Talk:Recoil/Archive March 07

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I don't think this article is accurate. In most small arms, most of the recoil does not come from conservation of momentum with the bullet itself, as the article asserts. Instead, it comes from the rapid escape of gasses after the bullet exits the muzzle. The recoil due to the opposing force of the bullet itself is usually quite small. However, I am not an expert on recoil (as I am attempting to learn more about it myself), so I do not think I can correct the article. AaronWL 00:52, 19 February 2006 (UTC)

You are partially correct. The recoil energy of a small and large arms are determined by the total chemical energy held within the powder charge. The muzzle energy of a small arm as calculated by the projectile motion is about one-third the total chemical energy of the powder charge. There are two different equations for figuring muzzle energy and recoil energy. The equation that calculates muzzle, remaing and terminal energy of a projectile is the Transitional kinetic energy equation. There are actuly three equations for calculating recoil. however for the purposes of this discussion, calculating recoil is a seperat and complex exersize. Edited (thought expanded) Greg Glover 20:15, 3 October 2006 (UTC)

Depends on the caliber. A .45 ACP uses a 230 grain bullet, and about 7 grains of powder; in this case the bullet generates nearly all the recoil. In a .22-250 Remington varmint load, the bullet is about 40 grains and 35-40 grains of powder, so the recoil is as much powder as bullet. scot 18:13, 2 October 2006 (UTC)

1 Stuff to be done
2 Responce to the Above
3 Response to the Author and Protest
4 Expert assistance needed

[edit] Stuff to be done

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)

This artical is completly inaccurate (wrong). I will rewrite the artical this week.

The major problem with this artical is the the author incorectly believes recoil is momentum. Recoil is not momentum. Momentum is a measurement of velocity. Recoil is a measurement of energy: From Newton's second law; F=ma. edited Greg Glover 16:09, 3 October 2006 (UTC)

On the contrary, recoil IS momentum, pure and simple. You push a bullet (and powder mass) one way, the gun goes the other way, that's conservation of momentum, period. How do you think that gun gets its velocity?

You stated, “you push a bullet (and powder mass) one way, the gun goes the other way, that’s conservation of momentum, period.” I would answer you by pointing out you have answered your own question. You are correct, in that “...bullet (and powder mass) ...” and “gun” are equal and therefor are the theory of conservation of momentum, “period”. I would like to point out as you have so generously reminded me that conservation of momentum is the theory of both Newton’s first and third Axiom: PE = KE and ma = ma.

The theory of conservation of momentum as I understand it, is; that all actions ( bodies in motion; the third law) and non-actions (bodies at rest; the first law) within the universe will never change; unless acted upon by and outside force. So, a cartridge in a gun that has not been fired is also a function of the theory of conservation of momentum. Do you concur? Therefore if we follow your logic through to its conclusion (as demanded by the theory of conservation of momentum), a gun with an unfired cartridge will also have “velocity”. Which is incorrect.

On the contrary, you cannot say the gun has no velocity. It is moving around the center of the Earth at anywhere from 0 to 1000 miles per hour, it is moving around the Sun at a velocity of 66,600 miles per hour, the Sun is moving about the center of the Milky Way at about 140 miles per second, and who knows how fast the galaxy is moving away from the center of the universe. The concept if "zero velocity" is not, according to Eintstein, a valid concept unless you define a frame of reference. Relativity aside, you're not considering the system as a whole.

Consider the gun, cartridge and bullet as a system. When you fire them, they don't move. The center of mass of the system remains exactly where it started, relative to the given frame of reference. Consider a cork gun. You compress the piston far enough, and the cork pops out (ignore the air for now--consider it a magnetic replusion cork gun). The gun recoils in reaction to the cork, until the cork hits the end of the string. Suddenly, everything stops. If you measure the mass of the cork, and how far it went, compared to the mass of the gun, and how far it went, then you'll discover, summing everything up, that the center of mass moved not at all. A firearm is exactly the same, the center of mass of propellant, gun, and bullet does not move, rather the parts just get further away. from each other. scot 20:58, 5 October 2006 (UTC)

You’re argument is a nosequitur.Greg Glover 21:34, 5 October 2006 (UTC)

Let me answer your question of “How do you think that the gun gets its velocity.” by saying this. A gun with an unfired cartridge gets its momentum (velocity) by an outside force as demanded by the theory of conservation of momentum. That outside force is the chemical energy held within the powder charge as it transitions form a solid to a gas. The pressure of the expanding gas within the chamber is what causes movement (velocity) of both the bullet and gun. In short, it is the energy of the powder that “gets the gun its velocity".

The force is not outside the gun; that is a fallacy that goes back to the "you can't fly a rocket in a vacuum because there's nothing to push against". If you consider the bullet and gun a system, then their total momentum does not change. If you consider just the bullet, then the momentum does change, and the "outside force" is the gun, or vise versa. scot 20:58, 5 October 2006 (UTC)

The last time I reload anything I did not fail to place a powder charge. Had I failed to place a powder charge, I am sure that when the firing pin struck the primer there would be no recoil.Greg Glover 21:34, 5 October 2006 (UTC)

Sir, I would kindly ask you not to reply to my replies with anymore hyperbole. You are obviously quite learned. I understand I am challenging you belief system. But try to use your skills in a more cognitive manner.Greg Glover 21:34, 5 October 2006 (UTC)

Reference: see Momentum; conservation of momentum.Greg Glover 20:35, 5 October 2006 (UTC)

Measurements of energy are stated in units of foot-pound force (ft-lb_f). Therefor all measurements in "free recoil" are staed in foot-pound force. As in: a Winchester Model 70 that weighs 9 pounds including scope and chambered for a Spingfield .30-06' loaded with a 180 grain bullet will impart to your shoulder 20.6ft-lb_f of "free recoil" to your shoulder.

But again, energy is not the whole equation, otherwise a 40 grain bullet at 4000 fps would interchangeable with a 500 grain bullet at 1100 fps. It's not, because momentum and energy interact in entirely different ways. Recoil is generated by conservation of momentum; how you choose to express it, whether in momentum, energy, velocity, or "Marcus Inches", is up to you. (The Marcus Inch, BTW, was a measure of how far back Marcus' shoulder was kicked when firing a given handgun. A .45-70 T/C generated a good 8 Marcus Inches of recoil.)

