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deriving a matrix

Started by me_here_me Jul 31, 2007 at 6:52 AM 4 replies 800+ views

Original Post

Jul 31, 2007 at 6:52 AM

According to a paper that I am trying to implement, the following expression returns a 4x4 matrix.

[Source]
/*
                      Np                                T  
SumPY = (1/Np) { SUMMATION_SIGN [ (pi - MueP) (yi - MueY) ] }
                    i = 1

"T" stands for transpose and is on top of (yi - MueY)
"p", "y", "MueP", "MueY" are all 3D points {x, y, z}
"Np" is the number of points, currently I have 4.
*/
[/Source]

[/source][/source][/source] I am not able to understand that how should I get a 4x4 matrix out of the above expression? Help needed in correctly understanding the equation. thanks in advance

ToohrVyk

238

ToohrVyk

Jul 31, 2007 at 6:59 AM

If X and Y are 3D vectors, then XY^T is a 3x3 matrix. If you want a 4x4 matrix, I assume you're working with projective geometry, so you can add a fourth w=1 component to your vectors.

Kambiz

100

Kambiz

Jul 31, 2007 at 7:19 AM

me_here_me

Author

me_here_me

Jul 31, 2007 at 7:26 AM

Thanks for the help.

and there is also a summation in question. Which of the following approaches is correct:

1.

[Source]/*                      Np                     p = (1/Np) { SUMMATION_SIGN [ (pi - MueP)  ] }                    i = 1                      Np                     y = (1/Np) { SUMMATION_SIGN [ (yi - MueY)  ] }                    i = 1          TSumPY = py*/[/Source]

[/source]

2. Or I generate a 4x4 matrix on each iteration on "i" and then add the matrices together at the end?

thanks for the help

Brother Bob

Mod

218

Brother Bob

Jul 31, 2007 at 8:20 AM

Second one is correct. Expand for a few iterations and you will see that first is not correct, but second is.

sum i = 0 to 2  ==  (a0*b0) + (a1*b1) + (a2*b2) !=                !=  (a0+a1+a2) * (b0+b1+b2)

Exactly what the formula says. For each i, make a matrix and sum them for all i. The summation is over (p_i-MueP)(y_i-MueY)^T, so that's what you sum.

me_here_me

Author

me_here_me

Jul 31, 2007 at 8:48 AM

thanks all for your help

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