Tape corrections for 86 m, 15 kg pull

Method. A tape standardised in catenary has the sag of the standardisation span built into its stated length. We first add this sag back to get the tape's true length, then apply the field sag for supports every 10 m.

Data

• $w = 30$ g/m $= 0.03$ kg/m; pull $P = 15$ kg (same as standard); $L = 86$ m
• Standard: 20 m tape $= 20.006$ m in catenary (one 20 m span)
• Field spans: 8 spans of 10 m + 1 span of 6 m

1. Tape length

Sag in standardisation (20 m span):

$$\frac{w^{2}\,l^{3}}{24\,P^{2}} = \frac{0.03^{2}\times 20^{3}}{24\times 15^{2}} = 0.00133\ \text{m}$$

True (flat) tape length $= 20.006 + 0.00133 = 20.00733$ m.

$$C_a = \frac{86\,(20.00733 - 20)}{20} = +0.0315\ \text{m}$$

2. Sag correction in the field

Each 10 m span: $\dfrac{0.03^{2}\times 10^{3}}{24\times 15^{2}} = 0.000167$ m; for 8 spans $= 0.001333$ m. The 6 m span: $\dfrac{0.03^{2}\times 6^{3}}{24\times 15^{2}} = 0.000036$ m.

$$C_s = -(0.001333 + 0.000036) = -0.0014\ \text{m}$$

Temperature and pull corrections are zero (same as standard, no data).

3. Correct length

$$L_{true} = 86 + 0.0315 - 0.0014 = 86.0302\ \text{m}$$

Answer: corrected length of the line $= 86.0302$ m.