2
$\begingroup$

Let's say we have a simple hollow cube(chosen for simplicity) with outside side length=10cm, thickness=5mm (or another arbitrary shape with arbitrary dimensions).

Clarification: I am NOT talking about a helicopter. It has no tail rotor, no forward speed, no controllers etc.

It has a passive rotor(no power, no engine) on top of it with 4 blades(total diameter of rotor l=30cm) (with arbitrary foil shape). We drop it from height of H.

What would be free-body diagram on this rotor-cube system? I am imagining that there would be 3 forces. First one is downward and the other two is upward. Both velocities are descent velocities, the area S in both of them is $\frac{\pi l^2}{4}$, l being diameter of circle created by rotating blades.

  1. Weight Force $W=mg$
  2. Lift force $L=½ρv²SC_l$
  3. Drag force $F_d=½ρv²SC_d$

Is this correct? Also please post your source, whether the name of the book from or site link which you learned how to find free-body diagram of these kinds of aviation systems.

Also will all kinds of foils work in this scenario? What is the important parameter in foils that will make it start spinning and descent with constant speed?

$\endgroup$
3
$\begingroup$

There's some info on autorotation, including some references, in this answer. And also a pertinent question on Physics SE. Free body diagrams included.

In short:

  • Autorotation only works in a satisfactory way if the vehicle has forward velocity: in a vertical descent, in optimal circumstances, autorotation works as well as a leaky parachute does.
  • Getting the rotor to actually rotate from standstill is a hair raising kettle of fish, the first thing about autorotation that helicopter pilots learn is that they must maintain rotor rotation from when it was driven by an engine. During any time of rotor standstill the aircraft is simply falling, with no real means of aligning the rotor with the airflow.
  • Autogyro's start their flight by starting the rotor to rotate. Can be done on ground by driving fast and making use of the horizontal velocity, or by a flywheel mechanism attached to the propeller engine. Again, the first thing to do is to rotate the rotor.

The equations in OP are valid for fixed wing aeroplanes, and for the rotor blades. But since with rotor blades the velocity over the blade is a function of the blade radius, plus faster over the forward moving blade than over the retreating blade, they are not defined relative to airspeed V but to tip speed of the rotor blades $\Omega R$. Impulse theory using non dimensional coefficients, from Prouty Helicopter Performance, Stability and Control page 25:

$$ T = C_T \cdot \rho A (\Omega R)^2$$

with

  • T = rotor thrust
  • $\rho$ = air density
  • A = disk area
  • $\Omega R$ = tip speed

Note that Prouty uses the USA definition for $C_T$, which incorporates the factor ½ of the dynamic pressure $½ \rho V^2$ and is therefore twice as much as $C_T$ values used in Europe.

For the thrust to drive the rotor, analogous to aeroplane drag:

$$Q = C_Q \cdot \rho A (\Omega R)^2 R$$

And for the power:

$$P = C_P \cdot \rho A (\Omega R)^3$$

Also will all kinds of foils work in this scenario? What is the important parameter in foils that will make it start spinning and descent with constant speed?

Airfoil shape is of much less importance than blade twist, solidity ratio and a whole lot more parameters, which would make the answer very broad. Best looked up in Prouty and Leishman.

| improve this answer | |
$\endgroup$
  • $\begingroup$ "Getting the rotor to actually rotate from standstill is a hair raising kettle of fish" - Can't we think of it as a wind turbine? Wind turbines are standing still until air is flowing through them, so can't it also work for a rotor? $\endgroup$ – photinaomo Mar 1 at 16:16
  • 1
    $\begingroup$ Still-air vertical descents are quite feasible if you design the gear to handle the vertical velocity of around 500 fpm youtube.com/watch?v=GlyR-aSEuig. $\endgroup$ – John K Mar 2 at 3:45
  • $\begingroup$ Vertical descents are fine, the hard bit is stopping at the ground, as you only have stored energy in the blades. If you have some forward speed, you can use that energy too, and there's usually quite a lot of it, as forward speed is greater than vertical speed. $\endgroup$ – Robin Bennett Mar 2 at 12:15
  • 1
    $\begingroup$ @photinaomo - helicopters typically only have a few degrees of negative pitch, so if they stop, they're deeply stalled and there's not much force turning them. Wind turbines can usually feather their blades to face the on-coming air. A typical paper helicopter starts from stationary, but drops rapidly for several feet while it does. $\endgroup$ – Robin Bennett Mar 2 at 12:21
1
$\begingroup$

What is the important parameter in foils that will make it start spinning

The aerofoil needs to create lift in a direction that causes rotation. Typically aerofoils stall at around 15 degrees angle of attack, so you would need to rotate your blades to around -75 degrees to get them started, and gradually reduce this as they speed up.

and descent with constant speed?

During descent, the blades have slight negative pitch, so their lift is tilted slightly in the direction of rotation, and provides the energy input to overcome drag. The angle of attack is still positive, as the aircraft is descending. If the blade speed and angle of attack are the same as for normal flight, so is the lift, and there is no vertical acceleration (i.e. you have constant speed descent).

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.