θ

θ2i  75.842 deg

θ2f  θ2i  β

θ2f  30.842 deg


δp  ϕ  θ

rp  3.000

δp  23.413 deg

10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  atan2 P2x P2y R1 

2

ρ  35.292 deg

2

P2x  P2y

R1  2.331

 

O2x  0.948

 

O2y  3.414

 

O4x  3.323

 

O4y  2.337




θrot  126.601  deg



Condition( g u v w)  "non-Grashof"




w  2.333

θ  50.759 deg

Link 3:

v  4.359

θ  123.413  deg

Link 4:

u  3.360

σ  5.216  deg

Link 1:

g  7.163

θ  126.601  deg

Coupler:

rp  3.000

δp  23.413 deg

Crank angles:

θ2i  75.842 deg θ2f  30.842 deg




Design a linkage to carry the body in Figure P5-2 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown.

Given:


P1y  0.0

P2x  1.903

P3x  1.389

P3y  1.830

P2y  1.347


θP1  101  deg

θP2  62.0 deg

θP3  39.0 deg


β  75.0 deg

Free choices for the US dyad : γ  0.0 deg Solution: 1.

2.

3.

γ  30.0 deg



2

p 21  2.331

δ  atan2 P2x P2y

δ  35.292 deg

2

2

p 31  2.297

δ  atan2 P3x P3y

δ  52.801 deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  39.000 deg

α  θP3  θP1

α  62.000 deg


 

B  sin β

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1 D  sin α G  sin β

 

L  p 31  cos δ

 A F AA   B G 

B C D 

 G H K  A D C   F K H 

The components of the W and Z vectors are:

 

 

C  cos α  1

 

 

 

M  p 21  sin δ

 E  L CC    M  N   

 

F  cos β  1

 

K  sin α

 

N  p 31  sin δ

 W1x  W   1y   AA  1 CC  Z1x     Z1y 



W1x  3.110

W1y  1.061

Z1x  0.297

Z1y  3.201


θ  18.843 deg


ϕ  84.698 deg

 W1x2  W1y2 , w  3.286  


 Z1x2  Z1y2 , z  3.215  



 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

F'  cos γ  1

 

M  p 21  sin δ

B' C D 

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

N  p 31  sin δ

 U1x  U   1y   AA  1 CC  S1x     S1y 

The components of the U and S vectors are: U1x  1.658

U1y  3.361

S 1x  2.853

S 1y  2.013


σ  63.740 deg

ψ  atan2 S 1x S 1y

ψ  144.792  deg


 U1x2  U1y2 , u  3.748  


 S 1x2  S 1y2 , s  3.492  


V1x  Z1x  S 1x

V1x  3.150

V1y  Z1y  S 1y

V1y  1.188

θ  atan2 V1x V1y

θ  20.657 deg

v  Link 1:

2

2

V1x  V1y

v  3.367

G1x  W1x  V1x  U1x

G1x  4.602

G1y  W1y  V1y  U1y

G1y  3.234



θ  atan2 G1x G1y

g  6.

7.

8.

9.

2

θ  35.099 deg

2

G1x  G1y

g  5.625


θ2i  θ  θ

θ2i  16.256 deg

θ2f  θ2i  β

θ2f  91.256 deg


δp  ϕ  θ

rp  3.215

δp  64.041 deg


O2x  3.407

O2y  z sin ϕ  w sin θ

O2y  2.140

O4x  s cos ψ  u  cos σ

O4x  1.195

O4y  s sin ψ  u  sin σ

O4y  5.374



θrot  35.099 deg




w  3.286

θ  18.843 deg

Link 3:

v  3.367

θ  20.657 deg

Link 4:

u  3.748

σ  63.740 deg

Link 1:

g  5.625

θ  35.099 deg

Coupler:

rp  3.215

δp  64.041 deg



Crank angles:

θ2i  16.256 deg

θ2f  91.256 deg


Y

P3 P2

Z3 A3 Z2 P1

V3 W3

X

S3

A2

S2

S1

75.0°

W2

B3

Z1

V2

B1 , B2

40.0° O2

W1 V1 30.0°

A1 G1

U1 U2

U3

O4




Given:

Solution: 1.

2.

Design a linkage to carry the body in Figure P5-2 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis and design it for the fixed pivots shown. P21x  1.903

P21y  1.347

P31x  1.389

P31y  1.830

O2x  0.884

O2y  1.251

O4x  3.062

O4y  1.251

Body angles:

θP1  101  deg

θP2  62 deg

θP3  39 deg



α  39.000 deg

α  θP3  θP1

α  62.000 deg


3.

4.

R1x  0.884

R1y  O2y

R2x  R1x  P21x

R2x  2.787

R2y  R1y  P21y

R2y  2.598

R3x  R1x  P31x

R3x  2.273

R3y  R1y  P31y

R3y  3.081

2

2

R1  1.532

2

2

R2  3.810

2

2

R3  3.829

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  1.251


ζ  54.754 deg

ζ  atan2 R2x R2y

ζ  42.990 deg

ζ  atan2 R3x R3y

ζ  53.582 deg






















 

C3  0.753

 

C4  3.274

 

C5  1.313






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ

C1  0.103 C2  2.205







 

C6  R1 sin α  ζ  R2 sin ζ 2

C6  2.182

2

A1  C3  C4

A1  11.288

A2  C3 C6  C4 C5

A2  2.654

A3  C4 C6  C3 C5

A3  8.134

A4  C2 C3  C1 C4

A4  1.324

A5  C4 C5  C3 C6

A5  2.654

A6  C1 C3  C2 C4

A6  7.297

K1  A2  A4  A3  A6

K1  55.842

K2  A3  A4  A5  A6

K2  30.136

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

K3  0.392

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  118.708  deg

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  62.000 deg


β  β

 A5  sin β  A3  cos β  A6   A1  

β  59.564 deg

 A3  sin β  A2  cos β  A4   A1  

β  59.564 deg



β  β



R1x  3.062

R1y  O4y

R2x  R1x  P21x

R2x  1.159

R2y  R1y  P21y

R2y  2.598

R3x  R1x  P31x

R3x  1.673

R3y  R1y  P31y

R3y  3.081

2

2

R1  3.308

2

2

R2  2.845

R1 

R1x  R1y

R2 

R2x  R2y

R1y  1.251


R3  6.

7.


2

2

R3x  R3y

R3  3.506


ζ  157.777  deg

ζ  atan2 R2x R2y

ζ  114.042  deg

ζ  atan2 R3x R3y

ζ  118.502  deg






















 

C3  1.340

 

C4  0.210

 

C5  0.433

 

C6  0.301






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

2

C1  1.111 C2  1.204

A1  C3  C4

A1  1.840

A2  C3 C6  C4 C5

A2  0.495

A3  C4 C6  C3 C5

A3  0.517

A4  C2 C3  C1 C4

A4  1.847

A5  C4 C5  C3 C6

A5  0.495

A6  C1 C3  C2 C4

A6  1.236

K1  A2  A4  A3  A6

K1  1.553

K2  A3  A4  A5  A6

K2  0.344

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

2

K3  1.033

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  62.000 deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  36.991 deg


γ  



 A5  sin γ  A3  cos γ  A6   A1  

  73.415 deg

 A3  sin γ  A2  cos γ  A4   A1  

  73.415 deg

  acos

  asin


γ  


p 21 

2

P21x  P21y

p 21  2.331

δ  atan2 P21x P21y 2

p 31 

δ  35.292 deg

2

P31x  P31y

p 31  2.297

δ  atan2 P31x P31y 9.

