From: Sandy King (sanking@clemson.edu)
Date: 06/30/03-11:13:32 AM Z
Chris,
>Sandy,
> Thanks for the info below--I haven't step wedged since I started
>photography (I've always said I am not a scientist!) I had to dig mine out
>of storage (and then ordered 2 more--but no directions!).
The step wedge is a very useful tool. You can learn more about a
process with it in two days with it than blindly experimenting for
weeks.
>
> With what you are saying, then the first step where I no longer see
>the black number in the black square, and when there is not a definable step
>between two numbered squares, would equate to maximum black--maximum red, or
>blue, or pink, or green in this case, correct?
The first maximum black would be my speed point.
>If that is true, then my
>results still stand that the am di reaches this sooner. I find the maximum
>black happens sooner (toward white, higher number) on the am di than the sod
>or pot. How I was (erroneously) reading the "speed" was that if the number
>of steps went further down on the scale toward the 21, then the solution was
>faster.
At a given percentage solution ammonium dichromate gives more
sensitivity than potassium dichromate. You will have a longer
exposure scale, and the expansion is at both ends of the scale, i.e.
in the shadows and in the highlights. Most of the expansion is in
the highlights, however.
Assume the following.
A. With a specific process you print a step wedge and the result give
you the first solid black (or red or whatever) at Step #3, and the
last step before solid paper white is Step #13. You know from this
that the speed point is at Step #3 and that the process has en
exposure scale of 12 steps, or six stops, or log 1.80.
B. With another process you print a step wedge and the result gives
you a first solid black (or blue, red, etc.) at Step #5, and the last
step before solid paper white is Step #13. What you know is that, 1)
the second process is twice as fast as the first, and 2) it has a
shorter exposure scale, i.e. is more contrasty.
>
> One thing I haven't done, which you say below, is compare the three at
>non-saturated solutions, or take all three at, let's say, 7% so that I can
>be sure to be below the pot di's saturation point. Do you think that that
>test would be more valid to do than the one I am doing now (using all at
>their saturation point)?
I do tests all the time like this with carbon. What I find is that at
a given percentage solution ammonium dichromate gives more
sensitivity than potassium dichromate. If you reduce the percentage
solution of either you will get a more contrasty image and less
sensitivity. If you increase the percentage solution you will less
contrast and more sensitivity. If you do this in gum I would bet a
lot of money that you would observe the same relationship.
>
> It seems from all the literature that the pot di is the most commonly
>used form. I always use am di because I learned on it. I personally find
>that the fact it is so quick to expose and prints more steps, or midtones or
>whatever the terminology is, makes it my first choice to reach for in gum
>printing. Then pot di, as Dave Rose says, for the final coat. Sodium--I
>still can't see why to use it, except as Judy says that you could almost
>infinitely dilute it, and it would be realllly slow. But it seems to me you
>could handle that with timing under the UV.
With carbon you can get the same results with either ammonium
dichromate or potassium dichromate, so long as you adjust the amount
to account for the greater sensitivity of ammonium dichromate. I find
that a 3% solution of ammonium dichromate will give almost the same
results in terms of speed point and number of printable steps as a
4.5% solution of potassium dichromate.
Sandy King
Sandy
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