I'm looking for a good source on recoil data, my main interest is the difference between longer and shorter barrels. With a longer barrel there's more complete combustion which should create more ft/lbs of actual recoil, but yet shorter barrels seem to recoil more, especially in large caliber revolvers. Does anyone have knowledge of the science of recoil?

Schmidt_Rubin (what is your first name, anyway?):

I found these formulas on:
http://www2.rpa.net/~bologna/formulas.htm

* Recoil Energy

1. Calculate recoil impulse:
I=((WB * VB) + (4000 * WC))/225400
Where WB=wt of bullet in grains, VB=velocity of bullet in fps, and WC is the wt of the powder charge in grains. (I is in lbs/sec)
2. Calculate the velocity of the recoiling gun:
VG=(32.2 * I) / WG
Where WG is the wt of the gun in lbs., 32.2 is the acceleration due to gravity, and I is the recoil impulse from calculation #1
3. Calculate the free recoil energy of the gun in ft lbs.
EG=(WG * VG^2) / 64.4

Unfortunately the calculator there doesn't work, so let's go through an example:

example: 120 grain bullet with muzzle velocity of 2750 fps using a powder charge of 41 grains. Gun wt is 6 lbs.

1. I=(120*2750 + 4000*41)/225400=2.1917 lbs/sec
2. VG=32.2*2.1917/6=11.7621 fps
3. EG=6*11.7621^2 / 64.4=12.8895 ft lbs

Let's do another example, based on data at http://7.62x54r.net/MosinID/MosinAmmo004.htm , for Hungarian light ball 7.62x54r, out of a Russian M44 carbine:

Rifle weighs 8.25 pounds, a 148.8gr bullet, powder charge of 46.8gr, and a muzzle velocity of 2701 fps.

I = ((148.8 * 2701) + (4000 * 46.8))/225400
=> 2.6136 lb/secs

VG = (32.2 * 2.6136)/8.25
=> 10.201 ft/sec

EG = (8.25 * 10.201 * 10.201)/64.4
=> 13.33 ft-lbs.

This is close to his measured (I think it is measured, anyway) recoil of 14.9 ft-lbs.

Now let's do the same calculations for the same round out of an M91 (Finnish) weighing 9.50 lbs, and generating a muzzle velocity due to the longer barrel of 2861 ft/sec:

I = ((148.8 * 2861) + (4000 * 46.8))/225400
=> 2.7192 lb/secs

VG = (32.2 * 2.7192)/9.50
=> 9.217 ft/sec

EG = (9.50 * 9.217 * 9.217)/64.4
=> 12.53 ft-lbs.

This result doesn't agree nearly as well with the value at 7.62x54r.net, but it illustrates my point. Some reasons it doesn't agree are a different impulse value for the powder charge (4000 instead of 4700) and the use of a more exact value for the acceleration due to gravity (32.2 vs. 32).

About a pound lighter recoil. This is due to the higher weight of the rifle itself. This is also why Flea likes lots of weight in his tactical rifles. Less recoil, so you can better spot your own shots.

I'll leave other examples to the reader, as all good instructors do
Cheers,
--Hawk

Update:
While the above formulae give you an idea what is going on, they seem to be appoximate and have some problems. The calculators below will give you more precise (consistent to the nearest 0.1 foot pounds) answers:

http://www.rfgc.org/reload/recoil_calc.htm

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

http://www.shortmags.org/shortmags/smo-recoil.htm

Here is the Lyman formula that many of these are based on:

E = 0.5 x (((Wr / 32) x (((Wb x MV) + (4700 x Wp)) / (7000 x Wr))^2)

Where E = recoil Energy in ft. lbs., Wr = Weight of rifle in pounds, Wb = Weight of bullet in grains, MV = Muzzle Velocity of bullet in feet-per-second, Wp = Weight of powder in grains.

Cheers,
--Hawk

Hey Guys,
Excellent topic and great replies . I pinned it , so it'll be easier to find in the future.
Respectfully,
Harry

Now what does that mean in english Is there anything on these sites about why there's more felt recoil in shorter barrels?
Jeff

These formulas show the recoil impulse of the bullet and powder. From there you can calculate free recoil velocity of the gun, and free recoil energy.

What the formulas don't show is the time/impulse curve for a given load and gun. With a shorter barrel the bullet velocity is a little less (and the term WB * VB is a bit smaller), but the term 4000 * WC is about the same (note that different formulas use from 4000 to 4700 fps as the effective velocity of the powder charge, it depends on the type of powder, barrel leght, caliber, etc.). Even though the total recoil impulse may be a little smaller, it feels bigger because the effect of "velocity of powder * podwer mass" in the time/impulse curve is more abrupt, and because of the larger muzzle blast.

And of course revolver with shorter barrels have less weight, so the free recoil velocity is larger, as is the recoil energy. This combined with the sharper (almost the same recoil impulse in less time) recoil, and the larger blast make short barrelled magnums very umpleasant.

Yes. In addition to what Tiro said above, reread my posts, completely. If you take an M44 and an M91, the felt recoil will be higher on the M44 (shorter barrel) because the gun is lighter. This allows the impulse of firing to accelerate the gun backwards toward your shoulder faster.

It has nothing to do with the length of the barrel per say, except in that a shorter barrel results in a lighter rifle, all other things being equal. If you add 5 pounds to a M44, the felt recoil will be much less than the felt recoil on a stock M91. Let's do that for the case above, using one of the calculators (bullet weight, powder charge, and muzzle velocity stay the same, but rifle weight changes from 8.25 pounds to 13.25 pounds). Using the first calculator I list, the felt recoil goes from 14.91 ft-lbs (stock rifle) to 9.286 ft-lbs (heavy rifle), a decrease of almost 5 ft-lbs. So you see, it has nothing to do with the shorter barrel, and everything to do with the weight of the rifle.

Cheers,
--Hawk

Hi all:
Just to clarify things a bit more, here are the two forms of the recoil formulae, reworked to use the same powder impulse (4700) and the same acceleration due to gravity (32.2). As an exercise for the reader, verify that the two versions give the same result, and report back :

First form:


1. Calculate recoil impulse:
I=((WB * VB) + (4700 * WC))/225400
Where WB=wt of bullet in grains, VB=velocity of bullet in fps, and WC is the wt of the powder charge in grains. (I is in lbs/sec)
2. Calculate the velocity of the recoiling gun:
VG=(32.2 * I) / WG
Where WG is the wt of the gun in lbs., 32.2 is the acceleration due to gravity, and I is the recoil impulse from calculation #1
3. Calculate the free recoil energy of the gun in ft lbs.
EG=(WG * VG^2) / 64.4


Second (Lyman) Form (Edited so the variables mean the same things as in the above equations):
E = 0.5 x (((WG / 32.2) x (((WB x VB) + (4700 x WC)) / (7000 x WG))^2)


Cheers,
--Hawk

On the site sst.benchrest.com there is a link to a recoil calculator. It's simple, easy & fast, regarding calculating recoil.