Modern Rodinal Substitutes Part II

gainer · Sep 19, 2009

Photo Engineer said:
Patrick;

Would it be too much to ask if the two photos matched in contrast and density? That is my only reservation.

PE

At the present time it would in fact be more to ask for than I am able to provide. As you know, there is a lot involved in finding the contrast versus developing time for a particular film-developer combination. My equipment for doing the necessary measurements is limited, and my equipment for walking from one end of the darkroom to the other is also limited. Those appendages I used to walk, run and even dance on occasionally (I seem to remember they are called "legs") are causing me great pain.
No one else seems to be doing anything but theorizing. I am feeling very much put upon because when I try to get some testing of theories started, the criticism of my work seems to be the most important business at hand.

If I were to change contrast and density, it would be by digital means. If I had done that with perfection, you would never have known the difference. Believe me, the amount of sulfite up to as much as 380 grams per liter of stock has little or nothing to do with grain or sharpness in a 1+50 dilution. The effects will be on storage life of the stock, working life and activity of the working solution.

Kirk Keyes · Sep 20, 2009

gainer said:
That BS as you stupidly call it is and always has been very important in almost any branch of science, except possibly the one that defines BS.

It seems to me it's simply an issue of mismatched contrast indices, and not an issue of scanning...

I guess I don't see the point of posting scans if pushing the Nyquist limit invalidates the scans. IF that's what is going on...

gainer · Sep 20, 2009

Kirk Keyes said:
It seems to me it's simply an issue of mismatched contrast indices, and not an issue of scanning...

I guess I don't see the point of posting scans if pushing the Nyquist limit invalidates the scans. IF that's what is going on...

All you need to do to see what happens to apparent granularity is to scan the same print or negative at a number of different scanning rates. Examine a light gray area such as the frame of the window screen behind the easy chair.

You really ought to repeat the experiment and then chastise me for not having the same CI. After all, these are two different developers, and the best you can do with matching contrast is to get the characteristic curves to cross at some 2 points. If, then, you see that the lines coincide, you're fortunate. Otherwise, especially when comparing small areas between Dmax and Dmin, you will have different values of CI due to differences in curve shape. Certainly, such difference in curve shape need to be investigated, but if the grain, acuity, resolution are not comparable, and the curve shape of the one with better values of these criteria is known and acceptable, why bother?

So, I say chastise me for not presenting the individual characteristic curves, but not for mismatch of CI. Unfortunately, the phototransisor on my densitometer appears to have died. Since that is also my easel photometer, it is very difficult to match density values at a single point.

gainer · Sep 20, 2009

The sampling theorem for band-limited signals says that any 2fm independent samples per second will completely characterize a band limied signal of band limit Fm. Our domain is space instead of time, so we have a signal that is made up of grains and expressed in bits per millimeter or other spatial unit. By the sampling theorem, it seems that if you want to characterize fully the real print or negative, you must scan it at high enough bits per mm to see the finest grain.

gainer · Sep 20, 2009

I should add that when I sample a small part of the image at 2400 dpi I see finer grain in the printout than at any lower rate. As I said before, the size of that file coupled with my 26.4 K internet speed might not leave me much time to live after it got delivered. APUG might not even be around. It might be like that aerodynamic problem in the early days of mainframe digital that took longer to solve on the IBM 7094 than the mean time between failures.

gainer · Sep 22, 2009

Here is an example of two different ways of showing the same negative. One is a scan directly from the negative at 2400 dpi. The other is scanned at 2400 dpi from a portion of an analog 8x10 print. Finally, I resampled each one to have the same height in pixels.
As you may see, the scan from the print has much higher resolution and very much finer apparent grain than the scan directly from the negative. It is not going to be easy to compare grain and resolution in digitized photos.
There will be those among you who will not believe what I just showed. You will have to repeat the process with a negative of your own.

DutchShooter · Sep 23, 2009

I completely believe that (apparent) sharpness and grain are totally different for a scanned negative and a scanned print of the same negative.
Maybe it may be more fair that you reduce the scan resolution from the print with the enlargement factor used for making the print from the negative: you then have at least a comparable pixel density to start from. But off course this is only ignoring the real problem, that a digitized version of reality will have a difficult time competing with the real thing

gainer · Sep 23, 2009

DutchShooter said:
I completely believe that (apparent) sharpness and grain are totally different for a scanned negative and a scanned print of the same negative.
Maybe it may be more fair that you reduce the scan resolution from the print with the enlargement factor used for making the print from the negative: you then have at least a comparable pixel density to start from. But off course this is only ignoring the real problem, that a digitized version of reality will have a difficult time competing with the real thing

Perhaps you noted that although each image is the same size on screen as the other, the sizes of the .jpg files are 140.4 for the scan from print and 77.0 for the scan directly from the negative. No matter how I go about it, I will either not get a fair comparison of two negatives or a fair assessment of grain and resolution of a single negative that I can transmit over the Internet.

