Short answer: You can't really. Representing a point-light requires an infinite number of coefficients. Long answer: You can approximate it using a circular shape. The paper "Algorithms for Spherical Harmonic-Lighting" By Ian G. Lisle and S.-L. Tracy Huang gives a method for calculating the coefficients for this directly.
You'd also have to know the distance to the light.
Short answer: You can't really. Representing a point-light requires an infinite number of coefficients. Long answer: You can approximate it using a circular shape. The paper "Algorithms for Spherical Harmonic-Lighting" By Ian G. Lisle and S.-L. Tracy Huang gives a method for calculating the coefficients for this directly.
You'd also have to know the distance to the light.
Ah ok. The confusion was because an infinite/directional and a point/omni light are different things.
Well, as I mentioned, let's take the direction vector [0 0 1], with an infinite point light source.
Ah ok. The confusion was because an infinite/directional and a point/omni light are different things.
[quote name='opengl_beginner' timestamp='1353232912' post='5001996']
Well, as I mentioned, let's take the direction vector [0 0 1], with an infinite point light source.
Directional lights are trivial to compute, you simply evaluate the SH basis functions in the given direction and scale appropriately (see Normalization section.) Spherical Light sources can be efficiently evaluated using zonal harmonics. Below is a diagram showing an example scene, we want to compute the incident radiance, a spherical function, at the receiver point P. Given a spherical light source with
center C, radius r, what is the radiance arriving at a point P d units away? The sin of the half-angle subtended by the light source is r/d, so you just need to compute a light source that subtends an appropriate part of the sphere. The ZH coefficients can be computed in closed form as a function of this angle: ? = integral ( integral (y_l, \theta, 0, 2Pi ) ,\theta, 0, a ) where a is the half-angle d subtended. See Appendix A3 ZH Coefficients for Spherical Light Source for the expressions through order 6.
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I assume a = r/d, where d is infinite and thus a = 0.
I've looked into the Appendix for L = 1, ...., 6:
L=0: ?sqrt(?)(?1+cos(a))
L =1: 1/2 sqrt(3) sqrt(?) sin(a)2
L=2: ?1/2 * sqrt(5) * sqrt(?) * cos(a) (?1 + cos(a)) (cos(a) + 1)
L=3 ?1/8 *sqrt(7)* sqrt(?) (?1 + cos(a)) (cos(a) + 1) (5 cos(a)^2 ? 1)
L=4 ?3/8*sqrt(?)*cos(a) (?1 + cos(a)) (cos(a) + 1) (7 cos(a)2 ? 3)
L=5 ? 1/16* sqrt(11)*sqrt(?)* (?1 + cos(a)) (cos(a) + 1) (21 cos(a)4 ? 14 cos(a)2 + 1)
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So, clearly, all coefficients will be 0 due to the term (-1 + cos(a)), which is 0 for a = 0.
Where is the flaw in my derivation?
Hey ginkgo! Well, yes - but isn't that exactly the idea behind spherical harmonics - approximating a function using a polynomial of a finite degree (e.g. degree 2 already gives an error rate less than 1%).
I expect the equivalent situation in spherical harmonics will also have funny ripples that it would take many terms to make small.
