Simple Curve

NATIONAL UNIVERSITY College of Engineering Department of Civil Engineering

Engineering surveys Field work (fsrvy2l)

Plate no. 1 Lay out simple curve by Tangent and offset (Title)

GUERZON, GREGORIO JR. R. Submitted by:

ENGR. JOSELITO BATAC Submitted to:

I.

OBJECTIVES

 To gain familiarity with the theory, design and layout of Horizontal curves for most types of transportation routes, such as highways, railroads, pipelines, etc.,  To be able to lay a simple curve by using the tape alone.

II. EQUIPMENTS/MATERIALS Surveying pin Steel tape Chalk Nylon string

III. PROCEDURE 1. With the instrument placed at the PI, the instrument man sights on the preceding PI or at a distant station and keeps the chainman on the line while the tangent distance is measured to locate the PC. After the PC has been staked out, the instrument man then trains the instrument on the forward PI to locate the PT. 2. String line PC-PI or the tangent line to have a better reference. 3. Then, compute the value of X and Y by a given formula. 4. After that, layout the X’s (along tangent line) and Y’s (perpendicular to the tangent line) to locate each station.

*note: X’s are the distance form PC along tangent line and Y’s are the offset to determine the location of each stations. *note: To have a perpendicular distance must have two equidistance measurements from each side of point X (along tangent line). Then, swing an arc with equidistance measurement form the two points. Connect point X to the intersection of the arcs.

IV. DIAGRAM

V.

DATA AND COMPUTATION FIRST CURVE (R=50,I=20)` STATION

I

X =RsinI

Y =R−RcosI

∆1/8

2o 30l

2.181

0.048

∆1/4

5o

4.361

0.190

∆3/8

7 30

6.53

0.428

Mc

10o

8.68

0.76

∆5/8

12 30

10.82

1.185

∆3/4

15o

12.94

1.704

∆7/8

17o 30l

15.06

2.314

PT

20o

17.101

3.015

o

o

l

l

X ∆ 1/ 8=RsinI=50 sin 2o 30l=2.181 Y ∆ 1 /8=R−RcosI =50−50 cos 2o 30 l=0.048 X ∆ 1/ 4=RsinI =50sin 5 o=4.361 Y ∆ 1 / 4=R−RcosI =50−50 cos 5 o=0.190 X ∆ 3/ 8=RsinI=50 sin 7o 30l =6.53Y ∆ 3/ 8=R−RcosI =50−50 cos 7 o 30l=0.428 X Mc =RsinI=50 sin 10o=8.68Y Mc =R−RcosI =50−50 cos 10 o=0.76

X ∆ 5/ 8=RsinI=50 sin 12o 30l=10.82 Y ∆ 5 /8=R−RcosI =50−50 cos 12o 30l =1.185 X ∆ 3/ 4=RsinI =50 sin15 o=12.94 Y ∆ 3/ 4=R−RcosI =50−50 cos 15o=1.704 X ∆ 7/ 8=RsinI=50 sin 17 o 30l =15.06Y ∆ 7 /8=R−RcosI =50−50 cos 17o 30l=2.314 X PT =RsinI =50 sin20 o=17.101 Y PT =R−RcosI =50−50 cos 20o =3.015

VI. PICTURES

VII. OBSERVATION I observe that laying out the curve by tangent and offset is not that hard but because of our schedule that we have In our field work which is at night time and also there is a limited space in our school that makes us struggle in some aspect of laying it out. But fortunately we manage to layout it. Some errors may occur if we misread the steel tape and in rounding off the computed value of X and Y.

VIII. CONCLUSION I therefore conclude that we can layout the curve just using steel tape. And only by computing the the offset in certain point in the tangent line. The tangent distance is the distance the tangents from the PI to the PC. The value of Y increases as the value of X also increases. And also be careful on the angle that you are going to use because it has a lot of bearing in computing the tangent offset.



Plate no. 2A Lay out simple curve by chord And mid-ordinate (Title)



I.

