Tape standardized in catenary: 20 m span

Reading of the data. A tape standardised in catenary reads 29.895 m between its end graduations when hanging in catenary at 50 N, so it is too short by 0.105 m as a hanging tape. The sag effect included in this standard must be accounted for to get the true length of the tape, then the 20 m span is corrected for its own sag. The field tension is taken equal to the standard tension, 50 N (not stated).

Data

• Mass per metre $= 0.015$ kg/m, so weight per metre $w = 0.015 \times 9.81 = 0.14715$ N/m
• Pull $P = 50$ N, standard catenary length of 30 m tape $= 29.895$ m
• Recorded span $L = 20$ m

Step 1: Sag of the 30 m tape at standardisation

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

Length of tape fully supported (flat):

$$l_0 = 29.895 + 0.009744 = 29.9047\ \text{m}$$

Step 2: Correction for tape length (for 20 m)

$$C_a = 20\times\frac{29.9047 - 30}{30} = -0.0635\ \text{m}$$

Step 3: Sag correction for the 20 m span

$$C_s = -\frac{w^{2}\,l^{3}}{24\,P^{2}} = -\frac{0.14715^{2}\times 20^{3}}{24\times 50^{2}} = -0.0029\ \text{m}$$

Step 4: Horizontal length

$$D = 20 - 0.0635 - 0.0029 = 19.9336\ \text{m}$$

Answer: horizontal length of the span $= 19.9336$ m.