Does Momentum Equal Stopping Power? Let’s Find Out!

PHOTO_20160430_1516532

In a recent post, I discussed four ballistics myths that I’ve heard over the years, and why they are just that – myths. One of these was the myth that the momentum of a projectile is equivalent or otherwise indicative of the stopping power of that projectile. I have for several years now been arguing that it is not, and the subject comes up again and again in my comments, so it’s high time I directly addressed it in a post. The short answer is “no”, as I wrote in “4 Ballistics Myths”:

4. Momentum = Stopping Power

Momentum, that is the product of mass times velocity, is a very easy-to-grasp physical quantity. A bigger man bumping into you on the street at the same walking speed will push you aside more violently than a smaller man. A larger rock or stone thrown with the same velocity will make a bigger splash in water. It’s a simple quantity that’s easily calculated and readily understood. The bigger something is, and the faster moving it is, the more oomph it has.

That’s why it’s very natural for momentum to be used as a rough metric for stopping power. This connection is made all across the gun world, from gun reviews noting that a bigger round rings steel more loudly than a smaller one, to the Taylor Knock-Out Index which combines momentum with the diameter of the bullet in an attempt to quantify stopping power against big game. However, although momentum is an important quantity in ballistics, it does not imply the degree of terminal effectiveness, or “stopping power”, that a projectile has.

Momentum is a conserved quantity, which means that as a bullet is driven forward by the force of expanding gases, the gun firing that bullet is also driven rearward with the same momentum as both the bullet and propellant gases combined. This means that the momentum produced by any bullet fired from a shoulder or hand fired weapon is not enough to significantly wound a human being, much less kill. The momentum of a bullet, when it strikes a target, does not do more than perhaps somewhat bruise the surrounding tissues and very slightly accelerate the target rearward.

This should be self-evident: Since momentum is conserved, if it were directly correlated to “stopping power”, then guns should be just as deadly to the shooter as they are to the target. Guns are not deadly to their shooters, Q.E.D., momentum is not correlated with stopping power.

Is it as simple as that, though? Several of my commenters thought that other factors might be in play, and one of the most common objections to this simple proof is that bullets have a much smaller frontal area than gunstocks do. Surely if you shot a rifle that had a buttplate surface only as wide as the bullet it fired, you’d severely injure yourself shooting it, right?

Yesterday at the range, I tested this problem to settle it for good. I took my Ruger 10/22 rifle* and modified it with a little field gunsmithing tape, a block of foam and a .22 caliber cleaning rod. The foam served as a solid base for the cleaning rod, and was taped firmly to the stock of the rifle.

*I should note that I chose this rifle not because it’s a puny .22 LR, but because it has a thick, synthetic stock that would give good purchase for the tape, and which wouldn’t be potentially damaged by it. I mostly own wood-stocked milsurps, so this was really the only rifle I own that fit the bill.

The cleaning rod was stuffed in the rear until it contacted the rifle stock, so that all forces produced by the gun’s recoil bore directly onto the .22 caliber cleaning rod. I also taped the action of the rifle shut, so that the force of recoil wouldn’t be absorbed by the rifle’s action spring. Finally, because experimentation is nothing without documentation, I filmed the whole thing:

So what happened? Well, I’ll disappoint quite a few people by saying this, but, not a whole lot. Despite getting the result I expected, even I was surprised, as I hardly even felt the cleaning rod against my shoulder during the shot. The magnitude of the effect of the rifle’s recoil was so low, in fact, that I actually feel confident that I could perform this test with much larger weapons (such as the No. 4 Lee-Enfield hiding in the brown case on the table in the video) without serious discomfort.

So this test proves to my satisfaction that momentum isn’t correlated with stopping power (such as that term is meaningful). One of the sources of confusion here, I think, is that kinetic energy – which is clearly a very important factor in terminal effectiveness – can be described as one half times momentum times velocity, but this does not mean that momentum itself, when not multiplied by those other factors, directly has an effect on the “stopping power” of a bullet. In fact, this test not only refutes that idea, but supports the idea of energy being the most relevant physical quantity, as the major physical difference between the projectile flying downrange and the cleaning rod up against my shoulder was their velocity. So the velocity seems to be the more important quantity here, although mass is not unimportant.

Hmmm, if only there were a physical quantity like momentum, but one that emphasized velocity more, maybe by, like, having it in there twice?

kineticEnergy

Though, energy by itself is not a direct indicator of terminal effectiveness, either, as I have said many times, but it – and especially its transference to the target and how it is applied in a gunshot as work – is a key ingredient to the “stopping power” recipe. Although, that is a topic for one of my Ballistics 101 posts.

