Tape standardized in catenary: 24 m span

Method. A tape standardised in catenary reads 29.8940 m for its nominal 30 m when hanging at 50 N. To find the true tape length, the sag included in this standard value is added back. Then the 24 m span is corrected for tape length and for its own sag. The field pull is taken as 50 N (the standard pull).

Data

$w = 0.015 \times 9.81 = 0.14715$ N/m, $P = 50$ N, $L = 24$ m.

1. Sag at standardisation (30 m)

$$\frac{w^{2}\,l^{3}}{24\,P^{2}} = \frac{0.14715^{2}\times 30^{3}}{24\times 50^{2}} = 0.009744\ \text{m}$$

True tape length $l_0 = 29.8940 + 0.009744 = 29.9037$ m.

2. Tape length correction

$$C_a = 24\times\frac{29.9037 - 30}{30} = -0.0770\ \text{m}$$

3. Sag correction for the 24 m span

$$C_s = -\frac{0.14715^{2}\times 24^{3}}{24\times 50^{2}} = -0.0050\ \text{m}$$

4. Horizontal length

$$D = 24 - 0.0770 - 0.0050 = 23.9180\ \text{m}$$

Answer: horizontal length of the span $= 23.9180$ m.