Film curve plotting and fitting

[Photo Engineer]

Nicholas is correct.

I would like to add again that the flat portion of the "bell" curve above, or dD/dLogE, the derivative, effectively describes the maximum usable latitude of a film. That is the flat or horizontal portion. The value in the above case appears to be incorrect on the y axis though as it should actually equal the gamma. Mees shows this in Fig 299 of his Revised Edition on P 870. He uses 2 y Axes, one is density and the other is gamma on the left and right respectively to remove the ambiguity.

In the above curve, the rise is about 0.2 units per 0.1 speed (if I assume log E) for a gamma of 2.0. If I have misinterpreted this, please accept my apologies. I would, of course, be wrong if the X axis were not in units of Log E.

And, the print gamma would be dD/dLogE film x dD/dLogE paper for each point allowing us to construct a "system" response assuming that the film and paper respond on a 1:1 basis.

PE

 

[RalphLambrecht]

Nicholas

I like to stick to real data and not some made-up idealized curves. I have proven to you that polynomials do what you said they can't. Also, I have proven that single equations can do what you said cannot be done.

 

[RalphLambrecht]

Nicholas is correct.

I would like to add again that the flat portion of the "bell" curve above, or dD/dLogE, the derivative, effectively describes the maximum usable latitude of a film. That is the flat or horizontal portion. The value in the above case appears to be incorrect on the y axis though as it should actually equal the gamma. Mees shows this in Fig 299 of his Revised Edition on P 870. He uses 2 y Axes, one is density and the other is gamma on the left and right respectively to remove the ambiguity.

In the above curve, the rise is about 0.2 units per 0.1 speed (if I assume log E) for a gamma of 2.0. If I have misinterpreted this, please accept my apologies. I would, of course, be wrong if the X axis were not in units of Log E.

And, the print gamma would be dD/dLogE film x dD/dLogE paper for each point allowing us to construct a "system" response assuming that the film and paper respond on a 1:1 basis.

PE

Sorry PE, but I've never seen curves like this in my film or paper testing. We need to get our heads out of the books and into the data!

 

[Photo Engineer]

Sorry PE, but I've never seen curves like this in my film or paper testing. We need to get our heads out of the books and into the data!

Ralph;

I agree. That is why you find these curves only in text books and being used by the engineers that design films. In fact, these curves were so tedious to gather, that we used to just draw the curve with a French curve to best fit the points and then draw a straight line along the curve and use the line to approximate the gamma and latitude. This worked until computers took over.

The user should not need to curve fit or anything more than just one or two sensi exposures to be used for test purposes. From there on out it is photo time (or should be).

I didn't spend weeks in the lab beforehand when I knew I had to go take some aerial photos. I took my trusty (well, kinda old and battered) USAF Speed Graphic and some Super XX, went out, and shot. Then 9 minutes in D-76 and good old F-5.

Yeah, I knew how to derive the speed, and how to run curves and plot process control, but too much is "Paralysis by Analysis" as Nicholas said.

So, for the most part, we don't even need the data! Sorry.

PE

 

[Nicholas Lindan]

The value in the above case appears to be incorrect ...

The curve isn't from any film - it is a demonstration from a spreadsheet for calculating HD curves. The linear portion is longer than usual and the curves are normalized and dimensionless.

 

[Kirk Keyes]

The user should not need to curve fit or anything more than just one or two sensi exposures to be used for test purposes. From there on out it is photo time (or should be).

That's why I gave up looking for curves or even think about spining when I made my own spreadsheet to calculate film speed and contrast. I decided that linear interpolation between adjacent points was going to be accurate enough to find the info I was looking for - speed point, 0.1+B+F, and contrast index. It's simple and easy to get a spreadsheet to do, with a little bit of macro programming.

But if someone figures out an easy solution, I'm here to see it!

 

[Photo Engineer]

I used points on a spreadsheet to do my curves as well. Good enough!

Here is the example from Mees (revised edition) for anyone interested. This is a real example.

