How this instrument works
A hydraulic jump is the abrupt, turbulent thickening that happens when a fast, shallow sheet of water — moving faster than its own surface waves can travel, a supercritical flow — is forced to slow down and can no longer hold that thin, racing state. Instead of settling gradually, the flow jumps: within a channel length of a few metres it turns from a smooth, glassy sheet a few centimetres deep into a churning, aerated roller several times as thick. A small version is visible in any kitchen sink, where the thin fast disc spreading out from under the tap suddenly thickens into a slower ring; engineers watch the same event, at a much larger scale, at the foot of a spillway chute or below a sluice gate.
The formula behind this instrument, y₂ ⁄ y₁ = ½(√(1 + 8Fr₁²) − 1), is not built from Bernoulli's equation, even though the jump looks like the kind of problem Bernoulli usually solves. That is deliberate: a hydraulic jump destroys mechanical energy on purpose, turning it into the heat and noise of turbulence, so energy is exactly the quantity not conserved here. What does hold, across the short length of the roller, is momentum — bed friction and the channel's weight component act over too short a reach to matter, so the momentum function upstream equals the momentum function downstream. Setting those two expressions equal and solving for the depth ratio is how the French engineer Jean-Baptiste Bélanger reached this result in 1828, and the formula still carries his name.
The ratio only describes a real event when Fr₁ is genuinely greater than 1. Feed in a subcritical value and the algebra still returns a number, but it describes a spontaneous deepening running backwards — subcritical flow thinning on its own into supercritical flow — which never happens, because that direction needs energy added rather than destroyed. The formula also assumes a horizontal, prismatic, rectangular channel with hydrostatic pressure either side of the jump; it says nothing about where the jump sits, how long its roller extends (roughly four to six times the downstream depth, from separate laboratory correlations), or how violently it churns — that last question is answered by Fr₁ alone, through a classification used to design spillway stilling basins.
- Enter the Upstream Froude number — flow speed divided by wave speed just before the jump; a real jump needs it above 1.
- Read the Sequent depth ratio (y₂ ⁄ y₁) — how many times deeper the subcritical flow becomes immediately after the jump.
- Multiply that ratio by your known upstream depth, y₁, to convert it into an actual downstream depth in metres or feet.
- Compare the Upstream Froude number to the standard jump classes — 1.7 to 2.5 weak, 2.5 to 4.5 oscillating, 4.5 to 9 steady, above 9 strong — to judge how rough the jump will be.
Worked example — Froude number 3 on a spillway apron
Take a spillway apron running 0.5 m deep at the toe, with the chute accelerating the sheet to 6.643 m/s — an upstream Froude number of exactly 3, since Fr₁ = 6.643 ⁄ √(9.80665 × 0.5) = 3. Feed Fr1 = 3 into the formula and the sequent depth ratio comes out to 3.772001872658765, which this instrument displays as 3.772002: the flow leaving the jump ends up very nearly three and three-quarters times as deep as the flow that entered it.
Multiply that ratio by the 0.5 m upstream depth and the downstream depth works out to 1.886 m, while continuity slows the current to about 1.761 m/s — a downstream Froude number near 0.410, comfortably subcritical. The specific energy head falls from 2.75 m to about 2.044 m across the jump, a loss near 25.7 percent, spent on the turbulence and spray of the roller rather than carried on downstream — precisely the energy a stilling basin is built to absorb instead of the unlined riverbed below it.
Questions
Why is this formula derived from momentum instead of energy?
Because energy is exactly what a hydraulic jump destroys — it is not conserved across the roller, so Bernoulli's equation cannot be applied over it. Momentum is conserved instead, since bed friction and the channel's weight component act over too short a distance to matter. Setting the upstream and downstream momentum function equal, then solving for the depth ratio, is how Jean-Baptiste Bélanger derived this exact formula in 1828.
What happens if I enter a Froude number below 1?
The algebra still returns a number, but it no longer describes a real hydraulic jump. A jump only forms when the approach flow is genuinely supercritical, Fr1 greater than 1; below that, the formula would describe subcritical flow spontaneously thinning into supercritical flow, which never happens on its own because it would require adding energy rather than losing it.
Why does the jump look different at different Froude numbers?
Because the turbulence scales with Fr1. Standard hydraulic references classify the roller as undular from about 1 to 1.7, weak and smooth from 1.7 to 2.5, oscillating and wave-prone from 2.5 to 4.5, and steady — the range engineers favour for stilling-basin design — from 4.5 to 9. A Fr1 of 3, like the worked example here, sits in that unsettled oscillating band, one reason designers often push a basin's Froude number higher rather than leave it there.
How much energy does a hydraulic jump actually destroy?
It depends on Fr1, but it can be substantial: at Fr1 = 3, the specific energy head drops from 2.75 m to about 2.04 m for a 0.5 m upstream depth, a loss near 25.7 percent. That destroyed energy becomes heat and the noise and spray of the turbulent roller — the entire reason engineers place a jump inside a concrete stilling basin below a spillway, so the energy is spent there instead of scouring the natural riverbed downstream.
Does this formula work for any channel shape?
No — it assumes a horizontal, prismatic, rectangular channel with hydrostatic pressure on both sides of the jump, the case Bélanger originally solved. A sloped chute or a trapezoidal canal section changes the momentum balance and needs a modified form with extra terms for the weight and friction this simple ratio ignores; treat this result as the flat-bed baseline the more complex cases build on.