A. CASES OF OMITTED MEASUREMENTS INVOLVING ADJOINING SIDES
1. Length of One Side and Bearing of Another Side Unknown 2. Lengths of Two Sides Unknown 3. Bearing of two Sides Unknown
B.
ILLUSTRATIVE PROBLEMS
1. Omitted Measurements 2. Omitted Measurements 3. Omitted Measurements
Lesson 41
41-1. CASES OF OMITTED MEASUREMENTS INVOLVING ADJOINING SIDES
The following are the usual cases of omitted quantities involving adjoining sides: 1ST CASE: Length 2ND CASE: 3RD CASE:
of one side and Bearing of another side Unknown Length of Two Sides Unknown Bearing of Two Sides Unknown
For the traverse shown in Figure 41-1, assume that the omitted or missing data occur in DE and EA which are adjacent course. co urse. The conventional approach to this th is problem is to first connect station A and D by a straight line thus making the length and bearing of AD, in the traverse ABCDA.
Figure 41-1 Traverse with omitted measurements.
As the initially unknown quantities, the unknown data of line AD are determined d etermined by described earlier in Article 40-2, and then the triangle ADE is solved for the missing angles and /or len gths of courses. The solutions usually require only basic participation of geometry and trigonometry
1. Length of one Side and Bearing of another Side Unknown. Figure 41-2 shows a closed traverse for which the length of side CD and the bearing of side DE are Unknown. The
Lengths and Bearing of the other othe r side’s AB, BC, and EA are known. kn own. Since the Latitudes and Departures of the known sides may be calculated, the length and Bearing of the 2. Closing line EC may in turn determined.
Fig. 41-2
Redrawn in Figure 41-3 is triangle CDE which contains the rising quantities. By simple Geometry, the angle at station C can be computed from the known bearings of the sides of the triangle are already known, the sine law may be used to find the angle at D. Thus
41-3 triangle CDE
Find the above Equation the angle at D can be determined, Thus Giving two of the three angles of triangle CDE. Angle E equals 180 Dergrees – (angle C = angle D) and the Length of Side CD Can be determined by sine law or
CD =
CE and CD = CE sin E
Sin E
Sin D
Sin D
For this particular case, however, if the angle between the side of unknown bearing and the side of unknown length is close to 90 degrees, the solution may be very weak and will yield unsatisfactory answer. This is because the value of the sine near 90 degrees change s very slowly and a small variation in the calculated values can cause a relatively large error in the angle calculated by the sine law.
Fig.41-4
2.) Length of Two Sides Unknown .In the figure 41-4 the lengths of side BC and CD are the unknown quantities. Since the latitudes and departures of sides AB , DE , EF , and FA can be computed and summed, the length the bearing of the closing line BD may in turn be determined by the sine law as follows.
FiG.41-5 By Sine Law: BC = CD = BD SinD SinB SinC BC = BD SinD SinC CD = BD SinB Sinc
3.) Bearings of Two Sides Unknown. In this particular rase it is assumed that the bearings of sides DE and EA (pig. 41-8) are the unknown quantities. Since the lengths.
Fig. 41-6
And the bearings of sides AB, BC, and CD are known, their respective latitudes and departments can be computed and; and from the results the length and bearing of the line (AD) is determined. Then the lengths of all sides of triangle ADE (Fig.41-7) are known as well as the bearing of side AD.
Triangle ADE:
By Cosine Law:
2 Let: a = DE, b = AD, c = EA, and Angle C = Angle A
() () () 2( )()
The angle at A may also be computed by the following equation.
Sin =
(−)(−) ()()
Where: S = (1/2) (DE + AD + EA) After the angle at A is determined the sine law may to applied to obtain the other missing angles in triangle ADE. Thus
and
With the above equations the remaining two angles of triangle ADE can be computed and the unknown bearings of sides DE and EA correspondingly determined, As a check , the sum if the three computed angles should equal to 180 degrees.
41- 2 ILLUSTRATIVE PROBLEMS I. OMITTED MEASUREENTS. Given the following sketch and tabulated data for closed traverse in which the length of DE and the bearing of AE have not been observed in the field. Determine these unknown quantities.
