Kyle,
(it
looks like you just figured it out, but I was about to send this response
anyway; let me know if it is consistent with your newly found
wisdom)
It looks like you want to be able to describe an
object in world coordinates, and view that object from a camera
positioned in world coordinates.
Let's say that the camera origin is at (Px, Py, Pz),
the camera's "up" direction is VUP = (VUPx, VUPy, VUPz), and the camera is
looking in the direction DIR = (DIRx, DIRy, DIRz). These are all defined
in world coordinates, which we will call W-space. This position and two
directions ("up" and "forward") define a "frame" or coordinate
system, which we will call C-space. You probably want to have the
VUP vector align with the +Y axis of C-space, and the DIR direction align with
the -Z axis of C-space.
We
are missing a vector to align with the +X axis of C-space, the "right" vector:
RT = (RTx, RTy, RTz). So, we need a new "right" vector (in
world space) which is perpendicular to VUP and DIR. This can be computed
with a cross product. Assuming DIR and VUP are normalized (unit
length):
RT =
DIR x VUP
Note
that RT is normalized, perpendicular to VUP and DIR, and points to the "right"
as required. If you do the cross product in the opposite order, RT will
point "left" which is not what we want.
With
these three vectors, and the camera position (all in W-space) we can easily
build a matrix that will take points in C-space and tell us where they are in
W-space. This matrix is:
RTx VUPx -DIRx
Px
RTy VUPy -DIRy
Py
RTz VUPz -DIRz
Pz
0 0 0
1
Notice that the first column, the "X column",
contains the RT vector, and similarly with the second and third columns.
(DIR is negated because we want to look down the -Z axis, not the
+Z)
However, this matrix is not quite what you need: this
matrix take points in C-space and expresses them in W-space, i.e. a "C-to-W"
matrix. I think you really want a matrix that will take points in
W-space and express them in C-space, a "W-to-C" matrix. This is simply
the inverse of the C-to-W matrix:
(corrected matrix)
RTx RTy RTz (-Px*RTx -
Py*RTy - Pz*RTz)
VUPx VUPy VUPz (-Px*VUPx -
Py*VUPy - Pz*VUPz)
-DIRx -DIRy -DIRz (Px*DIRx +
Py*DIRy + Pz*DIRz)
0
0 0 1
(sorry for the bad formatting) Using the
terminology for most graphics APIs, this is a view
matrix.
This
explanation assumes that your are using the "typical" transformation matrix
orientation, where the translation parameters are in the fourth column
(not the fourth row), i.e.
R R
R T
R R
R T
R R
R T
0 0
0 1
Clarification: I can easily translate the
vrp to the world coordinate origin. I just want to rotate the viewer's
coordinate system so that the axis coincides with the world coordinate
system. Does it have to do with calculating the projection, and then
doing the cross product? I am in dire straits. Any help is
greatly appreciated.
/**
* Kyle Wayne Kelly
* Computer Science
Student
* University of New Orleans
*
504-391-3985
*
http://www.cs.uno.edu/~kkelly
* Glory to God!
* El temor del hombre pondra lazo;
* Mas
el que confia en Jehova sera exaltado. (Proverbios 29:25)
*/
----- Original Message -----
Sent: Sunday, April 22, 2001 2:33
PM
Subject: [JAVA3D] Let me try to say
this more clearly
I am implementing a 3D system. I want
to shift the viewers reference point to the origin of the
world coordinate system. Second, I want to rotate the viewers
coordinate system, so that the users axis lie on top of the world
coordinate system. I want to use properties of special orthogonal
matrices. Can anyone explain how to use special orthogonal matrices
to rotate the viewers system to the world coordinate system? I am
implementing this in Java, but I am not using Java3D. Does anyone
know of any web pages that might have good information about this or can
explain how this rotation works? If you can explain using vectors
without any actual coding then that would be perfect.
/**
* Kyle Wayne
Kelly
* Computer Science Student
* University of New
Orleans
* 504-391-3985
* http://www.cs.uno.edu/~kkelly
* Glory to God!
* El temor del hombre pondra lazo;
*
Mas el que confia en Jehova sera exaltado. (Proverbios 29:25)
*/
