Mathematic equations (help)

WhiteLightning · Post by **WhiteLightning** » Sun Mar 04, 2007 10:52 pm

So to calculate how much I need to dilute my distillate to provide a lower % of alc. I use this formula:
(amount of distillate)x(% of distilliate)= (amount of 100% alc. in sample)
(amount of 100% alc. in sample) / (desired % of alc.) = (amount of distillate at that % which will be left after dilution)
(amount of diluted spirit) - (original amoutn of alc.) = (How much water needs to be added to dilute original sample)

ex. (750ml) x (.70%) = (525ml) / (.40%) = (1312.5ml) - (750ml) = (562.5ml)

How do you figure out the % of alc. after dilution.

ex. If I were to add 2 bottles of sprite* to 1 bottle of everclear and then add 1lb of sugar.

Thanks

Froggy · Post by **Froggy** » Mon Mar 05, 2007 1:07 am

I ended up making a spread sheet.

Try this: http://members.iinet.net.au/~shanej/Alc ... 0Table.xls" onclick="window.open(this.href);return false;" rel="nofollow

rangaz · Post by **rangaz** » Mon Mar 05, 2007 4:45 am

just remember that volumes are not additive. ie. 1L alcohol + 1L water doesn't equal 2L. To get around this you measure the initial volume of alcohol and its strength, then measure the final volume of the liquids combined instead of volume of liquid added. The simple equation is a ratio

initialABV*volume=finalABV*finalvolume

rearranged

initialABV*volume=finalABV
finalvolume

Matt · Post by **Matt** » Sun Mar 11, 2007 4:32 pm

Even that equation is only true if the initial alcohol concentration is 100%ABV to begin with.

rangaz · Post by **rangaz** » Sun Mar 11, 2007 11:52 pm

no because %ABV is a measure of how many units of pure alcohol in in a set number of units. ie the volume of pure alcohol in a sample if it were to be separated completely from the other constituents

Matt · Post by **Matt** » Sun Mar 18, 2007 4:10 pm

Oops ofcourse, my bad.