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histogram logarithm

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Anonymous
December 6, 2004 2:04:15 PM

Archived from groups: rec.photo.digital (More info?)

I am new to reading digital photo histograms. I've been told that the
horizontal scale of luminosity is logarithmic. That the histogram is
divided into 8 segments (not visually differentiated) of equal width.
The first segment at the right side contains twice the information as
the next segment, and that segment contains twice the info as its
neighbor to the left, and so on. Meaning that the distribution image
isn't quite what it might appear at first glance, its being on a
logrithmic rather than linear scale. Apparently, the advantage of a
log scale is that since the dark (left) end of the histogram
represents images with far fewer photons, a linear scale histogram
would be pretty thin on that end. Can anyone direct me to a better
explanation of this? One that doesn't have a degree in physics as a
prerequisite?

More about : histogram logarithm

Anonymous
December 6, 2004 8:57:22 PM

Archived from groups: rec.photo.digital (More info?)

<drs@canby.com> wrote in message
news:uga9r050mvvlnna2mvh7vculf4v7pksu88@4ax.com...
>I am new to reading digital photo histograms. I've been told that the
> horizontal scale of luminosity is logarithmic.

Right, just like film curves and all the other exposure scales you've seen.
Equal distances correspond to equal numbers of f-stops.

> That the histogram is
> divided into 8 segments (not visually differentiated) of equal width.

No. It is more or less continuous, or divided into hundreds of segments.

The reason for the log scale is that we perceive light logarithmically. The
shutter speeds on your camera are logarithmic (1/1000, 1/500, 1/250... not
1/1000, 2/1000, 3/1000...). The f-stops on your camera lens produce
logarithmically scaled intensities. One "stop" is a factor of 2, not an
increment of 2.

In short: Don't panic. Photographic measurements have been logarithmic all
along. What you think of as "midtones" are halfway along a logarithmic
scale, not halfway along a linear scale.

If the histogram were not logarithmic, the highlights would take up far too
much of it.
Anonymous
December 7, 2004 2:03:40 AM

Archived from groups: rec.photo.digital (More info?)

In article <uga9r050mvvlnna2mvh7vculf4v7pksu88@4ax.com>, drs@canby.com wrote:

> I am new to reading digital photo histograms.
>...
>...Can anyone direct me to a better explanation of this? One that doesn't
> have a degree in physics as a prerequisite?

You might want to start with:

http://www.luminous-landscape.com/tutorials/understandi...

ron

P.S. Didn't this question come up not too long ago - like within the past
couple of weeks?
Related resources
Can't find your answer ? Ask !
Anonymous
December 7, 2004 11:17:29 AM

Archived from groups: rec.photo.digital (More info?)

On Mon, 06 Dec 2004 23:03:40 -0800, ronwong@purgethis.inreach.com (Ron
Wong) wrote:

>http://www.luminous-landscape.com/tutorials/understandi...

I've read that article and quite a few others. None mention the
logarithmic distribution and from what I was recently told,
understanding the difference between a log scale and a linear scale is
very helpful in using histograms. He was saying that the histogram of
a properly exposed black-and-white checkerboard (exactly same amount
of blacks and whites) would not show a symmetrical curve. The curve
would be weighted toward the dark end of the scale. A useful thing to
know, apparently. Not that being unaware of that means a person can't
use a histogram to improve exposure.
Anonymous
December 7, 2004 3:51:18 PM

Archived from groups: rec.photo.digital (More info?)

Timo Autiokari wrote:

> The histogram that digital cameras show is not logarithmic nor
linear.

??? I take a picture with my 10D. I observe histogram. I say "it
needs another stop". So I adjust the exposure to add 2x the light,
take a new picture and look at the histogram. The whole thing shifts
over one horizontal "bar" on the display, just as one would expect.

So it seems fairly log-ish to me. And really, if the histogram didn't
behave this way, the feature would be rather useless.
Anonymous
December 7, 2004 7:07:11 PM

Archived from groups: rec.photo.digital (More info?)

> ??? I take a picture with my 10D. I observe histogram. I say "it
> needs another stop". So I adjust the exposure to add 2x the light,
> take a new picture and look at the histogram. The whole thing shifts
> over one horizontal "bar" on the display, just as one would expect.
>
> So it seems fairly log-ish to me. And really, if the histogram didn't
> behave this way, the feature would be rather useless.

