Tape corrections for 86 m, 10 kg pull

Method. The tape was standardised in catenary, so the sag in the standard length is added back to find its true length. Then the sag for the supports at every 10 m is applied.

Data

• $w = 30$ g/m $= 0.03$ kg/m; $P = 10$ kg (same as standard); $L = 86$ m
• Standard: 20 m tape $= 20.006$ m in catenary (20 m span)
• Field spans: 8 spans of 10 m + 1 span of 6 m

1. Tape length

Sag at standardisation: $\dfrac{0.03^{2}\times 20^{3}}{24\times 10^{2}} = 0.0030$ m. True tape length $= 20.006 + 0.0030 = 20.009$ m.

$$C_a = \frac{86\,(20.009 - 20)}{20} = +0.0387\ \text{m}$$

2. Sag correction

10 m span: $\dfrac{0.03^{2}\times 10^{3}}{24\times 10^{2}} = 0.000375$ m; for 8 spans $= 0.003000$ m. 6 m span: $\dfrac{0.03^{2}\times 6^{3}}{24\times 10^{2}} = 0.000081$ m.

$$C_s = -(0.003000 + 0.000081) = -0.0031\ \text{m}$$

3. Correct length

$$L_{true} = 86 + 0.0387 - 0.0031 = 86.0356\ \text{m}$$

Answer: corrected length of the line $= 86.0356$ m.