# Chinook Helicopter Torque

The Chinook Helicopter has 2 rotors to counteract the torque generated by spinning the blade.

Theoretically, could you use a smaller "back" rotor that is farther away from the main rotor to achieve the same result, ie no twisting?

• Unless I'm missing something, this is the point of the tail rotor (besides steering). – lionelbrits Nov 11 '13 at 2:48
• I want the tail rotor to point up – Mike Buck Nov 11 '13 at 2:56
• It is the torque from the air resistance that is going to be a problem to match with a small back rotor. – ja72 Nov 11 '13 at 15:12
• @ja72 Yes, that is my impression. You could compensate, though, with a steeper angle of attack and a higher speed, but I don't know how far that can go. – Emilio Pisanty Nov 11 '13 at 15:15
• You also going to create a torque imbalance along the main shaft, if the torque on the far sides of the gearboxes are going to be equal, then the torques on the near side (closest to the engine) are going to be un-equal, unless the output speeds are the same. – ja72 Nov 11 '13 at 15:36

The separation between the rotors does not actually matter. What matters is that the torque exerted by each of the motors on the respective rotor be the same and in opposite directions. Those torques then add vectorially and cancel out to give zero net torque on the helicopter.

It's not clear to me exactly what you mean by "small". It is indeed possible to have a rotor with shorter blades, but it will need to be driven faster to provide the same torque. This will typically be associated with a smaller power consumption and less lift coming from that rotor.

In general, though, it is the symmetry between the rotors which makes the configuration work. While in practice they won't be exactly the same, they will be very similar to each other, and straying far from this configuration will be very impractical unless you have some stringent requirements on your 'secondary' motor, as well as radical new engines for it to match.

• Is there a formula that can be used to calculate the diameter/rpm of the second rotor, ie if the main rotor is 24" at 1000 rpm how fast does a 11" second have to spin. obviously ignoring thrust & other things, this should just be for keeping the heli from yawing – Mike Buck Nov 11 '13 at 2:05
• That I can't answer. Your guiding principle, though, should be that the torque applied on both rotors should be the same. The rotor's shape and angle of attack would impact how that torque translates into rotational speed and lift. That's on you to figure out anyway ;). – Emilio Pisanty Nov 11 '13 at 2:08
• @MikeBuck it makes a different if the two motors are geared together or not. If there is a gearbox the when the speed of the small rotor is double for example, the torque supplied to it would be half. – ja72 Nov 11 '13 at 15:17

Assuming a mass moment of inertia of $I_1$ for the main rotor and $I_2$ for the secondary rotor, and a coefficient of drag of $\beta_1$ and $\beta_2$ respectively then the torque on the rotor shafts are

$$T_1 = I_1 \dot \Omega + \beta_1 \Omega^2 \\ T_2 = I_2 (\gamma \dot \Omega) + \beta_2 (\gamma \Omega)^2$$ where $\Omega$ is the main rotor speed, and $\gamma$ the gearing ratio for the small rotor.

To make those equal you need gearing of $$\gamma = \frac{I_1}{I_2}$$ and coefficient of drag $$\beta_2 = \beta_1 \frac{I_2^2}{I_1^2}$$

Since $I_1 > I_2$ this requires the drag to be $\beta_2 \gg \beta_1$ which is a) hard to do with a small rotor, and b) very inefficient.

FYI - The total torque on the motor is going to be

$$T_E = \left( I_1 + I_2 \gamma^2\right) \dot \Omega + \left(\beta_1 + \beta_2 \gamma^3\right) \Omega \\ = \left(I_1 + \frac{I_1^2}{I_2} \right) \dot \Omega + \left( 1 + \frac{I_1}{I_2} \right) \beta_1 \Omega^2$$

The interchangeability of rotors for tourque might be nearly impossible because of the ma in f=ma.The rotor speeds would have to continuously vary ideally.Now if it were just a question of lift it would be in aronautical engineering. But then again as they say ,nothing is really impossible,you just have to find a way of doing it.