I wish I could get the credit for creating a set of Units of Measure (Marcus Inches of recoil) but I have to give that credit to Sir Isaac Newton. I think you are mixing Apples and Oranges. Bullet velocity to firearm velocity is not comparable and is not Newton third law.

56 grains of H4350 has about 10,000ft-lbf of chemical energy. Only about one third of that chemical energy is imparted to the base of a bullet. However all 10,000ft-lbf are imparted to the firearm. Also that said bullet weighs 180 grains, said rifle weighs 70,000 grains. Can you visualize the difference?

Greg Glover 3:16 (PDT)

"Felt recoil" has no known equation that I am awear of.

This is true, yet you claim that recoil energy is the only issue of consequence; I claim that energy, momentum, and duration all count in how the shooter perceives recoil. Read last month's Dillon Precision catalog, on the 255 grain .45 ACP load; it has a 170 power factor, yet has a far lower perceived recoil than the 200 grain PF 170 the author was previously using. With identical power factors (which is a measure of momentum), the gun is going to recoil at exactly the same velocity, so why is there a difference in felt recoil?

When I get home tonight I will run the numbers and tell what the differences are tomarrow. Also the "Power Factor" you sight was an equation created by a group of competitive handgun shoots (the name escape me) circa 1978 for classifying different calibers of handguns to make the competitive fair and equitable (eg a .380 ACP vs .357 Bains & Davis ).Greg Glover: 4:05 2 OCT 06 (PDT)

Yeah, well forgive me if I ask for references to back that one up. I've written both internal and external ballistics simulation sofware (for personal use, not sale), so I'm pretty well aquainted with the physics involved. scot 21:38, 2 October 2006 (UTC)

The law determing recoil is conservation of momentum, which is mass times velocity, not kinetic energy, which is the integral with respect to velocity of momentum (hence mass times 0.5 times velocity squared). Conservation of momentum determines the free recoil velocity of the firearm; from this the recoil energy can be calculated. Perception of recoil has much to do with the free recoil velocity, but the mass of the recoiling object still counts, because that determines how much the recoiling firearm can accelerate the shooter's body as it transfers the momentum on. A 1 lb. pistol recoiling at 10 f/s is going to disturb the shooter much less than a 10 lb. rifle recoiling at 10 f/s. The recoil energy, which is more tied to velocity than mass, determines how whether the recoil is "soft" or "sharp". A .38 Spl firing a 158 grain bullet at 700 fps beats a .221 Remington Fireball firing 40 grains at 2500 fps, but the feel of the recoil is far, far different since the .221 is accelerating the gun in a small fraction of the time the .38 is. IF the gun was placed a few inches from the shooter upon firing, so that it reached full velocity by the time it hit, it wouldn't matter, but it is firmly attached to the shooter at firing, so it does matter. If you don't beleive me, try shooting a 12 gauge heavy field load with the shotgun held a couple of inches from your shoulder--that will really sharpen up the recoil. Ouchie! scot 21:25, 2 October 2006 (UTC)

Feel free to add to the list. scot 18:13, 2 October 2006 (UTC)

I am the author of the only book ever writen on kinetic energy. The title is Terminal Performance. You can down load or buy a soft cover addition fromm LuLu.com

The velocity of a gun under recoil is as you have stated "A 1 lb. pistol recoiling at 10 f/s is going to disturb the shooter much less than a 10 lb. rifle recoiling at 10 f/s." Momentum is equal to velocity. Recoil is equal energy. To figure the recoil energy of a firearm a person can use the momentum equation to calculate the velocity of that firearm. However the energy imparted to the shooter is calculated by mass times velocity squared or foot-pound force.

You are correct, a 1 lb. pistol recoiling at 10 f/s is going to disturb the shooter much less than a 10 lb. rifle recoiling at 10 f/s. But my question is this: how many foot-pounds force dose the 1 lb. pistol create and how many foot-pounds force dose the 10 lb. rifle create?

Greg Glover 2:50 2 OCT 06 (PDT)

"The map is not the territory." I think the issue here is that we're talking about different things. I'm talking about the concept of recoil--recoil is the reaction of any device that launches a projectile, and recoil is the result of conservation of momentum. You are talking about measurements of recoil; you can measure recoil in terms of kinetic energy, momentum, or some improviced measurement like the aforementioned "Marcus inch". Now, if we're concerned about how the object upon which the recoil acts will deal with that, then we need to pick the appropriate measurement. If the gun is mounted solidly, say to a tank, then the question is how many pounds of force will be applied to the mounting brackets, and is that within the acceptable load? If the gun is mounted to a spring, as in a recoil operated firearm, then the question is how fast will the slide be moving, and is that slide velocity within the acceptable range for correct operation? If we have a gun on a hydraulic recoil cylinder, then the question might be how much heat will be generated by the agitation of the hydraulic fluid? When we're dealing with a person, then we're in the realm of felt recoil, and "sharp" and "soft" recoil start to have an impact, and measurement becomes a subjective combination of velocity, mass, and duration of the impulse.

Internal ballistics is based on Greenhills formula which is proportional to the cube; volume. Hydraulics is based on the square; area. You are mixing apples and oranges again.

I am sorry sir recoil is about energy and the soft or sharp recoil you are talking about was started by Col. Arthur Alphin of A-Square “Any Shot You Want” copywrite 1996. This is junk science. Buy my book and you will learn everything. My book has many reference for you to check.

Does that explanation make sense? If so, then lets see if we can use that as a guide for re-working the current contents of the article, and also laying the groundwork for some more expansion. scot 21:51, 2 October 2006 (UTC)

I believe you and I see recoil from two different perspectives. I see recoil from the prospective as a hunter and shoot that wants to understand what I feel and why. Also how it is a function of my firearm. I perceive your explanation of recoil as an explanation of what happens mechanically with a wide variety of mechanical devices. That would be either a heat engine its self or a device in conjunction with it to change the recoil.

I wish to collaborate with you on both our understandings of recoil. May be you can write the technical side of the mechanics of recoil and I can write the physical (Newtonian mechanics) side of recoil. Maybe there needs to be two different articles: Mechanical Recoil and Free Recoil. Greg Glover 2 OCT 06 3:52 (PDT)

I don't think a split is a good idea, I'm sure we can deal with everything needed in one article. Let's start by assembling a list of topics that we want to see covered in the article, and also a few definitions to make sure we're talking about the same things. Then we can haggle over definitions and arrange the topics into an outline. At that point we can divide the outline up and start writing, and then review each other's sections to come to a consensus. To keep things here neat, I'm starting a new section now...