To conclude, this test may be disappointing to those of my readers who expected a result more like:

PHOTO_20160430_151554

but I also hope that you feel like you learned something by reading this article and watching the video. And of course, I hope this finally puts to rest the matter of momentum and its relationship to stopping power… Although that doesn’t mean I don’t still want to hear what you all think in the comments!



Nathaniel F

Nathaniel is a history enthusiast and firearms hobbyist whose primary interest lies in military small arms technological developments beginning with the smokeless powder era. In addition to contributing to The Firearm Blog, he runs 196,800 Revolutions Per Minute, a blog devoted to modern small arms design and theory. He can be reached via email at nathaniel.f@staff.thefirearmblog.com.


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  • Anonymoose

    Try that with a Mosin-Nagant and get back to me (when you get out of the hospital).

    • marathag

      Well, without energy transfer all you have is a 5.7mm hole, and the top half of prairie dog isn’t reduced to pink scraps, like other 22 caliber rounds tend to do when they hit a small bag of meat at 2300 fps

      • MarkVShaney

        That’s a function of friction.

        • marathag

          friction heats, no denying that, but its not a steam explosion.
          You don’t set telephone books on fire when shot, either.

          it’s around 2,300,000 joules of energy to make one kilo of water to steam

          .223 has around 1600 joules. So even if it dumps all it’s energy from KE to heat, it’s not even enough to make one gram of water flash to steam.

    • How is energy “nebulous”? It’s .5 * M * V^2. Pretty straightforward.

      And woulda lookit that, that equation for energy seems to perfectly describe this quantity of “momentum times velocity” you hint at!

      • gyrfalcon

        So repeat your experiment with a mosin.

        • Naw, I’m not man enough. As you can plainly see, I’m one of those doughy sub-male millenial types who writes lengthy Tumblr posts about how hard my life is because I got a hangnail.

        • Jwedel1231

          Since the mass of the gun vs the bullet is so much bigger, and the momentum would be the same (see article above), that means the velocity would be a lot slower. Plug that near-zero (compared to the bullet) velocity into .5*mass*v^s and the whole energy equation for the rifle approaches zero.

          *edit* I really would like to see someone shoot a decent rifle caliber w/ a protrusion (cleaning rod, for example) the same diameter as the bullet using a gel block instead of a shoulder, then compare penetration of the cleaning rod vs the penetration of the bullet into gel.

      • Anonymoose

        It’s nebulous because no one seems to be able to explain how this “energy” kills a living thing. Hydrostatic shock doesn’t real. .22LR is just as lethal (on humans and small game) as 9mm given good shot placement, yet people harp on and on about foot-pounds as if it’s this extremely significant number that you absolutely have to have to have an effective shot. Shot placement is king, penetration is queen. All else is just nitpicking.

        • Oh, that’s easy. It has to do with work; I’ll explain when I get home and am not on mobile.

        • marathag

          Blood is a slightly compressible fluid.
          Impact causes displacement. And incompressible matter is moved into matter that is more compressible.

          This continues till the projectile decelerates to where it has no more KE to shed, or exits the body.

          That moving around not good for that compressible matter, will cause loss of cohesion.
          That loss of cohesion in tissue is what is your gunshot wound damage.

          Damaged Tissues leak out that red stuff as the circulatory system is degraded, and all the synapses in that vicinity fire off charges that end up in the brain saying ‘Houston, We have a problem’

          Since most brains aren’t as well ordered as Jim Lovell, most do the equivalent of bouncing off the walls, pure panic, and the Brain just starts shutting systems down.

        • I think the real issue is that lethality is different than “stopping power.” An airgun pellet filled with Ricin (think the ol KGB umbrella gun) is more lethal than a .308, but its stopping power is 0.

          The vast majority of “stops” have nothing to do at all with lethality; 82% of the people shot with handguns survive according to one stat I read. Yet we’re not having 82% failures to stop.

          According to “Handgun Wounding Factors and Effectiveness,” the seminal work that has driven pretty much all handgun ammunition design for the last 30 years, psychological factors are far and away the most important aspect of a “stop.”

          Lethality and Stopping Power are almost unrelated in terms of a “meaningful combat interval” for defensive handgun use; even with a shot through the heart, there remains enough oxygen in the brain to continue fighting for 10-15 seconds. Blood loss / “big hole” theory also falls short because you have to lose 1/3 of the blood in your body (about 1.3 liters) before passing out. This takes a lot longer than the typical gunfight lasts.

          In short, unless the round strikes the central nervous system, then lethality/wounding is largely irrelevant in a 3-10 second, close range handgun fight.

