Kinematics of a landing gear
In this blog post, I explore the calculation of the velocity of a point on a landing gear wheel using multiple reference frames. I will show how to break down the complex motion into simpler components using intermediate frames. I start by analyzing the motion of the wheel with respect to the landing gear arm, and then I analyze the motion of the arm with respect to the aircraft. This approach will allow me to find the final velocity of the point using vector algebra. I’ll go through the entire derivation, clarifying each step.
Let’s consider a landing gear with a wheel of radius r_3 rotating at angular velocity \omega_3 and an arm of length r_2 rotating at angular velocity \omega_2. I want to find the velocity of a point P on the wheel relative to the fixed frame of the aircraft.
I will define three reference frames:
- frame 1: fixed frame (aircraft body),
- frame 2: frame attached to the landing gear arm,
- frame 3: frame attached to the wheel
First, I consider frames 2 and 3, with frame 2 as the fixed frame. The velocity of point P with respect to frame 2 is given by:
\mathbf v_{P/2} = \mathbf v_{O_3/2} + \mathbf v_{P/3} + \boldsymbol{\omega}_{3/2} \times \mathbf r_{O_3 P/3}
Since the origin of frame 3 is fixed in frame 2, \mathbf v_{O_3/2} = \mathbf 0. Also, point P is not moving in frame 3, so \mathbf v_{P/3} = \mathbf 0. The angular velocity of frame 3 relative to 2 is \boldsymbol{\omega}_{3/2} = \omega_3 \mathbf i_2, and the position vector from the origin of frame 3 to P is \mathbf r_{O_3P/3} = r_3 \mathbf j_2.
Substituting, I get:
\mathbf v_{P/2} = r_3\omega_3 \mathbf k_2
Now, I consider frame 1 as fixed and frame 2 moving. The velocity of point P with respect to frame 1 is:
\mathbf v_{P/1} = \mathbf v_{O_2/1} + \mathbf v_{P/2} + \boldsymbol{\omega}_{2/1} \times \mathbf r_{O_2P/2}
Again, the origin of frame 2 is fixed relative to frame 1, so \mathbf v_{O_2/1} = \mathbf 0. I know the value of \mathbf v_{P/2} from the previous calculation. The angular velocity of frame 2 relative to 1 is \boldsymbol{\omega}_{2/1} = \omega_2 \mathbf j_2 and \mathbf r_{O_2P/2} = r_3 \mathbf j_2 + r_2 \mathbf k_2.
Substituting, and simplifying I obtain:
\mathbf v_{P/1} = r_2 \omega_2 \mathbf i_2 + r_3\omega_3 \mathbf k_2
Thus, the velocity of point P with respect to the fixed frame is the sum of two terms: one due to the rotation of the arm and one due to the rotation of the wheel itself.
This is expressed in the frame 2, but it can be transformed to other frames with geometric and kinematic considerations.
For more insights into this topic, you can find the details here.