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Distance measurement on sloping ground

Short answer

Horizontal distance on sloping ground is obtained by stepping (holding the tape horizontal with a plumb bob) or by measuring slope length L and reducing it: D = √(L² − h²) or D = L cosθ. Hypotenuse allowance adds an extra length so that each chain length equals one horizontal chain.

Explain distance measurement in sloping ground.

Horizontal distance is required for plotting, but on sloping ground the tape follows the slope, so the measurement must be reduced to the horizontal. There are two ways: direct (by stepping) and indirect (measuring along the slope and applying a correction).

1. Direct method: stepping

The line is measured in short horizontal steps, as shown. The tape is held horizontal, with one end resting on the ground and the other end transferred to the ground by a plumb bob. The line is measured downhill when possible, so that the plumbed end is held at a convenient height. The steps are added to get the total horizontal distance.

  A o------- step 1 (tape horizontal)
            | plumb bob
            o------- step 2
                    |
                    o------- step 3  -> B

Limitations: it is slow, plumbing errors build up, and it is suited only to short steps on steep ground.

2. Indirect method: measuring along the slope and calculating

The slope length LL is measured and the horizontal length DD is found by one of the following:

  • Height difference hh from levelling: D=L2−h2D = \sqrt{L^{2} - h^{2}}, or the slope correction Ch=−h22LC_h = -\dfrac{h^{2}}{2L} (approx.).
  • Slope angle θ\theta measured with a clinometer or theodolite: D=Lcos⁡θD = L\cos\theta, correction Ch=−L(1−cos⁡θ)C_h = -L(1-\cos\theta).
  • Hypotenuse allowance: used in chain surveying on slopes. At the end of each chain length laid along the slope, the arrow is set forward by an extra length (the allowance), so that each chain length then corresponds to exactly one chain length (30 m) horizontally. For a slope θ\theta and a 30 m chain, the allowance is 30sec⁡θ−30=30(sec⁡θ−1)30\sec\theta - 30 = 30(\sec\theta - 1). For example, at θ=10∘\theta = 10^\circ the allowance is 0.46 m.

Example

Slope length L=30L = 30 m and h=2h = 2 m. Then D=900−4=29.933D = \sqrt{900 - 4} = 29.933 m (correction −0.067-0.067 m). The approximate formula gives −4/60=−0.067-4/60 = -0.067 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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