          And thats where energy and temporary stretch cavity is so important, as it leads to a higher probability of a psychological stop. The more energy/stretch cavity, the more pain neurons are activated, and the more likely the attacker will realize they have been shot, and that getting shot sucks.

          My belief is that “i’ve been shot” psychological factor due to energy transfer is why JHP’s have proven so much more effective than FMJ, rather than their enhanced lethality.

          And now I realize I’ve written an essay, sorry!

          • ostiariusalpha

            Kind of short for an essay, but you are not wrong. Marathag has a bit more insight though, most stopping power (that is not due to direct damage to the CNC) comes from a baroreceptor reflex that decreases blood flow to the brain and causes near instant unconsciousness. The baroreceptors monitor and regulate blood pressure and if they detect a sudden spike in pressure, like from the temporary stretch cavity of projectile entering at a high velocity, they will try to compensate by dilating your arteries to drop pressure… on everything. Whether it’s a human attacker or deer in the woods, even if you blow a baseball sized hole in them, they won’t immediately drop without a baroreceptor reflex being triggered; and certain conditions, such as an adrenaline surge which raises blood pressure, can counteract the reflex temporarily. Likewise, you can trigger it without any actual blood loss, such as by stimulating the baroreceptor nerve clusters in the neck with a sharp blow, or by swallowing a chunk of pretzel that you didn’t chew properly.

          • randomswede

            “baroreceptor reflex” is that related to the “hydrostatic shock” that I used to hear thrown about?

            To the immediate drop, I was of the understanding that destroying the oldest parts of the brain or severing the nervous system above the shoulders are “lightswitch” kills (The FBI “T” I’ve heard it called). Would that be triggering the reflex or did you omit that as you were mainly referencing body hits?

            I’ve also heard of a heart shot dropping someone like a sack of potatoes, I assume that would be along these lines as what little biology schooling I have tells me the person should be “alive” for a second or so before the lack of blood flow starves the brain of oxygen.

          • ostiariusalpha

            The “lightswitch” kill you’re referring to is the direct CNS (central nervous system) strike that I mentioned as separate from a baroreceptor reflex. It is the heart shot that causes a DRT (dead right there) which is a baroreflex.

          • randomswede

            Sorry, I must not have been awake and completely missed the parenthesis. -_-

          • Thank you, that’s very interesting. Do you have any suggestions for where I could read up more on the barorecptor in relation to stopping power? Google is mostly pulling up web-md style articles on the subject.

          • That’s very interesting, can I see the paper?

        • Max Popenker

          living kill is made by disrupting tissue; this is a physical work, and amount of work the bullet can do is function of its kinetic energy
          Also, there’s no such thing as “Hydrostatic shock”, because we talk dynamics.

          • ostiariusalpha

            No, hydrostatic shock is very much a real thing proven by modern medical science; Fackler and other skeptics have been proven wrong pretty conclusively. If you mean that it gets mythologized by gun rags & ammunition marketing to exaggerate its ability to cause incapacitating wounds, you would be on more solid ground, but hydrostatic shock is an important component of inducing a baroreceptor reflex that can and will cause immediate incapacitation through unconsciousness.

          • Ostiarius, from what I can tell, the ARL and the goat labs would disagree with that assessment. Do you have any documentation to back it up?

          • ostiariusalpha
          • Kudos for having the sauce. I will read it.

          • ostiariusalpha

            Now for the conditional statement; as I alluded to before, nothing I’ve read demonstrates that the remote hemorrhaging caused by a ballistic pressure wave in any way supercedes the effect of the direct laceration from the bullet, even on a more energetic rifle round like the 5.56. But the hemorrhaging itself is quite real. Fackler wouldn’t even admit that a rifle bullet could damage bone that it hadn’t directly struck, despite that there was considerable evidence to the contrary.

          • Well, in Fackler’s defense, his original premise was that the tissue damaged by temporary cavitation can recover and doesn’t need to be excised as necrotic. As time went on, he seems to have become a victim more and more of his own navel-gazing, but much of what he wrote is still worthwhile.

            Just not gospel.

          • Max Popenker

            man, have you ever attended basic physics class in school?
            hydro-STATIC means static (non-changing, constant) pressure in liquid medium, and there’s nothing static about bullet passing through the body; the hydro-DYNAMIC shock (damage created by shock wave passing through liquid-filled internal organs), while a real thing, works only for really fast bullets.

