1. Adjustment of Angles
Lecture – 07

2. Adjustment of Angles
After completion of field work of measurements of angles, it is necessary to adjust the angles. Generally the angles of a triangle and chain of triangles are adjusted under two heads.
Station Adjustment
Figure Adjustment

3. 1. Station Adjustment
Sum of the angles about a station should be 360o. If not, find the difference and adjust the difference equally to all the angles algebraically to make their sum equal to 360o.
Suppose; for a station B. 1 2 3 4 5 6 7

4. Station Adjustment Angles Observed Value Correction Corrected Value ∟3
---
-12'' ∟4 ∟5 ∟6
∑ = 360o 00' 48''
∑ = 360o 00' 00''

5. 2. Figure Adjustment:
The determination of most probable values of angles involved in any geometrical figure so as to fulfill the geometrical conditions is called the figure adjustment. All cases of figure adjustment necessarily involve one or more conditional equations. The geometrical figures used in a triangulation system are:
Triangles
Quadrilaterals
Polygons with central stations

6. a. Triangle Adjustment
Triangulation of ordinary precision, the sum of the angles of a triangle is equal to 180o. For large triangles, covering big area, correction for spherical excess is to be applied because sum of angles of a spherical triangle will be more than 180o and the correction to be applied is as an addition for 01'' for every 75 square miles.
For triangle ABC
Angles
Observed Value
Correction
Corrected Value ∟1
---
-4'' ∟5 ∟6
∑ = 180o 00' 12''
∑ = 180o 00' 00''

7. b. Adjustment of Braced Quadrilateral
1. Geometric Condition:
(a) Sum of all the angles should be equal to 360o.
(b) Sum of equal pair of angle should be equal.
∟2 + ∟3 = ∟6 + ∟7
∟1 + ∟8 = ∟4 + ∟3
Suppose L.H.S > R.H.S by 12''
Divide this 12” by 4; correction = 12''/4 = 3''
Add 3'' to the angles of R.H.S and subtract 3'' from the angles of L.H.S.

8. b. Adjustment of Braced Quadrilateral

9. b. Adjustment of Braced Quadrilateral
2. Trigonometric Condition:
log (sin 1) + log (sin 3) + log (sin 5) + log (sin 7) = log (sin 2) + log (sin 4) + log (sin 6) + log (sin 8)
For this adjustment following procedure is adopted.

10. Procedure:
Record ‘log (sinϑ)’ of each angle obtained after geometric adjustment.
For each angle record the ‘log (sinϑ)’ difference for 01'' i.e., the difference between the previous value in step (1) & the value obtained after adding 01'' to actual angle.
Find the average required change (α) in ‘log (sinϑ)’ by dividing the difference of sum of odd and even angle values by 8.

11. Procedure
Find the average difference “β” for difference for 01''. i.e., total ‘log (sin ϑ)’ difference for 01'' of all angles divided by 8.
The ratio α / β gives the no. of seconds to apply a correction.
Add the correction to each of four angles where ‘log (sin ϑ)’ is smaller & subtract the correction from the other four angles.

12. Problems:
Given the observed angles of a braced quadrilateral shown in figure. Adjust it geometrically as well as trigonometrically.
Angle
Observed Value a
38°44'06" b
23°44'38" c
42°19'09" d
44°52'01" e
69°04'21" f
39°37'48" g
26°25'51" h
75°12'14"

13. Log (sinϑ) differnece for 01"
Solution:
Angle
Observed Value
Figure Adjustment
Log (sinϑ)+10
Log [sin(ϑ+01")]+10
Log (sinϑ) differnece for 01"
Corrected Values
Condition 1
Condition 2 a
38°44'06"
 38°44’05"
38° 44`5.5`` b
23°44'38"
23° 44‘37"
23°44`35``       c
42°19'09"
42° 19‘08"
 42°19`06``       d
44°52'01"
44° 52‘00"
 44°51`59.5``     e
69°04'21"
69° 04‘20"
 69°04`19.5`       f
39°37'48"
39° 37‘47"
 39°37`49``     g
26°25'51"
26° 25‘50"
 26°25`52``       h
75°12'14"
75° 12‘13"
 75°12`13.5``  
360°00`08``
360°00`
= //8=6.25×10^-6
2.5*10^-6

