# 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

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

More about : characteristic curve function digital cam

Marvin

January 30, 2005 4:59:01 PM

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

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

"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

"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

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

"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

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

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

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

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