Recoil Equation?


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Auburn1992
May 6, 2008, 11:41 PM
I remember coming across an equation to find the recoil of a firearm by getting the bullet weight, gun weight, powder, and velocity. The problem is now I can't find it. Also, I remember when I tried it out once (before I lost the equation) I came up with an absurd number that couldn't be true.

If anyone knows where this can be found please let me know. Or, if you're feeling generous and want to work it out for me :rolleyes:, it would be greatly appreciated.

Gun weight = 4.6lbs Muzzle Velocity = 2050 Bullet Weight = 325gr Barrel = 10 inch

*I am not sure of the powder weight as I am not a reloader. It would be however many grains is in a .45-70 cartridge.

Thanks for any help!

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SaMx
May 6, 2008, 11:56 PM
just find the momentum of the bullet. That should give you the momentum of the gun in recoil, and because you know the weight you can find the velocity. From that you can find the recoil energy, which is what you want I think.

things get complicated because it's not just the gun that recoiling, the shooter does too.

Gun Slinger
May 7, 2008, 12:42 AM
Auburn,

Here you go:


Free recoil energy (in f.p.e.) = ([MV + 4700 P] 7000) (64.34809711) (W)

where:

M = weight of the projectile in grains
V = velocity of the projectile in feet per second
P = weight of the powder charge in grains
W = weight of the forearm in pounds

An example just to go by so you are not haunted by those pesky absurd numbers:

A Glock 17, weighing 1.55 pounds (24.8 oz.) fires a 147 grain bullet at a muzzle velocity of 1000 f.p.s. using a powder charge of 4.5 grains:

M = 147 grains
V = 1000 f.p.s.
P = 4.5 grains
W = 1.55 pounds


([147 grs. x 1000 f.p.s. + 4700 x 4.5 gr. ] 7000) (64.34809711) (1.55) = 5.78535 f.p.e. of recoil energy.

jakk280rem
May 7, 2008, 12:44 AM
pick up a copy of lymans reloading guide #48. in the back are all kinds of handy-dandy equations. btw here it is:
free recoil energy
(ft lbs.)=(W1 Vp+4700 W2)squared/64.348 Wg
where:
W1=weight of bullet in pounds
W2=weight of powder charge in pounds
Wg=weight of gun in pounds
Vp=muzzle velocity

jakk280rem
May 7, 2008, 12:47 AM
btw there are 8000 grains in one pound

HammerBite
May 7, 2008, 12:53 AM
btw there are 8000 grains in one pound
7000

Gun Slinger
May 7, 2008, 12:57 AM
Originally posted by jakk280rem:
free recoil energy
(ft lbs.)=(W1 Vp+4700 W2)squared/64.348 Wg

jakk,

Your equation is missing the conversion value for pounds to grains (7000) and will give an erroneous (very high) value for the value of free recoil energy.

In my example above, your equation minus the conversion value for grains to pounds would yield a value of 283,482,553.8 foot pounds of free recoil energy instead of ~5.8 foot pounds. I wouldn't want to fire a gun that generated in excess of 283 million foot pounds of recoil. Yikes! :what: Talk about recoil headache! :eek:

Without proper conversion factor in the equation:
([147 grs. x 1000 f.p.s. + 4700 x 4.5 gr. ]) (64.34809711) (1.55) = 283,482,553.8 f.p.e. of recoil energy.

With proper conversion factor in the equation:
([147 grs. x 1000 f.p.s. + 4700 x 4.5 gr. ] 7000) (64.34809711) (1.55) = 5.78535 f.p.e. of recoil energy.




ETA: jakk, you have my apologies. I didn't see that you had converted the weights from grains to pounds prior to inserting them into the equation as variables so no conversion factor would be needed to be added to the equation that you supplied above. I left the example above though so that everyone can see what happens when you fail to convert units and the resulting unimaginable numbers that can occur. Up too late, not paying attention as I type away.....Guess I need more coffee! :D

peck1234
May 7, 2008, 01:02 AM
my god i hate math, deal with the dam recoil!!! : )


pretty intresting actually....

Gun Slinger
May 7, 2008, 01:07 AM
peck1234,

I'd rather be shooting than crunching numbers, too.