Linear would be next to useless. My camera works the same way. Many log
graphs are powers of ten but digital camera histograms (the horizontal axes)
are powers of two.
Anonymous
December 7, 2004 11:44:10 PM

Archived from groups: rec.photo.digital (More info?)

On Mon, 06 Dec 2004 11:04:15 -0800, drs@canby.com wrote:

>reading digital photo histograms.

The histogram that digital cameras show is not logarithmic nor linear.

With current digital cameras the histogram shows the distribution of
the linear captured data after the data has been a) automatically
edited (finalized) by the firmware of the camera and b) compensated
for CRT viewing. So, in short, with most if not all digital cameras
the histogram is calculated from the camera finalized JPEG image. This
unfortunately is so even if one is shooting in the linear RAW mode.

In case the following three requirements are in effect:

1) the camera outputs the image data in the sadRGB color space (not
many do so even if the manufacturers may so indicate in the manual)

2) the camera does behave colorimetricly, in other words it does not
apply non-colorimetric "enhancements" (most if not all point&shoot
cameras are non-colorimetric and some dSLRs are also).

3) the coding system of the camera is such that it maps the captured
data up to level 255 (most do so)

then you can estimate your exposure value (EV) from the histogram
using the following table:

Histogram ends at 255 == EV 0
Histogram ends at 230 == EV -0.33
Histogram ends at 207 == EV -0.67
Histogram ends at 186 == EV -1
Histogram ends at 168 == EV -1.33
Histogram ends at 151 == EV -1.67
Histogram ends at 136 == EV -2
Histogram ends at 122 == EV -2.33
Histogram ends at 110 == EV -2.67
Histogram ends at 99 == EV -3
Histogram ends at 89 == EV -3.33
Histogram ends at 80 == EV -3.67
Histogram ends at 72 == EV -4

So, e.g. if the histo of your photo ends at level 207 it means that
you'd need to overexpose by +2/3 stops in order to capture hull
histogram (in other words to utilize the capability of the camera in
the best possible way).

Most cameras do not apply the sadRGB transfer function so the
following table that is calculated for gamma 1.72 space (native gamma
space of Mac systems) will often give more accurate estimation:

Histogram ends at 255 == EV 0
Histogram ends at 223 == EV -0.33
Histogram ends at 195 == EV -0.67
Histogram ends at 170 == EV -1
Histogram ends at 149 == EV -1.33
Histogram ends at 130 == EV -1.67
Histogram ends at 114 == EV -2
Histogram ends at 100 == EV -2.33
Histogram ends at 87 == EV -2.67
Histogram ends at 76 == EV -3
Histogram ends at 67 == EV -3.33
Histogram ends at 58 == EV -3.67
Histogram ends at 51 == EV -4

Now, a totally another issue then is the way how you perceive the
effect of the f/stops (aperture) when you look through the viewfinder
and change the aperture setting of the lens (while keeping the DOF
button pressed). In this situation the vision adapts (light
adaptation, it behaves about logarithmicly) so the change of the
aperture (that also has logarithmic scale) will be perceived as a
change that is equal in "effect" or in "amount", stop after stop, so
the perceived effect is about linear. That is the desired effect and
infarct the very reason why the aperture scaling was chosen to be
logarithmic.

Timo Autiokari http://www.aim-dtp.net
Anonymous
December 7, 2004 11:44:11 PM

Archived from groups: rec.photo.digital (More info?)

On Tue, 07 Dec 2004 20:44:10 +0200, Timo Autiokari
<timo.autiokari@aim-dtp.net> wrote:

>On Mon, 06 Dec 2004 11:04:15 -0800, drs@canby.com wrote:
>
>>reading digital photo histograms.
>
>The histogram that digital cameras show is not logarithmic nor linear.

Thanks. I'm quickly out of my element. I'll keep reading.
Anonymous
December 8, 2004 1:03:37 PM

Archived from groups: rec.photo.digital (More info?)

Timo Autiokari wrote:

>>So it seems fairly log-ish to me. And really, if the histogram
didn't
>>behave this way, the feature would be rather useless.
>
> Well, it does not behave like that.

???

It does behave like I described. If it didn't, I wouldn't even bother
with the thing, as it takes a fair bit of time to call it up.
Anonymous
December 8, 2004 8:48:47 PM

Archived from groups: rec.photo.digital (More info?)

eawckyegcy@yahoo.com wrote:

>So it seems fairly log-ish to me. And really, if the histogram didn't
>behave this way, the feature would be rather useless.

