gradient at occluding boundary
Posted 23 September 2010 - 11:06 AM
Posted 23 September 2010 - 06:00 PM
Posted 25 September 2010 - 04:53 AM
He's doing this because, like me, he has no idea what you are actually asking because you didn't give enough information in your original post.
Posted 25 September 2010 - 09:41 AM
Trying to explain my problem in details.............
Let a surface is defined as z=f(x,y), (z is height)
then gradient of this surface is defined by (p,q)
where p is the partial derivative of z with respect to x
and q is the partial derivative of z with respect to y
gradient becomes unbounded at boundary means p or q (the partial derivatives of height) becomes infinite on the boundary.
I do not understand what makes p or q infinite at the boundary.
Posted 25 September 2010 - 06:11 PM
Posted 27 September 2010 - 12:31 PM
udvat, we're still missing information as to the concrete context of the question. What or why are you saying that something is becoming unbounded?
Your gradient's value, z, is determined by your function f(x,y). Likely, your function works correctly over a finite, defined x-y plane, such that all z-values are bounded within that plane, meaning they do not exceed a certain value, M. Once you go beyond that plane, your function is unbounded, ie. it doesn't constrain itself to be below some M.
Posted 28 September 2010 - 05:29 AM
We suppose only positive z = sqrt(1 - x^2 - y^2)
and we can calculate gradient p = del_z / del_x = -x / sqrt(1 - x^2 - y^2),
q = del_z / del_y = -y / sqrt(1 - x^2 - y^2).
Here we change variables from x, y, z to r, A such as
x = r cosA,
y = r sinA,
z = sqrt(1 - r^2), (0 <= r <= 1, 0 <= A < 2pi).
Then p = -r cosA / sqrt(1 - r^2),
q = -r sinA / sqrt(1 - r^2).
At the occluding boundary, r equals 1. So p and / or q will become infinite.
Posted 28 September 2010 - 06:20 AM
One way to picture this is by planar projecting a texture straight down on the Z axis onto this sphere. The gradients p, q then measure how much the texture has to be stretched to fit the sphere. It's not stretched too much in the middle, but is severely stretched near the edges of the unit circle.
Posted 28 September 2010 - 12:31 PM
Correct. (Although, like Reedbeta, I am not sure why you are talking about "occlusion".)
Anyways, you seem to have a grasp on the math. You know yourself that p or q can swing to infinity, ie. is unbounded. So, I still don't see the actual question?
Posted 29 September 2010 - 07:55 AM
occluding boundary is meaningful. But we can use the word boundary to deal with the partial dreivative issue.
@Alphadog, I could not understand why gradient becomes infinite at the boundary and so that was my question. In my last post I myself tried to answer my question (why gradient becomes infinite) with the help of your (Reedbeta and you) insightful discussions. So that one is solved.
But now I am confused with the point, if partial derivative becomes infine at the boundary of the sphere, can we call the spehere surface differentiable?
Posted 29 September 2010 - 03:26 PM
Posted 29 September 2010 - 04:49 PM
Moreover, the sphere itself is still a completely regular, smooth surface with no singularities or anything. The breakdown at the boundary is really an artifact of the choice of coordinates, not anything to do with the geometry of the sphere.
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