Recoil of .45 ACP vs. 9mm Luger

[TMan]

MLB,

handgun mass & CG

I think I'm brain-dead. What is CG?

You are right about the vector, I never really thought of it in those terms. It is very noticeable in the derringer between shooting the bottom barrel and shooting the top barrel.

I took the little Taurus ultra-recoil to the range yesterday, and noticed there isn't much muzzle flip with it. The part that is bothering me is the part that is coming back. Another thing I noticed is that when I fire it, the trigger guard comes back and hits against the front of my middle finger. That is also adding to the discomfort of shooting this thing.

What is the effect of the compensator on recoil? Is it actually reducing it by releasing some of the energy and slowing down the bullet as it leaves the muzzle, or does it only change the direction of the vector?

[klmhq]

CG = Center of Gravity

The way I understand it, a compensator takes some of the gases used for propelling the bullet and uses that gas for pushing the muzzle back down. So basically it completely cancels the upward vector of the recoil.

I totally agree that any firearm's recoil will be a complex vector. Between the slide moving backward and the extractor flinging the case sideways (yes, that will affect the gun too), it's a hideously complex thing.

Revolvers would be a lot less complex, plus their center of gravity is more centered on the actual gun (because most of the weight is in the cylinder).

[MLB]

Sorry TMan. A bit exuberant in the abbreviation department. That actually came from your comments on the balance or weight distribution of the gun.

While the compensator will slow the bullet by a negligible amount (the bullet is almost out of the barrel by the time any gas can vent), its the downward thrust of the gasses pushing the muzzle down that helps control recoil (the upward part anyway, it should have little to no effect on backward recoil.) That's also why they are at the top of the barrel.

KLM, I hadn't thought of the spent brass ejection. Now that's what my father-in-law would call "picking the fly sh*t out of pepper"! ;D

[klmhq]

mlb said:

KLM, I hadn't thought of the spent brass ejection. Now that's what my father-in-law would call "picking the fly sh*t out of pepper"! ;D

(laughs)
True. I read somewhere that when Nasa sent the Cassini Probe to Saturn is was the equivalent of a person in Houston shooting a .22 rifle and hitting a dime... in France!

Small details matter... sometimes.

[TMan]

Well back to Newton: if we slow down the velocity of the bullet because of the compensator, then we loose energy, and must loose the same amount of energy in the gun - yes?

As far as shooting brass is concerned: the Ruger P90 has hit me in the forehead so hard that on two occasions it drew blood. The Accelerator is spitting it about 10 feet away (.22 WMR ). That takes some energy.

Incidentally, when my buddy shot the Accelerator yesterday for the first time, as soon as he shot the first round he turned with a grin and said: "Hey, it is humming at me." So the internal vibration of the recoil spring is also absorbing energy.

At the gun store this afternoon I was talking to the owner/gunsmith about the S&W .357 Magnum revolvers that weigh 12 oz. My Taurus weighs 21 oz and is only a .38 Special. I don't even want to think about the recoil of shooting one of those little S&W light weights.

While I was there I picked up, along with some other things, a book about John M. Browning by Curt Gentry and another book about the Model 1911 by Robert Campbell. Like guns, you can't have too many books, and you don't have to lock them in a safe.

["DoubleAction"]

I think everyone is beginning to confuse compensators or muzzle brakes with barrel porting or Hybra Porting.

[MLB]

You never get something for nothing. If some force pushes the muzzle down, that energy is coming from somewhere. All of the energy starts in the cartridge case...

Muzzle brakes, compensators and the like only complicate our original question. Simple is good. ;D

[TMan]

admin said:

I think everyone is beginning to confuse compensators or muzzle brakes with barrel porting or Hybra Porting.

Whoa, what is the difference between a barrel port and a compensator? I thought the only difference was in spelling.

Modified to add:
Well that was nice DA: pull the pin, toss in the grenade and run. My understanding of a muzzle brake is that it directs part of the energy back toward the shooter to really reduce the recoil (and increase the noise). I still don't know the difference between porting and compensator's unless it is that the compensator is external to the gun like it is on the Sig P220 Sport, and porting is having the holes drilled into the barrel like my Taurus 608

["DoubleAction"]

What some perceives as recoil is often modified with compensators, which uses the external gases as the bullet exits the muzzle of the barrel.
With ported barrels, the recoil is modified by using the internal gases which does reduce the velosity of the bullet before it exits the barrel.
Both methods diverts the gases upward to reduce the efffect of muzzle rise as the bullet exits the barrel.
With the muzzle brakes, the exiting gases are vented in such a way as to apply a forward force on the barrel's muzzle.
I think everyone already knew these things; I'm just following up on my reply.

[TMan]

Well, I'm not included in "everyone", but I remember that on the P220 Sport the "compensator" was external and had to be removed prior to field stripping the gun. On the Taurus it is just holes in the barrel. Therefore, from your reply, my deduction was correct. However, it never dawned on me that with the compensator it was using external vs. internal gases.

I think the really big guns, like on tanks, have muzzle brakes and from what I remember in the pictures are external so they wouldn't slow down the velocity of the bullet.

