Solved

Need clarification for computing  normal of a sphere

Posted on 2008-11-01
8
1,181 Views
Last Modified: 2013-12-26
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)
0
Comment
Question by:sph2105
  • 3
  • 2
  • 2
  • +1
8 Comments
 
LVL 19

Expert Comment

by:BrianGEFF719
ID: 22860805
I've read this question a few times and am not sure what you're asking. But why are you working in Cartesian coordinates anyway? This problem has spherical symmetry you should consider working with spherical coordinates.

Brian
0
 
LVL 19

Expert Comment

by:BrianGEFF719
ID: 22860822
>>(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

0
 
LVL 44

Expert Comment

by:Arthur_Wood
ID: 22861288
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
0
 

Author Comment

by:sph2105
ID: 22861358
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) ?
0
Is Your Active Directory as Secure as You Think?

More than 75% of all records are compromised because of the loss or theft of a privileged credential. Experts have been exploring Active Directory infrastructure to identify key threats and establish best practices for keeping data safe. Attend this month’s webinar to learn more.

 

Author Comment

by:sph2105
ID: 22861460
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)?
0
 
LVL 27

Accepted Solution

by:
aburr earned 500 total points
ID: 22861717
yes
grad f, evaluated at a point P: (x1, y1, z1  is normal the the surface f(x, y, z) = c at P,
0
 
LVL 27

Expert Comment

by:aburr
ID: 22861769
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-z2)
where x1,y1,z1 and x2,y2,z2 are two points (center and surface points)
0
 

Author Comment

by:sph2105
ID: 22861792
Thanks aburr.
0

Featured Post

Is Your Active Directory as Secure as You Think?

More than 75% of all records are compromised because of the loss or theft of a privileged credential. Experts have been exploring Active Directory infrastructure to identify key threats and establish best practices for keeping data safe. Attend this month’s webinar to learn more.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

As game developers, we quickly learn that Artificial Intelligence (AI) doesn’t need to be so tough.  To reference Space Ghost: “Moltar, I have a giant brain that is able to reduce any complex machine into a simple yes or no answer. (http://www.youtu…
Foreword (May 2015) This web page has appeared at Google.  It's definitely worth considering! https://www.google.com/about/careers/students/guide-to-technical-development.html How to Know You are Making a Difference at EE In August, 2013, one …
This is a video describing the growing solar energy use in Utah. This is a topic that greatly interests me and so I decided to produce a video about it.
With Secure Portal Encryption, the recipient is sent a link to their email address directing them to the email laundry delivery page. From there, the recipient will be required to enter a user name and password to enter the page. Once the recipient …

948 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

18 Experts available now in Live!

Get 1:1 Help Now