True length of 1020 m line, 20 m tape

Assumptions. Standardised fully supported at $20^\circ$C and 100 N, so no standard sag is added back. $g = 9.81$ m/s$^2$.

Data

• $L = 1020$ m, supports every 10 m (102 spans of 10 m)
• $P_0 = 100$ N, $P = 160$ N; $T_0 = 20^\circ$C, $T = 30^\circ$C
• $A = 0.03$ cm$^2$, $E = 2.1\times 10^{6}$ kg/cm$^2$, so $AE = 0.03\times 2.1\times 10^{6}\times 9.81 = 618{,}030$ N
• $\alpha = 11\times 10^{-6}$/$^\circ$C; $w = 24$ g/m $= 0.024\times 9.81 = 0.23544$ N/m

1. Temperature correction

$$C_t = 11\times 10^{-6}\times(30 - 20)\times 1020 = +0.1122\ \text{m}$$

2. Pull correction

$$C_p = \frac{(160 - 100)\times 1020}{618{,}030} = +0.0990\ \text{m}$$

3. Sag correction

Per 10 m span: $\dfrac{w^{2}\,l^{3}}{24\,P^{2}} = \dfrac{0.23544^{2}\times 10^{3}}{24\times 160^{2}} = 0.0000902$ m.

$$C_s = -102\times 0.0000902 = -0.0092\ \text{m}$$

4. True length

$$L_{true} = 1020 + 0.1122 + 0.0990 - 0.0092 = 1020.2020\ \text{m}$$

Answer: true length of the line $= 1020.202$ m.