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Chasles' theorem:
"Any rigid-body motion can be specified by a rotation about a screw axis and a translation parallel to that axis."

For a rolling wheel, the axis of rotation is perpendicular to the direction of translation. How do we reconcile these facts ?

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All velocities are relative to a specific frame.

You can describe the wheels motion relative to the vehicle body as purely rotation and no parallel translation.

You can alternatively describe the motion of the vehicle to the ground (assuming a basic cart going straight forward) as purely translation along a forward facing axis of the vehicle without rotation.

If you want the velocity of the wheel with respect to the ground. The vehicle is moving forward at velocity $v$ then the top of the wheel is moving at $2*v$ and the bottom of the wheel is moving at zero velocity then the axis is at the contact point on the ground. (The place where it has zero velocity.) And the rotational velocity is $v/r$ where $r$ is the radius of the wheel. And there's no translational component.

One critic part of the above theorem is that the axis of choice of the representation is not tied to the physical mechanisms causing the motion. In fact it may diverge significantly and also change moment to moment for a more complex trajectory.

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  • $\begingroup$ That's not what I'm asking. If a rolling wheel is a rigid motion, then how can you describe that motion using a screw ? The translational part can be described by a screw axis parallel to the axis of rotation, but then you need infinite pitch. $\endgroup$
    – drC1Ron
    Commented Sep 5, 2022 at 23:30
  • $\begingroup$ You want to describe the rolling wheel with respect to the ground? You always need to clarify the reference frame. I've updated the answer to explain that. There will be no translational part for that. In screw theory it's fine to have both zero pitch and infinite pitch. That means it's either purely rotational or purely translational. When a car rolls down the street you think of the wheels as both translating and rotating. But as I said this reduced representation does not necessarily reflect the behavior it's just a way to ensure a reduced representation for mathematical manipulation. $\endgroup$
    – Tully ♦
    Commented Sep 6, 2022 at 16:14
  • $\begingroup$ I want the inertial reference frame to be ground. Otherwise you just have a rotating disk. My claim is that you need two screws to describe the motion. Otherwise, the screw axis must translate along the translation of the wheel. $\endgroup$
    – drC1Ron
    Commented Sep 6, 2022 at 17:07
  • $\begingroup$ Yes, as I said above the representation is only valid for the moment that you're representing. It does not guarantee that the axis does not move for the next moment. If you want a wheel that rolls 90 degrees forward the axis of rotation to represent it as a pure rotation will be at the point that is 45 degrees in front of the first wheel and 45 degrees behind the second wheel some distance under the ground. If you were to trace the route between those two points the wheel would arc up into the air. But again this is not about representing the trajectory but just the pure displacement. $\endgroup$
    – Tully ♦
    Commented Sep 6, 2022 at 19:37

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