Talk:Recoil

message: *Talk:Recoil/Archive March 07
:Greg Glover 22:45, 21 March 2007 (UTC)
thread_title: Archives

message: *Tie in muzzle brakes, along with examples above of powder/bullet masses.  
*Add issues of recoil and controllability with rapid fire
*Bore height and muzzle rise
*Shotgun recoil absorbing stocks
*Recoil absorbing materials (sorbothane, neoprene)
The contents of this subheading has been archived (March 07)Greg Glover 22:31, 21 March 2007 (UTC)
thread_title: Stuff to be done

message: Frankly, folks, I feel the whole argument is kind of silly. f=ma is very important here. You could also define recoil as "work". So here's the simple answer. If a firearm produces work in recoiling (my hand weighs three pounds with the gun, it moved three inches up, and gravity and my triceps brought it back), recoil can and should be expressed in terms of energy. I believe this is Chuckhawks' way of expressing recoil as well -- in joules. This allows you to do comparitive analysis of recoil, such as back-of-the-envelope estimations of "a punch in the arm" "a kick from a mule" "a bowling ball dropped on your toe" and such.
And really, the technical detail here, while relevant, is entirely silly. It's starting to look like Loop quantum gravity which has had its own set of "my physics is better than yours" wars, too.
Yawn. Sorry I can't be more... erudite. ... aa:talk 09:42, 30 October 2006 (UTC)
replies: text: I completely agree with you. However it seems many folks view firearms from a prospective of quantum physics and conservation of momentum. I like many people cannot read or write in calculus. So I cannot view the physics of firearms from the prospective of quantum physics. All I know is that a bullet goes forward and I go backwards. Then I use a simple calculation that tells me what I felt while shooting. I find much of this argument the same as the one in my office, "It feels like an icebox in here. No, its to hot in here."  And the thermostat reads 20c (68f) degrees.Greg Glover 23:55, 2 January 2007 (UTC)
thread_title:  a very small comment on the recoil dilemma 

message: According to my sources (http://www.intuitor.com/moviephysics/ , under the box entitled It's Not Newtons 3rd Law, about 2/5 of the way down) the recoil has little to do with Newtons third law and it can not be used to explain why victims in real life do not fall backwards.
Therefore I think the section on recoil, like many have said does need to be rewritten, or at least written off for the moment. EvilGuru 23:14, 11 March 2007 (UTC)
replies: text: Well, one of the unwritten rules of firearms is "Anything you've seen about firearms on TV or in a movie is almost certainly false".  Recoil is the phenomenon experienced by the shooter, not the target, and has everything to do with action and reaction.  The movement of a live target, when hit with small arms fire, has more to do with the target's response than any transfer of momentum to the target.  In the case of movies, the target is generally attached to a wire that is use to pull them rapidly backwards...  A metallic silhouette target, on the other hand, reacts only to the momentum of the bullet, so that is governed by the 3rd law.  scot 15:22, 12 March 2007 (UTC)
replies: text: Recoil vs Energy was rewriten by agreemnet (see my wiki page) and posted today. EvilGuru I second your thought. Are you ready for a "Perception of Recoil" rewrite.Greg Glover 00:11, 14 March 2007 (UTC)
thread_title:  Its Not Newtons Third Law 

message: Newton's laws of motion:
   1. An object in motion will remain in motion unless acted upon by another force.
   2. Force equals mass times acceleration.
   3. For every action there is an equal but opposite reaction.
Let's start with JUST law 2:  F = m * a
A closed form solution to a ballistics problem would require more calculus than I'm comfortable with, so let's do things the old fashioned, pre-calculus, iterative way.  It works better anyway, since computers can do the huge number of calculations for us, and doing iteratively means we don't have to worry if our functions can't be expressed in nice neat forms.
Lets start with a good, solid example; a 2.5 lb. .45 ACP with a 230 grain bullet, a 5 inch barrel, a chamber volume .45 inches in diameter and .5 inches long, and 20,000 psi of chamber pressure.  We'll ignore bore friction for right now, since it's a constant and negligable compared to the resistance provided by the inertia of the bullet, and also ignore powder burn rate (we assume it's all burned by the time the bullet has moved significantly--this will give us higher than realistic velocities) and any adiabatic cooling and heat loss.  All of these things are secondary to the main concern, which is calculating acceleration--the pressure simulation just gives us a rough approximation of a chamber pressure curve, and any f(t) would work just as well, but this at least keeps us working with real numbers.
Here's the source code for the simulation, written in BASIC:
 ' constants
 pi = 3.14       ' ratio of circumference to diameter
 g = 32.2        ' gravity in feet per second per second
 ' units in feet, seconds, and slugs
 '  EXCEPT bore and pressure, which are in pounds and inches
 distB = .5 / 12              ' distance bullet has travelled down barrel
 distG = 0                    ' distance gun has recoiled
 length = 5 / 12              ' length of barrel
 diameter = .45               ' bore diameter
 pressure = 20000             ' chamber pressure (psi)
 massB = 230 / 7000 / g       ' mass of bullet (230 grains)
 massG = 2.5 / g              ' mass of gun (2.5 pounds mass)
 velG = 0                     ' velocity of gun
 velB = 0                     ' velocity of bullet
 time = 0                     ' elapsed time of simulation     
 ' simulation time step, seconds
 dt = .00001
                        
 ' calculate inital chamber volume, cubic inches
 vol = distB * pi * (diameter / 2) * (diameter / 2)
 CLS
 WHILE distB + distG < length
        ' calculate volume of gas
        volNew = (distB + distG) * pi * (diameter / 2) * (diameter / 2)
        