δ  52.801 deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 

M  p 21  sin δ

B C D 

 A F AA   B G 

 

C  cos α  1

 E  L CC    M  N   

 G H K  A D C   F K H 

N  p 31  sin δ

 W1x   W1y   AA  1 CC  Z1x   Z1y   

10. The components of the W and Z vectors are: W1x  1.262 w 


W1y  1.109 2

Z1x  0.378

2

W1x  W1y

Z1y  2.360

w  1.680


 

A'  cos γ  1

 

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

H  cos α  1

D  sin α

G'  sin γ

 

L  p 31  cos δ

 

 

 

M  p 21  sin δ

 

F'  cos γ  1

 

K  sin α

 

N  p 31  sin δ


 A' F' AA    B'  G' 


B' C D 

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   



U1y  0.830 2

u 

U1x  U1y

S1x  2.736

2

S1y  0.421

u  0.892


V1x  2.359

V1y  Z1y  S1y

V1y  1.939 v 


2

2

V1x  V1y

v  3.054

G1x  W1x  V1x  U1x

G1x  3.946

G1y  W1y  V1y  U1y

G1y  1.110  10

g 


2

G1x  G1y

2

 15

g  3.946


O2x  0.884

O2y  Z1y  W1y

O2y  1.251

O4x  S1x  U1x

O4x  3.062

O4y  S1y  U1y

O4y  1.251


2

z  2.390

2

2

s  2.769


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y


ψ  171.262  deg

ϕ  atan2( Z1x Z1y )

ϕ  99.095 deg

rP  z

θ  atan2 z cos ϕ  s cos ψ z sin ϕ  s sin ψ  θ  39.430 deg



δp  ϕ  θ

δp  59.666 deg


g  3.946

Link 2:

w  1.680

Link 3:

v  3.054

Link 4:

u  0.892

Coupler point:

rP  2.390

δp  59.666 deg





Design a linkage to carry the body in Figure P5-3 through the two positions P1 and P2 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown.

Given:


P1y  0.0

P2x  0.907

P2y  0.0


θP1  111.8  deg

θP2  191.1  deg


β  44.0 deg

ϕ  50.0 deg

γ  55.0 deg

ψ  20.0 deg



1.


2.


 P1x     P1y 

R2 

 P2x     P2y 

 P21x     R2  R1  P21y 

P21x  0.907 P21y  0.000

p 21  3.

4.

2

2

P21x  P21y

p 21  0.907


α  79.300 deg

δ  atan2 P21x P21y

δ  180.000  deg


  B  sin β

A  cos β  1

 

C  cos α  1 W1x 

W1y 

Z1x  0.964


 

A  0.281

D  sin α

B  0.695

E  p 21  cos δ

C  0.814

F  p 21  sin δ

A   C Z1x  D Z1y  E  B  C Z1y  D Z1x  F  2  A A   C Z1y  D Z1x  F   B  C Z1x  D Z1y  E 2  A

Z1y  1.149 D  0.983

 

E  0.907

 

F  0.000 W1x  1.705

W1y  2.490



2

w 

2

W1x  W1y

w  3.018

θ  atan2 W1x W1y 5.

θ  124.405  deg


S 1x  2.349

S 1y  s sin ψ

 

A  0.426

D  sin α

 

B  0.819

E  p 21  cos δ

 

C  0.814

F  p 21  sin δ

A  cos γ  1 B  sin γ

C  cos α  1

 

D  0.983

 

E  0.907

 

F  0.000


U1x 


U1y 

2

2

U1x  U1y

u  2.654

σ  atan2 U1x U1y 6.

σ  76.373 deg



V1x  1.385


V1y  2.004

θ  atan2 V1x V1y

θ  124.648  deg

v  Link 1:

2

2

V1x  V1y

v  2.436

 


 

8.

G1x  3.715


G1y  2.094

θ  atan2 G1x G1y

θ  150.596  deg

g  7.

U1x  0.625

U1y  2.579

2  A

u 

S 1y  0.855

2

2

G1x  G1y

g  4.265


θ2i  θ  θ

θ2i  275.001  deg

θ2f  θ2i  β

θ2f  319.001  deg


δp  ϕ  θ

rp  1.500

δp  74.648 deg


9.



ρ  0.000  deg

2

2

P1x  P1y

R1  0.000

 

O2x  0.741

 

O2y  1.341

 

O4x  2.975

 

O4y  3.434




θrot  150.596  deg




w  3.018

θ  124.405  deg

Link 3:

v  2.436

θ  124.648  deg

Link 4:

u  2.654

σ  76.373 deg

Link 1:

g  4.265

θ  150.596  deg

Coupler:

rp  1.500

δp  74.648 deg

Crank angles:

θ2i  275.001  deg

θ2f  319.001  deg




Y

A1 Z1 P2

V1

P1

Z2 A2

X S1

44.0°

B1

W2 V2

S2

55.0° U1

G1

B2 U2

O4

W1 O2




Design a linkage to carry the body in Figure P5-3 through the two positions P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown.

Given:

Coordinates of the points P2 and P3 with respect to P1: P2x  0.907

P2y  0.0

P3x  1.447

P3y  0.0


θP2  191.1  deg

θP3  237.4  deg


β  66.0 deg

ϕ  60.0 deg

γ  44.0 deg

ψ  30.0 deg



1.


2.


 P2x     P2y 

R2 

 P3x     P3y 

 P21x     R2  R1  P21y 

P21x  0.540 P21y  0.000

p 21  3.

4.

2

2

P21x  P21y

p 21  0.540


α  46.300 deg

δ  atan2 P21x P21y

δ  180.000  deg


A  0.593

D  sin α

 

B  0.914

E  p 21  cos δ

 

C  0.309

F  p 21  sin δ

B  sin β

C  cos α  1

W1y 


 

A  cos β  1

W1x 

Z1x  1.000

 

A   C Z1x  D Z1y  E  B  C Z1y  D Z1x  F  2  A A   C Z1y  D Z1x  F   B  C Z1x  D Z1y  E 2  A

Z1y  1.732 D  0.723

 

E  0.540

 

F  0.000 W1x  0.227

W1y  1.771



2

w 

2

W1x  W1y

w  1.786

θ  atan2 W1x W1y 5.

θ  97.314 deg


S 1x  2.598

S 1y  s sin ψ

 

A  0.281

D  sin α

 

B  0.695

E  p 21  cos δ

 

C  0.309

F  p 21  sin δ

A  cos γ  1 B  sin γ

C  cos α  1

 

D  0.723

 

E  0.540

 

F  0.000


U1x 


U1y 

2

2

U1x  U1y

u  2.608

σ  atan2 U1x U1y 6.

