transformations question
Started by davidko, Oct 07 2001 05:46 PM
14 replies to this topic
#1 Members  Reputation: 122
Posted 07 October 2001  05:46 PM
This question is pretty basic, but I just wanted to make sure I''m not making a mistake.
To transform a point from model space to camera space is simple this:
X'' = world_to_camera * model_to_world * X;
Now, if we want to transform a point in camera space to model space, we simple do this:
X = world_to_camera(T) * model_to_world(T) * X''
where the (T) means a transpose of the matrix (we are assuming just rotations and translations here).
Let''s call world_to_camera(T) * model_to_world(T) = H
Now, since we are dealing with orthogonal matricies, then we should be able to use the rotational part of matrix H to transform a normal vector from camera space to model space, right?
#2 Members  Reputation: 122
Posted 07 October 2001  11:28 PM
quote:
Now, if we want to transform a point in camera space to model space, we simple do this:
X = world_to_camera(T) * model_to_world(T) * X''
Is that really true!?
I think the correct transformation is this:
X = model_to_world(T) * world_to_camera(T) * X''
regards
/Mankind gave birth to God.
#6 Members  Reputation: 113
Posted 09 October 2001  02:11 PM
hrm, well, the way i remember the order of multiplication is that you multiply the parent transforms first, and local object transform comes last.. so,
world_transform = root_node * node1 * node2 * node3.. * local_object_transform
So, the parent transforms are on the left, just as you''d expect.
bart
world_transform = root_node * node1 * node2 * node3.. * local_object_transform
So, the parent transforms are on the left, just as you''d expect.
bart
#7 Members  Reputation: 122
Posted 09 October 2001  09:03 PM
bpj1138:
I''m sorry but I don''t know what you mean. I just corrected davidko''s second transformation, which ought to be right now. So, what do you mean?
I haven''t looked up any formulas (no need to) but I''m pretty confident that I have reasoned correctly.
Could you explain a bit more?
/Mankind gave birth to God.
I''m sorry but I don''t know what you mean. I just corrected davidko''s second transformation, which ought to be right now. So, what do you mean?
I haven''t looked up any formulas (no need to) but I''m pretty confident that I have reasoned correctly.
Could you explain a bit more?
/Mankind gave birth to God.
#8 Members  Reputation: 491
Posted 09 October 2001  11:55 PM
I take it you mean rotations only and not rotations and translations since I don''t think the transpose and inverse are the same when translations are included. I''m not particularly strong with matrices but when I tried multiplying a matrix that included translation by its transpose I didn''t get the identity matrix.
#9 Members  Reputation: 122
Posted 10 October 2001  12:23 AM
quote:
Original post by LilBudyWizer
I take it you mean rotations only and not rotations and translations since I don't think the transpose and inverse are the same when translations are included. I'm not particularly strong with matrices but when I tried multiplying a matrix that included translation by its transpose I didn't get the identity matrix.
You're right. Ortogonal matrices' inverses are simply the transposes. Rotation matrices are ortogonal, and translation matrices are not ortogonal.
/Mankind gave birth to God.
Edited by  silvren on October 10, 2001 7:26:38 AM
#10 Members  Reputation: 1373
Posted 10 October 2001  04:55 AM
(Edit  Actually, ignore this post. As silvren pointed out, my statement is wrong  Graham)
That's not exactly correct. For example, a scaling transformation matrix is orthogonal:
But its transpose and inverse are the same only if sfx = sfy = sfz = 1.
Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.
Edited by  grhodes_at_work on October 10, 2001 11:55:42 AM
Edited by  grhodes_at_work on October 10, 2001 7:14:12 PM
quote:
Original post by silvren
You're right. Ortogonal matrices' inverses are simply the transposes.
That's not exactly correct. For example, a scaling transformation matrix is orthogonal:

But its transpose and inverse are the same only if sfx = sfy = sfz = 1.
Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.
Edited by  grhodes_at_work on October 10, 2001 11:55:42 AM
Edited by  grhodes_at_work on October 10, 2001 7:14:12 PM
#11 Members  Reputation: 122
Posted 10 October 2001  10:28 AM
quote:
Original post by grhodes_at_work
That's not exactly correct. For example, a scaling transformation matrix is orthogonal:
[sfx 0 0 ]
S = 0 sfy 0 
[ 0 0 sfz]
But its transpose and inverse are the same only if sfx = sfy = sfz = 1.
First, how can you determine that your matrix is orthogonal?
