Issue 011 · November 2026

Fourteen inches or twenty-eight: three riprap methods at the end of one barrel

Last issue ended at the outfall of a 24-inch barrel running full at 13.4 ft/s. I promised three riprap sizing methods that disagree about what that jet needs. They do, by a factor of two. But two of the three turn out to be the same equation, and most of the disagreement comes down to one input that's usually guessed: the tailwater.

TL;DR. At stage 107 the 24-inch RCP from issue 010 carries 42.2 cfs at 13.4 ft/s onto a free outfall. FHWA HEC-14's apron equation gives D50 = 17.3 in, using the minimum tailwater it allows (0.4D = 0.8 ft). The "0.02 / TW" formula in several state erosion-control manuals gives 17.5 in, because it's the same equation with g folded into the coefficient. Isbash, applied to the jet velocity, gives 14.2 in with the low-turbulence coefficient and 27.6 in with the high-turbulence one. Raise the tailwater from 0.8 ft to 2.0 ft and the HEC-14 stone drops from 17 in to 7 in. Which number to use depends on knowing the tailwater, more than on which manual you pick.

The jet

The outlet is the one from issue 010: a 4 ft × 4 ft riser draining through 80 ft of 24-inch RCP at 1 percent onto a free outfall. At pond stage 107 the barrel is in control and passes 42.2 cfs. Flowing full, that's

$$V = \frac{Q}{A} = \frac{42.2}{\pi (2.0)^2/4} = 13.4\ \text{ft/s}$$

That's fast enough to pull apart most soils within a few storms. The question is how big the stone at the outfall must be.

Method 1: the HEC-14 apron equation

FHWA HEC-14 (Hydraulic Design of Energy Dissipators for Culverts and Channels, 3rd ed.) sizes a riprap apron at a circular outlet with

$$D_{50} = 0.2\,D\left(\frac{Q}{\sqrt{g}\,D^{2.5}}\right)^{4/3}\frac{D}{TW}$$

with D the barrel diameter and TW the tailwater depth. When the tailwater is unknown or below 0.4D, the manuals that adopt it use TW = 0.4D. A free outfall has no tailwater to speak of, so TW = 0.8 ft:

$$D_{50} = 0.2(2.0)\left(\frac{42.2}{5.67 \times 5.66}\right)^{4/3}\frac{2.0}{0.8} = 1.44\ \text{ft} = 17.3\ \text{in}$$

Method 2: the "0.02 / TW" formula

Several state erosion-control standards (New Jersey's and DelDOT's, for two) write the apron equation as

$$d_{50} = \frac{0.02}{TW}\left(\frac{Q}{D}\right)^{4/3}\qquad \text{(ft, cfs)}$$

which gives 0.02/0.8 × (21.1)4/3 = 1.46 ft = 17.5 in. Expand the HEC-14 form and you get 0.2/g2/3 × (Q/D)4/3/TW. With g = 32.2, the coefficient is 0.0198. So methods 1 and 2 are the same equation: one keeps the units in the formula and the other folds them into the coefficient. The 0.2 in-difference is just 0.02 against 0.0198.

That's worth knowing when a reviewer asks why your calc doesn't follow the local manual. If the manual uses the 0.02 form, it agrees with HEC-14.

Method 3: Isbash on the jet velocity

Isbash's relation sizes a stone that a stream of given velocity won't move:

$$D = \frac{V^2}{2g\,C^2\,(S_s - 1)}$$

with Ss = 2.65 for typical stone. It comes with two coefficients: C = 1.20 for stones embedded in a bed with low turbulence, and C = 0.86 for exposed stones in high turbulence. Here D is read as D50, as most manuals do.

Isbash coefficientD50
C = 1.20 (low turbulence)14.2 in
C = 0.86 (high turbulence)27.6 in

The jet coming out of a barrel onto a flat apron is close to the definition of high turbulence, at least for the first few diameters. If you take Isbash at face value, the 0.86 coefficient governs and the stone is nearly twice the HEC-14 size.

Why they disagree

The apron equation is empirical. It was fitted to laboratory tests of riprap aprons downstream of culverts, so the spreading of the jet over the apron is built into the coefficient. Isbash knows nothing about spreading: it takes the full outlet velocity as the velocity on the stone. For the first stones at the barrel lip that's roughly right. Two or three diameters downstream it's conservative. So the two methods answer different questions. HEC-14 sizes the apron as a whole, and Isbash with C = 0.86 sizes the stone right at the jet.

The bigger lever is the tailwater, which sits in the denominator of the apron equation:

Tailwater TWHEC-14 D50"0.02/TW" d50
0.8 ft (0.4D, the floor)17.3 in17.5 in
1.0 ft13.8 in14.0 in
1.5 ft9.2 in9.3 in
2.0 ft (= D)6.9 in7.0 in

Going from no tailwater to a full pipe's depth of it cuts the stone by a factor of 2.5. On a real job the tailwater is the normal depth in the receiving channel at the same flow, and it can be computed. A channel with that normal depth needs about 7-in stone. The same outfall dumping onto a flat field needs 17-in stone, or 28-in by Isbash. That's the actual design decision. The choice of manual barely matters next to it.

What I'd put on the plans

Running it yourself

New: PDH courses written the same way

I've started writing continuing-education courses in the style of this newsletter: worked examples where every number is computed, a graded quiz, and a certificate of completion. The first is free: a 1-PDH ethics course built on eight case studies from everyday civil and water-resources practice. Many states require ethics hours each renewal. Three 2-PDH technical courses are $29 each: detention pond routing (it includes the barrel check from issue 010), open channel design (it includes riprap linings), and storm sewer design. Most state boards don't pre-approve providers, so check your own board's rules (Florida and New York require approved providers).

— Michael Flynn, PE

More from the Hydraulic Notebook

← Previous Issue 010 — Nine feet of headwater nobody drew: when the barrel takes over a pond outlet

← Issue 006 Your low-flow orifice is a weir half the time

← Issue 003 Why your Manning's n estimate is probably wrong (and why it matters more than you think)

Size the pieces: Riprap sizing calculator · Culvert calculator

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