σ  65.609 deg



V1x  1.598


V1y  3.232

θ  atan2 V1x V1y

θ  116.310  deg

v  Link 1:

2

2

V1x  V1y

v  3.606

 


 

8.

G1x  2.902


G1y  3.836

θ  atan2 G1x G1y

θ  127.111  deg

g  7.

U1x  1.077

U1y  2.375

2  A

u 

S 1y  1.500

2

2

G1x  G1y

g  4.810


θ2i  θ  θ

θ2i  224.425  deg

θ2f  θ2i  β

θ2f  290.425  deg


δp  ϕ  θ



rp  2.000 9.

δp  56.310 deg

Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  atan2 P2x P2y R1 

2

ρ  180.000  deg

2

P2x  P2y

R1  0.907

 

O2x  1.680

 

O2y  0.039

 

O4x  4.582

 

O4y  3.875




θrot  127.111  deg




w  1.786

θ  97.314 deg

Link 3:

v  3.606

θ  116.310  deg

Link 4:

u  2.608

σ  65.609 deg

Link 1:

g  4.810

θ  127.111  deg

Coupler:

rp  2.000

δp  56.310 deg

Crank angles:

θ2i  224.425  deg

θ2f  290.425  deg




Y

A2 66.0° V1

Z1

A3

W1 Z2 W2

P3

O2

P2

S1

B2

U1

S2

V2

G1 44.0° B1 U2

O4

P1 X




Design a linkage to carry the body in Figure P5-3 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis without regard for the fixed pivots shown.

Given:


P1y  0.0

P3x  1.447

P3y  0.0

P2x  0.907

P2y  0.0


θP1  111.8  deg

θP2  191.1  deg

θP3  237.4  deg


β  80.0 deg

Free choices for the US dyad : γ  20.0 deg Solution: 1.

2.

3.

γ  50.0 deg



2

p 21  0.907

δ  atan2 P2x P2y

δ  180.000  deg

2

2

p 31  1.447

δ  atan2 P3x P3y

δ  180.000  deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  79.300 deg

α  θP3  θP1

α  125.600  deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

 

K  sin α

 

L  p 31  cos δ

 

M  p 21  sin δ

B C D 

 E  L CC    M  N   



G H K 

  F K H  A

 

F  cos β  1

 

G  sin β

 A F AA   B G 

 

C  cos α  1

D C

N  p 31  sin δ

 W1x  W   1y   AA  1 CC  Z1x     Z1y 

The components of the W and Z vectors are: W1x  1.696

W1y  0.038

Z1x  0.396

Z1y  0.872




ϕ  114.406  deg

 W1x2  W1y2 , w  1.696  


 Z1x2  Z1y2 , z  0.958  



θ  1.280  deg


 

 

A'  cos γ  1

 

B'  sin γ

C  cos α  1

 

E  p 21  cos δ

 

 

H  cos α  1

D  sin α

 

G'  sin γ

 

 

M  p 21  sin δ

B' C D 

 E  L CC    M  N   



G' H K 

  F' K H 

A'

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

F'  cos γ  1

D C

N  p 31  sin δ

 U1x  U   1y   AA  1 CC  S1x     S1y 

The components of the U and S vectors are: U1x  1.483

U1y  0.409

S 1x  0.048

S 1y  0.650


σ  15.412 deg

ψ  atan2 S 1x S 1y

ψ  85.790 deg


 U1x2  U1y2 , u  1.538  


 S 1x2  S 1y2 , s  0.652  


V1x  Z1x  S 1x

V1x  0.444

V1y  Z1y  S 1y

V1y  0.222

θ  atan2 V1x V1y

θ  153.426  deg

v  Link 1:

2

2

V1x  V1y

v  0.496

G1x  W1x  V1x  U1x

G1x  0.230

G1y  W1y  V1y  U1y

G1y  0.225

θ  atan2 G1x G1y

θ  135.700  deg


g  6.

7.

8.

9.

2


2

G1x  G1y

g  0.322


θ2i  θ  θ

θ2i  134.419  deg

θ2f  θ2i  β

θ2f  214.419  deg


δp  ϕ  θ

rp  0.958

δp  39.020 deg


O2x  1.300

O2y  z sin ϕ  w sin θ

O2y  0.834

O4x  s cos ψ  u  cos σ

O4x  1.530

O4y  s sin ψ  u  sin σ

O4y  1.059



θrot  135.700  deg




w  1.696

θ  1.280  deg

Link 3:

v  0.496

θ  153.426  deg

Link 4:

u  1.538

σ  15.412 deg

Link 1:

g  0.322

θ  135.700  deg

Coupler:

rp  0.958

δp  39.020 deg

Crank angles:

θ2i  134.419  deg θ2f  214.419  deg




Y A3 B3

A2

P3 P2 W2 O2 O4

U1

B2 U2

B1 W1

X

P1 S1 A1



PROBLEM 5-19 Statement: Given:

Solution: 1.

2.

Design a linkage to carry the body in Figure P5-3 through the three positions P1, P2 and P3 at the angles shown in the figure. Use analytical synthesis and design it for the fixed pivots shown. P21x  0.907

P21y  0.0

P31x  1.447

P31y  0.0

O2x  1.788

O2y  1.994

O4x  0.212

O4y  1.994

Body angles:

θP1  111.8  deg

θP2  191.1  deg

θP3  237.4  deg



α  79.300 deg

α  θP3  θP1

α  125.600  deg


3.

4.

R1x  1.788

R1y  O2y

R2x  R1x  P21x

R2x  0.881

R2y  R1y  P21y

R2y  1.994

R3x  R1x  P31x

R3x  0.341

R3y  R1y  P31y

R3y  1.994

2

2

R1  2.678

2

2

R2  2.180

2

2

R3  2.023

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  1.994


ζ  48.118 deg

ζ  atan2 R2x R2y

ζ  66.163 deg

ζ  atan2 R3x R3y

ζ  80.296 deg






















 

C3  3.003

 

C4  1.701

 

C5  2.508






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ

C1  0.238 C2  1.150







 

C6  R1 sin α  ζ  R2 sin ζ 2

C6  0.133

2

A1  C3  C4

A1  11.912

A2  C3 C6  C4 C5

A2  4.666

A3  C4 C6  C3 C5

A3  7.307

A4  C2 C3  C1 C4

A4  3.048

A5  C4 C5  C3 C6

A5  4.666

A6  C1 C3  C2 C4

A6  2.671

K1  A2  A4  A3  A6

K1  5.293

K2  A3  A4  A5  A6

K2  34.731

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

K3  25.158

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  125.600  deg

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  37.070 deg


β  β

 A5  sin β  A3  cos β  A6   A1  

β  18.241 deg

 A3  sin β  A2  cos β  A4   A1  

β  18.241 deg



β  β



R1x  0.212

R1y  O4y

R2x  R1x  P21x

R2x  1.119

R2y  R1y  P21y

R2y  1.994

R3x  R1x  P31x

R3x  1.659

R3y  R1y  P31y

R3y  1.994

R1 

2

2

R1x  R1y

R1  2.005

R1y  1.994


6.