[edit] Responce to the Above

As I promised. Here are the numbers and a bit of commentary.

To start with, check Wikipedia for Foot-pound force and check the references. All ballistics; internal, exterior and terminal are based on Sir Isaac Newton work. All calculations and definitions for energy are derived from Newton’s second law of motion F=ma including the greatest kinetic energy equation of them all: E=mc*2

F=ma is has nothing to do with third law of motion. The laws are:

First Law: Objects in motion tend to stay in motion, and objects at rest tend to stay at rest unless an outside force acts upon them.

Second law: The rate of change of the momentum of a body is directly proportional to the net force acting on it, and the direction of the change in momentum takes place in the direction of the net force.

Third law: To every action (force applied) there is an equal but opposite reaction (equal force applied in the opposite direction).

Recoil as a concept has to do with the third law, and is all about FORCE, as the third law states, not energy. If recoil were all about energy, then you would expect a 2000 ft. lb. cartidge to generate 2000 ft. lb. of recoil energy; in a 10 lb. rifle that works out to 113 fps, which is obviously not the case. What happens is the same force that pushes the bullet forward is pushing the rifle backwards, and you get:

F1 = m1 * a1

F2 = m2 * a2

Since the forces are equal, you can do this:

m1 * a1 = m2 * a2

Since velocity is equal to acceleration times time, and since the time is the same for both sides, multiply by time:

m1 * (a1 * t) = m2 * (a2 * t)

m1 * v1 = m2 * v2

Your fundamental understanding of what happens in a small arms to create recoil is incorrect. There are no “forces” as in F1 and F2. Therefor your mathematical argument is yielding a false value. There is only one “force”. That force is the energy deliver to the open system by the energy of expanding gases from the powder charge. The bullet movement, muzzle blast, hot barrel and gun recoil are all sum parts of the total energy released. Bullet velocity (momentum) and gun velocity are not equal.Greg Glover 21:15, 5 October 2006 (UTC)

If there is only one force, then why don't bullet and gun end up going the same way? Force is expressed as a vector, with magnitude and direction, so there must be two forces inolved to get acceleration in different directions. Now it is true that there is one pressure involved, the pressure produced by the combustion of the powder. The integral of the chamber pressure over the surface of the interior of the cartridge yeilds the sum force pushing back on the gun, and the integral of the chamber pressure on the surface of the rear of the bullet yeilds the sum force pressing on the bullet. It just so happens that due to the geometry that these end up being equal in magnitude and opposite in direction (which makes sense, the force integral over any closed surface will always be 0, and when you cut a hole in it, the force on the hole is equal and opposite to the force on the remainder). scot 21:32, 5 October 2006 (UTC)

I'm not sure if that was an answer or you where thinking out load. But that was a respectfull and cognitive responce. I may be able to responed to your responce later today. But in short if I understand your understanding is that, pressure exerts its self in all directions. That is why I say and mathimatical (F/d²) it says there is one force.Greg Glover 21:46, 5 October 2006 (UTC)

Please abandon your explanation of recoil.Greg Glover 21:01 3 October 06

Not until you show me your math for calculating recoil energy, and show me that you are NOT using momentum. scot 21:51, 3 October 2006 (UTC)

And mass times velocity is momentum, which is a more fundamental force than Newtownian physics, as it holds up at under relativisitc conditions where Newtonian physics breaks down. Now one thing we've both been ignoring is that powder mass and velocity plays into recoil as well; the powder gas column is going to accelerate to an average velocity of about half the bullet velocity at the point the bullet exits the barrel. At that point, the powder gas will escape, accelerating further, but the gas will expand more or less radially from the muzzle at that point. Just adding the powder mass to the bullet mass should get you a good upper bound on the momentum, but without empirical measurement or some truly icky fluid dynamics calculations, you're not going to get exact numbers.

To respond to the article in Dillon Precision catalog. The 255gr. Bullet has a far lower perceived felt recoil then the 200gr. Bullet when fired from a .45ACP. Let us assuming the handgun is a M1911-A1 because we need a gun weight. I checked several reloading books to get an approximation of the powder charge for the same powder (Apples and Apples). Here are the numbers and why a 200gr. .45cal bullet has a greater perceived felt recoil.

200gr. At the at the muzzle is 1013fps and 456ft-lbf respectively for the bullet and 8.41ft-lbf for the 2.25lb handgun.

255gr. At the at the muzzle is 806fps and 366ft-lbf respectively for the bullet and 7.87ft-lbf for the 2.25lb handgun.

Ah, but you aren't using the right loads. In both cases the loads had an IPSC power factor 170, which equates to 850 and 666 fps respectively; this yeilds a recoil velocity of 10.79 fps in the 2.25 lb gun,, and a recoil energy of 4.07 foot pounds of energy.

There are two separate equations for calculating bullet energy and recoil energy. Bullet weight or velocity is not directly connected to felt recoil. When all things are equal a the bullet that has the greatest amount of calculable transitional kinetic energy at the muzzle will recoil harder than a slower bullet. However when all things are not equal bullet weight or velocity is not an indicator of recoil energy; see the list below

It is Newton’s Second Law of motion that dictates this; F=ma

I believe you are writing about recoil as definition as a verb in the #1 (depending on dictionary)definition; the mechanics of recoil. I am writing about recoil as definition as a noun in the #3 (depending on dictionary) definition; Newtonian mechanics (physics). A definition specific to small arms.

To respond to the article in Dillon Precision catalog. “The 255gr bullet has a far lower perceived felt recoil then the 200gr bullet when fired from a .45ACP.” That is absolutely correct and the calculation bare this out. Let us assuming the handgun is a M1911-A1 because we need a gun weight. I checked several reloading books to get an approximation of the powder charge for the same powder (Apples and Apples). Here are the numbers and why a 200gr. .45cal bullet has a greater perceived felt recoil.

Muzzle energy and Recoil calculations: 40gr. Bullet at 4000fps; 1421ft-lbf and 5.47ft-lbf respectively. 500gr. Bullet at 1100fps; 1344ft-lbf and 12.67ft-lbf respectively.