          • ostiariusalpha

            Except that’s not what hydrostatic means, the static refers to the fluid’s non-flowing behavior. An interesting branch of Fluid Static (Hydrostatic) studies is concerned with differentiable pressure gradients and densities in a body of fluid while at equilibrium. In this case, the hydrostatic shock involves the movement of a pressure wave through through body fluids and fluid-filled tissues in a manner unrelated to whatever flow dynamics that those fluids might or might not be undergoing at any time. So calling the phenomenon “hydrostatic shock” is not really inaccurate, but it is not particularly elegant either, since many body fluids, especially blood, alternate between flowing and being at relative equilibrium from moment to moment. That’s probably why that term isn’t very popular with researchers any more.

    • Max Popenker

      wounding process is, basically, a job damaging or moving aside the tissue or shattering the bone; in physics, amount of work is equal to ENERGY used to make this work. So, capability of bullet to do some damage is based on an amount of kinetic energy bullet has at the moment of striking the target.
      How this energy would be used during the bullet passage through the target, depends on bullet diameter, design and target properties.

      • Wham bam, thank you ma’am, Max is on the case!

  • iksnilol

    I think these experiments are stupid. Why risk your life like that?

    You’re no better than those people who wear a bulletproof vest only to shoot themselves to prove that it works.

    • Anonymoose

      Do it underwater. It’s okay. 😀

      • iksnilol

        Are you referring to that grotesque suicide attempt by that Norwegian guy? I am still kinda disgusted they let the film on youtube.

        Here’s the video I am thinking about:

        • Bill

          Hey, it’s cool – he was wearing eye pro

    • MPWS

      Why bullet-proof west (up to class III) is not good against knife? Hmmm…

    • In what way did I risk my life, or are you joking?

      • Giolli Joker

        He was clearly joking.😆

        • Kevin Harron

          With iksnilol, it is a good thing to make sure. He’s said some pretty retarded things in the comments here.

          • MPWS

            I believe he’s ex East European. Different mind set; but he’s jovial and has good English. It’s challenge for you which does not hurt. 🙂

          • iksnilol

            ex?

            ONLY THING EX ABOUT ME… is a list of women I really should not have been seeing, that and my parents were born in a country which technically doesn’t exist anymore.

            Nah, but really, be careful around short blondes. And Norwegian money is not bad either. Those are life lessons I feel I can teach so far in life.

          • randomswede

            A language that makes everyone sound like they eat “happy pills” for breakfast cereal is a nice bonus too.
            Seriously I swear it’s the difference in between the mental health of the Swedes versus the Norwegians is language harmonics(?) to a major part.

          • iksnilol

            I never really thought about that. Then again, I live far from the Swedish border. Though there’s one guy who pretends to be Swedish.

          • iksnilol

            Wut?

            Every marine a machinegunner is a good doctrine. It’s worked for the Orks in W40K for decades.

      • iksnilol

        Well, if your math was wrong or something you’d have been killed.

        I mean, many peeps get killed with .22 LR, don’t underestimate its 100-150 or so joules of energy.

        • Jwedel1231

          That’s the problem with sarcasm. The better at it you get, the harder it is to see. I’m a big fan of sarcasm, so I know that pain all too well.

          • iksnilol

            Yup, we really need some irony punctuation.

  • Giolli Joker

    “The momentum of a bullet, when it strikes a target, does not do more
    than perhaps somewhat bruise the surrounding tissues and very slightly
    accelerate the target rearward.”
    I will repeat it: this sentence hurts you.
    If less informed people give such a value to momentum and they decouple it from kinetic energy, you’re not supposed to go on their ground, let them climb up on yours.
    Let’s define stopping power first… oh, wait, we can’t, as it’s not a measurable quantity.

    • Everything I write hurts my credibility with somebody. I’m not terribly worried about it.

      • Giolli Joker

        I see what you mean and I agree… however I know you can do much better than that sentence and I prefer to tell you.
        I hope the next topic will go in the depth of “what is stopping power” questioning the very meaning and existence of the same. 😉

        • I am using “stopping power” as a shorthand, only, for the sum issue of lethality. Yes, we can have a long discussion on the meaninglessness of the term “stopping power”, but on the other hand it’s still very clear what I mean, I think.

          • Giolli Joker

            I’m pretty sure most your readership has always had a pretty clear idea about what you meant with the word “caliber” as well.😉

          • Right, but I am starting from the ground up for that series.

          • georgesteele

            I think, actually, that the terms “stopping power” and “lethality” are NOT synonymous. I’ve maintained that what a hunter wants is lethality and good stopping power, but what a person defending him or herself wants is stopping power, and lethality be damned.

            The hunter needs to have the prey dead, ideally without having to track it all over creation. The defender needs to stop the fight NOW, so that he or she can safely beat feet – without taking another person’s life. Very different objectives – unless you’re a cannibal.

