I Don't Understand the Luminous Landscape "Expose Right" Article

I Don't Understand Luminous Landscape "Expose Right" Article

Thomas Knoll and Michael Reichmann coauthored the "Expose Right"
article on Michael's Luminous Landscape website.

Last fall I emailed Thomas Knoll this question. After he did not
reply, early this year I emailed the question to Michael Reichmann.
He did not reply either. I have posted a copy of the question I sent
Reichmann at http://factsfacts.com/LuminousLandscape.htm.

Since I have not received a reply from either of the authors of the
article in question, I am posting my question here.

I will try to state my question simply here. This is intended only as
a summary statement. Please read the complete question.

The table in the article shows the tonal values of an image shot with
a digital camera as a geometric distribution, and implies that the
tonal values of film are distributed linearly. Are the tonal values
of film really distributed linearly in this context?

If you reply please do so as requested in my question to Reichmann.
That is, please either agree that my "Film" column in the table is
correct, or give the 5 values that should be in the Film column.

Thanks.

jim

 

I Don't Understand Luminous Landscape "Expose Right" Article

Anyone wishing to understand the article could run this simple test ...
pick a fairly monotone subject (sky, gray card, etc) and shoot it in
RAW mode at ISO 400 metered properly, then set your exposure
compensation +2 and shoot it again, then -2 and shoot it again. When
you convert the three RAW files don't make any exposure adjustments to
the first image, adjust the +2 image by -2 stops, adjust the -2 image
by +2 stops. You should have roughly equivalent exposures by now if
you convert these ... check the noise levels of the three "equivalent"
exposures" and you'll see what he means.

You could have gotten the same improvement that the +2 image gives by
shooting as metered at ISO 100.

Try it and see.

 

The article is discussing the behavior of digital cameras not film. How
many levels you have in film is not so clear cut, you can deal with the
SN of film and get some sense of it, Roger Clark has done a lot of work
on this, see this link

http://clarkvision.com/imagedetail/digital.signal.to.noise

Scott

 

In message <>,

This should be true, but some RAW converters treat RAW data based upon
their absolute values, and not the values after a purely linear
"exposure" adjustment. ACR, for example, nails the highest RAW values
to 255 in the output, no matter if you use -4 "exposure".

Not exactly; ISO 400 at +2 EC is usually better quality than ISO 100 at
0 EC, if the RAW data doesn't clip.

--

 

And figure 5 on this page:
http://clarkvision.com/imagedetail/dynamicrange2
which does discuss film/digital

I still find that one shocking, particularly given the implication that
digital has less info in the shadows than digital highlights. But still
digital has more information in the shadows than film.

But that is not really what the luminous landscape article is discussing
I think. They are talking about how RAW digital files carry more info in
the highlights due to their being linear data of simple intensity.
That's the best I can understand anyways. Brighter means bigger number
in that linear scheme so there are more significant digits to assign
finer gradations of tonal range. Once it's converted from RAW, that
extra info is dropped. During the RAW conversion process, there is an
opportunity to sneak up & expand that highlight information before
dropping it.


One result of this effect (if I understand correctly -and I'm unsure) is
that an unmanipulated RAW image looks too dark & muddy. Most of the
information is crammed in the highlights. A curve needs to be applied to
the image to increase contrast back to what our non-linear eyes expect
to see.

Maybe someone can comment on whether the diagrams below correctly
explain the situation. Normal scenes appear to our eyes with most of the
complexity in the middle ranges and not a lot going on in the highlights
or shadows. Digital RAW format contains extra highlight data which is
generally useless unless you stretch it to the left before clipping to a
normally curved format. So yes RAW has less info in the shadows but it
has all it needs as long as you don't need to adjust the contrast later.
I think film is in the same boat on the shadow end of the chart, in fact
according to the figure 5 mentioned above, film has less shadow detail
and less room for adjusting highlight detail.

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shadow highlight

1. LINEAR RAW DATA

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shadow highlight

2. NORMAL (FILM, EYE, TIF, JPEG) -Bell Curve


Or I could be misunderstanding. I'm just trying to learn as I go.

 

The raw file holds the light values for the pixel in a linear format,
the normal jpg format is going to hold the data in a non-linear format
to fit more dynamic range in the 8 bits. This is done with the gamma of
the image, the light level is proportional to the pixel level raised to
the gamma power. A common gamma that is used is 2.2, this is what
photos that are in the sRGB color space use. This also turns out to be
the gamma of most PC monitors, MACs use a lower number of 1.8.

So it has nothing to do with our non-linear eye and everything to do
with our non-linear monitors.

Scott

 

John Sheehy writes ...