Actually, there are three measures that are all valid ways of measuring the "recoil phenomoenon": force, which is what Newton's laws deal with; momentum, which is the measure I prefer as it is calculable from just cartridge data; and energy, which you prefer but requires both the cartridge data and the firearm mass. Now that I think of it, I think that Newton puts it best; both the energy and momentum measures fail to consider the time element, which I think contributes significantly to the shooter's perception of recoil. You can call "sharp" and "soft" recoil junk science, but it's not--it's an attempt to quantify an qualitative perception. For example, I can say that two objects reflect light at wavelengths of 590nm and 480nm; that's a quantitative measure. However, I could also say "Purlpe and orange clash", and that's a qualitative property that cannot be measured. Does that mean fashion, and all those "pick your colors" tables are "junk science"? I think it's just a case of trying to come up with an approximate quantification for a qualitative property. In both cases, it's a matter of looking at empirical data (what people think "kicks harder" or "looks icky") and trying to tie that to some measurable physical quantity. In the case of the IPSC, I think the use of a momentum based measure came from the use of steel plates as targets; the bullet impact is an inelastic collision, and so what matters in knocking the target down is how much momentum is transferred to the target; hitting it with a high energy, low momentum round like, say, a .17 Remington will just convert all that energy into liquid lead and copper, and still not knock the target down. scot 17:06, 3 October 2006 (UTC)

I had to laugh when I read your challenge. I thought to my self, “is this a trick question.” “Should I unleash the long version (a quadratic based recoil energy computation) on this guy or what?” Anyway that would not be fair. Both the long and short versions are equations for recoil energy that are derived in part by using the “dreaded” quadratic equation. Neither of these equations employs momentum (mv). However as you know most if not all kinetic energy equations use mass (m) and velocity (v) as factors within them. Also I was thinking of using your 10lbs rifle moving at a rate of 10fps and plugging it directly into the transitional kinetic energy equation. I certainly would have met your criteria. But that would a bit of deception too.

To be total transparent, I have not used the short or long versions in 25 years. Both those equations are in a box somewhere in my storage unit. It would take me weeks to find them. And I know quite well your mathimatical skills are up to the task of use both the long or short versions.

Anyway I use the easy two step method, which is half of what you are doing less the powder mass. I first set up Newton’s third law. I put the gun mass on one side and the bullet/powder mass on the other side. I then find for the gun velocity. That is the first part.

The second part is to plug the gun velocity into the Tke equation (you know as the impotent classic statement) and it yields a proper measurement for recoil energy.

So, I ask you to please consider this before dismissing me. Momentum is a quality of velocity and energy is a quality of acceleration. Momentum as in mv is proportionate and energy is exponential as in mv^2. That is why your .17 Remington with a 40gr bullet will knock the pigs over at 200yds even though the momentum values are so low. By the way a steel pig is made of a non-yielding material so even with a liquid impact all the energy is transferred to the target. The computations for semi-yielding material such as the broad side of an Elk are different than steel plates.

If I may make another assumption, I would guess you are a handgun enthusiast. You have reference the IPSC power factor several times. You are correct. The power factor is a function of momentum and in my papers has proven it an impotent measurement for energy. There for you can’t use the power factor to consistently predictor knockdown power; steel plates or animals. However you can use transitional kinetic energy values to predict the knockdown power of steel plates ever time. As for animals maybe someday my published Impact Penetration Factor will get scientifically tested in a controlled environment.

Recoil calculations: 6.75lb .45-70 T/C with a 500gr bullet at 1800fps equals 62ft-lbf 1lb pistol at 10fps equals 2ft-lbf 10lb rifle at 10fps equals 16ft-lbf 2.25lb handgun chambered in .38 Special equals 3.05ft-lbf 2.25lb handgun chambered in .221 Rem. Fireball equals 9.24ft-lbf 6.75lb 12ga. Shotgun chambered in Heavy Field load equals 46ft-lbf

Greg Glover16:17, 3 October 2006 (UTC)

[edit] Response to the Author and Protest

Mr. Scot, with respect and humbleness. If you wish to redefine established science as predicated on Sir Isaac Newton’s work; the father of modern physics and author of Newtonian Mechanics: I must protest.

Actually, Newton was the father of classical physics, Einstein is generally given the title of Father of Modern Physics, in part because he proved Newton wrong. E=mc^2 has no place in Newtonian physics, because Newtonian physics does not recognize any velocity limits, gravitational bending of space, the fact that a 10 pound object approaching the speed of light weights many tons, or the fact that if it's moving fast enough a 5 mile long train will fit completely in a 1 mile long tunnel.

No Sir, Einstine did not prove Newton wrong. Einstein expanded on on the Newtons work (see Mechanics, Eneinsteinia vs Newtonian). This inturn solved the problem of Murcury's orbit. I belive Einstein was proven right about 1915 during an eclips. All of Newton's work is still alive and well today. Where do you think the kinetic energy tables come from. The KE equation "classic statement" and Tke equation "spacific statement" are derived directly for F=ma. I have done all the computations. (Edited by self to add references)Greg Glover 16:58, 5 October 2006 (UTC)

I ask you, please do not attempt to redefine "recoil" as it is known in small and large arms ballistics. Established science as predicated on Newtonian Mechanics, states clearly and concisely that in physics, recoil is a function of the second law; F=ma. The Units of Measure as established by science for Force (F) is the foot-pound force (ft-lb_f) and comes from the English Engineering System. Its counterpart in the Metric System (SI) is the erg or more commonly used is the Joule or Newton meter

Right, foot-pounds is a measure of energy, pounds is a measure of force, and slugs is the unit for mass; 32.2 pounds is exerted by 1 slug at 1 gravity, 7000 grains per pound-mass. Impulse is pound-seconds, acceleration is feet per second squared. Got the units down. As for relationships, velocity is the integral with respect to time of distance, acceleration is the integral w/r/ time of velocity, jerk is the integral w/r/ time of acceleration, and no one can decide what the integral w/r/ time of jerk should be called.