A matrix, if my memory's still with me, is only called orthogonal if its inverse is the same as its transpose, and in your case the orthogonality depends on the variables' values.
Second, the following properties must be true for your matrix to be orthogonal:
sfx = +1 sfy = +1 sfz = +1
So you can in fact get N=2^3=8 different matrices that are orthogonal, not only one.
To sum it up:
A*A(T)=I iff square matrix A is orthogonal.
regards
/Mankind gave birth to God.
Edited by  silvren on October 10, 2001 5:29:39 PM
Edited by  silvren on October 10, 2001 5:30:31 PM
Edited by  silvren on October 10, 2001 5:31:35 PM
#12 Members  Reputation: 113
Posted 10 October 2001  10:31 AM
OOPS.. sorry, I realized my message had nothing to do with the discussion. What I was talking about is computing the world transform for a particular node in a scene graph structure, and the order of multiplication involved there, but what this post is talking about is "change of coordinates", between two separate nodes in the scene graph.
It's also better not to talk about this with the word "camera", because it may become confused with a camera transform for a particular coordinate system/device.
So "change of coordinates" is changing between two local coordinate systems, say A, and B. You have a coordinate in A's local system, and you want to interpret it in B's local system.
In this case, you have to multiply A's coordinate first by A's world transform, to get the coordinate into world coordinates. Then multiply it by the inverse of B's world transform. So it's actually A * B(T)? heh, or it might be B(T) * A..
bart
Edited by  bpj1138 on October 10, 2001 5:34:34 PM
It's also better not to talk about this with the word "camera", because it may become confused with a camera transform for a particular coordinate system/device.
So "change of coordinates" is changing between two local coordinate systems, say A, and B. You have a coordinate in A's local system, and you want to interpret it in B's local system.
In this case, you have to multiply A's coordinate first by A's world transform, to get the coordinate into world coordinates. Then multiply it by the inverse of B's world transform. So it's actually A * B(T)? heh, or it might be B(T) * A..
bart
Edited by  bpj1138 on October 10, 2001 5:34:34 PM
#13 Members  Reputation: 1373
Posted 10 October 2001  12:13 PM
quote:
Original post by silvren
First, how can you determine that your matrix is orthogonal?
A matrix, if my memory''s still with me, is only called orthogonal if its inverse is the same as its transpose, and in your case the orthogonality depends on the variables'' values.
Second, the following properties must be true for your matrix to be orthogonal:
sfx = +1 sfy = +1 sfz = +1
So you can in fact get N=2^3=8 different matrices that are orthogonal, not only one.
To sum it up:
A*A(T)=I iff square matrix A is orthogonal.
silvren, you are absolutely correct on both points. Thanks for calling me on my error. What the hell was I thinking? (Ever hear of a brain fart?)
Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.
#14 Members  Reputation: 113
Posted 10 October 2001  12:55 PM
It''s true that orientation matrix with a position vector is not "orthogonal square matrix", so you cannot transpose it. To compute the inverse, you can still transpose the 3x3 axis vector matrix, but then you have to negate the position vector and dot it with the three axis vectors.
so if you have the orientation matrix (R = right, U = up, D = direction, P = position):
Rx Ry Rz 0
Ux Uy Uz 0
Dx Dy Dz 0
Px Py Pz 1
the inverse orientation matrix is:
Rx Ux Dx 0
Ry Uy Dy 0
Rz Uz Dz 0
P*R P*U P*D 1
notice the three axis vectors are still simply transposed from columns to rows. the position vector has to be negated and dotted with the three axis (in the original matrix, not the inverse)
bart
so if you have the orientation matrix (R = right, U = up, D = direction, P = position):
Rx Ry Rz 0
Ux Uy Uz 0
Dx Dy Dz 0
Px Py Pz 1
the inverse orientation matrix is:
Rx Ux Dx 0
Ry Uy Dy 0
Rz Uz Dz 0
P*R P*U P*D 1
notice the three axis vectors are still simply transposed from columns to rows. the position vector has to be negated and dotted with the three axis (in the original matrix, not the inverse)
bart
#15 Members  Reputation: 122
Posted 10 October 2001  03:53 PM
Normally, for graphics stuff, the interest in having an orthogonal matrix stems from wanting to efficiently transform normals, in which case translation doesn''t even factor in (i.e. how do you tranlate a normal)? So the usual textbook answer is, yes, if you have a rotation, translation, and uniform scaling, then your matrix is orthogonal. Of course, they usually say this in the context of normals, since translating a model will not affect its normals.