Well, it does not behave like that. The 4 vertical bars in the 10D
histogram are not f/stops, if they were the 10D would only have a 5
f/stop dynamic range. But is has about 8 stops of dynamic range.

You can easily test this, with the camera in manual, take macrophotos
of white evenly illuminated paper (set the lens un-focussed so paper
surface will be optically blurred) at dfferent exposures by 1/3 f/stop
stepping. The histo will show a nice peak for each shot and you can
verify how the EV adjustment affect to the data (how the peak is moved
as the function of EV).

The histogram is calculated as I explained earlier. For the 10D and
other dcameras that have equally spaced 4 vertical divisor bars in the
histogram display one of the following tables will give a decent
approximation (the bars are counted from right to left):

Assuming gamma 1.72 space, if the histogram ends:
at the location of Whitepoint: EV = 0
in the middle of Whitepoint and 1st bar: EV = -0.26
at the location of 1st bar: EV = -0.55
in the middle of 1st bar and 2nd bar: EV = -0.89
at the location of 2nd bar: EV = -1.27
in the middle of 2nd bar and 3rd bar: EV = -1.72
at the location of 3rd bar: EV = -2.27
in the middle of 3rd bar and 4th bar: EV = -2.99
at the location of 4th bar: EV = -3.99
in the middle of 4th bar and Blackpoint: EV = -5.71

Assuming gamma 2 space, if the histogram ends:
at the location of Whitepoint: EV = 0
in the middle of Whitepoint and 1st bar: EV = -0.3
at the location of 1st bar: EV = -0.64
in the middle of 1st bar and 2nd bar: EV = -1.03
at the location of 2nd bar: EV = -1.47
in the middle of 2nd bar and 3rd bar: EV = -2
at the location of 3rd bar: EV = -2.64
in the middle of 3rd bar and 4th bar: EV = -3.47
at the location of 4th bar: EV = -4.64
in the middle of 4th bar and Blackpoint: EV = -6.64

Assuming gamma 2.2 space, if the histogram ends:
at the location of Whitepoint: EV = 0
in the middle of Whitepoint and 1st bar: EV = -0.33
at the location of 1st bar: EV = -0.71
in the middle of 1st bar and 2nd bar: EV = -1.13
at the location of 2nd bar: EV = -1.62
in the middle of 2nd bar and 3rd bar: EV = -2.2
at the location of 3rd bar: EV = -2.91
in the middle of 3rd bar and 4th bar: EV = -3.82
at the location of 4th bar: EV = -5.11
in the middle of 4th bar and Blackpoint: EV = -7.31

Timo Autiokari http://www.aim-dtp.net
Anonymous
December 8, 2004 8:48:48 PM

Archived from groups: rec.photo.digital (More info?)

"Timo Autiokari" <timo.autiokari@aim-dtp.net> wrote in message
news:9g7er0h1gico8llu5ja30sapofukfsu701@4ax.com...
> eawckyegcy@yahoo.com wrote:
>
> >So it seems fairly log-ish to me. And really, if the histogram didn't
> >behave this way, the feature would be rather useless.
>
> Well, it does not behave like that. The 4 vertical bars in the 10D
> histogram are not f/stops, if they were the 10D would only have a 5
> f/stop dynamic range. But is has about 8 stops of dynamic range.
>
> You can easily test this, with the camera in manual, take macrophotos
> of white evenly illuminated paper (set the lens un-focussed so paper
> surface will be optically blurred) at dfferent exposures by 1/3 f/stop
> stepping. The histo will show a nice peak for each shot and you can
> verify how the EV adjustment affect to the data (how the peak is moved
> as the function of EV).
>
> The histogram is calculated as I explained earlier. For the 10D and
> other dcameras that have equally spaced 4 vertical divisor bars in the
> histogram display one of the following tables will give a decent
> approximation (the bars are counted from right to left):
>
> Assuming gamma 1.72 space, if the histogram ends:
> at the location of Whitepoint: EV = 0
> in the middle of Whitepoint and 1st bar: EV = -0.26
> at the location of 1st bar: EV = -0.55
> in the middle of 1st bar and 2nd bar: EV = -0.89
> at the location of 2nd bar: EV = -1.27
> in the middle of 2nd bar and 3rd bar: EV = -1.72
> at the location of 3rd bar: EV = -2.27
> in the middle of 3rd bar and 4th bar: EV = -2.99
> at the location of 4th bar: EV = -3.99
> in the middle of 4th bar and Blackpoint: EV = -5.71
>
> Assuming gamma 2 space, if the histogram ends:
> at the location of Whitepoint: EV = 0
> in the middle of Whitepoint and 1st bar: EV = -0.3
> at the location of 1st bar: EV = -0.64
> in the middle of 1st bar and 2nd bar: EV = -1.03
> at the location of 2nd bar: EV = -1.47
> in the middle of 2nd bar and 3rd bar: EV = -2
> at the location of 3rd bar: EV = -2.64
> in the middle of 3rd bar and 4th bar: EV = -3.47
> at the location of 4th bar: EV = -4.64
> in the middle of 4th bar and Blackpoint: EV = -6.64
>
> Assuming gamma 2.2 space, if the histogram ends:
> at the location of Whitepoint: EV = 0
> in the middle of Whitepoint and 1st bar: EV = -0.33
> at the location of 1st bar: EV = -0.71
> in the middle of 1st bar and 2nd bar: EV = -1.13
> at the location of 2nd bar: EV = -1.62
> in the middle of 2nd bar and 3rd bar: EV = -2.2
> at the location of 3rd bar: EV = -2.91
> in the middle of 3rd bar and 4th bar: EV = -3.82
> at the location of 4th bar: EV = -5.11
> in the middle of 4th bar and Blackpoint: EV = -7.31
>
> Timo Autiokari http://www.aim-dtp.net