No Sir, you have not got the units down as I have already shown. I think you are accustomed to using the metric system which is impotent as a system of values within the small arms world (The world I speak of is the United States). Maybe in engineering you can use the metric system but not in the small arms world. As I have posted elsewhere we use the English Engineering System with its true and correct 4 UOM; pound force (F), pound mass (m), time (t)and distance (d). That is why all computations of bullet energy and recoil energy are in foot-pound force (Fd).

Your inference to the “jerk” is junk science made up by the (and with great respect) sporting arms writer Jon Sundra (I hope I spelled his name right). There is no way to make acceleration fast or slow. By introducing time back into acceleration, is to render acceleration mathimatically impotent and equal to velocity. And there you are, back to momentum again.

Jerk is not junk science, it's an estabilshed scientific and engineering term--very important in engineering since it describes the momentary loads components are subjected to when acceleration changes rapidly. To deny the existance of jerk is to deny that acceleration can change. Jerk is a pretty rarely encountered unit, but it's there, and you use it every day when you push the throttle down to increase your rate of acceleration to get through a stoplight before it changes--or alternately when you push harder on the brake because the person in front of you stopped quickly. In both those cases you are changing the rate at which you are accelerating, which is the third derivative of change in position, f/s^3. As for Jon Sundra, never heard of 'em. scot 19:04, 5 October 2006 (UTC)

Correction noted. Mr. Sundra invented the "Zerk", with a Z. However I beleive F/S³ still has no place within the narrow deffinition of recoil as we apply it to small arms. The "jerk" may have a place within the engineering of a firearm and its componnets to ensure proper functioning, but I don't think it apples to the gun/shooter system.Greg Glover 21:58, 5 October 2006 (UTC)

I'll have to take your word on zerk, the only references I see are to a movie Zerk the Jerk and to a type of grease fitting. As for jerk and perceived recoil, I'm not sure it's of any consideration or not on the timescale involved. I'm sure that the duration of the recoil force does have an impact on the shooter's perception; however that duration is more a function of the gun interacting with the shooter than the chamber pressure interacting with the gun. This goes back to the issue of shooting a shotgun with the stock held an inch from your shoulder, vs. correctly snugged into the "pocket". The first gives you all the recoil at once, because the gun is moving at full speed when it hits you, while the latter gives you a longer period of time to absorb the recoil--or maybe it just makes the gun effectively heavier, meaning less energy ends up in the equation. I'll have to think on that one, and calculate the compressions required to make a difference... At any rate, this is going off into original research, so is not suitable for the article, but there is plenty of literature out there on making a "softer" load. scot 22:15, 5 October 2006 (UTC)

Also, I ask you, please do not attempt to define “felt recoil” as it is known in small arms ballistics. Science may have created a computer model for felt recoil using a Cray computer but I am not aware that any simple, algebraic formulae that exists to express this function.

Nope, not even a Cray will help you, because felt recoil is subjective. All you can do is observe and form rules of thumb that say "for most shooters increasing factor X increases felt recoil." The terms "hot", "warm", "cool", and "cold" are also completely subjective and therefore undefinable, but that doesn't mean they're not useful, and that there aren't rules of thumb for them.

I understand you wish to apply Newton’s first law of motion to recoil which is what conceives the law of Conservation of Momentum. Newton’s First Axiom is stated as “Every body continues in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by force impressed upon it.” Newton’s first law is stated mathematically as PE = KE.

Thus you are lead to believe (incorrectly) that recoil is a function of the third law because the third law is derived from the first (as mathematics dictates). Newton’s Third Axiom is stated as “To every action there is always opposed an equal reaction: or, the mutual actions of two bodies upon each other are always equal, and in the direction to the contrary parts.” Newton’s third law is stated mathematically as ma = ma.

Isn't that exactly what happens when you fire a gun? Bullet goes one way, gun goes opposite way? And if you plug time into ma = ma, to get the final velocity, don't you get momentum?

However, a definition in English (be it the first or third law) dose not make Conservation of Momentum any way shape or form a part of recoil.

Then how are you calculating the recoil energy of a gun? Start with something simple, the 2.25 lb. 1911 pistol firing a 230 grain bullet at 850 fps. Here's how I do it. First, conserve momentum using mv = mv

I will show you my math just below yours in italics. Maybe then we can come to a concensus.

230gr / 7000gr/lb / 32.2f/s^2 * 850f/s = 2.25lb / 32.2f/s^2 * v

(230gr * 850f/s / 7000gr/lb) + (7gr * 5200f/s / 7000gr/lb) = 2.25lb * V

Your units don't match up--you need to divide by the acceleration of gravity to get from pounds (a unit of force) to slugs (a unit of mass). Also, you assume all the powder gasses go straight forward, an assumption that can easily be disproven by placing an empty 50 round .22 LR cartridge box over the muzzle and firing the gun. Much of the powder is diverted perpendicular to the muzzle as the bullet leaves, as evidenced by the .22 LR box disentegrating (assuming you're firing a .45 ACP, which has significant residual chamber pressure at the muzzle). If you consider only the action that happens before the bullet exits the muzzle, then the average velocity of the powder gas column will be 1/2 the muzzle velocity of the bullet; the powder at the breech is not moving, the powder at the base of the bullet is moving at bullet velocity, and the powder in between averages out. scot 19:04, 5 October 2006 (UTC)

Rearrange and cancel stuff:

(0.0329lb * 850f/s) / 2.25lb = v

33.13f/s / 2.25lb = V

12.4f/s = v

14.72f/s = v

Now, given that we have conserved momentum, we can then solve for kinetic energy with ke = .5 * m * v ^2:

Now given that we have found the velcity, we can solve for the transitional kinetic energy equation:

tke = mv² / 2g_c .

g_c is equal to 32.163 lb_m * ft / lb_f * s² .

ke = .5 * 2.25lb / 32.2f/s^2 * (12.4f/s)^2

tke = 2.25lb * 14.72²f/s / 64.326 lb_m * ft / lb_f * s²

ke = 5.37 ft-lbs

tke = 7.58ft-lb_f

Now, what are you doing differently?

“Now what are you doing differently?” What I do, is to take into consideration the powder charge weight and use a mean standard for the powder charge velocity. I use 5200fps. In theory 100% of the powder charge when detonated is given up to the system. If I had the actual energy per grain for the powder its self I could give an exact measurement of the recoil energy. However, only Vita Vorhi gives out that information so its not worth my time. I also use the correct equations for finding both velocity and energy. I use 32.163 as the dimensional constant (g_c) rather than 32.2.