One would think that a direct scan from the negative would eliminate the losses due to enlarger lens and paper. The film grain was enlarged in the print as well as the other characteristics of the film image, and then scanned at 2400 dpi. In my opinion, it would be better to judge the qualities of film and development on the enlargement of the paper print rather than to reduce its quality to match the direct scan. Either that, or find a way to scan at 19200 dpi from the neg.

Alan Johnson · Sep 24, 2009

Posting scans of prints every time a formula is tested seems a bit like hard work for you to have to do and slows the whole process down.
In my case I would be happy to rely on your Mk 1 calibrated eyeball to assess grain unless a formula you consider of special interest is found.

gainer · Sep 24, 2009

I do have 19200 dpi on my scanner, but my system memory will not handle the resulting file when I try to scan only the area I showed above. I got up to 7200 dpi, but found that the resolution was worse than at 2400. I'm beginning to think that the resolution of my scanner's optical system is not as good as that of my enlarger, There is a point at which the scanner shows very clearly the lack of clarity of its own optical system.

Kirk Keyes · Sep 24, 2009

gainer said:
There will be those among you who will not believe what I just showed. You will have to repeat the process with a negative of your own.

I'm not surprised the print looks better than the neg. It's hard to get that much magnification with the scanner alone as you can with an enlarger in the middle of the process.

Anon Ymous · Sep 24, 2009

gainer said:
I do have 19200 dpi on my scanner,...I got up to 7200 dpi, but found that the resolution was worse than at 2400...

Which is interpolated, not real, hence the better resolution at 2400 vs 7200. Your enlarger's resolution is much better.

Photo Engineer · Sep 24, 2009

I have to say that I can get far better resolution from direct scans of negatives and transparencies than I can get from prints and I make prints up to 16x20 regularly, both B&W and color.

PE

gainer · Sep 24, 2009

Photo Engineer said:
I have to say that I can get far better resolution from direct scans of negatives and transparencies than I can get from prints and I make prints up to 16x20 regularly, both B&W and color.

PE

But can I afford the scanner you have?
I don't remember seeing anything in the instructions about interpolating, and HP is usually pretty good about such things.
I have 6 Super B sized prints scanned from 35 mm negatives hanging on my wall, and they look pretty good to me and all others who have seen them. Of course, I have them arranged so that grain-sniffers can't get close enough to comment on that. They can, however, get close enough to see the proper perspective.
The only reason I went to such high dpi was to see what would happen. Now I know.

Ray Rogers · Sep 25, 2009

gainer said:
I have them arranged so that grain-sniffers can't get close enough to comment on that.

grain-sniffers ?

gainer · Sep 25, 2009

Ray Rogers said:
grain-sniffers ?

That is what we old timers call viewers who get close enough to the photo to smell the grain and too close to see the picture.

gainer · Sep 28, 2009

You guys can keep looking for the perfect Rodinal (if that's what anyone is actually doing). I have found it. I tested it last night and left the used developer in an open 500 ml pitcher. I saw that the color had not changed appreciably, so I tested it again. The negatives are identical. The batch that I mixed a week ago has not changed color. It may be that the 2 week seasoning period imposed by AGFA was to give it a chance to get a little color.

I follow the developing instructions on my old AGFA Rodinal bottle, not glass but the white plastic kind. The only change I make is to agitate on the minute instead of the half minute. There is something about European scenery or at least European photographers that drives them on average to a slightly higher contrast than I like.

I used mostly the recipe proposed by Ian, but left bromide and the other stuff out for now. I don't miss it so far. I used somewhat less K2SO3, but still more than most other recipes dictate.
To make 500 ml:
Take 225 ml of the 45% K3SO3 solution and add to it 21 grams of paraminophenol base. This, of course forms a slurry that is protected by the sulfite from immediate oxidation by the KOH you are about to add. It will take at least 11 grams of KOH to dissolve the precipitate. I do not know how to tell you to keep KOH, even dry, from forming K2CO3 with atmospheric CO2.

