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I would like to provide a more thorough answer to this question here

https://aviation.stackexchange.com/q/3709

but I realized I don't know enough about angular momentum. If an airplane wheel is rotating at 100 rpm, and the wheel weighs 10kg, with a diameter of 50cm and a uniform mass (approximations applicable to a standard small aircraft), what is the difference in force necessary to bring the plane to 20 degrees of bank as opposed to when the wheels are stopped?

I know that this involves calculating angular momentum, which I have at 5kgm/s per wheel, so 10kgm/s total, I'm just not sure how I would quantify the affect of this angular momentum when trying to bank the aircraft 20 degrees over a course of 5 seconds (replicating first turn in the airport pattern).

I bet the following terms are involved: $\sin(20), 5s, 10kgm/s.$

Not sure if it's relevant, but we can assume the aircraft wheels are suspended 1 meter below the aircraft.

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  • $\begingroup$ The link is dead for me. $\endgroup$ – John Alexiou May 5 '14 at 17:06
  • $\begingroup$ BTW, angular momentum and even change in angular momentum can not and will not exert any force. $\endgroup$ – John Alexiou May 5 '14 at 17:08
  • $\begingroup$ @ja72 I fixed the link. I realize that the terminology isn't quite right, maybe you could edit the post? $\endgroup$ – OneChillDude May 5 '14 at 18:30
  • $\begingroup$ I suggest "What is the gyroscopic effect of a fast spinning wheel during an standard airplane turn?" $\endgroup$ – John Alexiou May 5 '14 at 19:56
  • $\begingroup$ 3D dynamics are really complicated, and I am not sure you would understand an analytical solution. In layman terms, there is an additional overturning torque along the axis of the plane when the wheel is spinning and the plane is turning. $\endgroup$ – John Alexiou May 5 '14 at 20:15
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You asked for it.

Here is how you find the forces and moments acting on the wheel from the plane to gauge the effect of spin. Reverse the sign to find the forces acting to the plane by the wheel.

  1. Consider a coordinate system with $z$ axis along the plane axis, and $y$ up (opposite gravity). The plane precession angle is $\varphi$ (as it turns about $y$), the plane bank angle $\psi$ and the wheel spin angle $\theta$. The turn radius is $R$, the wheel is hanging a distance $\ell$ from the center of mass of the plane and it has a radius $r$.
  2. The 3×3 orientation matrix of the wheel spindle is $E = {\rm Ry}(\psi){\rm Rz}(\varphi)$, but for evaluation purposes we will always use $\psi=0$, $\dot{\psi} = \Omega$ and $\ddot{\psi}=0$ for a steady normal turn. $E=\begin{bmatrix}\cos\varphi & -\sin\varphi & 0 \\ \sin\varphi & \cos\varphi & 0 \\ 0 & 0 & 1 \end{bmatrix}$
  3. The mass moment of inertia tensor along the spindle and wheel is $I_{body} = \begin{bmatrix} I_{disk} & 0 & 0 \\ 0 & \frac{1}{2} I_{disk} & 0 \\ 0 & 0 & \frac{1}{2} I_{disk} \end{bmatrix}$ where $I_{disk}=\frac{m}{2} r^2$ for a uniform disk. The mass moment of inertia on the world coordinates is $I = E I_{body} E^\top$ yielding $I= I_{disk} \begin{bmatrix} \frac{\cos^2(\varphi)+1}{2} & \frac{\sin\varphi\cos\varphi}{2} & 0 \\ \frac{\sin\varphi\cos\varphi}{2} & \frac{\sin^2(\varphi)+1}{2} & 0 \\ 0 & 0 & \frac{1}{2} \end{bmatrix}$
  4. The position of the wheel center of mass is $\vec{r} = {\rm Ry}(\psi) \left( \hat{i} R - {\rm Rz}(\varphi) \hat{j} \ell \right) = (R+\ell \sin\varphi, -\ell \cos\varphi, 0)$. In general a varying bank has $\dot\varphi \ne 0$.
  5. The velocity of the wheel center of mass is $\vec{v} = \dot{\vec{r}} = (\ell \dot\varphi \cos\varphi,\ell \dot\varphi \sin\varphi, -(\ell \sin\varphi+R) \Omega )$ and the rotational velocity $\vec\omega = \hat{j} \Omega + {\rm Ry}(\psi) ( \hat{k} \dot\varphi + {\rm Rz}(\varphi) \hat{i}\dot\theta) = (\dot\theta \cos\varphi, \Omega+\dot\theta \sin\varphi,\dot\varphi)$. Remember that wheel spin speed is $\dot\theta$ and turn rate is $\Omega$.
  6. Similarly the acceleration of the wheel center of mass is $\vec{a} = (-\ell (\Omega^2 +\dot\varphi^2)\sin\varphi - \Omega^2 R, \ell\dot\varphi^2\cos\varphi,-2\ell\Omega\dot\varphi\cos\varphi)$ and the rotational acceleration $\vec\alpha = (\dot\varphi (\Omega-\dot\theta \sin\varphi), \dot\varphi\dot\theta \cos\varphi,-\Omega \dot\theta \cos\varphi)$
  7. The force needed to keep the center of mass in motion is $\vec{F} = m \vec{a} = (-m \ell (\Omega^2 +\dot\varphi^2)\sin\varphi - m \Omega^2 R, m \ell\dot\varphi^2\cos\varphi,-2 m \ell\Omega\dot\varphi\cos\varphi)$ and not related to wheel spin (of course).
  8. The moment applied to the spindle to keep the wheel in motion is $\vec{M} = I \vec{\alpha} + \vec{\omega} \times I \vec{\omega} = $ $$\boxed{ \vec{M} = \begin{pmatrix} I_{disk} \dot\varphi (\Omega \cos^2\varphi - \dot\theta\sin\varphi) \\ I_{disk} \dot\varphi \cos\varphi ( \Omega \sin\varphi + \dot\theta) \\ - I_{disk} \Omega \left(\dot\theta + \frac{\sin\varphi}{2} \Omega \right) \cos\varphi)\end{pmatrix}}$$