Greetings Timo,
you seem to be an expert on the histogram.
I've never used mine in the field. I have a Pentex spot meter that I use
for evaluating highlights and shadows. Perhaps I no longer need the spot
meter. Do you use the histogram much in the field? I would like to learn
more on how to use it.
-tom
Anonymous
December 9, 2004 12:27:30 AM

Archived from groups: rec.photo.digital (More info?)

"Michael A. Covington" <look@www.covingtoninnovations.com.for.address&gt; wrote
in message news:41b4e82b$1@mustang.speedfactory.net...
>
> <drs@canby.com> wrote in message
> news:uga9r050mvvlnna2mvh7vculf4v7pksu88@4ax.com...
> >I am new to reading digital photo histograms. I've been told that the
> > horizontal scale of luminosity is logarithmic.
>
> Right, just like film curves and all the other exposure scales you've
seen.
> Equal distances correspond to equal numbers of f-stops.
>
> > That the histogram is
> > divided into 8 segments (not visually differentiated) of equal width.
>
> No. It is more or less continuous, or divided into hundreds of segments.
>
> The reason for the log scale is that we perceive light logarithmically.
The
> shutter speeds on your camera are logarithmic (1/1000, 1/500, 1/250... not
> 1/1000, 2/1000, 3/1000...). The f-stops on your camera lens produce
> logarithmically scaled intensities. One "stop" is a factor of 2, not an
> increment of 2.
>
> In short: Don't panic. Photographic measurements have been logarithmic
all
> along. What you think of as "midtones" are halfway along a logarithmic
> scale, not halfway along a linear scale.
>
> If the histogram were not logarithmic, the highlights would take up far
too
> much of it.


First off, I like to thank all of those who refrained from calling me an
idiot for a previous post-- my delicate ego can't take too many more shocks.
Secondly...

I trained my camera on an indoor scene under constant illumination and took
twelve manually-focussed exposures at f5.6 with shutter speeds ranging from
1/25s to 1/2s in roughtly 1/3EV increments. Next, I made twelve histograms
and recorded the numbers for the peak luminance readings over the specified
exposure range, obtaining a data set which I normalized to 8 bits, i.e., the
values range from 0 to 255, corresponding to the bins of the camera
histogram. So what did I find? Clearly, the position for peak luminance
varied linearly as the natural logarithms of the exposure readings. Here's
the data

Shutters speeds: 1/25, 1/20, 1/15, 1/13, 1/10, 1/8, 1/6, 1/5, 1/4, 1/3,
1/2.5, 1/2
Measured corresponding peak luminance bin numbers: 39, 52, 66, 80, 104,
123, 143, 170, 187, 204, 228, 244

Data were plotted using my ancient version of Mathcad.

It was a pain in the neck to measure the position of the peak luminance
values in Photoshop, but hey, science requires sacrifice.

