February 6th, 2012 08:59 AM
Remember me saying the G22 had about 50% more recoil than the G17, well....
Years ago I asked a fellow shooter that happened to be a physicist to calculate the recoil force of a gun based on bullet mass, bullet energy and weight of the gun. Well what I didn't tell him was that I would be working on the same problem from a engineering perspective.
He used an entirely different approach than I did. I nor he at no time knew how the other was approaching the problem. When he emailed me his solution I compared it to mine - the formula was exactly the same! Exactly!
1- Neither of us took the reciprocation of the slide into account; unquestionably valid for revolvers since they have no slides. And, I'm not sure the slide reciprocation would make much difference, but even if it did, if we are comparing semi-to semi, i.e. a G22 to a G17 the slide effects would be a wash anyway.
2- Resistance between the bullet and the barrel was not included. Could make some difference but how would you ever determine what the friction was?
The formula for recoil force we both came up with using different approaches is:
Er = Mb * Eb / Mg
Er = recoil energy
Mb = bullet mass
Eb = bullet muzzle energy
Mg = mass of the gun
Realizing that mass is proportional to weight and that in the formula the mass of the bullet and the mass of the gun is expressed as a ratio, the formula can be simplified to:
Fr = Wb * Eb / Wg
Wb is the weight, instead of mass, of the bullet in pounds, i.e. bullet grains divided by 7000
Wg is the weight of the gun in lbs.
Here's an example
A Federal .40 cal 180 gn vs a Federal 9mm 115 gn bullet. One fired from a G22, the other from a G17
The weight of the guns:
G22 = 22.9 oz or 1.43 lbs
G17 = 22 oz or 1.357 lbs
muzzle energy = 400 ft-lbs
bullet weight = 180 / 7000 = 0.0257 lbs
muzzle energy = This is a bit problematic, the specific spec says 356 ft-lbs, the chart says 400 ft-lbs, let's use 400.
bullet weight = 115 / 7000 = 0.0164 lbs
Fr(40) = 0.0257 lbs * 400 ft-lbs / 1.43 lbs = 7.18 lbs
Er(9mm) = 0.0164 lbs * 400 ft-lbs / 1.357 lbs = 4.83 lbs
% difference = ( (7.18 - 4.83) / 4.83) * 100% = 48.7%
Interestingly, simply assuming the two bullets to have the same energy, dividing 180 by 115 gives 1.56 which would represent a 56% increase in recoil. However there is a significant difference in the weight of the G22 and G17 and that's why the formula gave a lower percentage.
This is theoretical, but the bullet weights happened to be the very ones I was shooting when I perceived felt recoil to be about 50% more and I had no idea about the math at the time, I was simply going by feel.
And, the M1V1 = M2V2 relationship suggests that a 20 oz slide could move 0.2" by the time the bullet exits the barrel. That assumes a motionless frame and that the slide unlocks immediately.
I'm too young to be this old!
Getting old isn't good for you!
February 6th, 2012 10:21 AM
February 6th, 2012 10:37 AM
"If you make something idiot proof, someone will make a better idiot."
February 6th, 2012 10:45 AM
I'll take ya'lls word for it.
February 6th, 2012 10:57 AM
February 6th, 2012 11:20 AM
A solution hunting for the question. Actually, that is quite interesting that you both ended up at exactly the same place given the disciplines. I guess that's why they call math the pure science? It makes perfect sense, actually, so I like it! Ya'll will need to add friction coefficients in there someday.
Thanks! Neat post!
February 6th, 2012 11:23 AM
I think you're on the right track, but dimensional analysis shows your result (Fr) should be in foot-pounds, not pounds. Your approach is calculating recoil energy, then scaling it to the weight of the gun, which makes sense.
The NRA Firearms Fact Book devotes nearly 10 pages to the calculation of free recoil... it's worth a look. Very handy book to have in one's firearms library.
NRA Endowment Member
February 6th, 2012 11:30 AM
Ahhh shoot! I was doing a different calculation in force and let that carry over into this. The formulas are correct, but as you point out give the results in ft-lbs. I hadn't shifted out of my force gear yet.
Originally Posted by gasmitty
But then that's why I posted so much detail, for checks exactly like this - for peer review - the scientific method and all.
I corrected the units in the OP. And many thanks for the catch!
I'm too young to be this old!
Getting old isn't good for you!
February 6th, 2012 11:44 AM
No charge. Professional courtesy among geeks.
