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Tape corrections for 86 m, supports at 15 m

Short answer

Add standard sag 0.0045 m to 30.006 m to obtain true tape 30.0105 m, apply tape length correction +0.0301 m, subtract field sag -0.0030 m. Corrected length is 86.027 m.

A 30 m steel tape was standardized in catenary conditions under a pull of 15 kg and found to be 30.006 m. This tape was used to measure a distance of 86 m in catenary conditions at a pull of 15 kg. Supports were provided at every 15 m. The weight of the tape was 30 gm/m. Apply necessary tape corrections for the measured length of line.

Method. The tape was standardised in catenary, so first the sag included in its standard length is added back to find its true length. Then the sag for the supports at every 15 m is applied.

Data

  • w=30w = 30 g/m =0.03= 0.03 kg/m; P=15P = 15 kg (same as standard); L=86L = 86 m
  • Standard: 30 m tape =30.006= 30.006 m in catenary
  • Field spans: 5 spans of 15 m + 1 span of 11 m

1. Tape length

Sag in standardisation (30 m span):

w2 l324 P2=0.032×30324×152=0.0045 m\frac{w^{2}\,l^{3}}{24\,P^{2}} = \frac{0.03^{2}\times 30^{3}}{24\times 15^{2}} = 0.0045\ \text{m}

True tape length =30.006+0.0045=30.0105= 30.006 + 0.0045 = 30.0105 m.

Ca=86 (30.0105−30)30=+0.0301 mC_a = \frac{86\,(30.0105 - 30)}{30} = +0.0301\ \text{m}

2. Sag correction

15 m span: 0.032×15324×152=0.000563\dfrac{0.03^{2}\times 15^{3}}{24\times 15^{2}} = 0.000563 m; for 5 spans =0.002813= 0.002813 m. 11 m span: 0.032×11324×152=0.000222\dfrac{0.03^{2}\times 11^{3}}{24\times 15^{2}} = 0.000222 m.

Cs=−(0.002813+0.000222)=−0.0030 mC_s = -(0.002813 + 0.000222) = -0.0030\ \text{m}

3. Correct length

Ltrue=86+0.0301−0.0030=86.0271 mL_{true} = 86 + 0.0301 - 0.0030 = 86.0271\ \text{m}

Answer: corrected length of the line =86.0271= 86.0271 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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