        ' calculate new pressure based on expansion
        p = pressure * vol / volNew
        ' calculate force based on bore diameter and pressure
        force = p * pi * (diameter / 2) * (diameter / 2)
        ' calculate acceleration, f=ma -> a = f/m
        accelB = force / massB
        accelG = force / massG
        ' calculate velocities
        velB = velB + accelB * dt
        velG = velG + accelG * dt
        ' calculate travels
        distB = distB + velB * dt
        distG = distG + velG * dt
        time = time + dt
 WEND
                                        
 PRINT "Bullet has exited at time "; time; "s"
 PRINT "Residual pressure         "; p; "psi"
 PRINT "Bullet velocity           "; velB; "fps"
 PRINT "Gun velocity              "; velG; "fps"
 PRINT "Bullet energy             "; velB * velB * massB / 2; "ft-lbs"
 PRINT "Gun energy                "; velG * velG * massG / 2; "ft-lbs"
 PRINT "Bullet momentum           "; velB * massB; "lb-s"
 PRINT "Gun momentum              "; velG * massG; "lb-s"
 PRINT "Bullet travel             "; distB; "feet"
 PRINT "Gun travel                "; distG; "feet"
And here is the output:
 Bullet has exited at time  6.900001E-04 s
 Residual pressure          2009.819 psi
 Bullet velocity            769.0192 fps
 Gun velocity               10.10711 fps 
 Bullet energy              301.7299 ft-lbs
 Gun energy                 3.965591 ft-lbs
 Bullet momentum            .7847134 lb-s
 Gun momentum               .7847134 lb-s
 Bullet travel              .4174829 feet
 Gun travel                 4.939299E-03 feet
The gun is 76 times heavier than the bullet, and the velocities reflect that ratio, as do the energies.  Note that the momentums, defined as mass times velocity, match exactly.  I am ignoring powder mass, but as it's only 5 grains and will be accelerated by only 2000 psi for a very short span, that's not a lot compared to the 230 grain bullet at 20000 psi.  scot 22:14, 22 March 2007 (UTC)
replies: text: This is wrong. You did your calculation for an isothermal gas (p V = const) where you should have used an isentropic process. 87.157.195.25 (talk) 18:08, 4 October 2017 (UTC)
replies: text: Good point.  That would reduce the velocity somewhat, as the Y value would be, at a wild guess, 1.35, for a mix of largely N2 and CO2.  Of course, the simulation is already off because it assumes an infinitely fast burning rate; the optimal rate for the .45 ACP with real-world powders would put the pressure peak somewhere around an inch or two of the bullet's travel down the barrel.  This is why an actual 20kpsi load for the .45 ACP generates significantly higher velocities, peaking over 950 fps at burn rates around that of Accurate Arms #5.  38.109.137.2 (talk) 23:04, 15 December 2017 (UTC)
thread_title: Internal ballistics simulation and recoil

message: The backward momentum is equal to the mass of the gun times its reverse velocity. This backward momentum
is equal to the sums of the two forward momentums by the law of conservation of momentum and is due to and equal to the projectile's mass times its velocity added to the mass of the escaping gases, burnt propellant, unburnt propellant, and other gases, times their respective velocities.
newton third law governs  recoil, conservation of momentum is relevant at the point of impact has little to do with recoil
:What?  So the interaction between bullet and target is subject to the laws of physics, but the interaction between firearm and bullet is not?  ALL particle interactions— elastic and  inelastic—are subject to conservation of energy AND conservation of momentum.  Newton's third law describes the effects of conservation of momentum.  In the case of recoil, the third law requires that the momentum of the ejecta (the action) be balanced by the momentum of the firearm (the reaction).  Even rotational inertia must balance out, so the firearm twists a tiny bit in the direction opposite the bullet, though this force is tiny enough that the shooter is generally unaware of it.  scot 22:37, 26 March 2007 (UTC)
thread_title:  Newtons laws 

message: The recoil impulse is incorrect.  Recoil impulse is calculated in lb.s
For example: A .308 Winchester fires a 150 gr. bullet at 2700 ft/s in a 8 lb gun using 44 gr. H4895. Recoil impulse formula states:
Ir=massbullet*velocitybullet+1.75*masscharge*velocitybullet
So the units will look like this:
(150 gr./(7000 gr./lb)/32.17 ft/s2*2700 ft/s+1.75*2700 ft/s*(44 gr./(7000 gr./lb)/32.17 ft/s2
when reducing units we get a value in lb.s not a value in milliseconds.
The section pertaining to values calculated provides an impulse in ms (milliseconds) which does not compute unless you are going to divide the impulse value by the mass of the gun which would be 8lb/32.17ft/s2. DeusImperator (talk) 19:46, 5 December 2010 (UTC)
thread_title:  Recoil Impulse Incorrect 

message: The second photo purports to show recoil from firing a S&W Model 500.  What the photo shows is not recoil, but muzzle flip, which is not the same thing.  96.35.164.102 (talk) 06:32, 19 January 2012 (UTC)
thread_title:  Photo incorrect 

message: The perceived recoil does not seem to be a function of merely recoil momentum of the firearm. A cartridge fired from a light-weight firearm has a stronger perceived recoil than the same cartridge fired from a heavier firearm, although both firearms have the same recoil momentum equal to the momentum of the projectile.
One could stipulate that the perceived recoil is a function of recoil energy of the firearm, although such a function may also depend on other variables. It seems very unlikely that such a stipulation was not exhaustively tested. This article would greatly benefit from a reference of references to such tests.  — Preceding unsigned comment added by 76.95.183.31 (talk) 07:13, 27 December 2013 (UTC)
thread_title:  Momentum or energy?