GOVERNMENT ENGINEERING COLLAGE - BHAVNAGAR SUBJECT : SURVEYING

TRIGONOMETRIC LEVELLING CURVATURE Prepared by: Amipara Priyank (130210106001) Bharay Vijay (130210106008) Jadav Bhavesh (130210106024) Vala Akshit (130210106062)

Content Introduction Correction For Curvature 3. Axis – Signal Correction Observed Angle Is Angle Of Elevation Observed Angle Is Angle Of Depression 4. Determining Differences In Elevation Elevation From Single Observation Angle Of Depression Correction In Linear Measure 5. Elevation From Reciprocal Observation

Introduction Trigonometric levelling is the process of determining the differences of elevation of stations from observed vertical angles and known distances. The vertical angles are measured by means of theodolite. The horizontal distances may either be measured (in case of plane surveying) or computed (in case of geodetic surveying). Relative heights are calculated using trigonometric functions. If the distance between instrument station and is small, correction for earth’s curvature and refraction is not required. If the vertical angle is (+ve), the correction is taken as (+ve). If the vertical angle is (-ve), the correction is taken as (-ve).

Correction for Curvature The effect of the curvature is to make the object appear lover than they really are. In spirit levelling, the effect is to increase the staff reading and the correction is, there fore, subtracted from the staff reading. The effect of refraction is in the opposite direction to that of curvature. In trigonometrical levelling employed for determining the elevation of widely distributed points the correction for curvature is applied directly to the observed angles.

Axis – Signal Correction

Axis – Signal Correction The height of the signal is not same as that of the height of the instrument axis above the station, a correction known as the axis signal correction or eye and object correction is to be applied. Let, h₁= height of instrument at P, for observation to Q h₂= height of instrument at Q, for observation to P S₁= height of the signal at P, instrument being at Q s₂= height of the signal at Q, instrument being at P d= horizontal distance between P and Q α= observed of elevation uncorrected for the axis signal β= observed of depression uncorrected for the axis signal α₁= angle of elevation corrected for axis signal β₁= angle of depression corrected for axis signal

For observations from P to Q : tanδ₁ = ( s₂ - h₁ )cos²α/D ………………(1) The correction δ₁ is negative for the angle of elevation so, corrected angle of elevation (α₁) α₁= α-δ₁ …………………(2) If α is very small, tanδ₁ ≈δ₂=(s₂-h₁)/D radians =( s₂-h₁)/Dsin1” seconds ……………..(3) For observation from Q to P : tanδ₂ = (s₁-h₂)cos²β/D ……………..(4) The correction δ₂ is positive for the angle of depression so, corrected angle of depression {β₁}, β₁ = β + δ₂ ………………(5) If β is very small, Tanδ₂≈δ₁=(s₁-h₂)/D radians = (s₁-h₂)/D sin1” seconds ………………(6)

EXAMPLES The following observations were made in a trigonometric levelling. Observed altitude = + 3ᵒ10’49” Height of instrument = 1.24 m Height of signal = 5.32 m Horizontal distance = 4935 m Coefficient of refraction = 0.07 If R sin1” = 30.88 m. Correct the observed angle for refraction, curvature and axis signal. SOLUTION: α = 3ᵒ10’49” h1 = 1.24 m , S2 = 5.32 m , D =4935 m , m = 0.07 Rsin1” = 30.88 m , Central angle = 4935/30.88 = 159.81”

1. Correction for refraction(r), r = mθ = 0.07(159.81) =11.19” ...(-ve) 2. Correction for curvature, (cc) : cc = θ/2 = 159.81/2 =79.90 sec .....(+ve) 3. Correction for axis signal : δ1 = (S2 – h1)/Dsin1” = (5.32 – 1.24)/4935(1/206265) = 170.53 sec ..........( -ve) sin1” = 1/206265 = 0.000004848 Here the angle is (+ve) So, total correction = - r + θ/2 – δ1 = - 11.19” + 79.90” – 170.5353 = - 101.82” = - 0ᵒ1’ 42” correct altitude = + 3ᵒ10’49” – 0ᵒ1’42” = 3ᵒ9’7”

Difference in elevation The difference in elevation between the two points P and Q can be determined by 1. single observation 2. reciprocal observation 1. by single observation Case 1 : Angle of elevation H = difference in elevation = D sin(α₁- mƟ + Ɵ/2)/cos(α₁ - mƟ + Ɵ) ………..(1) Case 2 : Angle of depression H= Dsin ( β₁+ mƟ – Ɵ/2)/cos ( β₁ + mƟ –Ɵ)

SOLUTION: So, axis signal correction = δ1 =(S2-h1)/Dsin1” Find R.L. Of Q from the following observation: Horizontal distance between P and Q= 9290 m Angle of elevation from P to Q = 2ᵒ06’18” Height of signal at Q = 3.96 m Height of instrument at P =1.25 m Coefficient of refraction = 0.07 R.L. Of P =396.58 m Rsin1”= 30.88 m SOLUTION: D=9092 m , α = 2ᵒ06’18” , S2 =3.96 m, h1 = 1.25 m , Rsin1” = 30.88 m M = 0.07 Central angle (θ) =D/Rsin1”= 9290/30.88=300.84” So, axis signal correction = δ1 =(S2-h1)/Dsin1” = (3.96 -1.25)/9290(1/206265)=60.17” ....(-ve) = 0ᵒ1’0.017” So, α1=α - δ1= 2ᵒ06’18”- 0ᵒ1’0.017” = 2ᵒ5’17.83” θ/2 = 300.84”/2 = 150.42” r= mθ = 0.07×300.84” = 21.06” so, difference in elevation (H) =D sin(α1-mθ + θ/2)/cos (α1-mθ + θ)

H = 9290sin(2ᵒ5’17. 83” – 21. 06” + 150. 42”) / cos (2ᵒ5’17. 83” – 21 = 344.34/0.999 = 344.68 m So, R.L. Of Q =R.L. Of P + H = 396.58 + 344.668 = 741.26 m

By reciprocal observation : The difference of elevations of P and Q may be determined more accurately by reciprocal observations. the observation are made simultaneously from both the stations P and Q . Case 1: when α₁ is angle of elevation and β₁ is angle of depression. H = D sin(α₁ + β₁/2)/cos[α₁ + β₁/2 + Ɵ/2] If H obtained is +ve, Q is higher than P, but if H is negative, Q is lower than P. Case 2 : If both α₁ and β₁ are the angles of depression. so, H = D sin[(β₁ -α₁)/2]/cos [(β₁ - α₁)/2 + Ɵ/2] In general , H = D sin[(β₁ ±α₁)/2]/cos[(β₁±α₁)/2 + Ɵ/2] Use + sign when α₁ is the angle of elevation and use – sign when α₁ is the angle of depression.

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