            Ideally, a defensive round would slam the attacker to the floor like a heavyweight’s haymaker, keeping the person unconscious for a long enough time for the defender to leave (or cuff the attacker), but with no permanent injury or risk of death. So if we are restricted to using a firearm for defense (and we aren’t) the real issue is: what kind of firearm round achieves a CNS or kinetic shock sufficient to accomplish that goal.

            I don’t know the answer to that, and there may not be one.

          • “Stopping power” doesn’t really exist as such, and it’s become common in military circles to substitute the word “lethality” for those basic concepts to avoid confusion.

            But I see your point, anyway.

          • Bill

            Excellent points. Interpreted one way, a TASER may have more “stopping power” than many, if not most, man-portable firearms.

          • georgesteele

            Yes – and now, thanks to the Caetano decision in MA, we may all be able to get them included under 2A right to carry (some states don’t). That would benefit everyone – state, attacker, and victim alike.

          • Marcus D.

            I agree that there are a large number of variables in determining the “lethality” of any given projectile. Lethality as best I could define it, would be the probability that any given projectile, when fired from a specified cartridge, will cause sufficient disruption of tissue to cause death, whether immediately from disruption of critical tissues, or through bleeding. Lethality can be accomplished in different ways. For example, the projectile can be lead of various densities, FMJ, HPs, ballistic tips, etc., each of which varies in the amount of projectile expansion–and the size of the resulting hole–after impact. Second is penetration, which for military rifles is also more or less meaningless, since most have enough “juice” to fully penetrate the human body, but which is of critical importance for hand gun rounds. Since military rounds generally must be FMJ, lethality is a function of velocity, bullet diameter, and for the .223/5.56, the ability of the round to yaw (tumble) after impact. The .223, for example, is so small in diameter that it causes very little tissue damage unless it yaws, but quite a lot when it does.

            I surmise that there really is no such thing as “knock down power,” aka “stopping power,” which is more likely a function of a projectile of sufficient kinetic energy striking solid tissue (bones) (or perhaps a ballistic vest), and transferring its energy to that solid tissue. When people use this terminology, they are almost always talking about lethality, and the (often presumed) power of a particular (handgun) round to cause fatal wounds.Of course, since there is virtually no difference in the wound characteristics of any fmj round between .38 and .45, there being after all only a negligible difference in diameter (.09″ to be precise), the difference in lethality has much more to do with penetration and bullet design, i.e., the ability of a given round to expand and make a bigger hole. Obviously, the more the expansion of a particular projectile, the more power is needed to shove it through tissue

            I suggest that one rarely hears these terms discussed with rifle rounds; for the latter, the discussion is always penetration, range, accuracy and wound channel. I recognize that some modern hunters believe they need to take a .300 Win Mag to the forest to hunt white tail for its greater “knock down power,” but truth be told, the .243 Winchester is more than sufficient for such game, it one’s aim is true. All the more powerful round does is decrease the amount of meat to harvest. But that’s just me.

      • MPWS

        You don’t need to. You brought this web to much higher standard academically that it has been before. People owe you respect.

  • Isaac Newton

    Perhaps a Fermi problem is in order: calculate the momentum of the projectile and rifle (should be the same)…then compare it against the calculation of the energy of the projectile and energy of the rifle (should not be the same). This may prove your point better.

    • I’ve tried the math approach before in my comments, it tends to make people’s eyes glaze over.

      • tts

        Its good that you try though. Even if most people’s eyes glaze over some will try to give it a chance. Personally I’m crap at math but I can usually worth through a algorithm or formula with a calculator.

        • I think I have to keep it to a minimum, though.

      • Ah, the MEGO effect: My Eyes Glaze Over

    • ExMachina1

      “Perhaps a Fermi problem is in order: calculate the momentum of the projectile versus the rifle (should be the same).”

      I agree, however he’s arguing in the video that they are not equal. Somehow, he thinks the momentum of the rifle is greater because the mass of the propellant gasses somehow only get added to the bullet side of the equation. That’s simply wrong and doesn’t make his “explanation” any easier to follow…

      • Hi ExMachina,

        The momentum of the rifle’s recoil is equal to the momentum of the ejecta, that means the projectile plus the propellant. The projectile is moving at about 1,100-1,200 ft/s, and so momentum equals that times its mass (34 gr, IIRC). The propellant is moving much faster, in the case of .22 LR, probably on the order of 2,000 ft/s, although I can’t be sure, and the charge weight is probably about 2 grains.

        So then we have a total momentum for the ejecta of 34 grains times 1,100 ft/s plus 2 grains times 2,000 feet per second, which gives us 41,400 grains-feet/sec, or about 5.9 poundsfeet per second.

        The rifle’s momentum should be equal to this.

