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Correct length of base from sections

Short answer

Total measured length 231.624 m. Corrections: standardization -0.0185 m, temperature +0.0167 m, sag per section totalling -0.1604 m. Correct length = 231.462 m.

A steel tape weighing 0.68 kg was standardised on the flat and found to have length 49.996 m at 20°C, tension 5 kg. It was used in catenary at the same tension to measure a horizontal base, at average temperature 26°C, and the readings on the successive sections were 49.105, 49.373, 48.976, 49.817 and 34.353. What was the correct length of the line? Take α=12×10−6\alpha = 12\times10^{-6} per °C.

Assumptions. The tape (50 m) was standardised on the flat, and is used in catenary at the same tension of 5 kg, so there is no pull correction. Each reading is a separate suspended span. The total measured length is 49.105+49.373+48.976+49.817+34.353=231.62449.105 + 49.373 + 48.976 + 49.817 + 34.353 = 231.624 m.

Data

W=0.68W = 0.68 kg for 50 m, so w=0.0136w = 0.0136 kg/m; P=5P = 5 kg; standard length =49.996= 49.996 m at T0=20∘T_0 = 20^\circC; T=26∘T = 26^\circC; α=12×10−6\alpha = 12\times 10^{-6}/∘^\circC.

1. Tape length

Ca=231.624×(49.996−50)50=−0.0185 mC_a = \frac{231.624\times(49.996 - 50)}{50} = -0.0185\ \text{m}

2. Temperature

Ct=12×10−6×6×231.624=+0.0167 mC_t = 12\times 10^{-6}\times 6\times 231.624 = +0.0167\ \text{m}

3. Sag (for each section, Cs=−w2 l324 P2C_s = -\dfrac{w^{2}\,l^{3}}{24\,P^{2}})

Section length (m)Sag correction (m)
49.105−0.03650-0.03650
49.373−0.03710-0.03710
48.976−0.03621-0.03621
49.817−0.03811-0.03811
34.353−0.01250-0.01250

Total Cs=−0.1604C_s = -0.1604 m.

4. Correct length

L=231.624−0.0185+0.0167−0.1604=231.4617 mL = 231.624 - 0.0185 + 0.0167 - 0.1604 = 231.4617\ \text{m}

Answer: correct length of the line =231.462= 231.462 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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