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Old   September 3, 2002, 23:25
Default curvature
  #1
x.zhang
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Hi, there:

Does anybody know how to calculate the curvature of a convolution surface?

Thanks.

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Old   September 3, 2002, 23:32
Default sorry ,I post it again!
  #2
x.zhang
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sorry,

The question should be:

Does anybody know how to calculate the curvature of a "revolution" surface?

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Old   September 4, 2002, 03:39
Default Re: sorry ,I post it again!
  #3
versi
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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.
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Old   September 4, 2002, 07:04
Default Re: sorry ,I post it again!
  #4
Student.
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Try using the following in Google as search terms

"curvature of a revolution surface" ......leave out the quotes when using Google.
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Old   September 4, 2002, 23:24
Default Re: sorry ,I post it again!
  #5
x.zhang
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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.
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Old   September 4, 2002, 23:59
Default Re: sorry ,I post it again!
  #6
versi
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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.
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Old   September 24, 2002, 11:28
Default Re: sorry ,I post it again!
  #7
Tom
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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 ).
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Old   September 24, 2002, 12:17
Default Re: sorry ,I post it again!
  #8
Tom
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Oops!! the square root should by a power 3/2; i.e. the cude of the square root.
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