# Again transmissivity for color filter / glass

Hi Greg,

thank you again and YES it is confusing!

Actually, the Y value of transmittance can change for different light sources.

Forget my other question, at first i want to make sure that i model a simple filter in the right way.
And that is why i want to stick to the values LEE gives for a 3200K lamp.
These are: X= 58.8 , Y=62.4 , Z=37.2
with xyz_srgb.cal i get: R= 76.088002 G=61.6146944 B= 29.8622268, ( i want to calculate this with the D65 white point)
As the filter won't have a bigger transmission than 1, i divide through 100.
Now i get values of: R= .761 G= .616 B= .299;
If i apply these to glass polygon i get a yellow/orange filter with my tungsten 3200 K source in the scene.
Then is subtract 1 from RGB = .239 , .384, .701;
These values should now represent my blue filter, and it looks quite good!

void glass LeeGlass
0
3 .239 .384 .701

Am i right up to now?

These values would then be my normal transmittance. In order to model a disc of 'real'colored glass (e.g. n=1.52) i would apply these values to 'trans.cal' how you suggested an get the transmissivity.
rcalc -f trans.cal -e '\$1=tn(\$1)'
.239
0.260699013
.384
0.418782103
.701
0.763929466

Am i right here too?

Now i want to triple the effect of the filter, that means have 3 blue filters in a row. My Y=.624 would then be Y� = .243, can i simply multiply my normal transmittance of RGB with this factor to get the triple effect?

I know these are pretty simple questions, but as i plan to use the 'trans' material later i want to make sure to be on the right way!

Christian

Hi Christian,

I'm afraid you lost me, here.

Forget my other question, at first i want to make sure that i model a simple filter in the right way.
And that is why i want to stick to the values LEE gives for a 3200K lamp.
These are: X= 58.8 , Y=62.4 , Z=37.2

OK, I see these values on the URL you sent earlier <http://www.leefilters.com/LPFD.asp?PageID=193>.

with xyz_srgb.cal i get: R= 76.088002 G=61.6146944 B= 29.8622268, ( i want to calculate this with the D65 white point)

Since the XYZ values were measured using a 3200 K tungsten illuminant, converting to sRGB only gives you back the same orange color. You haven't really changed illuminants, because xyz_srgb.cal doesn't do a white-point conversion.

As the filter won't have a bigger transmission than 1, i divide through 100.
Now i get values of: R= .761 G= .616 B= .299;
If i apply these to glass polygon i get a yellow/orange filter with my tungsten 3200 K source in the scene.

Does the scene your simulating use a tungsten source or not? If it does, you should model it as a pure white light, because you have incorporated the illuminant color into your filter. (This isn't a good idea for transmission as the angular function will be incorrect.)

Then is subtract 1 from RGB = .239 , .384, .701;

This where you lost me completely. What made you think to subtract each value from 1? It doesn't work that way, even if it gets you what you were expecting.

These values should now represent my blue filter, and it looks quite good!

If you want blue out of your filter from a tungsten source, you won't get it. You'll get something bluer than the source, but not truly blue. This all gets involved in the white balance issue, which I cover in depth in the paper I keep mentioning:

I know it's a difficult read, but until you understand white balancing, colors are going to look consistently wrong to your eyes.

-Greg