So there you go, I guess.
Anonymous
December 9, 2004 4:48:27 AM

Archived from groups: rec.photo.digital (More info?)

Thanks for the data. It's reassuring to know that the histograms we
actually see are plotted the way the traditions of photographic theory led
me to expect.

"Paul H." <xxpaulhtck@zzcomcast.yycom> wrote in message
news:u4WdnTxt9LC3fyrcRVn-oQ@comcast.com...
>
> "Michael A. Covington" <look@www.covingtoninnovations.com.for.address&gt;
> wrote
> in message news:41b4e82b$1@mustang.speedfactory.net...
>>
>> <drs@canby.com> wrote in message
>> news:uga9r050mvvlnna2mvh7vculf4v7pksu88@4ax.com...
>> >I am new to reading digital photo histograms. I've been told that the
>> > horizontal scale of luminosity is logarithmic.
>>
>> Right, just like film curves and all the other exposure scales you've
> seen.
>> Equal distances correspond to equal numbers of f-stops.
>>
>> > That the histogram is
>> > divided into 8 segments (not visually differentiated) of equal width.
>>
>> No. It is more or less continuous, or divided into hundreds of segments.
>>
>> The reason for the log scale is that we perceive light logarithmically.
> The
>> shutter speeds on your camera are logarithmic (1/1000, 1/500, 1/250...
>> not
>> 1/1000, 2/1000, 3/1000...). The f-stops on your camera lens produce
>> logarithmically scaled intensities. One "stop" is a factor of 2, not an
>> increment of 2.
>>
>> In short: Don't panic. Photographic measurements have been logarithmic
> all
>> along. What you think of as "midtones" are halfway along a logarithmic
>> scale, not halfway along a linear scale.
>>
>> If the histogram were not logarithmic, the highlights would take up far
> too
>> much of it.
>
>
> First off, I like to thank all of those who refrained from calling me an
> idiot for a previous post-- my delicate ego can't take too many more
> shocks.
> Secondly...
>
> I trained my camera on an indoor scene under constant illumination and
> took
> twelve manually-focussed exposures at f5.6 with shutter speeds ranging
> from
> 1/25s to 1/2s in roughtly 1/3EV increments. Next, I made twelve
> histograms
> and recorded the numbers for the peak luminance readings over the
> specified
> exposure range, obtaining a data set which I normalized to 8 bits, i.e.,
> the
> values range from 0 to 255, corresponding to the bins of the camera
> histogram. So what did I find? Clearly, the position for peak
> luminance
> varied linearly as the natural logarithms of the exposure readings.
> Here's
> the data
>
> Shutters speeds: 1/25, 1/20, 1/15, 1/13, 1/10, 1/8, 1/6, 1/5, 1/4, 1/3,
> 1/2.5, 1/2
> Measured corresponding peak luminance bin numbers: 39, 52, 66, 80, 104,
> 123, 143, 170, 187, 204, 228, 244
>
> Data were plotted using my ancient version of Mathcad.
>
> It was a pain in the neck to measure the position of the peak luminance
> values in Photoshop, but hey, science requires sacrifice.
>
> So there you go, I guess.
>
>
Anonymous
December 9, 2004 2:34:30 PM

Archived from groups: rec.photo.digital (More info?)

Tom Nakashima wrote:
> I've never used mine in the field. I have a Pentex spot meter that I
use
> for evaluating highlights and shadows. Perhaps I no longer need the
spot
> meter. Do you use the histogram much in the field?

I do not carry a spot meter at all! I have the D60 and am using both
the histogram and the overexposure warning (blinking pixels on the LCD
view) to evaluate the exposure level. These both tools are not very
accurate, e.g. the pixel blinking only happens when all the three color
components of a pixel result 255,255,255 in the JPEG version, the
histogram is binned according to some weighted average of the R, G and
B values of the pixels of the JPEG version, and the histogram is
neither lin log or log scaled so it is difficult to see what amount of
EV correction is needed.

The biggest problem of all however is that both the blinking and the
histogram are taken from the camera finalized JPEG. I only shoot RAW
and I convert to linear PSD. The automatic editing algorithms that the
camera use to create the JPEG version clips the data somewhat so even
if the histogram shows clipping and there are pixels blinking on the
LCD view, even then the raw data still can be underexposed.

Anyhow these tools are useful once one is familiar with the above
mentioned faults. I measure using the evaluative or center-weighted
average metering modes of the camera and then I use the histogram and
pixel blinking to assess the exposure level.

