Recoil Formula?

F

FUD

Moderator
I remember reading about and hearing about (at the gun range) about a "recoil formula" where you take the weight of the bullet, the velocity and the weight of the gun ... factor it all together and if you come up with a number greater than 200, then your ability to shoot that bullet from that gun accurately is questionable. However, I can't seem to find any more details as to what the exact formula is.

Has anyone ever heard of this? Can somebody post the formula?

With the long week-end coming up, I wanted to do the math to see where my various gun/ammo combinations fall using this formula.

FUD
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I know I've seen this, and can probably find it if I look through all my stuff. I think I may even remember the article in Shooting Times where I saw it quoted, because it was dealing with small, light handguns.

IIRC, it's basically just a momemtum balance, since momentum is conserved. In other words,

V(gun) * M(gun) = V(bullet) * M(bullet)

where V is velocity and M is mass.

But the formula that is used to give a "recoil index" does something to estimate force or it isn't unit consistent so the number that comes out is not a true value, but an "index".

It's similar to the "power factor" used in lots of pistol shooting competition these days. M(bullet in grains) * V(bullet in ft/sec). This doesn't really give a normal momentum value (because of non-standard units), but it is consistent or proportional to a momemtum value. So it's just as good, as long as you get used to it.

I'll see if I can find that recoil index formula if somebody else doesn't come up with it first.
 
free recoil=

(weight of the bullet X muzzle velocity + 4700 X weight of powder charge)squared
_____________________________________________

64.348 X weight of gun in pounds


with this formula you can figure free recoil of any gun/load combo per Lyman manual 46th ed.

------------------
Democracy is two wolves and a lamb voting on what is for lunch.
Liberty is a well armed lamb contesting the outcome of the vote.
Let he that hath no sword sell his garment and buy one. Luke 22-36
They all hold swords, being expert in war: every man hath his sword upon his thigh because of fear in the night. Song of Solomon 3-8
The man that can keep his head and aims carefully when the situation has gone bad and lead is flying usually wins the fight.
 
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.
 
riddleofsteel, your formula is missing a factor of 7000 squared in the denominator. That's necessary to convert bullet and powder charge weight from grains to pounds (7000gr/lb). Other than that, it's right.

HumpMan, the units are ft-lb. Powder charge has to be included because the muzzle blast contributes a "rocket effect", propelling the gun to the rear. The powder is assumed to be 100% converted to gas, and the 4700 is the assumed velocity of the gas. You are right about recoil being partially powder-dependant. By the way, that's the theory behind a muzzle brake: redirect the gas 90 degrees to the bore, and you eliminate its contribution to recoil. Redirect it more than 90 degrees (slightly rearward), and it actually provides some forward force.
 
Jeff,CA,

Thanks for the clarifications. I did some more searching and found at least three references that tried to explain (or at least showed the formula) the recoil energy calculation. They were all confusing, however, due to the way they worked in the conversion from grains to pounds(force) and the conversions needed to convert from pounds(force) to pounds(mass).

And yes, I am convinced that the article I started from was printing ft./lbs. when they should have been printing ft-lbs. Actually I saw it listed using the "/" in a couple of places, so it's either a common mistake, or they don't intend it to be taken as a mathematical symbol.

Ok, so at least I am pleased to see that it's all based on the physics of momentum conservation. And now I understand that the powder weight is significant since it gets expelled as high speed gas and particles and has to be accounted for in the momentum balance. And if you can do the momentum balance, you can calculate the speed of the recoiling gun and using this, you can calculate the recoil of the gun using the simple formula that related a moving mass and the energy. (E = 1/2 * M V^2)

The problem, it seems, has to do in how to know what exact momentum the powder generates. Using a constant effective velocity could be used, but others introduce a constant fudge factor (k) which can be seen as an adjustment to the nominal bullet velocity. I have seen this factor as being between 1.0 and 2.0, with 1.5 being assumed for handguns.

OK, the math isn't too bad, if you don't have to introduce the conversion factors. Depending on where and how you do that, it can certainly muddy things up.

I derived it, and then using a spreadsheet, and an assumed k of 1.5, calculated what powder charge was being used to calculate that table I first posted. I came up with a power weight of 8.15 grains which doesn't seem too out of line for a .38 Special load.

So my derivation goes like this:

M(gun) = Mass of gun (not weight)
V(gun) = Velocity of gun (ft/sec)
M(bullet) = Mass of bullet (not weight)
V(bullet) = Velocity of bullet (ft/sec)
M(powder) = Mass of powder (not weight)
V(powder) = Velocity of powder.

