badapplebrewing wrote:
My question is: Is there a formula or maybe even a calculator (i.e. 5.5 gal. at 1.064 + 1.75 liters at 1.035 = 1.0??) that I can use to determine what the actual OG is POST-pitch without physically pulling a sample out of the carboy? I know there are alot of variables, so I'm just looking for some general advice.
You have 5.5*3.7854 liters of wort at 1.064 SG which weighs 5.5*3.7854*1.064*0.998203 = 22.1124 kg. SG 1.064 is 15.672 °P so the wort contains 0.15672*22.1124 = 3.4654 kg extract and 22.1124 - 3.4654 = 18.647 kg of water. The starter is 1.75L at 1.035 SG so it weighs 1.808 kg and, as 1.035 SG is 8.7811°P so it contains 0.087811*1.75*1.035/.998203 = 0.1593 kg of extract and 1.808 - 0.1593 = 1.6487 kg water. Thus you have a total of 3.4654 + 0.1593 = 3.6247 kg extract in 22.1124 + 1.808 = 23.9204 kg of mixed wort/starter so the mixture is
100*3.6247/23.9024 = 15.1646 °P which is 1.06182 SG.
To do this you must convert 2 SGs to Plato which you can do from Plato = -616.868 + 1111.14*S-630.272*S*S +135.997*S*S*S. To convert from Plato to SG you must put the SG -> P conversion into a spreasheed and use trial SG's until you match the P you are converting. Or, in Excel, let the Solver do it for you automatically.
So yes, there is a procedure. You can compare the "accurate" calculation to the approximate one in the post which precedes this one and see if you think it's worth the trouble.
What you have here is a sort of "effective" OG because you are treating the calculation as if all the sugar went in at once. That seems the only reasonable thing to do. But what about things like evaporation in the fermenter or water you use to sparge hops (don't know if you do that but I'm not leaving any extract behind if I can help it) or extract you lost during a boil over or left in the hops if you don't sparge? Should those be counted in calculation of OG? Because those things are hard to quantify brewers sometimes calculate OG after the beer is finished by analyzing for alcohol content and then sticking that back into the Balling equation to get an OG estimate. That is, to my way of thinking, the ultimate effective OG beacuse it accounts for any losses or gains of water or extract throughout the entire process, including starters. The overall efficiency of the process is the effective OG (in Plato/100) times the mass of the beer you got into kegs divided by the weight of grain/fermentables you used.