        • ExMachina1

          “The momentum of the rifle’s recoil is equal to the momentum of the ejecta, that means the projectile plus the propellant. ”

          No. The momentum of the **system** before must equal the momentum of the **system** after (system=bullet, rifle and propellant mass) There is no reason to conclude (or even suspect) that the propellant mass contributes a vector in only the direction of the bullet In fact the propellant gas should contribute vectors of equal velocity in every direction so will essentially have zero net momentum after firing.

          My only point was that you are playing loose with the physics here (both the math and the terminology) so it’s not surprising that people are confused.

          • Yes, some of the propellant mass remains in the firearm, but the vast majority is ejected. That is how a gun works, it’s not a PV=rRT problem.

          • Martin

            First up I agree with your contention that momentum and stopping power are not directly related but the way you’re going about “proving” this is very strange. (The biggest problem you have first up is that you cannot produce a formula for stopping power – it doesn’t exist because stopping power is largely an undefined quantity.)

            To be honest it is starting to look more and more like you only know enough about momentum to sound credible but to actually be dangerous.

            If you really want to make a difference to this conversation contact the physics professor at your local university and involve him/her in the discussion. A lot of physics guys love a conversation like that and love being involved in real world problems. With his/her help run a scientific experiment and produce some numbers – that will do much more to your credibility than the unscientific/contrived experiment you performed already.

            If you really want to put this issue to bed like you claim that is my suggestion. And don’t try to please the guys who’s “eyes glaze over” you won’t get through to them anyway, prove it to all the well educated and capable people that are on on this site already and that still disagree with you.

            Martin.

          • Martin,

            I have to write for an audience that isn’t always as saturated with the technical details as people directly involved in the cutting scientific and technical edges of these problems. That means, I often have to use terminology that is imprecise but descriptive. “Stopping power” is technically meaningless, but vulgarly very meaningful. So it is a tool for me to communicate with quite a lot of people very succinctly. Not everyone likes it (in fact, I don’t), but it’s still useful. Without going into an extended several-thousand-word discourse, I am able to describe to people that momentum is not correlated to “killingness”, in such a way that is (I hope) informative, entertaining, and mostly correct.

            Whether I know enough about momentum to be “dangerous”, well, I dunno. I’ve got college-level engineering and physics courses under my belt, and I do my best when writing these articles to consult the literature beforehand, but I cannot be responsible for everything my readership may conceivably do when they read my writing. I simply can’t predict that.

            Was this experiment unscientific? You betcha, but frankly, such a problem doesn’t deserve much more. We’re talking introductory level physics, here, fundamentally, so conducting a state-funded study with all the bells and whistles would be simply gratuitous.

          • Mark

            The gas will act equally across the interior surfaces of the barrel at each instant as it expands and drives the bullet forward. Thus the net contribution from the expansion is essentially ZERO until the bullet clears the muzzle and the propellant gas suddenly has an escape route.

            By then the bullet has gained most of the chemical energy produced by the burnt propellant. We know that the gaseous mass can’t travel faster than the bullet.

            Perhaps firing a blank cartridge would illustrate this better???

          • Actually, the gas escape velocity is always higher than the bullet, so you’ve got that backwards.

            Finally, you’re making the mistake that this is a PV=nRT problem, just like the fellow above. It’s not, because it’s a gas flow problem, which is different.

          • Mark

            Apologies, I was trying to illustrate the initial near-zero contribution from the expanding gas. I don’t think this analysis will get very far without the mathematics behind it. Math don’t lie!

          • M40

            “For every action, there is an equal and opposite reaction.”
            That doesn’t always mean what people think it does.

            In this case we have a 40gr. bullet, and a 40 THOUSAND gr. rifle.
            The rifle therefore ‘kicks’ at 1/1000th the speed of the bullet.
            If the bullet is propelled to 1000fps, then the rifle is propelled to 1fps.

            Simply reducing the surface area of the buttstock with the cleaning rod doesn’t change the fact that you’re experiencing 1/1000th of the velocity.

            PS – If NASA fired an (unrested) rifle in space, the bullet would speed away at 1000fps, and the rifle would drift in the opposite direction at 1fps. Here on earth, if you take 2 identical rifles and shoot one at the other… the rifle that gets hit should “kick” at nearly the same amount as the rifle that fired the projectile (minus energy lost to wind resistance).

          • MarkVShaney

            Not confused at all. If the rifle mass = bullet mass then their momentum is the same. Here’s a thought experiment. Please do not try this at home. Take a rifle barrel. Insert 2 projectiles, one from each end and a power charge between them at the center of the barrel. 1 projectile represents the “bullet” – lets call it 50 grains. The other projectile represents the “rifle” and also 50 grains. Put the tube in your mouth and ignite the charge. You have just demonstrated two masses with equal momentum.

