True length of AB: 114.095 m, 12°C

Assumptions. The tape is standardised fully supported; used suspended in 3 full tapes of 30 m and a last part of 24.095 m. The slope is uniform at 2.5 m per 100 m.

Data

• $W = 9$ N per 30 m, so $w = 0.3$ N/m
• $A = 0.028$ cm$^2$, $E = 2.11\times 10^{6}$ N/cm$^2$, so $AE = 59{,}080$ N
• Standard: 30.015 m at $20^\circ$C, $P_0 = 70$ N; field: $T = 12^\circ$C, $P = 100$ N; $\alpha = 1.15\times 10^{-5}$/$^\circ$C
• $L = 114.095$ m

1. Tape length

$$C_a = \frac{114.095\times 0.015}{30} = +0.0570\ \text{m}$$

2. Temperature

$$C_t = 1.15\times 10^{-5}\times(12 - 20)\times 114.095 = -0.0105\ \text{m}$$

3. Pull

$$C_p = \frac{(100 - 70)\times 114.095}{59{,}080} = +0.0579\ \text{m}$$

4. Sag

Full tape: $\dfrac{9^{2}\times 30}{24\times 100^{2}} = 0.010125$ m; three tapes $= 0.030375$ m. Last part: $\dfrac{0.3^{2}\times 24.095^{3}}{24\times 100^{2}} = 0.005246$ m.

$$C_s = -(0.030375 + 0.005246) = -0.0356\ \text{m}$$

5. Slope

$$C_h = -\frac{0.025^{2}\times 114.095}{2} = -0.0357\ \text{m}$$

6. True length

$$L_{true} = 114.095 + 0.0570 - 0.0105 + 0.0579 - 0.0356 - 0.0357 = 114.1282\ \text{m}$$

Answer: true horizontal length of AB $= 114.128$ m.