Need clarification for computing normal of a sphere
I am trying to compute the 3D surface normal for a given point on a sphere, in order to find the direction of a light source. I am given the sphere's orthographic projection on the image plane, which is basically a circle, and I have found the center, radius, and area of this circle.
Given that the equation of a sphere is (x-xc)^2 + (y-yc)^2 + (z-zc)^2 = r^2, where xc,yc, and zc are the coordinates of the centroid of the sphere. I have solved for (z-zc) using this equation. I just want to make sure: if I compute the gradient (2(x-xc), 2(y-yc), 2(z-zc)), it safe to assume that the sphere's gradient will give me the normal vector TO the sphere's surface at the point (x,yz)? (the normal originates at the sphere's center)
Given that the equation of a sphere is (x-xc)^2 + (y-yc)^2 + (z-zc)^2 = r^2, where xc,yc, and zc are the coordinates of the centroid of the sphere. I have solved for (z-zc) using this equation. I just want to make sure: if I compute the gradient (2(x-xc), 2(y-yc), 2(z-zc)), it safe to assume that the sphere's gradient will give me the normal vector TO the sphere's surface at the point (x,yz)? (the normal originates at the sphere's center)
>>(x-xc)^2 + (y-yc)^2 + (z-zc)^2
>>Gradient (2(x-xc), 2(y-yc), 2(z-zc))
That's not correct.
If your function is f(x,y,z) = (x-xc)^2 + (y-yc)^2 + (z-zc)^2
Then you have:
f(x,y,z) = (x(1-c))^2 + (y(1-c))^2 + (z(1-c))^2
grad f(x,y,z) = (1-c)^2 * (x^2 + y^2 + z^2)
= (1-c)^2 + (2x a_x + 2y a_y + 2z a_z)
Another way to look at it:
f(x,y,z) = (x(1-c))^2 + (y(1-c))^2 + (z(1-c))^2
grad f = 2(x(1-c)) * (1 - c) a_x + 2(y(1-c)) * (1 - c) a_y + 2(z(1-c)) * (1-c) a_z
= (1-c)^2 +(2x a_x + 2y a_y + 2z a_z)
Brian
>>Gradient (2(x-xc), 2(y-yc), 2(z-zc))
That's not correct.
If your function is f(x,y,z) = (x-xc)^2 + (y-yc)^2 + (z-zc)^2
Then you have:
f(x,y,z) = (x(1-c))^2 + (y(1-c))^2 + (z(1-c))^2
grad f(x,y,z) = (1-c)^2 * (x^2 + y^2 + z^2)
= (1-c)^2 + (2x a_x + 2y a_y + 2z a_z)
Another way to look at it:
f(x,y,z) = (x(1-c))^2 + (y(1-c))^2 + (z(1-c))^2
grad f = 2(x(1-c)) * (1 - c) a_x + 2(y(1-c)) * (1 - c) a_y + 2(z(1-c)) * (1-c) a_z
= (1-c)^2 +(2x a_x + 2y a_y + 2z a_z)
Brian
BrianGEFF719>> xc is NOT x*c, but x(center) etc so his formula for the coordinates of the surface of the sphere, in Cartesian coordinates (x,y,z) is correct. The comment about using Spherical coordinates is also the proper way to approach this problem.
AW
AW
ASKER
Sorry for being unclear, as I typed this question in a rush. Let me try to clarify:
xc, yc, and zc are indeed x(center), y(center), and z(center). as suggested by Arthur.
The reason I want to stick with Cartesian coordinates is because I am working with an image which I will have to process later, so rather than switching back and forth between Cartesian and polar, I would rather just stay in Cartesian.
What I am basically asking is: does using the gradient formula I stated give me the normal from originating from the center of the sphere to the point (x,y,z) ?
xc, yc, and zc are indeed x(center), y(center), and z(center). as suggested by Arthur.
The reason I want to stick with Cartesian coordinates is because I am working with an image which I will have to process later, so rather than switching back and forth between Cartesian and polar, I would rather just stay in Cartesian.
What I am basically asking is: does using the gradient formula I stated give me the normal from originating from the center of the sphere to the point (x,y,z) ?
ASKER
wow, let mer retype that, since I cant edit posts
does that gradient 2*(x-xc), 2*(y-yc) 2*(z-zc) give the normal originating at the center of the sphere to point (x,y,z)?
does that gradient 2*(x-xc), 2*(y-yc) 2*(z-zc) give the normal originating at the center of the sphere to point (x,y,z)?
ASKER CERTIFIED SOLUTION
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you can also calculte the equation of a line from the center to (and through) your point from
(x-x1)/(x1-x2) = (y-y1)/y1-y2)=(z-z1)/(z1-z
where x1,y1,z1 and x2,y2,z2 are two points (center and surface points)
(x-x1)/(x1-x2) = (y-y1)/y1-y2)=(z-z1)/(z1-z
where x1,y1,z1 and x2,y2,z2 are two points (center and surface points)
ASKER
Thanks aburr.
Brian