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#ActualSébastien Lagarde

Posted 29 September 2012 - 01:23 PM

Hey,

I don't understand your thread but for this part:
"and they decide the coefficients are:
(Pi, 2 Pi / 3, Pi / 4) for 3 bands of Zonal Harmonics. (so simply 0 on the non zonal terms)"
I join you a shot of mathematica showing the derivation. It is just a projection of max(cos(theta), 0) into SH then an application of the convolution term.

Also, the D3DX SH math library is now open source and containt the cubemap projection, so enjoy:
http://blogs.msdn.co...onics-math.aspx

Cheers

#1Sébastien Lagarde

Posted 29 September 2012 - 01:21 PM

Hey,

I don't understand your thread but for this part:
"and they decide the coefficients are:
(Pi, 2 Pi / 3, Pi / 4) for 3 bands of Zonal Harmonics. (so simply 0 on the non zonal terms)"
I join you a shot of mathematica showing the derivation. It is just a projection of max(cos(theta), 0) into SH then an application of the convolution term.

Also, the D3DX SH math library us now open source and containt the cubemap projection, so enjoy:
http://blogs.msdn.com/b/chuckw/archive/2012/07/28/spherical-harmonics-math.aspx

Cheers

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