-3
\$\begingroup\$

I understand the basic logic of providing histograms of JPEGs. The sensor pixel values get combined, processed, interpreted as a color, then broken out into new RGB values. So, if you were to give the red histogram of the sensor, it would be different than the red histogram of the JPEG, so it would not represent the color mix accurately. I get that.

However, most of the time, I am not using the histogram to interpret color, I want to use it to determine whether my sensor is being saturated (or underexposed). The histogram for the sensor would be useful for that.

In other words, let's say the max value for the sensor pixel is 256, and one of the pixels registers a 256, but ends up getting a value of 254 in the resulting JPEG. So, the camera is saying "oh look the pixel is not saturated anymore". Well, I don't care that the camera decreased the value after the fact. I want to know if the sensor pixel was saturated originally. I want to know if the "sensor well" is full or not.

This would seem to be an obvious feature, but for some reason even the most expensive cameras seem to only provide JPEG histograms. Am I missing something here, or is this just a blind spot in current functionality designs?

\$\endgroup\$
8

1 Answer 1

1
\$\begingroup\$

For one thing, raw images are not RGB, and are not usable "as is". Mostly though, raw image data also has no white balance or contrast or profiles added yet. These modifications of course can (and often do) change the histogram significantly. Since raw data cannot be shown on RGB LCD monitors, cameras shooting raw do also embed a JPG image (with full camera processing) into the raw file, and this JPG image and its histogram is shown on the camera rear LCD preview. So you are not very bad off. :)

As to checking clipping and exposure in that histogram, this does then assume the camera JPG white balance and processing will more or less match the subsequent raw processing done later (so poor as Auto WB is, it's a good plan when shooting raw... NOT for the image data, but to help the rear LCD preview).

\$\endgroup\$
6
  • \$\begingroup\$ As I said in my question I understand this. I don't care about the raw image being "usable" or not. I want to know if the sensor is getting saturated (or underexposed) and I don't see any reason why cameras should not be providing this information. Your answer does not address this question. \$\endgroup\$ Aug 29, 2017 at 16:58
  • \$\begingroup\$ Sorry, I thought it did. You know if there were any useful purpose for showing it, it would be shown. Here is a sample white balance multiplier suggested for a raw image from Exif: WB RB Levels: 1.8984375 1.44140625 1 1 ... Ignoring for a second that WB is multiplied in 12 bit raw data values (not in raw data though), if a red value 200 is multiplied by 1.898, it will exceed 255. So it seems pedantic asking if red and blue were saturated before, the right question is, are they saturated after this suggested WB multiplier is applied? \$\endgroup\$
    – WayneF
    Aug 29, 2017 at 17:27
  • \$\begingroup\$ And of course, the embedded JPG and its histogram does EXACTLY SHOW that multiplication result (assuming a similar WB will be applied to raw later). Gamma is applied later too, but gamma does not shift the end points. \$\endgroup\$
    – WayneF
    Aug 29, 2017 at 17:27
  • \$\begingroup\$ @WayneF I agree with Tyler. The optimal exposure when the workflow uses RAW files is to expose such that no significant portion of the image exceeds the max well value. A histogram that can show the CFA grouped well values would be extremely useful as, in most circumstances, it captures materially more of the dynamic range. An optimally exposed RAW file generally produces a washed out in camera jpeg but will generate excellent results in post processing by stopping down the "exposure." The histogram is really all that's necessary for optimizing exposure. \$\endgroup\$
    – doug
    Aug 30, 2017 at 4:28
  • \$\begingroup\$ This same conversation has been made over several existing questions/answers. \$\endgroup\$
    – Michael C
    Aug 30, 2017 at 7:50

Not the answer you're looking for? Browse other questions tagged or ask your own question.