So if the spin is $\dot\theta \ge \frac{\sin\varphi}{2} \Omega$ then the effects along the axis of the plane of the wheel spinning are more that the effect of the wheel orbiting around the turn. Otherwise it is less. There is component of the reaction moment along $\ell$ which is equal to $I_{disk} \dot\varphi \dot\theta$ and a reaction moment along the spin axis of the wheel equal to $I_{disk} \Omega \dot\varphi \cos\varphi$. The former will cause the plane to yaw when a bank is introduced $\dot\varphi \ne 0$ and the wheel is spinning $\dot\theta$.

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    $\begingroup$ Thanks for the great response! Although I'm not sure I fully explained my question, because this should in fact be a 2d mechanics problem. You seem to be assuming that the plane will be turning around the Y axis, but that should not be the case. I simply want to know what the effect of the angular momentum would be when rolling 20 degrees of bank, making a coordinated turn. I'll edit my question for clarity. $\endgroup$ – OneChillDude May 6 '14 at 13:50
  • $\begingroup$ The coordinated turn is about the Y axis. The plane traces a circle if viewed from above. $\endgroup$ – John Alexiou May 6 '14 at 13:56
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    $\begingroup$ Right, the question was about 20 degrees of bank. Bank alone in a small aircraft will not bring you into a turn, you will also need rudder motion to enter into a full turn. $\endgroup$ – OneChillDude May 6 '14 at 13:58
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    $\begingroup$ This is what a coordinated turn is. Bank + Rudder to get turning. The rudder causes $\Omega$ about the vertical. $\endgroup$ – John Alexiou May 6 '14 at 13:58
  • $\begingroup$ " how I would quantify the affect of this angular momentum when trying to bank the aircraft 20 degrees over a course of 5 seconds". I realize that the most true effects of this are more complicated when considering the 3rd axes of rotation, but I think we can confidently put this into terms of approximated difference in aileron pressure of the flight control in terms of theta, where theta is the speed of the wheels spin $\endgroup$ – OneChillDude May 6 '14 at 13:59

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