Difficulty: Medium
Correct Answer: 28.28 m
Explanation:
Introduction / Context:
Offsets from a chain line carry two principal error sources: (i) error in setting the perpendicular direction and (ii) error in measuring the offset length. For a given drawing scale and allowable displacement on the plan, the maximum offset length must be limited so the combined positional error does not exceed the plotting tolerance.
Given Data / Assumptions:
Concept / Approach:
Let L be the offset length (ground). With 1 in 40 accuracy, the standard deviation/limit of both the directional (from setting perpendicular) and the linear measurement components can be taken as ≈ L/40 each (conventional exam assumption when only a single accuracy figure is given). The resultant ground-position error ≈ √[(L/40)² + (L/40)²] = (L/40)√2. This must be ≤ 1 m (ground equivalent of 0.05 cm at the stated scale). Solve for L.
Step-by-Step Solution:
Verification / Alternative check:
Even if the two components do not exactly equal each other, using the combined error as above yields a conservative limit close to tabulated values in surveying texts for similar scales and accuracies.
Why Other Options Are Wrong:
14.14 m — overly conservative (half of the allowable limit).
200 m — far exceeds the limit; plan displacement would be much larger than permitted.
None of these — incorrect because 28.28 m satisfies the conditions.
Common Pitfalls:
Forgetting to convert the plotting tolerance to ground units; adding errors arithmetically instead of by root-sum-square when they are independent; ignoring the angular component altogether.
Final Answer:
28.28 m
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