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What is the Characteristic Curve function of a digital cam..

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Anonymous
January 30, 2005 1:33:46 PM

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

Hi,
I'm looking for the equation for the conversion of
the linear intensity data from a digital camera sensor
to the image output. Please see the measured
curve in Figure 7 at:
http://clarkvision.com/imagedetail/dynamicrange2

This is not a standard Gamma function. Using a
Gamma function, the Gamma appears to be around 8
at the high end, dropping to 1 at the low end.

Does anyone know the mathematical formula for
curves like this one? I've tried several, but
can't quite match the data over the whole range.

Roger
January 30, 2005 4:59:01 PM

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

Roger N. Clark (change username to rnclark) wrote:
> Hi,
> I'm looking for the equation for the conversion of
> the linear intensity data from a digital camera sensor
> to the image output. Please see the measured
> curve in Figure 7 at:
> http://clarkvision.com/imagedetail/dynamicrange2
>
> This is not a standard Gamma function. Using a
> Gamma function, the Gamma appears to be around 8
> at the high end, dropping to 1 at the low end.
>
> Does anyone know the mathematical formula for
> curves like this one? I've tried several, but
> can't quite match the data over the whole range.
>
> Roger
>
There are several linearizing functions for p[hotographic emulsions that have been known for many years. I published a
little note on the subject:
Remarks on Linearization of Characteristic Curves in Photographic Photometry.
M. Margoshes
Appl. Optics 8, 818 (1969)

The one that worked best for me was the Kaiser transform. I incorporated it into a Basic computer program, published in
Application of Digital Computers in Spectrochemical Analysis;
Computational Methods in Photographic Photometry.
M. Margoshes and S. D. Rasberry
Spectrochim. Acta 23B, 497-513 (1969)

The functions were developed by trial, not based on emulsion theory. It would be suprising if they worked equally well for
digital camera sensor. I can't tell from the curves in the reference you cited whether they do or don't come close to the
curves for photographic emulsions. The conventional Hurter-Driffield characteristic curve for photo emulsions is logarithmic
in both axes, and the one in the Web page you cited is linear in both axes.

Get back to me if you can't find the papers. I don't have time now to dig out my copies.
Anonymous
January 30, 2005 4:59:02 PM

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

Marvin wrote:

> Roger N. Clark (change username to rnclark) wrote:
>
>> Hi,
>> I'm looking for the equation for the conversion of
>> the linear intensity data from a digital camera sensor
>> to the image output. Please see the measured
>> curve in Figure 7 at:
>> http://clarkvision.com/imagedetail/dynamicrange2
>>
>> This is not a standard Gamma function. Using a
>> Gamma function, the Gamma appears to be around 8
>> at the high end, dropping to 1 at the low end.
>>
>> Does anyone know the mathematical formula for
>> curves like this one? I've tried several, but
>> can't quite match the data over the whole range.
>>
>> Roger
>>
> There are several linearizing functions for p[hotographic emulsions that
> have been known for many years. I published a little note on the subject:
> Remarks on Linearization of Characteristic Curves in Photographic
> Photometry.
> M. Margoshes
> Appl. Optics 8, 818 (1969)
>
> The one that worked best for me was the Kaiser transform. I
> incorporated it into a Basic computer program, published in Application
> of Digital Computers in Spectrochemical Analysis;
> Computational Methods in Photographic Photometry.
> M. Margoshes and S. D. Rasberry
> Spectrochim. Acta 23B, 497-513 (1969)
>
> The functions were developed by trial, not based on emulsion theory. It
> would be suprising if they worked equally well for digital camera
> sensor. I can't tell from the curves in the reference you cited whether
> they do or don't come close to the curves for photographic emulsions.
> The conventional Hurter-Driffield characteristic curve for photo
> emulsions is logarithmic in both axes, and the one in the Web page you
> cited is linear in both axes.
>
> Get back to me if you can't find the papers. I don't have time now to
> dig out my copies.

Marvin,
Thanks! This does help, sort of. I started web searches
for papers online. I am an Optical Society of America member,
and have subscribed to Applied Optics since 1975. So I am
shocked to find when I want a paper online, I must
pay $15 for it. This is simply not acceptable. Other journals
I subscribe to give you access, for example, as an AAS member,
it costs $12/year for online access to all papers,
so $15 for one paper for a member is just ridiculous.
Scientific journals are really becoming a hindrance
to scientific information and research.

Sorry for the rant--this is simply becoming a larger problem
community wide. The National Institute of Health is putting
all health articles online for free access. Other sciences need
to do the same. I do realize it does cost to create and maintain
such databases, but times are changing, and things should get
easier, not harder. More and more libraries are not stocking
journals because they are becoming too expensive and too many.