Gotta go now. Talk later.

alanrockwood · Oct 2, 2009

Just a few random comments:

First, I have only skimmed through about half the comments on this thread, but I want to congratulate PE and Patrick for keeping a civil tone so far!!! Thank you both.

Next, a few comments and questions about hydroxide chemistry. First, a well-know quirk about hydroxide is the ability to absorb atmospheric CO2 and be converted to carbonate. This would obviously change the pH of the hydroxide solution. Are there special considerations that photographic chemists take to minimize this effect? I suppose that keeping bottles capped would go a long way toward controlling this.

Second, hydroxide solutions are known to attack glass, converting the silica in the glass to silicate. This could have two effects. The first is to damage glass bottles. The second is, as with the CO2 chemistry noted above, a shift in pH of the solution. In fact, the chemistry of the conversion of SiO2 to SiO3-- is very analogous to the conversion of CO2 to CO3--, given the relative positions of C and Si in the periodic table. Is the silica-silicate chemistry noted here something of concern in photographic chemistry?

alanrockwood · Oct 2, 2009

Oops, what do you know, Patrick's post just ahead of mine (and possibly some others as well) mentioned the hydroxide, CO2, carbonate issue. I hadn't gotten that far in the threads when I made my post, so I apologize if my post seems out of sequence or deals with something already covered.

Photo Engineer · Oct 2, 2009

Alan;

I mentioned somewhere the reaction of glass with hydroxide. In fact, my 50% hydroxide solution is starting to etch the glass dropper bottle that it is in. I use a dropper for addition of the last bit of hydroxide. (well, at least I did! )

It takes a fair amount of time to etch even the cheaper glass in the dropper bottles and it goes after the clear glass dropper first.

PE

Ian Grant · Oct 2, 2009

Following on from Ron's (PE) comment. Early Rodinal and modern Calbe R09 has no free Hydroxide anyway so is not going to etch glass. Even modern Agfa/A&O Rodinal only has a small amount of free Hydroxide so even when it was still supplied in a glass bottle it would have taken a considerable number of years longer etch the glass than a 50% solution

Ian

Photo Engineer · Oct 2, 2009

Ian;

If the pH is above 7, there is free hydroxide. As the pH increases, the amount of OH increases. So, by the reports that the pH of the stock is about 11 or 12, this is pretty strong stuff.

PE

Kirk Keyes · Oct 2, 2009

gainer said:
I do not know how to tell you to keep KOH, even dry, from forming K2CO3 with atmospheric CO2.

You can apply tape (electrical tape works well) around the lid of the bottle to help the lid seal better. You may have noticed that many chemicals come packaged that way. It not only keeps the air/moisture/CO2 from getting into the bottle, it also helps keep the contents of the bottle in the bottle.

Ian Grant · Oct 2, 2009

Photo Engineer said:
Ian;
If the pH is above 7, there is free hydroxide. As the pH increases, the amount of OH increases. So, by the reports that the pH of the stock is about 11 or 12, this is pretty strong stuff.
PE

Strong as in a 50% soloution of Potassium Htdroxide, quite the opposite

My Jacobson & Jacobson lists Potassium Hydroxide at 0.05% as giving a pH of 12, 0.5% a ph of 13, and then the Potassium Sulphite level on it's own would give a pH of 9-10.

Calbe R09 with no free Hydroxide does indeed have a pH of 11.8, all version of the formulae until Agfa's 1964 re-formulation contained no excess hydroxide - just enough to react with the free base. If this wasn't the case the pH would be higher we know that the amount of KOH needed to reactwith 50g of Free base is 24g. At 2.4% that should give a pH of somewhere around 13.5, but it doesn't - that's they key to what's going on with Rodinal.

Now take into account that modern Agfa/A&O Rodinal has a pH of nearer 14 but has 20% less (approx) p-Aminophenol it also contains more Potassium Hydroxide 2.7% that when it reacts with the Free Base leave 0.73% free hydroxide which gives a pH of over 13 exactly as we'd expect.

This hold the key to Rodinal, it starts life as the p-Aminophenol free base then there's a reaction with the Hydroxide, which may be slow.

If the hydroxide doesn't react then the Calbe R09 would have a pH close to the Agfa/A&O version - which it doesn't.

Ian

Photo Engineer · Oct 2, 2009

Ian;

I have to disagree. If the pH is above 7, then the OH concentration goes up one order of magnitude for every pH unit from 8 - 14. It is as simple as that. So, at pH 11 or 12, there is excess free OH present. There is no question. This is basic chemistry!

PE

Modern Rodinal Substitutes Part II

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