I'm not asking about the mechanics of acquiring images or using any particular HDR program. Nor am I asking about tone mapping of the 32-bit HDR image. I'm just after some basic intuition of how the input photos are used so that the HDR algorithm is more than a "black box". Most of what I've found online just discusses how to acquire the individual photos and use them in a particular program without explaining how the photos are used.

So how is the 32-bit image put together from the individual photos? The brightest unclipped regions in the scene and the darkest unclipped regions in the scene are retained for the final result, but those are represented by 8 bits (or maybe slightly more if raw photos are used).


  • \$\begingroup\$ en.wikipedia.org/wiki/Tone_mapping \$\endgroup\$
    – xenoid
    Commented Dec 28, 2023 at 10:10
  • \$\begingroup\$ "Nor am I asking about tone mapping of the 32-bit HDR image. I'm just after some basic intuition of how the input photos are used so that the HDR algorithm is more than a "black box"." Tone mapping is the algorithm. \$\endgroup\$
    – Philip Kendall
    Commented Dec 28, 2023 at 10:22
  • \$\begingroup\$ As per the answer by @xenoid, tone mapping is the final step. \$\endgroup\$ Commented Dec 28, 2023 at 17:42
  • \$\begingroup\$ @PhilipKendall, tonemapping is only the output/export algorithm to translate the HDR data into a viewable LDR format. Not_Einstein is asking about what's done on the input set of image files to create the HDR file. \$\endgroup\$
    – inkista
    Commented Dec 29, 2023 at 19:48

1 Answer 1


Assume you take a picture with 3 different exposures: balanced, under-exposed by 3EV, and over-exposed by 3EV. In a 8-bit image, you can only have 8EV of range(*). So

  • The "balanced" image contains the "middle gray" in the middle, and everything -4EV to +4EV. Pixels darker that -4EV are blacked out, pixels lighter than 4EV are whitened out.
  • In the under-exposed image, average gray is 1EV, anything more than 1EV darker that average gray is blacked out, but at the other end you have pixels that are +7EV from the average gray.
  • In the over-exposed image, it is the opposite, the average gray pixels are are at +7EV, and everything light than 1EV from that is whitened out. But in the shadows, you have the data for pixel that are up to -7EV from middle gray.

So between the three pictures, you go from -7EV darker than middle gray to +7EV lighter, so in total you have 14EV of range.

                <----- range of EV ----------->
                  <--- captured -->  
Over-exposed    --[++++++---------]------------
Balanced        ------[--++++++++++++--]-------
Under exposed   -----------[---------+++++++]--
Extracted       --[+++++++++++++++++++++++++]--

The computer that does the processing isn't itself limited to 8 bit integer values, and in practice the luminosity of pixels is encoded as a decimal value between 0.0 (black) and 1.0 (white) so middle gray is 0.5 and the darkest non-black pixel of a picture with 14EV of dynamic is 1/2^14=0.000061035. Values can be extracted from the images above like this:

  • (value/255)/8 (because 8=2³) of over-exposed pixels in the dark parts (**)
  • (value/255) of balanced pixels in the mid-tones
  • (value/255)*8 of the under exposed pixels in the highlights

In practice this could be a weighed mix if the images, so have a smooth transition.

You can of course take even more shots (-6,-3,0,+3,+6) to extend the range (20EV in this case).

The final step if transforming this 14EV image into something encoded on 8-bit (so much lower dynamics) which is usually done with tone-mapping.

(*) This is a bit simplified (at least if working from JPEG) because the pixels values are gamma-corrected, and in that case

  • The processing must first convert the 8-bit values to "linear" luminosity.
  • The EV aren't linearly mapped to the recorded values, the gamma correction gives more importance to the darker parts of the image.

(**) if the image is gamma-corrected, value/255 is actually (value/255)^(1/2.2)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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