So with a compensator, the redirection of the gases is basically free i.e. no loss in velocity of the bullet, but with porting you do lose velocity.

Wow, I feel like I just crawled out from under a rock.

["DoubleAction"]

TMan; Porting and Compensators are designed primarily to work in reducing the recoil on the higher velosity cartridges with lighter bullets to aid in rapid fire or reduce the perceived recoil of the larger bore guns.
One thing is the swap out, or give and take; With some porting the loss in muzzle velosity is only 5%, where the perceived recoil might be reduced as much as 20 %.

[TMan]

Just when I convinced that velocity had more to do with recoil than mass of the bullet, I picked up a copy of the July/August American Handgunner. In it the author does a review of the .460 S&W Magnum. He shoots several different loads. I'll list the CorBon ones

Weight	Velocity	Energy
200	2259	2266
250	1827	1853
325	1560	1756
395	1485	1934

A little quote from the author, Charles E. Petty talking about the factory load: "I would refer to as prototypes, with bullets from 200 all the way up to 395 gr. At that level the .460 becomes obnoxious --- just like the .500 --- but the 200 and 250 gr. loads are spectacular enough without being punishing."

A little bit further he states: "The 200 gr load is blistering fast but the light bullet keeps recoil down. The built-in muzzle break is also very effective, so that first shot was really anti-climatic. It's more of a push, and muzzle rise is modest. There is none of the wrist-wrenching twist of the 500. Actually that comes later with the heaviest bullets."

So much for theory.

[MLB]

Just a 15% change in energy from the smallest to the largest. The biggest surprise is that the smallest energy was reported to have the most recoil.

[TMan]

Just when you thought this thread was dead...

I'm still bothered by experience vs. theory, especially what I read concerning the new S&W 460 XVR. So, I bought a new book Understanding Firearm Ballistics by Robert A. Rinker ISBN: 0-9645598-4-6.

A few excerpts from the book:

"Recoil increases as the square of the velocity". Well, we knew that.

"Recoil also increases as the square of the bullet's weight."Whoa, this is big news. I thought it increased only to the first power. Now what people have said that they felt shooting the heavier bullets makes a lot of sense.

"But, it decreases only to the first power with the gun's weight so a change in the gun's weight has much less effect."

The formula he gives for Recoil Energy is:


RE = 1/2 GW (bw*bv +cw*C/7000)²
Where:
RE = Recoil Energy in ft.lbs.
W= Weight of gun in lbs.
C= Constant (4000 f.p.s)
G = Gravitations constant (32.17 ft./sec./sec.)
bw = bullet weight in grams
bv = bullet velocity in f.p.s.
cw = power charge weight in grams


I also thought it was interesting that he said the weight of the powder charge has to be considered in the recoil energy.

"High velocity .22 centerfire cartridges, such as the .22-250 , will receive as much recoil from accelerating the powder charge as from accelerating the bullet."

He also mentions that the average person can handle 15 ft. lbs of recoil without discomfort, which is what we get from shooting a .44 Magnum with a 240 gr bullet and 22.5 grs. of charge out of a 3 pound gun.

What I find interesting is the .357 Magnum with 158 gr bullet in a 2.1 lb gun has recoil energy of 10.1. Okay, doing the math on the S&W 340PD with its weight of 12 oz. gives us a RE of 27 ft.lbs. Now I understand the remarks people have made when shooting this gun that usually start with the word "mutha".

Well, I'm not sure I understand all this stuff, but it is making more sense, and follows in-line with what some people have experienced shooting heavier bullets.

[klmhq]

tman said:

J
"Recoil also increases as the square of the bullet's weight."Whoa, this is big news. I thought it increased only to the first power. Now what people have said that they felt shooting the heavier bullets makes a lot of sense.

I would want to know how he comes to that conclusion.

The formula he gives for Recoil Energy is:


RE = 1/2 GW (bw*bv +cw*C/7000)²
Where:
RE = Recoil Energy in ft.lbs.
W= Weight of gun in lbs.
C= Constant (4000 f.p.s)
G = Gravitations constant (32.17 ft./sec./sec.)
bw = bullet weight in grams
bv = bullet velocity in f.p.s.
cw = power charge weight in grams

Huh? Where did this formula come from. Does he describe how he derived it? What assumptions he used?

I also thought it was interesting that he said the weight of the powder charge has to be considered in the recoil energy.

"High velocity .22 centerfire cartridges, such as the .22-250 , will receive as much recoil from accelerating the powder charge as from accelerating the bullet."

Again, huh? What does he base that on? The power charge is converted into hot gases.

I'll have to seen if I can find it at Barnes and Noble and see his formulas.

[TMan]

KLMHQ,

I would want to know how he comes to that conclusion.

I believe it is based on the formula he gave, which by the way he didn't say how it was derived, but states that it is given without explanation for those that like math.

Again, huh? What does he base that on? The power charge is converted into hot gases.

That is true, but the heavier the charge the more recoil from it. He goes into this stuff rather deeply, and I'm not that great a typist... This is a 400 page book, and Chapter 6 is all about recoil.

