How to correctly calculate the Dynamic Range of an image

Question

I'm trying to compare the dynamic range of a correctly exposed LDR picture with the dynamic range of the HDR image obtained with 3 LDR pictures.

Right now I'm calculating the DR as:

 stops = log2(max) - log2(min);

Where

max = maximum pixel value found in the image
min = minimum pixel value found in the image

The reference correctly exposed image is an 8 bits Jpeg, so the dynamic range is always 8 stops, since the values goes from 0 to 255. The HDR image is an RGBe radiance map, so the values are 32 bits floating point.

I was wondering a few things:

1. Is this the correct way to calculate the dynamic range from a picture?
2. All jpg files have pixel values from 0 to 255, so every correctly exposed jpg will have the same dynamic range. Is this right?
3. Sometimes the calculated dynamic range of the HDR image is lower than the original. How is this possible?

Thank you for your help

EDIT: adding an image to better explain my reasoning

Left: standard correcly exposed image - Right: HDR image blended from 3 bracketed photos

Answer 1

First, regardless of the particular dynamic range definition, calculating "anything vs zero" luminance is not useful, because then the ratio becomes infinite. Your:

log2(max)-log2(0)

is another way of saying:

log(max/0)

and we can't divide by zero (and log2(0) is undefined)

So let's forget the 0 for now (zero means the image is not well exposed anyway because you can't have any detail at level 0), then, indeed:

log2(255)-log2(1) = ~8 (EV)

But this is not the dynamic range either, because in JPEG (and sRGB in general) pixel values are not pixel luminance.

Taking the proper gamma formula into account, we finally get:

log2(1)-log2(1/3294.6) = ~12 (EV)

(See also: 8bit monitor theoretical contrast)

Answer 2

The end result of High Dynamic Range Imaging (HDRI) is not to produce an image with dynamic range as high as the scene it attempts to reproduce. We do not have display technology available that can do that. At least not practical ones.

The aim of HDRI is to take a very high dynamic range scene and reproduce it in a way that we can see the very bright and the very dark details in the actual scene using our limited display technologies that have a much more limited dynamic range.

It's not a whole lot different than when Ansel Adams used the zone system to capture up to 14 stops of dynamic range on his negatives and then squeeze all of that information onto papers that were capable of only 6-8 stops of dynamic range.

In other words, we're fitting the high dynamic range of a scene into our low dynamic range display technologies. Thus the end result of HDRI is a low dynamic range image that includes as many details as possible from a high dynamic range scene.

Answer 3

Firstly, an 8bit jpeg is 24bit color; 8bits per color. Your 32bit HDR is also 8bit per color plus an additional 8bit shared channel (Alpha).

Secondly, not every correctly exposed image will have values from min (black/0) to max (white/255). HDR is rewriting values to fit w/in the display/reproduction capability... if in that process it eliminates any true blacks or whites that exist in the original then the result will be a reduced DR.

You are not wanting to measure the difference between min/max, that will always be the same due to the limitations of reproduction/display. What you want to measure is the difference in the number of steps/values that exist in between min and max.

I.e. an image that only has black and white values only contains two stops of dynamic range. Whereas an image that has black, white, and mid grey has three stops (values) of DR displayed within the image; even though black and white are equally as far apart. The greatest increases in displayed DR occur in the darkest parts of an image, where small differences in tonal values still equals a stop of luminance.

Answer 4

Ok I think I found a solution by myself. I'll try to explain it here, let me know what you people think about it.

INPUT: 3 grayscale 8 bit images

OUTPUT: 1 HDR radiance map (single precision)

So, to compute the HDR image I use the Devebec algorithm

This algorithm outputs the radiance map computed from the 3 (or more) input photos. Every pixel value represent the relative illuminance (irradiance actually) of the corresponding point in the scene. I can then calculate the dynamic range simply by using the max and min pixel value from this matrix:

stops = log2(max) - log2(min);

It's not important that the irradiance values are not the absolute one, because we are computing a range:

absolute_value = scale_factor + relative_value

range = absolute_max - absolute_min = (relative_max + scale_factor) - (relative_min + scale_factor) = relative_max - relative_min

To blend the images the algorithm need to recover the camera response function. This function relate the pixel values to the relative illuminance of the scene. This means that once I have the camera response curve, and the maximum and minimum pixel value in the original image (in this case 0 and 255) I can retrieve the relative irradiance values of the original image too:

exp(g(255)-ln⁡(dt) ) = E_max
exp(g(0)-ln⁡(dt) ) = E_min

where

g = inverse of camera response function (calculated by the script)
dt = shutter speed used on the original image

I can then compute the actual dynamic range of the original image and compare it to the HDR image dynamic range.

Yet, seems like some images lose DR when blended together. So I'm a little confused. For example, these images once blended have narrower DR than before.