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What are the Corrections Applied in Surveying?

For surveying, we need to have some prerequisite conditions. If these conditions are not met we can have a huge variation in result. Therefore we have to apply corrections to get the true result.

Ideal Conditions
1) A tape accurate to 0.00305m or 0.01 ft should be used.
2) Tension of the tape should be about 66.7N or 15 lb.
3) Temperature should be determined within 5.56°C or 10°F
4) The slope of the ground, should be within 2 percent

On ground these are nearly impossible to achieve and thus corrections need to be applied.

Corrections Applied for Temperature
The correction applied on steel tape is Ct=0.0000065s(T-T0)
where
Ct= temperature correction to measured length, ft (m)
T=temperature at which measurements are made, F ( C)
T0= temperature at which tape is standardized, F ( C)
s= measured length, ft (m)

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Correction Applied to Measurements on Slope
Ch= s ([email protected]) [exact]
or = [email protected]2[approximate]
or = h2/2s
where
Ch= correction to be subtracted from slope distance, ft (m)
s= measured length, ft (m)
@ =slope angle, degree
h= difference in elevation at ends of measured length, ft (m)

Correction Applied for Tension
Cp=s[Pm-Ps]/SE

Correction Applied for Sag when not Fully Supported
Cs=w2L3/24Pm2

where
Cp= tension correction to measured length, ft (m)
Cs = sag correction to measured length for each section of unsupported tape, ft (m)
Pm actual tension, lb (N)
Ps tension at which tape is standardized, lb (N) (usually 10 lb) (44.4 N)
S=cross-sectional area of tape, in2 (mm2)
E= modulus of elasticity of tape, lb/in2 (MPa) (29 million lb/in2(MPa) for steel) (199,955 MPa)
w= weight of tape, lb/ft (kg/m)
L= unsupported length, ft (m)

What are Slope Corrections?
We know that the horizontal distance H= [email protected], where L slope distance and @=vertical angle, measured from the horizontal.

For slopes of 10 percent or less
Cs=d2/2L

For a slope greater than 10 percent
Cs=d2/2L+d4/8Ld3

What are Temperature Corrections in terms of length?
Ct=(actual tape length-nominal tape length)L/nominal tape length

For nonstandard tension:
Ct(applied pull-standard tension)L/AE
where A= cross-sectional area of tape, in2 (mm2)
E=modulus of elasticity29,000,00 lb / in2 for steel (199,955 MPa).

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For sag correction between points of support, ft (m):
C= -w2L3s/24P2
where
w = weight of tape per foot, lb (N)
Ls= unsupported length of tape, ft (m)
P=pull on tape, lb (N)

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Kanwarjot Singh

Kanwarjot Singh is the founder of Civil Engineering Portal, a leading civil engineering website which has been awarded as the best online publication by CIDC. He did his BE civil from Thapar University, Patiala and has been working on this website with his team of Civil Engineers.

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3 comments on "What are the Corrections Applied in Surveying?"

anand says:

I hope this is more useful..

AZHARUDDIN SK says:

For a slope greater than 10 percent
Cs=d2/2L+d4/8Ld3
remove d3 from the above formula
corrected one is as below
Cs=d2/2L+d4/8L3

md hasnat says:

On a map covering an area of 2.5 sq. miles,the scale is by mistake recorded to be 12in =1 mile with the result that the area comes to to be 4 sq.mile and 2844.44 sq.chain.Determine what will be the correct scale of the map ?
PLEASE SOLVE IT

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