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

On 28 Mar 2005 14:32:18 GMT, Roland Karlsson
wrote:

Owamanga wrote in :

Yes, 118/255 is no where near black, even if you store the file in
Native American Smoke Signal format. Storage makes no difference to
what 118/255 means inside Photoshop once the file has been decoded.

118/255 within Photoshop has no meaning without a coding.

It is entirely separated from the compression (file coding) used to
store the image.

Fill a 100x100 pixel image with 118/255 RGB in photoshop and save it
as a JPEG, BMP, TIFF, PSD even GIF, choose any of the formats, lossy
or lossless. Now open the file again, and you'll see data that shows
118/255.

When you do that coding the RAW data is gone. There might
be zillions of unlinearities before arriving att this number.

Agreed, that wasn't my point. I was trying to explain the way Alan was
measuring - eg the meaning of 118 is 118/255 in photoshop's pixel
level 'info' tool.

I can give you an example. Some time ago there was som discussions
in this forum ragarding native JPEG in the D10. It was shown that
you lost high lights if you let the camera do its own conversion
to JPEG. It was therefore shown that you shall NEVER let D10 do
the JPEG compression - always use RAW.

Agreed, a good idea on any camera. But Alan isn't attempting to
measure DMin or DMax here, so the blown highlights isn't really going
to be an issue.

I think this was shown withot a doubt.

Certainly.

As I have written in another reply - there are only three ISO values
that are meaningful to compute for the camera:

1. The ISO the light meter in the camera assumes.
2. Saturation based ISO.
3. Noise based ISO.

All three can be found in this paper:
http://www.kodak.com/global/plugins/...asurements.pdf

None of those three meassurements are based upon measuring
a grey tone in the resulting picture. A grey tone in the
resulting picture depends on the linearity of the system.
Therefore - you don't use a grey tone when defining
the ISO of the system.

OK - a solid state sensor is very linear - but not 100%.
And there is nothing that says that the A/D conversion
is linear. And there is nothing that says that the
RAW import must be linear.

Yes. There is more than one way to cook an egg.

An what about the numbers
shown in Photoshop. What do 118/255 mean for a 16 bit
image? For an 8 bit image? Don't forget that Photoshop
do color management.

Your gamma settings will make no difference to the info tool's ability
to point to a pixel and give you a reading of 118/255. All it would do
is make that gray appear to be darker or lighter than another system
displays it. Alan wasn't metering off his screen, so PS's color
management doesn't really come in to it.

Are the number before or after
converting to the view color space?

Presumably the color space consistently remains sRGB from the camera
to Photoshop. Or if a color space change was involved, it was the
*same* color space change from each source.

So - in short. I don't think you can use a grey tone to
determine the ISO sensitivity for a complex system.

Okay, but to compare two sources, when Photoshop's configuration
remains the same between the two, I think still has some validity.

--
Owamanga!
http://www.pbase.com/owamanga

Roland Karlsson wrote:

Alan Browne wrote in news:d27daj$oqj$1
@inews.gazeta.pl:


Again, the measurement _I_ did was off of the RAW first then JPG. With
(as I stated) all parameters set to '0' for conversion to JPG, the
values in grey remained about (+/- a few) the same. (Only color temp
was set to flash temp of 5500K).


The values in RAW are 12 bit linear - the values in JPEG are 8 bit
_non_ linear. How can you be meassuring the value 118 on the RAW data?
That would be almost totally black. There must be some misunderstanding
here.

Try READING the post. I've stated several times that the Adobe RAW
import utility in 16 bit/colr mode displays the R,G,B data in the range
of 0..255. Hence, 118 / 255 is the 18% grey level.

I also posted a link to a screenshot:
http://www.aliasimages.com/images/RAWSS.jpg whiuch you seem to have
ignored for the sake of being a difficult curmedgeon.


Moreover - look at my other reply that shows what ISOs you
can meassure. There are actually several ones, depending on
what you want to accomplish. And if you don't believe me or
my argumentation - maybe you believe Kodak. The link I give
in the other reply is to a Kodak document. Please read it.

Been there, done that. However the method use dby C d'I is different
(yet in reference to the same standards and processes). The method I
used results in a similar outcome.

Reading for middle grey is a legitiamte means of confirming the ISO of
the cameras. They all have different toe and shoulders, but they should
all have the same 18% grey if their ISO sensitivity is correct.

