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

In message ,
Alan Browne wrote:

Regardless of the conversions:

-the grey level was maintained exactly if the RAW conversion was held to
no changes other than color temperature. eg: the 8 bit values in the
RAW and JPG were the same.

What do you mean by "the 8 bit values in the RAW"?

The exposure of a grey-card in the Canon 20D is in the 300s in the green
channel, out of up to 3975 RAW levels above blackpoint.

The whole idea of characterizing cameras that produce RAW files by their
JPEG renders makes me cringe. JPEG is bull****; it's just a side effect
of the camera; an extra feature with limited value. It makes me sick
that my $1500 DSLRs have green-channel-only histograms that represent a
"JPEG", and tell me nothing about the state of the real exposure, as it
is digitized in the RAW "RGB" data.
--


John P Sheehy

wrote:
In message ,
Alan Browne wrote:


Regardless of the conversions:

-the grey level was maintained exactly if the RAW conversion was held to
no changes other than color temperature. eg: the 8 bit values in the
RAW and JPG were the same.


What do you mean by "the 8 bit values in the RAW"?

In the RAW import utility of PS E 3, it displays R,G,B as 0..255 (using
the info pointer) even if in 16/bit / color mode. Here's a screen shot.
http://www.aliasimages.com/images/RAWSS.jpg (the pointer was on the grey
back of one of the gulls).



The exposure of a grey-card in the Canon 20D is in the 300s in the green
channel, out of up to 3975 RAW levels above blackpoint.

4095 minus blackpoint = 3975 ?

300 / 3975 = 0.075
399 / 3975 = 0.1

Seems well below 18% grey. But I suppose this is non-linear.


The whole idea of characterizing cameras that produce RAW files by their
JPEG renders makes me cringe. JPEG is bull****; it's just a side effect
of the camera; an extra feature with limited value. It makes me sick
that my $1500 DSLRs have green-channel-only histograms that represent a
"JPEG", and tell me nothing about the state of the real exposure, as it
is digitized in the RAW "RGB" data.

First off, in my test of this, the R,G,B's are those reported in the
Adobe RAW converter (set at 16 bit/channel, but the RGB info is
displayed in the range 0..255 for each color).

Regarding your "green channel only histos" I suggest you write a latter
to the manufs. By the way, do you have a reference for it being
green-channel only?

The C d'I article did not state how they were getting the RGB info (what
software), just that they were. When I imported the RAW into PS E 3., I
had to set all conversion sliders to 0 (except color temp which I set at
5500K).

In PS, of course it's not JPG, it's native PS. The info window/pointer
gives the values at that point.

Cheers,
Alan.


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Alan Browne wrote in news:d26dqa$5jl
:

Hmmm ... I assure you ... I am not trying to be difficult.

Hmmm yourself.



Nope - I don't buy it.

The cameras we are talking about use RAW as their native
format. Any meassurements on ISO sensitivity must be made
on the original RAW data.

The procedure you describe means that this RAW data is
modified and then converted to JPEG before making the
ISO sensitivity meassurements. This conversion involvs
both color correction and also an unknown non linear
gamma type conversion.

The strobe method for exposure is also somewhat peculiar.
This method avoids any inaccuracys in the light
meassuring system of the camera and also any inaccuracies
in the exposure time. But - you still have the accuracy
in the aperture and lens attenuation to take into
account.

I would say that the method (although involved and
rather scientific executed) is rather meaningless.


/Roland

Alan Browne wrote in :

...

There are three ISO values you can meassure for a 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

To meassure 1. you use a large uniformly lit area. You then meassure
this area with a callibrated light meter. Then you meassure the same
with the camera's light meter. The actual value can then be deduced.
E.g. if you set the camera to ISO 100 and the callibrated light meter
gives you the same exposure at ISO 125 - then the camera actually
is meassuring ISO 125.

How you can meassure both 2. and 3. is hinted at in the Kodak paper.
It is rather involved - and it does not look the slightest like
the one you described was made in the journal.

The three methods will probabaly give you three different values.
And this is OK. You can expose more or less and still get a
useful image. The actual ISO for the system depends on how you
want to use it.


/Roland

Roland Karlsson wrote:

Alan Browne wrote in news:d26dqa$5jl
:


Hmmm ... I assure you ... I am not trying to be difficult.

Hmmm yourself.




Nope - I don't buy it.

The cameras we are talking about use RAW as their native
format. Any meassurements on ISO sensitivity must be made
on the original RAW data.

As I did. I assume that C d'I did as neutral a test as possible, but
they did not state how they did that. They do state various things such
as their dark room, their light source (2000 lux), etc.


The procedure you describe means that this RAW data is
modified and then converted to JPEG before making the
ISO sensitivity meassurements. This conversion involvs
both color correction and also an unknown non linear
gamma type conversion.

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 strobe method for exposure is also somewhat peculiar.
This method avoids any inaccuracys in the light
meassuring system of the camera and also any inaccuracies
in the exposure time. But - you still have the accuracy
in the aperture and lens attenuation to take into
account.

I chose the flash as I could accurately meter it seperately, it's a very
repeatable test, I could do with the studio lights down to avoid other
color sources.

As to the aperture I'm already satisfied from other experience that it
is accurate enough.

As to lens attenuation, that is also fairly insignificant. (Or why
would incident metering be useful at all).


I would say that the method (although involved and
rather scientific executed) is rather meaningless.

I would say that their method and mine beat your conjecture.

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.

Alan Browne wrote:

300 / 3975 = 0.075
399 / 3975 = 0.1

Seems well below 18% grey. But I suppose this is non-linear.

You may be misinterpreting what "linear" means.

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.

The 0-255 scale you're thinking in is non-linear in terms of the amount of
light represented -- it represents levels after a gamma correction.

--
Jeremy |

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.

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.



/Roland

On 28 Mar 2005 13:27:20 GMT, Roland Karlsson
wrote:

Alan Browne wrote in news:d27daj$oqj$1
:

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.

If you took a look at the binary file itself (after some
decompression, and a good idea of JPEG format) this is true.

How can you be meassuring the value 118 on the RAW data?

He's keeping the same method of measurement across both formats, 118
means 118/255 on each of the RGB channels. Photoshop has already
interpreted the RAW file at this point and moved it into these 3 bytes
per channel (or 24 bits per pixel) representation that Alan is
quoting.

Even in the case of a non-linear file format such as JPEG, the
software (Photoshop or whatever) has already decoded it back into a
linear representative workspace. Inside Photoshop, fully saturated
green is R=0,G=255,B=0 regardless of the compression format used to
store the file onto the disk.

If you want Alan to change his measurement units based on the storage
compression, his results or statements would be entirely meaningless
to anyone wanting to compare the two (which was the point).

That [118] would be almost totally black.

He didn't use a binary editor to look at the RAW file itself, he's
quoting Photoshop numbers.

There must be some misunderstanding here.

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.

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

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.

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.
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. 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?

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


/Roland

Jeremy Nixon wrote:

Alan Browne wrote:


300 / 3975 = 0.075
399 / 3975 = 0.1

Seems well below 18% grey. But I suppose this is non-linear.


You may be misinterpreting what "linear" means.

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?


The 0-255 scale you're thinking in is non-linear in terms of the amount of
light represented -- it represents levels after a gamma correction.

This is all very unclear to me. Not your post. I just haven't taken
the time to look at the math.



--
-- 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.