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!
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!