We must be talking about something totally different, because this is
total nonsense ... in the test shots I mentioned (monotone subject --
in my case a heavily overcast sky -- at 0, +2 and -2) a straight
conversion using ACR of each image leaves me with these RGB values for
the brightest part of the sky ... 60/67/79 for the -2 shot, 129/140/165
for 0, 223/229/246 for the +2. Think about it ... if what you say is
correct (it isn't) then every monotone image would get mapped to white,
which obviously doesn't happen.

Run the test and compare the noise levels. That's what I did. ISO 100
looks the same as ISO 400 at +2 corrected for exposure with Capture One
....

Bill

 

First, the digital camera response isn't a "geometric distribution".
The response is linear. But the terms in the left-hand column of your
chart are logarithmic. The top row is labelled "within the first
f/stop", which by definition contains all tones between 1/2 of maximum
and maximum. Naturally, because the sensor and A/D converter are
linear, half of the available codes are assigned to this half of the
brightness range.

The problem with this approach is that the *eye* isn't linear. It sees
the step from full brightness to 1/2 brightness as being about the same
size as the step from 1/2 to 1/4, which is the same as from 1/4 to
1/8. So if you have a limited number of bits available (e.g. 24 bit
colour, with only 256 codes per colour), you're better off to use a
nonlinear transformation to spread the available codes so that each stop
of range gets something like the same number of codes. This is in fact
done.

Second, what do you mean by "the tonal values of film are distributed
linearly"? Film itself is analog; it has no levels or steps. Digitized
film *does* have levels, but the distribution of them depends on the
software running the film scanner. Some film scanners output data that
is proportional to film density, which is the logarithm of
transmittance. In this case, there will be about the same number of
code values per each one-stop range of input brightness. Some scanners
output linear data - but it's linear in film transmittance, not original
scene brightness. And there are other possibilities.

Basically, your question makes no sense unless the right column is about
*some specific film, scanner, and software* combination. Film itself
has no levels.

Your question is unanswerable as posed. There is no answer that would
apply to all film scanners.

Dave

 

Jim:

I don't understand why you are trying to make a comparison to film.
That article is strictly about digital photography, one of the quirks
of using an electronic sensor, and how to compensate for it.

It has nothing to do with using film, it implies nothing about the
behavior of film. There is no connection to using film.


Drifter
"I've been here, I've been there..."

 

I didn't get all of what you are saying but I was reading some other
articles about a way of converting RAW in a linear format to 16 bit TIF.
Done that way, the image looks almost pure black but when curves are
applied, it can be adjusted to get more info out of high contrast
images, less blown highlights. The thing is it usually messes up the
colors so the technique is to make a B&W image with the linear
conversion then do another normal one for color & merge the two.

I think Dave is on to something about the logarithmic scale of stops
versus linear digital capture though that seems such an extreme
difference. Probably it has more to do with that than my bell curve
histogram idea.

 

Versus what? If it's a quirk, what doesn't have this "quirk?" Film
maybe?

jim

 

Okay that's fine, have it your way. My question still remains. In
the full luminosity range of *film*, is half of the tonal range in the
first F/stop or is the range more geometrically distributed such that
1/5 of the tonal range is in the first F/stop and 1/5 in each of the
other five F/stops. Said another way, if the luminosity range were
divided and in the table would an equal amount go in each row. And,
if it is not linearly distributed (like the digital image) and it is
not geometrically distributed, what is the distribution?

I'm talking about film. I'm not talking about film scanners or the
results of scanning film.

jim

 

It's a tip on how to take advantage of 'bonus' data hidden in RAW
digital files.

 

The same technique could be used when scanning film theoretically, if
you wanted to make adjustments.

 

Okay, obviously you want to have a discussion about film. In that
case maybe Rec.Photo.Digital isn't the place for you.


Drifter
"I've been here, I've been there..."

 

The problem with this question is that you are trying to measure
something that doesn't exist. You want to know how many gradations
there are allocated to each stop of exposure range, but there aren't any
gradations in film in the first place.

Suppose you have a scene where the intensity of objects ranges from 0 to
1 in some particular units. So the light intensity is 0.5 for one stop
down from maximum, 0.25 for two stops down, 0.125 for 3 stops down, and
so on. It does not make sense to ask whether half the tonal range is
between 0.5 and 1.0 (the brightest f/stop). There are infinitely many
values between 0.5 and 1, just as there are infinitely many values
between 0.25 and 0.5, and betweeen 0.125 and 0.25. There is no
meaningful way of "counting" or "measuring" the number of values
assigned to each stop of brightness range.

