there is a spreadsheet somewhere that does this calculation (in one of the groups or inkjetmall or somewhere)
However, it goes like this.
The answer you want is 1.29
the question is: what % of the pure black ink gives a density of 1.29
you find the answer is somewhere between 40% (density 1.22) and 45% (density 1.33).
without any maths you can see that 1.29 is not too far away from halfway between these. Halfway would be at 42.5% with a density of (1.22+1.33)/2 ie 2.55/2 = 1.275 so the exact answer will be a wee bit more.
;
Using the maths:
from 40% to 45% represents a 5% change
while the density grows from 1.22 to 1.33: a change of .11
So: THIS IS THE MATHS, .11/5 = .022
Which means each 1% extra ink gives .022 extra density.
So if you start at 40% ink with 1.22 density add .022 repeatedly until you arrive at 1.29
ie 1.22+n x .022 = 1.29
subtracting 1.22 gives n x .022 = .07
solve and you get a touch more than 3 lots of adding;
Final answer: 40% + 3 lots of adding = 43%
Hope you can follow the explanation
To: QuadtoneRIP@yahoogroups.com
Date: Tue, 13 Oct 2015 19:59:22 -0700
Subject: [QuadtoneRIP] Help with a "bit of simple math".
With a densitometer:
1. measure the 100% Light Black patch: Density 1.29
2. measure the 40% Black patch: Density 1.22. I selected 40% by eye,
but you can use the densitometer to find the darkest patch with
density less than the density of 100% Light Black
3. Measure the density of the next darker patch – 45% Black: Density
1.33.
Then it’s bit of simple math to calculate the percentage of black needed to
produce the same density as 100% Light Black. In this case we get a value
of 43.2% Black is equivalent to 100% Light Black.
I'm apparently math impaired because for the life of me I can't figure out a formula that will give me the answer of 43.2%Black. I can come close (43%) by trial and error but I think a formula would be better and in my case more accurate.
So, I guess my question is: can any one provide the formula for determining that 43.2% result.
Thanks