Ideal Hover Power (Momentum Theory)

The power is the easy part. The rotor pushes with thrust $T$ on air moving through the disc at $v_i$, so the work rate is $P = T v_i$, and substituting the induced velocity gives $P = T^{3/2}/\sqrt{2\rho A}$. That is the whole derivation. What comes out of it is one of the more consequential exponents in engineering.

Hover power goes as thrust to the three-halves power, and that is why a helicopter cannot be scaled up gracefully. Double the weight and the power required does not double, it rises by $2^{3/2} = 2.83$. Now put that beside the square-cube law that governs any structure: make an aircraft twice as long in every dimension and its weight grows eightfold while its disc area grows only fourfold, so its disc loading doubles, and the power per unit of weight rises by another $\sqrt{2}$ on top of the weight increase itself. The penalties compound. A fixed-wing aircraft does not suffer this — in cruise its lift is nearly free and the engine only pays for drag — which is why airliners are the size they are and heavy-lift helicopters are not.

Rearrange the same relation per unit of thrust and the design lever appears in one line: $P/T = \sqrt{DL/2\rho}$. The power it costs to hold up each newton depends on nothing except the disc loading and the air density, each under a square root. Low disc loading is the only lever that helps, and it helps slowly. Four times the disc area halves the power per unit of weight. That single sentence explains a great deal that otherwise looks like styling: it is why a Chinook carries its load on two enormous rotors rather than one compact one, why every heavy-lift helicopter has the longest blades that can be built and transported, why the Mi-26 has an eight-blade rotor thirty-two metres across, and why a quadcopter — four small discs, very high loading for its size — is aerodynamically dreadful, flying for twenty minutes on a battery that would keep an equivalent fixed-wing model up for two hours. The quadcopter is not badly designed. It is trading hover efficiency for mechanical simplicity and control authority, and this equation is the bill.

The density term works the same way and in the cruellest possible direction. Thin air on a hot, high day raises the ideal power by $1/\sqrt{\rho}$ at exactly the moment a turbine engine has least to give, which is why hover ceilings are quoted in and out of ground effect and why density altitude, not elevation, is what the performance chart is indexed on.

Now the honesty, because it matters more than the arithmetic. This number is a FLOOR, not a prediction. It is induced power alone — the energy that ends up in the downwash. A real rotor also drags its blades edgewise through the air and pays profile power for it whether it is lifting or not; loses the outer few percent of the disc to tip loss; leaves swirl in the wake; and distributes thrust unevenly along the radius, which costs more induced power than the uniform ideal. None of that is in the equation, because none of it was in the actuator disc. To get a real number, divide by a figure of merit — see the next page — of roughly 0.70 to 0.80. Then add seven to twelve percent for a conventional tail rotor, another two to four for the transmission, and leave the engine something in reserve, because a hover with no excess power is a hover you cannot leave.

One reading tip for the inverse direction. Thrust goes as $P^{2/3}$, so doubling the installed power buys only 59% more lift. That is the same law read backwards, and it is why re-engining a helicopter never delivers the payload gain the horsepower figures seem to promise.