[Fwd: 3D Rotation Transformation]

[Ron Boisvert <boisvert@nist.gov>]

-------- Original Message --------
Subject: 	3D Rotation Transformation
Date: 	Wed, 10 Sep 2008 11:34:42 -0400
From: 	Sione <sionep@xtra.co.nz>
Reply-To: 	sionep@xtra.co.nz
To: 	boisvert@nist.gov
References: 	<4836D9F0.9010401@nist.gov>




Dear List,

Perhaps, my problem is not related directly to JAMA, however I am using 
JAMA for a  3D ray-tracing visualization.

Does anyone know  the  cartesian  representation of  a  cone (in  3D)   
after it has been rotated anti-clockwise along the  x-axis  by  an  
angle   t  ?     The  cartesian equation for the original cone   is :

x^2 + y^2  =  (k*z)^2 

where   k   is  a  constant.   The [3x3]  transformation matrix for the 
rotation by angle  t  is :

T = [ 1,     0,     0*;*    0,      cos(t),     -sin(t)*;*   0,     
sin(t),      cos(t) ]

If  a  point   P  , ie,  (x,y,z)  on the cone is rotated by  T,   it is 
maps into   say   point  P',  ie,  (x', y', z')   which  results in:

x'  =  x
y'  = cos(t)*y - sin(t)*z
z'  =  sin(t)*y + cos(t)*z

Can my rotated (transformed) cone be now represented by:

x*'*^2 + y*'*^2  =  (k*z*'*)^2

The cartesian equation of the transformed  cone is really needed, since 
it is easy to compute the normal  vector to any point on the transformed 
cone's surface via  Jacobian.  Once the normal vector is calculated for 
a specific point on the transformed cone, then  Snell's Law (from 
optics)  is applied to the incident  ray vector that is refracted by the 
transparent  cone.

Is my interpretation correct here that the cartesian of the transformed 
cone is  x*'*^2 + y*'*^2  =  (k*z*'*)^2   ?

Any hints would be appreciated.

Thanks,
Sione.