OBJECTIVES  To gain familiarity with the theory, design and layout of Horizontal curves for most types of transportation routes, such as highways, railroads, pipelines, etc.,  To be able to lay a simple curve by using the tape alone.  To be able to lay out a simple curve chord and mid-ordinate


III. PROCEDURE 1. With the instrument placed at the PI, the instrument man sights on the preceding PI or at a distant station and keeps the chainman on the line while the tangent distance is measured to locate the PC. After the PC has been staked out, the instrument man then trains the instrument on the forward PI to locate the PT. 2. Compute and string line the chord PC-PT then compute the mid ordinate and plot it perpendicular to the chord to locate the station Mc. The middle ordinate is located from the middle of the chord 3. String line PC-Mc to determine its chord length. Then compute the mid ordinate plot it perpendicular to the chord to locate station ∆1/4. Again, the middle ordinate is located in the center of PC-Mc. 4. String line PC-∆1/4 to determine its chord length. Compute the mid ordinate plot it perpendicular to the chord to locate station ∆1/8. 5. String line ∆1/4-Mc to determine its chord length. Compute the mid ordinate plot it perpendicular to the chord to locate station ∆3/8.

6. String line PT-Mc to determine its chord length. Compute the mid ordinate plot it perpendicular to the chord to locate station ∆3/4. 7. String line PT-∆3/4 to determine its chord length. Compute the mid ordinate plot it perpendicular to the chord to locate station ∆7/8. 8. String line ∆3/4-Mc to determine its chord length. Compute the mid ordinate plot it perpendicular to the chord to locate station ∆5/8.

*note: To have a perpendicular distance from chord must have two equidistance measurements in each side of the middle of the chord (along the chord). Then, swing an arc with equidistance measurement from the two points. Connect the middle of the chord to the intersections of two points.

IV. DIAGRAM

V.

DATA AND COMPUTATION FIRST CURVE (R=50,I=20) CHORD MIDDLE ORDINATE=R(1-cosI/2) 17.365m 0.76m

Station P.C.-P.T.

I 20o

P.C.-Mc

10

8.85m

0.19m

P.C.-∆1/4

5o

4.47m

0.048m

P.T.-Mc

10

9.02m

0.19m

P.T-∆3/4

5o

4.49m

0.048m

∆3/4-Mc

5o

4.47m

0.048m

∆1/4-Mc

5o

4.48m

0.048m

o

o

( ))

(

( ))

Mo PC−Mc =R 1−cos

( ( )) ( ( )) ( ( )) ( ( )) ( ( )) ( ( )) ( ( ))

I 20 o =50− 1−cos =0.76 2 2

(

Mo PC−PT =R 1−cos

o

I 10 =50− 1−cos 2 2

I 5o =50− 1−cos =0.048 2 2

(

( ))

(

I 10o =50− 1−cos =0.19 2 2

Mo PC−∆ 1 /4 =R 1−cos

Mo PT− Mc =R 1−cos

( ))

(

( ))

o

I 5 Mo PT−∆ 3 / 4=R 1−cos =50− 1−cos 2 2

( ))

(

( ))

Mo ∆ 3/ 4− Mc =R 1−cos

=0.048

I 5o =50− 1−cos =0.048 2 2

(

Mo ∆ 1/ 4− Mc =R 1−cos

VI. PICTURES

=0.19

o

I 5 =50− 1−cos 2 2

=0.048

VII. OBSERVATION I observe in this field work that some errors may occur if angle that being used is wrong. In laying out, if the mid ordinate is not perpendicular to the chord have a huge effect in the curve.

VIII. CONCLUSION I therefore conclude that the curve can be lay out using chord and mid ordinate. Also that as the angle become smaller, the value of mid ordinate decreases. The curve is symmetric thats why the procedure in the left side of the Mo should be the same in the right side.



Plate no. 3 Lay out simple curve by

Intersecting chords (Title)



I.

