Chain surveying—alignment error: If a 30 m chain (tape) is laid with its line diverging by a constant perpendicular offset d from the correct alignment, what is the error in the measured length along the intended line?

Introduction / Context:

In chain (tape) surveying, accurate linear measurement assumes the tape is laid exactly along the intended straight alignment. If, however, the tape is displaced sideways by a constant perpendicular distance d, the measured length becomes the hypotenuse of a small right triangle rather than the required horizontal projection along the true line. This creates a systematic alignment (sagitta)-type error even when the tape is otherwise correct and straight.

Given Data / Assumptions:

• Tape length L = 30 m.
• Perpendicular divergence from true line = d (constant over the measured length).
• Flat ground and negligible slope, sag, or temperature errors so we isolate alignment error only.

Concept / Approach:

If the tape is offset by d, the measured straight line is the hypotenuse, while the true length along alignment is the adjacent side. For a short constant offset, the true projection is D = √(L² − d²). The error in length (measured minus true) is therefore c = L − √(L² − d²). Using a binomial expansion, this simplifies to the well-known approximation c ≈ d²/(2L) for small d/L.

Step-by-Step Solution:

Verification / Alternative check:

Numerical illustration: With d = 0.3 m, c ≈ 0.3²/60 = 0.09/60 = 0.0015 m (1.5 mm), which matches calculator results to within rounding for such small offsets.

Why Other Options Are Wrong:

• d²/30: doubles the correct error; it ignores the 1/2 factor from the binomial expansion.
• d/30 and 30 d²: dimensionally inconsistent or grossly incorrect magnitude.
• Zero: alignment divergence always produces a positive error.

Common Pitfalls:

• Treating the offset as a random error; it is systematic if the divergence is constant.
• Confusing alignment error with slope or sag corrections, which require different formulas.

Final Answer:

d² / 60