The Firing Line Forums

Go Back   The Firing Line Forums > The Skunkworks > Handloading, Reloading, and Bullet Casting

Thread Tools Search this Thread
Old September 11, 2010, 11:14 PM   #1
Senior Member
Join Date: July 11, 2009
Posts: 295
powder charge/ OAL/ Perceived Recoil relationship

im mainly inquiring about pistol cartridges but this probably pertains to rifles too i guess

my question is kind of general so please if you can help me with a pertinent but general answer.

If you take 2 different cartridges that have been pre-tested and both have their bullets going at the same velocity out of a certain barrel:

1. Has a shorter OAL but a lighter powder charge

2. Has a longer OAL but a heavier powder charge

lets assume the powder is a fast burning one like Titegroup, Bullseye or CLays

If there is going to be any difference at all, which one do you beleive would have more perceived recoil/flip?

I am going to assume that the cartridge with the shorter OAL is going to snap more because the powder has more time to burn in the case before the bullet has left into the barrel.//??
Field is offline  
Old September 12, 2010, 12:10 AM   #2
Senior Member
Join Date: January 5, 2009
Location: Just off Route 66
Posts: 5,067
Technically speaking if both cartrages have the exact same max pressure then the recoil should be exactly the same. However the real world very seldom resembles the scientific one and I would guess that the one with the longer burn time (more powder) will seem to be a heavier recoil. I am not sure how you could measure that.

Jim243 is offline  
Old September 12, 2010, 10:28 AM   #3
Senior Member
Join Date: June 17, 2010
Location: Virginia
Posts: 4,820
Recoil (acceleration of the rifle backward) has a direct relationship with the projectile's acceleration forward. Over a fixed barrel distance you can (for the most part) equate that with muzzle velocity.

So within the bounds of perceived recoil, any realistic powder/projectile combo attaining the same muzzle velocity has the same recoil.
mehavey is offline  
Old September 12, 2010, 12:05 PM   #4
Join Date: March 4, 2005
Location: Ohio
Posts: 14,058
Recoil is caused by three things:

1. Newton's equal and opposite reaction force to accelerating the mass of the bullet.

2. Newton's equal and opposite reaction force to accelerating an average of about half the powder mass down the barrel along with the bullet. (Note: this is not a nuclear reaction, so mass is neither created nor destroyed, so the mass of the propellant gas is the same as the mass of the powder that burned to make it.) Gas at the breech stays put while gas at the bullet base accelerates with it. Hence, the 50% powder mass average.

3. Newton's equal and opposite reaction force to accelerating the mass of the propellant gas out of the muzzle after the bullet clears the muzzle, effectively uncorking it. The gas in the muzzle blast is accelerated to much higher velocity than the bullet, so this contribution is disproportionate to the gas mass. This is called "rocket effect" or sometimes "post-acceleration effect". It can contribute over half of the total recoil impulse in magnum rifles with heavy powder charges. This is why venting the gas laterally through holes near the muzzle before the bullet clears the muzzle (a muzzle brake) reduces recoil. The jets can be straight out in all directions and don't have to be biased up or down or back to pull the gun forward in order to reduce recoil. They just have to drop the barrel pressure before the uncorking. This is also why brakes make less difference with light loads than with heavy ones. There isn't so much pressure or powder mass to vent.

Since your longer COL cartridge has more powder mass, it will get more recoil contribution from causes 2. and 3., above, and therefore have the larger total recoil impulse.

Perceived recoil is more complicated, having to do not only with the actual recoil impulse, but how hard a slide slams into the frame in counter-battery in a self-loader, how much muzzle rise there is, how loud the muzzle blast is, etc. But those all tend to be increased for the longer COL cartridge, too. Because the shorter COL load with less powder burns at a higher pressure it produces higher peak acceleration, but this tends to be countered by the fact the greater early acceleration gets the bullet moving faster,sooner, so its total barrel time is shorter. That means it applies the cause type 1. recoil, above, for a shorter period of time. This is supposed to make the perceived recoil sharper, if lower.

Can your nerve endings tell the difference in "sharpness"? That's debatable, but I've heard people claim it. My own experience with the 1911 suggests otherwise. Target loads with with a really fast powder, like N310 or Clays, seem to recoil less than getting the same bullet to the same velocity with Power Pistol. I can only say for sure that in rifles, the same velocity load fired with a higher peak pressure load of a faster powder recoils less than using a heavier charge of slower burning powder.
Gunsite Orange Hat Family Member
CMP Certified GSM Master Instructor
NRA Certified Rifle Instructor
NRA Benefactor Member
Unclenick is offline  
Old September 12, 2010, 07:14 PM   #5
Senior Member
Join Date: July 11, 2009
Posts: 295
thank you for a very informative replies
Field is offline  
Old September 14, 2010, 10:07 PM   #6
Senior Member
Join Date: January 20, 2009
Location: already given
Posts: 105
Part of your question can be computed almost exactly. Recoil relates to Conservation of Momentum and "free recoil" relates to the energy associated with the gun when it is fired. It can be derived from the gun's momentum using several methods but I have found the method known as the "Momentum Short form" to be the easiest to use. It will give you the translational kinetic energy imparted to the gun, which is the "free recoil" of the gun.

