Chaining Errors – Situations Causing ~1 in 1000 Relative Error with a 20 m Chain Which of the following situations typically introduces an error on the order of 1 part in 1000 when using a 20 m chain for distance measurement?

Difficulty: Medium

Correct Answer: All of the above

Explanation:


Introduction:
Understanding the magnitude of common field errors helps surveyors judge when corrections are necessary. With a 20 m chain, small mishandlings can create significant relative errors near 1 in 1000, impacting layout and quantity computations if not controlled.


Given Data / Assumptions:

  • Nominal chain length = 20 m.
  • Small deviations in length, alignment, or slope are considered.
  • Relative error is approximated as error / measured length.


Concept / Approach:

Each listed situation yields an error roughly 20 mm over 20 m, which is 1/1000. Length error is direct. Alignment and slope errors can be estimated using small-angle approximations: the excess due to using a sloping or misaligned line rather than true horizontal/straight alignment is about d^2/(2L), which with the given numbers is near 0.02 m (20 mm) over 20 m.


Step-by-Step Solution:

Length error: 20 mm over 20 m ⇒ 0.02/20 = 1/1000.Alignment (one end 0.9 m off): error ≈ d^2/(2L) = 0.9^2/(40) ≈ 0.020 m ⇒ ~1/1000.Slope (0.9 m vertical over 20 m): horizontal shortfall ≈ h^2/(2L) = 0.9^2/40 ≈ 0.020 m ⇒ ~1/1000.Middle offset 0.45 m: similar magnitude by geometry of bowing/offsets.


Verification / Alternative check:

Exact computations using Pythagoras or alignment triangles confirm each is about 20 mm over 20 m, matching the 1 in 1000 scale.


Why Other Options Are Wrong:

Each single case indeed produces the specified magnitude; hence selecting any one alone would be incomplete. The correct holistic choice is “All of the above.”


Common Pitfalls:

Ignoring small alignment errors as “negligible”; forgetting to reduce slope distances to horizontal; overlooking periodic standardization of chains.


Final Answer:

All of the above

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