--
-- 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:

Owamanga wrote in :


Yes, 118/255 is no where near black, even if you store the file in
Native American Smoke Signal format. Storage makes no difference to
what 118/255 means inside Photoshop once the file has been decoded.


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.

Thence, converting to JPG without any other changes, it is not
surprising that the JPG shows pretty much the same number (within a few).

If this does not satisfy you, then please explain the reverse: Why am I
getting these numbers? Coincidence perhaps? No.


I can give you an example. Some time ago there was som discussions
in this forum ragarding native JPEG in the D10. It was shown that
you lost high lights if you let the camera do its own conversion
to JPEG. It was therefore shown that you shall NEVER let D10 do
the JPEG compression - always use RAW.

I think this was shown withot a doubt.

As I have written in another reply - there are only three ISO values
that are meaningful to compute for the camera:

1. The ISO the light meter in the camera assumes.

I was not using the light meter.

2. Saturation based ISO.

I was using RAW to record.

3. Noise based ISO.

All three can be found in this paper:
http://www.kodak.com/global/plugins/...asurements.pdf

None of those three meassurements are based upon measuring
a grey tone in the resulting picture. A grey tone in the
resulting picture depends on the linearity of the system.
Therefore - you don't use a grey tone when defining
the ISO of the system.

Perhaps not to define it. However, if 18% grey is what it is supposed
to be, then a shot taken of such an 18% target under conditions that
eliminate as many uncertainties as possible (as I did) will reveal if
the sensor exposed for a particular ISO is recording for that ISO. In
the 7D case it is thus. In the A200 test (C d'I) it is thus.


OK - a solid state sensor is very linear - but not 100%.
And there is nothing that says that the A/D conversion
is linear. And there is nothing that says that the
RAW import must be linear. An what about the numbers
shown in Photoshop. What do 118/255 mean for a 16 bit
image? For an 8 bit image? Don't forget that Photoshop
do color management. Are the number before or after
converting to the view color space?

Before. It is completely independant of colorspace. It is pre-color
management (display/printer).



So - in short. I don't think you can use a grey tone to
determine the ISO sensitivity for a complex system.

You're steadfastly missing the point. The 18% grey test is not to
DEFINE the ISO, but rahter to confirm the exposures taken with a given
IOS setting. If the 18% comes out as 18% then the ISO is correct in the
mid tone. Otherwise, as people use their meters to measure light
relative to an 18% mid tone, they will end up with over or under
expsosure of the mid tone.

--
-- 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
:

Presumably the color space consistently remains sRGB from the camera
to Photoshop. Or if a color space change was involved, it was the
*same* color space change from each source.

The camera might output Adobe RGB. The screen is probably sRGB.
The working color space inside Photoshop can be either or something
totally different. The 118/255 is meassured within the working color
space.

Moreover - even if the working color space and the camera's color
space are equal - the picture has to be converted from the linear
representation to the non linear 8 bit representation before it
can be representaed as 118/255.

So - in short. I don't think you can use a grey tone to
determine the ISO sensitivity for a complex system.

Okay, but to compare two sources, when Photoshop's configuration
remains the same between the two, I think still has some validity.

Yes - it has some validity. But only as long as you plan
on using the same conversion utility when you are going to
make the pictures as you used when you did the measurements.


/Roland

Roland Karlsson wrote:

Owamanga wrote in
:


Presumably the color space consistently remains sRGB from the camera
to Photoshop. Or if a color space change was involved, it was the
*same* color space change from each source.


The camera might output Adobe RGB. The screen is probably sRGB.
The working color space inside Photoshop can be either or something
totally different. The 118/255 is meassured within the working color
space.

How many times have I stated: RAW for my tests?


Moreover - even if the working color space and the camera's color
space are equal - the picture has to be converted from the linear
representation to the non linear 8 bit representation before it
can be representaed as 118/255.

Again, try reading. I stated "in the Adobe RAW converter."


Okay, but to compare two sources, when Photoshop's configuration
remains the same between the two, I think still has some validity.


Yes - it has some validity. But only as long as you plan
on using the same conversion utility when you are going to
make the pictures as you used when you did the measurements.

Yes and no. First off I am a 100% RAW man (since I finally found out
that PS E 3.0 does 16 bits/color). My images will always come in to PS
via that path.

No, in the sense that once the ISO offset or error is known, then the
photog can make further tests to determine his own exposure preferences
for his work path.


