• FEATURED

View more

View more

View more

### Image of the Day Submit

IOTD | Top Screenshots

### The latest, straight to your Inbox.

Subscribe to GameDev.net Direct to receive the latest updates and exclusive content.

# Bottom Row of 4x4 Matrix

Old topic!

Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.

17 replies to this topic

### #1caibbor  Members

Posted 15 December 2012 - 09:40 PM

I'm writing a matrix library.

In an opengl matrix, you have:

R R R T
R R R T
R R R T
A A A B

where R is rotation and T is translation.

what is A and B?

how would I apply A and B to a Vec3? ( I know how to translate and rotate )

how would I invert A and B?

### #2Yours3!f  Members

Posted 16 December 2012 - 03:03 AM

well first take a look at this:
http://www.songho.ca/opengl/gl_transform.html
(the other pages are very useful too!!! so look around)

I'm not really sure if A is used at all, I did this a long time ago, but B is supposed to be the perspective divide element.
Or at least B and the projection matrix modify the 4th element of the position, so that in clip space if you divide by it, you get the ndc coordinates.
like when you do the transformations:
-Model (world) space (apply T here):
model_space_pos = model_mat * vertex
-View (camera) space (apply R here):
view_space_pos = view_mat * model_space_pos
-Clip space (apply projection matrix here):
clip_space_pos = projection_mat * view_space_pos
-normalized device coordinates (do the perspective divide, apply B here):
ndc_pos = clip_space_pos.xyz / clip_space_pos.w
-viewport coordinates (scale & bias to get texture coordinates, scale by window coordinates)
viewport_pos = (ndc_pos.xy * 0.5 + 0.5) * vec2( screen_width, screen_height )

and to the other questions:
you would apply the 4x4 matrix to a 4 component vector (ie vec4)
and you would invert A and B by inverting the 4x4 matrix.

### #3Brother Bob  Moderators

Posted 16 December 2012 - 04:14 AM

I'm not really sure if A is used at all, I did this a long time ago, but B is supposed to be the perspective divide element.
Or at least B and the projection matrix modify the 4th element of the position, so that in clip space if you divide by it, you get the ndc coordinates.

The entire bottom row is used for the perspective division, not just the B-element.

For example, for an orthographic projection, the bottom row is [0, 0, 0, x] where x is some value depending on the parameters of the projection matrix. That means that the fourth component of the resulting vector after multiplication is the fourth component of the in multiplying vector, to some scale factor. For glOrtho, x=1 at all times though, which means that as long as the Z-coordinate of the input is 1.0, which is often the case, there fourth component of the resulting vector is also 1.0, and there is effectively no perspective division, hence no perspective since it is an orthographic projection.

On the other hand, for a perspective matrix, the bottom row is [0, 0, x, 0]. That means that the fourth component of the resulting vector is the third component of the multiplying vector, to some scale factor x. The third component is the depth, and hence the perspective division is now dependent on the depth; you now get a perspective effect.

Likewise, the first two elements of the bottom row can also be non-zero to get a perspective effect along the X and/or the Y-axis instead.

how would I apply A and B to a Vec3? ( I know how to translate and rotate )

It doesn't make much sense to talk about how to apply these elements to a 3-element vector. You simply cannot multiply a 3-element vector by a 4-by-4 matrix in the first place. What you do when adding multiplying the 3-element vector by the rotation part and then adding the translation part as a separate step is really just assuming that the missing fourth component of the 3-element vector is unity. In order to handle the fourth row of the matrix correctly, you have to do the same assumption again, carry out the multiplication, and see how the bottom row affects the other three elements, as well as performing the final perspective division to ensure that the assumption that the fourth component really is unity even after the multiplication.

In the end, you really have to carry out a full 4-vector times 4x4-matrix multiplication, although you can assume that one element is unity and eliminate its multiplication with the corresponding elements of the matrix, and just add them.

Edited by Brother Bob, 16 December 2012 - 04:14 AM.

