Mathfilter - Gaussian Curvature in Cylindrical Coordinates
December 28, 2010 12:15 PM Subscribe
Mathfilter - Need help with Gaussian curvature in cylindrical coordinates
We have done analysis in the past where we've computed an approximation for Gaussian curvature of a surface in Cartesian coordinates.
What we've been doing for Cartesian (in Matlab) is
[fu,fv] = gradient(Z)
[fuu, fuv] = gradient(fu)
[fvu,fvv] = gradient(fv)
GC = (fuu*fvv - fuv*fuv)/(1 + fu^2 + fv^2)
(more or less, trying to simplify the equations for AMF)
So now I have a surface that I'm modeling in cylindrical coordinates, and I can do the same thing as above for R as a function of Theta and Z. The problem is that it's only taking into account the change in R, not the fact that there is curvature inherent in it being a cylinder.
Looking on Wolfram (equations 27, 32 and 37 and thereabouts), it seems like there's a centripetal component that I don't know how to apply. Dividing by the (constant?) radius doesn't seem like it would work, so I think I'm missing something.
Any help would be appreciated, either explaining how to modify these equations to work correctly, or some other approximation that has worked for you.
Thank you.
We have done analysis in the past where we've computed an approximation for Gaussian curvature of a surface in Cartesian coordinates.
What we've been doing for Cartesian (in Matlab) is
[fu,fv] = gradient(Z)
[fuu, fuv] = gradient(fu)
[fvu,fvv] = gradient(fv)
GC = (fuu*fvv - fuv*fuv)/(1 + fu^2 + fv^2)
(more or less, trying to simplify the equations for AMF)
So now I have a surface that I'm modeling in cylindrical coordinates, and I can do the same thing as above for R as a function of Theta and Z. The problem is that it's only taking into account the change in R, not the fact that there is curvature inherent in it being a cylinder.
Looking on Wolfram (equations 27, 32 and 37 and thereabouts), it seems like there's a centripetal component that I don't know how to apply. Dividing by the (constant?) radius doesn't seem like it would work, so I think I'm missing something.
Any help would be appreciated, either explaining how to modify these equations to work correctly, or some other approximation that has worked for you.
Thank you.
Well, curvature is coordinate-independent, so you could just convert to standard coordinates and compute it that way.
posted by number9dream at 6:37 PM on December 28, 2010
posted by number9dream at 6:37 PM on December 28, 2010
This thread is closed to new comments.
posted by lex mercatoria at 2:01 PM on December 28, 2010