No--the laws of thermodynamics say that you cannot get 100% conversion of energy, something is always lost--otherwise rocket engines would bet 100% efficient. Much of the energy of combustion is lost in the form of heat transferred to the bullet, cartridge case, and barrelm, the light of the muzzle flash, the deformation of chamber, case, and bullet, and the sound of firing. As for 32.163 vs. 32.2, since the bullet weight will vary by a couple of percent, you're just making more work for no additional precision. Also, Earth's gravity varies from 32.215f/s^2 in Oslo to 32.083f/s^3 in Mexico City, so technically 32.2 is overly precise. scot 19:04, 5 October 2006 (UTC)

As I understand your perception of recoil is based in momentum. I perceive recoil from the basis of energy. Work with me here for a moment; please. The definition of energy is the quality to do work. So with that in mind let me use this analogy: When a Remington 1100, cycles its shotgun shells, who dose the work? And how dose the bolt recoil? Well it's not by me, because the shotgun is an autoloader. I don’t think it the velocity of the expanding gases that are cycling the shells and recoiling the bolt either; which I believe is mv. I perceive the work of cycleing shotgun shells and recoiling the bolt, as being done by the pressure (p) from the expanding gases being exerted on the lower receiver piston; which I believe is F/d². I beleive force though distance (whether a product or quotiant)is a true and correct use of energy (see foot-pound force ). Therefor, recoil is a function of energy. Do you concur?

Does the energy taken out to cycle the bolt change the recoil? Assume for a moment a closed gas system, so all the gas used to cycle the bolt keeps going out the end of the barrel, rather than being vented after pushing back the piston. Another thing to consider is wet suppressors; look at how much water or grease can be vaporized by the energy of a shot; does that decrease teh recoil?

Look at the mechanics of elastic collisions vs. inelastic collisions; one conserves energy and momentum, the other only conserves momentum, with the excess energy being dissipated as deformation of the objects, or heat, or sound, or some other form. For example, consider the ballistic pendulum, the poor man's chronograph. Take a 5 lb. lump of clay hanging from a string, and shoot it. Momentum is conserved, therefore you can calculate the velocity of the bullet from the height (which is related to the speed) of the pendulum swing. Energy is not conserved, but rather is dissipated by the forming of the large crater in the clay. scot 19:04, 5 October 2006 (UTC)

Also if you notice the expanding gases were both propotional as in mv which is equal to velocity and exponential as in F/d² which is equal to energy.

Greg Glover 04 OCT 06

To all wiki, community members, who may have read the forgoing exercise I did in bullet-points. I ask for your peer review. I understand, that some of you may think to your selves, “big deal, the guy added the mass for the powder charge and came up with a couple of extra foot-pound force.” I humbly ask you this, how can an equation yield a value of ft-lb_f when all the other values are: f/s, lb and gr? This is the crux of my mathematical argument. Please consider the true and correct use of the dimensional constant as proof of my knowledge of the subject. I ask you to consider that recoil is a function of energy not momentum? I also ask you if I may add to the article and if so how? Finally if the community feels I have not met the burden of proof I will withdraw from this discussion.

Greg Glover 04 OCT 06

We seem to be arriving at the same answers, but I have no clue how you are getting there if you aren't conserving momentum. You cannot be using straight f=ma because you don't know the force--it's not at all constant, but looks like the curves shown in internal ballistics--I know this, because I based my internal ballistics simulation on emprical pressure transducer evidence from 7.62 and 5.56 NATO testing. The only way to guesstimate force is to reverse engineer it from f = ma; knowing the mass and the velocity, you can calculate the average force--however, this average force is completely ficticious, because it assumes a flat acceleration curve, which will never happen (unless you know how to solve the rail erosion problems with rain guns).

I would say, that I can’t go straight to F=ma using the two part method because I don’t know the “velocity”. “F” is what we are figuring for (in reference to KE). However when employing the quadratic equation, one can go strait to F=ma without “conserving momentum” because ma is know in the projectile and powder charge.

As for the rest of you paragraph, I’m lost to the connection between recoil and internal ballistics. Recoil would fall into the category of exterior ballistics, such as the projectile that just came out of the other end of the gun; Newton’s third law: ma = ma.Greg Glover 18:59, 5 October 2006 (UTC)

It is clear to me after reading your rewrite proposal and response to my response, you lack the basic theoretical understanding Newton’s laws of motion in general and to specifically to comment on recoil. Again, this is in the area of physic in general and concerning ballistics specifically. If you wish to formulate an explanation of recoil in general or as a mechanical function (eg a spring under tension); I have no complaints. However, if you wish to continue to express the “backward movement in firearms” as a function of Conservation of Momentum, I will have to make a formal complaint.

So why are firearms not subject to the same laws of physics as, say, a crossbow, or an airgun, or a railgun, or a rocket? These all work by pushing some mass in one direction, and recoiling in the other direction. In the case of the firearm, you also have to account for the powder mass, and in the case of recoil operated firearms the recoil acts on a component that is loosely coupled to the frame by a spring at the moment of firing, but physics is physics, it applies equally to all.

I don’t think you understand, I always take into account for the powder charge. All of the items you have mentioned are covered under physics, yes. However I am trying to keep you focused on the transitional energy of motion: chemical, muzzle, remaining, terminal and recoil. I can not have a meaningful discuss of recoil if you keep introducing things such as electromagnetism, propulsion and mechanical operations.

I feel at this time you have no scientific basis in which to write a definition. If you do, show me your work and references.

Fine, go read internal ballistics, external ballistics, recoil operation, muzzle brake, iron sights, hull speed, wave making resistance, point blank range, and wildcat cartridge, all of have me as a major contributing author. I have had college level physics courses, I can explain the relativistic effects on mass, time and length, I can tell you how the kinetic energy formula is derived from F = m * a, and I have written both internal and external ballistics simulation software.

My references are: Sir Isaac Newton, F=ma; Christiaan Huygens; Robert Hooke; Loard William Thomas Kelvin and Marquise du Chatelet Gabrielle-Emilie le Tonnelier ee Breteuil . My work is published in the book Terminal Performance. I show all the mathematical proof for Kinetic Energy, Transitional kinetic energy, foot-pound force and recoil. All the math in the book is complete and all computation were done by myself. All references are included. Also I am the only person that I am aware of since “du Chatelet” to publish all the mathematical computation to prove Kinetic Energy.