### #4caibbor  Members

Posted 16 December 2012 - 05:09 PM

thank you for the insight. I'm still re-reading this a few times to let it sink in.

one problem I'm having is trying to create a frustum from a unit cube by multiplying each of the cube's vertices by the inverse of a projection matrix (as I'm told that is how it's done) so I can do basic view-frustum culling with world objects. currently, the result I'm getting is not a frustum.

### #5caibbor  Members

Posted 16 December 2012 - 05:26 PM

okay, I think I understand everything you've said.

doing this wihtout inverting for the first try:

I've got a vector (1,2,3) and I want to multiply it by my perspective matrix:

0.75 0 0 0
0 1 0 0
0 0 -1.0001 -0.20001
0 0 -1 0

after applying rotation and translation, I have this:

0.75,2,-3.20031

and then to apply the projection division... this is where I'm not totally sure how to apply it.. but I think I would end up with a Vec4:

0.75,2,-3.20031,-1

### #6Brother Bob  Moderators

Posted 16 December 2012 - 05:42 PM

I addressed that in my last paragraph; you have to perform a full 4-dimensional multiplication because you cannot multiply a 3-dimensional by a 4x4 matrix. You have to expand the multiplication using a full 4-dimensional vector.

Here's your matrix and vector and the product of the two.
>> M
M =
0.7500    0         0         0
0         1.0000    0         0
0         0        -1.0001   -0.2000
0         0        -1.0000    0
>> v
v =
1
2
3
1
>> p = M*v
p =
0.7500
2.0000
-3.2003
-3.0000

See how the fourth element of the vector p is -3.0? That's because your 3-dimensional vector is actually a 4-dimensional vector with an implicit element at the end with a value of 1.0.

The perspective division is just dividing all elements of the vector by the fourth element:
>> p./p(4)
ans =
-0.2500
-0.6667
1.0668
1.0000

So your resulting projected 3-dimensional vector is (-0.2500, -0.6667, 1.0668).

Edited by Brother Bob, 16 December 2012 - 05:46 PM.

### #7caibbor  Members

Posted 16 December 2012 - 05:45 PM

oh wow, that's easy. thank you very much! sometimes it takes baby-speak and sock puppets to explain things to me

### #8caibbor  Members

Posted 16 December 2012 - 09:26 PM

I'm curious... what program are you using there in the last post? has some interesting notation

### #9BitMaster  Members

Posted 17 December 2012 - 02:38 AM

My guess would be MatLab or Octave.

### #10Brother Bob  Moderators

Posted 17 December 2012 - 03:44 AM

My guess would be MatLab or Octave.

It's Matlab, to be specific.

### #11larspensjo  Members

Posted 17 December 2012 - 05:15 AM

I am using Octave for this, and I can recommend it. I attached a zip file to this post that contains 4 simple macros to help you experiment with matrices. It helped me to quickly experiment and gain understanding.
rot: Create a rotation matrix.
scale(x,y,z): Create a scaling matrix
transl(x,y,z): Create a translation matrix
Shearz: Create a shear matrix in z.

Examples:
octave:3> scale(1,2)
ans =
1   0   0   0
0   2   0   0
0   0   1   0
0   0   0   1
octave:4> v=[1;2;3;1]
v =
1
2
3
1
octave:6> scale(1,2)*v
ans =
1
4
3
1
octave:7> rot(pi/2)
ans =
1.00000   0.00000   0.00000   0.00000
0.00000   0.00000  -1.00000   0.00000
-0.00000   1.00000   0.00000   0.00000
0.00000   0.00000   0.00000   1.00000
octave:8> rot(pi/2)*v
ans =
1
-3
2
1
octave:9> inverse(rot(pi/2))
ans =
1.00000   0.00000  -0.00000  -0.00000
0.00000   0.00000   1.00000  -0.00000
0.00000  -1.00000   0.00000  -0.00000
0.00000   0.00000   0.00000   1.00000
octave:11> T=transl(1);
octave:19> S=scale(2);
octave:20> S*T*v
ans =
4
2
3
1
octave:21> T*S*v
ans =
3
2
3
1

#### Attached Files

Current project: Ephenation.
Sharing OpenGL experiences: http://ephenationopengl.blogspot.com/

### #12caibbor  Members

Posted 17 December 2012 - 03:08 PM

thank you, larspensjo. I was delighted to find octave available for my linux distro. but how about projection/perspective matrices?

