Applied physics · classical mechanics · thermodynamics
A 2-quart bottle, some tap water, and a bicycle pump can fly higher than a person is tall in under half a second. Underneath that is a genuine rocket-propulsion problem — Newton's laws, an ideal gas, and Bernoulli's equation, in a package simple enough to hold in your hand. This page derives the model, then lets you fly one: tune the water volume, launch pressure, and nozzle size and watch the optimum emerge.
01 · Anatomy
A water rocket is a sealed pressure vessel that happens to also be a projectile. Below the neck sits a pocket of compressed air; below that, the water that the air pushes out through a narrow orifice — and because the airframe is sealed all the way to the tip, that water fills the nose cone's interior too, not just the cylindrical body. The fins are the one part that carries no propellant at all — they exist only to keep the ejected water pointed straight down for as long as possible.
02 · The governing equations
Five relationships fully describe the flight: how the trapped air's pressure falls as it expands, how fast water leaves the orifice, how that flow becomes a force, how that force accelerates a rocket that is losing mass as it flies, and — underlying all of it — the momentum bookkeeping that explains why "just add more water" isn't the answer.
Step 1 — the compressed-air spring
The water leaves in a few tenths of a second — far too fast for heat to flow across the bottle wall — so the air's expansion is adiabatic rather than isothermal. As it fills the volume vacated by departing water, its pressure falls along:
where:
P air pressure at any instant during expansionV air volume at that same instantγ adiabatic index of air (1.4)P₀ initial absolute pressure (pump gauge + atmospheric)V₀ initial air volume = bottle − water volumeA smaller starting air pocket (i.e. more water) means the same volume of ejected water expands the gas by a larger relative amount — so pressure, and thrust, collapse faster.
Where γ comes from — Siméon Poisson, 1823
The exponent γ in Step 1 is named for the relation French mathematician Siméon Poisson derived for gases undergoing rapid, heat-free compression or expansion. For air — a diatomic gas, mostly N₂ and O₂ — kinetic theory puts γ at 7/5 = 1.4, the same value used throughout this page. It's a direct consequence of air molecules having 5 active degrees of freedom (3 translational, 2 rotational) to store energy in as they're squeezed.
Step 2 — the jet
Treat the water as inviscid and incompressible, and the falling air-water interface as far slower than the jet squeezing through the small orifice. The unsteady Bernoulli equation between interface and exit then collapses to a Torricelli-like form:
where:
v water exit (jet) velocityCd discharge coefficient (≈ 0.85)P instantaneous air pressure inside the bottlePatm atmospheric pressureρ water densityCd is a discharge coefficient — a real jet narrows slightly just past the opening (the "vena contracta") and loses a little more to viscosity, so it never quite matches the frictionless ideal.
The trade Bernoulli found — Daniel Bernoulli, 1738
Daniel Bernoulli published the pressure-speed relationship behind Step 2 in his 1738 book Hydrodynamica — the same principle that explains airplane lift and why a pinched garden hose sprays farther. It's really a statement of energy conservation for a moving fluid: speed up the flow, and pressure must drop to pay for it. Here that trade runs in reverse — the bottle's trapped pressure pays to speed the water up as it escapes.
Step 3 — Newton's third law
The rocket feels a reaction force equal to the rate of momentum the jet carries away. Because the water has already reached v by the time it exits — its pressure energy is already converted to kinetic energy — thrust reduces to:
where:
F thrust forceṁ mass flow rate of water leaving the orificev water exit velocity, from Step 2ρ water densityA orifice (nozzle) areaCd discharge coefficientP instantaneous air pressurePatm atmospheric pressureThrust scales directly with orifice area and with pressure — but not with how much water is left. A wider nozzle doesn't add momentum, it just delivers the same total momentum faster.
The law doing the work — Isaac Newton, 1687
Step 3 is Newton's third law — for every action, an equal and opposite reaction — applied to a stream of departing water instead of a cannonball or a collision. The rocket doesn't need anything to "push against" outside itself; it pushes against the water, and the water pushes back. That's also why rockets work in the vacuum of space, where a propeller or a jet engine's intake air have nothing to grip.
Step 4 — variable-mass dynamics
The rocket's own mass is falling as fuel — water — leaves, so Newton's second law needs its variable-mass (Meshchersky) form, with aerodynamic drag opposing motion:
where:
m instantaneous rocket mass (dry mass + remaining water)v rocket velocityF thrust, from Step 3 — drops to zero once the water is goneg gravitational accelerationρ air densityCd airframe drag coefficientA frontal area for dragm(t) itself is mdry (empty rocket mass — bottle + fins + nose) plus ρwVwater(t), the water still on board. Once the water is gone, F drops to zero and the rocket simply coasts under gravity and drag to apogee — the highest point of its trajectory, where vertical velocity passes through zero.
