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Drop Spillway Calculator

Straight-drop principal spillway and grade-control structure per NRCS NEH-11 and Rand (1955). Computes discharge over the crest, impinging-jet geometry from the drop number, conjugate depth required at the apron, and total stilling-basin length. Use this for screening sizing of small embankment-dam principal spillways.

ft
ft
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cfs
cfs/ft
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Rand 1955 / NRCS NEH-11 power-law fits for straight-drop geometry. End sill, blocks, and rip-rap downstream are sized separately. Apply at d < ~8 ft for principal spillways; for larger drops use a Type II/III stilling basin.

$$ q = \frac{Q}{L} = C_d H^{3/2}, \quad y_c = (q^2/g)^{1/3}, \quad D_N = \frac{q^2}{g d^3} $$
$$ \frac{x_a}{d} = 4.30\, D_N^{0.27}, \quad \frac{y_2}{d} = 1.00\, D_N^{0.22} $$
q unit discharge · yc brink (critical) depth · DN drop number · xa distance to plunge point · y2 conjugate (sequent) depth required for the jump.

How to use this calculator

Enter the crest length L, head over crest H during the design event, and drop height d. The calculator computes free-fall discharge over the crest using the chosen weir formula, then applies Rand's drop-number relations to size the stilling basin downstream. The stilling-basin length Lbasin is the minimum apron required to contain the impinging jet plus the hydraulic jump that follows it. Confirm tailwater depth in the receiving channel exceeds y2 — if it doesn't, the jump will sweep out and the basin won't function. Add an end sill at Lbasin sized at 0.07·d (NRCS NEH-11 detail).

Drop-number scaling

Rand 1955 / NRCS NEH-11 — straight-drop hydraulic jump geometry
DNxa/dy2/dLbasin/d
0.0011.040.302.8
0.0051.510.424.0
0.011.820.494.7
0.052.660.686.7
0.103.210.797.9
0.504.711.1011.3
1.005.681.2813.4

Computed from xa/d = 4.30·DN0.27, y2/d = 1.00·DN0.22, Lbasin = xa + 6·y2. Source: Rand, W. (1955) Flow Geometry at Straight Drop Spillways, ASCE Paper 791. NRCS NEH-11 (Drop Spillways).

Worked example

Example — small NRCS sediment-dam principal spillway

Given: L = 20 ft, H = 2 ft over a sharp crest, d = 6 ft drop to apron.
q = 3.33 · 21.5 = 9.42 cfs/ft → Q = 188 cfs
yc = (9.42² / 32.2)1/3 = 1.37 ft
DN = 9.42² / (32.2 · 6³) = 0.0128
xa/d = 4.30 · 0.01280.27 = 1.94 → xa = 11.6 ft
y2/d = 1.00 · 0.01280.22 = 0.51 → y2 = 3.1 ft (required tailwater)
Lbasin = 11.6 + 6 · 3.1 ≈ 30 ft total apron length

References: Rand, W. (1955). Flow Geometry at Straight Drop Spillways. ASCE Proceedings 81 (Paper 791). USDA NRCS, NEH-11 (Drop Spillways). Chow, V.T. (1959), Open-Channel Hydraulics, §15-9.

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