curvature

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 September 3, 2002, 22:25 curvature #1 x.zhang Guest   Posts: n/a Hi, there: Does anybody know how to calculate the curvature of a convolution surface? Thanks.

 September 3, 2002, 22:32 sorry ,I post it again! #2 x.zhang Guest   Posts: n/a sorry, The question should be: Does anybody know how to calculate the curvature of a "revolution" surface?

 September 4, 2002, 02:39 Re: sorry ,I post it again! #3 versi Guest   Posts: n/a curvature in the rotational direction: 1/r, where r is the radius of the body (the distance from body surface to the rotational axis); curvature in the azimutal plane: Know the first and second derivatives first, then compute curvature according to a standard formula; curvature = div ( N ), where is the unit normal vector.

 September 4, 2002, 06:04 Re: sorry ,I post it again! #4 Student. Guest   Posts: n/a Try using the following in Google as search terms "curvature of a revolution surface" ......leave out the quotes when using Google.

 September 4, 2002, 22:24 Re: sorry ,I post it again! #5 x.zhang Guest   Posts: n/a I think you are wrong. I have thought it over before I come here for help! 1/r is not the curvature in any direction!!! e.g. When a hemi-circle rotate, a sphere is formed. I think now you know what I mean.

 September 4, 2002, 22:59 Re: sorry ,I post it again! #6 versi Guest   Posts: n/a r is not necessarily radius of revolution. There are two principal radi of cuvature for a general surface in 3D space. Refer to a book on geometry.

 September 24, 2002, 10:28 Re: sorry ,I post it again! #7 Tom Guest   Posts: n/a If you write your surface as z = f(r) where r is the radius, then the curvature is f''/sqrt(1 + f'^2) where ' = d/dr. ( If you have r = g(z) then the formulas just the same; g''/sqrt(1 + g'^2) '=d/dz ).

 September 24, 2002, 11:17 Re: sorry ,I post it again! #8 Tom Guest   Posts: n/a Oops!! the square root should by a power 3/2; i.e. the cude of the square root.

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