7.


2

2

R2  2.287

2

2

R3  2.594

R2 

R2x  R2y

R3 

R3x  R3y


ζ  96.069 deg

ζ  atan2 R2x R2y

ζ  119.300  deg

ζ  atan2 R3x R3y

ζ  129.760  deg






















 

C3  0.161

 

C4  3.327

 

C5  0.880

 

C6  1.832






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

2

C1  1.297 C2  0.811

A1  C3  C4

A1  11.096

A2  C3 C6  C4 C5

A2  3.222

A3  C4 C6  C3 C5

A3  5.954

A4  C2 C3  C1 C4

A4  4.186

A5  C4 C5  C3 C6

A5  3.222

A6  C1 C3  C2 C4

A6  2.906

K1  A2  A4  A3  A6

K1  3.816

K2  A3  A4  A5  A6

K2  34.288

2

K3 

2

2

2

A1  A2  A3  A4  A6

2

2

K3  25.658

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3  

γ  125.600  deg

 K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3  

  41.699 deg




γ  

 A5  sin γ  A3  cos γ  A6   A1  

  31.159 deg

 A3  sin γ  A2  cos γ  A4   A1  

  31.159 deg

  acos

  asin

γ  

Since both values are the same , 8.


p 21 

2

P21x  P21y

p 21  0.907

δ  atan2 P21x P21y 2

p 31 

δ  180.000  deg

2

P31x  P31y

p 31  1.447

δ  atan2 P31x P31y 9.

δ  180.000  deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 A F AA   B G 

 

C  cos α  1

 

M  p 21  sin δ

 E  L CC    M  N   

B C D 

 G H K  A D C   F K H 

N  p 31  sin δ

 W1x   W1y   AA  1 CC  Z1x   Z1y   

10. The components of the W and Z vectors are: W1x  1.943


w 

W1y  1.529 2

Z1x  0.155

2

W1x  W1y

Z1y  0.465

w  2.472


 

A'  cos γ  1

 

D  sin α

 

B'  sin γ

 

E  p 21  cos δ

 

C  cos α  1

 

F'  cos γ  1



 

 

G'  sin γ

 

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

H  cos α  1

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   

13. The components of the W and Z vectors are: U1x  0.395 14. The length of link 4 is:

U1y  2.460 2

u 

U1x  U1y

S1x  0.183

2

S1y  0.466

u  2.491


V1x  0.338

V1y  Z1y  S1y

V1y  0.931 v 


2

2

V1x  V1y

v  0.991

G1x  W1x  V1x  U1x

G1x  2.000

G1y  W1y  V1y  U1y

G1y  0.000

g 


2

G1x  G1y

2

g  2.000


O2x  1.788

O2y  Z1y  W1y

O2y  1.994

O4x  S1x  U1x

O4x  0.212

O4y  S1y  U1y

O4y  1.994


2

z  0.491

2

2

s  0.501


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y


ψ  68.558 deg

ϕ  atan2( Z1x Z1y )

ϕ  108.446  deg

rP  z

θ  atan2 z cos ϕ  s cos ψ z sin ϕ  s sin ψ  θ  109.959  deg



δp  ϕ  θ

δp  1.513  deg


g  2.000

Link 2:

w  2.472

Link 3:

v  0.991

Link 4:

u  2.491

Coupler point:

rP  0.491

δp  1.513  deg





Write a program to generate and plot the circle-point and center-point circles for Problem 5-19 using an equation solver or any program language.

Given: P21x  0.907

P21y  0.0

P31x  1.447

P31y  0.0

O2x  1.788

O2y  1.994

O4x  0.212

O4y  1.994

Body angles:

θP1  111.8  deg

θP2  191.1  deg

θP3  237.4  deg

Assumptions: Let the position 1 to position 2 rotation angles be: β  18.241 deg and γ  31.159 deg Let the position 1 to position 2 coupler rotation angle be: α  79.3 deg Solution: 1.


Use the method of Section 5.6 to synthesize the linkage. Start by determining the magnitudes of the vectors P21 and P31 and their angles with respect to the X axis. p 21 

2

2

P21x  P21y

p 21  0.907

δ  atan2 P21x P21y p 31 

2

δ  180.000  deg

2

P31x  P31y

p 31  1.447

δ  atan2 P31x P31y 2.

δ  180.000  deg

Evaluate terms in the WZ coefficient matrix and constant vector from equations (5.25) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α  125.6  deg

β  0  deg 1  deg  360  deg

 

B  sin β

 

 

E  p 21  cos δ

A  cos β  1 D  sin α

 

 

 

C  cos α  1

 

F β  cos β  1

 

K α  sin α

 

N  p 31  sin δ

 

H α  cos α  1

 

M  p 21  sin δ

G β  sin β L  p 31  cos δ

 

 

 

   

B C D   A  F β G β H α K α           AA  α β    B A D C  G β F β K α H α           

 E  L CC    M  N   

 1 CC















W1y α β  DD α β 2









Z1y α β  DD α β 4

DD α β  AA α β

W1x α β  DD α β 1 Z1x α β  DD α β 3


















3.


Check this against the solutions in Problem 5-19:





W1y α 37.070 deg  1.529





Z1y α 37.070 deg  0.465

W1x α 37.070 deg  1.943 Z1x α 37.070 deg  0.155









These are the same as the values calculated in Problem 5-19. 4.

Form the vector N, whose tip describes the center-point circle for the WZ dyad.

























Nx α β  W1x α β  Z1x α β Ny α β  W1y α β  Z1y α β 5.

Plot the center-point circle for the WZ dyad. Center-Point Circle for WZ Dyad 0

1





Ny α β   2

3

4 2

1



0

1



2

Nx α β 

4.

Form the vector Z, whose tip describes the center-point circle for the WZ dyad. β  37.070 deg





α  0  deg 1  deg  360  deg





Zx α β  Z1x α β

5.









Zy α β  Z1y α β

Plot the circle-point circle for the WZ dyad (see next page).



Circle-Point Circle for WZ Dyad  0.1

 0.2





Zy α β   0.3

 0.4

 0.5  0.2

 0.1

0





0.1

0.2

0.3

Zx α β 

6.

Evaluate terms in the US coefficient matrix and constant vector from equations (5.31) and form the matrix and vector to get the center-point and circle-point circles for the left dyad. α  125.6  deg

γ  0  deg 1  deg  360  deg

 

B  sin γ

 

 

E  p 21  cos δ

A  cos γ  1 D  sin α

 

 

 

C  cos α  1

 

F γ  cos γ  1

 

K α  sin α

 

N  p 31  sin δ

 

H α  cos α  1

 

M  p 21  sin δ

G γ  sin γ L  p 31  cos δ

 

 

 

   

B C D   A  F γ G γ H α K α           AA  α γ    B A D C  G γ F γ K α H α           

 E  L CC    M  N   

 1 CC















U1y α γ  DD α γ 2









S1y α γ  DD α γ 4

DD α γ  AA α γ

U1x α γ  DD α γ 1 S1x α γ  DD α γ 3 7.

















