ghank15
Well-Known Member
I was rereading The Joy of Home Brewing(Charlie Papazian), and I came across the formulas in the book that are used to find the alcohol content of beer. The one based on specific gravity (starting gravity-final gravity)X(105), when used for recipes in the back of the book yield alcohol contents that seem rather low.
For example, the IPA recipe in the back of the book shows an OG of 1.052-1.056 and an FG of 1.014-1.018.
Even using the highest OG and the lowest FG, the total ABV is 4.41%.(this is an IPA, remember)
If I am doing my math and interpreting the formula correctly, this seems awfully low. Don't get me wrong, there is more to beer than alcohol content. But when homebrewing, I would like to get results that are comparable/better to/than commercial beers.
What are the reasons for this disparity? Is it possible to homebrew an IPA that is stronger than even the basic BMC products(which are between 4.2 and 5% ABV)?
For example, the IPA recipe in the back of the book shows an OG of 1.052-1.056 and an FG of 1.014-1.018.
Even using the highest OG and the lowest FG, the total ABV is 4.41%.(this is an IPA, remember)
If I am doing my math and interpreting the formula correctly, this seems awfully low. Don't get me wrong, there is more to beer than alcohol content. But when homebrewing, I would like to get results that are comparable/better to/than commercial beers.
What are the reasons for this disparity? Is it possible to homebrew an IPA that is stronger than even the basic BMC products(which are between 4.2 and 5% ABV)?