Spot metering is very important with chrome, especially when the chomes
are to be projected on the whitescreen. In this situation one is
totally at the mercy of the tonal reproduction curve of the chrome, one
simply must expose the main subject well/correctly so that it will then
also reproduce correctly. The drawback is that there can often be
obtrusive strong clipping at the bright end.

In my opinion, with digital imaging (and when chrome is scanned) it is
not a good approach to force the main subject to some hypothetical
optimal level using exposure adjustment. There is no fixed tonal
reproduction curve that needs to be taken into account, and forcing the
exposure according to the main subject will usually cause the clipping
at the bright end (or can cause un-necessary underexposure that
increase the noise and cause clipping at the dark end).

With digital workflow it is much better to expose so that nothing is
clipped at the bright end (less of course e.g. direct and speculars
rays from light sources that almost always need to be clipped) and then
in the post-process adjust the tonality (e.g. using Curves) so that the
main subject is reproduced well.

So this is called as the "expose to the right" -method. I always try to
expose for full histogram without major clipping. That way the small
dynamic range of the camera is taken fully into use, noise is minimized
and the most important part of the dynamic range of the scene is
captured. In the post-process I then have the best possible starting
point for editing the image in case the main subject needs to be
presented differently than how it was recorded, or in case the scene
had very large dynamic range etc.

Timo Autiokari
Anonymous
December 11, 2004 8:32:20 AM

Archived from groups: rec.photo.digital (More info?)

Paul H. wrote:
>Clearly, the position for peak luminance varied linearly as the
natural
>logarithms of the exposure readings.

Interesting results! So from your data what amount of RGB levels
equals one f/stop?

I also did an evaluation, very cood topic, I now have much better grip
on the histogram of my Canon D60. I took 22 exposures of a Q-60 color
input target, stepping each exposure by 1/3 f/stop. I started from an
obvious overexposure in order to see how the histogram show that
situation. I do not discuss the form of the histogram as the most
important information that the histogram gives is the exposure and it
is the location where the histogram end on the right that shows this
information.

Like it is with most dSLRs the D60 histogram is divided into 5 equal
sized segments by vertical lines. I have named these as segments 5, 4,
3, 2 and 1. So the segment 1 is the _rightmost_ segment. I do not list
the RGB values since at the picture taking time these are not
available, the only things that are available there are the histogram
and the blinking pixels on the Camera LCD. I have tried to characterize
the appearance of the histogram as accurately as possible.

Results:

+1 1/3 EV: the histogram draws a thint vertical white line on right
edge of the histogram all the way from the bottom to the top. Both the
PEG and RAW data are severely overexposed.

+1 EV: the histogram draws a vertical white line on right edge of the
histogram all the way from bottom to the top. Both the PEG and RAW data
are severely overexposed.

+2/3 EV: the histogram draws a vertical white line on right edge of the
histogram all the way from bottom to the top. JPEG data is severely
overexposed, RAW data is clipped somewhat but quite usable.

+1/3 EV: the histogram draws a vertical white line on right edge of the
histogram about, 80% of full height. Plenty of color patches (of the
target) are still blinking fully. JPEG data is severely overexposed,
RAW data is fully usable, only a little clipped.

0 EV: The histogram ends just about at 255, there is one empty pixel in
the histogram on the right, (and no vertical white line on right edge).
Blinking pixels on the most light patches. Jpeg data is clipped
according to the blinking, RAW data is about correctly exposed.

-1/3 EV: Both the histogram and the blinking appear *very* much the
same (verified by a loupe) as in the previous 0 EV shot (still the same
single pixel on the right is empty on the histogram). Jpeg data is
clipped according to the blinking pixels, RAW data is -1/3 EV
underexposed. And I really did not mess up with the manual controls nor
did the illumination level change, the illumination was from DC 12V
driven Tungsten lamp with dichroic filter for daylight simulation in
anothervice a totally dark room.

-2/3 EV: Histo ends at about 80% of the segment 1 filled. No blinking
pixels anymore, no clipping on JPEG data either.

-1 EV: Histo ends at about 60% of the segment 1 filled.

-1 1/3 EV: Histo ends at about 20% of the segment 1 filled.

-1 2/3 EV: Histo ends at about 80% of the segment 2 filled.

-2 EV: Histo ends at about 40% of the segment 2 filled.