Another thing I thought that was interesting is that under certain conditions unburned hydrogen can burn in a muzzle blast, and this also results in a secondary recoil. Having seen the horrific muzzle blast on a .223 Bushmaster pistol, I can see how that could happen. I'm told that there isn't any appreciable recoil on the pistol, but it does have what I would consider a significant muzzle brake.

Amazon.com has it, but they don't discount it. I paid MSRP for the book, $24.95, at Cabala's in Fort Worth.

Modified to add: I just ran some numbers and the formula does not give me meaningful results. I could be doing something wrong. Like that would be a first...

[TMan]

Okay, the guy's formula isn't written correctly. He probably had it right and the publishers screwed it up.

The correct formula that I derived is:

RE = ((bw*bv + 4000*cw)/7000)²/(2*G*W)

Where:
bw = weight of bullet in grains
bv = velocity of bullet in f.p.s.
4000 is the velocity of the gasses
cw = weight of charge in grains
G = 32.17 the gravity acceleration constant
W = weight of gun in pounds.

I'm not going to go through all the steps, but his other formula on page 76 for I is wrong too, and conflicts with the formula on page 78 for I, but I think he was just showing concepts with it and doesn't have the adjustments for unit of measure and gravity that are shown with the second formula.

It is possible to have a formula to demonstrate concepts without being mathematically correct.

I don't know why I can't get it in my head to use Google, but I did this morning and found an interesting write-up on recoil. I checked his formula's against the ones in the book, and after I figured out the correct formula, I checked it against his example figures.

www.loadammo.com/Topics/August01.htm

I apologize for not checking the accuracy of the author's formula prior to posting it. I was mesmerized with his facts and figures by the time I got to the recoil chapter.

The main thing is: the effect of doubling the mass of the bullet is the same as doubling the velocity. With this information, and the fact that 15 ft-lbs of recoil energy is all most people are comfortable with, I can pretty much determine how much a gun will recoil, at least what the maximum would be minus the effect of recoil springs, compensators, muzzle-brakes, super-grips, etc.

Modified formula to make it easier to enter into spreadsheet. KLMHQ, it is amazing how much the weight of the charge makes a difference in the Recoil Energy.

[TMan]

OK, I ran some numbers. All were using Sellier & Bellot ammo, which I still have a lot of; guess I need to do more shooting and less buying. For the .44 Mag, it is Magtech 240gr SJSP (44A)

Since I have a 9mm Kimber and a .45 Kimber, I used their weights to do my calculations. The weight of the charge wasn't readily available, so I used www.ammoguide.com and found loads that matched for velocity and muzzle energy. From their corresponding entries I obtained the weight of the charge. Just for grins, I also calculated the Recoil Energy of the little Taurus, which by now you are tired of hearing me whine about.

Gun	Recoil Energy
.45 Kimber	6.4
9mm Kimber	4.0
.38 Spc Taurus	7.8
.44 Mag S&W	9.96

Wonderful, except for one thing: I have noticed that I tire easily. My first few shots from a magazine are usually right in the center, and then they kind of meander off looking for clean paper. I took six rounds of the .44M that I shot and weighted them on my postal scale. Right about 1/2 a pound. Adding that into the equation gives me a recoil energy for my first shot of: 8.64, a reduction of 1.32 ft-lbs of recoil energy.

Back to my original post of eons ago: you can't determine what the recoil is going to be from the muzzle energy; KLMHQ was correct in his assertion. If you recall on the S&B ammo it has a ME of 371 and 421 for the .45 and 9mm respectively, but we have experience and mathematically have shown that there is more recoil from a .45 than from a 9mm if the weights of the guns are the same.

Having said that, I will still assert that muzzle energy can still give you an indication of what is about to happen when you pull the trigger. A .460 Weatherby Magnum with a 500 gr bullet, a velocity of 2530, and a 116 grain charge, fired from a 10 pound rifle will have a muzzle energy of 7110 and a recoil energy of 94.4. Should I ever experience this, I hope I am on the receiving side of the bullet vs. the sending side. ;D

[MLB]

Very good stuff TMan. I'm not surprised that the powder charge is in there (without it there would be no recoil ;D ), but that it can be generalized that way. So many grains of one powder is not the same as so many grains as another. That would mess up the "constant". The charge, bullet mass and bullet velocity could be encompassed in ME.

I don't understand the recoil energy changing as the square of the bullet mass though.

[TMan]

MLB, the constant doesn't relate to the energy in the powder. A powder with greater energy will result in a higher velocity and that is taken care of by the bw*bv portion of the formula.

I suggest doing what I did on Sunday and go to the web site I pointed to and print off the contents. He shows three formulas. What you can do is substitute the values in one formula from the results in the previous one. You will get a formula that will yield the same results as the final one that I posted.

It makes sense to me about the increase in recoil from the heavier bullets because it coincides with what I've heard that people have experienced.

I picked up a new book at the local Barnes and Noble yesterday too. In it is a review of the S&W 340PD. They convinced me: I don't want one. Running the parameters through the formula also convinced me that I don't want one. Can you believe it? A gun that I don't want. ;D