


Rodinal formula from Wolfen via British postWWII intelligence
Silverprint has an ad/info in the December 2006 B&W Magazine with a version of Rodinal taken from post WWII military intelligence, modified to use the currently more commonly avaliable sodium salts. That page is available online:
http://www.silverprint.co.uk/PDF/Rodinal4.pdf
Lee

The Wolfen recipe is difficult to convert for modern use. The solutions are measured by weight and not by volume.
Dissolve 34 kg of paraaminophenol in 340 litres of water. Add 558 kg of a 30% solution of potassium sulphite at 55C followed by 50 kg of a 34% potassium hydroxide solution, then 5.52 kg of potassium bromide in a little water. Add 42 g P.1347 (an Agfaspecific antifoggant). Filter and allow to stand for 14 days.
One must know the density of the solutions used (potassium sulfite and potassium hydroxide) and account for them in determining the total amount of water. I would like to see the calculations used to arrive at the formula given in the Silverprint article.
The use of sodium salts rather than potassium ones results in a softer working developer. This can be seen by comparing the two negatives in the article.

The example of the neg they show in the Silverprint ad looks to me completely different from the true Rodinal one.

Isn't this no longer an issue, since Rodinal is back in production, or did I missread that post from a few days ago?
Harry Lime

It looks softer working to me as well, but it's difficult to judge as the pattern of the brighter sky (denser part of the neg) and clouds, and presumably the pattern of sunlight on the foreground objects, has changed a lot between exposures.
You read the earlier post about A&O putting the current Rodinal back into production correctly Harry, although I don't see why it would necessarily lessen curiosity about the topic.
Lee

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If the original solution strengths are in grams/gram of solution, it's easy enough at least to get the weights. I then calculate:
34 grams paminophenol
167 grams K2SO3 or 133 grams Na2SO3
17 grams KOH or 12 grams NaOH
and I'm assuming that the total volume at the factory will be 500 liters, but my calculated weights would work for 500 ml. These are close to the proportions I have been using in my homebrew. I'll bet there would have been much less difference between the illustrating photos if these proportions had been used. The 12 grams of NaOH with 34 grams of paminophenol will leave a little precipitate. If you leave it for 2 weeks, most of that may be used up by aerial oxidation as all the air in the original mixture takes a while to work. The oxidation of the aminophenol in the presence of sulfite makes a monosulfonate and the hydroxide, which make some of the precipitate soluble and active as developer. The solution will also darken somewhat over time as a little oxidized aminophenol makes a lot of dark.

Note: if you use paminophenol hydrochloride, you will need about twice as much NaOH. For every mole of paminophenol.HCl you will need one mole of NaOH to neutralize the HCl and a little less than one more mole to make the sodium aminophenolate salt, leaving some aminphenol base precipitated. This solution will contain some sodium chloride, but I doubt it will make much difference in performance or grain size.

Originally Posted by gainer
If the original solution strengths are in grams/gram of solution, it's easy enough at least to get the weights. I then calculate:
34 grams paminophenol
167 grams K2SO3 or 133 grams Na2SO3
17 grams KOH or 12 grams NaOH
and I'm assuming that the total volume at the factory will be 500 liters, but my calculated weights would work for 500 ml.
I shouldn't do these things late at night. I think the weights are right, but I should have calculated the volume of water. It starts with 340 liters. 167.4 kg of K2SO3 comes in 30% of 558 kg od solution, leaving 390.6 kg of water, or 390.6 liters. 17 kg of KOH comes in 34% of 50 kg of solution leaving 33 kg of water. Thus, the total water in the mixture is 763.6 liters. What are the odds that the final volume was topped off at 1000 liters? Pretty good I think. If that is the case, the solution would be about half the strength I have been using in my homebrew. Sorry about that. I'll stick with mine, though, for my use.

The Wolfen recipe does not indicate that the final solution should be diluted to a set volume. Rather I get the impression that the amounts were arrived at by the size of the reaction vessel.

When I get some more paminophenol I shall see what the volume is without adding any more water. It may be that the volume of solids added to the 763 or so liters comes out to be about 1000 liters. It is difficult to tell because some solids seem to fit between water molecules with practically no change in volume. It is true that the conversion to sodium makes it tricky, and perhaps an excercise in futility. Would converting to moles of the potassium salts/ gram of solution before attempting the conversion help?
My opinion is that the sulfite content is not critical but should probably be sufficient to make the monosulfonate from the available aminophenol; that the developing agent ends up being the sodium or potassium aminophenolate, and that the presence of some precipitated aminophenol, at least initially, is important. The ratio of hydroxide to aminophenol seems to be such as to provide more aminophenol than can be converted to the phenolate by the amount of hydroxide provided. Anyway, it's fun to think about and play with.