I presume the way to invert the perspective section of a matrix is to just negate the values, just the same as negating the translation parts. is this correct?

### #13Brother Bob  Moderators

Posted 17 December 2012 - 04:01 PM

thank you, larspensjo. I was delighted to find octave available for my linux distro. but how about projection/perspective matrices?

I presume the way to invert the perspective section of a matrix is to just negate the values, just the same as negating the translation parts. is this correct?

You cannot invert the individual sections of a matrix just like that. It just happens to work for translation under very specific circumstances. Likewise, you can invert other sections under other specific conditions.

If you have a matrix consisting of the perspective division component only, then you can invert it by negating the bottom row just like you do with translation. But if you have other components of the projection matrix, and a usual projection matrix does have both translation and non-uniform scaling, then you cannot do that anymore.

But take a look at how OpenGL's matrices are commonly designed here in appendix F. It lists their inverse also.

### #14caibbor  Members

Posted 18 December 2012 - 09:51 AM

But take a look at how OpenGL's matrices are commonly designed here in appendix F. It lists their inverse also.

I tried implementing my own fucntions to generate a matrix based on those equations, as well as an inverse:

Multiplying a matrix together with it's own inverse should result in an identity matrix. I'm getting mostly identity, except col 3 row 4 is always -5.9, and col 4 row 3 is always -1.2. I've tripple-checked that my matrices match the ones in the image.

I think the problem is in my matrix multiplication. I know how to multiply the 3x3 rotation section of the matrix and then add the translation, but what do I do with the perspective section?

Edit: I've been googling and even looked into a few CG programming books. nobody seems to say how to handle the bottom row of a GL 4x4 when multiplying two matrices together...

Edited by caibbor, 18 December 2012 - 12:19 PM.

### #15Brother Bob  Moderators

Posted 18 December 2012 - 01:08 PM

You have two 4x4 matrices so you have to multiply them as 4x4 matrices. The bottom row is just like any other row in the matrix, there is nothing special about it at all.

### #16caibbor  Members

Posted 18 December 2012 - 06:07 PM

alright, then riddle me this.

(all in column-major order)

The projection matrix I created is:
0.074978 0.000000 0.000000 0.000000 0.000000 0.099970 0.000000 0.000000 0.000000 0.000000 -1.000100 -1.000000 0.000000 0.000000 -0.200010 0.000000

It's inverse I calculated is:
13.337285 0.000000 0.000000 0.000000 0.000000 10.002963 0.000000 0.000000 0.000000 0.000000 0.000000 -4.999750 0.000000 0.000000 -1.000000 20001000.000000

an multiplying those matrices together should be identy, but instead I get:
1.000000 0.000000 0.000000 0.000000 0.000000 1.000000 0.000000 0.000000 0.000000 0.000000 1.000000 0.000000 0.000000 0.000000 -20000994.000000 1.000000

