Tape standardized in catenary: 23 m span

Method. A tape standardised in catenary has the sag effect included in its stated length. We add this sag back to find the true tape length, then correct the 23 m span for tape length and for its own sag. The field pull is taken equal to the standard pull, 50 N.

Data

$w = 0.015\times 9.81 = 0.14715$ N/m, $P = 50$ N, $L = 23$ m, standard length $= 29.8950$ m.

1. Sag at standardisation (30 m)

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

True tape length $= 29.8950 + 0.009744 = 29.9047$ m.

2. Tape length correction

$$C_a = 23\times\frac{29.9047 - 30}{30} = -0.0730\ \text{m}$$

3. Sag correction for the 23 m span

$$C_s = -\frac{0.14715^{2}\times 23^{3}}{24\times 50^{2}} = -0.0044\ \text{m}$$

4. Horizontal length

$$D = 23 - 0.0730 - 0.0044 = 22.9226\ \text{m}$$

Answer: horizontal length of the span $= 22.9226$ m.