Check this against the solutions in Problem 5-19:





U1x α 41.699 deg  0.395





U1y α 41.699 deg  2.460









S1x α 41.699 deg  0.183



S1y α 41.699 deg  0.466

These are the same as the values calculated in Problem 5-19. 8.

Form the vector M, whose tip describes the center-point circle for the US dyad.

























Mx α γ  U1x α γ  S1x α γ My α γ  U1y α γ  S1y α γ 9.

Plot the center-point circle for the US dyad. Center-Point Circle for US Dyad 0

 0.5

1



My α γ

  1.5

2

 2.5  1.5

1

 0.5



Mx α γ

0



0.5

10. Form the vector S, whose tip describes the center-point circle for the US dyad. γ  41.699 deg





α  0  deg 1  deg  360  deg





Sx α γ  S1x α γ

11. Plot the circle-point circle for the WZ dyad (see next page).









Sy α γ  S1y α γ



Circle-Point Circle for the US Dyad 1.5

1



Sy α γ



0.5

0  0.5

0





Sx α γ

0.5

1




Design a fourbar linkage to carry the box in Figure P5-4 from position 1 to 2 without regard for the fixed pivots shown. Use points A and B for your attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base.

Given:


P1y  0.0

P2x  184.0

P2y  17.0


θP1  90.0 deg

θP2  45.0 deg

Coordinates of the points A1 and B1 with respect to P1: A1x  17.0

A1y  43.0

B1x  69.0

B1y  43.0

Free choice for the WZ dyad : β  44.0 deg Free choice for the US dyad : γ  55.0 deg Solution:


1.


2.


 P1x     P1y 

R2 

 P2x     P2y 

 P21x     R2  R1  P21y 

P21x  184.000 P21y  17.000

p 21  3.

4.

2

2

P21x  P21y


α  45.000 deg

δ  atan2 P21x P21y

δ  5.279  deg

Using Figure P5-4, the given data, and the law of cosines, determine z, s, , and . z 

P1x  A1x2   P1y  A1y 2

z  46.239

s 

P1x  B1x 2  P1y  B1y 2

s  81.302

v 

 A1x  B1x 2  A1y  B1y2

v  52.000

 v2  z2  s2    2  v z 

ϕ  acos

 v2  s2  z2    2  v s 

ψ  π  acos 5.

p 21  184.784

Solve for the WZ dyad using equations 5.8.

ϕ  111.571  deg

ψ  148.069  deg



Z1x  z cos ϕ

Z1x  17.000 A  0.281

D  sin α

 

B  0.695

E  p 21  cos δ

 

E  184.000

 

C  0.293

F  p 21  sin δ

 

F  17.000

B  sin β

C  cos α  1

W1y  w 

 

D  0.707


W1x  53.979


W1y  192.131

2  A 2

2

W1x  W1y

w  199.570

θ  atan2 W1x W1y 6.

θ  105.693  deg


S 1x  69.000

D  sin α

 

B  0.819

E  p 21  cos δ

 

E  184.000

 

C  0.293

F  p 21  sin δ

 

F  17.000

C  cos α  1

U1y  u 

S 1y  43.000

A  0.426

B  sin γ

U1x 

S 1y  s sin ψ

 

A  cos γ  1

 

D  0.707


2

U1x  15.598

U1y  154.713

2  A 2

U1x  U1y

u  155.497

σ  atan2 U1x U1y 7.

Z1y  43.000

 

A  cos β  1

W1x 


σ  95.757 deg



V1x  52.000


V1y  0.000

θ  atan2 V1x V1y

θ  0.000  deg

v  Link 1:

2

2

V1x  V1y

v  52.000

 


G1x  13.619



 


G1y  37.418

θ  atan2 G1x G1y

θ  70.000 deg

g  8.

9.

2

2

G1x  G1y

g  39.819


θ2i  θ  θ

θ2i  35.692 deg

θ2f  θ2i  β

θ2f  8.308  deg


δp  ϕ  θ

rp  46.239

δp  111.571  deg

10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  0  deg R1 

ρ  0.000  deg

2

2

P1x  P1y

R1  0.000

 

O2x  70.979

 

O2y  235.131

 

O4x  84.598

 

O4y  197.713


These fixed pivot points fall on the base and are, therefore, acceptable. 11. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis to the line O2O4.


θrot  70.000 deg



Condition( g u v w)  "non-Grashof"




w  199.570

θ  105.693  deg

Link 3:

v  52.000

θ  0.000  deg

Link 4:

u  155.497

σ  95.757 deg

Link 1:

g  39.819

θ  70.000 deg

Coupler:

rp  46.239

δp  111.571  deg

Crank angles:

θ2i  35.692 deg

θ2f  8.308  deg


Y

P1

X Z1 A1

P2

S1 B1

Z2

V1 A2

V2 55.0°

U1 44.0° W1

W2

O4

O2

U2

S2

B2





Given:


P1y  0.0

P3x  211.0

P3y  180.0


θP1  90.0 deg

θP3  0.0 deg


A1y  43.0

B1x  69.0

B1y  43.0



1.


2.


p 21  3.

4.

 P1x     P1y 

R2 

2

 P3x     P3y 

 P21x     R2  R1  P21y 

2

P21x  P21y

P21y  180.000 p 21  277.346


α  90.000 deg

δ  atan2 P21x P21y

δ  40.467 deg


P1x  A1x2   P1y  A1y 2

z  46.239

s 

P1x  B1x 2  P1y  B1y 2

s  81.302

v 

 A1x  B1x 2  A1y  B1y2

v  52.000

 v2  z2  s2    2  v z 

ϕ  acos

ϕ  111.571  deg

 v2  s2  z2    2  v s 

ψ  π  acos 5.

P21x  211.000

ψ  148.069  deg


Z1x  17.000


Z1y  43.000



 

A  0.658

D  sin α

 

B  0.940

E  p 21  cos δ

 

C  1.000

F  p 21  sin δ

A  cos β  1 B  sin β

C  cos α  1 W1x 

W1y  w 

 

D  1.000

 

E  211.000

 

F  180.000


W1x  34.467


W1y  184.825

2  A 2

2

W1x  W1y

w  188.012

θ  atan2 W1x W1y 6.

θ  79.436 deg


S 1x  69.000

 

A  1.087

D  sin α

 

B  0.996

E  p 21  cos δ

 

C  1.000

F  p 21  sin δ

A  cos γ  1 B  sin γ

C  cos α  1 U1x 

U1y  u 

S 1y  s sin ψ

 

D  1.000

 

E  211.000

 

F  180.000


2

U1x  U1y

u  154.999

σ  atan2 U1x U1y 7.