OBJECTIVE

 To gain familiarity with the theory, design and layout of Horizontal curves for most types of transportation routes, such as highways, railroads, pipelines, etc.,  To be able to lay a simple curve by using the tape alone.  To be able to lay out a simple curve by intersecting chords


III. PROCEDURE

1. With the instrument placed at the PI, the instrument man sights on the preceding PI or at a distant station and keeps the chainman on the line while the tangent distance is measured to locate the PC. After the PC has been staked out, the instrument man then trains the instrument on the forward PI to locate the PT. 2. Compute the chord PC-∆7/8 and PT-∆7/8 and swing an arc to determine their intersection. Now you have the station ∆7/8. 3. Next is to compute the chord PC-∆3/4 and PT-∆3/4. Swing an arc with the computed measures; the intersection is the station ∆3/4. 4. Then compute the chord PC-∆5/8 and PT-∆5/8. Swing an arc with the computed measures; the intersection is the station ∆5/8. 5. Compute the chord PC-Mc and PT-Mc. Swing an arc with the computed measures; the intersection is the station Mc. 6. Compute the chord PC-∆3/8 and PT-∆3/8. Swing an arc with the computed measures; the intersection is the station ∆3/8. 7. Compute the chord PC-∆1/4 and PT-∆1/4. Swing an arc with the computed measures; the intersection is the station ∆1/4. 8. Lastly, compute the chord PC-∆1/8 and PT-∆1/8. Swing an arc with the computed measures; the intersection is the station ∆1/8.

IV. DIAGRAM

V.

DATA AND COMPUTATION FIRST CURVE (R=50,I=20)

Station

I

PC-PT

20

17.365

PC-∆7/8

17o 30l

15.21

PC-∆3/4

15 o

13.05

PC-∆5/8

12 30

PC-Mc

10o

PC-∆3/8

7 30

PC-∆1/4

5o

4.36

PC-∆1/8

2o 30l

2.18

PT-∆1/8

17o 30l

15.21

PT-∆1/4

15 o

13.05

PT-∆3/8

12 30

PT-Mc

10o

PT-∆5/8

7 30

PT-∆3/4

5o

4.36

PT-∆7/8

2o 30l

2.18

C PC−PT =2 Rsin

C=2 Rsin o

o

o

l

6.54

l

l

l

( ) I 15 =2 Rsin ( )=2 ( 50 ) sin ( =13.05 2 2 ) I 12 30 =2 Rsin ( )=2 (50 ) sin ( =10.89 2 2 ) I 10 =2 Rsin ( )=2 ( 50 ) sin ( =8.72 2 2 )

I 17 30 C PC−∆ 7 /8=C PT−∆ 1 /8=2 Rsin =2 (50 ) sin =15.21 2 2

()

o

C PC−∆ 3 /4 =C PT−∆ 1 / 4

o

C PC−∆ 5 /8=C PT −∆ 3 /8

o

C PC−Mc =C PT −Mc

l

10.89 8.72

( 2I )=2 ( 50) sin ( 202 )=17.365 o

10.89 8.72

o

o

l

6.54

( 2I )

I 7 o 30 l C PC−∆ 3 /8=C PT −∆ 5 /8=2 Rsin =2 (50 ) sin =6.54 2 2

( ) I 5 =2 Rsin ( )=2 ( 50 ) sin ( )=4.36 2 2 I 2 30 =2 Rsin ( )=2 (50 ) sin ( =2.18 2 2 ) ()

o

C PC−∆ 1 /4 =C PT−∆ 3 / 4

o

C PC−∆ 1 /8=C PT −∆ 7 /8

l

VI. PICTURES

VII. OBSERVATION I observe that laying out curve using intersecting chords is one of the hardest because we need to swing an arc with the chord that computed with the largest amount of 15.21. An error may occur if the swing is incorrect or the measurement is wrong.

VIII. CONCLUSION I therefore conclude that the intersection of two chords is the location of the specific station. We just need to compute the value of each chord of each station.



Plate no. 5 Lay out simple curve by Chord and offset (Title)



I.