Fig. 41- 8
LINE
LENGTH
BEARING
LATITUDE +N
S 75˚48’ E S 15˚ 18’ W S 83˚06’ W
AB 1,094.32 m BC 1,590.51 m CD 1,294.74 m DE Unknown EA 1,739.92 m SUMS SOLUTION:
N28˚39’ W Unknown
----+0.00
DEPARTURE +E 1,051.19
-S 265.99 1,534.14 482.92
-W 419.89 1,201.31 ---
-2,263.05
--+1,051.19
-1,621.00
Determining length and Bearing of closing Line.
= ∑+∑= +.+(−,. )=−,. = ∑+∑= +,.+(−,. )= −.
= √ √ (2,283.08) (569.81) = 2,353.08 (Length of closing line DA) (569.81) 569.81 0.249583 œ (2,283.05) 2,283.05 œ 14˚01′ (Therefore the bearing of the closing line DA N 14 01’ E) ˚
Determining Bearing of Line DE and length of Line EA. Fig.41-10 Angle D = 28˚39’ + 14˚01’ = 42˚40’ By sine law: DA Sin E
= EA Sin D
Sin E = (DA) Sin D (EA) Sin E = (2,353.08)Sin 42˚40’ 1,738.96 Sin E = 0.91076
;
Angle E = 66˚30’
Since the sum of interior angles of ADE should equal to 180˚ Angle A = 180˚ - {Angle D + Angle E} = 180˚ - {42˚40’ + 86˚30’} Angle A = 70˚50’ About station E
Ǿ = 180˚ - {Angle E + 28˚39’} By Sine Law:
;
()
70˚50’ (1,738.96) 42˚ 40’ 2,432.62 (Length of line De) 2. OMITTED MEASUREMENTS. Given
the following sketch and tabulated data for closed Iraverse in which the length of BC and CD have not been measured in the field. Determine these unknown quantities. Fig.41-10
LINE AB BC CD DE EA SUMS
LENGTH 639.32 m Unknown Unknown 570.53 m 1,082.71
BEARING N 09˚30’ W N 56˚ 55’ W S 56˚13’ W S 02˚02’ E S29˚31’ E
LATITUDE +N 630.55 ---
+630.55
-S
DEPARTURE +E
--570.17 9.13 -579.30
20.24 1,092.67 +1,102.91
SOLUTION: Determining length and Bearing of closing Line.
= ∑+∑= +.+(−. ) = . = ∑+∑= +,.+(−. )= −.
= 998.71
= √ √ (51.25) (∓997.39) (Length of closing line BD)
− −(+. ) −. − −(+.) −. 19.461268
œ
œ 87˚04′
(Therefore the bearing of the closing line DA N
Determining Length of lines BC and CD.
Fig.41-11
87˚04′E)
-W 105.52 -----
-105.52
Angle B = 180˚ - (87˚04’ + 56˚55’)
= 36˚01’
Angle C = 56˚13’ + 56˚55’
= 113˚18’
Angle D = 87˚04’ - 56˚13’
= 30˚51’
SUMS
= 180˚00’ (Check!)
By Sine Law:
30˚50′ 36˚01′ 113˚08′ 30˚51′ 998.71113˚08′ 556.91 (Length of Line BC) 36˚01′ 998.71113˚08′ 638.61 (Length of Line CD)
I. OMITTED MEASUREMENTS. Given the following sketch and tabulated data for a closed iraverse in which the bearings of DE and EA have been measured in the field determine these unknown directions.
Fig.41-12
LINE AB BC CD DE EA SUMS
LENGTH 1,081.35 m 1,589.50 m 1,293.72 m 2,506.94 m 1,737.89 m
BEARING S 73˚47’ E S 13˚14’ W S 67˚07’ W Unknown Unknown
LATITUDE +N
----+0.00
-S 301.99 1,533.65 503.07
-2,338.71
DEPARTURE +E 1,030.99
-W 417.64 1,191.90 ---
--+1,038.33
-1,609.54
SOLUTION:
= ∑+∑= +.+(−,. )= −,. = ∑+∑= +,. +(−,. )= −. = √ √ (2,338.71) (571.21) = 2,407.46 (Length of closing line DA)
(2,338.71 ) 2,338.71 œ (571.21 ) 571.21 0.244241 œ 13˚44′ (Therefore the bearing of the closing line DA N 13˚44 E) Determining bearings of lines DE and EA.
