What you see there is not any kind of clipping, is just that the Y-axis scale in which LR decided to show you the histogram doesn't reach its maximum, so it shows you a truncated version of the complete histogram. But it's just a matter of plotting, nothing related to the content of your image gamut.
The following image:
has the following histograms (left Photoshop, right complete histogram):
As can be seen, PS truncates the Y axis of the graph in order to make the plot more representative.
To analyze gamut clipping in the histogram, just look at histogram ends: in my picture, the Glencoe landscape in the Highlands was so intensely green that sRGB didn't manage to encode it, clipping the blue channel in 0.
For photographers who never heard about histograms, I think it's a good idea to look at the statistical concept of histogram
, much wider that its particular application to photography.
Okay, thank you for explaining. What you described "a truncated version of the complete histogram" I was interpreting to mean "can't represent all of the colors" when was looking at the histogram.
But, in other words, I am not seeing color clipping at all here, what I am seeing is the limitatons of that histogram(?).
Histograms aren't helpful for gamut or gamut clipping info. Photoshop's soft proofing capabilities are designed for help analyze gamut clipping. View>ProofSetup>Custom... and View>GamutWarning are the two tools worth getting to know if you're not familiar with them already.
Okay, thank you.
Quite true, but the histogram indicates that the scene is of rather low contrast, and setting of better black and white points would drastically improve the image. With ACR one can see gamut clipping in the chosen color space, but with LR one would have to export to ProPhotoRGB and use Photoshop's soft proofing tools.
Thank you. I know how to set the white point, but how do I set the black point?
I’d agree with you too. The colors seen in the histogram are useful to see the gamut clipping of one, two (saturation) or more channels (white/black all three). This tool is more useful in ACR because you can see the effect of gamut clipping as you change the RGB encoding options (toggle from sRGB to ProPhoto RGB as an example). Unfortunately, in LR, you get to see clipping based on ProPhoto primaries and if you export in anything but ProPhoto, the resulting histogram is different and you probably clipped saturation “blindly”.
Thank you for that. I always use ProPhoto.
To answer the OP’s question about all three channels clipping, I would suggest he alter the various rendering controls (Exposure, Blacks) to push the histogram out and see the results of the image. IOW, having a fixed histogram appearance can often make the image appearance ugly and awful! Edit images to appear as you desire on a calibrated and profiled display, not to produce a certain appearance of a histogram. Also using the clipping indicators (Alt/Option key) as you drag various sliders (again, Exposure, Blacks, Recovery) can be useful to see not only what is and isn’t clipping but where you might want to set the end of the tone scale and then viewing the image. Do you have and do you want a highlight that’s close to clipping (a specular?). Do you want to block up shadow detail as part of an artistic expression of the image? Look at the work of Greg Gorman and see how his style has no regard for shadow detail by design. There is no rule that says because you have shadow detail, you need to render and express it in your image. Clip it to death if you like that rendering.
You see, this is how I always do edit: to "my eye" not
any histogram or need to conform. However, I just wanted to try to understand the histogram more, to have it help me. In other words (as Michael likes to say), "The difference between science and art." Recognizing this difference, I will always side with what looks good to me, but I just want to try to understand the science more.
I have many images that I edit, where my histogram is all off, but I like the way it looks. And yet, I have other images where the histogram is "perfect" but the image lacks punch to my eyes. And yet, I am sure in many other ways my being able to fully understand the histogram can only help me.
So, good point, and thanks for your time.