What's so hard about kinetic energy? Start with f = m * a, and integrate with respect to time; that's a part of all freshman level physics courses. Here, I'll do it for you...

It is obvious you know your mathematics. In fact you know it better than me. Your style is reminiscent to my fathers. He was an aerospace engineer. I am assuming you are an engineer too. I think this is where the problem lies. Your level of understanding is way above the subject. Kinetic energy is an easy subject. May be you can figure KE straight from the old classic statement but I don’t use the metric system and its three Units of Measure (UOM). I like everyone else that picks up a reloading book or shooting magazine uses the English Engineering System with its true and correct 4 UOM. That is why all computations of bullet energy and recoil energy are in foot-pound force.

f = m * a

Since it's an equivalence, we just have to look at one side. Start with the units for force; we'll use SI KMS units:

kg * m / s^2

Integrate with respect to time

1/2(kg * m / s)

I ask you sir, (and this is rhetorical, I already know the answer) how did you get from F=ma to ma then say “Ingrate with respect to time”, without violating am myriad of orders of operation and get to ½(mv)? By the way, where did that ½ come from (that’s rhetorical too)? The answers are in my book and you can check Foot-pound force within Wikipedia to see that I do know the answer. But maybe you can answer those questions in your own words?

I know you say it's rhetorical, but I'm going to go over this, because I'm not sure you're following the math. I'm going to leave out all the derivations of the fundamental theorums of calculus (it's been too many years, and it's spread over many pages of my calc book), but it goes something like this:

The derivitive of the function y = x^2 with respect to x is a function which describes the slope of x^2 at any given point. To calculate the derivative, you multiply the equation by the power of the variable in question, and reduce that power by one. So, y = x^2 becomes y' = 2x (technically y=x^2dx, but the dx is just a mathematical reminder to account for the fact that derivatives drop constants; i.e. the derivative of y = 10 + x^2 is also y' = 2x, because both curves have the same slope at any given x value).

The integral of a function, which describes the area under the curve of that function, is the reverse of the derivative; that is, integrating the function y' = 2x should give us y = x^2. This means that if we integrate the function y = x with respect to x, we have to increase the power of x, and divide by the new power, hence we get y' = x^2 / 2.

As for how did I get from "F = ma" to "ma", if F and ma are equal, then anything I do to one side of the equation will have exactly the same result on the other side. Force IS mass times acceleration, if I express them as units, then I get this:

kg * m/s^2 = kg * m/s^2

The point is, if I take the area under the force curve, by integrating with respect to time, I get (kg * m / s) / 2. Another thing that you might not have noticed about momentum is that the units work out to be pound-seconds or newton-seconds, the same units used for determining rocket thrust. Since a gun is just a really short duration rocket, with really dense exhaust, the unit does make sense. scot 19:29, 5 October 2006 (UTC)

And you get kilogram-meters per second, which is the unit for momentum, look it up. Conceputally what we just did by integrating with respect to time is to take the area under the momentum curve.

The first two sentences in “recoil” are: “The recoil is the backward momentum produce by a gun when fired. It is equal to and opposes the forward momentum of the projectile.” Both these sentences are patently false.

OK, I will correct that. It should read, straight from Newton, "The recoil is the backward force produced by a gun when it is fired. It is equal to and opposes the force acting on the projectile. It may be calculated using the law of conservation of momentum".

So at this time, I implore you to remove your explanation of recoil. If you wish to argue with the above preeminent scientist and mathematicians concerning energy and force, then you can take up my protest with the powers to be, within Wikipedia.

Well, let me just point out that all of said classical physics scientists are long dead, and their work has been shown to apply only to narrow realms. The universe is not continuous, time and space are not constant, and there are limits that were never suspected by the classical physicists. scot 21:46, 3 October 2006 (UTC)

No Sir, classic physics is alive and well here on earth. The concept is just too simple for today engineers and scientist to be concerned with.

No, it's not too simple, it's just limited to low velocities and low accelerations. You can use the relatavistic functions on firearms ballistic problems, and the Lorentz factor will always give you a value of .99999999 or some other value so close to one as to make no difference whatsoever. My point is that Newton et. al. were not the end-all and be-all of physics; they got hte answers right--or close enough--for the situations they were dealing with, but there were still a few minor issues that didn't quite work out, like Mercury's orbit. scot 19:04, 5 October 2006 (UTC)

Greg Glover 20:00, 3 October 2003 (UTC)

Have made an attempt at a clean-up of the physics involved, and also have tightened up the article at the same time. And yes, classical mechanics is alive and well today. Now, if we want to write about felt recoil in a shooting range located at the edge of a black hole, or near another intense gravitational field, then it would really get interesting :-) Yaf 04:24, 4 October 2006 (UTC)

Actually, reguarding relatavistic momentum effects, there are a few devices where that actually can have an impact (forgive the pun). Particle accelerators are currently avaialable, and do move projectiles at relatavistic speeds, and the ion drive is basically a particle accelerator used as a rocket engine. If you can move the particles at velocities where the mass dilation becomes significant, you can get away with far, far less propellant mass than would otherwise be needed. Granted, this should go in a special section, since there's no need in calculating the infantesimal values yielded by the Lorentz transformation for anything less than a particle accelerator or the like. scot 14:11, 4 October 2006 (UTC)

I like what I see, very good. Not perfict but doable. I'm not sure if I can but in the spirit of good wikietiqutt I will remove the dispiut flag I put up.

Well I geuss someone eles did ti for me.

Good morning community. You know, I love a good heated discussion. If you read though this discussion, you will see that I believe recoil is a function of energy. The mathematical proof is the original kinetic energy equation of KE=wz which is found on the Foot-pound force page. Is it not true that any body in motion contains energy because of that motion? How you find for that motion is up to you. That motion can be measured in two forms: velocity; feet per second and energy; foot-pound force. Once you know the velocity of a firearm during recoil, you can then find for its transitional kinetic energy (velocity of the fire arm is not need when employing the quadratic equation). However the community over at the kinetic energy article never bothered to have a discussion (mathematical proof and references provided) they just cooped my article. I hope that do not do the same to this article since it is linked over there.

