Discussion
If the width of bench terrace is W, drop D and existing land slope S; then for 150% batter slope, the drop D will be
Answer: D
The drop is \(D = \frac{3\,W\,S}{(300 - 2\,S)}\).
Take the batter slope as 150%, a rise of 1.5 for each 1 horizontal, so the riser of height \(D\) takes up a horizontal width of \(D/1.5 = 2D/3\). On the original land slope \(S\)% the horizontal distance between terraces is \(100D/S\), and it equals the terrace width plus the riser width:
\[ \begin{aligned} \frac{100D}{S} &= W + \frac{2D}{3} \\ D\left(\frac{300 - 2S}{3S}\right) &= W \\ D &= \frac{3WS}{300 - 2S} \end{aligned} \]
In general \(D = \frac{WS}{100 - SU}\), where \(U\) is the horizontal run per unit rise of the riser. \(\frac{WS}{100 - S}\) is the case \(U = 1\), a 100% (1:1) batter, and \(\frac{2WS}{200 - S}\) is \(U = 1/2\), a 200% batter. \(\frac{WS}{100}\) ignores the riser entirely, which only holds for a vertical riser.
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