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Point light attenuation's off..


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#1 cozzie   Members   -  Reputation: 1584

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Posted 16 July 2013 - 02:29 PM

Hi,

I've managed to implement a reasonally flexible lighting shader, including both directional and point lights. With various numbers of max. lights (different copies of the effect/shader).

 

But there's something strange with the light attenuation of my point lights.
For testing purposes I've set up the following scene:

 

- place a point light at position x (-5, 1, 5)

- give the point light a max. range of 5.0, fullpower (full bright) range of 3.0

- place a sphere with blended material at position x (-5, 1, 5)

 

And then simply check if the point light stops shining where the sphere begins.

Unfortunately... it doesn't.

I've double checked the range/scale of the sphere, it matches the 5 'units' in the same scale as the pointlight range of '5'.

 

Here's a screenshot of the result and the actual attenuation calculation.

Any idea where it's going wrong?

 

plightatt.jpg

// PER PIXEL ATTENUATION
float dist = length(PointLightPos[i] - input.wPos);
float att = saturate(1 - (dist - PointLightFPRange[0]) / (PointLightRange[0] - PointLightFPRange[0]));
att *= att;


Edited by cozzie, 16 July 2013 - 02:30 PM.


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#2 cozzie   Members   -  Reputation: 1584

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Posted 16 July 2013 - 02:49 PM

Updated: when I set the full power range to zero, the sphere matches the light range.
And similar, when I scale the sphere to 8 instead of 5 it's also matching.
So definately something wrong in combination with the fullpower range (seems to be added to the max range).

#3 cozzie   Members   -  Reputation: 1584

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Posted 18 July 2013 - 02:01 PM

Anyone?

#4 unbird   Crossbones+   -  Reputation: 4973

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Posted 18 July 2013 - 02:17 PM

What is PointLightFPRange ? Can't you just use one falloff range ?

Edit: Sorry, reading helps. How about:
saturate(1 - (dist - PointLightFPRange[0]) / PointLightRange[0]);

Wait, pen and paper helps even more. You want a linear function f(x) = a * x + b (lets first ignore the saturation) where.
f(r) = 1
f(R) = 0
Where - for notational convenience - R is the full falloff range and r the (smaller) full power range. So
f(r) = a*r + b = 1
f(R) = a*R + b = 0
Subtracting these equations gives.
a*r + b - a*R - b = 1 - 0 <=>
a*(r - R) = 1 <=>
a = 1 / (r-R)
Using in the above second equation:
R / (r-R) + b = 0 <=>
b = - R / (r - R)
So, now with saturation:
saturate(dist / (r-R) - R / (r-R) =
saturate((dist - R )/ (r-R)) =
saturate((dist - PointLightRange[0])/ (PointLightFPRange[0] - PointLightRange[0]))
You almost had it wink.png

Edit: graph with range 3 and 5

Edited by unbird, 18 July 2013 - 03:19 PM.


#5 cozzie   Members   -  Reputation: 1584

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Posted 18 July 2013 - 04:12 PM

Thanks, I'll have to rewrite that exercise myself :)
So basically what I'm deviding by was wrong, will test it out tomorrow and let you know

#6 cozzie   Members   -  Reputation: 1584

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Posted 20 July 2013 - 04:37 AM

Hi unbird,

I've got it, also wrote it down with some examples on paper.

My dummy explanation is that I take fully lit (1.0) minus the range that's attenuated divided by the total attenuation range.

 

Here are 2 screenshots with a point light with range 5, one with fullpower range also 5 and one with FP range 2.

There's still some more brighter light in the center in the 5 / 2 example, but that's probably because of the Y position of the light, above the terrain (the light being a sphere).

 

I'm also trying to find out why I do the attenuaton in the power of 2 (att = att * att).

Any idea?

 

plight5-2.jpg

 

plight5-5.jpg


Edited by cozzie, 20 July 2013 - 04:38 AM.


#7 unbird   Crossbones+   -  Reputation: 4973

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Posted 22 July 2013 - 06:08 AM

Here are 2 screenshots with a point light with range 5, one with fullpower range also 5 and one with FP range 2.
There's still some more brighter light in the center in the 5 / 2 example, but that's probably because of the Y position of the light, above the terrain (the light being a sphere).

Well, if that isn't just a debug/test output visualizing your attenuation, then yes, that's expected. It's the "cosine term" (N dot L, normal dot light direction) of your point light calculation. The shallower the angle at which the light hits, the less the surface gets illuminated.
 

I'm also trying to find out why I do the attenuaton in the power of 2 (att = att * att).
Any idea?


Why you do that ? How would I know ? wink.png

In this case it smoothes the fade out at the full range, see here

Question is, if this is desired. A more physically correct attenuation would even respect the inverse square law (1/dist2). But that would make your light's bounding sphere infinite. IIRC the fixed function pipeline does support different formulas (linear, 1/r or 1/r^2 attenuation). Just use something that looks good enough.

#8 cozzie   Members   -  Reputation: 1584

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Posted 22 July 2013 - 03:25 PM

Thanks.

I've now added the "att *= att" because indeed it smoothens things.

It also looks good for my goal now and the range is as I expect.






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