I read through the article which seemed very descriptive but since you asked for feedback scott ... I wanted to comment on something in particular, only because I dealt with the same "math" problem recently.
Quote:
When you are starting out, it would be a good idea to get some additional shot glasses, and mark them with the percentage dilution they will contain. The stock solution you get when you buy it is a 20% solution. Mark your shot glasses as 10%, 5%, 2.5% and 1.25%. Put one drop of NA2 in each glass. Using a different dropper, add 1 drop distilled water to the 10% glass, 2 drops to the 5%, three drops to the 2.5% and so on. If you need that much contrast, you can always pull the 20% direct from the bottle. You will find with use and practice that good negatives require very little contrast agent to make a good print, and that a very little contrast agent goes a very long way.
With my example print here, I am adding 1 drop of 2.5% NA2.
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your dilutions are incorrect Scott.
you mention putting 1 drop of 20% in each shot glass that are marked 10%, 5%, 2.5%, 1.25%. and then sequentially adding 1 drop, 2 drops, 3 drops etc of dist water.
My understanding for the correct manner to dilute is as follows:
10%solution:
add 1 drop of distilled water to the 1 drop of 20% solution = 2 drops of 10% solution
5% solution:
add 3 drops of distilled to 1 drop of 20%= 4 drops of 5%
2.5% solution:
add 7 drops of distilled to 1 drop of 20% = 8 drops of 2.5%
1.25% solution:
add 15 drops of distilled to 1 drop of 20% = 16 drops of 2.5%
you have to double your total quantities, it is an exponential relationship not just an adding relations... not just add one drop to get different dilutions.
I went through a similar process although I am taking small bottles and making up solutions and storing each of my dilutions in seperate bottles marked 10%, 5%, 2.5%
10% solution:
5ml of 20% Na2 added to 5ml dist water = 10ml of 10% Na2
5% solution:
5ml of 20% Na2 added to 15ml dist water = 20ml of 5% Na2
2.5% Solution:
5ml of Na2 added to 35ml of dist water = 40ml of 2.5% Na2
Others can comment on this, or correct me, but Im pretty sure as I made a mistake at first and was corrected (by Kerik).
When people are taking notes about their prints using your calculations theyd actually not be listing the correct %Na2 drop.