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## equation for bowling ball on a trampoline

i´m searching for the displacement of the surface of a elastic rectangle for a given x and y and a force at a position.

like a bowling ball on a trampoline

the equation should include a var for the elasticity of the surface and the position and strength of the point-shaped force. the return should be a delta-z to any given xy-pair.

I tried a lot of sin cos stuff but could not figure it out. One of the problems is the surface should be fixed at the edges (like a trampoline). so the influence of the force is getting weaker (or the surface is getting stiffer) near the edge.

the is no need for a physically correct model. A function that gives me a roughly such a curve would help me alot

Any suggestions???

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Your problem is one of solving Laplace's equation subject to Neumann boundary conditions. This problem is equivalent to a problem in electrostatics. For small displacements, the height of the trampoline is equivalent to the electric potential, and the bowling ball is eqivalent to a point charge. The fixed boundary of the trampoline is a conducting, grounded box. You should use the method of image charges1. Your original charge maps into a two-dimensional lattice of image charges, as reflected in the sides of the rectangle. The sign of the image charge alternates in a checkerboard pattern. To find the resulting potential, You could either sum the two-dimensional infinite series analytically, or just add up the closest charges numerically. I think the analytical solution involves an elliptic integral, which I prefer to avoid, so if I had to do this I'd do it numerically.

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The cartoon image of a bowling ball on a trampoline is quite far from a small-displacement linear approximation, but I suppose if the trampoline can handle several jumping humans then the bowling ball is light enough. The series you suggest converges rather slowly as it stands; the elliptic integrals aren't that bad, and at least they have more rapidly converging series expansions. – Noam D. Elkies Jan 13 at 0:39

that was more than i expected.

After all i used just a normal distribution to get the look. To be more accurate 2 normal distriputions one for the force and one for the flexibility of the surface.

But thanks anyway...

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 If Carl Feynman's answer was "more than what you expected," then consider asking at math.stackexchange.com next time. – Joel Reyes Noche Jan 17 at 9:20