Helicopter hover flight question I'm an avionics technician and I work on helicopters. This week I was trying to explain where comes the lift force to some friends. I'm a self-learner but I have a poor physics school background.
I told them that the helicopter sustains by providing the same acceleration the earth pull him, to the air. But thinking about this at home I tried to computed this to see if its plausible:
My helicopter is an EC135 that can weight up to 3000kg. The rotor diameter is 10.2m, that's a rotor surface of 81.7 m², but since 1 cubic meter of air is 1.225kg does that mean that it needs to "suck" (accelerates from 0 to +9.8m/s) 3000/1.225=2449 cubic meter of air each seconds? Or accelerates half this value, twice the quantity of air? That's a column of 30 meter of air above the rotor. That's a theoretical wind of 108km/h under the rotor, I often stand very close to the helicopter when it takes off and that number seems big to me. Is my reasoning correct or is there some coefficients that lower the thing?
 A: For this kind of problem, I find it easier to think about momentum than about mass and acceleration.  Remember that momentum is mass times velocity,
$$
\vec p = m\vec v
$$
and so Newton's second law is secretly about the rate of change of momentum just as much as about acceleration:
$$
\vec F = m\vec a = \frac{\mathrm d}{\mathrm dt}(m\vec v) 
= \frac{\mathrm d}{\mathrm dt} \vec p
$$
If your helicopter with mass $3000\,\rm kg$ were to suddenly fall out of the sky with acceleration $g = 10\rm\,m/s^2$, after each second it would gain momentum
$$
\begin{align}
\frac{\mathrm d}{\mathrm dt} \vec p &= m\vec g \\
&= 3000\,\mathrm{kg} \cdot 10\,\mathrm{m/s^2}
\\ &= 3\times10^4\,\frac{\rm kg}{\rm s}\,\frac{\rm m}{\rm s}
\end{align}
$$
I've done something a little sneaky with the units here: I've split up the $\rm s^{-2}$, to suggest that you can think of this rate of change of momentum as the product of a mass flow rate (usually $\dot m$, with a dot to mean it's a rate, in $\rm kg/s$) and an exhaust velocity ($\vec v_\text{out}$, in $\rm m/s$).
The mass flow rate is also proportional to the air flow rate: the faster the air is flowing, the more mass flow there is.  If the density of the air is $\rho \approx 1\rm\,kg/m^3$, and the rotor blades have area $A \approx 100\rm\,m^2$ (which is just to keep the arithmetic simple, but winds up having roughly the same product $\rho A$ as your values) then the air flow rate would be
$$
\dot m = \rho A v
$$
If we want to find the rate of change of the momentum, we have to do a bit of calculus to get the flowing air from being at rest (above the rotor) to flowing at the exhaust speed:
$$
\frac{\mathrm d}{\mathrm dt} \vec p = \int_0^{v_\text{out}} \dot m \ \mathrm dv
= \rho A\cdot \frac12 v_\text{out}^2
$$
You might remember that the definition of pressure is force per unit area, $P = F/A$.  So we can take this force $F = \mathrm dp/\mathrm dt$ and relate it to the pressure difference between the upper and lower edges of the helicopter's rotors:
$$
P_\text{below} - P_\text{above} = \frac{F_\text{net}}{A} = \frac12\rho \left(v_\text{below}^2 - v_\text{above}^2\right)
$$
This is actually just the Bernoulli equation,
$$
P + \rho gh + \frac12\rho v^2 = \text{constant}
$$
So you can say, with complete equivalence,

*

*your helicopter hovers by transferring momentum to the air at the same rate that gravity would otherwise transfer momentum to the helicopter


*your helicopter hovers by making the pressure below the rotor higher than the pressure above the rotor.
You can figure out the downdraft speed from these relationships, too:
\begin{align}
v_\text{out}^2 &= \frac{2 F_\text{net}}{\rho A}
\\
v_\text{out} &=
\sqrt{
\frac{2\cdot 3\times10^4\,\rm kg\,m\,s^{-2}}{1\rm\,kg\,m^{-3} \cdot 100\rm\,m^2}
}
\sim 25\rm\,m/s
\end{align}
This corresponds to a mass flow rate
$$
\dot m = \rho A v_\text{out} \approx 2500\rm\,kg/s
$$
just like you calculated.
I wouldn't think of this as a thirty-meter column of air above the rotor.  Instead, I'd think of it as an order-of-magnitude estimate for how far below the helicopter you would expect to feel the downdraft.  When the helicopter is on the ground (or hovering just above ground level), you can unquestionably feel the downdraft.  As the helicopter rises, turbulence mixes the downdraft into a larger area $A$, so the average velocity of the downdraft decreases.  If you have experience standing near helicopters as they take off or land, you might be able to estimate the shape of the cone where the downdraft is strong.  If that cone were narrow, you might expect to feel the downdraft even when the helicopter is thirty meters up, as you say.  But if that cone is wide, you would expect the turbulent downdraft from the helicopter to have dissipated by thirty meters below.
If you were to put a bigger rotor on your helicopter, giving a bigger flow area $A$, you would have larger mass flow rate, even at a smaller downdraft velocity.
A: When analysing a fan, propeller or engine which isn't moving, the air which is being sucked through the rotor starts far away and not moving. Air is drawn in from all around, so the area far away is big so the speed is slow.
We normally assume it exits in a 'free jet', which is a column of air that's all moving at the same direction and same speed, which is equal to the speed passing through the rotor itself. This is simple to analyse, and is close to reality, although in reality there'll be a vortex at the rotor tips and swirl downstream, but it's good enough for an estimate to understand.
So if we say that at the rotor, the air is moving straight down with speed $V_r$, and starts far away at speed $0$, then $m$ kg of air changes in momentum by $mV_r$. Force is the rate of change of momentum, $\frac{\Delta(mV)}{\Delta t}$ so the force upwards on the helicopter (which must be equal to its weight) is equal to the rate of change of momentum of the air.
In this case, $\Delta V$ isn't related to time (so not an acceleration) - the velocity changes gradually and non-uniformly between the far-away air that isn't moving, and the rotor speed $V_r$. Instead, the time-derivative is linked to the mass - there's not a fixed mass of air, there's a mass flow rate of $\dot{m}$ kg/s. So here $\frac{\Delta(mV)}{\Delta t}=\Delta V\frac{m}{\Delta t}=V_r\dot{m}$
The upwards force is 3000kg*$g$, so 29430N. The mass going through the rotor per second is therefore 29430/$V_r$.
The mass flow is also equal to density * area * velocity, so $\rho*81.7*V$ with an area of 81.7$m^2$. So $1.225*81.7*V_r=29430/V_r$, $V=17.1ms^{-1}=$61.6kph=38.3mph
So there are 1711 kilograms of air passing through per second, each of which accelerates from 0 to $17.1ms^{-1}$
