scm@onion.UUCP (01/26/87)
Can anyone help me with this problem? I am trying to
represent some solid objects by using quadric surfaces which
seem to be a favourite amongst geometric modellers and to
then move them around in a world model.
If I have a surface defined by
Q(x,y,z) = ax^2 + by^2 + cz^2 + dxy + eyz + fxz
+ gx + hy + iz + k
(ie sphere, cone etc)
does anyone know if there is a straight-forward way of
transforming this surface in my world model using the usual
4x4 scale, rotate and translate matrices?
I know that you can scale a sphere by altering k and translate
it by adding values to a,b or c, but this doesn't seem
to be a very logical approach. Will I have to re-express
these quadric solids as a lot of polygons, surface patches etc etc?
If you know of some mathematical way of re-expressing the
above equation so that the usual transformations can be
applied, I would be pleased to hear from you.
Name: Steve Marsh
E-mail: scm@onion.cs.reading.ac.uk
Establishment: University of Reading, UK.
Research: Systems Architecture Group.
Department: Department of Computer Science
Phone me at: (0734) 875123 ext 7881 from inside UKstolfi@jumbo.UUCP (01/29/87)
oxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxo Stephen Marsh asked: > If I have a surface defined by > > Q(x,y,z) = ax^2 + by^2 + cz^2 + dxy + eyz + fxz + gx + hy + iz + k > > (ie sphere, cone etc) does anyone know if there is a > straightforward way of transforming this surface using the > usual 4x4 scale, rotate and translate matrices? Let the coordinates of a point p be written as a 4-element row vector [x y z 1] (or [X Y Z W] in homogeneous coordinates, where x=X/W, y=Y/W, z=Z/W). A projective transformation (one that preserves straight lines) of R^3 is determined by a 4x4 matrix M, such that the image of a point p is simply the product p M of the 4-vector p by the matrix M. Linear, Euclidean, and affine transformations or R^3 (rotations, translations, scalings, reflections, etc.) are all special cases of projective maps. For example, a translation by (dx, dy, dz) is given by the matrix 1 0 0 0 0 1 0 0 0 0 1 0 dx dy dz 1 A quadratic surface can be represented by a symmetric 4x4 matrix Q, such that a point p lies on the surface if and only if the matrix product p Q p' = 0, where p' is the transpose of p, i.e. the 4-element COLUMN vector <x y z 1>. Specifically, the quadratic surface with equation ax^2 + by^2 + cz^2 + dxy + eyz + fxz + gx + hy + iz + k = 0 is given by the matrix a d/2 f/2 g/2 d/2 b e/2 h/2 f/2 e/2 c i/2 g/2 h/2 i/2 k So, how do we map a quadratic surface Q by a map M? We want a surface K such that p is on Q if and only if (p M) is on K; that is, we want a matrix K such that p Q p' = 0 iff (p M) K (p M)' = 0 Since (p M)' = M' p', we conclude that K = M~ Q M~' will do the trick, where M~ is the inverse of M, and M~' its transpose. j. oxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxo
ph@pixar.UUCP (01/31/87)
Represent the point (x,y,z) by the homogeneous row vector P = [x, y, z, 1]
and plane (or normal vector) ax+by+cz+d=0 by column vector N = [a, b, c, d]^t,
where M^t means transpose of matrix M. The equation P*N=0 (where '*' denotes
matrix multiplication) determines all the planes N passing through point P, or
all the points P which lie on plane N. Note the duality of points and planes.
Represent the quadric surface axx+bxy+cxz+dx+eyy+fyz+gy+hzz+iz+j=0 by the matrix
[ a b/2 c/2 d/2 ]
Q = [ b/2 e f/2 g/2 ]
[ c/2 f/2 h i/2 ]
[ d/2 g/2 i/2 j ]
Points on the quadric surface satisfy P*Q*P^t=0 (substitute P and Q above
to check) and planes tangent to the surface satisfy N^t*Q^-1*N=0,
where M^-1 means matrix inverse or adjoint. Note the duality once again.
If T is a 4x4 transformation matrix (perspective, rotation, scaling,
translation, etc) then to transform point P by T: P' = P*T and
to transform plane N by T: N' = T*N, where M*N is matrix multiplication.
To transform a quadric: Q' = T^-1*Q*T^-1^t. You can substitute the
expressions above into P'*Q'*P'^t=0 to check that it works.
A non-intuitive consequence of this is that any perspective transformation
of a quadric is a quadric! A perspective transformation of a sphere,
for example, can be a hyperboloid of two sheets.
The normal vector or tangent plane can be found by taking the gradient of
the quadratic form: N = grad(P*Q*P^t)/2 = Q*P^t for a point P on the surface.
If P is not on the surface, this plane N is still geometrically meaningful;
it is called the "polar plane" of P, and dually, the "pole" of the plane N
is P = N^t*Q^-1.
References:
E. A. Maxwell, General Homogeneous Coordinates in Space of Three Dimensions,
Cambridge U. Press, London, 1951.
Paul Heckbert, The Mathematics of Quadric Surface Rendering and SOID,
New York Inst. of Tech. Computer Graphics Lab TM4, July 1984.
Jim Blinn, The Algebraic Properties of Homogeneous Second Order Surfaces,
The Mathematics of Computer Graphics seminar notes, ACM SIGGRAPH '84,
July 1984.
Paul Heckbert
Pixar 415-499-3600
P.O. Box 13719 UUCP: {sun,ucbvax}!pixar!ph
San Rafael, CA 94913 ARPA: ph%pixar.uucp@ucbvax.berkeley.edusteve@thumper.UUCP (01/31/87)
>So, how do we map a quadratic surface Q by a map M? We want a surface K >such that p is on Q if and only if (p M) is on K; that is, we want a matrix >K such that > > p Q p' = 0 iff (p M) K (p M)' = 0 > >Since (p M)' = M' p', we conclude that > > K = M~ Q M~' > >will do the trick, where M~ is the inverse of M, and M~' its transpose. This will also work if M~ is replaced by M* (the adjoint of M), which always exists (the inverse doesn't), and is slightly cheaper to compute. BTW: A very nice description of some complicated stuff! -Steve Miller ihnp4!bellcore!thumper!steve
ph@pixar.UUCP (01/31/87)
Correction to a typo in my previous posting: to transform a plane N by matrix T: N' = T^-1*N, not N' = T*N. It makes perfect sense if you remember the correspondence between plane equations in the two coordinate systems: P'*N'=0 <==> P*N=0 : P'*N' = (P*T)*(T^-1*N) = P*(T*T^-1)*N = P*N Interesting note: some computer graphicists mistakenly transform normals (planes) using N' = T^t*N, but this is correct only for rotations.