Originally Posted by Tangle
NRA Endowment Member
February 6th, 2012 11:46 AM
I am not a math whiz so many years a go I just took a G19 and a G23 out to the range and shot them. My conclusion at that time was that the G23 recoiled more then the G19. I tried different bullet weights and got the same results.
February 6th, 2012 11:49 AM
Physics: what you take when your constipated.
Ignorance is a long way from stupid, but left unchecked, can get there real fast.
February 6th, 2012 11:50 AM
February 6th, 2012 11:52 AM
February 6th, 2012 12:16 PM
When an energy force is suddenly generated between two objects (e.g. propellant ignition between bullet and firearm) all the calculations are correct when arriving at a reasonably accurate estimate of recoil force by comparing the ratio of total firearm weight to total bullet weight. Since the firearm and bullet each have their particular degrees of "rest inertia" that's directly proportional to the mass (weight) of each object, each object will momentarily nully any force that tries to instantly set it into motion with an opposing force equal to its particular mass/inertia (weight) of rest. Because the bullet's weight is generally thousands of times less than that of the firearm, the instantaneous force generated between them will send the bullet moving forward (and the firearm moving to the rear as "recoil") with the same proportional degree of both force and speed (i.e. firearm moves to the rear with thousands of less joules of force and speed as the bullet moving in the opposite direction does).
Unless someone comes up with some exotic type of large counter-weight mechanism that would be moved forward within the firearm as other parts of it were moved to the rear by a sudden applied energy force, there's little else that will "reduce" the actual recoil force of any weapon when the weight of the weapon, bullet, and applied force remain constant. However, as mentioned in other threads, the total recoil force can be spread out over a longer duration of time to make it "feel" less intense like a hard push instead of a sharp, tooth-rattling jolt.
Because of this, most SA handguns will probably "feel" like they have a bit less recoil than that of an identical weight revolver when both are shooting the exact same bullet weight with an identical power charge because the SA inherently spreads the total recoil force out over a longer period of time. Since a revolver is rigid in nature, it will instantly transfer the total recoil force into the shooter's hand.
Going back to the mass/inertia thing, even though the slide is an integral part of an SA and a portion of its total weight, the slide weighs less (and has less mass/inertia) than the rest of the firearm - which is also being braced in the shooter's hand. When the initial energy force is applied (as recoil), much of the initial recoil energy is transfered to the lighter weight slide (which sends it moving to the rear much faster than the heavier frame & barrel which is also braced in the shooter's hand) and only the remaining portion of the initial recoil force is applied to the shooter's hand. Once the slide bottoms out against the frame on it's rearward travel, the portion of initial recoil energy applied to the slide is then transfered to the frame and on into the shooter's hand as a "second jolt".
Even though your hand is receiving the same total amount of recoil force with either the revolver or SA (when weapon weight, bullet weight, and power charge all remain identical), you'll generally feel a sharper jolt from the revolver since it's delivering the total recoil force all at once where the SA is dividing the total force into a couple of less-intense bumps even though it happens so fast they're hard to distinguish.
All of the above ramble dealing with actual recoil force still doesn't consider that many other factors can also greatly influence how that recoil force is perceived or "felt" because I've fired a number of handguns which delivered a considerably higher amount of recoil force but were still more comfortable and manageable to shoot than ones with less total recoil force. Even though a revolver doesn't spread the total recoil force over a longer duration of time, its weight distribution remains constant and the recoil-induced weapon movement is over in a hurry.
However, upon firing, the weight distribution of an SA starts moving to the rear, and the slide transfering the rest of the recoil force to the frame of a now more butt-heavy weapon tends to snap the weapon and shooter's wrist upward to make the recoil "feel" more uncomfortable and create a greater muzzle rise off target. An SA with a heavier slide to frame weight ratio would increase this upward jolt to the wrist when it bottomed out, and an SA with a shorter slide travel distance would put the two recoil bumps closer together to possibly "feel" a little more intense than the wider spacing of recoil force from a longer slide travel. Barrel length between two identical weight weapons (with same bullet weight and power charge) will also make a noticeable difference in the "feel" of identical recoil force amounts.
How un-cool is it to waste so much time writing a lot of babble that still doesn't come close to being able to predict how the total recoil is perceived, felt, or the manner in which it's applied to the hand and wrist between different handguns even though they may be of the same weight and shooting identical weight bullets of the same caliber and powder charge. Some just feel good, and some are simply uncomfortable to shoot regardless of all the math and physics
February 6th, 2012 01:32 PM
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