S URVEYING
12.7 TAPE CORRECTIONS
The following five corrections may be found for the measured lengths of tape:
(i) Corrections for absolute length (ii) Corrections for pull
(iii) Corrections for temperature (iv) Corrections for slope and
(v) Corrections for sag.
12.7.1 Corrections for Absolute Length
Let, l = designated length of tape, la = actual length of tape.
Then correction for chain length c = la – l
Hence, if the total length measured is L, the total correction Ca = c
l L ...(12.2)
∴ Corrected length = L + Ca
If A is the measured area with incorrect tape, the correct area
= 1 + 2
F H I
c
K
l A = 1 2
F
+H I
c
K
l A, since C is small. ...(12.3)
12.7.2 Corrections for Pull
If pull applied while standardising the length of tape and pull applied in the field are different, this correction is required.
Let, P0 = Standard pull
P = Pull applied in the field
A = Cross-sectional area of the tape L = Measured length of line.
E = Young’s modulus of the material of tape, then Cp = (P P ) L
AE
− 0 ...(12.4)
The above expression takes care of sign of the correction also.
12.7.3 Correction for Temperature
Let T0 = Temperature at which tape is standardised Tm = Mean temperature during measurement
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α = Coefficient of thermal expansion of the material of the tape and L = Measured length,
Then the temperature correction Ct is given by,
Ct = Lα (Tm – T0) ...(12.5)
The above expression takes care of sign of the correction also.
12.7.4 Correction for Slope
If the length measured is ‘L’ and the difference in the levels of first and the last point is ‘h’, then slope correction
Csl = L – L2 −h2
= L [1 – 1−( / )h 2 2]
= L 1 1
2 8
2 4
−
F
− −HG
I KJ L
N M M
O Q P P
h h
L2 L4 ....
= h2
2L ...(12.6)
If measured length is L and slope is ‘θ’, then
Csl = L – L cos θ = L (1 – cos θ) ...(12.7)
This correction is always –ve.
12.7.5 Correction for Sag
While measuring on unevenly sloping ground, tapes are suspended at shorter length and horizontal distances are measured. This technique eliminates errors due to measurement along slopes, but necessitates correction for sag [Fig. 12.23]. Hence, measured length is more than actual length. Thus the correction is –ve. The correction, which is difference between the length of catenary and true length is given by
Cs = 1 24
W
P L
F
2H I
K
...(12.8)where, W = the weight of the tape of span length P = the pull applied
and L = measured length
It may be noted that if pull is more than standard pull, the correction for pull is +ve, while correction for sag is always –ve. The pull for which these two corrections neutralise each other is called
‘normal tension’. Hence normal tension Pn may be found as, Cp = Cs
True length True length
P P
Catenary
Fig. 12.23
(P P L) AE
n − 0 = 1
24 W 2
Pn
F HG
I KJ
L(Pn – P0) Pn2 = 1
24 W2 AE
or Pn = 0 204
0
. W AE
Pn −P ...(12.9)
The value of Pn is to be determined by trial and error method.
Example 12.2: A distance of 1500 m was measured with a 20 m chain. After the measurement chain was found to be 80 mm longer. If the length of chain was perfectly correct while starting measurement, what is the true length of the line measured?
Solution: Average correction per chain length c = 0 80
2
+ = 40 mm = 0.04 m Chain length = 20 m
∴ Correction for measured length Ca = L c
l = 1500 0 04 20
. = 3.0 m
∴ True length = 1500 + 3.0 = 1503 m Ans.
Example 12.3: A survey was conducted with a 20 m chain and plan of the field was drawn to a scale of 1 cm = 5 m. The area of the plan was found to be 62.8 cm2. However when the chain was tested at the end of work, it was wound to be 20.10 m. Assuming the length was exactly 20.0 m in the beginning of survey work, determine the true area of the field.
Solution: Initial length of the chain = 20 m Length at the end of work = 20.1 m
∴ Average length of chain = 20 20 1 2 + .
= 20.05 m
∴ Average correction per chain length = 20.05 – 20.00 = 0.05 m Measured area on plan = 62.8 cm2
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Scale 1 cm = 5 m
∴ Measured area on field = 62.8 × 52
= 1570 m2
∴ Corrected area on the ground = A l l
F
1 2HG I
KJ
= 157020 05 20 00
. 2
.
