How is histogram calibrated?

[Chris Carlen]

Is it in "stops" or factors of 2, ie the rightmost edge represents 100%, the next one to the left represents 50%, next to the left 25%, 12.5%, 6.25%, 0 ?

I think it must be in stops because a shot of a uniform white wall produces a peak just a little above the 12.5% mark. This might argue for the metering being set for 13%, though an internal flash shot of a white wall is just a little more to the right, at almost exactly 2.5 stops below 100%. Since the log to the base 2 of 0.18 is -2.47, this makes sense. Perhaps it means my camera is calibrated a little underexposing with ambient light, but perfect with flash.

But that doesn't mean it doesn't underexpose flash shots, which is a lack of FEC problem, not a metering problem.

Your thoughts welcome.

Good day!

 

[Max Leung]

I'm thinking that if you use the internal flash, it meters assuming 18%, but if you use an external flash like the Canon 420EX it uses 13% instead.

I have to do more testing...although I hate testing and would rather take pictures.

Is it in "stops" or factors of 2, ie the rightmost edge represents
100%, the next one to the left represents 50%, next to the left
25%, 12.5%, 6.25%, 0 ?

I think it must be in stops because a shot of a uniform white wall
produces a peak just a little above the 12.5% mark. This might
argue for the metering being set for 13%, though an internal flash
shot of a white wall is just a little more to the right, at almost
exactly 2.5 stops below 100%. Since the log to the base 2 of 0.18
is -2.47, this makes sense. Perhaps it means my camera is
calibrated a little underexposing with ambient light, but perfect
with flash.

But that doesn't mean it doesn't underexpose flash shots, which is
a lack of FEC problem, not a metering problem.

Your thoughts welcome.

Good day!

 

[Doug Kerr]

I don't know, but it might work in terms of gamma precompensated luminance. That is, the luminance represented by some point on the scale is equal to the scale position (on a 0-1 scale) to the 2.2 power. (That is what we usually see in the R, G, B scales in photo editors.)

If so, then the beginning of the scale, the four marker lines, and the top would represent these luminance values (on a scale of 0-100%)

0%
3%
13%
32%
61%
100%

In that case, the last marker line represents 0.7 stop from the top; the next to theh last maeker line would represent 1.6 stop from the top.

But I don't know if that's how it works.

Best regards,

Doug

 

[Chris Carlen]

I don't know, but it might work in terms of gamma precompensated
luminance. That is, the luminance represented by some point on the
scale is equal to the scale position (on a 0-1 scale) to the 2.2
power. (That is what we usually see in the R, G, B scales in photo
editors.)

If so, then the beginning of the scale, the four marker lines, and
the top would represent these luminance values (on a scale of
0-100%)

0%
3%
13%
32%
61%
100%

In that case, the last marker line represents 0.7 stop from the
top; the next to theh last maeker line would represent 1.6 stop
from the top.

But I don't know if that's how it works.

Best regards,

Doug

Interesting. Seems kind of funny if we all don't really know how it's calibrated yet we use it all the time. It should be possible to test. Thanks for this theory, though, because if I test for "stops" and the peaks don't line up, I'd be at a loss for a new theory. This one sounds like it has just as much likelyhood of being correct as mine.

But where might we go to find out for sure, other than by experimenting?

 

[DRG]

Is it in "stops" or factors of 2, ie the rightmost edge represents
100%, the next one to the left represents 50%, next to the left
25%, 12.5%, 6.25%, 0 ?

I think it must be in stops because a shot of a uniform white wall
produces a peak just a little above the 12.5% mark. This might
argue for the metering being set for 13%, though an internal flash
shot of a white wall is just a little more to the right, at almost
exactly 2.5 stops below 100%. Since the log to the base 2 of 0.18
is -2.47, this makes sense. Perhaps it means my camera is
calibrated a little underexposing with ambient light, but perfect
with flash.

But that doesn't mean it doesn't underexpose flash shots, which is
a lack of FEC problem, not a metering problem.

Your thoughts welcome.

Good day!

Stops ARE factors of two. I think you mean is it linear or in stops (which is actually a logarithmic scale). The answer is that each tick mark is one stop, equal to a factor of two.

I shot the same exact scene increasing shutter speed by 2x each time and the histogram moved rigidly leftward by one tick with each shot.

This is very different from the histogram in Photoshop, which is linear. Half scale on that histogram represents 128 out of 255 tone values, or just 1 stop down from full value.

David

 

[Chris Carlen]

Is it in "stops" or factors of 2, ie the rightmost edge represents
100%, the next one to the left represents 50%, next to the left
25%, 12.5%, 6.25%, 0 ?

Stops ARE factors of two. I think you mean is it linear or in stops
(which is actually a logarithmic scale). The answer is that each
tick mark is one stop, equal to a factor of two.

Yes, I shouldn't have used the word "or" when I meant "in effect."

I shot the same exact scene increasing shutter speed by 2x each
time and the histogram moved rigidly leftward by one tick with each
shot.

This is a good test to prove the case. I have been interested in doing a test in which grey blocks of varying levels according to a white, 50%, 25%, etc. pattern are imaged. These should line up right on the marks if metering is done on an 18%.

This is very different from the histogram in Photoshop, which is
linear. Half scale on that histogram represents 128 out of 255 tone
values, or just 1 stop down from full value.

Yes, I have noticed that it is linear in PS as well. Having the camera in stops is sensible, because it is the most useful info for determining and setting required exposure compensation.

Thanks for the reply.

 

[Max Leung]

Another test I was thinking of trying, was creating canned JPEG images in photoshop, where the whole image is of varying grays: 13%, 18%, etc. Then upload them into the flash and preview them on the camera and see how it creates the histograms.

Is it in "stops" or factors of 2, ie the rightmost edge represents
100%, the next one to the left represents 50%, next to the left
25%, 12.5%, 6.25%, 0 ?

Stops ARE factors of two. I think you mean is it linear or in stops
(which is actually a logarithmic scale). The answer is that each
tick mark is one stop, equal to a factor of two.

Yes, I shouldn't have used the word "or" when I meant "in effect."

I shot the same exact scene increasing shutter speed by 2x each
time and the histogram moved rigidly leftward by one tick with each
shot.

This is a good test to prove the case. I have been interested in
doing a test in which grey blocks of varying levels according to a
white, 50%, 25%, etc. pattern are imaged. These should line up
right on the marks if metering is done on an 18%.

This is very different from the histogram in Photoshop, which is
linear. Half scale on that histogram represents 128 out of 255 tone
values, or just 1 stop down from full value.

Yes, I have noticed that it is linear in PS as well. Having the
camera in stops is sensible, because it is the most useful info for
determining and setting required exposure compensation.

Thanks for the reply.

 

[Doug Kerr]

http://forums.dpreview.com/forums/read.asp?forum=1031&message=6427296

Best regards,

DOug