Hi Dad,
Here is my answer to your question about "torque roll".
The roll of the plane upon launch is due to conservation of angular momentum. To understand the effect, imagine that your plane is freely falling, ie. "weightless", in the vacuum of space. If you were to turn on the throttle, the propeller in the tail will rotate in one direction, but the body of the plane must rotate in the opposite direction to ensure that the system (body + propeller) has net zero angular momentum. Intuitively, at the same time that the motor in the plane rotates the propeller, the motor simultaneously rotates the body of the plane in the opposite direction.
If the propeller rotates at a given rpm, does the body of the plane rotate at the same rpm? No. The rate of rotation for the body is determined as follows.
The angular momentum L is the product L = I omega, where omega is the angular velocity/rate of rotation in rpm, and I is the moment of intertia. Roughly speaking, I measures how the mass of the rotating body is distributed with respect to the axis of rotation. The greater the mass and the further the mass is from the axis of rotation, the greater I.
For your plane, the propeller and the body both have some fixed moment of inertia. The propeller is small and lightweight, so it will have a small I. The body of the plane is more massive, and the wings stick out from the body, so the moment of inertia of the body will be greater. Note that a swept wing plane will have a smaller moment of inertia than a plane with straight wings, since the moment of inertia increases the further the mass is from the axis of the plane. So we have a conservation equation
L = I_body omega_body = I_propeller omega_propeller.
This equation determines how fast the body rotates for a given throttle speed on the propeller. Since I_body is larger than I_propeller, the angular velocity omega_body will be proportionally smaller than omega_propeller.
All this discussion has been for the weightless, freely-falling plane. When you are holding the plane before launch, your hand is applying a torque to the body of the plane to prevent it from rotating, even though the propeller is rotating. But as soon as you release the plane, the plane starts to fall freely, and the body begins to rotate with velocity omega_body in the equation above. Eventually, when the airflow is established over the wings, the trim of the flaps will supply the necessary torque to prevent the body from rotating.
Here is my recommendation to prevent "torque roll" on the basis of the preceding analysis.
> 1. Attitude of launch.
Attitude of launch is irrelevant.
> 2. Throttle magnitude at launch.
Throttle magnitute is directly relevant. To prevent "torque roll", you want the throttle magnitude to be as small as possible. The throttle magnitude is the same as omega_propeller in my equation, and omega_body is "torque roll". They are proportional. I suspect that you are noticing the effect more for your jet plane because you are launching with a higher throttle, and the moment of inertia I_body is smaller than for your other hand-launched planes.
> 3. Weight distribution (More weight forward or aft on the aircraft).
Weight distribution is relevant, but not forward/aft distribution. The relevant weight distribution is wings vs body. To decrease "torque roll", you want to increase the moment of inertia I_body of the plane, ie. you want more weight on the wings, further from the roll axis. You might try taping pennies on the wingtips, or something like that. The effect would also be smaller with a lighter propeller, since then I_propeller is smaller.
> As an aside, when flying horizontally gliding
at no throttle, an abrupt increase of throttle to full
throttle causes the plane to rotate left and down. This
obviously cannot be prevented. But it may be diminished
using the factors mentioned above hopefully.
Yes, this effect is exactly the same as the "torque roll" effect, due to conservation of angular momentum.
Here is my answer to your question about "torque roll".
The roll of the plane upon launch is due to conservation of angular momentum. To understand the effect, imagine that your plane is freely falling, ie. "weightless", in the vacuum of space. If you were to turn on the throttle, the propeller in the tail will rotate in one direction, but the body of the plane must rotate in the opposite direction to ensure that the system (body + propeller) has net zero angular momentum. Intuitively, at the same time that the motor in the plane rotates the propeller, the motor simultaneously rotates the body of the plane in the opposite direction.
If the propeller rotates at a given rpm, does the body of the plane rotate at the same rpm? No. The rate of rotation for the body is determined as follows.
The angular momentum L is the product L = I omega, where omega is the angular velocity/rate of rotation in rpm, and I is the moment of intertia. Roughly speaking, I measures how the mass of the rotating body is distributed with respect to the axis of rotation. The greater the mass and the further the mass is from the axis of rotation, the greater I.
For your plane, the propeller and the body both have some fixed moment of inertia. The propeller is small and lightweight, so it will have a small I. The body of the plane is more massive, and the wings stick out from the body, so the moment of inertia of the body will be greater. Note that a swept wing plane will have a smaller moment of inertia than a plane with straight wings, since the moment of inertia increases the further the mass is from the axis of the plane. So we have a conservation equation
L = I_body omega_body = I_propeller omega_propeller.
This equation determines how fast the body rotates for a given throttle speed on the propeller. Since I_body is larger than I_propeller, the angular velocity omega_body will be proportionally smaller than omega_propeller.
All this discussion has been for the weightless, freely-falling plane. When you are holding the plane before launch, your hand is applying a torque to the body of the plane to prevent it from rotating, even though the propeller is rotating. But as soon as you release the plane, the plane starts to fall freely, and the body begins to rotate with velocity omega_body in the equation above. Eventually, when the airflow is established over the wings, the trim of the flaps will supply the necessary torque to prevent the body from rotating.
Here is my recommendation to prevent "torque roll" on the basis of the preceding analysis.
> 1. Attitude of launch.
Attitude of launch is irrelevant.
> 2. Throttle magnitude at launch.
Throttle magnitute is directly relevant. To prevent "torque roll", you want the throttle magnitude to be as small as possible. The throttle magnitude is the same as omega_propeller in my equation, and omega_body is "torque roll". They are proportional. I suspect that you are noticing the effect more for your jet plane because you are launching with a higher throttle, and the moment of inertia I_body is smaller than for your other hand-launched planes.
> 3. Weight distribution (More weight forward or aft on the aircraft).
Weight distribution is relevant, but not forward/aft distribution. The relevant weight distribution is wings vs body. To decrease "torque roll", you want to increase the moment of inertia I_body of the plane, ie. you want more weight on the wings, further from the roll axis. You might try taping pennies on the wingtips, or something like that. The effect would also be smaller with a lighter propeller, since then I_propeller is smaller.
> As an aside, when flying horizontally gliding
at no throttle, an abrupt increase of throttle to full
throttle causes the plane to rotate left and down. This
obviously cannot be prevented. But it may be diminished
using the factors mentioned above hopefully.
Yes, this effect is exactly the same as the "torque roll" effect, due to conservation of angular momentum.
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