A little more information is needed here... what exactly do you mean by "fall off"? I'm assuming you're talking about a dynamic case where the sphere picks up enough horizontal momentum to no longer have contact with the elipse despite gravitational acceleration? Or are you talking about some other scenario?

If the former is the case, it's going to be a function of the geometery of the elipse... specifically the derivative of the slope of the contact surface (2nd derivative of the ellipse surface itself). The point at which gravitational acceleration downward is lesser in magnitude than the derivative of the slope of the ellipse (in relation to the horizontal speed of the sphere) is where the sphere will being free-fall.... (if I'm making the correct assumptions about your problem).

A little more information is needed here... what exactly do you mean by "fall off"? I'm assuming you're talking about a dynamic case where the sphere picks up enough horizontal momentum to no longer have contact with the elipse despite gravitational acceleration? Or are you talking about some other scenario?

If the former is the case, it's going to be a function of the geometery of the elipse... specifically the derivative of the slope of the contact surface (2nd derivative of the ellipse surface itself). The point at which gravitational acceleration downward is lesser in magnitude than the derivative of the slope of the ellipse (in relation to the horizontal speed of the sphere) is where the sphere will being free-fall.... (if I'm making the correct assumptions about your problem).

Thank you for taking the trouble to think about the problem.

I didn't have another scenario in mind - your interpretation is correct. By falling off, I mean the sphere loses contact with the ellipse.

The traditional way to think about the problem of a sphere rolling off a slope of circular cross-section is to resolve the forces normal to the point of contact between the spheres and surface.

You get something along the lines...

mg sin(theta) - N = m v**2 /(R+r)

where R and r are the radii of the surface and sphere, m is the mass of the sphere, N is the reaction force exerted by the surface on the sphere etc.

Conservation of energy is used to find v in terms of theta and there is a bit of geometry related to non-slippage of the rolling sphere. Then it is just a matter of letting N=0, which is the condition for the sphere to part company with the surface. It works out well.

In the case where a sphere is falling off an elliptical surface I was using the same approach as for falling off a circular surface. The R+r term above gets replaced by the radius of curvature of the elliptical path taken by the center of the sphere. This is a rather complicated formula.

The condition for non-slippage is complicated also.

Your idea of using the fall-off rate of the slope (2nd derivative) as compared to g is intriguing. I will see where that leads.

Thanks Dave.

-- Cheers, Brad

Give me a fast ship... for I intend to go in harm's way.

I just figured an easier way would be to determine at what point the velocity vector has a slope lesser in magnitude than the slope of the ellipse surface. (ex. the ball is travelling at 3i - 4j m/s = slope of -4/3... if the slope of the ellipse at that point is -5/3, then the ball will loose contact).

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