:)

mnrivrat
May 7, 2008, 01:14 AM
My recoil formula :

The heavier the slug, the higher the velocity, the lighter the gun = more recoil .

The lighter the slug, the slower the velocity, the heavier the gun = less recoil .

Now ain't that better than algebra ? :D:D

Gun Slinger
May 7, 2008, 01:16 AM
mnrivrat,

Undeniably so! And much easier, too.

:D

230RN
May 7, 2008, 01:34 AM
Can you use the whole mass of the gun if it's a recoil-operated locked breech firearm?

I thought you could only figure in the mass of the recoiling parts, say the slide of a pistol, in that situation.

Someone correct me if I'm wrong.

Gun Slinger
May 7, 2008, 01:49 AM
230RN,

Yep.

Momentum must be conserved: mv = mv

Recoil begins with the bullet's first movement and the 'whole' gun must recoil.

McCall911
May 7, 2008, 03:46 AM
Mathematically challenged, but would like/need to know anyway?

No worries!

http://www.handloads.com/calc/recoil.asp

1911Tuner
May 7, 2008, 06:58 AM
Can you use the whole mass of the gun if it's a recoil-operated locked breech firearm?

I thought you could only figure in the mass of the recoiling parts, say the slide of a pistol, in that situation.

Someone correct me if I'm wrong.

You are correct...and that's where it gets a little tricky with an autoloader...be it locked breech/recoil operated OR straight blowback.

The slide and barrel assembly is the actual gun. The frame is basically a gun mount, and the gun is moving on rails. There is very little disturbance that comes from the explosion of the powder charge.

What you feel as recoil comes in two phases. One, from the compression of the spring...which starts out light and increases with compression. The second is from the slide striking the impact abutment in the frame...which provides you with the greatest percentage of muzzle flip.

Because the spring's influence changes over the length and duration of the slide's travel...so does the perception of recoil. It happens so quickly that you can't really put your finger on it...but "kick" from a spring-loaded gun and mount arrangement is more of a push than a punch.

Lastly...The absolute muzzle velocity has little to do with what you feel as recoil...especially in a pistol. By the time the bullet reaches the point of release...about all you're getting is a light push. In some large capacity bottlenecked rifle rounds, the gasses exiting after the bullet's release can generate a fairly substantial amount of recoil...or "aftershock" if you will...but the actual recoil from the action/reaction of the bullet and breechblock is pretty much over by that point, and all you're feeling is the momentum that was conserved BY the action/reaction event.

Dismantler
May 7, 2008, 07:04 AM
I Google recoil tables. :)

1911Tuner
May 7, 2008, 07:16 AM
I Google recoil tables

Standard recoil tables are good as long as you're aware that they assume a fixed-breech weapon...like a bolt-action rifle or a revolver. When the "gun" part of the machine slides back and forth on rails, with a spring providing the brake and the accelerator...it changes things.

230RN
May 7, 2008, 10:23 AM
Thanks, 1911Tuner. I seem to recall that Hatcher noted that the recoil of the 1911 was much milder if the slide were locked to the frame.

I also recall that the recoil of the 1917 revolver was much milder than the same round in the 1911, although there are confuting factors involved, like the gas escaping between the cylinder and the barrel.

With the revolver, the whole mass of the gun must be calculated in, whereas with the auto, basically, only the mass of the slide and barrel should be included in "opposing" the "kick" rather than the whole gun.

(Thanks for the correction on blowback-operated arms, also. I guess I just phrased that wrong.)

I remember being surprised at how mild the recoil was in the revolver compared to the auto when I first fired the 1917.

In fact, I loaded balloon-headed .45 AR (Auto Rim) cases with the 255 gr .45 Colt ("Long Colt") bullets without a big increase in the revolver's "felt recoil."

Boy, I wish I hadn't traded in that revolver for something else. Butt-kick, butt-kick, butt-kick. I think I bought it for about forty bucks and got something like $70 when I traded it in, thinking what a smart feller I was.

Auburn1992
May 7, 2008, 10:36 PM
Does anyone think they could give me a good idea of what felt recoil would be like in a BFR 10" .45/70?

