Horizontal length of 30 m span, 50 m tape

Assumptions. The tape is standardised in catenary (50 m span), so the sag in its standard length is added back. The 30 m span is measured in catenary at the same tension, 5 kg. The coefficient of expansion is not given, so the steel value $\alpha = 12\times 10^{-6}$/$^\circ$C is assumed.

Data

$W = 0.68$ kg for 50 m, so $w = 0.0136$ kg/m; $P = 5$ kg; standard length $= 49.996$ m at $20^\circ$C; $T = 26^\circ$C; span $L = 30$ m.

1. Tape length

Sag at standardisation: $\dfrac{0.0136^{2}\times 50^{3}}{24\times 5^{2}} = 0.03853$ m. True tape length $= 49.996 + 0.03853 = 50.03453$ m.

$$C_a = 30\times\frac{50.03453 - 50}{50} = +0.0207\ \text{m}$$

2. Temperature

$$C_t = 12\times 10^{-6}\times 6\times 30 = +0.0022\ \text{m}$$

3. Sag for 30 m

$$C_s = -\frac{0.0136^{2}\times 30^{3}}{24\times 5^{2}} = -0.0083\ \text{m}$$

4. Horizontal length

$$D = 30 + 0.0207 + 0.0022 - 0.0083 = 30.0146\ \text{m}$$

Answer: horizontal length of the 30 m span $= 30.0146$ m.