Any way to calculate ABV?

I just bottled a honey bourbon stout yesterday and have a series if readings. Wondering if there's a way to calculate the alcohol percentage. Here are the readings I took throughout my brewing process.

Original gravity: 1.044
Transfer to secondary added honey(SG: 1.425). Gravity before transfer 1.020. After adding honey, still 1.020. Didn't stir it but it did start to ferment again rapidly.
Gravity before transfer to bourbon barrel: 1.014
Gravity at bottling time: 1.011

All readings were taken at 72-75 degrees F.

I don't know the calculations off the top of my head, but if you know how much honey you added you can definitely figure it out. I believe honey adds 35 ppg (Points per pound per gallon). So, assuming you had a 5 gallon batch take (.035 * lbs of honey added)/5. Add that to your OG. Then do the usual calculation of OG-FG for ABV.

ABV = (Starting SG - Final SG) * 131

(1.044-1.011) * 131 = 3.93%

Edit: This assumes your readings posted were already temperature corrected

You can start by noting that your original wort with specific gravity 1.044 was 10.956% extract by weight. A liter of water weighs 998.2 grams, a liter of wort with specific gravity 1.044 weighs 1.044*998.2 grams = 1042.12 grams and 10.956% of that is 114.175 grams. You don't say how much honey you added to how much fermenting beer so I can't go further but honey of specific gravity 1.425 is 81.5% sugar by weight. The process from this point on is figure out how much honey you added to each liter of wort and get the effective orriginal extract per liter, and subtract the final extract (1.011 SG ~ 2.82 °P) from this: OriginalExtract - 2.82. Then this gets multiplied by a factor which depends on the effective original gravity but is about 0.421. That's an estimate of the ABW. Convert to ABV and you are done.

I added 1 lb of honey

Assuming that was to 5 gal of beer (19 L) that would be 454*0.815*1/19 = 19.5 grams of extract per liter. Obviously, this isn't much compared to the 114.2 grams from the malts but in any case you have a total of 114.2 + 19.5 = 133.7 grams per liter. Before adding the honey each liter had a mass of 1042.12 grams and now it has mass 1042.12 + 19.5 = 1061.62 grams of which 133.7 is extract so the effective wort strength (what you would have gotten had you added the honey in the kettle) is 100*133.7/1061.62 = 12.6 °P. So ABW = 0.421*(12.6 - 2.82)= 4.12 % and dividing this by 0.791 gives ABV = 5.21%

This is all pretty iffy, of course. It is possible to refine the accuracy a little by using the correct multiplier for the difference. 0.421 is about right for 12 °P beers and it varies little with OG. I'm not in McLean so I don't have access to my library and can't look up the 'correct' factor but all of this is dependent on Balling's assumptions about how much sugar gets consumed in the production of 1 gram of alcohol and how much yeast biomass goes with that.

1 lb of honey = 35 pts per pound per gallon. Assuming a 5 gallon batch that gives you 7 points added to your specific gravity (35 ppg/ 5). This is 1.051. This becomes 1.052 when corrected for your temp. Your Final gravity is 1.012 when corrected for temp. Running that through an ABV calculator in Beer Alchemy. gives you 5.3%.

Yes, their is a way to calculate ABV but you need to take into account the dilution to the ABV that adding in the honey caused.

I have made a short write up on this topic in this thread.

https://www.homebrewtalk.com/f13/seths-combined-sg-dilution-abv-formula-362212/

You do, of course, have to take into account the mass of water added but you also have to take into account the mass of water lost through evaporation (be it in boil or fermenter). This is where the idea of 'effective' original extract comes into play. You measure the volume of the final product and its specific gravity. This you convert to apparent extract (it's really better to measure true extract). You then sum all the extract added at various points in the process (mash, kettle, fermenter) and divide by the volume you have now whatever it may be. This can be converted to an estimated effective original extract. Subtract true extract from this and use the Balling formula with the true extract multiplier or subtract apparent extract and use the apparent extract multiplier. See #6. The shortcoming with this is that the multiplier depends (though weakly) on the OG and as you are assuming the original effective volume is the same as the final volume the multiplier will be off a little. If you really want to account for everything you should total the added extract and determine the final total true extract by multiplying TE times the mass of the finished beer (Volume*SG*0.998203). Taking the difference gives the number of grams of extract consumed. Dividing this by 2.0665 gives the grams of alcohol produced. Dividing this by the mass of the beer gives ABW and dividing that by 0.791 gives ABW. Given Ballings assumptions about yeast mass this is all, as I noted in #6 a bit iffy - more accurate for lagers than ales I would think. Also note that formulas like 125*(OG-FG) are an approximation to Ballings approximation. The multiplier is a function of OG.