Kentak
May 7, 2008, 11:24 PM
What does the weight of the powder charge have to do with it?

Unless you're going to get really, really technical and figure that 5 grains of powder becomes 5 grains of gas projected out of the gun along with the bullet. But, then you'd have to know the velocity of the gas, which is not the same as the velocity of the bullet.

Just shoot the freaking gun. The perception of recoil is subjective anyway. Formulas won't tell you how a particular gun will feel when you shoot it. There are all kinds of other factors such as the type of action, where the center of mass of the gun is, height of bore axis, blah, blah, blah.

K

Vicious-Peanut
May 7, 2008, 11:30 PM
I ran your load through the link someone posted above and got 55 lbs. Now that is rifle velocity, if I go to 1800 it is about 45 lbs.

230RN
May 7, 2008, 11:31 PM
See Hatcher's Notebook, Chapter XII, "The theory of recoil," where all this stuff is discussed, with some practical examples. (Like how come a handgun generally shoots higher with heavier, slower bullets.)

These questions are important to ballisticians and designers.

The recoil developed by the acceleration of the powder gases down the bore is calculated by using the mass of the powder at about half the muzzle velocity of the bullet. --the "half" being complicated by the fact that if one is talking about a bottlenecked cartridge case (as opposed to a straight case like pistol cases), the volume and length of the cartridge must be included in the calculations. (op cit,pp 286-287.)

The recoil due to the rocket-like muzzle blast (as opposed to just accelerating the powder down the bore) is one of those things that apparently has to be determined empirically, since the rocket thrust (which will ultimately determine the recoil energy due to the thrust) is equal to the Net Gas pressure at exit x the area of the exit x the mass rate of discharge.

The mass rate of discharge is the stumbling block here, and must be determined experimentally for each gun and load.

However, Hatcher gives a couple of reasonable approximations, which should be pursued by the reasonable student, on pages 287-289 op cit.

Everyone who shoots should have a copy of Hatcher's Notebook. While it was written in the middle of the last centruy, so many questions I have seen asked on gun boards are readily answered by this book:

How far will your gun shoot? How high will the bullet go? How dangerous is it when it comes down? What happens if a cartridge goes off in the open air? What happens when an ammunition dump blows up? What happens when you build a fire under a can of propellant powder?

...And... how do you figure out recoil?

AZAndy
May 7, 2008, 11:31 PM
I'm fascinated by this formula and would enjoy setting up a spreadsheet to run the numbers on various guns. My problem is that I don't know where to find the weight of the powder charge for factory ammunition-- is that something that the various manufacturers are willing to share? (Not reloading at this time, you see.)

Thanks,

Andy

Kentak
May 7, 2008, 11:40 PM
I thought you could only figure in the mass of the recoiling parts, say the slide of a pistol, in that situation.

And what happens to the momentum of the moving slide? It gets transfered to the frame by the recoil spring and by hitting whatever part of the frame stops its movement. Generally, a self-loader will stretch out the time recoil forces act in the system and this affects felt recoil.

K

1911Tuner
May 7, 2008, 11:49 PM
whereas with the auto, basically, only the mass of the slide and barrel should be included in "opposing" the "kick" rather than the whole gun.


Yep...and the mass of the recoil spring...and the effect of the spring itself...which provides at least half of the felt recoil in an autopistol.

Many fail to consider that the recoil spring is a part of a separate closed system that operates on Newton's principles governing action and reaction, just like the bullet, breechblock, and the force vector provided by the expanding gasses.

The spring provides a force vector between the slide and frame. This, of course, doesn't exist in a fixed-breech revolver or single-shot pistol.

As the spring is compressed, it works to return the slide back to its starting point. Because it's butted up to the frame...whatever force it exerts against the slide is exerted on the frame in equal measure...but in the opposite direction. So, while the spring pushes forward on the slide...it pushes just as hard backward on the frame. Change the spring's rate or its maximum loading...often referred to in "pounds"...and recoil changes. Speed the slide up with hotter ammo with a given spring...and it changes.

