True length of 358.650 m line

Reading of the doubtful value. The scanned tape weight "0.05 kg/cm" is taken as 0.05 kg per metre (a 30 m tape then weighs 1.5 kg, a realistic value for steel tape of 0.04 cm$^2$ section). The other values are as printed.

Data

• $l = 30$ m; standard pull $P_0 = 5.5$ kg (tape fully supported); $A = 0.04$ cm$^2$; $E = 2.1\times 10^{6}$ kg/cm$^2$
• Measured distance $L = 358.650$ m; field pull $P = 8$ kg; tape supported at the ends only (catenary)
• $w = 0.05$ kg/m, so $W = 0.05\times 30 = 1.5$ kg
• Number of tapes: 11 full tapes (330 m) + a last part of 28.650 m. No temperature or tape-length error is given, so only pull and sag are applied.

1. Correction for pull

$$C_p = \frac{(P - P_0)\,L}{A\,E} = \frac{(8 - 5.5)\times 358.650}{0.04\times 2.1\times 10^{6}} = +0.0107\ \text{m}$$

2. Correction for sag

Full tape: $C_s = -\dfrac{W^{2}\,l}{24\,P^{2}} = -\dfrac{1.5^{2}\times 30}{24\times 8^{2}} = -0.043945$ m per tape.

11 full tapes: $-11\times 0.043945 = -0.48340$ m.

Last length of 28.650 m: $C_s = -\dfrac{w^{2}\,l^{3}}{24\,P^{2}} = -\dfrac{0.05^{2}\times 28.65^{3}}{24\times 8^{2}} = -0.03828$ m.

Total sag correction $= -0.48340 - 0.03828 = -0.5217$ m.

3. True length

$$L_{true} = 358.650 + 0.0107 - 0.5217 = 358.1390\ \text{m}$$

Answer: correct length of the line $= 358.139$ m.