Our hypothesis here is once again that the statement is literally true. However, it contradicts the intuitive sense of many photographers, who feel that a cloudy day yields more saturated colors. We try to show that the increased saturation from “flat” surfaces, which do not produce a shiny reflection of the light source, is defeated by glare from “shiny” surfaces, such as certain foliage, giving the overall impression that the image is less saturated. This hypothesis can be further tested by introducing a polarizing filter to reduce the glare of reflection from “shiny” objects, thus yielding an image that is more saturated overall.
What is saturation?
Our first challenge is to understand what saturation is. Webopedia (http://www.webopedia.com/TERM/C/color_saturation.html) has the following definition:
“In graphics and imaging, color saturation is used to describe the intensity of color in the image. A saturated image has overly bright colors. Using a graphics editing programyou can increase saturation on under-exposed images, or vise versa.”
Beyond the obvious typographical and spelling errors, this definition is seriously flawed. It fails to distinguish luminance (“intensity,” “overly bright”) from saturation. Based strictly on this definition, one could conclude that a bright pastel color had more “saturation” than a deep red. This type of definition may be at the root of the misconception that photos taken in bright sun are more saturated than those taken in shade.
To see that this definition is questionable, if not mistaken, we can examine the color picker in Adobe Photoshop CC. (Prior versions of Adobe Photoshop will show similar results.) In the color picker, we see that each possible color can be understood in five ways: LAB, CMYK, as a hexadecimal number (#), RGB and HSB (for hue, saturation and brightness). For example, a blue sky might appear as L: 49, A:-6, B: -42 in the LAB color scheme. (Details of the interpretation of these schemes is beyond the scope of this document and can be found elsewhere.) The same exact color is C: 82%, M: 47%, Y: 2% K: 0% in the CMYK interpretation and has a hexadecimal number representation of 4e78b8. The RGB components are R: 78, G: 120, B: 184. For this color, the HSB hue is 216, the saturation is 58% and the brightness is 72%.
Staying with this same hue of 216, we can move the selection to the far right and observe that the brightness remains the same (72%), but the saturation rises toward 100% as we move the selection. When the saturation is 100%, we can observe that the R value (in RGB terms) is 0, the G value is 72 and the B value, as we might expect, is the highest, 184. By the same token, if we slide the selection to the far left, the saturation goes to zero without affecting hue or brightness, and the RGB values become 184, 184 and 184. The LAB representation becomes 75, 0, 0. In other words, it is a light gray, the very opposite of what we usually consider saturation.
A better definition of saturation can be found in Adobe help (https://helpx.adobe.com/photoshop/using/color.html):
“Strength or purity of the color (sometimes called chroma). Saturation represents the amount of gray in proportion to the hue, measured as a percentage from 0% (gray) to 100% (fully saturated). On the standard color wheel, saturation increases from the center to the edge.”
Finally, we turn to the Art Institute of Chicago (http://www.artic.edu/aic/education/sciarttech/2c1.html) for what is probably the simplest and most straightforward definition of saturation:
“the relative intensity of a hue when compared to gray.”
Experiments
We took two sets of photographs. One was a close-up (macro) of a small flower with a yellow center, white petals and green stem. The other was of a tree in full leaf. In both cases, we took photographs in bright sun and in obscured or overcast sun, with a lens shade and without, and with a polarizing filter and without.
For each photograph, we selected 3x3 pixel samples of several areas. For the flower, we sampled the yellow center, the white petals and the green stem. For the, we sampled leaves in full exposure to the light, in partial shade and in deep shade. For each photograph and for each sample, we used the Photoshop Color Sampler Tool, which gave us the RGB values for each sample. From these, we used the Color Picker to calculate the saturation from the RGB values.
Results
In nearly cases, the samples from photographs taken in the bright sun were more saturated than from those taken in overcast or obscured sun conditions. The one exception was that in the photographs of the flower taken in bright sun, the white petals were more saturated when a polarizer was used than not. But even in these photographs, the yellow flower center and the green stem were more saturated without the polarizer.
Conclusion
It turns out that this myth is, simply, TRUE! Colors are more saturated in bright sun than in overcast conditions. A polarizer can be used to “darken” the blue sky and to reduce reflections off of water and some foliage. When it comes to the polarizer’s effects, your own visual perception is the best guide. But in general, bright days are best for saturated colors.