By Cosine Law:
2 Let: a = EA, b = DE, c = DA, and Angle C = Angle C Therefore:
() () () 2()() (2,506.94) (2,407.46) (1, 7 37. 9 8) 2(1,737.98)(2,506.94) Angle E = 0.402737 Angle E = 66˚15 By Sine Law:
1,737.98 2,506.94 2,407.46 66˚15′ 1,737.2,98(Sin66˚15′) 407.46 0.660777 41˚22′ 2,506.2,94(Sin66˚15′) 407.46 0.953134 72˚23′
Solution Check: Angle A + Angle D + Angle E = 180˚00’ 72˚23’ + 41˚22’ + 56˚15’ = 180˚00’ 180˚00’ = 180˚00’ (check)
About Station D:
Ǿ = Angle D - 13˚44’ = 41˚22’ -13˚44’ Ǿ = 24˚38’ About Station E:
Ǿ = 180˚00’ – Angle E - Ǿ = 180˚00’ - 66˚15’ - 27˚38’ Ǿ = 86˚07’ (Therefore the Bearing of Line EA is N 86˚07’)
C. CASES OF OMITTED MEASUREMENTS INVOLVING ADJOINING SIDES
a. Length of One Side and Bearing of Another Side Omitted b. Lengths of Two Sides Unknown c. Bearing of two Sides Unknown
D.
ILLUSTRATIVE PROBLEMS
a. Omitted Measurements (non-Adjoining Sides)
Lesson 41
LESSON 42
41-1 CASES OF OMMITED MESUREMENTS INVOLVING NON-ADJOINING SIDES
In the previous lessons it was pointed out tha t the solutions of the different cases of omitted measurements may be classified according to whether the two unknown or omitted quantities (length and/or bearings) are in the same side , in any two adjoining sides , or in two non-adjoining sides. This lesson will cover solution to the following typical cases involving omitted quantities in two non-adjoining sides. 1ST CASE: Length of One Side and Bearing of another Side Unknown 2ND CASE: Length of Two Sides Unknown 3RD CASE: Bearings of Two Sides Unknown With an added graphical solution , the determination of missing data as explained in the previous two lessons are also applicable even though the sides the sides with two unknown quantities are non-adjoining . For the different cases of o mitted measurement involving non-adjoining sides the following principles are adapted: 1ST PRINCIPLE: A line may be moved from one location to a second location parallel with the first, and its latitude and departure will remain un -changed. 2ND PRINCIPLE: The algebraic sum of the Latitude and the Algebraic sum of the Departures of any system of lines forming a closed figure must be zero, regardless of the order in which the lines are placed.
A solution by simultaneous equations will give the missing parts. In practice, however , this method is seldom applied since geometric solutions are preferred b y most engineers with unknown quantities are not adjoining , one of the sides is considered as moved from its location to a second location parallel with the first , such that the two unknown sides adjoin .The solution then becomes identical with those described for adjoining sides. Problems of this t ype are simplified by first
plotting the given traverse data to scale, then creating a “transformed” figure prior to undertaking
the actual computations .Aside from the cases me ntioned above there are some special cases of omitted measurements which can be solved b y simple geometrical analysis. Other complicated cases of omitted measurements which will involve the use of simultaneous equation are not discussed here As a check on the calculations made on earth of the three cases, the latitudes and departures of the sides with unknown quantities must be computed. The algebraic sum of the latitudes and the algebraic sum of the departures for all sides should bo th be zero if the computation made for determining the unknown quantities are correct.
1. ) Length of One Sided and Bearing of another Side omitted. In figure 42-1 are given the directions and lengths of sides AB, CD, EF, and FA; length of side BC; and direction of side DE. The unknown quantities are the bearing of side BC, and length of side DE which are nonadjoining. The traverses first plotted to scale based on given data; with the sides shown in the order a, b, c, d, e, and f. The sides with unknown quantities are shown is dashed lines and sides with known quantities in solid lines
Fig.42-1
By rearranging the order of the sides into c, e, f, a, b, and d the figure is transformed as shown in figure 42-2. A closing line divides the figure into two parts and its length and bearing are determined from the algebraic sum of the departures and latitudes of the known sides. The first part is Formed by the known sides (a, c, e, and f) and the closing line; the second part by the unknown sides (b and d and the closing line.)
Simple geometrical constructions are involved in plotting out the transformed figure. In the process of rearranging the sides to arrive at the desire figure all the sides are drawn parallel to the corresponding sides of the previously plotted traverse.
In this example the locations of sides C and D have been interchanged. The transformed figure now shows that the sides with unknown quantities are made adjoining.
The problem of determining the unknown quantities is now seem to be identical with that of a cases whose sides are adjoining as explained in lesson number 41 2.) Lengths of Two Sides Omitted. Given in figure 42-3 are the lengths and directions of sides AB, CD, DE, and direction of the non-adjoining sides BC and EA are he unknown quantities.
As in the previous case, the traverse is first plotted to scale b ased on the given data; with the sides shown in the order a, b, c, d, and e. The transformed figure (Fig., 42-1) is formed by rearranging the order of the sides into c, d, a, e, and b.
The closing line divides the figure into two p arts. The first part id formed by the known sides (c, d, and a) and the closing line, and the second part by the unknown sides (b and a) and the closing line.
The transformed figure now shows that the sides with unknown quantities are made adjoining. The problem of determining these unknown quantities is now seen to be identical with that of a case whose sides are adjoining as explained in lesson number 41.
For this particular case, it must be noted that as the angle between the unknown sides approaches
90○, the solution becomes strong, and as the angle approaches 0○ or 180○, the solution becomes weak. The problem becomes indeterminate when the lines are parallel.
3. Bearings of Two Sides Omitted. Given in figure 42-5 are the lengths and directions of sides BC, DE, and EA; and the length of sides AB and CD. The bearing of the non-adjoining sides AB and CD are the unknown quantities.
Following the same procedure outlined in the first two cases, the traverse is plotted to scale based on the given data and with the sides drawn in the order a, b, c, d, and e. The transformed figure is arrived at by rearranging the order of the sides into b, d, e, a, and c as shown in Figure42-6.
Again, the closing line divides the figure into two parts. The first part is formed by the known sides (b, d, and e) and the closing line; the second part by the unknown sides (a and c) and the closing line. Since the transformed figure has the u nknown quantities now adjoining , the problem of determining the unknown quantities is again seem to be identical with one of the cases as explained in lesson number 41.
It will be noted that when the unknown sides are parallel but are not of the same length , their direction is that of the closing line. However, wh en the unknown sides are parallel and are also of the same length the solution for the problem becomes indeterminate since the length o f the closing line become a point.
42-2 ILLUSTRATIVE PROBLEM 1. OMITTED MEASUREMENTS
(Non-Adjoining Sides).Given the following data for a closed
traverse for which the lengths of the non-adjoining sides BC and EF have been measured in the field. Determine the unknown quantities.
LATITUDES COURSE
BEARING
DISTANCE
AB
S 92˚360’ E
620.08
BC
N 30˚30’ E
Unknown
CD
S 42˚45’ E
591.55
507.93
DE
S 60˚40’ W
588.48
337.28
EF
N 74˚05’ W
Unknown
---
FA
N 07˚02’ E
412.35
409.35
SUMS
+N
DEPARTURES -S
+E
27.05
619.47
---
+409.35
-W
--469.43 600.21 --50.49 -872.16
+1,139.39
+600.21
Note: 1. The first step Is to plot the traverse to scale bases on the given data. 2. The next step is to draw an equipment transformed figure such that the sides with unk nown quantities are made adjoining.
SOLUTION: a. Plotting the traverse to scale
Fig 42-7
b. Drawing an equivalent transformed figure. Fig.42-8
c. Determining the Length and the Bearing of the closing line.
∑ NL = +409.25 m
∑ED = +1,139.39 m
∑SL = -872.16 m
∑WD = -600.21 m
= ∑ + ∑ = + . + (−.) = − . (Difference between north and south latitudes) = ∑ + ∑ = +,. +(−. ) = +. (Difference between east and west departures) d.Determining Lengths of sides BC and EF. Fig. 42-9
Angle 1 = 30 ˚30’ + 49˚21’ = 79˚51’ Angle 2 = 74 ˚05’ + 49˚21’ = 24˚44’ Angle 3 = 180 ˚00’ + (30˚30’ + 74˚05’) = 75˚25’
Solution check: Angle 1 + Angle 2 + Angle 3 = 180 ˚00’ 79˚51 + 24˚44’ + 75˚23’ = 180˚00’ 180˚00’ = 180˚00’ (check)
By Sine Law:
. ( . ) ˚ ( . ) ˚
307.22 (Length of side BC)
( . ) ˚ ( . ) ˚
722.90 (Length of side EF)
