Emitter Sampling Rates
Posted: Thu Feb 27, 2014 7:30 am
I have a question about emitter sampling rates.
Say you have a studio lights setup with three lights (please see attached image)...One light is red at 400W, the second is blue also at 400W, and the third is green but at only 10W. These three lights are equally spaced around a cube which sits in the center of the lights. Assume for argument sakes, that all three light colors behave equally and are not wavelength dependent.
Now, according to how I understand how Octane works, since the green light has such a low power in the scene, the cube surface that faces the green light will take longer to clear up noise than the other surfaces of the cube, since the green light is throwing out less "photons" than the other two lights. As I understand it, this is true if the Sampling rate of all three emitters are equal. So let's say that the rates for all three lights was "1". I end up with a grainy looking green side of the cube...Or at least it takes longer for it to clear up than the other sides. But if I raise the sampling rate of the green light to, say, "6", then theoretically, even at 40x less lamp wattage, the green side of the cube would clear up faster and not be so grainy in the end, correct? If this is so, is this faster rate of green cleanup at the expense of the rate that noise clears up for the other sides of the cube. In other words, since more effort is spent in Octane handling the green light, and the GPU only has limited resources, is that effort taken from the available resources given to the red and blue lights? I am assuming that all warps in the GPU are already full of tasks.
Say you have a studio lights setup with three lights (please see attached image)...One light is red at 400W, the second is blue also at 400W, and the third is green but at only 10W. These three lights are equally spaced around a cube which sits in the center of the lights. Assume for argument sakes, that all three light colors behave equally and are not wavelength dependent.
Now, according to how I understand how Octane works, since the green light has such a low power in the scene, the cube surface that faces the green light will take longer to clear up noise than the other surfaces of the cube, since the green light is throwing out less "photons" than the other two lights. As I understand it, this is true if the Sampling rate of all three emitters are equal. So let's say that the rates for all three lights was "1". I end up with a grainy looking green side of the cube...Or at least it takes longer for it to clear up than the other sides. But if I raise the sampling rate of the green light to, say, "6", then theoretically, even at 40x less lamp wattage, the green side of the cube would clear up faster and not be so grainy in the end, correct? If this is so, is this faster rate of green cleanup at the expense of the rate that noise clears up for the other sides of the cube. In other words, since more effort is spent in Octane handling the green light, and the GPU only has limited resources, is that effort taken from the available resources given to the red and blue lights? I am assuming that all warps in the GPU are already full of tasks.