Several other searches for things like
Hurter-Driffield characteristic curve resulted
in the same must pay a subscription issue.

Anyway, if you could find a reprint and post the equation,
I would be greatfull. Please also include the full reference
for me to cite. In the meantime, I'll keep searching.

Roger
Anonymous
January 30, 2005 9:43:03 PM

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

"Roger N. Clark (change username to rnclark)" <username@qwest.net> wrote
in news:41FD1A7A.5030408@qwest.net:

> I'm looking for the equation for the conversion of
> the linear intensity data from a digital camera sensor
> to the image output. Please see the measured
> curve in Figure 7 at:
> http://clarkvision.com/imagedetail/dynamicrange2
>
> This is not a standard Gamma function. Using a
> Gamma function, the Gamma appears to be around 8
> at the high end, dropping to 1 at the low end.
>
> Does anyone know the mathematical formula for
> curves like this one? I've tried several, but
> can't quite match the data over the whole range.

It sure looks like the curve is linear from 100-2000,
logarithmic from 2000-10000 with one constant, logarithmic
from 10000-30000 with another constant and then it
saturates.

I would be surprised if you find one simple formula for this.

BTW - did I get you right that the RAW values are linear?
I.e. that this non linear behaviour is only an artefact of the
conversion?

In that case - why destroy the initial linear response?


/Roland
Anonymous
January 30, 2005 11:22:11 PM

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

"Roger N. Clark (change username to rnclark)" <username@qwest.net>
wrote in message news:41FD1A7A.5030408@qwest.net...
SNIP
> Does anyone know the mathematical formula for
> curves like this one? I've tried several, but
> can't quite match the data over the whole range.

It looks like a Rodbard function fits rather well (although I don't
have the full dataset to fully check). "Rodbard" is a four parameter
general curve fit function proposed by David Rodbard at NIH:
y = d + (a-d) / (1 - (x/c)^b) ,

with the following approximated Parameters:
a = 0.018343439862814965
b = 1.0643295289322523
c = 6262.359656339619
d = 72032.53123412121

Bart
Anonymous
January 30, 2005 11:22:12 PM

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

Bart van der Wolf wrote:

>
> "Roger N. Clark (change username to rnclark)" <username@qwest.net> wrote
> in message news:41FD1A7A.5030408@qwest.net...
> SNIP
>
>> Does anyone know the mathematical formula for
>> curves like this one? I've tried several, but
>> can't quite match the data over the whole range.
>
>
> It looks like a Rodbard function fits rather well (although I don't have
> the full dataset to fully check). "Rodbard" is a four parameter general
> curve fit function proposed by David Rodbard at NIH:
> y = d + (a-d) / (1 - (x/c)^b) ,
>
> with the following approximated Parameters:
> a = 0.018343439862814965
> b = 1.0643295289322523
> c = 6262.359656339619
> d = 72032.53123412121
>
> Bart

Thanks, I'll play with this.

Here is some raw data from my 1D Mark II
(from Figure 7 at
http://clarkvision.com/imagedetail/dynamicrange2 )
This, by the way, shows that the jpeg and raw are
saturating at about the same point and below the
linear sensor saturation. Noise in the linear data
are dominated by read noise at the low end
(~ 10 DN) and photon statistics at the high end
(~ sqrt(DN*(52300/65535)). The data were random
lines extracted from the data in figure 7 above.
Why the raw saturates at 65499 I do not know.

Scene
Inteni- scaled
sity Raw JPEG
DN camera image
Value Value Value
16-bit 16-bit 16-bit
64 710 1028
103 1162 1285
180 1731 1542
245 2273 1542
297 2761 2056
348 3266 3341
478 4437 3855
736 6732 5911
1007 9158 8481
1356 12107 10280
2105 17493 16448
3153 23694 23387
4311 28998 29555
5499 33421 33667
7967 40916 41120
12366 48857 47802
17866 54049 54227
22868 57562 57568
27723 60346 59624
32204 61927 61166
36840 63391 62194
41378 64293 63222
47811 65135 64507
52789 65450 65278
55180 65499 65278
57366 65499 65535
58254 65499 65535
59034 65499 65535
59559 65499 65535
65535 65499 65535

Roger
Anonymous
January 31, 2005 2:45:30 AM

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

"Roger N. Clark (change username to rnclark)" <username@qwest.net>
wrote in message news:41FD455B.2090403@qwest.net...
> Bart van der Wolf wrote:
SNIP
>> y = d + (a-d) / (1 - (x/c)^b)

SNIP
> Here is some raw data from my 1D Mark II
SNIP
> The data were random lines extracted from the data in figure 7
> above.
> Why the raw saturates at 65499 I do not know.
SNIP

Using the above Rodbard function, those scaled Raw camera values, as a
function of DN, are best fitted with parameters:
a = 307.5908474937861
b = 1.091885854762039
c = 6101.1148040713415
d = 71370.43124044045

The scaled JPEG values, as a function of DN, are best fitted with
parameters:
a = 143.93060451108522
b = 1.1247148320126283
c = 6038.784118954135
d = 70505.46445427524

The reducing slope at the top is obviously the hardest part to
accurately fit in a simple function. A piecewise set of functions
would be more accurate.