-2 1/3 EV: Histo ends at about 30% of the segment 2 filled.

-2 2/3 EV: Histo fills the segments 5,4 and 3 fully. (So the two
leftmost segments are empty).

Not consistent behavioud in any way. So my new general rules for
evaluation of the histogram are:
1) is overexposed in case there is the faint white line, of any height,
at the right edge of the histogram.
2) is underexposed if more than 20% of the segment 1 is empty.

Extremely difficult tool for accurate control/evaluation, when 50% of
the segment 1 is empty the image data is already underexposed over 1
f/stop.

Linear histogram would be perfect (in other words, taken directly from
the raw data). This would allow very accurate evaluation of the
exposure: When such histogram is half full the exposure was -1EV, so
there is ample of room for accurate adjustment and by a quick glance
one could immediately see if the exposure is good. The way the
histogram is now presented maps the topmost 1 f/stop to a very very
narrow range at the right edge of the histogram, on the tiny LCD of the
cameras there are only very small number of pixels representing the top
1 f/stop. One f/stop underexposure is _quite_ bad with the digicams due
to their small dynamic range and due to the highly increased noise
level in the finalized image.

Now, there are many ways the feedback could be improved (in addition to
the histogram and blinking pixels):

1) Option of linear x-axis. This gives much better resolution to where
it is really needed.

2) Option for logarithmic y-axis. The form of the histogram is not at
all important, what is important is where it ends at the right.

3) In case of underexposure show the amount of the underexposure in EV,
with a user settable warning level, if exeeded emits a beep. And in
case of overexposure show the amount of pixels that are overexposed
e.g. in per mille, with a user settable warning level, if exeeded emits
another kind of a beep.

Timo Autiokari
Anonymous
December 24, 2004 1:47:04 AM

Archived from groups: rec.photo.digital (More info?)

In message <uga9r050mvvlnna2mvh7vculf4v7pksu88@4ax.com>,
drs@canby.com wrote:

>I am new to reading digital photo histograms. I've been told that the
>horizontal scale of luminosity is logarithmic. That the histogram is
>divided into 8 segments (not visually differentiated) of equal width.
>The first segment at the right side contains twice the information as
>the next segment, and that segment contains twice the info as its
>neighbor to the left, and so on. Meaning that the distribution image
>isn't quite what it might appear at first glance, its being on a
>logrithmic rather than linear scale. Apparently, the advantage of a
>log scale is that since the dark (left) end of the histogram
>represents images with far fewer photons, a linear scale histogram
>would be pretty thin on that end. Can anyone direct me to a better
>explanation of this? One that doesn't have a degree in physics as a
>prerequisite?

It isn't exactly logarithmic; logarithmic would be where the actual
intensity of the light were proportional to a fixed, constant number
raised to a variable power. The way image histograms work is that a
variable base is rasied to a fixed power.

In other words, where x is the position in the histogram, and the
histogram goes from 0 to 255, the range of intensities represented are
0^2.2 to 255^2.2. This is called "gamma-adjusted" data. The midpoint
(50% intensity) is somewhere around 186 out of 255.

A logarithmic range would be 2.2^0 to 2.2^255, and the midpoint (50%
intensity) would be about 254! 0 intensity would be negative infinity,
and all intensities below 1 would be negative values.
--

<>>< ><<> ><<> <>>< ><<> <>>< <>>< ><<>
John P Sheehy <JPS@no.komm>
><<> <>>< <>>< ><<> <>>< ><<> ><<> <>><
Anonymous
December 24, 2004 2:16:34 AM

Archived from groups: rec.photo.digital (More info?)

In message <1102452678.422467.132160@z14g2000cwz.googlegroups.com>,
eawckyegcy@yahoo.com wrote:

>Timo Autiokari wrote:
>
>> The histogram that digital cameras show is not logarithmic nor
>linear.
>
>??? I take a picture with my 10D. I observe histogram. I say "it
>needs another stop". So I adjust the exposure to add 2x the light,
>take a new picture and look at the histogram. The whole thing shifts
>over one horizontal "bar" on the display, just as one would expect.
>
>So it seems fairly log-ish to me. And really, if the histogram didn't
>behave this way, the feature would be rather useless.

