Interesting. What are the units of that Free Recoil formula. I also don't really know what the weight of the powder charge has to do with it, and wouldn't it then be powder specific?
Well, I found the article I was thinking of, but it doesn't exactly sound like the thing that FUD first asked about. From the Combat Shooting column of the Nov/Dec 1999 issue of American Handgunner, columnist Dave Anderson discussed something he called "Recoil Energy". Unfortunately, he did not give a formula, although he did present the following table of values, using a hypothetical .38 Special firing a 158gr bullet at 900fps:
<BLOCKQUOTE><font size="1" face="Verdana, Arial">code:</font><HR><pre>
Gun Weight Recoil Energy
40 oz. 2.98 ft./lbs.
30 oz. 3.97 ft./lbs.
20 oz. 5.95 ft./lbs.
16 oz. 7.44 ft./lbs.
12 oz. 9.92 ft./lbs.
[/code]
I haven't yet been able to figure out the formula or even if the units ft./lbs. has any physical significance or it's just a form of index. It was the basic thesis of the author that going above 5 ft./lbs. leads to recoil levels that are hard to train with.
The author stated:
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR> "Experience with literally hundreds of thousands of shooters has shown that most can be trained to an acceptable level of skill with handguns having around 5 ft./lbs. of recoil energy. Police instructors say that as recoil levels start going over 6 ft./lbs., scores start to drop of." [/quote]
He goes on to state that he shoots a lot of .45 ACP (200gr at 925fps) and .38 Super (135gr at 1350fps) and that these loads, out of a 40 oz 1911-style handgun gives about 5 ft./lbs.
Now if only I can find out how to calculate this.