Bart
Anonymous
January 31, 2005 5:42:30 AM

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

In message <41FD455B.2090403@qwest.net>,
"Roger N. Clark (change username to rnclark)" <username@qwest.net>
wrote:

>Why the raw saturates at 65499 I do not know.

Blackpoint subtracted (36)?

The following chart is messed up here, because you used TAB characters.
TAB characters are useless on usenet.

>Scene
>Inteni- scaled
>sity Raw JPEG
>DN camera image
>Value Value Value
>16-bit 16-bit 16-bit
>64 710 1028
>103 1162 1285
>180 1731 1542
....
>59034 65499 65535
>59559 65499 65535
>65535 65499 65535
>
>Roger

--

<>>< ><<> ><<> <>>< ><<> <>>< <>>< ><<>
John P Sheehy <JPS@no.komm>
><<> <>>< <>>< ><<> <>>< ><<> ><<> <>><
Anonymous
January 31, 2005 12:50:34 PM

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

In article <41FD1A7A.5030408@qwest.net>, username@qwest.net says...
> Hi,
> I'm looking for the equation for the conversion of
> the linear intensity data from a digital camera sensor
> to the image output. Please see the measured
> curve in Figure 7 at:
> http://clarkvision.com/imagedetail/dynamicrange2
>
> This is not a standard Gamma function. Using a
> Gamma function, the Gamma appears to be around 8
> at the high end, dropping to 1 at the low end.
>
> Does anyone know the mathematical formula for
> curves like this one? I've tried several, but
> can't quite match the data over the whole range.
>
> Roger
>
>
Just a question about your experimental protocol.
I'm not sure how you established the correct exposure for
the scene. One of the well-known characteristics of color
negative film is its ability to handle overexposure well.
So I think it might be instructive to see the results for
this class of film with a variety of different exposures.
My personal (non-scientific) experience is that there is a
lot more detail in the shadows than is commonly expected if
the film is exposed at a lower nominal ISO. I commonly shoot
200 ISO film at 100-125 to compensate for lens vignetting and
have no problems.
Many pictures taken with ultra wide angle lenses show the sun
in the frame and it is surprising how close to the sun there
is still sky detail.

--
Robert D Feinman
Landscapes, Cityscapes and Panoramic Photographs
http://robertdfeinman.com
mail: robertdfeinman@netscape.net
Anonymous
January 31, 2005 12:50:35 PM

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

Robert Feinman wrote:
> In article <41FD1A7A.5030408@qwest.net>, username@qwest.net says...
>
>>Hi,
>>I'm looking for the equation for the conversion of
>>the linear intensity data from a digital camera sensor
>>to the image output. Please see the measured
>>curve in Figure 7 at:
>>http://clarkvision.com/imagedetail/dynamicrange2
>>
>>This is not a standard Gamma function. Using a
>>Gamma function, the Gamma appears to be around 8
>>at the high end, dropping to 1 at the low end.
>>
>>Does anyone know the mathematical formula for
>>curves like this one? I've tried several, but
>>can't quite match the data over the whole range.
>>
>>Roger
>>
>>
>
> Just a question about your experimental protocol.
> I'm not sure how you established the correct exposure for
> the scene. One of the well-known characteristics of color
> negative film is its ability to handle overexposure well.
> So I think it might be instructive to see the results for
> this class of film with a variety of different exposures.
> My personal (non-scientific) experience is that there is a
> lot more detail in the shadows than is commonly expected if
> the film is exposed at a lower nominal ISO. I commonly shoot
> 200 ISO film at 100-125 to compensate for lens vignetting and
> have no problems.
> Many pictures taken with ultra wide angle lenses show the sun
> in the frame and it is surprising how close to the sun there
> is still sky detail.
>

In my tests on the above page, I did a series of exposures
from -3 to +3 stops. The one I chose saturates the
highlights and was over exposed compared to the meter
reading. Si I am quite sure I captured the entire
latitude of the film.

Roger
!