All code related to this is shown below. I've been up and down and all the methods, as far as I can tell, look correct.
[source lang="cpp"]// Mat4f class is a 4x4 matrix of floats.// Mat4f::data is a float[16], and is column-major// returns float& intentionally, method is privateinline float& Mat4f::M( const int col, const int row ) { assert( col < dim && row < dim && col >= 0 && row >= 0); return data[row+col*4];}void Mat4f::Multiply( const Mat4f& mat, Mat4f& out ) const { assert ( &mat != &out ); int r,c; for ( c=0; c<4; c++ ) { for ( r=0; r<4; r++ ) { out.M(c,r) = M(r,0) * mat.M(0,c) + M(r,1) * mat.M(1,c) + M(r,2) * mat.M(2,c) + M(r,3) * mat.M(3,c); } }}// calculation from: http://www.glprogramming.com/red/images/Image23.gifvoid my_glFrustum( float l, float r, float b, float t, float n, float f, Mat4f& out ) { float n2 = n * 2.0f; out.data[0]=n2/(r-l); out.data[1]=0.0f; out.data[2]=0.0f; out.data[3]=0.0f; out.data[4]=0.0f; out.data[5]=n2/(t-b); out.data[6]=0.0f; out.data[7]=0.0f; out.data[8]=(r+l)/(r-l); out.data[9]=(t+b)/(t-b); out.data[10]=-(f+n)/(f-n); out.data[11]=-1.0f; out.data[12]=0.0f; out.data[13]=0.0f; out.data[14]=-(n2*f)/(f-n); out.data[15]=0.0f;}// calculation from: http://www.glprogramming.com/red/images/Image23.gifvoid my_glFrustum_inv( float l, float r, float b, float t, float n, float f, Mat4f& out ) { float n2 = n * 2.0f; out.data[0]=(r-l)/n2; out.data[1]=0.0f; out.data[2]=0.0f; out.data[3]=0.0f; out.data[4]=0.0f; out.data[5]=(t-b)/n2; out.data[6]=0.0f; out.data[7]=0.0f; out.data[8]=0.0f; out.data[9]=0.0f; out.data[10]=0.0f; out.data[11]=-(f-n)/(n2*f); out.data[12]=(r+l)/n2; out.data[13]=(t+b)/n2; out.data[14]=-1.0f; out.data[15]=(f+n)/n2*f;}void my_gluPerspective( float fov, float x,float y,float near,float far, Mat4f& out, bool invert = false );void my_gluPerspective( float fov, float x,float y,float near,float far, Mat4f& out, bool invert ) { // inspired by [url="http://www.cse.unsw.edu.au/~cs3421/labs/nate/frustsrc.html"]http://www.cse.unsw....e/frustsrc.html[/url] float aspect = x / y; float t = 1.0f/tanf(fov * 3.141f / 360.0f); if ( invert ) { my_glFrustum_inv(-t * aspect, t * aspect, -t, t, near, far, out); } else { my_glFrustum(-t * aspect, t * aspect, -t, t, near, far, out); }}inline void my_gluPerspective_inv( float fov, float x,float y,float near,float far, Mat4f& out ) { my_gluPerspective( fov, x, y, near, far, out, true );}int main ( void ) { Mat4f should_be_identity; Mat4f projection; Mat4f projection_inverse; my_gluPerspective(90,800,600,0.1f,2000,projection); my_gluPerspective_inv(90,800,600,0.1f,2000,projection_inverse); projection.Multiply(projection_inverse, should_be_identity); for ( int i=0; i<16; i++ ) printf("%f ", should_be_identity.data[i] ); /* expected output: 1.000000 0.000000 0.000000 0.000000 0.000000 1.000000 0.000000 0.000000 0.000000 0.000000 1.000000 0.000000 0.000000 0.000000 0.000000 1.000000 actual output: 1.000000 0.000000 0.000000 0.000000 0.000000 1.000000 0.000000 0.000000 0.000000 0.000000 1.000000 0.000000 0.000000 0.000000 -20000994.000000 1.000000 */ return 0;}[/source]

Edited by caibbor, 18 December 2012 - 06:09 PM.

### #17Brother Bob  Moderators

Posted 18 December 2012 - 06:19 PM

Your element 15 of the inverse is incorrect. The denominator is missing a parenthesis to ensure that the multiplication by f doesn't end up in the numerator.

### #18caibbor  Members

Posted 18 December 2012 - 06:52 PM

"I always do that. I always mess up some mundane detail." - Michael Bolton, Office Space

thanks.

Old topic!

Guest, the last post of this topic is over 60 days old and at this point you may not reply in this topic. If you wish to continue this conversation start a new topic.