--
-- 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 :

The 18% grey test is not to
DEFINE the ISO, but rahter to confirm the exposures taken with a given
IOS setting. If the 18% comes out as 18% then the ISO is correct in the
mid tone. Otherwise, as people use their meters to measure light
relative to an 18% mid tone, they will end up with over or under
expsosure of the mid tone.

OK - your test meassures the fidelity of mapping an 18%
grey card to the value 118 in the controlled environment
and using the given work flow.

But - personally I fail to se the relevance of such a test.
At least I always apply levels to my pictures. Then the 118
value will move. I also apply some color balance. Then it will
move even further.

The main purpose of the choice of exposure is to keep the
important parts of the picture within e.g. 10-90% of max
exposure before clipping. Where the 18% grey is to be found
in an 8 bit conversion of the data I find rather uninteresting.
If the picture is rather flat I over expose some to move
the exposure to parts with less noise - or under expose some
to avoid blurriness.


/Roland

Alan Browne wrote in news:d29ami$7gu$1
@inews.gazeta.pl:

Try READING the post. I've stated several times that the Adobe RAW
import utility in 16 bit/colr mode displays the R,G,B data in the range
of 0..255. Hence, 118 / 255 is the 18% grey level.

The 118 cannot be a linear value. It must be a predicted value
by the Adobe RAW utility. And this is exactly how Photoshop
displays the values whan working with 16 bit pictures. Photoshop
(and I assume Adobe RAW) shows the predicted value in the assumed
working color space using 8 bit representation. This is a non
linear mapping of the 16 bit linear data. And this non linear
prediction might vary between conversion utilities.

I also posted a link to a screenshot:
http://www.aliasimages.com/images/RAWSS.jpg whiuch you seem to have
ignored for the sake of being a difficult curmedgeon.

I have looked at it. And I cannot see that it changes a thing.


/Roland

Roland Karlsson wrote:

I have looked at it. And I cannot see that it changes a thing.

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

--
-- 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.

On 28 Mar 2005 16:39:55 GMT, Roland Karlsson
wrote:

Alan Browne wrote in :

The 18% grey test is not to
DEFINE the ISO, but rahter to confirm the exposures taken with a given
IOS setting. If the 18% comes out as 18% then the ISO is correct in the
mid tone. Otherwise, as people use their meters to measure light
relative to an 18% mid tone, they will end up with over or under
expsosure of the mid tone.

OK - your test meassures the fidelity of mapping an 18%
grey card to the value 118 in the controlled environment
and using the given work flow.

But - personally I fail to se the relevance of such a test.
At least I always apply levels to my pictures. Then the 118
value will move. I also apply some color balance. Then it will
move even further.

The main purpose of the choice of exposure is to keep the
important parts of the picture within e.g. 10-90% of max
exposure before clipping. Where the 18% grey is to be found
in an 8 bit conversion of the data I find rather uninteresting.
If the picture is rather flat I over expose some to move
the exposure to parts with less noise - or under expose some
to avoid blurriness.

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

It all makes sense now.

--
Owamanga!
http://www.pbase.com/owamanga

Alan Browne wrote in news:d29bvh$br2
:

How many times have I stated: RAW for my tests?

Yes you have - but repeating it does not make it
more valid

This is what really happens when you read 118.

I am 100% sure about it. If you have some arguments
I find valid - telling hat I am wrong I will of
course listen. But don't repeat that you are working
with RAW data - as I find that remark being irrelevant.

The 118 value is _not_ taken from the RAW data, it is
_computed_ from the RAW data. The Adobe RAW conversion
utility does exactly like Photoshop when working with
linear input. Photoshop maps the 16 bit PSD file to
the choosen working color space. This color space might
be sRGB and is almost always non linear. It might also be
Adobe RGB. You can check what you have choosen in the
Adobe RGB utility. It is not a surprise that the converted
file has the same values - as the Adobe RAW conversion
utility uses the same computation when doing the conversion
as it uses when computing the predicted value.

So - if Photoshop tells me 118/255 it then means the
value that the 16 bit linear pixel gets when it is converted
to the working color space represented as 8 bit.

If you take a linear 16 bit PSD picture and work in another
color space - then the 118 will be displayed as something
else.

The 118 does not say anything about the greyness of the pixel.
It only tells what the pixel will be represented as when
saved in an 8 bit file using the working color space.


/Roland