This document contains a lecture outline on trigonometric leveling with examples of numerical problems. It includes: - An introduction to trigonometric leveling and the lecturer's contact information. - Four example problems demonstrating calculations for corrected vertical angles, central angles, curvature and refraction corrections, and determining height differences or reduced levels between stations using trigonometric leveling methods and accounting for instrument heights, signal heights, and refraction.

1. Advance Surveying
Trigonometric Leveling
Prof. Rajguru R.S.
Civil Engineering Department
(rajgururajeshcivil@sanjivani.org.in)
Sanjivani College of Engineering,
Kopargaon,MH,India

2. Lecture Outline
 Numerical on Trigonometric Leveling

3.  Numerical:
• Ex.5.1: A vertical angle of elevation was observed from a station P as 2°
32’ 25’’. Determine its true value if the height of instruments at P is 1.2m
& height of signal at the other station Q is 5.2m. & Two station P & Q are
5200m apart. Take the value of Rsin1’’ as 30.88m. Find also the true value
of the angle observed if it was an angle of depression.
• Solution:
• Case I Given: α = 2° 32’ 25’’ ℎ1 = 1.2𝑚 𝑆2 = 5.2𝑚 D =
5200m , RSin1’’ = 30.88 Assume m =0.07 R = 6370 km =6370
000m
• Sin1’’ =
30.88 𝑚
6370000 𝑚
= 4.85 × 10−6
sec
• Axis Signal Correction, δ =
𝑺 𝟐 −𝒉 𝟏
𝑫𝒔𝒊𝒏𝟏"
=
𝟓.𝟐 −𝟏.𝟐
𝟓𝟐𝟎𝟎 ×𝟒.𝟖𝟓 ×𝟏𝟎−𝟔
• =158.67 ” sec = 0 ° 2’ 38.67’’ (-ve)

4.  Numerical:
• Central angle, 𝞱 =
𝐷
𝑅𝑆𝑖𝑛1"
=
5200
30.88
= 168.39” =2’ 48.39’’
• Curvature correction, C =
𝞱
2
=
168.39
2
= 84.19’’ = 0 ° 1’ 24.19’’ (+ve)
• Refraction correction , 𝛾 = m 𝞱 = 0.07 × 2’ 48.39’’ = 0 ° 0’ 11.79’’ (-ve)
• Corrected angle of elevation, ∝1 = α – δ + C -𝛾
• = 2° 32’ 25’’- 0° 2’ 38.67’’ + 0° 1’ 24.19’’ – 0° 0’ 11.79’’ = 2° 30’ 58.73’

5. • Corrected angle of depression:
• Corrected angle of depression, 𝛃1 = 𝛃 + δ - C +𝛾
• = 2° 32’ 25’’ + 0 ° 2’ 38.67’’ - 0 ° 1’ 24.19’’ + 0 ° 0’ 11.79’’ = 2 ° 33’
51.27’’

6.  Numerical:
• Ex.5.2 Two triangulation station A & B are 3200.65 m apart. Find the
difference of elevation of two stations for the following data.
• Angle of depression at B to A = 2° 18’ 16’’.
• Height of signal at A = 4.23m
• Height of instrument at B = 1.24m
• Coefficient of refraction at B = 0.07
• Rsin1’’ = 30.88 m
• R.L. of B = 242.6 m
•
Solution: Sin1’’ =
30.88 𝑚
6370000 𝑚
= 4.85 × 10−6
sec
• Axis Signal Correction, δ =
𝑺 𝟐 −𝒉 𝟏
𝑫𝒔𝒊𝒏𝟏"
=
𝟒.𝟐𝟑 −𝟏.𝟐𝟒
𝟑𝟐𝟎𝟎.𝟔𝟓 ×𝟒.𝟖𝟓 ×𝟏𝟎−𝟔
• = 192.62 ” sec = 0 ° 3’ 12.62’’ (+ve)
•

7.  Numerical:
• Vertical angle corrected for axis signal correction, 𝛃1 = 𝛃 + δ = 2 ° 18’ 16’’
+ 0 ° 3’ 12.62’’ = 2 ° 21’ 28.69’’
• Central angle, 𝞱 =
𝐷
𝑅𝑆𝑖𝑛1"
=
3200.65
30.88
= 103.64” = 1’ 43.64’’
• Curvature correction, C =
𝞱
2
=
103.64”
2
= 51.82’’ (-ve)
• Refraction correction , 𝛾 = m 𝞱 = 0.07 × 103.64’’ = 7.25’’ (+ve)
• H = D
𝑠𝑖𝑛(𝛃1− 𝞱
2
+ m 𝞱)
𝐶𝑂𝑆(𝛃1− 𝞱 + m 𝞱)
= 3200.65
𝑠𝑖𝑛(2 ° 21’ 28.69’’ −51.82′′ + 7.25′′)
𝐶𝑂𝑆(2 ° 21’ 28.69’−103.64’’+7.25’’)
= 131.12m
• R.L. of A = 242.6-131.12 = 111.48 m

