# How to calculate amount of water to add to lower OG to target

I have been using less water than needed in my partial mash recipes. For a target of 8 gal into 2 (5 gal) fermenters, I have been using 3 (+1 gal sparge) in the mash step and 2 gal for boiling bittering hops, then adding the mash and sparge water to the boil kettle and mixing in the extract and raising to boiling and adding late addition hops. Then I chill the concentrated wort to about 90 degr. F, measure the sp. gr. and add water to bring to target volume. (I know this is not the classic schedule, but I cannot think if a good reason why it shouldn't give quite similar results and it is faster.) Yesterday I did this and got to a significantly higher OG than my target. I wanted to calculate the amount of water to add to the fermenters after splitting the volumes, since I was near the top of my kettle after adding water. My target was 1.060 so I set up an equation to solve

``````8 gal * 1.076 = x_vol * 1.060
``````

I got a ridiculously low number (about 2 cups) and eventually figured how to do it correctly, but I thought I would just ask a question here so it could be on the record to the next person to benefit from my mistake. (I did a search to see if this had been addressed earlier but failed to find a duplicate.)

## 1 Answer

Yeah, so the formula for this is C1V1 = C2V2 i.e. concentration of liquid 1 multiplied by the volume of liquid 1 is the same as the concentration of liquid 2 multiplied by the volume of liquid 2. Liquid 1 is your start point, liquid 2 is your end point. The premise is that whatever is expressed by the C variable can't change quantity by dilution or evaporation, it can only change concentration.

You can rearrange this into: (C1V1)/C2 = V2 or (C1V1)/V2 = C2

Units don't matter as long as they're the same on both sides.

Examples: 10L x 15*P, but you want the volume to be 15L: (10Lx15*P)/15L = 10*P will be the new gravity.

100 Gal at 1.060, but you want the gravity to be 1.050: (100*60)/50 = 120Gal will be the new volume.

1 pound of sugar in 1 gallon of water yields a sp.gr. of 1.046. (A gallon of malt extract weighs about 10 pounds.) And the relationship of sp.gr to "sugar" is linear, so you could figure out what the concentration was originally in weight per volume if you wanted to. The important point to get is that the amount of dissolved solids in water is measured, not by the specific gravity, but rather by its difference from 1.000, which for solution with SG of 1.076 is 0.076. So it's the ratio of the deviations from 1 that matter in the calculations.

• Just as a side note: if using SG for concentration, it's best to use just the "points", i.e. for 1.068, use 68. My maths isn't awesome, so I can't explain why (someone help?), but if you use 1.068, you get the wrong answer. – Frazbro Nov 30 '18 at 1:41
• Exactly. That was the point that I wanted someone to make "on the record." I think you should to explain that the "points" represent a value proportional to the total dissolved solutes. The actual solute concentration is something on the order of 0.076 and the target is 0.060, but since we a using ratios (because of the cancellation of units as you say) you could just as accurately be using 76 and 60 in my case. I'd like to have words to that effect in the main answer body before giving the checkmark. – 42- Nov 30 '18 at 4:59
• Sorry, late reply. I omitted specific mention of solutes because this theory can be applied to more than just solutions - it just describes situations where one of two variables is limited, and you want to modify the other. – Frazbro Dec 12 '18 at 23:06