I have done some rather extensive comparisons of pistol calibers, less so for rifles. These comparisons are "free recoil" comparisons and not what you are really seeking. What you are asking to compare would be the "net energy" of the gun, sometimes referred to as "felt recoil". Even there it gets fun. Unclenick has added "perceived recoil" to the mix. Even if we set up a device to measure "felt recoil", we have the subjective issue of "perceived" to deal with. Our measuring device could very well show the same net energy from two different cartridges, thus they would have the same felt recoil value, yet we are likely to perceive them differently. Our measuring device would be set up and calibrated to measure the net energy. Our personal built-in measuring device (perception) does not work that way. We are not necessarily feeling the net energy, but rather the net energy transfer over a slight, but perceptible time period. Energy over time is power. We are capable of perceiving the recoil over very short time intervals, thus we are not really feeling the net energy, but rather how that net energy is spread to our hand over time, i.e., a power transfer.

What we can compute very accurately is the energy values. What we actually feel is that energy value over time. That is subjective because we all have uncalibrated and variable power measurement ability, with no direct net energy measurement ability like our instruments.
highrolls is offline  
Old September 14, 2010, 10:36 PM   #7
Senior Member
Join Date: April 15, 2009
Location: Wyoming
Posts: 1,717
Thank you Uncle Nick and highrolls for attempting to apply science to something that I have wondered about sometimes. My first inclination would be that if the mass and velocity of the bullet were equal, then the opposing recoil would also be equal, regardless of how those velocities were attained. But Uncle Nick points out some complexities of that assumption.

I do know that I have compared my subjective perceived recoil from my 7 mm Rem mag firing the same weight bullet at the same (approximate) velocity as a friend's 7 MM WSM. The WSM recoil seemed to be noticeably milder to me. I know it is a shorter cartridge and uses less powder than the Rem mag, but I figured since the mass and velocities were equal that the recoil difference was due to differences in the rifle or recoil pads or something. But maybe, the recoil really WAS different in the WSM cartridge.
Doodlebugger45 is offline  
Old September 15, 2010, 10:37 PM   #8
Senior Member
Join Date: January 20, 2009
Location: already given
Posts: 105
I think Unclenick likes to make things more complicated than they need to be. Sure, a lot of things like muzzle brakes and auto slide motions and etc. will complicate the calculation of "net energy" but it is easier to avoid that and stay with the basic "free recoil". A good buddy recently asked me to give him some idea of what to expect when firing his recently acquired Mosin M1891/30 as compared to two other rifles he was familiar with. Here was the answer I gave him.

The above reference gave me the weight without bayonett of 4 Kilograms. One popular safe load I found for it would be 150 grain jacketed boat tail spitzer, over a charge weight of 43 grains, which drives that bullet at an average velocity of 2772 fps. That's all that we need to calculate it, but we need to convert to metric since I like Joules.

150 grains = 9.71 grams and 43 grains = 2.79 grams

2772 fps = 844.9 mps and charge velocity of 5200 fps = 1585 mps

Now Free recoil = (Total Momentum) squared and the result divided by twice the gun mass (8 Kg in this case)

Here goes: Bullet mass 150 grains (9.71 grams) times velocity 2772 fps (844.9 meters per second)
Plus powder mass 43 grains (2.79 grams) times charge velocity 5200 fps (1585 meters per sec)
using the metric conversions in parens gives a numerator of 12626.13
Divide numerator by 1000 (to convert the grams to kilograms) is 12.63
Square the result is total momentum squared 159.42
Divide total momentum squared by twice the gun mass (twice 4 Kg)

gives 19.93 Joules

So your rifle will have a free recoil of about 20 Joules and should compare favorably with
the Marlin model 1894 carbine at 2.9 kg (6.5 lbs) in .30-30 winchester at 17 Joules or
the Remington Sendero 4.5 Kg (10 lbs) in .308 winchester at 20 Joules. So fire either of those
two and the M1891/30 should feel somewhere in the same ballpark.

Notice I avoided the perception argument entirely.
highrolls is offline  

Thread Tools Search this Thread
Search this Thread:

Advanced Search

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump

All times are GMT -5. The time now is 11:04 AM.

Powered by vBulletin® Version 3.8.7
Copyright ©2000 - 2018, vBulletin Solutions, Inc.
This site and contents, including all posts, Copyright © 1998-2018 S.W.A.T. Magazine
Copyright Complaints: Please direct DMCA Takedown Notices to the registered agent:
Contact Us
Page generated in 0.10390 seconds with 7 queries