Horizontal length of 16 m, 29.8850 m tape

Method. The tape was standardised in catenary at 40 N, so its true (fully supported) length is found by adding back the sag in the standard. It is then used fully supported at the same tension (40 N), so there is no sag in the field.

Data

$w = 0.015\times 9.81 = 0.14715$ N/m, $P = 40$ N, standard length $= 29.8850$ m, $L = 16$ m.

1. Sag at standardisation (30 m)

$$\frac{w^{2}\,l^{3}}{24\,P^{2}} = \frac{0.14715^{2}\times 30^{3}}{24\times 40^{2}} = 0.015225\ \text{m}$$

True tape length $= 29.8850 + 0.015225 = 29.9002$ m.

2. Tape length correction

$$C_a = 16\times\frac{29.9002 - 30}{30} = -0.0532\ \text{m}$$

3. Horizontal length

Fully supported, so the sag correction is zero:

$$D = 16 - 0.0532 = 15.9468\ \text{m}$$

Answer: horizontal length of the 16 m $= 15.9468$ m.