Talk:Recoil

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)

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?

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)

Greg Glover: Formost expert on transitional kinetic energy 2:16 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?

Feel free to continue this discussion here or IM greg_glover12@yahoo.com

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...

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

With all due respect Sir, I don’t know where to start with you. You have just violated two mathematical rules, the rule of associative properties and the rules regarding orders of operation. Also Acceleration is equal to velocity divide by time or distance divided by time squared; as in feet per second per second. Acceleration is stated mathematically as: a = v/t or d/t*2.

Since when is multiplication not associative? And if a = v / t then is not also v = a * t?

!!!!!!!!!!!!!!!!!Please abandon your explanation of recoil!!!!!!!!!!!!!!!!!!

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)

Greg Glover 21:01 3 October 06

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)

Rewrite proposal

Here are the topics I listed above, plus the existing article topics, in a possible outline:

• Definition of recoil
• Recoil vs. energy (probably change to "momentum vs. energy")
  • Recoil is the reaction, and is a function of momentum
  • Ways to quantify recoil
  • Recoil energy can be calculated if the firearm mass is known
• Perception of recoil
  • Psychological effects
  • Controllability
  • "sharp" or "snappy" recoil vs. "soft" recoil
  • Bore axis and torque
• Recoil in mounted guns
  • Rigid vs. soft mounts
• Controlling recoil
  • Tie in muzzle brakes, along with examples above of powder/bullet masses.
  • Shotgun recoil absorbing stocks
  • Recoil absorbing materials (sorbothane, neoprene)
• Myths

As for definitions and units: I think "recoil" should be defined as the phenomenon that results from conservation of momentum when a projectile is launched. Since recoil is a function of momentum, I think that if we're going to pick a unit for "recoil", it should be mass times velocity. "Free recoil" is a measure of the recoil of a particular firearm/cartridge combination. This can be, and most often is, expressed as kinetic energy, or mass times velocity squared, with velocity being the velocity the firearm would reach if hung from string and fired.

As for perception issues: I think the difference between "soft" and "sharp" recoil, for two loads with the same free recoil energy from the same firearm, has to do with acceleration. A light, fast bullet will accelerate the gun faster than a heavy, slow bullet, and this means that the sharp load will accelerate the gun to a higher percentage of the free recoil velocity before the shooter's body starts to slow it down (since people are squishy, it's going to have room to accelerate). A soft load accelerates slower, and therefore will have compressed the shooter's body before it reaches such a high velocity, and will therefore won't reach as high a percentage of the free recoil velocity. Because of this, neither momentum nor energy is sufficient to quantify a shooter's perception of recoil, but the duration of the acceleration has to factor in there as well. scot 15:00, 3 October 2006 (UTC)

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. 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.

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. 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 rendere acceleration mathimatically impotent and equal to velocity. And there you are back to momentum again.

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

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

Rearrange and cancel stuff:

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

12.4f/s = v

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

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

ke = 5.37 ft-lbs

Now, what are you doing differently? 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).

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)

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.

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)