# Math question

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#### pherball

##### Well-Known Member
ok i'm hoping someone can tell me how to do this math:

1. figuring out original gravity given x amount of dilution: for example i have 3 gallons of wort @ 1.080 and i want to know what the gravity will drop to with each gallon of water i add to it.

2. my goal here is to brew a concentrated wort and then dilute it into two different strength batches: so if someone could tell me how to calculate the following:

say i have 3 gallons of 1.080 wort and i want to split that into two batches, say one batch of 1.040 and one batch of 1.050 and lets assume that i split the original 3 gallons in half, so 1.5 gallons in each carboy. Then all i need to know is how much water to dilute each carboy with in order to achieve the desired strengths.

Sorry if this is complicated, i know it is possible, probably basic high school algebra, but i am retarded. thanks

#### Beerthoven

##### Well-Known Member
3 gallons at 1.080 is 240 total gravity points (3 X 80 = 240)

If you pour 1.5 gallons into two carboys you will have transfered 120 gravity points into each carboy:

1.5 gallons X 80 pts/gallon = 120 points

So, given that you have 120 gravity points in a carboy, how many gallons of water do you need to get to a 1.040 wort? Note that water has 0 gravity points.

Y gallons X 40 pts/gallon = 120 points => Y = 120/40 = 3 gallons. So, you need to add 1.5 (= 3 - 1.5) gallons of water to the first carboy to get a 1.040 wort. This can also been seen intuitively. If you have a 1.080 wort, just double the volume to get to a wort that is half strength.

Same for the next carboy. You have 120 gravity points in the carboy. How many gallons of water to you need to add to get a 1.050 wort?

Z gallons X 50 pts/gallon = 120 => Z = 120/50 = 2.4 gallons. So you need to add 0.9 (= 2.4 - 1.5) gallons to the second carboy to get a 1.050 wort.

#### Dr_Deathweed

##### Well-Known Member
Concentration1 x volume 1 = concentration2 x volume 2

that should get you in the ball park. Oh, SG of H20 is 1.000

#### mrkristofo

##### Well-Known Member
Just thinking out loud here...

Density is proportional to solutes dissolved, and therefore solute concentration, so the following dilution equation should work:

(conc1)(volume1)=(conc2)(volume2)
(Grav1)(volume1)=(Grav2)(volume2)

soo...
(80)(1.5gal)=(40)(x gal)
x= 3 gal, so add 1.5 gal water

(80)(1.5gal)=(50)(x gal)
x= 2.4gal, so add 0.9gal water

Intuitively that seems right, because you're diluting with water (density 1.000), so if you want to go from 1.080 to 1.040 you need to cut the density in half, so double the volume.

Edit: Not quick enough on the draw.

#### Beerthoven

##### Well-Known Member
mrkristofo said:
Just thinking out loud here...

Density is proportional to solutes dissolved, and therefore solute concentration, so the following dilution equation should work:

(conc1)(volume1)=(conc2)(volume2)
(Grav1)(volume1)=(Grav2)(volume2)

soo...
(80)(1.5gal)=(40)(x gal)
x= 3 gal, so add 1.5 gal water

(80)(1.5gal)=(50)(x gal)
x= 2.4gal, so add 0.9gal water

Intuitively that seems right, because you're diluting with water (density 1.000), so if you want to go from 1.080 to 1.040 you need to cut the density in half, so double the volume.

Edit: Not quick enough on the draw.
I like your solution better than mine; its more elegant.

#### homebrewer_99

##### Well-Known Member
Nothing else beats out just doing it...

Sometimes I'll brew a batch and thing the OG is high so I'll just add more top off water until I get to a gravity I like...makes more bew that way...