PE

 

[Nicholas Lindan]

not some made-up idealized curves. I have proven to you that polynomials do what you said they can't.

Ummm. It's not 'some made up idealized curve'. It has all the characteristics of a real HD curve. If this can't be fit, then a real-real curve hasn't a chance of getting fit as this is a much easier case. And this is what I say a single polynomial can not do.

A real HD curve has a toe and shoulder asymptote where it goes to and stays at either DMin or Dmax. Fitting just the center part of the curve is, well, trivial -- that it can be done isn't under debate. What is at issue is that a polynomial can't do the whole shebang and is a poor choice if one is going to pick a do-it-all modeling function for an 'S' curve. Additionally, a polynomial does not derive from any understanding of the physics involved and provides no additional insight of its own.

A polynomial isn't the real curve - it's just something that has been bent to look like the curve - at least a bit of it. Change the parameters and it now looks like the Jungfraujoch. It's just a mathematical pencil, a programmable french curve, as it were. OK, it draws a picture of the graph - but in the end it leaves the question: So what?

For practical exposure of negative materials all that is needed is the straight line section which specifies the shadow speed and the highlight speed/range of exposure. Any sort of erf or polynomial is overkill. If one is going to go into the form of the curve one may as well use the real thing instead of ersatz.

 

[ic-racer]

I used points on a spreadsheet to do my curves as well. Good enough!

Here is the example from Mees (revised edition) for anyone interested. This is a real example.

PE

So, I see a film curve then its first derivitave. It indeed does look alot like what Nicholas was showing.

If I took the G to be 1.1 and then 0.3G would be 0.33 which would put the speed point on the x-axis around 0.8.

Is that how it was being used?

 

[Kirk Keyes]

Ummm. It's not 'some made up idealized curve'. It has all the characteristics of a real HD curve.

It's too smooth. I've never measured real data that smooth. It just will not work with multigrade papers very well...

 

[Kirk Keyes]

Additionally, a polynomial does not derive from any understanding of the physics involved and provides no additional insight of its own.

How does it fit with the physics of a multiemulsion paper or film?

 

[Nicholas Lindan]

It's too smooth.

Yes, in that it is idealized: the curve of a perfect film. A real film is the addition of a series of such curves each with different parameters - and hence is the characteristic of your multi-emulsion film. For a multi-emulsion paper see http://www.darkroomautomation.com/support/appnotevcworkings.pdf

 

[Photo Engineer]

So, I see a film curve then its first derivitave. It indeed does look alot like what Nicholas was showing.

If I took the G to be 1.1 and then 0.3G would be 0.33 which would put the speed point on the x-axis around 0.8.

Is that how it was being used?

Actually, the speed point is closer to point "A" or about 1.5 on this curve which is the beginning of the straight line portion of the curve and is just a tad overexposed as I have said before. This H&D curve is identical to the one previously posted as the idealized negative curve.

A similar curve was derived from a real film and used to make pictures in my previous post and shows how far up the curve the best pictures are regardless of where the speed point might be.

On average, that is gained by about 1/3 stop overexposure. With many modern, sharp toed films, this overexposure is redundant but not harmful.

PE

 

[Photo Engineer]

How does it fit with the physics of a multiemulsion paper or film?

Kirk, with films and papers the curves of individual emulsions are additive, so if you coat 1/3 of the silver of each of 3 emulsions, equally spaced apart for toe and shoulder and with identical latitudes, you get a modern 'straight line" center part to the H&D curve. In practice this is often difficult to do but it can be done. I've done it as have many dozens of Kodak, Fuji and Ilford engineers.

Basically, the key is having equally spaced threshold speeds, toe curves and size frequency distributions.

PE

 

[Stephen Benskin]

Ron,

My copy of the 3rd edition of Theory of the Photographic Process has almost that exact curve, but they don't go into detail as to the various lettered points other than "A" and "B". I'm curious what "O" and "P" designate. It sounds like you have the 2nd edition. Does it define those other points?

