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SpinningWing > Helicopters > Helicopter Rotors > Momentum Theory for a Hovering Helicopter

Momentum Theory for a Hovering Helicopter

Momentum theory provides a relatively simple estimate of how a helicopter rotor changes/induces airflow. We’ll focus here on a helicopter main rotor in hover. The rotor is powered and accelerates air that is otherwise at rest relative to the rotor. The analysis can be generalized to other flight regimes and even wind turbines that are powered by the air, but we will save that for another day. Even in the simplest context, this concept can be difficult to understand at first.

You may also want to check out our momentum theory calculator.

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We’ll start by listing the assumptions required for this analysis. These assumptions are not satisfied by practical helicopter rotors, but the results provide convenient approximations widely used in the helicopter industry. The most useful equations derived from these assumptions will be highlighted at the bottom of this article.

  1. The rotor provides uniform thrust over its area.
  2. The rotor is powered and accelerates air that is otherwise at rest (e.g. a helicopter rotor in hover).
  3. The flow is steady, inviscid, irrotational and incompressible.
  4. Swirl (air tending to rotate with the rotor) is ignored in this analysis. It can be included with a little additional complication.

Big Picture

The rotor attains thrust \(T\) by pushing air downward. Newton’s equation \(F=ma\) is used to equate the thrust (\(F\)) to the rate of momentum change of the air \(ma=\frac{dp}{dt}\) (\(p=mv\)). The change in momentum is achieved by accelerating the air downward. The acceleration begins at some distance \(d_0\) above the rotor and continues to some distance \(d_3\) below the rotor (see diagram down below). Since the thrust is finite, the accelerated air is confined to a finite volume called the streamtube. The radii of this tube at different heights can be related using conservation of mass—the rate at which air mass enters any volume of the streamtube (mass flux) must equal the rate that air mass flows out. The mass flux through a cross-section \(s_x\) from the diagram down below, with area \(A_x\) is just \( \rho A_x v_x\). Since the volume enclosed by \(s_x\) and some other cross-section \(s_y\) must have 0 net mass flux, we arrive at this useful equation (after dividing by \(\rho \)) $$ \begin{equation} A_x v_x = A_y v_y \label{eq:massflux} \end{equation}.$$

We’ll use the conservation of energy to equate the power applied by the rotor to the rate that energy is added to the air. We’ll also use Bernoulli’s equation above and below the rotor to generate to eliminate many variables and reach relatively simple equations for the induced velocity and power as a function of thrust.

For convenience, we’ll mark 4 cross-sections of the streamtube as shown in the diagram: \(s_0,s_1,s_2,s_3\). Section \(s_0\) is where the air starts to accelerate, \(s_1\) is just above the rotor, \(s_2\) just below the rotor and \(s_3\) below the rotor, where the flow reaches maximum speed. The induced velocity, area and pressure at section \(s_x\) are denoted \(v_x\), \(A_x\) and \(p_x\). Since \(s_1\) and \(s_2\) are arbitrarily close, \(v_1=v_2\) and \(A_1=A_2\).

Diagram of the streamtube with labeled cross-sections for momentum theory analysis.


We'll start with \(F=ma=\frac{dp}{dt}\), where \(F\) is the rotor thrust and \(p\) is the net momentum of all the air. Since the bottom of the streamtube lengthens as time passes (airflow in other regions is constant), we'll compute \(\frac{dp}{dt}\) from the lengthening of the bottom of the streamtube. In a period of time \(dt\), the streamtube lengthens by \( v_3dt \) and adds momentum \(\rho Vv_3\) to the system, where the volume \(V\) of this new section of the tube is \(V= A_3 v_3 dt\). Hence $$\begin{equation} T=\frac{dp}{dt}=\rho A_3 v_3^2 \label{eq:fma} \end{equation}$$.

Similar to above, we compute the kinetic energy added to the air from this new volume of streamtube as \(\frac{1}{2}\rho Vv_3^2\), which gives an energy per unit time (power) of \( \frac{1}{2}\rho A_3 v_3^3 \). Since the power applied to the air by the rotor is \(Tv_1\), we get $$\begin{equation} Tv_1 = \frac{1}{2}\rho A_3 v_3^3 \label{eq:conserve} \end{equation}$$.

Dividing \eqref{eq:conserve} by \eqref{eq:fma} and doing some arithmetic reveals that the airspeed is twice as large in the far wake, compared to the rotor $$\begin{equation} v_1=\frac{1}{2}v_3 \label{eq:veleq} \end{equation}.$$

Using the mass conservation equation \eqref{eq:massflux}, it follows that the area \(A_3\) is half as large as \(A_1\): $$\begin {equation} A_1 v_1 = A_3 2v_1 \rightarrow A_1=2A_3. \label{eq:arear3} \end{equation}$$.

Using Bernoulli’s equation, we can relate quantities at \(s_0\) to that at \(s_1\) via \( p_0-(p_1+\frac{1}{2}\rho v_1^2)=0\). Likewise, we can relate \(s_2\) to \(s_3\) via \( p_2+\frac{1}{2}\rho v_2^2-(p_3+\frac{1}{2}\rho v_3^2)=0 \). Adding those two equations and solving for the pressure difference \(p_2-p_1\) gives equation \eqref{eq:pressdiff} using the following facts in the arithmetic: (1) the pressures \(p_0\) and \(p_3\) must be the ambient pressure \(p_a\), (2) \(v_1=v_2\) and (3) \(v_3=2v_1\). $$\begin{equation} p_2 - p_1=\frac{1}{2}\rho v_3^2 = 2\rho v_1^2. \label{eq:pressdiff} \end{equation}$$

The thrust on the rotor comes from the pressure drop across the rotor multiplied by the rotor area, i.e. \(T = (p_2-p_1)A\). Combining this with \eqref{eq:pressdiff} and solving for \(v_1\) we get a very useful equation for the induced velocity at the rotor, i.e. $$\begin{equation} v_1= \sqrt{ \frac{T}{2\rho A} } \label{eq:v1} \end{equation}$$

This allows us to compute the power required by the rotor \(P=Tv_1\) as $$\begin{equation} P=T \sqrt{ \frac{T}{2\rho A}}. \label{eq:pt} \end{equation}$$

We can also get the power as a function of the induced velocity \(v_1\) by substituting \(T=2\rho v_1^2 A\) (from rearranging \eqref{eq:v1}) into \eqref{eq:pt} which gives $$\begin{equation}P= 2\rho v_1^3 A. \label{eq:pvi} \end{equation}$$.

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