OBJECTIVE  To gain familiarity with the theory, design and layout of Horizontal curves for most types of transportation routes, such as highways, railroads, pipelines, etc.,  To be able to lay a simple curve by using the tape alone.  To be able to lay out a simple curve by chord and offset

II. EQUIPMENTS/MATREIALS Surveying pin Steel tape Chalk Nylon string

III. PROCEDURE 1. With the instrument placed at the PI, the instrument man sights on the preceding PI or at a distant station and keeps the chainman on the line while the tangent distance is measured to locate the PC. After the PC has been staked out, the instrument man then trains the instrument on the forward PI to locate the PT. 2. String line PC-PT to have better reference. 3. Compute the mid ordinate of chord PC-PT. That is the offset of station Mc from the chord. 4. Compute the horizontal distance of each station from Mo (X). 5. Then compute the offset (Y) of each station from the chord by the given formula. 6. Layout the computed Y’s perpendicular to the X’s. *note: To have a perpendicular distance from chord must have two equidistance measurements from point X (along the chord). Then, swing an arc with another equidistance measurement from the two points. Connect point X to the intersection of the two arcs.

IV. DIAGRAM

V.

DATA AND RESULT Station

X(distance from Mo)

Y =Mo + √ R 2−X 2 −R

∆1/8 ∆1/4 ∆3/8 Mc ∆5/8 ∆3/4 ∆7/8

6.51 4.34 2.17 2.17 4.34 6.51

0.33 0.57 0.71 0.76 0.71 0.57 0.33

Distance of PC-Mo and PT-Mo= 8.68 X ∆ 1/ 8=X ∆ 7/ 8=8.68

( 34 )=6.51

X ∆ 1/ 4= X ∆ 3 /4 =8.68

( 12 )=4.34

X ∆ 3/ 8=X ∆ 5/ 8=8.68

( 14 )=2.17

(

Mo=R 1−cos

(

( ))

I 20 o =50− 1−cos =0.76 2 2

( ))

Y ∆ 1 /8=Y ∆ 7 /8=Mo + √ R2−X 2−R=0.76 + √50 2−6.512−50=0.33 Y ∆ 1 /4 =Y ∆ 3 / 4=Mo + √ R 2−X 2−R=0.76+ √ 502 −4.34−50=0.57 Y ∆ 3 /8 =Y ∆ 5 /8=Mo + √ R2−X 2−R=0.76+ √ 502−2.172−50=0.71

VI. PICTURES

VII. OBSERVATION I observe that this is the easiest way to lay out a curve. Errors may occur only if the offset is not perpendicular to the chord. I hope that this formula that we used is valid.

VIII. CONCLUSION

I manage to take a research in this method on how to compute the offset from the chord. The formula that I got is very useful because all we need to do is to get the perpendicular distance from Mo and substitute it to the formula. This is also the easiest way to lay out the curve. The curve is symmetric that’s why some value are identical.



Plate no. 2B Lay out simple curve By moving chords (Title)



I.

OBJECTIVE  To gain familiarity with the theory, design and layout of Horizontal curves for most types of transportation routes, such as highways, railroads, pipelines, etc.,  To be able to lay a simple curve by using the tape alone.  To be able to lay out a simple curve by intersecting chords


III. PROCEDURE 1 Locate your PI and establish a line using your backward tangent to locate the PC of the curve. 2 From PI, Swing an arc using your forward tangent. 3 From PC, Swing an arc using the chord length. The intersection of the arcs is the PT 4 Locate the midpoint of the long chord. 5 From its midpoint, measure 2 equal distances and swing another 2 equal distances to establish a perpendicular line. 6 Locate the MO of the curve, the endpoint of the MO is the MC. 7 Establish the chord of 1/2Δ that connects PC and MC. 8 From that chord, establish a perpendicular line and locate the MO, the endpoint of that MO is 1/4Δ station. 9 Establish the chord of 1/4Δ that connects PC and 1/4Δ. 10 From that chord, establish a perpendicular line and locate its MO, the endpoint of the MO of that station is your 1/8Δ. 11 To locate your 3/8Δ, get the chord of 3/4Δ, endpoint of the MO of that chord is your 3/8Δ station. 12 To locate your 3/4Δ, get the chord of 1/2Δ that connects PT and MC. The endpoint of MO is your station. 13 To locate your 7/8Δ, get the chord length that connects 3/4Δ and PT. establish a perpendicular line to locate the MO, the endpoint of the MO is your 7/8Δ station.

14 To locate your 5/8Δ, establish a chord that connects PC and 1/4Δ station, establish a perpendicular line then locate your MO, the endpoint of your MO is your 5/8Δ station.

Simple Curve

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