Tape correction for 24.726 m distance

Method. The tape was standardised in catenary (30 m span), so the sag in its stated length is added back to give the true tape length. The 24.726 m is measured as one catenary span at the same pull.

Data

$w = 0.03$ kg/m, $P = 5$ kg, standard length $= 30.015$ m (catenary), $L = 24.726$ m.

1. Tape length

Sag at standardisation: $\dfrac{0.03^{2}\times 30^{3}}{24\times 5^{2}} = 0.0405$ m. True tape length $= 30.015 + 0.0405 = 30.0555$ m.

$$C_a = 24.726\times\frac{30.0555 - 30}{30} = +0.0457\ \text{m}$$

2. Sag correction

$$C_s = -\frac{0.03^{2}\times 24.726^{3}}{24\times 5^{2}} = -0.0227\ \text{m}$$

3. Corrected length

$$L_{true} = 24.726 + 0.0457 - 0.0227 = 24.7491\ \text{m}$$

Answer: corrected length $= 24.7491$ m.