I think Gmax, or as it is often called, gamma infinity, is most easily obtained by estimating it from two gradients measured at the same temperature. It will be advantageous to have semi-log graph paper at hand. I found a web site from which one can download pdf files which allow one to print out on demand a number of different special graph papers. Use http://incompetech.com/graphpaper/logarithmic/.
Start with an estimate of Gmax. I can be pretty wild. I use 2.0 in my program. Using this value and two known values of CI at the same temperature, calculate 2.0 - CI for each CI, plot these values against development time (time on the linear axis), draw the line connecting these two points and use the value where it intersects the log axis as a new estimate of Gmax. Use that value to calculate Gmax - CI for each point, plot a new line, and repeat this process until two consecutive values of Gmax are close enough for government work.
What I was trying to do is see if I can get "k" for each of my films. I already have boatloads of "G at temp X" data. If I had G-Max, then I could solve for "k". Then I would be set. After years of computer use, I don't think I know how to use graph paper and pencil anymore I'm going to have to re-read you response a few times to get it.
Now that I have digested you response I see I can use two curves at the same time but different temperatures (two G's) to obtain "k" mathematically, rather than graphing.
I believe the following is true of the exponential equation listed previously:
k = ln (1+ (r/100))
Where:
k = constant from the exponential equation I am seeking
r = percent increase in G per unit temperature
So, once I have 'k' from the above relationship, I believe I can solve for G-max.
This would really be great if it works.
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Now that I have digested you response I see I can use two curves at the same time but different temperatures (two G's) to obtain "k" mathematically, rather than graphing.
I believe the following is true of the exponential equation listed previously:
k = ln (1+ (r/100))
Where:
k = constant from the exponential equation I am seeking
r = percent increase in G per unit temperature
So, once I have 'k' from the above relationship, I believe I can solve for G-max.
This would really be great if it works.
I have a feeling that it will not work. The constant k is not constant with temperature, but Gmax is. There appears to be no slope information when constant temperature samples are used. I won't tell you not to try. I could be wrong. There's a first time for everything. Let me give you some data points for which I know the answer and see what you can come up with. G1= .55, t1 = 20 C; G2 = .65, t2 = 24 C. Time was 12 minutes for both. You should get a Gmax = .86, give or take a little for reading error. I have numerous other data points to check with if you can come up with a way to predict them.
For the numbers you give, based on what we have discussed, you have to give a film and developer to match the data to. I too have reservations about the derivation of our fellow APUG member. I read it earlier and something bothered me so I did not post, but then you are better at math than I am I think.
Patrick, I truly wish we could meet or talk sometime. We have so much information to exchange.
Maybe I have misread something?? Is the little 't' in that original equation Temp or time??
I thought you had posted the formula to relate to temp. But I am thinking (by the way you are writing) that I have misinterpreted things. You are just relating any G to the G-max for given TIMES right. I thought TEMP.
The little t is for time. The k that goes with it is variable with temperature as well as the film-developer combination. Since my last post, I have found that it is theoretically not possible to define experimentally all the constants of that equation for even one film-developer combo without measuring some variations due to both time and temperature. Going back to the differential equation:
dG/dt = k( Gmax - G)
If indeed you measure a small increment of each of G and t at a known value of G you can get pretty close to dG/dt and if you do it for 2 values of G you will have a simple simultaneous set to solve. BUT the values you must measure are very much smaller than Gmax which makes the required precision of measurement much to high for practicality. You still must have at least enough information about variation with time to get the derivatives you need.
OTOH, you only need to do two characteristic curves at different development times for each film of interest to be able to choose your time and temperature to get a desired CI for any one film-developer combo in your stable. You already have the temperature data reduced.
I'm going to have to re-read you response a few times to get it.
Let me give you some data points for which I know the answer and see what you can come up with. G1= .55, t1 = 20 C; G2 = .65, t2 = 24 C. Time was 12 minutes for both. You should get a Gmax = .86, give or take a little for reading error. I have numerous other data points to check with if you can come up with a way to predict them.
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