Why is the pull correction positive for every bay?

Field pull (120 N) is greater than the standard pull (89 N), so the tape stretches and reads short

The tape was standardised in catenary

The temperature is above standard

Why is no tape-length correction applied?

The tape is standardised on the flat and taken as nominally 30 m, so it has no length error

Because temperature is standard

How is the true length of AB obtained?

Sum of the corrected horizontal bay lengths, about 117.962 m

Sum of measured lengths minus the sag only

Average of the four bays times 4 without slope correction

2078 Kartik · 6 marks

True length of base line AB (four bays)

Short answer

For each bay, apply slope correction h^2/2l, sag, pull and temperature corrections, then sum. True horizontal length of the base line is 117.962 m.

Four bays of base line AB were measured under a tension of 120 N and the data is given below. If the tape was standardised on the flat under a pull of 89 N and a temperature of 20°C, calculate the true length of the line.

Bay	Length (m)	Difference in level (m)
1	29.478	+0.294
2	29.208	-0.384
3	29.396	+0.923
4	29.916	-0.726

Field temperature = 31°C; cross-sectional area of tape = 3.24 mm²; density = 7700 kg/m³; coefficient of linear expansion = 0.000001/°C; Young modulus = $15.3\times10^{4}$ MN/m².

Assumptions. The tape is nominally 30 m long, standardised on the flat (fully supported), so it has no length error. Each bay is measured with the tape in catenary at 120 N. Slope corrections use $h^2/2l$ .

Data

Area $A = 3.24$ mm $^2 = 3.24\times 10^{-6}$ m $^2$ ; density $= 7700$ kg/m $^3$
Mass per metre $= 7700\times 3.24\times 10^{-6} = 0.024948$ kg/m, so $w = 0.24474$ N/m
$E = 15.3\times 10^{4}$ MN/m $^2 = 1.53\times 10^{11}$ N/m $^2$ , so $AE = 495{,}720$ N
$P = 120$ N, $P_0 = 89$ N, $T = 31^\circ$ C, $T_0 = 20^\circ$ C, $\alpha = 1\times 10^{-6}$ / $^\circ$ C

Formulas for each bay of length $l$ with level difference $h$

C_h = -\frac{h^{2}}{2l},\quad C_t = \alpha\,(T-T_0)\,l,\quad C_p = \frac{(P-P_0)\,l}{AE},\quad C_s = -\frac{w^{2}\,l^{3}}{24\,P^{2}}

Calculation

Bay	$l$ (m)	$h$ (m)	$C_h$	$C_s$	$C_p$	$C_t$	Horizontal length (m)
1	29.478	+0.294	$-0.00147$	$-0.00444$	$+0.00184$	$+0.00032$	29.4743
2	29.208	$-0.384$	$-0.00252$	$-0.00432$	$+0.00183$	$+0.00032$	29.2033
3	29.396	+0.923	$-0.01449$	$-0.00440$	$+0.00184$	$+0.00032$	29.3793
4	29.916	$-0.726$	$-0.00881$	$-0.00464$	$+0.00187$	$+0.00033$	29.9048

Sum of measured lengths $= 117.998$ m.

\text{True length} = 29.4743 + 29.2033 + 29.3793 + 29.9048 = 117.9617\ \text{m}

Answer: true horizontal length of the base line AB $\approx 117.962$ m (the sum of the bays computed with unrounded values is $117.9616$ m).

✓ Answer checked by Claude Opus 5.5 (high effort)

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True length of base line AB (four bays)

Data

Formulas for each bay of length lll with level difference hhh

Calculation

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Formulas for each bay of length $l$ with level difference $h$