Transcript Axis – Signal Correction Axis – Signal Correction

GOVERNMENT ENGINEERING
COLLAGE - BHAVNAGAR
SUBJECT : SURVEYING
TRIGONOMETRIC LEVELLING
CURVATURE
Prepared by:
Amipara Priyank
Bharay Vijay
Jadav Bhavesh
Vala Akshit
(130210106001)
(130210106008)
(130210106024)
(130210106062)
Content
1. Introduction
2. 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
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=( s₂-h₁)/Dsin1” seconds ……………..(3)
• For observation from Q to P :
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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
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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:
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α = 3ᵒ10’49”
h1 = 1.24 m , S2 = 5.32 m , D =4935 m , m = 0.07
Rsin1” = 30.88 m ,
Central angle  
•
D
Rsin1"
= 4935/30.88 = 159.81”
• 1. Correction for refraction(r),
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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”
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= (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
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= - 11.19” + 79.90” – 170.5353 = - 101.82”
= - 0ᵒ1’ 42”
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correct altitude = + 3ᵒ10’49” – 0ᵒ1’42”
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= 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
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H = difference in elevation
= D sin(α₁- mƟ + Ɵ/2)/cos(α₁ - mƟ + Ɵ)
• Case 2 : Angle of depression
• H= Dsin ( β₁+ mƟ – Ɵ/2)/cos ( β₁ + mƟ –Ɵ)
………..(1)
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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
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M = 0.07
Central angle (θ) =D/Rsin1”= 9290/30.88=300.84”
• So, axis signal correction = δ1 =(S2-h1)/Dsin1”
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= (3.96 -1.25)/9290(1/206265)=60.17” ....(-ve)
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= 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” –
21.06” +300.84”)
= 9290 sin (2ᵒ7’27.19”) / cos (2ᵒ9’57.06”)
= 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.
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so, H = D sin[(β₁ -α₁)/2]/cos [(β₁ - α₁)/2 + Ɵ/2]
• In general ,
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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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