          • That’s not how momentum works at all. Momentum is mass times velocity, two objects do not need to have the same mass or velocity to have the same product of their mass and velocity.

            Further, two objects with the same mass do not necessarily have the same momentum, such as for example if they are moving at different speeds.

    • MarkVShaney

      1 word. Inertia. Explains exactly why the rifle does not puncture his chest. The inertia of the bullet is much lower than the inertia of a rifle. Fire a 40 grain bullet from a 40 grain rifle (minus friction and gravity) and see what happens.

      • Jeff Edwards

        Bravo! This is exactly correct. The charge is absorbed by the mass of the rifle, the bolt, the spring (which sucks up energy and then redistributes it back FORWARD again) and the friction of the bullet against the lands inside the barrel as it pulls forward on them. If the energy of the bullet was equal to the mass of the rifle (less the attendant frictional losses) you’d have a little more blood on your hands!

      • me me

        also bullet does take a longer time to accelerate up barrel than its does decelerating at impact. Not quite as dramatic as “its not the fall off the roof that kills you, its the sudden impact with the ground” but its still a force multiplier. Some of that effect explaining why pistols tend to be less effective than rifles — long path allows higher velocity to be achieved while still being controllable.

        • MarkVShaney

          There were several mentions of ‘conservation of momentum’ in this thread- and you are describing that perfectly. I think in general the blog should have an appendix “Firearms not Physics” 🙂

    • cs

      Also the friction of sandbags the rifle is resting on is using a considerable amount of the guns energy. Perhaps if Nathan had the rifle mounted Newtons Cradle kind of swing, he would have seen better results.

  • PK

    “clearly it was my fat that saved me had I been a reasonable weight I would have surely died – I think that means I’m immune to bullets so maybe DARPA should start breeding really fat supersoldiers”

    Killing it with the tags once again, I’m in tears!

  • Thomas S

    I don’t know what to say. You are on the right track, momentum does not = stopping power. That as a general concept is completely correct.

    “Stopping power” has more to do with terminal ballistics than a given velocity or diameter of a projectile. Bullets designed to “dump” their energy entirely inside a target are more effective than a bullet designed to punch through a target when that target is made of something squishy. This is why hollow points are better defensive rounds than FMJ.

    So what role does momentum have? Well an increase in velocity yields a greater effect on energy of the projectile than an equivalent increase in mass (standard physics). This is why a lightweight rifle round can still vastly outperform a comparatively heavier pistol round.

    Why does a rifle not damage the shooter to the same extent as the bullet does the target? The energy transferred accelerates mass as a function of the design. On one hand it is accelerating a few grams of metal and on the other, a pound to several pounds of material.

    So while yes, the powder imparts the same amount of energy into the weapon as it does the bullet, that energy is simply not enough to accelerate the vastly heavier firearm to velocities that are dangerous to the user. The fact that the energy is dissipated over a larger area also helps.

    Granted that is still a bit of an oversimplification, there are still things like energy following path of least resistance. The short of it is, this helps accelerate the bullet even more than the firearm it is being shot from.

    Anyway, in general terms momentum does not equate to stopping power in the practical world. However, please do the math on say a 40gr .22 projectile accelerated to say 18k per second. A human would not survive a hit in anything approaching a vital are. Now this boys and girls is why the military wants toys like railguns or why things like project thor sort of become scary. Turns out dropping something with the mass of a volkswagen from space can have impact energy on the order of nuclear detonations…

    Let’s not right momentum completely off, it has a major effect (rifle vs pistol) but we really aren’t talking about vast differences in momentum in firearms.

    • Isaac Newton

      “So while yes, the powder imparts the same amount of energy into the weapon as it does the bullet, that energy is simply not enough to accelerate the vastly heavier firearm to velocities that are dangerous to the user.”

      This is not true, unless by energy you mean momentum.

    • You seem to be confusing momentum and energy; they are two different quantities.

  • Tassiebush

    It’s a cover up. That skewering picture was the real outcome and you’re too vain to admit it! 😉

    • ostiariusalpha

      Notice that he refused to show his grievous pectoral wound in the video. You could tell by the way he couldn’t keep his right hand up for more than a second at the end, and then the video ended abruptly before he started crying.
      #momentumconspiracy

      • Tassiebush

        And with the still it looked like every painting and print I have ever seen of a man being wounded by a firearm. You can’t fake that sort of look!

        • Don Ward

          Back… And to the left.
          Back… And to the left.
          Back… And to the left.

