So your "standard" quadratic attenuation function is just 1 / d^2. The problem with that is that it never actually hits 0, and so you can't bound it by a radius. The easiest way to do that is to just compute the attenuation value at the desired radius, and then subtract that from the attenuation factor:

float attenuation = max(1.0f / (distance * distance) - 1.0f / (radius * radius), 0.0f);

If you'd like you can pre-compute the 1 / r^2 part ahead of time on the CPU, since it's constant for that light source.

Epic outlined another formula here: Lighting Model
falloff = (saturate(1 ? (distance/lightRadius)4)2) ÷ (distance2 + 1)

Here is a test scene with all 3 equations. There are 3 point lights.

#1: Physically correct quadratic fall-off (scaled by light radius).

float Falloff( in float _fDistance, in float _fRadius ) {
	float fFalloff = 1.0 / Sqr( _fDistance );
	return fFalloff;
}

[attachment=25241:FalloffD2.png]
It indeed never reaches 0, but all 3 lights have approximately the correct visible ranges (you would still see the cut-off if you dropped to 0 at “range” units though).

#2: Epic’s fall-off.

float Falloff( in float _fDistance, in float _fRadius ) {
	float fFalloff = 0;
	float fDistOverRadius = _fDistance / _fRadius;
	float fDistOverRadius2 = Sqr( fDistOverRadius );
	float fDistOverRadius4 = Sqr( fDistOverRadius2 );
	fFalloff = Sqr( saturate( 1.0 - fDistOverRadius4 ) );
	fFalloff /= Sqr( _fDistance ) + 1.0;
	return fFalloff;
}

[attachment=25242:FalloffEpic.png]
Gives a very good fall-off. They outlined the artistic intent in the paper.

#3: The above formula.

float Falloff( in float _fDistance, in float _fRadius ) {
	float fFalloff = max( 1.0 / Sqr( _fDistance ) - 1.0 / Sqr( _fRadius ), 0.0 );
	return fFalloff;
}

[attachment=25240:FalloffMod.png]
Looks almost the same as Epic’s—only a few pixels in the specular around the headlights are different. The artistic intent is different however, and may give noticeably different results in other situations.


L. Spiro