It looks "log-ish" towards the right. It is not log-ish at all at the
black end. In fact, zero intensity would be at an infinite distance to
the left.

y=k^x - this is logarithmic

y=x^k - this is how gamma adjusted numbers work

A logarithmic value of zero always represents 1, as any number raised to
the zero power is one. The negative numbers represent the space between
linear 1 and linear infinitessimal values. Luminance starts at zero,
not at one.
--

<>>< ><<> ><<> <>>< ><<> <>>< <>>< ><<>
John P Sheehy <JPS@no.komm>
><<> <>>< <>>< ><<> <>>< ><<> ><<> <>><
Anonymous
December 24, 2004 2:31:42 AM

Archived from groups: rec.photo.digital (More info?)

In message <1102620870.134829.45560@z14g2000cwz.googlegroups.com>,
timo.autiokari@aim-dtp.net wrote:

>So this is called as the "expose to the right" -method. I always try to
>expose for full histogram without major clipping. That way the small
>dynamic range of the camera is taken fully into use, noise is minimized
>and the most important part of the dynamic range of the scene is
>captured. In the post-process I then have the best possible starting
>point for editing the image in case the main subject needs to be
>presented differently than how it was recorded, or in case the scene
>had very large dynamic range etc.

But alas, after every exposure the camera has all the information it
needs to tell you how you could optimize the exposure. Imagine if the
camera would report something like this to you, after the shot:

---------------------------------

% Clipped - R 0% G 8% B 0%

---------------------------------

REMAINING RAW HEADROOM (IN STOPS)

% Clipping Red Green Blue

0% 1.1 0 0.1
1% 1.2 0 0.3
2% 1.4 0 0.5
5% 1.7 0 0.8
10% 2.1 0.3 0.9

If it wasn't an action shot with changing lighting, and you could do it
again, you could perfect it ( or choose a suitable level of compromise).
--

<>>< ><<> ><<> <>>< ><<> <>>< <>>< ><<>
John P Sheehy <JPS@no.komm>
><<> <>>< <>>< ><<> <>>< ><<> ><<> <>><
Anonymous
December 24, 2004 2:31:43 AM

Archived from groups: rec.photo.digital (More info?)

JPS@no.komm wrote:
> In message <1102620870.134829.45560@z14g2000cwz.googlegroups.com>,
> timo.autiokari@aim-dtp.net wrote:
>
>> So this is called as the "expose to the right" -method. I always try
>> to expose for full histogram without major clipping. That way the
>> small dynamic range of the camera is taken fully into use, noise is
>> minimized and the most important part of the dynamic range of the
>> scene is captured. In the post-process I then have the best possible
>> starting point for editing the image in case the main subject needs
>> to be presented differently than how it was recorded, or in case the
>> scene had very large dynamic range etc.
>
> But alas, after every exposure the camera has all the information it
> needs to tell you how you could optimize the exposure. Imagine if the
> camera would report something like this to you, after the shot:
>
> ---------------------------------
>
> % Clipped - R 0% G 8% B 0%
>
> ---------------------------------
>
> REMAINING RAW HEADROOM (IN STOPS)
>
> % Clipping Red Green Blue
>
> 0% 1.1 0 0.1
> 1% 1.2 0 0.3
> 2% 1.4 0 0.5
> 5% 1.7 0 0.8
> 10% 2.1 0.3 0.9
>
> If it wasn't an action shot with changing lighting, and you could do
> it again, you could perfect it ( or choose a suitable level of
> compromise).
>

Wouldn't it be able to do the evaluation near-instantaneously and apply
the result to a second shot (if not a first) in milliseconds?


--
Frank ess
Anonymous
January 4, 2005 1:58:53 PM

Archived from groups: rec.photo.digital (More info?)

"Frank ess" <frank@fshe2fs.com> wrote in message
news:FZednVQ5o9Lx4lbcRVn-jA@giganews.com...
> JPS@no.komm wrote:
SNIP
>> If it wasn't an action shot with changing lighting, and you could
>> do it again, you could perfect it ( or choose a suitable level of
>> compromise).
>>
>
> Wouldn't it be able to do the evaluation near-instantaneously and
> apply the result to a second shot (if not a first) in milliseconds?

As a belated reaction, theoretically it could do this rather fast
(depends on image size and processor speed), especially when
implemented in an ASIC. However the results could drasticly reduce the
useful dynamic range. This could be caused by a single catchlight.
There is nothing wrong with clipping catchlights, because it allows
for better exposure in the shadows.
Of course one can apply some sort of heuristic to deal with the
exclusion of catchlights, but it would add to the time between shots.

Bart
!