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1. If you are using fiber paper, isnt glass needed to kept it flat?

If the paper curls, the negative will be much more than .5mm away from the paper. There is also a higher probability that the negative will shift, especially if you are waving your hands above it to dodge or burn.

You also mention the circles of confusion will be "10 times smaller than they need to be for most "8x10 at 'normal' viewing distance"" Isnt the entire point of making 8x10 contact prints to have the sharpest and most tonally beautiful print possible, not merely sufficient for a certain viewing distance? After all, when i view prints in a gallery i like to get as close as possible and study every detail, especially with a smaller print.

2. Originally Posted by mcfactor
If you are using fiber paper, isnt glass needed to kept it flat?
Yeah, that's why a contact printing frame becomes necessary.

Originally Posted by JBrunner
Use a good quality contact printing frame with clean scratch free glass, and strong springs, throw away the slide rule, and print away.

3. Paper, negative, glass, in a contact printing frame to make it easier, a bulb overhead and a watch or timer. Turn on the light, make the exposure, unload the frame and develope the paper.

Or get a degree in physics to prove mathematically that it's all impossible.

4. Just a couple of comments. First, I don't disagree with those of you who state the OP should use a contact printing frame - you have given good reasons such a holding the paper flat, avoiding newtons rings, etc.
However, I find it interesting that whenever someone posts some kind of mathematical derivation here, there a bunch of people who chime in saying "you're over thinking this". I just don't buy into this - there are actually some of us who enjoy carrying out a good derivation to see where it leads.

Jason, math does not break down when it hits the real world - it simply becomes more complicated. One of the truly remarkable things about this universe we live in is that it can in fact be very accurately described by relatively concise mathematical equations. Indeed, this leads into a great philosophical question. Is math something humans discover, or is it something we invent?

mfactor - I a COC of 0.022mm is smaller than you can detect no matter how close you get.

ic-racer - In your diagram, I see that you computed your COC for the special case of a point directly under the edge of the exit pupil. However, when you making a contact print - say 8x10, most of the image points will be further off to the side so you will no longer have right triangles to work with - I believe they will be obtuse triangles. Did you try computing the COC for these?

5. Originally Posted by dslater

Jason, math does not break down when it hits the real world - it simply becomes more complicated. One of the truly remarkable things about this universe we live in is that it can in fact be very accurately described by relatively concise mathematical equations. Indeed, this leads into a great philosophical question. Is math something humans discover, or is it something we invent?
I don't disagree with this at all. What should have said is that all variables are not accounted for. And yes, doing the math is fun for some persons (thank God, or we wouldn't have much in the way of photography, or other technology) so I'm not, and haven't tried to knock the OP in any way, just pointing out the variables, and that the combined experience of the contact printers here isn't "misinformation"

The real issue of the OP seems to be scratched/pitted/ or dirty glass. He has found that large source illumination gets under the imperfections in the flawed glass, and eliminates them, but this is of course at the expense of any further control over any specific area of the print. #2 solution is to simply lay the negative on the paper, sans the offending glass, and hope for the best, because the math, absent any paper curl, negative curl, humidity, vibration, bump, or draft, says it works. (and it very well might, now and again)

I say get some glass that doesn't suck.

Contemplating the vectors of the photons probably isn't much use in the real world pursuit of a fine contact print, which is, as I understand it, the goal. Time spent doing is the way, and plenty of more accomplished and eminent foreheads have beat the walls before us. Contact printing is as simple as it gets, and brutally reveals any deficiency in the negative or the process. You can't calculate around that. Get some good glass, and clean the printing area- that's the problem, and you can figure all the day long, but it won't change until you do something about it.

I don't think any photographic pursuit, mathematical or otherwise is a waste of time, but I honestly don't see the application here, when what's needed are the proper tools, and attention to the details that matter.

6. Ice-racer,
Are you attempting to produce a photographic work of art via the contact print process? If so, a contact printing frame with a scratch-free glass and an even light source should be all the tools you need. You do not need to engineer the process to death - you just need a good negative that has the printing characteristics that match the paper you are printing on and your personal vision as an artist to guide you to a finished print. You should also be having some fun in the process!

7. Whoa!!! You are making this so much more difficult then it needs to be. If you are contact printing all you need is a voltage regulator (a good suggestion when enlarging as well), a simple light source (ie. a light bulb), a frame, and some sort of timer, Ansel Adams recommended a metronome. This is a rudimentary setup and it is capable of producing the highest quality prints you will EVER see, look at the works Edward and Brett Weston. When printing this way you will need graded paper, which is no big deal.

always remember the words of Albert Einstein "everything should be made as simple as possible, without being made more simply." Words to live by.

Yours;

8. Originally Posted by mcfactor
If you are using fiber paper, isnt glass needed to kept it flat?

If the paper curls, the negative will be much more than .5mm away from the paper. There is also a higher probability that the negative will shift, especially if you are waving your hands above it to dodge or burn.

You also mention the circles of confusion will be "10 times smaller than they need to be for most "8x10 at 'normal' viewing distance"" Isnt the entire point of making 8x10 contact prints to have the sharpest and most tonally beautiful print possible, not merely sufficient for a certain viewing distance? After all, when i view prints in a gallery i like to get as close as possible and study every detail, especially with a smaller print.
For my tests I just put the negative/paper unit in a conventional 8x10 easel. (see results http://www.largeformatphotography.in...9&postcount=90)

In terms of the circles of confusion, you are so right. In fact I discount most dissertations on circle of confusion because it all depends on viewing distance (as you point out) and a good print (in my opinion) DRAWS the viewer IN to examine it closely.

Looks like I ran out of space to post prints, so check out the LF forum link above for the results. I need to think about it some before I decide the next step .

In case it wasn't clear in the first post, using a piece of glass over the paper and the enlarger as a light source produced too much artifact (ie dust and shadows from scratches) that my prints were a waste . Thus the need to sort things out and find a better way of doing it.

I don't WANT to be doing this testing , I'd rather make prints, but they are not turning out. The calculations in the initial post were just to get an idea of WHAT to test as the next step.

Anyway, check out the link above and see what you think.

9. Originally Posted by dslater

ic-racer - In your diagram, I see that you computed your COC for the special case of a point directly under the edge of the exit pupil. However, when you making a contact print - say 8x10, most of the image points will be further off to the side so you will no longer have right triangles to work with - I believe they will be obtuse triangles. Did you try computing the COC for these?
Excellent point, though I think they would probably still be "similar" triangles and the same A, A prime, B, B prime, C, C prime in the diagram would still be equal.

Anyway, the results were a little 'blurrier' then expected. Probably OK at safe viewing distance, but I want to make out all the leaves and pebbles

10. Originally Posted by ic-racer
Anyway, the results were a little 'blurrier' then expected. Probably OK at safe viewing distance, but I want to make out all the leaves and pebbles
:rolleyes:

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