True length of 780 m line, tape 0.693 kg

Assumption. The tape was standardised fully supported ($20^\circ$C, 10 kg), so there is no standard sag to add back.

Data

• $L = 780$ m; supports every 15 m, so $780/15 = 52$ spans
• $P_0 = 10$ kg, $P = 15$ kg; $T_0 = 20^\circ$C, $T = 32^\circ$C
• $A = 0.03$ cm$^2$, $E = 2.1\times 10^{6}$ kg/cm$^2$, $\alpha = 11\times 10^{-6}$/$^\circ$C
• $w = 0.693/30 = 0.0231$ kg/m

1. Temperature correction

$$C_t = 11\times 10^{-6}\times(32 - 20)\times 780 = +0.1030\ \text{m}$$

2. Pull correction

$$C_p = \frac{(15 - 10)\times 780}{0.03\times 2.1\times 10^{6}} = +0.0619\ \text{m}$$

3. Sag correction

Per span: $\dfrac{0.0231^{2}\times 15^{3}}{24\times 15^{2}} = 0.000334$ m.

$$C_s = -52\times 0.000334 = -0.0173\ \text{m}$$

4. True length

$$L_{true} = 780 + 0.1030 + 0.0619 - 0.0173 = 780.1475\ \text{m}$$

Answer: true length of the line $= 780.1475$ m.