Is .163178144 (.1632) the factor for deciding the ABV % from the percentage fermentability of any fermentable beer solid, in 5 gallons ? cheers. DD.

  • 1
    Please add more information to this question, formula you are using and your units that you are using (e.g. plato/brix etc) – jsolarski Sep 10 '18 at 15:39
  • Ok, for want of answering my own question, I was really looking for confirmation of that number or one like it. So my example would be a malt extract that is 64% fermentable ,is made up to a 1040 wort, so .64 x 40 points = 25.6 points, now x by .163178144,will give 4.177 % v/v. now divide this by .131 (you know what this factor is), and you end up with the gravity drop of 31.8844, take this from the OG and you have the FG 1008.1145, which works out at just over 20%,- a one fifth gravity beer. Sorry if you didnt understand my question,Yes it was a little brief. Close issue. – Custodian Sep 10 '18 at 19:59

The relationship is not linear so multiplying by 0.13 or 0.131 works for the majority of 4-7 percent beers roughly, but you need to factoring in the gravity drop and weigh this against the effect of the alcohol on your gravity.

ABV = (OG - FG) * factor

Here is a table of the factors to use:

(OG - FG)   % ABV        Factor

Up to 6.9    Up to 0.8   0.125
7.0 - 10.4   0.8-1.3     0.126
10.5 - 17.2  1.3-2.1     0.127
17.3 - 26.1  2.2-3.3     0.128
26.2 - 36.0  3.3-4.6     0.129
36.1 - 46.5  4.6-6.0     0.13
46.6 - 57.1  6.0-7.5     0.131
57.2 - 67.9  7.5-9.0     0.132
68.0 - 78.8  9.0-10.5    0.133
78.9 - 89.7  10.5-12.0   0.134
89.8 -100.7  12.0-13.6   0.135
  • Best answer of all, brilliant, I think that it must be to do with the alcohol 'curve' on graph. I did hear about this a few decades ago but thought it a myth.Now i am straight with this,- variable factors, after 46 years of ignorance, thank you.! – Custodian Sep 20 '18 at 16:35

What are you u doing to get your readings,I use a hydromete. For me it is the easiest.I recommend.Go the Brewers friend website,it has a ABV calculator.

  • Yes, Jack I know about 131, Its the percentage fermentability side that I am looking at. See note to Martin, Cheers. – Custodian Aug 11 '18 at 16:16

No, not to my understanding. I believe the hydrometer often used is a saccharometer. It’s calibrated to measure sugar content and it’s effect in boyency. As far as total dissolved solids (TDS) go it’s generally assumed these are fermentable sugars in wort. But this because increasingly untrue for darker beers. In my experience large amounts of dark and heavily roasted malts will produce more TDS that cannot be fermented.

I believe the common equation is: ABV = OG - FG * 131

Were OG and FG is original gravity and final gravity respectively.

Best of luck!

  • Hi Martin thanks for that, I know about the x .131, but on the saccharometer it has the potential alcohol of 16 ozs sucrose in 1 gallon H2O, Marked at just less than 6%, say 5.95%, so if sugar is 36.5 points per lb/gall then multiply by .1632, in order to reach the 5.95, as sugar is regarded as 100% fermentable. So, which of these three figures is correct? Cheers, Just trying to dot the i, and cross the t. – Custodian Aug 11 '18 at 13:15
  • I don’t follow, but here goes what I know, maybe it can help. Most formulas use weight calculate expected ABV. This is used when priming bottles for instance. Also, sucrose, dextrose, dry malt extracts all have different alcohol yields in terms of a weight unit (g or oz). In beer I believe maltose is the common wort sugar. The 131 constant is merely an approximation. I’ve seen anything in the range from 129 to 131 used. But for more accurate yield numbers there are other formulas. But in the range 1-7 % ABV its supposed to be accurate enough. Best of luck! – Martin Aug 11 '18 at 21:40
  • Thanks for that Martin, moving away from the saccharometer then, If you had a wort of 1040, with a known percentage fermentable maltose content say 64%, then 25.6 points of that 40 are going to ferment (the resultant aim will be a one fifth gravity beer, so we are looking at circa 4.1 % ABV, I have used a factor of .1632 to get the same result as using .1293, on the gravity drop method but is .1632 a relevant factor? – Custodian Aug 11 '18 at 22:43
  • Many thanks for your efforts Martin, but I have just found a question of it from 5 years ago from 'daveb' with an absolutely brill answer from Mdma . Take a look. All the best. – Custodian Aug 12 '18 at 11:08

I use plato, and the formula for that is (OG-FG)/(2.0665-0.010665*OG). You can find that in Fix

  • Multiply the gravity drop by .13 is best and easiest, if you know your 13 times table,even easier. Cheers. – Custodian Sep 10 '18 at 9:47
  • Not in Plato, I'm led to believe. Or at least, not if you want an accurate result. – Frazbro Sep 10 '18 at 9:57
  • i dont use plato – Custodian Sep 10 '18 at 23:05

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