14. angle
Observed
Cond 1
Cond 2
Log (sinϑ)+10
Log [sin(ϑ+01")]+10
Corrected Values 1
56°06'33" 2
31°56'58" 3
27°43'52" 4
63°30'12" 5
56°50'50" 6
25°23'45" 7
34°20'45" 8
64°08'54"
360°1'49"

15. Assignment:
Given the observed angles of a braced quadrilateral shown in figure. Adjust it geometrically as well as trigonometrically.
Angle
Observed Value 1
56°06'33" 2
31°56'58" 3
27°43'52" 4
63°30'12" 5
56°50'50" 6
25°23'45" 7
34°20'45" 8
64°08'54"

16. SATELLITE STATION
A satellite station is used when instrument can not be set up at the main station. The distance of the satellite from its station is usually very small as compared to the length of the sides of the triangulation.
In the figure ABC represent part of a triangulation station. Station A could be observed but it was not possible to set up an instrument there. A satellite station ‘S’ was therefore used and angles from S to A, B & C are measured. Suppose distance AS = S'.

18. sinδ1 = S' / l1 x sin θ1----------- (1) For very small value of δ1:
Satellite station
In the triangle ACS:
sinδ1 / S' = sinθ1 / l1
sinδ1 = S' / l1 x sin θ (1)
For very small value of δ1:
sinδ1 = δ1 (in radians)
= (δ1 x 180) / π (in degrees)
= (δ1 x 180 x 60 x 60) / π (seconds)
= x δ1 (seconds)

19. Satellite station
Equation (1) can be written as:
δ1 = (S' x sinθ1) / l (in radians)
δ1 = ( x S' x sinθ1) / l1 (in seconds)
Similarly:
δ2 = ( x S' x sinθ2) / l2 (in seconds)
BSC = θ and BAC =?
Since; α + β + θ = (α + δ1) + (β + δ2) + BAC.
Therefore;
BAC = θ - (δ1 + δ2) [For station inside the triangle]
BAC = θ + (δ1 + δ2) [For station outside the triangle]

20. PROBLEM
A, B & C were stations of a minor triangle ABC; C not being suitable for an instrument. A satellite station ‘S’ was therefore set up outside the triangle ABC in order to determine the angle at C. The distance of ‘S’ from C was m. The length of AC = m and of BC = m. The angle ASC was found to be 63o 48' 00'' and the angle ASB 71o 54' 32''. Calculate angle ACB?

21. Solution: Given Data: S' = 17.00 m Angle ASC = θ2 = 63o 48' 00''
Angle ASB = 71o 54' 32''
θ1 = CSB = ASB – θ2
= 71o 54' 32'' – 63o 48' 00''
θ1 = 08o 06' 32''
AC = l2 = m
BC = l1 = m
To Determine:
Angle ACB = θ =?

22. Calculations:
Using the relations:
sinδ2 = (S' x sin θ2) / l2
= [17 x sin (63o 48' 00'')] / 16479
= sin-1[9.256 x 10-4] radians
δ2 = 0o 3' 10.92'' degrees
Similarly:
Sinδ1 = (S' / l1) x sin θ2
= [17 x sin (08o 06' 32'')] / 21726
= sin-1[ x 10-4] radians
δ1 = 0o 0' 22.77'' degrees

23. Now:
Since,
Angle ACB = θ
α + β + θ = (α + δ1) + (β + δ2) + ASB
θ = δ1 + δ2 + ASB
= 0o 3' 10.92'' + 0o 0' 22.77'' + 71o 54' 32''
= 71o 58' 5.69'' (Answer)

24. Assignment
In a triangle ABC, Station C was a church spire & could not be occupied. A satellite station was selected at m from ‘C’ and inside the triangle ABC. From ‘S’ angle CSA = 135o 40' 30'' and ASB = 71o 29' 30'' were measured. And the lengths AC and BC were known to be approximately 2511 m and 1894 m respectively. Compute the angle ACB?