Two-peg test style check: computing the correct far staff reading Peg A and peg B are 80 m apart. With a dumpy level set at the midpoint, staff readings are A = 1.32 and B = 1.56. Then the level is set 10 m from A on BA produced; staff readings are A = 1.11 and B = 1.39. What should be the correct staff reading at B from this second setup (after removing collimation error)?

Difficulty: Medium

Correct Answer: 1.345

Explanation:


Introduction / Context:
This is a classic two-peg test scenario used to detect and correct the line of collimation error of a level. When the instrument is placed midway between pegs, the difference in staff readings gives the true difference in elevation between the pegs. When the instrument is placed near one peg, unequal sight lengths reveal the line-of-sight error, allowing computation of the corrected far staff reading.


Given Data / Assumptions:

  • Distance AB = 80 m.
  • Midpoint setup readings: A = 1.32, B = 1.56 → true difference Δ_true = 1.56 − 1.32 = 0.24 m (B lower).
  • Second setup 10 m beyond A: distances IA = 10 m, IB = 90 m.
  • Observed readings: A = 1.11, B = 1.39 → observed difference Δ_obs = 0.28 m.


Concept / Approach:

The line of collimation error produces a reading error proportional to sight length. Let e be the error per 100 m added to a reading (positive if it increases with distance). The difference in errors between B and A is e * (IB − IA) / 100. Matching observed and true differences determines e, which is then used to correct the far reading.


Step-by-Step Solution:

Compute true elevation difference: Δ_true = 0.24 m.Compute observed difference for second setup: Δ_obs = 1.39 − 1.11 = 0.28 m.Error difference = Δ_obs − Δ_true = 0.28 − 0.24 = 0.04 m.Let e be error per 100 m in a staff reading; then 0.04 = e * (90 − 10) / 100 = e * 0.8 → e = 0.05 m per 100 m.Far sight B error = e * (IB/100) = 0.05 * 0.9 = 0.045 m; since Δ_obs > Δ_true, B's observed reading is too large.Correct B reading = 1.39 − 0.045 = 1.345 m.


Verification / Alternative check:

Check using A's correction: near sight error = e * 0.1 = 0.005 m. Corrected difference = (1.39 − 0.045) − (1.11 − 0.005) = 1.345 − 1.105 = 0.240 m, matching Δ_true.


Why Other Options Are Wrong:

1.435 and 1.425 do not satisfy the true difference constraint when combined with the corrected near reading.

“None of these” is false because 1.345 is consistent with the two-peg correction.


Common Pitfalls:

Using distance difference 80 m incorrectly; forgetting to apply corrections to both near and far readings; sign confusion when deciding whether to add or subtract the error.


Final Answer:

1.345

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