We all know exposure warnings that a digital camera offers: If shows which parts are so bright or dark that there's no more distinction towards even brighter or darker areas possible. Meaning that those areas can't show any more detail. And that's a limitation of the recording feature of the hardware, i.e. of the sensitivity range of the sensor.

That's all clear.

Now, what puzzles me is that Lightroom keeps creating such warnings when I change an image's exposure. E.g. like this: I open the "Histogram" and "Basic" panes, turn on the overexposure warning in the histogram (click on the top right rectangle) and then move the Exposure slider to the right. That soon leads to areas becoming totally white, and the overexposure warning kicking in.

This makes no sense to me, as here those are just numbers now, and there's no limit to the dynamic range of the numbers, right? I would expect that the editing software can work with endless dynamic range, and only when I apply this to a output medium, i.e a printer, monitor or whatever, I need to choose the available range of that medium, and then either clip the range or compress it.

But here, Lightroom appears to tell me right away that I'm outside of some range, but which range is that, exactly?

Background: Let's imagine I have a display medium that can handle a wider EV range than what the camera had. Now imagine I have a picture that contains the sun, all overexposed, due to the camera's limits. I would now want to use an image editing software to manually put more "detail" into this img area of the sun, raising the brightness there for better effect, because my out medium could handle that additional range. How would I accomplish this, if a tool such as Lightroom apparently already would clip my range at that point?


3 Answers 3


There is going to be a limit to the numbers you can use, and absolute white is as far as you can go -- it doesn't matter whether that's a 255,255,255 value in an 8-bit image or 65535,65535,65535 in a 16-bit image (or the corresponding values in a 32-bit image). Once you've hit black at one end and white at the other, you've exhausted the available dynamic range, no matter what the output device is.

Screens, backlit transparecies and projectors can never get darker than "off" or brighter than "all the way on". Paper can never get brighter than the paper itself, nor darker than the darkest ink you can put on it. A high dynamic range might change the distance between black and white (as in a screen with a very high contrast ratio), or it may mean a less visible, smoother transition between levels, but black will still be black and white will still be white. So even though the original bits in the RAW file might still be available for Lightroom (or a Photoshop smart object or similar) to calculate with, it's the output values that matter as far as clipping goes. White and black are hard clipping limits, so the overexposure and underexposure clipping warnings are telling you something you ought to know.

  • 3
    \$\begingroup\$ It should also be noted that, from an artistic standpoint, clipped highlights and blacks are not necessarily a bad thing. Particularly in the case where high contrast is desirable, artistically, one may WANT clipping. \$\endgroup\$
    – jrista
    Dec 21, 2011 at 18:20
  • 1
    \$\begingroup\$ Indeed. And sometimes they are unavoidable for all practical purposes -- specular highlights can be made grey without gaining any detail, while deep unlit recesses and black velvet will rarely provide anything other than the read noise from the sensor. But one needs to be aware that there is clipping going on in order to make informed choices. \$\endgroup\$
    – user2719
    Dec 21, 2011 at 18:43
  • \$\begingroup\$ Yours and the answer from Miguel still do not really make me understand the effect. I am a software developer, so I very well understand the file format and meaning of number ranges. And because of this background of mine, I have trouble accepting these limitations I've described in my question. \$\endgroup\$ Jan 10, 2012 at 11:04
  • \$\begingroup\$ @Thomas: Consider only one channel to simplify. Any operation done on an image is a function taking an integer (8, 16, or 32-bit, depending on bit depth) to another integer. For example, a contrast adjustment may take the 8-bit value of 250 to 255 (i.e. bright values get brighter) and 5 to 0 (i.e. dark values get darker). But because 255 is the greatest value possible for an 8-bit channel, any initial value above 250 would also become 255. Likewise 0 is the smallest value possible, so values less than five would also become 0. These regions would be clipped in that particular channel. \$\endgroup\$ Jan 10, 2012 at 17:14
  • \$\begingroup\$ @ThomasTempelmann "And because of this background..."? That should make it easier to understand. You are creating a value that overflows a return type, and the transformation function has only two options: limit the return to the bounds of the type (a pixel is a struct of unsigned bytes|shorts|integers, where all 1s in any one member is channel saturation) or allow overflow and undefined behavior (not only would a single channel wrap back to zero, but its "carry" would affect the neighboring channel member -- which may even be in a different pixel). \$\endgroup\$
    – user2719
    Jan 11, 2012 at 1:32

You need to interpret those exposure warnings as this:

"If you export an LDR image with the current settings, these are the areas where you have lost detail that exists in the original unedited picture".

