Calculate OG from ABV and Attenuation?

[RobertRGeorge]

I am not sure if this is possible to do with algebra but here goes:

Coopers says to determine ABV use the formula ABV = (OG-FG)/7.46

Is there a way to use this to determine the OG required for a desired ABV knowing only the expected yeast attenuation?

FG would = OG x (1-A) where A = the decimal equivalent of percent attenuation of the yeast, right?

So OG = ABV x 7.46 + ((OG x (1-A))

Except now I have OG on both sides of the equation, making me think the solution is impossible. High school math failing me?

 

[jhoyda]

Yes, you need to expand the OG*(1-A) term. Doing it in my head, it looks like you get: OG= (7.46*ABV)/A

 

[ajdelange]

The formula for ABV is ABW*SG_Beer/0.791 and the formula for ABW is f(OG)*(OG -AE) where f(OG) is a function of the original gravity expressed as a simple second order polynomial and the apparent extract is AE = OG*(1 -ADF) where ADF is the apparent degree of fermentation (attenuation) expressed as a fraction. You can put these formulas into cells of a spreadsheet and calculate ABV from OG and ADF. Now all you need to do is try values of OG until you get the one that gives you the value of ABV you want. This is most easily done by using Excels SOLVER but you can grope for the answer manually as well. Hint: Guess. If the ABV us too high then make a much lower guess - one that gets you an ABV too low. Then substitute a guess half way between the high and low guesses for the high guess and if its result is too low restore the original high and use the halfway point for the low. In other words, after the bisection trap the desired ABV between the high and low guessed. Now bisect again, and again. You will zero in on the right answer pretty quickly.

A closed form solution is going to be a mess as f(OG) is second degree. Next time I'm at my other machine I'll look up F(OG).

All gravities here are in °P.

 

[passedpawn]

I am not sure if this is possible to do with algebra but here goes:

Coopers says to determine ABV use the formula ABV = (OG-FG)/7.46

Is there a way to use this to determine the OG required for a desired ABV knowing only the expected yeast attenuation?

FG would = OG x (1-A) where A = the decimal equivalent of percent attenuation of the yeast, right?

So OG = ABV x 7.46 + ((OG x (1-A))

Except now I have OG on both sides of the equation, making me think the solution is impossible. High school math failing me?

By FAR, the easiest formula is to just set your OG the same as the ABV you desire.

For example

• for a 5.5% beer, OG = 1.055
• for a 8.0% beer, OG = 1.080

This always works assuming attenuation of 76%, which is a great assumption. Do the math yourself if you don't believe me.

 

[RobertRGeorge]

Yes, you need to expand the OG*(1-A) term. Doing it in my head, it looks like you get: OG= (7.46*ABV)/A

Yep, that works thank you!

 

[RobertRGeorge]

By FAR, the easiest formula is to just set your OG the same as the ABV you desire.

For example

• for a 5.5% beer, OG = 1.055
• for a 8.0% beer, OG = 1.080

This always works assuming attenuation of 76%, which is a great assumption. Do the math yourself if you don't believe me.

That's a neat trick, thanks, I will certainly use it.

 

[RobertRGeorge]

All gravities here are in °P.

Which brings up the question I've had for a while, which is, what would it take to get the homebrewing community at large to start thinking in metric terms?

I posed my original question in points of gravity because it was most familiar.

Degrees Plato is so much easier to work with in calculating a brew and doing recipe conversions. And I understand it is in widespread use in commercial brewing.

 

[ajdelange]

Which brings up the question I've had for a while, which is, what would it take to get the homebrewing community at large to start thinking in metric terms?

A miracle

Degrees Plato is so much easier to work with in calculating a brew and doing recipe conversions. And I understand it is in widespread use in commercial brewing.

It is and it is but the 'points per pound' mentality is so engrained in home brewing that I don't think anything will ever get them to change. The fact is that it is impossible to do extract calculations without using the Plato scale which is, in fact, nothing more than concentration w/w. But one really needs specific gravity too. To calculate the pounds of extract in a gallon of wort we need to know how much the gallon of wort weights (which we get from the specific gravity) and the percentage of that weight which is extract (which is the °P).

 

[RobertRGeorge]

A miracle

It is and it is but the 'points per pound' mentality is so engrained in home brewing that I don't think anything will ever get them to change.

I guess I have my work cut out for me to convice the PPP people that there's an easier way!

But one really needs specific gravity too. To calculate the pounds of extract in a gallon of wort we need to know how much the gallon of wort weights (which we get from the specific gravity) and the percentage of that weight which is extract (which is the °P).

I don't follow you here. If I put 25 grams of sugar in enough water to make 250 ml, then I have a solution that reads 10 Plato. A 250 ml solution that reads 10 Plato would weigh 260 grams by definition, right? Therefore the water weighs 235 grams and by definition would be 235 ml in volume. What am I missing?

 

[ajdelange]

I don't follow you here. If I put 25 grams of sugar in enough water to make 250 ml, then I have a solution that reads 10 Plato.

No, you'd have a solution of 9.646 °P with a specific gravity of 1.03856 and a density of 0.998203 times that or 1.0367 grams/cc. As there are 250 cc the weight would be 250 times that or 259.17 grams. 0.09646 times that is 25 grams.

What am I missing?

I think what you are missing is that putting 25 grams of sugar into 250 mL of water is not the same as adding water to 25 grams of sugar and making up to 250 mL.

 

[RobertRGeorge]

No, what I did was put 25 grams of sugar in a glass measuring cup and topped it up to the 250 ml mark with water. Then I weighed the container and took the specific gravity of the solution.

Now since my scale does not read to hundredths of a gram, nor my saccarometer read to five decimal places, nor is my container a laboratory graduated cylinder, perhaps that is why I got a round answer.

Practically speaking though, why would the average home brewer with average home brewing equipment need to know anything other than degrees Plato considering that we're just talking about 2 different scales for measuring the same thing?

I know that points of gravity is very close to four times the degrees Plato, and within the measuring limits of my equipment that appears to be a reasonable approximation. I can imagine if I was brewing ten thousand liters at a time I might be concerned about accuracy to four decimal places, but for my purposes it seems to be false precision.

 

[ajdelange]

We aren't talking about two scales that measure the same thing. The specific gravity scale can be used to find the density of a solution. The Plato scale is a measure of the concentration of the solution w/w. The two are tied together by the Plato tables which list the specific gravity of sucrose solutions of known strength w/w (P). If you want to know efficiency (the pounds of extract you get from a pound of grain) then you must either weigh the wort (which is actually what I used to do) or measure its volume and specific gravity in order to get its weight in order to calculate the amount of extract produced. When analyzing the beer to determine true extract you multiply the Plato reading of the distillation residue by the specific gravity of the beer. To accurately solve any problem involving extract you need both (unless, as i noted earlier, you weigh the beer or wort). Because of the ASBC (modernized Plato) tables which are incorporated in the ASBC polynomial (which is third degree and can be inverted in closed form) it is easy to get Plato from SG and conversely.

Now if you don't care about accuracy then you can just use one system or the other or both using the 1 °P = 4 'points' approximation. Here are some numbers which give an idea as to how good that approximation is:
•Print SG(2)
1.0078
•print SG(4)
1.01569
•Print SG(6)
1.02369
•Print SG(8)
1.0318
•print SG(10)
1.04003
•Print SG(12)
1.04837
•Print SG(14)
1.05683
•Print SG(16)
1.06542
•Print SG(18)
1.07413
•Print SG(20)
1.08297