Specific Gravity Calculus

hey folks,

i've got a beer calculus question....

so i was silly and took my hydrometer reading using the wort *before* i topped up with cooled, boiled water. whoops.

however, i know exactly how much water i added, and therefore the % by which i diluted my wort. does that mean that i can calculate the actual OG of the wort that is fermenting?

i.e., i know that i added 13 cups (or .8125 gal), to bring my volume up to a full 5 gal. meaning that i diluted my original 4.1875 gal by 19.4%. since i know the SG of that 4.1875 gal, is there some fancy math i can do to calculate what the SG would be if it were diluted by 19.4%?

I've got a strong math, and decent physics background, but am not familiar enough with buoyancy/liquid density for this -- is it simply 19.4% less dense? i'm fairly sure it is not, but i really have no idea!

thanks for any thoughts!

Was this an extract kit? Did it give an OG? If so, and you used all the ingredients and ended up with the right volume you can be assured you were within a couple of points of that number and it will save you from a lot of agonizing.

Actually, unless you need to know the ABV, I would not worry too much about it.

It was a partial mash.

Thanks for the suggestion!

I am certainly comfortable just going with what my recipe tells me the OG should be (efficiency for my last 5 or 6 batches have all been pretty much exactly the same %). I'm certainly not worried about what the ABV is ultimately; I usually don't even bother pulling the hydrometer out in the first place, but I'm trying to be more consistent about taking better, more detailed notes/measurements/readings as i'm starting to revisit old recipes again (and have become generally a much more comfortable, experienced brewer).

Really this was one of those, I wonder if/what the math would be for this, more for my edification, than because it makes a big difference to me.

I'm sure someone will post a formula. In the meantime, you could use this: http://www.brewersfriend.com/dilution-and-boiloff-gravity-calculator/ and figure it out by playing the the desired gravity variable until you hit the right final volume. For instance if you OG was 1.070 at 4.1875 then at 5 gallons it would be approx 1.058 - 1.059.

dobie99 said:

hey folks,

i've got a beer calculus question....

so i was silly and took my hydrometer reading using the wort *before* i topped up with cooled, boiled water. whoops.

however, i know exactly how much water i added, and therefore the % by which i diluted my wort. does that mean that i can calculate the actual OG of the wort that is fermenting?

i.e., i know that i added 13 cups (or .8125 gal), to bring my volume up to a full 5 gal. meaning that i diluted my original 4.1875 gal by 19.4%. since i know the SG of that 4.1875 gal, is there some fancy math i can do to calculate what the SG would be if it were diluted by 19.4%?

I've got a strong math, and decent physics background, but am not familiar enough with buoyancy/liquid density for this -- is it simply 19.4% less dense? i'm fairly sure it is not, but i really have no idea!

thanks for any thoughts!

Click to expand...

You're making it too complicated. Let's say your SG (temp corrected, of course), is 50 points. We need to know the *total* gravity in your beer:

50 pts/gal * 4.1875 gal = 209.375 pts

Now we need to figured out what that would be in 5 gal:

209.375 pts / 5 gal = 41.875 pts/gal ~= 1.042

Disclaimer: this is assuming that your volume AFTER boiling is 4.1875. If this is the volume prior to boiling, then the numbers will be a bit different (b/c you have to account for evaporation). Is that pre- or post-boil?

someone might finish typing this out before i get it in... but it's just a simple ratio.

Specific gravity is (density fluid)/(density water). The higher specific gravity of wort than water is due to dissolved sugars increasing mass but not affecting volume. specific gravity may be unit-less but as it's a measure of density it sorta has units of mass/volume. You are doing your calculations though assuming you are working with a measure of concentration and not density. They can both be measured as mass/volume... but where density is Mass of solution/volume of solution, concentration is mass of solute/volume of solvent... so not equivalent. Basically you can't do a simple proportionality to work out the final density...