U1x  44.882

U1y  148.358

2  A 2

S 1y  43.000

σ  73.168 deg



V1x  52.000


V1y  0.000

θ  atan2 V1x V1y

θ  0.000  deg

v  Link 1:

2

2

V1x  V1y

v  52.000


G1x  41.585


G1y  36.467

θ  atan2 G1x G1y

θ  41.248 deg

 

 


g 

8.

9.

2


G1x  G1y

g  55.310


θ2i  θ  θ

θ2i  38.188 deg

θ2f  θ2i  β

θ2f  31.812 deg


δp  ϕ  θ

rp  46.239

δp  111.571  deg

10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  0  deg R1 

ρ  0.000  deg

2

2

P1x  P1y

R1  0.000

 

O2x  17.467

 

O2y  227.825

 

O4x  24.118

 

O4y  191.358




θrot  41.248 deg




w  188.012

θ  79.436 deg

Link 3:

v  52.000

θ  0.000  deg

Link 4:

u  154.999

σ  73.168 deg



Link 1:

g  55.310

θ  41.248 deg

Coupler:

rp  46.239

δp  111.571  deg

Crank angles:

θ2i  38.188 deg θ2f  31.812 deg 14. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design. Y

P1

X Z1

S1 B1

A1

V1

U1 W1 70.0°

95.0° P3

O4

O2

W2

A3

U2

Z2

V2

S2 B3





Given:


P2y  17.0

P3x  211.0

P3y  180.0


θP2  45.0 deg

θP3  0.0 deg


A1y  43.0

B1x  69.0

B1y  43.0



1.


2.


p 21  3.

4.

 P2x     P2y 

R2 

2

 P3x     P3y 

 P21x     R2  R1  P21y 

2

P21x  P21y

P21y  163.000 p 21  165.221


α  45.000 deg

δ  atan2 P21x P21y

δ  80.595 deg


0.0  A1x 2  0.0  A1y 2

z  46.239

s 

0.0  B1x2   0.0  B1y 2

s  81.302

v 

 A1x  B1x 2  A1y  B1y2

v  52.000

 v2  z2  s2    45 deg  2  v z 

ϕ  acos

ϕ  66.571 deg

 v2  s2  z2    45 deg  2  v s 

ψ  π  acos 5.

P21x  27.000

ψ  103.069  deg


Z1x  18.385


Z1y  42.426



 

A  0.500

D  sin α

 

B  0.866

E  p 21  cos δ

 

C  0.293

F  p 21  sin δ

A  cos β  1 B  sin β

C  cos α  1

W1x 

W1y  w 

 

D  0.707

 

E  27.000

 

F  163.000


W1x  117.950


W1y  70.852

2  A 2

2

W1x  W1y

w  137.594

θ  atan2 W1x W1y 6.

θ  30.993 deg


S 1x  18.385

 

A  0.293

D  sin α

 

B  0.707

E  p 21  cos δ

 

C  0.293

F  p 21  sin δ

A  cos γ  1 B  sin γ

C  cos α  1 U1x 

U1y  u 

S 1y  s sin ψ

 

D  0.707

 

E  27.000

 

F  163.000


2

U1x  U1y

u  204.640

σ  atan2 U1x U1y 7.

U1x  201.643

U1y  34.896

2  A 2

S 1y  79.196

σ  9.818  deg



V1x  36.770


V1y  36.770

θ  atan2 V1x V1y v  Link 1:

2

θ  45.000 deg

2

V1x  V1y

v  52.000

 


 


G1x  46.924 G1y  0.813



θ  atan2 G1x G1y g  8.

9.

2

θ  179.007  deg

2

G1x  G1y

g  46.931


θ2i  θ  θ

θ2i  210.000  deg

θ2f  θ2i  β

θ2f  150.000  deg


δp  ϕ  θ

rp  46.239

δp  111.571  deg

10. Locate the fixed pivots in the global frame using the vector definitions in Figure 5-2. ρ  atan2 P2x P2y R1 

2

ρ  5.279  deg

2

P2x  P2y

R1  184.784

 

O2x  47.665

 

O2y  130.278

 

O4x  0.742

 

O4y  131.092




θrot  179.007  deg




w  137.594

θ  30.993 deg

Link 3:

v  52.000

θ  45.000 deg

Link 4:

u  204.640

σ  9.818  deg



Link 1:

g  46.931

θ  179.007  deg

Coupler:

rp  46.239

δp  111.571  deg

Crank angles:

θ2i  210.000  deg θ2f  150.000  deg 14. Draw the linkage, using the link lengths, fixed pivot positions, and angles above, to verify the design. Y

P1

X P2

A2

Z2

S2

V2

B2

W1 O4

U1 O2

W2 P3 U2 A3 V2 B3

Z2 S2




Given:

Design a fourbar linkage to carry the box in Figure P5-4 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use any points on the object as attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base. Coordinates of the points P1 , P2 and P3 with respect to P1: P1x  0.0

P1y  0.0

P2x  184.0

P3x  211.0

P3y  180.0

P2y  17.0


θP1  90.0 deg

θP2  45.0 deg

θP3  0.0 deg

Free choices for the WZ dyad : β  80.0 deg

β  160.0  deg

Free choices for the US dyad : γ  80.0 deg Solution: 1.

2.

3.

γ  170.0  deg



2

p 21  184.784

δ  atan2 P2x P2y

δ  5.279  deg

2

2

p 31  277.346

δ  atan2 P3x P3y

δ  40.467 deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  45.000 deg

α  θP3  θP1

α  90.000 deg

Evaluate terms in the WZ coefficient matrix and constant vector from equations 5.26 and form the matrix and vector: A  cos β  1 B  sin β C  cos α  1

  D  sin α G  sin β L  p 31  cos δ  A F AA   B G 

 

  H  cos α  1 M  p 21  sin δ E  p 21  cos δ

B C D 

 E  L CC    M  N   



G H K 

  F K H  A

  F  cos β  1 K  sin α N  p 31  sin δ

D C

 W1x  W   1y   AA  1 CC  Z1x     Z1y 

The components of the W and Z vectors are: W1x  51.854

W1y  109.176

Z1x  43.555

Z1y  29.523


θ  115.406  deg


ϕ  145.869  deg


 W1x2  W1y2 , w  120.864  



 Z1x2  Z1y2 , z  52.618  



  D  sin α

  E  p 21  cos δ

A'  cos γ  1

 

C  cos α  1

 

G'  sin γ

 

H  cos α  1

 

 

M  p 21  sin δ

B' C D 

 E  L CC    M  N   



G' H K 

  F' K H 

A'

K  sin α

 

L  p 31  cos δ

 A' F' AA    B'  G' 

  F'  cos γ  1

B'  sin γ

D C

N  p 31  sin δ

 U1x  U   1y   AA  1 CC  S1x     S1y 


U1y  94.023

S 1x  62.812

S 1y  62.282


σ  110.448  deg

ψ  atan2 S 1x S 1y

ψ  135.242  deg


 U1x2  U1y2 , u  100.345  



V1x  Z1x  S 1x

V1x  19.256

V1y  Z1y  S 1y

V1y  32.759

θ  atan2 V1x V1y

θ  59.552 deg

v  Link 1:

2

2

V1x  V1y

v  38.000

G1x  W1x  V1x  U1x

G1x  2.458

G1y  W1y  V1y  U1y

G1y  17.606

θ  atan2 G1x G1y

θ  82.052 deg

g  7.

