Sag correction for 76.35 m line

Data

• Weight of 30 m tape $W = 0.7$ kg, so $w = 0.7/30 = 0.02333$ kg/m
• Pull $P = 5$ kg; recorded length $= 76.35$ m in three segments, taken as three equal spans of $76.35/3 = 25.45$ m

Sag correction per span

$$C_s = -\frac{w^{2}\,l^{3}}{24\,P^{2}} = -\frac{0.02333^{2}\times 25.45^{3}}{24\times 5^{2}} = -0.01496\ \text{m}$$

For the three spans

$$C_s = -3\times 0.01496 = -0.0449\ \text{m}$$

Corrected length

$$L = 76.35 - 0.0449 = 76.3051\ \text{m}$$

Answer: length of the line corrected for sag $= 76.3051$ m.

(If the three segments are instead read as two full 30 m tape lengths plus 16.35 m, the sag is $2\times\dfrac{0.7^{2}\times 30}{24\times 5^{2}} + \dfrac{0.02333^{2}\times 16.35^{3}}{24\times 5^{2}} = 0.0490 + 0.0040 = 0.0530$ m, and the corrected length is $76.297$ m.)