Horizontal length for 120.0 m recorded

Method. The tape is standardised in catenary, so the sag in its stated length is added back. In the field it is used in catenary in 30 m spans (120 m $=$ 4 tapes). $g = 9.81$ m/s$^2$.

Data

• $w = 0.0312\times 9.81 = 0.30607$ N/m
• Standard: 29.995 m at $15^\circ$C and $P_0 = 110$ N (catenary); field: $T = 22^\circ$C, $P = 90$ N, $L = 120.0$ m
• $A = 0.03$ 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

1. Tape length

Sag at standardisation: $\dfrac{0.30607^{2}\times 30^{3}}{24\times 110^{2}} = 0.008710$ m. True tape length $= 29.995 + 0.00871 = 30.00371$ m.

$$C_a = 120\times\frac{30.00371 - 30}{30} = +0.0148\ \text{m}$$

2. Temperature

$$C_t = 11\times 10^{-6}\times(22 - 15)\times 120 = +0.0092\ \text{m}$$

3. Pull

$$C_p = \frac{(90 - 110)\times 120}{618{,}030} = -0.0039\ \text{m}$$

4. Sag

Per tape: $\dfrac{0.30607^{2}\times 30^{3}}{24\times 90^{2}} = 0.013011$ m; $C_s = -4\times 0.013011 = -0.0520$ m.

5. Horizontal length

$$D = 120.0 + 0.0148 + 0.0092 - 0.0039 - 0.0520 = 119.9682\ \text{m}$$

Answer: horizontal length $= 119.968$ m.