8.  Numerical:
• Ex.5.3: Two triangulation station A& B are 2800m apart. Observations
were made for vertical angle of elevation from A to B and the mean angle
observed was 1° 28’ 32’’.The height of the instrument was 1.38 m and the
signal was 2.46m high. If the reduced level of station A was 125m and the
coefficient of refraction was 0.07, calculate the reduced level of B. The
radius of earth is 6372 km.
• Solution:
• Axis signal correction = δ =
𝑺 𝟐 −𝒉 𝟏
𝑫𝒔𝒊𝒏𝟏"
𝟐.𝟒𝟔−𝟏.𝟑𝟖
𝟐𝟖𝟎𝟎𝑺𝒊𝒏𝟏"
= 1’ 19.56’’
• Vertical angle corrected for axis signal correction, α1=α − δ = 1 ° 28’
32’’ -0° 1’19.56’’ = 1 ° 27’ 12.44’’

9.  Numerical:
• Central angle, 𝞱 =
𝐷
𝑅𝑆𝑖𝑛1"
=
2800
6372×103 𝑆𝑖𝑛1"
= 90.637”
• Curvature correction, C =
𝞱
2
=
90.637”
2
= 45.318’’ (+ve)
• Refraction correction , 𝛾 = m 𝞱 = 0.07 × 90.637 = 6.34’’ (-ve)
• H = D
𝑠𝑖𝑛(α1+ 𝞱
2
− m 𝞱)
𝐶𝑂𝑆(α1+ 𝞱 − m 𝞱)
• = 2800
𝑠𝑖𝑛(1 ° 27’ 12.44’’ + 45.318′′ − 6.34′′)
𝐶𝑂𝑆(1 ° 27’ 12.44’’ + 90.637′’− 6.34’’)
= 71.578 m
• R.L. of B = 125 + 71.578 = 196.578 m

10.  Numerical:
• Ex.5.4: Two station A & B are 1700m apart. The observation recorded
were as follows. Calculate the difference of level between A & B, the
coefficient of refraction and refraction correction.
Station A Station B
Height of Instrument 1.39 m 1.46 m Rsin1’’
Height of signal 2.2 m 2.00 m =30.88 m
Vertical Angle + 1° 08’ 05’’ - 1° 06’ 10’’

11.  Numerical:
• Solution: Vertical angle from A to B, α = + 1° 08’ 05’’
• Axis signal correction to α, δ=
𝑺 𝟐 −𝒉 𝟏
𝑫𝒔𝒊𝒏𝟏"
𝟐−𝟏.𝟑𝟗
𝟏𝟕𝟎𝟎𝑺𝒊𝒏𝟏"
= 74.012’’ (-ve)
• Vertical angle corrected for axis signal = α1=α − δ = 1° 08’ 05’’
• -0° 0’ 74.012’’ = 1° 06’ 50.98’’
• Vertical angle from B to A, 𝛃 = 1° 06’ 10’’
• Axis signal correction to 𝛃, δ =
𝑺 𝟏 −𝒉 𝟐
𝑫𝒔𝒊𝒏𝟏"
𝟐.𝟐−𝟏.𝟒𝟔
𝟏𝟕𝟎𝟎𝑺𝒊𝒏𝟏"
= 89.78’’ (+ve)
• Vertical angle corrected for axis signal = 𝛃1=𝛃 + δ = + 1° 06’ 10’+89.78’’ = 1°
07’ 39.78’’
• Central angle, 𝞱 =
𝐷
𝑅𝑆𝑖𝑛1"
=
1700
30.88
= 55.05’’
• H =D
𝑠𝑖𝑛
∝1+𝛃1
2
)
𝐶𝑂𝑆( ∝1+𝛃1
2
+ 𝞱
2
)
= 1700
sin(
1° 06’ 50.98’’ +1° 07’ 39.78’’
2
)
𝐶𝑂𝑆( 1° 06’ 50.98’’ +1° 07’ 39.78’’
2
+ 55.05′′
2
)
• = 33.263 m

12.  Numerical:
• Refraction correction, 𝛾 =
1
2
(𝞱 + α1 − ઺ 1)
• =
1
2
(55.05’’+1° 06’ 50.98’’ −1° 07’ 39.78’’) = 3.125’’
• Coefficient of refraction m =
𝛾
𝞱
=
3.125’’
55.05’’
=0.057

13. Thank you