I just had a look at @PabloImpallari
's stem weight calculator, and noticed something odd (I think).
So far, I've got a regular and an extrabold, but the only fields available are 'thin' and 'black', so I just use those instead.
Thin 92, Black 162, steps: 4 -> 92, 113, 137, 162 (Impallari Formula)
If I then pretend that I have a light weight at 42, and increase the steps to 7, this happens
Thin: 42, Black 162, steps 7 -> 42, 54, 71, 92, 116, 140, 162
So, in the second configuration, focussing on the last four values, the steps are different than in the first configuration (even though the regular is still at 92). Logically, I don't think this should be the case. The same problem would occur with the De Groot method. I may be wrong, though, so feel free to correct me.
If I am right, however, a better formula for calculating stem weights would be useful. I have used the following distribution for Proza and Proza Display, but only came up with the formula just now. Maybe the whole thing is nothing new, I don't know.
Assume that a light weight is 40, and a black weight is 200 (difference
is thus 160), and there are 4 steps
(in Pablo's calculator, this would be considered 5 steps. I say there are 5 values, thus 4 steps from value to value).
A linear approach looks like this:
1=40 2=80 3=120 4=160 5=200
+40 +40 +40 +40
My approach looks something like this:
1=40 2=80 3=125 4=175 5=230
+40 +45 +50 +55
Then, to make sure we stay within the 40 to 200 bounds, I multiply each of the figures in the line directly above this, by (160/190=0.84), (that's the original difference divided by the 'new' difference) which gives:
+34 +38 +42 +46
Thus, the instances become:
1=40 2=74 3=112 4=154 5=200
So far for the basic explanation. Now, to get to these instances right away, I created a formula. In the formula '#' means the number that is before the = sign. X is a value that can be freely chosen (in the example above it was 5). Y is the value of the previous (lighter) instance. Here goes:
2=Y+(difference/steps+X*(#-2)) * (difference/((steps-1)*steps)/2*X+difference)
40 + ( 40 +5 * 0 ) * ( 160 / 190 )) = 74
3=Y+(difference/steps+X*(#-2)) * (difference/((steps-1)*steps)/2*X+difference)
74 + ( 40 +5 * 1 ) * ( 160 / 190 )) = 112
Now, if wanted to add an even lighter weight, it would be set at 10, and X would still be 5, and all the other instances would remain the same. Even if I wanted to add a number of darker or lighter weights than I already have, the 'original' instances would remain the same. To illustrate:
2=10 + (190/5+0)*(190/240)=40
3=40 + 190/5+5)*(190/240)=74
I think this distribution has two advantages:
- It provides a flexible X, which determines how much you want to deviate from the linear distribution. The X can be varied for each typefamily.
- Adding lighter or darker weights does not change the distribution of the existing weights.
- A tool to do this automatically would be helpful, at least for me.
I hope my calculations and assumptions are all correct. If not, please explain in simple language. I'm somewhat brainfried. Sorry for the long post, but I did my best to explain it clearly.