ISO and actual sensitivity in DSLR's (D70, *istD, 20D, S3...)

Alan Browne wrote in news:d29cl5$dnq$1
@inews.gazeta.pl:

Explain then the quite improbable coincidence, given all your tripe,
that the results I get are so on the money?


I have written a long reply further down that explains what
happens when the Adobe RAW utility shows 118. I hope this
is the explanation you seeks.


/Roland

Alan Browne writes:

Assume for sake of simplification that the camera is using all 4096 levels
to represent black to white. Level 2048, the halfway point, is now one stop
down from maximum saturation, because one stop down is half as much light.
Level 1024 is, then, two stops down, 512 is three stops, and 256 is four
stops down.

Makes sense. But where is 18% grey?

Matte white objects like cloth and paper are about 90% reflectance. 18%
grey is 1/5 of that. So, *if* the camera exposed to put white at 4095,
you'd expect a grey card to be at about 820. Mid-grey is "mid" to our
eyes, but we see in an approximately logarithmic space, not a linear
one.

In practice, a camera isn't likely to put white at 4095 if it's setting
exposure automatically, because it wants some headroom for capturing
things brighter than white (e.g. specular reflections) without clipping
them abruptly. So white might be at 2048, grey at 410, and you'd have 1
stop of headroom above white.

Dave

Alan Browne writes:

118/255 within Photoshop has no meaning without a coding.
When you do that coding the RAW data is gone. There might
be zillions of unlinearities before arriving att this number.

It is as pure as the driven snow as it completely ignores display and
conversion issues. It is not JPEG. It is not TIF. It is not PSD. It
is not anything other than what the sensor read at a specified color
temperature. Period.

Well, no. The sensor read some voltage, which was probably converted
to a number in a range of 0-4095 or 0-16383 or something similar. Then
Photoshop's raw converter mapped that to the non-linearly coded (gamma
corrected) value 118 out of 255. 118 isn't raw data, and it also can't
be related back to an original sensor measurement without knowing what
the Adobe raw converter does internally.

And if the raw converter does any part of its processing adaptively,
depending on image content, you can never know exactly what the original
data was - even with a knowledge of internal operation of the converter.

Dave

Roland Karlsson wrote:

The 118 value is _not_ taken from the RAW data, it is
_computed_ from the RAW data. The Adobe RAW conversion

sigh As stated before all the data I posted was with all of the RAW
conversion parameters set to 0 with the exception of color temp. That
is to say no effect on the image. If I change parameters, such as
brightness, contrast, aturation, etc. you can be sure that the info
values change accordingly.

From the 7D manual: "Unlike the other image-quality modes, RAW image
data is unprocessed and requires image processing before it can be
used". Since the 'processing' I've done is _none_, there is no effect.

Secondly, due to the neutral conversion the JPG shows the same values
(close enough) at the same points in the image.


--
-- r.p.e.35mm user resource: http://www.aliasimages.com/rpe35mmur.htm
-- r.p.d.slr-systems: http://www.aliasimages.com/rpdslrsysur.htm
-- [SI] gallery & rulz: http://www.pbase.com/shootin
-- e-meil: there's no such thing as a FreeLunch.

Roland Karlsson wrote:

Alan Browne wrote in news:d29cl5$dnq$1
@inews.gazeta.pl:


Explain then the quite improbable coincidence, given all your tripe,
that the results I get are so on the money?



I have written a long reply further down that explains what
happens when the Adobe RAW utility shows 118. I hope this
is the explanation you seeks.

1) From the 7D manual
"Unlike the other image-quality modes, RAW image data is unprocessed and
requires image processing before it can be used"

2) As prev. stated, I had all 'adjustments' set to 0. Only the color
temp to match the light source was used.


--
-- r.p.e.35mm user resource: http://www.aliasimages.com/rpe35mmur.htm
-- r.p.d.slr-systems: http://www.aliasimages.com/rpdslrsysur.htm
-- [SI] gallery & rulz: http://www.pbase.com/shootin
-- e-meil: there's no such thing as a FreeLunch.

Owamanga wrote in
:

To summarize, a correct exposure isn't important in your workflow
because you have the levels sliders in Photoshop.

In a sense you are right. If you use your digital
camera (expensive or not) and takes a photo then
the most important thing is that the important
parts of the picture is within 10-90% of the exposure
range. Then you can make a picture that prints well.

In a sense you are not. It is important to
use a consitent exposure in your work flow. At least
if you are doing studio work. Pre visualisation
when taking the picture combined with a repeatable
work flow gives you better pictures.

But - I don't think that the meassurement where
you map 18% grey in an evenly lit scene to 118
in a sRGB picture is a useful method to get a
consistent work flow. I think it is a misunderstanding
of how exposure and color spaces work in digital
photography.



