I'm really confused. I'm not having a problem with efficiency. I'm having a problem with the fact that the software is (appears) using the wrong volume to calculate OG.

Quoting from your article:

â€ś(potential_pts) = (grain_pts) * (weight lbs) / volume_gals

Each grain has a dry grain potential, which you can find from our grain listing or from the malterâ€™s web site. The grain_pts is calculated from the grain potential by subtracting 1.000 and multiplying by 1000. For example, a grain with a potential of 1.035 becomes simply 35 points. 5 pounds of this grain in a 5 gallon batch would add 35*5/5 = 35 potential points to the beer. If we sum all of the potential points from the various grain additions we can get the overall potential. If we had no losses in the system, the 35 points above would give an ideal starting gravity for our beer of 1.035.

I mentioned that the potential points represents the gravity under ideal conditions. In practice one gets much less than this, usually around 70-80% for brewhouse efficiency overall. Therefore the actual original gravity is determined by the potential points times the gravity:

(batch_pts) = (potential_pts) * (brewhouse efficiency)

So if we consider a recipe with 40 potential points, and a 75% brewhouse efficiency we get 30 batch points or an original gravity of 1.030. This is how original gravity is estimated.â€ť

Letâ€™s do some math:

Using my example of a recipe with 100% pale malt with a (grain_pts)=36 and a efficiency of 75% with a target OG of 1.045.

If I put this recipe into Beer Smith, it tells me I need 17.5# of grain for a batch size of 10.5 gal in the fermentor. Under the water profile, it tells me, based on my equipment profile that my post boil volume will be 12.2 gal. The difference between the 12.2 gal post boil volume and the 10.5 gal fermentor volume is based on a 4% shrinkage (.49 gal) and 1.25 gal trub and chiller losses. These losses are due to a loss of volume and will have no affect on the gravity of the wort. (If I take a pot with 12.2 gallons of wort at 1.045 and dump 1.74 gallons of that wort on the ground, the gravity of the 10.46 gallons of wort left will still be 1.045).

Therefore: OG of 1.045 SHOULD be based on the 12.2 gallon post boil volume.

Using your equations:

(batch_pts)=[ (grain_pts) * (weight lbs) / volume_gals]* (brewhouse efficiency)

(batch_pts)=[(36*17.5lbs/12.2gal)*.75] = 1.039

Using the conventional method I used in my earlier post, I came up with 20.3 lbs of grain. Back to your equations:

(batch_pts)=[ (grain_pts) * (weight lbs) / volume_gals]* (brewhouse efficiency)

(batch_pts)=[(36*20.3bs/12.2gal)*.75] = 1.045

As far as the second point goes, I now see that the program adjust the pre boil volume to account for the longer boil. That makes sense. The point still remains that if I use Beer Smith 2, it will tell me that to get a gravity of 1.045 at 12.2 gal I need 17.5 lbs of Pale Malt. See the attached screen shot. Based on your own article and equations, this simply doesnâ€™t work.