It's only when you quantize the range to a number that you can ask this
question, and the answer depends entirely on the way you do the
quantization. If you map intensity linearly to an integer in (say) 0 to
4096, *then* there are 2048 values used for the first stop, 1024 for the
next stop, and so on. But if you quantize the logarithm of intensity,
then every stop gets exactly the same number of code values. Or if you
represent intensity using a computer floating-point number, you get
exactly the same number of code values for each stop.

Now, film's tonal range is compressed from the original scene. Every
time the original scene brightness increases by a factor of 2, the light
transmittance of the negative is reduced to about 0.66. It isn't 0.5
because of the lower contrast (gamma) of the negative. But film
transmittance is still a continuous thing. It doesn't make sense to ask
whether the first stop uses "the same amount" of the range as the second
stop of exposure range, because there's no meaning to "amount".

If you want, you can graph the response of film: the amount of light
that comes through the film vs. the amount of light the film was exposed
to. If you plot this graph on log-log scales (the usual way
characteristic curves are shown), you'll see that each stop of exposure
uses about the same fraction of the curve length. But that's precisely
because you chose a log-log scale in the first place, which shows each
stop of exposure as the same size on the X axis. If, instead, you drew
exactly the same data on a linear graph, the brightest one stop of the
exposure range would take up half the width of the graph. The
difference in apparent importance is purely due to the graph scale you
chose, not anything to do with the function you're graphing.

Both graphs will tell you exactly how much film density (or
transmittance) you can expect for a given amount of exposure. But
neither give you an objective answer to "how much of the range is used".
Each graph *suggests* an answer to that question, but both are lies.

Your question has no true answer at all when applied to film. There's
no unbiased way to ask "how much of the film range is between these two
points". If you use a linear definition of "how much", you'll find
that the brightest stop gets about half the range. If you use a
logarithmic measure of "how much" (and density is a logarithmic
measure), you'll conclude that each stop of exposure uses the same
fraction of film's range. But either way you'd be misleading yourself;
the answer is completely determined by the measuring scale you
choose.

It's only when you digitize film that you can meaningfully talk about
"how much" with respect to the pixel code range.

Dave

 

It is a "quirk" of any system that linearly digitizes a continuous quantity
(light) into a discrete representation. It applies to digitized audio
too. Since film doesn't quantize brightness, it does not apply to film.
That doesn't mean that film is better because of this.

The article compares two different approaches to working with a digital
camera, and shows that one produces better results than another. That's
all.

If you want to compare film to digital images, you need to take a
broader look at how various "quirks" of BOTH imaging methods influence
the quality of the final image. The quantization of brightness in a
digital camera has one sort of effect on the image, and it may or may
not be visible depending on the circumstances. The grain of film adds
another sort of error to film images, one that's not present in digital
images, and again it may or may not be visible.

But this article really doesn't address film at all. If you wanted, you
could publish the same sort of advice for film, but it would be for
different reasons. (It's generally better to overexpose negative film
slightly, but underexpose positive film. This could be called "expose
right" and "expose left" respectively).

Dave

 

I'm comparing film camera performance to digital camera performance.
Using your logic, if I were to post in rec.photo.film they would say,
"Obviously you want to have a discussion about digital cameras. In
that case maybe Rec.Photo.Film isn't the place for you. "

jim

 

Maybe they would. The whole "film vs digital" thing was thrashed to
death a long time ago.
If you like film, use film.
If you like digital, use digital.
If you like both, use both.
But for pete's sake stop trying to make a digital-to-film comparison
of an article that was written about how to best use a characteristic
of digital cameras. As several of us have tried to tell you, there is
no useful equal to this behavior in film just like there is no useful
equal to chemical mixes in digital. (could you imagine trying to set
up a written table equating developer mixes/times to Photoshop
settings? Jeeze, you'd go totally bonkers first).

Unfortunately when any of the responses here try to point out the
"apples/oranges" element out to you, your response has essentially
been a repeat of "but what about film" so I don't really know what
else to say in order to help you understand.


Drifter
"I've been here, I've been there..."

 

Actually, that would make somewhat more sense. All you'd need is data
showing how developer concentration and development time affects the
developed film (primarily the change in gamma) and then map it to
equivalent Photoshop settings. In fact, if you know that a certain
development change boosts gamma by a factor of 1.1, you can get almost
the same effect in Photoshop by setting the middle box in Levels to 0.9
(the reciprocal of 1.1).

Or if you have the complete characteristic curve of the film and
developer together, you could convert that into a Photoshop Curves
operation. This is a bit messier because the axes of a film response
graph are base-10 logarithms of light intensity and transmittance, while
in Photoshop the axes are pixel values which are roughly intensity to
the 0.45 power. But a spreadsheet will take care of the conversions
nicely.

Dave