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Tape corrections for 66 m in three spans

Short answer

Add sag at standardisation (0.0405 m) to 30.008 m to get true tape length 30.0485 m. Apply tape length correction (+0.1067 m) and sag correction for three 22 m spans (-0.0479 m) to 66 m, yielding corrected length 66.0588 m.

A 30 m steel tape was standardized in catenary condition under a pull of 5 kg and found to be 30.008 m. This tape was used to measure a distance of 66 m in three equal spans in catenary conditions at a pull of 5 kg. The weight of tape was 30 gm/m. Apply necessary tape correction for the measured length of line.

Method. The tape was standardised in catenary, so the sag in its standard length is added back to find the true tape length. Then the sag for the three 22 m spans is applied.

Data

  • w=0.03w = 0.03 kg/m; P=5P = 5 kg (same as standard); L=66L = 66 m in three equal spans of 22 m
  • Standard length of the 30 m tape in catenary =30.008= 30.008 m

1. Tape length

Sag at standardisation: 0.032×30324×52=0.0405\dfrac{0.03^{2}\times 30^{3}}{24\times 5^{2}} = 0.0405 m. True tape length =30.008+0.0405=30.0485= 30.008 + 0.0405 = 30.0485 m.

Ca=66×30.0485−3030=+0.1067 mC_a = 66\times\frac{30.0485 - 30}{30} = +0.1067\ \text{m}

2. Sag correction

Each span: 0.032×22324×52=0.015972\dfrac{0.03^{2}\times 22^{3}}{24\times 5^{2}} = 0.015972 m.

Cs=−3×0.015972=−0.0479 mC_s = -3\times 0.015972 = -0.0479\ \text{m}

3. Corrected length

Ltrue=66+0.1067−0.0479=66.0588 mL_{true} = 66 + 0.1067 - 0.0479 = 66.0588\ \text{m}

Answer: corrected length of the line =66.0588= 66.0588 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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