Tape sag correction (supported at 10 m intervals): A 50 m steel tape weighs 4 kg in total and is used with a pull of 5 kgf, supported at 10 m intervals. What is the approximate total sag correction for one 50 m tape length?

Difficulty: Medium

Correct Answer: 0.053 m

Explanation:


Introduction / Context:
When measuring distances with a tape carried between supports, the tape sags and the measured length exceeds the true horizontal distance. A sag correction must be subtracted. The magnitude depends on span length, tape weight per unit length, and applied pull. Field practice often supports a long tape at equal intervals to reduce the correction.


Given Data / Assumptions:

  • Total tape length L_total = 50 m; total weight W_total = 4 kgf → weight per meter w = 4 / 50 = 0.08 kgf/m.
  • Pull (tension) P = 5 kgf, assumed equal on each span.
  • Tape supported at equal 10 m intervals → number of spans n = 5; span length l = 10 m.
  • Use standard sag correction per span: c_span = (w^2 * l^3) / (24 * P^2); total correction c_total = n * c_span.


Concept / Approach:
Sag makes the measured distance too long. The correction is derived from the catenary approximation for small sag and uniform load. Supporting at intermediate points reduces l and thus the cubic dependence on span length, dramatically lowering the total correction compared to an end-supported 50 m span.


Step-by-Step Solution:

Compute w = 0.08 kgf/m.For each 10 m span: c_span = (0.08^2 * 10^3) / (24 * 5^2) = (0.0064 * 1000) / (24 * 25) = 6.4 / 600 = 0.010667 m.Total correction over 5 spans: c_total = 5 * 0.010667 m = 0.05333 m.Rounded to the nearest option: 0.053 m (to be subtracted from the measured distance).


Verification / Alternative check:
If the tape were supported only at its ends (one 50 m span), c = (0.08^2 * 50^3) / (24 * 25) ≈ 1.33 m, showing why intermediate supports are used; the 10 m support interval yields a much smaller, practical correction.


Why Other Options Are Wrong:

  • 0.043 m, 0.063 m, 0.073 m, 0.083 m: These do not match the computed 0.053 m for 10 m spans with the given w and P.


Common Pitfalls:
Applying the end-support formula to a multi-support setup; mixing SI units (N) with kgf without consistency; forgetting to subtract the correction from the measured length.


Final Answer:
0.053 m

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