Also...as the spring compresses, its resistance increases...and thus increases the equal force rearward. As the spring starts to slow the slide, the velocity side of the Mass X Velocity equation drops...so it's constantly changing as long as the slide moves.

Then, the slide hits the frame and yanks the gun back and up...and then the spring begins the return trip, and the action/reaction starts all over again.

mkonops
May 7, 2008, 11:51 PM
How does one factor in the "felt recoil" differences between different types of grips and such, if there really is any?

230RN
May 8, 2008, 12:23 AM
^^ Too much depends on the grips, the meat in the shooter's hand, the height of the bore above... let's say, the trigger. As someone above quite wisely said, "Just shoot the freakin' thing."

"Felt" recoil is an individual thing.

And what happens to the momentum of the moving slide? It gets transfered to the frame by the recoil spring and by hitting whatever part of the frame stops its movement. Generally, a self-loader will stretch out the time recoil forces act in the system and this affects felt recoil.

Don't forget that the relatively light slide will pick up much more energy than if the recoil just went to the whole gun.

Let's face it, if the slide weighed as much as the bullet, it would come back just as fast as the bullet went forward, picking up exactly the same amount of energy as the bullet --a hairy situation, no doubt.

(This, neglecting of course, the other recoiling parts, like the barrel and spring and the energy that goes into tipping the barrel down... if one really is inclined to pick at nits.)

jakk280rem
May 8, 2008, 03:13 AM
hammerbite, thanx for the save, one would think i would have commited that to memory by now but it was getting late. i took the equation staight out of the lymans #48 reloading manual pg.398.

1911Tuner
May 8, 2008, 06:30 AM
if the slide weighed as much as the bullet, it would come back just as fast as the bullet went forward, picking up exactly the same amount of energy as the bullet

Indeed. The momentum of the slide and the bullet are equal at the onset of action/reaction. Momentum must be conserved, and will be conserved...and objects in motion will remain in motion unless and until influenced by an outside force. The spring represents an outside force. Friction is another one. Even a fly sitting on top of the slide when the gun fires will constitute an outside force. Anything outside the system that acts to delay...reduce the speed and momentum...or block the movement will do just that.

If you fire a gun in the weightless void of outer space, the bullet will continue on at the exit velocity forever unless and until it either strikes something or enters the gravitational field of another object.

230RN
May 8, 2008, 12:30 PM
....

1911Tuner
May 8, 2008, 01:04 PM
230RN...Too late! It came in on the E-mail notification before ya whacked it!

:D

The response is...not unless the gun is fired by magic from afar. If your hand is on it...or even your finger and thumb in a pinch-fire...if the opposite velocity and momentum is effected for even a split nanosecond...the gun's rearward momentum is reduced, and it will conserve that momentum, and continue in motion AFTER the outside force influenced it.

If ya wanna get all technical...

:)

230RN
May 8, 2008, 06:54 PM
I whacked it 'cause I started to get too detailed in my Notepad composing for the post and found myself going into "This presumes the recoil forces are directed directly through the center of mass of the gun/shooter system... otherwise part of the energy would be converted to rotational energy and one would spin forever.... et cetera, et cetera, et cetera..."

But I decided the heck with that kind of stuff.

Wisely, if I may add.

I edit interminably, so one of my posts may change from moment to moment.

1911Tuner
May 8, 2008, 07:07 PM
Then, we may assume that you're not only a wise whacker, and a serious whacker...but also a surprise whacker.

Maybe your username should be...

BUSHWHACKER!

:D

McCall911
May 8, 2008, 07:11 PM
I guess I'm too old fashioned to use speadsheets for my more complex calculations. I still write my own BASIC and QBASIC programs, just like my ancestors did.

:D

230RN
May 12, 2008, 02:01 AM
McCall911,

Shhh... Top Seekrit:

I hate to mention this, but people are constantly amazed at how fast I can turn out results. Y' know why?

'Cause I frequently use good old Gee-Whiz Basic to analyze data.

One of the best and fastest ways to manipulate strings, 'cept for the fact that it doesn't handle Unicode or ASCII beyond character string 255 without a lot of fussin'.

Some complex algorithms using powers sometimes get screwed up though. Best to do a reality check on your results sometimes.

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