 S 1x2  S 1y2 , s  88.456  

2

2

G1x  G1y

g  17.777


θ2i  θ  θ

θ2i  197.458  deg

θ2f  θ2i  β

θ2f  37.458 deg


8.

9.



δp  ϕ  θ

rp  52.618

δp  205.422  deg


O2x  95.410

O2y  z sin ϕ  w sin θ

O2y  138.699

O4x  s cos ψ  u  cos σ

O4x  97.868

O4y  s sin ψ  u  sin σ

O4y  156.305



θrot  82.052 deg




w  120.864

θ  115.406  deg

Link 3:

v  38.000

θ  59.552 deg

Link 4:

u  100.345

σ  110.448  deg

Link 1:

g  17.777

θ  82.052 deg

Coupler:

rp  52.618

δp  205.422  deg

Crank angles:

θ2i  197.458  deg

θ2f  37.458 deg




Y

P1

X Z1 S1

P2

A1 V1 B1

Z2 S2

W1

A2 V2 B2

W2

U1

U2

O2 W3

O4

P3

U3

S3

V3 B3

Z3 A3




Design a fourbar linkage to carry the box in Figure P5-4 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use points A and B for your attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base.

Given:

Coordinates of the points P1 , P2 and P3 with respect to P1: P1x  0.0

P1y  0.0

P2x  184.0

P3x  211.0

P3y  180.0

P2y  17.0


θP1  90.0 deg

θP2  45.0 deg

θP3  0.0 deg

Coordinates of the points A1 and B1 with respect to P1: A1x  17.0 Solution: 1.

2.

3.

A1y  43.0

B1y  43.0



2

p 21  184.784

δ  atan2 P2x P2y

δ  5.279  deg

2

2

p 31  277.346

δ  atan2 P3x P3y

δ  40.467 deg

p 21 

P2x  P2y

p 31 

P3x  P3y


α  45.000 deg

α  θP3  θP1

α  90.000 deg


P1x  A1x2   P1y  A1y 2

z  46.239

s 

P1x  B1x 2  P1y  B1y 2

s  81.302

v 

 A1x  B1x 2  A1y  B1y2

v  52.000

 v2  z2  s2    2  v z 

ϕ  acos

 v2  s2  z2    2  v s 

4.

B1x  69.0

ϕ  111.571  deg

ψ  π  acos

ψ  148.069  deg

Z1x  z cos ϕ

Z1x  17.000


Z1y  43.000

S 1x  s cos ψ

S 1x  69.000

S 1y  s sin ψ

S 1y  43.000

Use equations 5.24 to solve for w, , 2, and 3. Since the points A and B are to be used as pivots, z and  are known from the calculations above.



Guess:

W1x  50

W1y  200

β  80 deg

Given

W1x cos β  1  W1y sin β  = p 21  cos δ  Z1x cos α  1  Z1y sin α

β  160  deg

       

   

 

       

   

 

       

   

 

       

   

 

W1x cos β  1  W1y sin β  = p 31  cos δ  Z1x cos α  1  Z1y sin α W1y cos β  1  W1x sin β  = p 21  sin δ  Z1y cos α  1  Z1x sin α W1y cos β  1  W1x sin β  = p 31  sin δ  Z1y cos α  1  Z1x sin α

 W1x   W1y     Find  W1x W1y β β  β   β  β  86.887 deg

β  165.399  deg

The components of the W vector are: W1x  65.636 The length of link 2 is: w  5.


W1y  86.672

θ  127.136  deg

 W1x2  W1y2 , w  108.720  

Use equations 5.28 to solve for u, , 2, and 3. Since the points A and B are to be used as pivots, s and  are known from the calculations above. Guess:

U1x  30

U1y  100

γ  80 deg

Given

U1x cos γ  1  U1y sin γ  = p 21  cos δ  S 1x cos α  1  S 1y sin α

       

   

 

       

   

 

       

   

 

       

   

 

U1x cos γ  1  U1y sin γ  = p 31  cos δ  S 1x cos α  1  S 1y sin α U1y cos γ  1  U1x sin γ  = p 21  sin δ  S 1y cos α  1  S 1x sin α U1y cos γ  1  U1x sin γ  = p 31  sin δ  S 1y cos α  1  S 1x sin α

γ  160  deg



 U1x   U1y     Find  U1x U1y γ γ  γ   γ  γ  76.700 deg

γ  161.878  deg

The components of the U vector are: U1x  33.074

U1y  110.894

The length of link 4 is: u  6.

Link 1:

V1x  52.000

V1y  Z1y  S 1y

V1y  0.000

θ  atan2 V1x V1y

θ  0.000  deg

2

2

V1x  V1y

v  52.000

G1x  W1x  V1x  U1x

G1x  19.438

G1y  W1y  V1y  U1y

G1y  24.222

θ  atan2 G1x G1y

θ  51.253 deg

g 

9.

 U1x2  U1y2 , u  115.721  

V1x  Z1x  S 1x

v 

8.

σ  106.607  deg


7.


2

2

G1x  G1y

g  31.057


θ2i  θ  θ

θ2i  178.389  deg

θ2f  θ2i  β

θ2f  12.990 deg


δp  ϕ  θ

rp  46.239

δp  111.571  deg


O2x  82.636

O2y  z sin ϕ  w sin θ

O2y  129.672

O4x  s cos ψ  u  cos σ

O4x  102.074

O4y  s sin ψ  u  sin σ

O4y  153.894

10. Determine the rotation angle of the fourbar frame with respect to the global frame (angle from the global X axis



to the line O2O4.


θrot  51.253 deg

11. Determine the Grashof condition.




w  108.720

θ  127.136  deg

Link 3:

v  52.000

θ  0.000  deg

Link 4:

u  115.721

σ  106.607  deg

Link 1:

g  31.057

θ  51.253 deg

Coupler:

rp  46.239

δp  111.571  deg

Crank angles:

θ2i  178.389  deg

Y

θ2f  12.990 deg P1


X Z1 A1

P2

S1 B1 V1

A2

U1

14. A driver dyad with a crank should be added to link 2 to control the motion of the fourbar so that it cannot move beyond positions 1 and 3.

Z2

S2

V2

B2

W2

W1

U2

O2 O4

W3

P3 A3

U3

Z3

V3

S3 B3




Given:

Solution: 1.

2.

Design a fourbar linkage to carry the box in Figure P5-4 through the three positions shown in their numbered order using the fixed pivots shown. Determine the range of the transmission angle. P21x  184.0

P21y  17.0

P31x  211.0

P31y  180.0

O2x  86.0

O2y  132.0

O4x  104.0

O4y  155.0

Body angles:

θP1  90 deg

θP2  45 deg

θP3  0  deg



α  45.000 deg

α  θP3  θP1

α  90.000 deg


3.

