Correct horizontal distance, 28.00 m measured

Method. The tape was standardised in catenary, so its stated length includes the effect of sag. When used fully supported there is no sag, so the sag in the standard must be added back to get the true tape length. $g$ is not needed (all in kg).

Data

• Standard: 30.005 m at $20^\circ$C and $P_0 = 5$ kg (catenary); field: $T = 25^\circ$C, $P = 12$ kg, fully supported; $L = 28.00$ m
• $A = 0.02$ cm$^2$, $E = 2.0\times 10^{6}$ kg/cm$^2$, $\alpha = 11\times 10^{-6}$/$^\circ$C, $w = 0.022$ kg/m

1. Tape length

Sag at standardisation: $\dfrac{0.022^{2}\times 30^{3}}{24\times 5^{2}} = 0.02178$ m. True tape length $= 30.005 + 0.02178 = 30.02678$ m.

$$C_a = 28.00\times\frac{30.02678 - 30}{30} = +0.0250\ \text{m}$$

2. Temperature

$$C_t = 11\times 10^{-6}\times(25 - 20)\times 28 = +0.0015\ \text{m}$$

3. Pull

$$C_p = \frac{(12 - 5)\times 28}{0.02\times 2.0\times 10^{6}} = +0.0049\ \text{m}$$

4. Sag

Zero (fully supported).

5. Correct horizontal distance

$$D = 28.00 + 0.0250 + 0.0015 + 0.0049 = 28.0314\ \text{m}$$

Answer: correct distance $= 28.0314$ m.