Basically it's not 19.4% less dense... it's has a 19.4% lower concentration of sugars. You can do this as a proportiality by using gravity points rather than SG. SO if your original SG is 1.050, you can remove the contribution of water and call that 50 gravity points.. then the proportionality works out

Final SGpoints = (initial SGpoints) * (Vinitial/Vfinal)

Edit: yup, beat me to it.

Blauhung said:

someone might finish typing this out before i get it in... but it's just a simple ratio.

Click to expand...

Slipped right in before ya

Blauhung's got a better description!

dobie99 said:

hey folks,

i've got a beer calculus question....

so i was silly and took my hydrometer reading using the wort *before* i topped up with cooled, boiled water. whoops.

however, i know exactly how much water i added, and therefore the % by which i diluted my wort. does that mean that i can calculate the actual OG of the wort that is fermenting?

i.e., i know that i added 13 cups (or .8125 gal), to bring my volume up to a full 5 gal. meaning that i diluted my original 4.1875 gal by 19.4%. since i know the SG of that 4.1875 gal, is there some fancy math i can do to calculate what the SG would be if it were diluted by 19.4%?

I've got a strong math, and decent physics background, but am not familiar enough with buoyancy/liquid density for this -- is it simply 19.4% less dense? i'm fairly sure it is not, but i really have no idea!

thanks for any thoughts!

Click to expand...

It would have helped if you gave us your starting OG before you added the water.

OG of wort before adding water = x
Remove the "1." off of the hydrometer reading and that gives you your gravity points.
Multiply that by your current volume (4.1875) and that gives you your total gravity units.

Let's say that x = 1.050

(x -1) * 1000 * 4.1875 = 209.375 TGU's

TGU's divided by your new volume (5 gallons) gives you your new gravity points. Add the "1." back onto the front and you have your new OG.

209.375 / 5 = 41.875 gravity points
41.875 / 1000 + 1 = an OG of 1.042

dobie99 said:

hey folks,



i.e., i know that i added 13 cups (or .8125 gal), to bring my volume up to a full 5 gal. meaning that i diluted my original 4.1875 gal by 19.4%. since i know the SG of that 4.1875 gal, is there some fancy math i can do to calculate what the SG would be if it were diluted by 19.4%?

Click to expand...

Yes but it's not that fancy.
i) Gravity Points is (gravity - 1)*1000. i.e. a gravity of 1.051 is 51 gravity points and so on.

ii) Using gravity points then for any solution if you boil water out or boild water in you Volume*gravity = constant *or* more commonly expressed v1*g1 = v2* g2.

And that's all you need to know:

So SG (in points)*4.1875 = x* 5.
x = SG * 5/4.1875


And your O.G. is 1 + x/1000.



Easy Peasy.

But then every one else got to it before I did, didn't they?

{Why? Because a gravity of 1-point-sumthin is a ratio. It's heavier than water because a unit of water will contain a specific amount of sugar. So a ratio of 1 + g/1000 means that a unit volume, say a gallon, of solution will have a specific amount of sugar in it. Let's call this amount g units where one unit is enough to make a one gallon of solution raise one gravity point. Then v gallons of this solution will have v*g total units of sugar. Well, if you boil down and or water this up to v2 gallons of solution, there will still be the same units, v*g units of sugar in it. So v2*g2 = v*g, where g2 is the gravity point of the modified solution. Ta-dah!}

woozy said:

No fancy math. Gravity Points is (gravity - 1)*1000. i.e. a gravity of 1.051 is 51 gravity points and so on.

Using gravity points then for any solution if you boil water out or boild water in you Volume*gravity = constant *or* more commonly expressed v1*g1 = v2* g2.

So SG*4.1875 = x* 5.
x = SG * 5/4.1875


And your O.G. is 1 + x/1000.



Easy Peasy.

But then every one else got to it before I did, didn't they?

Click to expand...

Close, but no.

x = (SG*4.1875)/5

passedpawn said:

Close, but no.

x = (SG*4.1875)/5

Click to expand...

D'OH!!!!

Yeah. Oops. Doing math on a keyboard....