Momentum balance says that:

M(gun)*V(gun) = M(bullet)*V(bullet) + M(powder)*V(powder)

Assume that V(powder) can be replaced with k*V(bullet)

Solve for V(gun) to get:

V(gun) = [M(bullet)*V(bullet) + M(powder)*k*V(bullet)] / M(gun)

The Recoil Energy of the moving gun is 0.5 * M(gun)* [V(gun)]^2 so:

E(gun) = 0.5 * M(gun) * [V(gun)]^2

or

R.E. = [0.5 / M(gun)] * [M(bullet)*V(bullet) + M(powder)*k*V(bullet)]^2

Simple, but you have to convert all pound weights into pound mass (or slugs, maybe, I forget the correct terminology) by dividing by the gravitational constant of 32.16 ft/sec-sec, and you of course have to convert any grain weights to pound weights by dividing by 7000.

Anyway, my goal was to reproduce the hypothetical table of the 158gr .38 Special 900fps loads being shot from various weight guns. If I assume a k of 1.5 and a powder weight of 8.15, I do get the original table results.

Now assuming that 1.5 is a reasonable fudge constant to use, the question now becomes for me, as a non-reloader, of how to get or estimate the powder charge weights of factory ammo.

But back to FUD's original question. All of the things I have seen are real kinetic energy calculations and yield number in the range of 3 to 20 ft-lbs for normal guns. FUD was looking for some formula that yielded 200 as some "index of controlability". I don't know if these Recoil Energy calculations are what he's looking for.

OTOH, the opening of the Dave Anderson article talks about this:
<BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>"Famed hunter and explorer Frederick Selous commenced his African hunting adventures in 1871 using a pair of Hollis 4-bore muzzleloaders, each weighing about 12.5 lbs. The projectiles weighed 1,750 grs. and, depending on the powder charge, recoil energy must have been in the range of 200 to 250 ft./lbs. [sic] By way of comparison, the .460 Wthby. devleops around 110 ft./lbs. [sic] of recoil energy, while rifles of the 30-'06 class have about 20 ft./lbs. [sic]" [/quote]

Maybe this is what FUD is looking for after all.


[This message has been edited by HumpMan (edited September 02, 2000).]
 
HumpMan, you're making my head swim. I'm having flashbacks to my physics classes.

Regarding what FUD was asking for, I think you may have gotten it right with your first post. Some of the action-shooting organizations use a "power factor", which, as you said, is an index of momentum. It's:

{W(bullet) * V(bullet)}/1000

The break between major and minor calibers is 180. Ever notice how the .40 has a 180 grain bullet at a nominal 1000 fps? PF = 180. No coincidence. For the .45 (230@850), PF = 195.5, close to FUD's 200.
 
to make it simple.....

gun goes bang
muzzle goes up and gun pushes back

bigger the bullet/faster the bullet=more flip and recoil

bigger the gun the more all this is controled

in a handgun almost everything else is mental masturbation

no commonly produced handgun recoils so much with standard loads that recoil figures are needed to calculate that big heavy calibers recoil more than little ones and heavy guns help all that.

rent one and shoot it. no calculations needed

LOL

------------------
Democracy is two wolves and a lamb voting on what is for lunch.
Liberty is a well armed lamb contesting the outcome of the vote.
Let he that hath no sword sell his garment and buy one. Luke 22-36
They all hold swords, being expert in war: every man hath his sword upon his thigh because of fear in the night. Song of Solomon 3-8
The man that can keep his head and aims carefully when the situation has gone bad and lead is flying usually wins the fight.
 
So if I gets me a cronygraph I kin tell if my .44 snubby kicks mor'n my .38 snubby. Kwel, alwais wondered bout that.

Left outta the formuli are the angular relationship tween the bbl n bones plus the grip to flesh interface. Hell of a lot more effect on accuracy than recoil energy.

Sam....foller me, I knows a shortcut.
 
OK, OK, I quit. :)

I was just curious about this "5 ft-lb" rule about what kind of force gets you into the zone where it is hard to control flinching and blinking.

Just for grins, I did the calculation for all the 9mm loads in the Remington catalog, assuming a 25 oz pistol and assuming the same k factor of 1.5 and a powder charge of 8.15 gr. What did I get? The standard loads in 115gr and 124gr were all a little less than 5 ft-lbs, the 147gr was just at 5 ft-lbs, and the +P loads were all a bit above 5 ft-lbs.

Maybe there's something to this 5 ft-lbs value.

Jeff,CA:
Maybe FUD was just asking about the Power Factor, but his original question mentioned the weight of the gun, which doesn't factor in. A Power Factor of 200 might well be shootable in a heavy enough gun, I would suppose.

[This message has been edited by HumpMan (edited September 03, 2000).]
 
Thanks to everyone for their comments. Now, does anyone know how to come up with the weight of powder charge when using factory ammo? A lot of the formulas present here ask for that number and I don't see it anywhere on the box.
 
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