'Advanced Water Calculator' when used in the 'SRM Color' mode

[Silver_Is_Money]

For those who actually prefer to use the "SRM Color" mode method offered by the Advanced Water Calculator' in BF, which predicts 'Mash pH' based solely upon a recipes estimate of final SRM beer color, as opposed to basing this upon calculated acidities for each of the recipes individual grist component entries, you may have noticed that rarely does the "SRM Color" mode's mash pH prediction output match the more sophisticated, yet much more time consuming and complicated "Malt Bill" mode. I believe that a good part of this discrepancy may often trace back to a misunderstanding with regard to what value one should enter within the "% Roasted Color" field found within the very easy to use "SRM Color" mode.

To noticeably improve the mash pH prediction output correlation between "SRM Color" mode and "Malt Bill" mode I've come up with some (quasi-empirical and quasi-mathematically derived ) nominal 'ballpark' guideline entries to be placed within the "% Roasted Color" field of the "SRM Color" mode as follows:

0.25% deep roasted as % Wt. of grist ~= 20% of finished beer's SRM color from roasted
0.5% deep roasted as % Wt. of grist ~= 30% of finished beer's SRM color from roasted
1% deep roasted as % Wt. of grist ~= 50% of finished beer's SRM color from roasted
2% deep roasted as % Wt. of grist ~= 65% of finished beer's SRM color from roasted
3.5% deep roasted as % Wt. of grist ~= 75% of finished beer's SRM color from roasted
5% deep roasted as % Wt. of grist ~= 80% of finished beer's SRM color from roasted
7.5% deep roasted as % Wt. of grist ~= 84% of finished beer's SRM color from roasted
10% deep roasted as % Wt. of grist ~= 87% of finished beer's SRM color from roasted
15% deep roasted as % Wt. of grist ~= 90% of finished beer's SRM color from roasted
20% deep roasted as % Wt. of grist ~= 92% of finished beer's SRM color from roasted

As to the reasoning here, all I can suggest is that if your entry into the "% Roasted Color" field is total garbage (such as it would be for a guess, as well as for placing the grists % of roasted by weight into this field), the pH adjustment suggestion given to you by the 'Advanced Water Calculator' when used in the "SRM Color" mode will also be total garbage. I.E., garbage in, garbage out.

 

[Silver_Is_Money]

I just logarithmically regressed the data as seen above and got a very nice correlation fit value of 0.9822 for this equation in the form of y=A+B*ln(x):

y = 47.5776 + 17.154 * ln(x)

Where:
'x' is the % by weight of grist that is deep roasted
'y' is the '% Roasted Color' value to be entered into BF's "SRM Mode"
ln = natural log

A series of examples:

Example for 0.25% deep roasted in the grist:

y = 47.5776 + 17.154 * ln(0.25)
y = 23.7971% of final beer SRM color from deep roasted

Example for 0.5% deep roasted in the grist:

y = 47.5776 + 17.154 * ln(0.5)
y = 35.6874% of final beer SRM color from deep roasted

Example for 3.5% deep roasted in the grist:

y = 47.5776 + 17.154 * ln(3.5)
y = 69.0675% of final beer SRM color from deep roasted

Example for 10% deep roasted in the grist:

y = 47.5776 + 17.154 * ln(10)
y = 87.0761% of final beer SRM color from deep roasted

Example for 15% deep roasted in the grist:

y = 47.5776 + 17.154 * ln(15)
y = 94.0315% of final beer SRM color from deep roasted

Example for 20% deep roasted in the grist:

y = 47.5776 + 17.154 * ln(20)
y = 98.9663% of final beer SRM color from deep roasted

BF might consider utilizing something along the lines of this regression formula and then asking the user to enter the % of grist that is deep roasted by weight, rather than requesting the user to enter the % of the final beers color that came from deep roasted.

 

[Silver_Is_Money]

Something closer to this revision may prove to work better overall.

y = 48 + 15 * ln(x)

Which yields:

 

[Silver_Is_Money]

Better yet. I doubt that I could get a much better 'overall' math model fit to the quasi-empirical values presented within in my first post to this thread than this:

y = 50 + 13.525 * ln(x)

Which yields: