Am 31.03.2016 um 04:28 schrieb Ben Goren <ben@trumpetpower.com>:
The sky is often north of 15,000 K, and horizon sunlight can be below 5,000 K. The most naive of math would suggest that, to pair with my 6500 K "sky" I need a "Sun" with a color temperature thousands of degrees below absolute zero --
I’m afraid that would be more naive than math. ;-) Wayne Bretl’s suggestion is extremely simple and probably sufficient for practical purposes, but it’s not really precise from a colorimetric POV. Using the values from Wayne Bretl’s example calculation (sky = 16,000 K, sun = 5,000 K, fake sky = 6,500 K), the precise calculation would be as follows: 1. Calculate the xy values and from them the XYZ values for daylight with 16,000 K (using the CIE Daylight formula): D160xy = (x = 0.2594, y = 0.2676) D160XYZ = (X = 0.9693, Y = 1, Z = 1.7670) The XYZ values for D50 and D65 are well known. 2. Calculate the Bradford chromatic adaptation matrix for a white point shift from D160 to D65 (= from the real to the fake sky) (A matrix is hard to print in an email, so I omit it here.) 3. Chromatically adapt D50XYZ (= the sun) with this matrix. You’ll get: adaptedD50XYZ = (X = 0.9961, Y = 1, Z = 0.5010) 4. Calculate the correlated CCT for the adapted XYZ value: adaptedD50CCT = 3718 K So your fake sun has 3718 K. This should be as precise as it gets in colorimetry. The value is roughly 10% higher than the value calculated with Wayne Bretl’s method; the chromaticity difference (in uv(1976)) is deltaC = 9.79, so it’s relatively high and certainly visible. It would be interesting to know which CCT value achieves the better result artistically. Bye Uli _________________________________________________________________________ Uli Zappe, Christian-Morgenstern-Straße 16, D-65201 Wiesbaden, Germany http://www.ritual.org Fon: +49-700-ULIZAPPE Fax: +49-700-ZAPPEFAX _________________________________________________________________________