Chaining on Slopes – Approximate Slope Correction per 100 Links for a Gradient of 1 in n If a 100-link chain (20 m) is laid along a uniform slope with a rise of 1 unit in n horizontal units (tan θ = 1/n), what is the approximate correction per chain length to reduce the sloping length to horizontal (in links)?

Introduction / Context:
When a chain is laid along a slope, the measured sloping distance exceeds the horizontal projection. For mapping and setting out, distances must be reduced to the horizontal. This question asks for the commonly used approximate correction per 100 links for a slope of 1 in n, a staple result in elementary surveying.

Given Data / Assumptions:

• Chain length along slope L = 100 links.
• Gradient 1 in n ⇒ tan θ = 1/n with small θ (n is reasonably large).
• We need the correction to subtract from the sloping length to get horizontal length.

Concept / Approach:

The exact horizontal length is H = L * cos θ, so the correction per chain is C_exact = L − H = L (1 − cos θ). For small θ, use cos θ ≈ 1 − θ^2/2 and tan θ ≈ θ (in radians). With tan θ = 1/n, take θ ≈ 1/n. Then 1 − cos θ ≈ θ^2/2 ≈ 1/(2 n^2). Therefore, C_approx ≈ L * (1/(2 n^2)). With L = 100 links, C_approx ≈ 100/(2 n^2) = 50/n^2 links. This is the standard approximate slope correction per 100 links to be subtracted from the measured sloping length.

Step-by-Step Solution:

Verification / Alternative check:

If n = 20, C_approx = 50/400 = 0.125 links (≈ 0.025 m), consistent with practical expectations for gentle slopes over 20 m.

Why Other Options Are Wrong:

100/n^2 doubles the correction; 100 (1 − cos θ) is exact form but the question asks for the approximate value in terms of n; 100 (sec θ − 1) relates to forward shift in stepping, not horizontal reduction; 25/n has wrong dependence on n.

Common Pitfalls:

Confusing the forward shift formula with reduction to horizontal; forgetting that the correction is to be subtracted; mixing degrees with radians in approximations.

Final Answer:

50 / n^2 links (to be subtracted)