4.

R1x  86.000

R1y  O2y

R2x  R1x  P21x

R2x  98.000

R2y  R1y  P21y

R2y  115.000

R3x  R1x  P31x

R3x  125.000

R3y  R1y  P31y

R3y  48.000

2

2

R1  157.544

2

2

R2  151.093

2

2

R3  133.899

R1 

R1x  R1y

R2 

R2x  R2y

R3 

R3x  R3y

R1y  132.000


ζ  123.085  deg

ζ  atan2 R2x R2y

ζ  49.563 deg

ζ  atan2 R3x R3y

ζ  21.007 deg


    C2  R3 sin α  ζ  R2 sin α  ζ C1  R3 cos α  ζ  R2 cos α  ζ







C3  7.000

 

C4  134.000

 

C5  65.473

 

C6  39.149

C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C2  24.329

 

C3  R1 cos α  ζ  R3 cos ζ



C1  60.553

C6  R1 sin α  ζ  R2 sin ζ



A1  C3  C4

2

A1  1.800  10

4

A2  C3 C6  C4 C5

A2  9.047  10

3

A3  C4 C6  C3 C5

A3  4.788  10

3

A4  C2 C3  C1 C4

A4  7.944  10

A5  C4 C5  C3 C6

A5  9.047  10

A6  C1 C3  C2 C4

A6  3.684  10

3

K1  A2  A4  A3  A6

K1  5.423  10

7

K2  A3  A4  A5  A6

K2  7.136  10

7

2

K3 

3 3

2

2

2

A1  A2  A3  A4  A6

2 7

K3  7.136  10

2

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  90.000 deg

 K  K 2  K 2  K 2  2 1 2 3  β  2  atan  K1  K3  

β  164.466  deg


β  β

 A5  sin β  A3  cos β  A6   A1  

β  85.240 deg

 A3  sin β  A2  cos β  A4   A1  

β  85.240 deg



Since 2 is not in the first quadrant, 5.

β  β


R1x  104.000

R1y  O4y

R2x  R1x  P21x

R2x  80.000

R2y  R1y  P21y

R2y  138.000

R3x  R1x  P31x

R3x  107.000

R3y  R1y  P31y

R3y  25.000

2

2

R1  186.657

2

2

R2  159.512

R1 

R1x  R1y

R2 

R2x  R2y

R1y  155.000


R3  6.

7.


2

2

R3x  R3y

R3  109.882


ζ  123.860  deg

ζ  atan2 R2x R2y

ζ  59.899 deg

ζ  atan2 R3x R3y

ζ  13.151 deg






















 

C3  48.000

 

C4  129.000

 

C5  43.938

 

C6  45.141






C4  R1 sin α  ζ  R3 sin ζ





C5  R1 cos α  ζ  R2 cos ζ





C6  R1 sin α  ζ  R2 sin ζ 2

C2  13.338

A1  C3  C4

A1  1.894  10

4

A2  C3 C6  C4 C5

A2  7.835  10

3

A3  C4 C6  C3 C5

A3  3.714  10

3

A4  C2 C3  C1 C4

A4  9.682  10

A5  C4 C5  C3 C6

A5  7.835  10

A6  C1 C3  C2 C4

A6  5.561  10

3

K1  A2  A4  A3  A6

K1  5.520  10

7

K2  A3  A4  A5  A6

K2  7.953  10

7

2

K3 

2

C1  80.017

3 3

2

2

2

A1  A2  A3  A4  A6

2

2

 K  K 2  K 2  K 2  2 1 2 3  γ  2  atan  K1  K3    K  K 2  K 2  K 2  2 1 2 3    2  atan  K1  K3   The first value is the same as 3, so use the second value

7

K3  7.953  10

γ  90.000 deg

  159.525  deg γ  



 A5  sin γ  A3  cos γ  A6   A1  

  75.253 deg

 A3  sin γ  A2  cos γ  A4   A1  

  75.253 deg

  acos

  asin


γ  


p 21 

2

P21x  P21y

p 21  184.784

δ  atan2 P21x P21y 2

p 31 

δ  5.279  deg

2

P31x  P31y

p 31  277.346

δ  atan2 P31x P31y 9.

δ  40.467 deg


 

B  sin β

 

 

E  p 21  cos δ

 

H  cos α  1

A  cos β  1

 

D  sin α

 

F  cos β  1

 

G  sin β

 

 

K  sin α

 

L  p 31  cos δ

 A F AA   B G 

 

C  cos α  1

 

M  p 21  sin δ

B C D 

 E  L CC    M  N   

 G H K  A D C   F K H 

N  p 31  sin δ

 W1x   W1y   AA  1 CC  Z1x   Z1y   

10. The components of the W and Z vectors are: W1x  62.394


w 

W1y  91.663 2

Z1x  23.606

2

W1x  W1y

Z1y  40.337

w  110.884


 

A'  cos γ  1

 

B'  sin γ

 

E  p 21  cos δ

 

H  cos α  1

D  sin α

G'  sin γ

 

 

 

C  cos α  1

 

F'  cos γ  1

 

K  sin α



 

 

L  p 31  cos δ

 A' F' AA    B'  G' 

 

M  p 21  sin δ

B' C D 

N  p 31  sin δ

 E  L CC    M  N   

 G' H K  A' D C   F' K H 

 U1x   U1y   AA  1 CC  S1x   S1y   

13. The components of the U and S vectors are: U1x  29.920


U1y  116.933 2

u 

U1x  U1y

2

S1x  74.080

S1y  38.067

u  120.700


V1x  50.474

V1y  Z1y  S1y

V1y  2.270 v 


2

2

V1x  V1y

v  50.525

G1x  W1x  V1x  U1x

G1x  18.000

G1y  W1y  V1y  U1y

G1y  23.000

g 


2

G1x  G1y

2

g  29.206


O2x  86.000

O2y  Z1y  W1y

O2y  132.000

O4x  S1x  U1x

O4x  104.000

O4y  S1y  U1y

O4y  155.000


2

z  46.736

2

2

s  83.288


z 

Z1x  Z1y

Angle BAP (p)

s 

S1x  S1y


ψ  152.803  deg

ϕ  atan2( Z1x Z1y )

ϕ  120.337  deg

rP  z

θ  atan2 z cos ϕ  s cos ψ z sin ϕ  s sin ψ  θ  2.575  deg



δp  ϕ  θ

δp  117.762  deg


g  29.206

Link 2:

w  110.884

Link 3:

v  50.525

Link 4:

u  120.700

Coupler point:

rP  46.736

δp  117.762  deg





Design a fourbar linkage to carry the box in Figure P5-5 through the three positions shown in their numbered order without regard for the fixed pivots shown. Use any points on the object as attachment points. Determine the range of the transmission angle. The fixed pivots should be on the base.

Given:

Coordinates of the points P1 , P2 and P3 with respect to P1: P1x  0.0

P1y  0.0

P2x  421.0

P3x  184.0

P3y �

Design of Machinery Solutions Manual Norton 5th Edition (1).pdf

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