ABV Adjustment, is my math right?

[IndyBlueprints]

I have a 5 gallon batch of mead that is 14.9% ABV, and I'd like to get it down around 10%. Am I thinking right here:

5 gallons is 640 ounces. If it's 14.9% ABV, then it is 95.36 ounces alcohol. If I add 1 gallon water to it, then the total is 768 ounces, with 95.36 Ounces alcohol, making it now 12.42% ABV (95.36/768=.124217) Add one additional gallon to it, total is now 896 Ounces. 95.36/896=0.1064, or 10.6% ABV.

It really is as easy as that, isn't it?

I know what you're all thinking..."WHY ON EARTH would I want to do that??" I am serving this at a beer festival, and I don't want to serve something out of "the norm" for most of their choices there.

I'm not completely convinced I want to do this, as I haven't done any bench testing to see how badly the flavor will be diluted, but I'm considering it.

 

[Silver_Is_Money]

That works. You could also try:

14.9/10 x 5 = 7.45 gallons

7.45 - 5 = 2.45 gallons of water to be added

 

[Silver_Is_Money]

And doing the above to end up at 10.6% I get:

14.9/10.6 x 5 = 7.028 gallons

7.028 - 5 = 2.028 gallons to be added.

 

[dmtaylor]

5 (14.9) = 10 x

x = 7.45

You need to add 7.45 - 5 = 2.45 gallons.

So Silver is right in my view.

I'm a degreed engineer, and I also play one at work, if it matters.