Step 5 — from acceleration to altitude
Step 4 gives acceleration, dv/dt — not height. Velocity is the integral of acceleration over time, and height is the integral of velocity over time, so getting from F(t) to a peak altitude means integrating twice:
where:
a(τ) instantaneous acceleration, from Step 4, at each moment τ between 0 and tv(τ) instantaneous velocity at each moment τv(0), h(0) velocity and height at the start of the intervalAn integral like this just means: slice the time from 0 to t into instants short enough that acceleration barely changes over any one of them, multiply each instant's acceleration by its sliver of time to get a tiny velocity bump, and add up every bump. Shrink the slices toward zero width and that sum becomes exact — that's what the ∫ symbol represents. Height comes from repeating the same slice-and-add process on velocity instead of acceleration, which is why reaching height costs two rounds of integrating rather than one.
During the powered phase, F(t), m(t), and drag are all changing together and the equation of motion has no closed-form solution — that's exactly why the live simulator above steps forward in tiny 0.5 ms increments (numerical integration) instead of evaluating a formula: it's literally carrying out the slice-and-add process above, one 0.5 ms sliver at a time. But once the water is gone, the equation simplifies enough to integrate by hand.
After burnout — the moment the water runs out — m(t) is just the constant mdry, and thrust is zero, so the equation of motion for the coast phase (rising, before apogee) reduces to:
where:
v rocket velocity during the coast phaseg gravitational accelerationk aggregate drag constantm dry rocket mass (constant after burnout)ρ air densityCd airframe drag coefficientA frontal area for dragThis is a separable differential equation in time — but height is what we actually want, and height and velocity are related by v = dh/dt. Swapping the independent variable with the chain rule, v dv/dh = dv/dt, turns it into a separable equation in height instead:
where:
dh an infinitesimal change in heightv rocket velocityg gravitational accelerationk aggregate drag constant, from abovem dry rocket massIntegrating from the burnout speed vb (where the coast begins) down to v = 0 (apogee) gives a closed-form coast-phase height gain:
where:
hcoast height gained during the unpowered coast phasem dry rocket massk aggregate drag constantg gravitational accelerationvb burnout velocity, read from Step 4's numerical integrationTotal apogee height is then hburnout (altitude already reached at burnout) plus this hcoast. The calculus only rescues the second, unpowered half of the flight — the first half stays a numerical problem, which is the same split the simulator makes internally.
03 · Live simulator
This numerically integrates the four equations above (explicit time-stepping, Δt = 0.5 ms) for a standard 2-quart bottle. Adjust the three variables this page is about — water fill, launch pressure, orifice diameter — and watch the flight profile respond in real time.
04 · Optimizing for height
Holding the other two sliders above fixed, here is how maximum altitude responds to sweeping each variable on its own. The shapes are not the same — and that difference is the whole design lesson.
Too little water and there's almost no reaction mass — the air just hisses out with nothing to push. Too much water and the air pocket starts too small, so pressure (and thrust) collapses within the first few inches of ejection, and the extra water's own weight has to be dragged along mostly unused. The optimum, computed live for your current pressure and orifice settings, sits at roughly a third of the bottle — the classic result reported in water-rocket literature.
Higher P0 raises exit velocity and thrust with no offsetting mass penalty — pressure is "free" energy density. The real limit is material: a standard 2-quart PET bottle typically bursts somewhere past 130 psi, and its safety margin erodes with every reuse, scratch, and UV-exposed hour. Competition and classroom launches generally stay at or below ~100–110 psi (shaded red above) for exactly that reason — this is a materials constraint, not a physics one.
Orifice area barely changes the total momentum delivered (that's set by pressure and water volume) — it changes how fast that momentum is delivered. A larger orifice empties the bottle quicker, which shortens the time gravity has to claw back velocity during the burn (a real "gravity loss," just as in a full-scale launch vehicle). Past a certain diameter the burn is already so short that this benefit saturates, while very large orifices strain the "orifice area ≪ chamber area" assumption the Bernoulli model depends on.
05 · Glossary
The equations above lean on a handful of terms that carry precise meaning. Here they are in plain English.
| Term | Plain-English meaning |
|---|---|
| Reaction mass | The stuff a rocket throws overboard to push itself forward — water, here. The more momentum it carries away, the more the rocket gains, per Newton's third law. |
| Dry mass | Everything on the rocket that isn't reaction mass: bottle, fins, nose cone. It never leaves, so it's dead weight the thrust has to haul the whole flight. |
| Gauge pressure | Pressure measured relative to the surrounding air, which is what a bike pump gauge reads. Add atmospheric pressure to get the absolute pressure the physics equations actually use. |
| Adiabatic | Happening too fast for heat to flow in or out. The trapped air's pressure drop is adiabatic because the whole flight is over in about a second — no time to exchange heat with the bottle wall. |
| Discharge coefficient (Cd) | A correction factor, a little under 1, for how a real jet of water falls short of a perfect, frictionless one — it narrows slightly right past the opening and loses a bit more to viscosity. |
| Numerical integration | Solving an equation by taking thousands of tiny time steps instead of one formula — what the live simulator does whenever no clean closed-form answer exists, as during the powered phase. |