 

[Photo Engineer]

Steve;

Sorry, but they appear to have no mention of "O", "P" or "Y". They do define Speed = k / inertia which in the graph is X.

K = a constant which is undefined in the text but appears to be unique to each film. Not sure. Sorry again.

PE

 

[Ray Rogers]

Ron,

My copy of the 3rd edition of Theory of the Photographic Process has almost that exact curve, but they don't go into detail as to the various lettered points other than "A" and "B". I'm curious what "O" and "P" designate. It sounds like you have the 2nd edition. Does it define those other points?

Steve,

I cannot find solid proof, but I am thinking
"p" is the lower point on the curve that corresponds to
the given exposure that created "the first excellent print".

[Steve: cf. graph top right pg. 441]
[PE: cf. pg. 878; Perhaps "p" of pg. 870 ~ "m" of pg. 878 (?)]

 

[Stephen Benskin]

Steve;
They do define Speed = k / inertia which in the graph is X.

I believe that was just a historical reference to Hurter and Driffield's Inertia speed method. I'm thinking point "P" is the Fractional Gradient speed point. That would make ic-racer's suggestion of 0.80 fall around today's fixed density speed point.

 

[Photo Engineer]

Steve, Ray;

Since the author of that chapter has not fully defined the points, I would have to say I have no solid evidence, but I tend to agree. OTOH, the position of the best prints indicate exposure above that point, as does Haist in his graphs and parts of his text.

Due to these ambiguities in texts, I use what works when I go out to take photos.

PE

 

[Stephen Benskin]

Ron,

Haist book(s) is one I don't have. I went online and found one used for $250.00 which is probably why I don't have them.

I looked at bib for the chapter in Theory of the Photographic Process, and except for the three papers published before the 30s, I have all but two. I bet one of those is where the graph in question came from.

Steve

 

[RalphLambrecht]

... So, for the most part, we don't even need the data! Sorry.

PE

You're right, but this thread has gone beyond photography. We are not talking about taking picture anymore. That said, I enjoy the occasional discussion about applied mathematics, and that's what we need the data for.

 

[Photo Engineer]

I would like to emphasize that the graph of just acceptable and first excellent prints shows overlap between the two regions. This is judgmental and also scene dependent. However, the region of first excellent that "stands alone" with no overlap is unambiguous and is on the upper part of the curve represented by the straight line. The derivative curve for that is curve "B" in the figure I posted and you see that this shows the majority of the entirely first excellent prints fall in the area at the top or on the flat portion of curve "B".

Even though his text partially contradicts his graphs, Grant Haist shows that the longest latitude is in this same region. These two sets of graphs in Mees and Haist clearly teach us something.

However, having run one of these tests myself with hundreds of observers, and having been a participant in several of these tests, I can say that there is a large amount of overlap just due to taste, opinion or whatever you may call it. After all, this is art and beauty (or excellence) is truly in the eye of the beholder.

PE

 

[RalphLambrecht]

Ummm. It's not 'some made up idealized curve'. It has all the characteristics of a real HD curve. If this can't be fit, then a real-real curve hasn't a chance of getting fit as this is a much easier case. And this is what I say a single polynomial can not do...

Nicholas

This is not correct. The horizontal parts of this curve are unrealistic. No HD curve looks like that. And when you ignore these unrealistic portions, then it can be fit with a simple polynominal. I have used them to fit all HD curves I've seen so far without problem. I don't use them because a 3rd order nonlinear function works even better. But you are making this tougher than it really is. I don't understand why.

 

[Ray Rogers]

...except for the three papers published before the 30s, I have all but two. I bet one of those is where the graph in question came from.

Steve

Steve,

If you PM me the titles of the ones you don't have,
I will see if they are available here.

 

[RalphLambrecht]

Ron,

My copy of the 3rd edition of Theory of the Photographic Process has almost that exact curve, but they don't go into detail as to the various lettered points other than "A" and "B". I'm curious what "O" and "P" designate. It sounds like you have the 2nd edition. Does it define those other points?

Stephen

What page are we looking at? I got the same edition as you have.