equation of a plane + point on plane and half width

I don't understand your question: A plane doesn't have a center point or an edge.

:-)

of course it doesn't.

I have a point on the plane.

Basically, I'm after getting the points of a circle from this equation.

so imagine a circle of known radius and centre is drawn onto that plane, so to speak.

its the radical plane formed by the circle made when two spheres intersect.

[A,B,C] is the normal of the plane,, let's assume it's normal for the sake of simplicity.

you can create two orthogonal tangents by projecting the normal onto it's maximal coordinate plane (for numerical stability) and normalising and rotating it 90 degrees in the plane, the cross product of this and the normal will then give you a second, orthogonal tangent

aka:


normal = (A,B,C)/sqrt(A^2+B^2+C^2)
m = min{|A| ,|B| ,|C|}
(m==|A|) -> tangent = (0,-C,<img src='http://public.gamedev.net/public/style_emoticons/<#EMO_DIR#>/cool.gif' class='bbc_emoticon' alt='B)' />/sqrt(C^2+B^2)
(m==|B|) -> tangent = (-C,0,A)/sqrt(C^2+A^2)
(m==|C|) -> tangent = (-B,A,0)/sqrt(B^2+A^2)

bitangent = cross(normal,tangent).


then each point on the circle of radius R in the plane centred at P would be:

P + R*(cos(theta)*tangent + sin(theta)*bitangent)

[A,B,C] is the normal of the plane,, let's assume it's normal for the sake of simplicity.

you can create two orthogonal tangents by projecting the normal onto it's maximal coordinate plane (for numerical stability) and normalising and rotating it 90 degrees in the plane, the cross product of this and the normal will then give you the second normal.

aka:


normal = (A,B,C)/sqrt(A^2+B^2+C^2)
m = min{|A| ,|B| ,|C|}
(m==|A|) -> tangent = (0,-C,<img src='http://public.gamedev.net/public/style_emoticons/<#EMO_DIR#>/cool.gif' class='bbc_emoticon' alt='B)' />/sqrt(C^2+B^2)
(m==|B|) -> tangent = (-C,0,A)/sqrt(C^2+A^2)
(m==|C|) -> tangent = (-B,A,0)/sqrt(B^2+A^2)

bitangent = cross(normal,tangent).


then each point on the circle of radius R in the plane centred at P would be:

P + R*(cos(theta)*tangent + sin(theta)*bitangent)

luca-deltodesco, thank you very much, that was the magic bit I couldn't quite recall and couldn't find reference to anywhere!!

(I should have been more clear in the original question).

thanks again.

out of interest, can anyone point me at a math tut or something where this is discussed?

out of interest, can anyone point me at a math tut or something where this is discussed?

Which part? The projection to 2D, the creation of a perpendicular vector, or the cross product?

Regardless Wikipedia has the answer...

[quote name='RavNaz' timestamp='1298937964' post='4780341']
out of interest, can anyone point me at a math tut or something where this is discussed?

Which part? The projection to 2D, the creation of a perpendicular vector, or the cross product?

Regardless Wikipedia has the answer...
[/quote]

the projection maths that luca uses to create the tangent



Vector n <3,4,5>

// Make unit length
normal = (A,B,C)/sqrt(A^2+B^2+C^2)

// for n<3,4,5> A is the minimum component.
m = min{|A| ,|B| ,|C|}

// the cross product of 2 vectors returns a vector perpendicular to the plane they exist on
// Since m = A, we use the X axis as the other vector. If m = B use <0,1,0> and if m=c use <0,0,1>

cross(n,<1,0,0>) = <0,-5,4> = <0,-C,B>

// Make unit length...since it is essentially a 2D vector, sqrt(b^2 + c^2) = magnitude

(m==|A|) -> tangent = (0,-C,<img src='http://public.gamedev.net/public/style_emoticons/<#EMO_DIR#>/cool.gif' class='bbc_emoticon' alt='B)' />/sqrt(C^2+B^2)
(m==|B|) -> tangent = (-C,0,A)/sqrt(C^2+A^2)
(m==|C|) -> tangent = (-B,A,0)/sqrt(B^2+A^2)


bitangent = cross(normal,tangent).


Hope this helps, cheers!