Transcript Surveying-II

Surveying-II
SURVEYING-II
Surveying-II
ADJUSTMENT OF ANGLES
IN
TRIANGULATION
Surveying-II
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.
(i).
(ii).
Station Adjustment.
Figure Adjustment.
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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.
Angles
Observed Value
Correction
Corrected Value
∟1
---
-12''
---
∟2
---
-12''
---
∟3
---
-12''
---
∟4
---
-12''
---
∑ = 360o 00' 48''
∑ = 360o 00' 00''
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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:
(a). Triangles.
(b). Quadrilaterals.
(c). Polygons with central stations.
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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
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TRIANGLE ADJUSTMENT
Angles
∟1
∟5
Observed Value
-----
Correction
-4''
-4''
Corrected Value
-----
∟6
---
-4''
---
∑ = 180o 00' 12''
∑ = 180o 00' 00''
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ADJUSTMENT OF BRACED
QUADRILATERAL
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 seconds 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.
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ADJUSTMENT OF BRACED
QUADRILATERAL
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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 :
PROCEDURE
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1. Record ‘log (sin)’ of each angle obtained after geometric
adjustment.
2. For each angle record the ‘log (sin)’ difference for 01'' i.e.,
the difference between the previous value in step (1) &
the value obtained after adding 01'' to actual angle.
3. Find the average required change (α) in ‘log (sin)’ by
dividing the difference of sum of odd and even angle
values by 8.
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PROCEDURE
4. Find the average difference “β” for difference for 01''. i.e.,
total ‘log (sin)’ difference for 01'' of all angles divided by
8.
5. The ratio α / β gives the no. of seconds to apply a
correction.
6. Add the correction to each of four angles where ‘log (sin)’
is smaller & subtract the correction from the other four
angles.
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SATELLITE STATION
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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'.
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In the triangle ACS:
sinδ1 / S' = sinθ1 / l1
sinδ1 = S' / l1 x sin θ1
----------- (1)
For very small value of δ1:
sinδ1 = δ1 (in radians)
= (δ1 x 180) / π
= (δ1 x 180 x 60 x 60) / π (seconds)
= 206265 x δ1
(seconds)
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Equation (1) can be written as:
δ1 = (S' x sinθ1) / l1
δ1 = (206265 x S' x sinθ1) / l1
(in radians)
(in seconds)
Similarly:
δ2 = (206265 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)
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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
17.00 m. The length of AC = 16479 m and of BC = 21726
m. The angle SAC was found to be 63o 48' 00'' and the
angle ASB 71o 54' 32''. Calculate angle ACB?
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Solution:
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Given Data:
S' = 17.00 m
Angle ASC = θ1 = 63o 48' 00''
Angle ASB = 71o 54' 32''
θ2 = CSB = ASB – θ
= 71o 54' 32'' – 63o 48' 00''
θ2 = 08o 06' 32''
AC = l1 = 16479 m
BC = l2 = 21726 m
To Determine:
Angle ACB =?
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Calculations:
Using the relations:
sinδ1 = (S' / l1) x sin θ1
= [17 x sin (63o 48' 00'')] / 16479
= sin-1[9.256 x 10-4]
radians
δ1 = 0o 3' 10.92''
degrees
Similarly:
Sinδ2 = (S' / l2) x sin θ2
= [17 x sin (08o 06' 32'')] / 21726
= sin-1[1.1037 x 10-4] radians
δ2 = 0o 0' 22.77''
degrees
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Now:
Since,
angle ASB = θ - (δ1 + δ2)
We have to find θ i.e., angle ACB.
Angle ACB = Angle ASB + (δ1 + δ2)
= 71o 54' 32'' + (0o 3' 10.92'' + 0o 0' 22.77'')
= 71o 58' 5.69'' (Answer)
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Problem
In a triangle ABC, was a church spire & could not be
occupied. A satellite station was selected at 12.10 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?
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SOLUTION
Given Data:
S' = 12.10 m
Angle ASC = θ1 = 135o 40' 30''
θ = 71o 29' 30''
Angle BSC = θ2 = 360o (θ1 + θ)
= 152o 50' 00''
AC = l1 = 2511 m
BC = l2 = 1894 m
To Determine:
Angle ACB =?
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Calculations:
Using the relations:
sinδ1 = (S' / l1) x sin θ1
= [12.10 x sin (135o 40' 30'')] / 2511
= sin-1[3.367 x 10-3] radians
δ1 = 0o 11' 34.5''
degrees
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Similarly:
Sinδ2 = (S' / l2) x sin θ2
= [12.10 x sin (152o 50' 00'')] / 1894
= sin-1[2.916 x 10-3]
radians
δ2 = 0o 10' 1.66''
degrees
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Now:
Since,
Angle ACB = θ - (δ1 + δ2)
We have to find ACB, so
Angle ACB = 71o 29' 30'' - (0o 11' 34.5'' + 0o 10' 1.66'')
= 71o 07' 53.84''
(Answer)
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Thanks