I see there were several posts in the time it took me to type mine up. Shows you how far away from college calculus I am. One difference that struck me was that I thought ABW would be more appropriate to use than ABV in this situation. Am I wrong there? Also I made an error in my work and updated my numbers. Gave me 4% sugar in the 32% and 1.7% sugar in the 10%. Those numbers seem pretty reasonable to me.

Originally posted by lithy
I see there were several posts in the time it took me to type mine up. Shows you how far away from college calculus I am. One difference that struck me was that I thought ABW would be more appropriate to use than ABV in this situation. Am I wrong there?
Yeah, you’re probably right. Makes sense using ABW when you’re calculating weights. Otherwise the result is the percentage sugar in liquid form.

Originally posted by VsXsV
Originally posted by lithy
I see there were several posts in the time it took me to type mine up. Shows you how far away from college calculus I am. One difference that struck me was that I thought ABW would be more appropriate to use than ABV in this situation. Am I wrong there?
Yeah, you’re probably right. Makes sense using ABW when you’re calculating weights. Otherwise the result is the percentage sugar in liquid form.
Yes, you are right, ABW is appropriate, another mistake on my part.


Ok, totally scrap what I said earlier. I started over and came up with this. SG = [Sum(V1 x SG1 + V2 x SG2...etc)] / V_total then SG = [(V_water x SG_water) + (V_alc x SG_alc) + (V_sug x SG_sug)] / V_total Assuming to total volume is 1 unit, we can leave that out. Similarly, the SG of water is 1, so there goes that. Note also that m=DV (the original equation is just a drawnout expression of that, in fact), so I’m going to substitute M_sug for that last expression. And V_alc is the ABV (NOT abw) and SG_alc is a known value. That gives me: SG = V_water + ABV(.789) + M_sug So V_water is unknown. We can redefine it as (1ABVV_sug). Then, using the m=DV substitution, plug in the density of sugar (I’m going to use 1.54, you can change this and redo the math if you want) to get V_sug = M_sug / 1.54. That gives: SG = 1  ABV  (M_sug/1.54) + ABV(.789) + M_sug Then just solve that equation for M_sug, expressed in grams (per mL of solution). Someone double check me on this, but I come up with: M_sug = 2.85 (SG + ABV(0.211)  1) So the OP’s examples: 2.85(1.050 + .1(0.211)  1) = 0.203 g/mL 2.85(1.050 + .32(0.211) 1) = 0.335 A more realistic example, 1.010 with 7%abv, gives 0.071 g/mL. These seem like fair numbers to me.

Someone please dissolve a measured amount of sugar in some vodka and measure the gravity.

Wow!
I just went out for a little Sunday Danish tasting and the thread had exploded with formulae ;)
I will read through tomorrow
By the way, to answer some concern, as some may know I’m a dedicated taster of Mead.
In the last year I tried to take a FG reading of what I opened at home, at least; I have about 75 readings now.
North American mead tends to be drier than European mead (I don’t think it’s a matter of taste, but just a matter of economics).
Therefore I have a few readings between 1050 and 1075.
Then I was wondering how mead compares to other drinks.
For example Porto, Marsala, Sauternes, and a number of digestives.
Friday I measured 2 bottles of Limoncello (that’s the one at 32% ABV); one had SG 1045, the other one 1092.
But I know that the ABV has an impact on the density, so I was wondering how much
impact it has

There are also ways to gently boil away the alcohol, compensate loss of volume with added water and then do a reading to compare with the first one. That way you’ve taken one parameter out of the equation and can figure out that particular value (which I understand was the one you wanted to measure here). However, I still figure not having the OG reading makes this a very inaccurate method. If ethanol makes it lighter than water and suger heavier, just having the sum of it all how can you really know what percentage of the volume holds what.


Originally posted by Danko
However, I still figure not having the OG reading makes this a very inaccurate method. If ethanol makes it lighter than water and suger heavier, just having the sum of it all how can you really know what percentage of the volume holds what.
We’re assuming that the percentage alcohol is given (correctly) on the product. So the only unknowns are water and sugar, and there should only be one combination of them that leads to the SG measured.

Originally posted by tronraner
Originally posted by Danko
However, I still figure not having the OG reading makes this a very inaccurate method. If ethanol makes it lighter than water and suger heavier, just having the sum of it all how can you really know what percentage of the volume holds what.
We’re assuming that the percentage alcohol is given (correctly) on the product. So the only unknowns are water and sugar, and there should only be one combination of them that leads to the SG measured.
Then I understand.
