Is there a relationship between dynamic range and white balance?
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\$\begingroup\$ Note that there is a significant difference in the answer depending on if you are looking for sensor or final image dynamic range. TFuto does a great job of explaining why it doesn't impact the sensor readings and I attempted to explain how adjusting white balance in post impacts the final dynamic range of the image in select color channels. \$\endgroup\$– AJ Henderson ♦Commented Jan 29, 2014 at 18:54
2 Answers
The white balance you select impacts the dynamic range for particular colors in the final output only. When you adjust the white balance of an image, you are not actually adjusting how the sensor picks up colors of light, but rather how each signal is weighted.
If, say, red is weighted more highly than green, then the limit to how high your red value can go is lower in terms of actual measures. A value of .85,1,1 would be white since the .85 red is being treated as 1. Any value above .85 would exceed the color space and so that amount of dynamic range of the sensor is effectively lost as it is impossible for the camera to sample enough green and blue to properly resolve a white object that reads a value of 1 for the red pixels.
The actual dynamic range the sensor picks up is unaltered, but later white balance adjustments render one color to have a wider range than the others, the dynamic range for those colors has to be limited to make a natural looking image.
This is why, for example, when a photo is taken underwater (where there is far more green and blue light than red) the colors end up looking muted and noisy if you try to properly white balance. Under those kinds of conditions, a white object will have blue and green fully exposed, but the red will only be at 10 or 20%. In order to achieve a white balance, that 20% range has to be expanded to fill the whole 100%. This expansion means that the effective dynamic range of the red color is much more limited. It may be expanded to match the level of intensity of the other colors, but far fewer intensity levels are represented.
(As a side note, one could argue this isn't really the dynamic range, but it depends on how you look at dynamic range. If you are looking for what is the brightest and darkest detectable value, then the dynamic range isn't altered by this. If you are looking for the number of distinguishable values within the image, then it is, since the usable range of the sensor is impacted by the white balance even if there wasn't any information in that range.)
No, there is no relation. (NOTE: I am describing dynamic range and white balance of the raw image captured by the sensor, not a post-processed image).
Dynamic range is related to the brightness (luminous flux amount) a pixel's R,G,B component can sense. Let's take R (red). Below a certain amount of light, that channel will receive a "0" value. Above certain amount of incoming light, it will always have a maximum value, for 8-bit RGB that is 255, for 16-bit RGB that is 65535.
Anything between R being 0 and its maximum value bidirectionally maps to a certain amount of light. That is the dynamic range the camera sensor can handle and can successfully map to color channel values. This range is fixed per camera sensor, and can be objectively measured.
White balance is color-related. You photograph a white or neutral gray object, and you would expect that to be represented with R,G and B channels having the same value on your photo. However, you will see differences as the light illuminating objects are colorized - the sun is yellowish or orange, the light from the sky is blue, etc. You correct white balance by telling your camera that a neutral object's RGB is what you call white or neutral gray, and the camera or post-processing algorithm adjusts each channel separately to arrive to a solution when what you call white or gray is actually white or gray for you, observing the final image.
So basically dynamic range is an interval of possible light intensities for a camera, white balance is the relative shift of RGB channels on a photo.
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\$\begingroup\$ Pulled my comments since we now understand each other and updated my answer to reinforce your point that it doesn't impact the sensor at all, because I do agree that is an important distinction to make. \$\endgroup\$– AJ Henderson ♦Commented Jan 29, 2014 at 18:51