Correct base line length 1120 m

Method. The tape is standardised in catenary, so its true length is found by adding back the sag at standardisation. The tape is used in catenary (20 m spans), 1120 m $= 56$ tapes.

Data

• $W = 7.85$ N for 20 m, $w = 0.3925$ N/m; weight density of steel $= 0.077$ N/cm$^3$
• Area: $A = \dfrac{W}{0.077\times 2000} = \dfrac{7.85}{154} = 0.05097$ cm$^2 = 5.097$ mm$^2$
• $E = 2.1\times 10^{5}$ N/mm$^2$, so $AE = 5.097\times 2.1\times 10^{5} = 1.0705\times 10^{6}$ N
• Standard: 19.978 m, $T_0 = 12.5^\circ$C, $P_0 = 100$ N; field: $T = 25^\circ$C, $P = 150$ N
• $\alpha = 12\times 10^{-6}$/$^\circ$C

1. Tape length

Sag at standardisation: $\dfrac{7.85^{2}\times 20}{24\times 100^{2}} = 0.005135$ m. True tape length $= 19.978 + 0.005135 = 19.98314$ m.

$$C_a = 1120\times\frac{19.98314 - 20}{20} = -0.9444\ \text{m}$$

2. Temperature

$$C_t = 12\times 10^{-6}\times(25 - 12.5)\times 1120 = +0.1680\ \text{m}$$

3. Pull

$$C_p = \frac{(150 - 100)\times 1120}{1.0705\times 10^{6}} = +0.0523\ \text{m}$$

4. Sag

Per tape: $\dfrac{7.85^{2}\times 20}{24\times 150^{2}} = 0.002282$ m; $C_s = -56\times 0.002282 = -0.1278$ m.

5. Correct length

$$L = 1120 - 0.9444 + 0.1680 + 0.0523 - 0.1278 = 1119.1481\ \text{m}$$

Answer: correct length of the base line $= 1119.148$ m.