This is very useful, because you can move sliders and immediately see the effect your change will have in a final exported LDR image.

Also note that Lightroom will certainly not have infinite dynamic range available to it. Computers are pretty bad at representing infinite, pictures are typically represented in the computer (Lightroom or otherwise) in one of just a few formats:

  • 8 bits per pixel: standard for LDR pictures. The range for each pixel value is 0-255.
  • 16 bits per pixel: standard for HDR pictures. The range for each pixel component is 0-65535. If you shoot raw, this is the most likely representation, even though raw files are typically 10 or 12 bits per pixel only.
  • 32-bit floating point: each pixel is a floating point number. In this mode typically the a pixel component value range from 0.0 (black) to 1.0 (white) and out of range values can be represented easily. While this may seem like a great format, floating point numbers also suffer from the problem that not all possible values can be represented, for any desired number you have to find the closest value that can be represented in this form, like with integers. Operations in floating point format are slower than with integers.
  • 64-bit floating point: same as above, but more accurate and slower to process.

Note that for all of the above, a color picture requires three value to represent a pixel, the red, green and blue components. So in reality, the 8 bits per pixel format uses 24 bits to represent a color picture. A grayscale picture has only one component per picture.

  • \$\begingroup\$ With integers, you also have to use the closest value that can be represented in the format. \$\endgroup\$
    – mattdm
    Dec 24, 2011 at 14:43
  • \$\begingroup\$ @mattdm: true, my answer didn't clearly indicate this. Hopefully it is clearer now. Thanks! \$\endgroup\$
    – Miguel
    Dec 24, 2011 at 17:59
  • \$\begingroup\$ I guess the other thing that's not clear is: are these increasing bits used to express more range, or more precision within the same overall range? How and why are the limits of the range chosen? \$\endgroup\$
    – mattdm
    Dec 24, 2011 at 18:27
  • \$\begingroup\$ In 32-bit floating point you can represent up to about 7 digits accurately, with the decimal point in any place within those 7 digits. 64-bit floating point increases that to 16 digits. The [0-1] range is good because it gives you the most precision in the decimal digits (i.e. 7 or 16 digits after the decimal point), while still leaving a lot of space for representing numbers that are outside of the range. \$\endgroup\$
    – Miguel
    Dec 24, 2011 at 19:21

I know it's lame to answer my own question, especially if others have also tried to provide an answer. Yet, the best explanation I've found just now is on this web page: photozone.de - Brightness & Exposure.

It reveals to me that I've had a misconception about the "Exposure" function in editing software.

One thing that Stan Rogers helped me to understand is that we're always only looking at a relative brightness range from 0 to 1.

What I expected "Exposure" to do is to shift all mid-gray values towards the extreme end (black or white), in a way that would compress the lowest or highest values proportionally to their position. I.e, that when I'd increase exposure, a mid-gray tone (0.5) would be increased more (e.g. by 0.2) than a tone that's already very bright (e.g. 0.9 only by 0.03). That way, especially if using high-precision numbers internally (i.e. floating point numbers), there'd be practically no loss in information.

However, it turns out that changing the Exposure is a direct imitation of what the camera does, i.e. it simply shifts the values of 0 to 1 by a constant offset, and that can lead to clipping if the shift exceeds the borders.

I was simply not expecting that a tool would offer a control that would change the information in such a way. In my opinion, this kind of Exposure control that unavoidably leads to clipping is hardly ever useful, whereas a "brightness" control as I've explained above, which would lighten everything up/down proportional to their current brightness is something that would alwmost always be the preferred operation because it avoids loss of detail better.

Yet, Lightroom and all other tools I know of all offer the Exposure control prominently instead of the - IMO - much better brightness control, thereby effectively misguiding the casual photographer into using tools that aren't giving him the best results.

Or do I (still) miss something here?

  • \$\begingroup\$ It's absolutely not lame to answer your own question. It's even encouraged and helpful. blog.stackoverflow.com/2011/07/… \$\endgroup\$
    – mattdm
    Jan 10, 2012 at 11:32
  • \$\begingroup\$ I don't know if I would say that Exposure is more prominent than brightness in Lightroom. It's listed first because (I believe) its the adjustment you usually want attempt first. I.e. if your exposure is off, adjust that first then do contrast and brightness adjustments, and then pull back on exposure adjustments if needed. Ideally your exposure will be spot on, and you won't need to mess with it. (I'm not that good yet, but I'm working on it.) \$\endgroup\$ Jan 10, 2012 at 17:29

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