Correct length of 360 m line

Assumptions. The tape was standardised in catenary at 7 kg with the same support spacing (20 m) as in the field, so its stated length is correct for that condition. Only the changes (pull and sag) are corrected. $g = 9.81$ m/s$^2$.

Data

• $L = 360$ m; field pull $P = 5$ kg; standard pull $P_0 = 7$ kg
• $w = 10$ g/m $= 0.01$ kg/m; $A = 0.03$ cm$^2$
• $E = 210\times 10^{3}$ N/mm$^2 = 2.1\times 10^{7}$ N/cm$^2 = 2.1407\times 10^{6}$ kg/cm$^2$
• Spans: $360/20 = 18$ spans of 20 m

1. Pull correction

$$C_p = \frac{(P - P_0)\,L}{A\,E} = \frac{(5 - 7)\times 360}{0.03\times 2.1407\times 10^{6}} = -0.0112\ \text{m}$$

2. Sag correction (difference between the sag at 5 kg and that already in the standard at 7 kg)

Sag at 5 kg: $18\times\dfrac{0.01^{2}\times 20^{3}}{24\times 5^{2}} = 0.02400$ m. Sag at 7 kg: $18\times\dfrac{0.01^{2}\times 20^{3}}{24\times 7^{2}} = 0.01224$ m.

$$C_s = -(0.02400 - 0.01224) = -0.0118\ \text{m}$$

3. Correct length

$$L_{true} = 360 - 0.0112 - 0.0118 = 359.9770\ \text{m}$$

(If the sag in standardisation is ignored, so that the full sag of $-0.0240$ m is applied, the length is 359.9648 m.)

Answer: correct length of the line $= 359.977$ m.