# Double exposure times for high-contrast filters? Not mine.

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• 03-13-2009, 05:43 PM
Nicholas Lindan
Quote:

Originally Posted by RalphLambrecht
The difference might be that I judge the intersection point(s) after applying a best-fit curve through the data points. Best-fit equations for s-shaped curves are not trivial.

The graphs aren't curve fits, but point-to-point graphs of the average of 3 independent runs. The run-to-run deviations were not large, the reason for the multiplicity was to remove the fliers one gets from operator error, defects in the paper, deviations in surface finish, etc. etc.

I did my master's in curve fitting algorithms and controls for NC machinery. Got a bookcase full of algorithms.

There is no reasonable way to fit VC paper to a curve - the stuff has 3 emulsions so you would need a summation of 3 emulsion curves. Most graded papers seem to be two emulsion VC-like papers - you can see the transition where the first emulsion shoulders out - so even there the situation gets complicated.

I found a power law fit works reasonably well, but the shoulder part of the curve is a bit of a problem. The physics behind the toe, 'linear' and shoulder regions are all different and if you use a function that is the same form as the underlying physics you end up with 3 functions for each emulsion.

In any case there is no gain in 'smoothing' the data. The deviation from the true curve, whatever it is, to the linear approximation is less than the within-run print to print variation. There is even quite a bit of variation across a sheet of photographic paper. For precision work I find I need to take a serpentine path across the paper to minimize the distance from patch to patch. You can see this in the difference between the densities produced by the two #16 patches on a 31 step tablet.

The manufacturer's data is very prettyfied, has no meaningful scale and is close to useless, though the Ilford curves at least do give an indication of the flat spot in the 00 curve.

The curves supplied on the web site can be trusted. I can read the light on the easel, go to the curve to find the exposure, expose, develop and dry and the density of the resulting print is what the curve indicates it should be.

What is very interesting isn't the HD curve but the derivative of the HD curve - a plot of local contrast. This really reveals the differences between papers and helps in picking the paper that has the right contrast in the right places.

If the equipment used assumes an ideal HD curve then there is no doubt you are going to be doing test strips... Even using the real curves still results in a need for test strips for some prints.

If you don't like ugly curves, then like most of reality, it is best not to look too close.
• 03-13-2009, 05:57 PM
MattKing
Quote:

Originally Posted by Nicholas Lindan
If you don't like ugly curves, then like most of reality, it is best not to look too close.

This would make a good website signature - and not just on APUG!

Matt
• 03-13-2009, 06:29 PM
RalphLambrecht
Quote:

Originally Posted by Nicholas Lindan
There is no reasonable way to fit VC paper to a curve - the stuff has 3 emulsions so you would need a summation of 3 emulsion curves. Most graded papers seem to be two emulsion VC-like papers - you can see the transition where the first emulsion shoulders out - so even there the situation gets complicated.

I found a power law fit works reasonably well, but the shoulder part of the curve is a bit of a problem. The physics behind the toe, 'linear' and shoulder regions are all different and if you use a function that is the same form as the underlying physics you end up with 3 functions for each emulsion.

Nicolas

I don't completely agree, because I get very good approximations with all types of photographic materials by using non-linear curve-fitting algorithms of 3rd to 5th order. The following equation was used to fit an ISO grade 4 contrast for MGIV-FB:

y=(3.365E+0+-5.329E+0*x+2.833E+0*x^2+-5.024E-1*x^3)/(3.471E+0+-8.779E+0*x+9.013E+0*x^2+-4.356E+0*x^3+8.218E-1*x^4)
R^2 = 9.999E-1

Feel free to plot this for x-values from 1.0 to 2.0 to see for yourself. As you can probably tell, I used the following equation format:

y=(a0+a1*x+a2*x^2+a3*x^3)/(b0+b1*x+b2*x^2+b3*x^3+b4*x^4)

Unfortunately, this is not helping the original question, but I'm very interested to continue this conversation via private eMails, if you like.
• 03-13-2009, 07:23 PM
BetterSense
Don't worry about drifting the thread. It's not a problem at all.
• 03-14-2009, 05:35 AM
RalphLambrecht
Quote:

Originally Posted by BetterSense
Don't worry about drifting the thread. It's not a problem at all.

Just so you know what you need and get, I attached two files. The first shows the step tablets you need, and how to use them. This will allow you to measure the exposure difference from highlight to shadows needed for each filter. You also need a table to turn the exposure differences into ISO contrast. That's what the 2nd attachment is for.
• 03-14-2009, 07:40 AM
BetterSense

I see how the 21-step wedge is used to measure the range of the underlying exposure, but is
this
all I need? Because I don't quite understand where the thirty-one-step exposed wedge underneath the 21-step wedge comes from. It seems like I need a 31-step wedge to contact-print onto my paper to start with, to duplicate what is depicted in that figure.
• 03-14-2009, 06:50 PM
RalphLambrecht
You need a 31-step transmission tablet in your typical negative format to enlarge it (mine is 4x5, which is also useful for other tests where contact printing is the way to go). If you don't have a densitometer, you also need a 21-step reflection tablet to measure your results.

In the example of my previous message, the target highlight and shadow densities (step 2 and 20 on the 21-step tablet) are produced at step 23 and 14 on the test print (31-step tablet). This makes a difference of 23-14=9, which is equal to 0.9 log exposure range, because each step on the tablet is 1/3 f/stop or 0.1 log exposure apart. In the attached table, you'll find that a log exposure range of 0.9 is equivalent to a paper contrast of roughly ISO 3.0 (0.88 would be exactly 3.0).
• 03-14-2009, 07:26 PM
Tom Kershaw
Quote:

You need a 31-step transmission tablet in your typical negative format to enlarge it (mine is 4x5, which is also useful for other tests where contact printing is the way to go). If you don't have a densitometer, you also need a 21-step reflection tablet to measure your results.
Could you comment on using a 4x5 transmission tablet as an in-between if your typical negatives are 6x6 medium format or 8"x10"?

Tom.
• 03-14-2009, 09:43 PM
RalphLambrecht
Quote:

Originally Posted by Tom Kershaw
Could you comment on using a 4x5 transmission tablet as an in-between if your typical negatives are 6x6 medium format or 8"x10"?

Tom.

I think the largest available are 4x5. Not quite sure what you mean with in-between. In your case, I would get one in 6x6 and 4x5.
• 03-15-2009, 01:15 PM
ic-racer
A quick and dirty way to read these 21 step contact prints is to just ignore the one that is 'almost white' and ignore the one that is 'almost black.' Then just count the number of gray steps between them and multiply by 15 to quickly approximate the ISO-R.
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