/Roland

Alan Browne wrote in news:d29fed$nlg
:

sigh As stated before all the data I posted was with all of the RAW
conversion parameters set to 0 with the exception of color temp. That
is to say no effect on the image. If I change parameters, such as
brightness, contrast, aturation, etc. you can be sure that the info
values change accordingly.

From the 7D manual: "Unlike the other image-quality modes, RAW image
data is unprocessed and requires image processing before it can be
used". Since the 'processing' I've done is _none_, there is no effect.

Secondly, due to the neutral conversion the JPG shows the same values
(close enough) at the same points in the image.

Please Alan.

Reread what I wrote. There is no 118 in the RAW data, the value
118 is computed. And it is the same computation made when
converting to JPEG. The 118 is the result of a non linear
computation.

That Adobe RAW says that it is working with the actual RAW
data is also correct. But it has nothing to do with the value
118, because that value is not what Adobe mwans whith image
data.

You can sigh how much you want. If you think I am a problem
when I am right and you don't understand - then it will soon
be your problem only - when I give up to explain it to you.



/Roland

Dave Martindale wrote:

Alan Browne writes:


118/255 within Photoshop has no meaning without a coding.
When you do that coding the RAW data is gone. There might
be zillions of unlinearities before arriving att this number.


It is as pure as the driven snow as it completely ignores display and
conversion issues. It is not JPEG. It is not TIF. It is not PSD. It
is not anything other than what the sensor read at a specified color
temperature. Period.


Well, no. The sensor read some voltage, which was probably converted
to a number in a range of 0-4095 or 0-16383 or something similar. Then

I agree to this point, I was a bit heavy in stating "the senor" above.
It should be, as stated in the 7D manual: "Unlike the other mage-quality
modes, RAW image data is unprocessed and requires image processing
before it can be used". It is such when I look at the pixel values,
except for the light source temp.

Photoshop's raw converter mapped that to the non-linearly coded (gamma
corrected) value 118 out of 255. 118 isn't raw data, and it also can't

At the RAW converter stage, I'm looking at the numbers for a given pixel
converted to the 0..255 scale, not the post conversion image/data where
gamma has been applied. There is, to put a point on it, no gamma
coefficent settable in the RAW converter and I have all settings set to 0.

be related back to an original sensor measurement without knowing what
the Adobe raw converter does internally.

And if the raw converter does any part of its processing adaptively,
depending on image content, you can never know exactly what the original
data was - even with a knowledge of internal operation of the converter.

Agree up to the point that I have the converter set (for this test) at
all '0', only the light source temperature is set.

If everyone else is correct about this, then I would like someone to
elaborate on the extraordinary coincidence that my values were very
close to 118 (except blue which was a little bit lower).

Cheers,
Alan.

--
-- r.p.e.35mm user resource: http://www.aliasimages.com/rpe35mmur.htm
-- r.p.d.slr-systems: http://www.aliasimages.com/rpdslrsysur.htm
-- [SI] gallery & rulz: http://www.pbase.com/shootin
-- e-meil: there's no such thing as a FreeLunch.

Roland Karlsson wrote:

of how exposure and color spaces work in digital
photography.

For tour own reasons you've completely misdirected the purpose of the OP.

--
-- r.p.e.35mm user resource: http://www.aliasimages.com/rpe35mmur.htm
-- r.p.d.slr-systems: http://www.aliasimages.com/rpdslrsysur.htm
-- [SI] gallery & rulz: http://www.pbase.com/shootin
-- e-meil: there's no such thing as a FreeLunch.

Alan Browne wrote in news:d29gr3$sve
:

If everyone else is correct about this, then I would like someone to
elaborate on the extraordinary coincidence that my values were very
close to 118 (except blue which was a little bit lower).

I hope you can bear with me - although you start to get
annoyed

If we assume that neither the camera nor the conversion
utilty finds any reason why it should do any compensations
and we also assume that the lens is without much flare and
we also assume that the non linear color space conversion
is well known (e.g. sRGB or Adobe RGB - well implemented) -
then a correctly exposed evenly lit 18% grey card shall
have a specific value in the picture.

I assume someone else have made similar meassurements like
you and found out that this value is 118. So - if you do the
same meassurements thoroughly you shall end up with 118.

But - and this is my point. Another camera or conversion
utility might end up with 132. And this value do not show
that the camera/utility has another sensitivity. It only
shows that the value is 132. To understand the sensitivity
of the acmera you must look at more values. And most important
are those near to 255 and 0.

In a way it seems like you want to use something like
the zone system. In the zone system, the picture is matched
into 10 zones in the print. But ... you are trying to
use only one zone.


/Roland