True length of 680 m line, 20 m tape

Assumption. Standardised fully supported at $10^\circ$C and 5 kg, so there is no standard sag to add back.

Data

• $L = 680$ m; supports every 10 m (68 spans)
• $P_0 = 5$ kg, $P = 16$ kg; $T_0 = 10^\circ$C, $T = 22^\circ$C
• $A = 0.03$ cm$^2$, $E = 2.1\times 10^{6}$ kg/cm$^2$, $\alpha = 11\times 10^{-6}$/$^\circ$C
• Weight per metre $= A\times\rho = 0.03\ \text{cm}^2\times 10\ \text{g/cm}^3 = 0.3$ g/cm $= 0.03$ kg/m

1. Temperature

$$C_t = 11\times 10^{-6}\times(22 - 10)\times 680 = +0.0898\ \text{m}$$

2. Pull

$$C_p = \frac{(16 - 5)\times 680}{0.03\times 2.1\times 10^{6}} = +0.1187\ \text{m}$$

3. Sag

Per 10 m span: $\dfrac{0.03^{2}\times 10^{3}}{24\times 16^{2}} = 0.000146$ m.

$$C_s = -68\times 0.000146 = -0.0100\ \text{m}$$

4. True length

$$L = 680 + 0.0898 + 0.1187 - 0.0100 = 680.1985\ \text{m}$$

Answer: true length of the line $= 680.1985$ m.