F H I
K
= 1577.860 m2 Ans.Example 12.4: A 20 m tape was used for measuring a line at an average temperature of 65º F. The measured distances on the ground and slope of the ground are as given below:
2º 18′ for 125 m
3º 30′ for 250 m
1º 42′ for a distance of 170 m
If the temperature at which tape was standardised is 80º F, find the true length of the line. Take α = 6.2 × 10–6/ºF.
Solution: Measured horizontal distance = ΣL cos θ
= 125 cos 2º 18′ + 250 cos 3º 30′ + 170 cos 1º 42′
= 544.367 m Temperature correction
Ct = Lα (Tm – T0)
= 544.367 × 6.2 × 10–6 (65 – 80)
= – 0.051 m Correct horizontal length
= 544.367 – 0.051 = 544.316 m Ans.
Example 12.5: To measure a base line, a steel tape 30 m long, standardised at 15º C with a pull of 80 N was used. Find the correction per tape length, if the temperature at the time of measurement is 25º C and the pull exerted is 150 N.
Take Young’s modulus E = 2 × 105 N/mm2 and coefficient of thermal expansion α = 11.2 × 10–6/°C. Cross-sectional area of tape is 8 mm2.
Solution: l = 30 m, α = 11.2 × 10–6/ºC, T0 = 15ºC, P0 = 80 N Tm = 25ºC, P = 150 N
∴ Correction for temperature
Ct = l α (Tm – T0)
= 30 × 11.2 × 10–6 (25 – 15)
= 3.360 × 10–3 m Correction for pull
Cp = (P P ) AE
− 0 l
= (150 80) 30 2 105 8
− ×
× ×
= 1.3125 × 10–3 m
[Note: Unit of AE comes out to be Newton only, if A is in cm2 or mm2 and E in N/cm2 or N/mm2].
∴ Total correction for temperature and pull
= Ct + Cp = 3.360 × 10–3 + 1.3125 × 10–3
= 4.6725 × 10–3 m per chain length
Example 12.6: Calculate sag correction for a 30 m steel tape under a pull of 80 N, if it is suspended in three equal spans. Unit weight of steel is 78.6 kN/m3. Area of cross-section of tape is 8 mm2. Solution: Length of each span = 10 m
W = wt. of taper per span length
= 78.6 × 10 × (8 × 10–6)
= 6288 × 10–6 kN = 6.288 N [Note: 1 mm2 = (0.001)2 m2 = 1 × 10–6 m2]
P = 80 N L = 10 m
∴ Correction for each span
= 1 24
6 288
80 10
. 2
F H I
K
×= 2.574 × 10–3 m
∴ Correction for three spans
= 3 × 2.574 × 10–3 m
= 7.722 × 10–3 m Ans.
Example 12.7: A 30 m steel tape was standardised under 60 N pull at 65º F. It was suspended in 5 equal span during measurement. The mean temperature during measurement was 90º F and the pull exerted was 100 N. The area of the cross-section of the tape was 8 mm2. Find the true length of the tape, if,
α = 6.3 × 10–6/ºF, E = 2 × 105 N/mm2 and unit weight of steel = 78.6 kN/m3. Solution: Correction for temperature:
Ct = l α (Tm – T0)
= 30 × 6.3 × 10–6 (90 – 65)
= 4.725 × 10–3 m Correction for pull:
Cp = (P P ) AE
− 0 l
= (100 60) 30 8 2 105
− ×
× × = 0.75 × 10–3 m
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Correction for sag:
Each span length = 30 5 = 6 m
Weight of tape per span, W = 78.6 × 6 × 8 × 10–6
= 3772.8 × 10–6 kN
= 3.7728 N
∴ Correction for sag per span
= 1 24
3 7728
100 6
. 2
F H I
K
×= 0.3559 × 10–3 m
∴ Corrections for 5 spans
Cs = 5 × 0.3559 × 10–3
= 1.779 × 10–3 m (– ve)
Noting that correction Ct and Cp are +ve, while correction Cs is –ve, we get Total correction = 4.725 × 10–3 + 0.75 × 10–3 – 1.779 × 10–3
= 3.696 × 10–3 m
= 3.696 mm
∴ True length of tape
= 30 + 3.696 × 10–3
= 30.003696 m Ans.