Again, I'm going to go back to "the map is not the territory". I see recoil as a concept, which is not dependent on any units. The units are merely a formatlity used to describe a phenomenon, and are chosen for the particular task. For example, if I ask how the recoil of a 240 grain .44 Mag at 1500 fps compares to a 115 grain 9mm at 1200 fps, how do you express it? If you don't know the firearms in question, you can't express the recoil in terms of energy. The logical unit choice in this case (in my opinion at least) is momentum--not having specified the firearm, the recoil energy cannot be calculated, but given the momentum and the firearm mass, you can calculate the speed and from there the energy. Giving someone recoil values in kinetic energy means that they need to have additional information, i.e. the mass of the firearm used for the measurement, to be able to convert that to a different gun.

Also there seems to be some annoyance to the use of English Units of Measure. Every time I put something up in English Units of Measure someone changes it to SI. Can someone here in the community explain why? I am dumfounded because all “kinetic energy” tables here in the U.S. are in English Units of Measure. I know this site is for everyone and most of the world uses SI, but I have never treaded in others areas of physics other than that which pertains to the open system of large and small arms. I am not even concerned with recoil suppression systems attached to small and large arms.

This website, while hosted in the US, is intended for and is used by the global English-speaking community, of which the US is the only member that has not officially switched to SI units. The official stance is that units should be given in SI units, which is the defacto global standard, in addition to any other units applicable. When working in ballistics, I just don't have a "feel" for SI units, so I can't intuit when I've made a mistake in calculations, so everything I write I write in English units. Ideally someone will then come along and add SI units (usually User:Bobblewik) for those who aren't conversant in English units. As for changing English to SI, that probably shouldn't be done with singificant justification. Now if there was an article that was all in SI and you added a bit that was in English, then it should be converted to make it consistent.

Thank you for that explanation. No I have never added English units to any site containing SI units. The explanations I am giving are specific to those that are English based. That is because the original English system (concerning energy) uses 4 units of measure not 3 units of measure found in SI. So for this article I can use SI but I will still argue (technical use of the word as in a mathematical argument) recoil energy in English units.

So maybe someone in the community can tell me what nerve I hit. Especially when I provide the math and referances?

Well, when discussions get sprited, then it's often best to just step back and wait a while, and maybe go find another editor who has contributed to similar subject matter to help arbitrate. Often all it takes is a bit of change in perspective to reach an agreeable compromise. scot 16:43, 4 October 2006 (UTC)

Greg, as a newcomer you probably aren't aware that the normal way sign and datestamp your talk page (not article, however) contributions is easy; all you need to do is add four tildes (like this ~~~~) at the end before you hit save. Gene Nygaard 09:17, 5 October 2006 (UTC)

Thank you, I'll try it.Greg Glover 15:00, 5 October 2006 (UTC)

[edit] Expert assistance needed

The article several times mixes up force and momentum.

If the momentum of the escaping gasses and propellant is significant, there is not enough information to "calculate" the strength of the recoil. The article Physics of firearms completely ignores this and considers only the bullet versus gun-shooter.

Force is clearly not a good measure since you must integrate it over time to get a meaningful result. What I don't know if, in discussing recoil, one should always take gun+shooter as one entity, or separate the effect on both. For the same momentum imparted on the bullet, the shooter will experience less of a "kick" if the gun is heavier. --Lambiam ^Talk 17:10, 5 October 2006 (UTC)

Yep, it did. Have attempted to clean up some of the confusion among F = ma, p = mv, a = (d(v(t))/dt), and F = dp/dt. Still am not entirely happy with this article, as I fear it is not really suitable for a general purpose introduction to recoil, and, as you point out, the force would have to be integrated over time to get a meaningful result. It may be better to simply start over on this article. Yaf 04:10, 6 October 2006 (UTC)

As I promised, I would find the long or short form for the free recoil energy equation. I found the short form. As I said, with either of these forms you don’t need to find the momentum of the gun to find the recoil energy. Again recoil is a function of energy (mv^2) as seen in this second term of the following equation. Again I have shown proof that free recoil is a function of energy.

Free Recoil is; RE = (1/2gcGw)×(((BwBv)+(CwCv))/7000)^2

Where as: RE is the free recoil energy in foot-pounds force, gc is the Dimensional constant (I use 32.163 and is not the acceleration of gravity), Gw is the gun weight in pounds, Bw is the bullet weight in grains, Bv is the bullet velocity in feet per second, Cw is the powder charge weight in grains and Cv is the powder charge velocity (I use 5200f/s)

Again most of what is written in the first, two subsections of this article is fundamentally incorrect. Also for use in this article I can convert the above equation into SI units of measure.Greg Glover 02:56, 17 October 2006 (UTC)

What are the units of gc? scot 14:27, 17 October 2006 (UTC)

OK, let me derive your formula for you. First, start with the formula for conservation of momentum; m is mass, v is velocity, _g for gun and _p for projectile. The rearward momentum of the gun is equal to the sum of the forward momentum of projectiles, powder gas, and any other mass discharged.

m g v g = \sum m p v p

Now, since we'll need this in a bit, solve for the gun velocity:

2. $v_g = \frac {\sum_{} m_p v_p} {m_g}$

Now we take the formula for kinetic energy, E_k, of the gun:

3. $E_k = 0.5 m_g v_g^2$

Now replace v_g with the equal quantity from equation 2.

4. $E_k = 0.5 m_g (\frac {\sum_{} m_p v_p} {m_g})^2$

Now simplify; (a/b)² equals a²/b², so...

5. $E_k = 0.5 m_g \frac {(\sum_{} m_p v_p)^2} {m_g ^2} = 0.5 \frac {(\sum_{} m_p v_p)^2} {m_g}$

And, since you started your equation with gun weight in pounds and projectile weight in grains, and you need mass in slugs to ge the units to work out, multiply the entire equation by 32.163 pounds per slug and divide the projectile weight by 7000 grains per pound:

6. $E_k = 32.163 \times 0.5 \frac {(\frac {\sum_{} m_p v_p} {7000})^2} {m_g}$

I think you'll see that your formula increases the free recoil energy as the gun mass increases, wheras it should, as my formula shows, decrease with increasing gun mass. scot 17:00, 17 October 2006 (UTC)

Talk:Recoil/Archive March 07

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