  • David B

    Whenever I scrolled down and saw the kenetic energy equation, I started having high school physics flashbacks.

  • Budogunner

    Doesn’t this rather ignore inertia? The mass of the firearm seems to be forgotten here.

    My .308 Savage is in an aluminum chassis system. It is a massive gun, by design, as I use it as a middle step in training with subsonic loads. Shooting subs, you feel a brief push at the shoulder, but can hear the loud smack of 208 gr. Of lead hitting the target.

    Because the gun has higher mass than the bullet, it is not propelled into my shoulder with the same velocity as the bullet leaving the barrel. Thus, my experience being “hit” by the butt of the stock is not at all analogous to being hit by the bullet.

    • That’s the point. The rifle has more momentum than the projectile, so if momentum were correlated to “stopping power”, we would expect the rifle to injure me to the same degree as the bullet would. But it didn’t, which disproves that theory.

      • MarkVShaney

        Greater mass (plus inertia) and less velocity does not “more momentum” make.

        • The rifle has more momentum than the projectile because it has momentum equal to all the ejecta, including the propellant.

          That is the law of conservation of momentum.

          • MarkVShaney

            You’re changing state.

          • How does a state change allow you to violate conservation of mass?

          • MarkVShaney

            Because you’re not throwing the loaded rifle at the target. You don’t understand the concepts you’re talking about.

          • Whether you’ve worked for NASA or not, it doesn’t seem like you understand how this equation is balanced. Because the momentum is conserved, that means we have an equation that looks like this:

            mass of the bullet * velocity of the bullet + mass of the propellant * velocity of the propellant = mass of the rifle * velocity of the rifle.

            If we balance this equation, we find that the velocity of the rifle is very low to equalize its momentum with that of the ejecta. As I stated before, the rifle has the same momentum as all of the ejecta. Think of it the same way as how a rocket engine works, where thrust is produced by the velocity and mass of the ejected mass. Same for a rifle.

    • MarkVShaney

      Excellent reply- and my point exactly. Someone must have been absent that day in Mechanics 101.

    • Conservation of momentum, means that the total momentum of the ejecta (propellant plus projectile) is equal to the momentum of the rifle.

      It does not mean the velocity is equal, and I never said any such thing.

      So here’s how it works out, roughly:

      34 grain projectile at approximately 1,100 ft/s = 37,400 grn-ft/s

      2 grain propellant at approximately 2,000 ft/s = 4,000 grn-ft/s

      Total momentum of the ejecta: 41,400 grn-ft/s

      Weight of rifle, approximately 5.5lbs, 38,500 grains

      41,400 grn-ft/s divided by 38,500 grains = approximately 1.1 ft/s recoil velocity.

  • Joel

    This myth will be hard to break. Perhaps this is because it is repeated so often. Many accept stopping power or “knock-down power.” I will note that the latter was explicitly called out in the 1989 FBI report on Miami.

    A 160 pound person walking at 3 miles per hour has 4,928,000 grain feet per second of momentum. A 230 grain 45 ACP round moving at 800 feet per second has 184,000 grain feet per second of momentum. That is, a 45 ACP round (under those conditions) has about 3.7% of the momentum of a medium or small sized person walking not very quickly. Thus, the 45 round will not knock them back or knock them down. If the bullet transferred all of it’s momentum to the person, they would decrease their speed only slightly (to about 2.9 MPH). In other words, they would continue moving forward but at a very slightly reduced speed.

  • rob in katy

    Wasn’t the ability of a bullet to knock a person backward just disproven with the gentle giant, ugh, thug, Michael Brown?

  • Major Tom

    Momentum is conserved yes, but you forget it goes both ways. The momentum imparted on the gun itself is vastly diminished compared to the momentum imparted onto the projectile owing to the mass of the gun and the absorption of the force into the shooter. That’s the big reason why bullets can fly for hundreds or even thousands of meters yet the gun only recoils a relative little bit.

    Secondly, momentum when talking about stopping power is directly correlated and equated to the bullet. When you use the stone example, how much does the stone throw you around when you chuck it into the lake? For many stone weights smaller than boulders, the effect is negligible on you but massive on the stone that was thrown. The same applies to bullets. The momentum of the bullet is vastly higher than the momentum of the gun it was fired from owing to the diminished mass of the bullet vs the gun. Just as a bigger and faster stone breaks more bones when you chuck it at someone yet doesn’t break you when you throw it. A bigger and faster bullet wounds and penetrates more effectively than a smaller one. That’s the reason why for example .22LR has vastly diminished kill rates for targets larger than a house cat or rabbit compared to bigger and faster calibers.

    • The physics you describe violate conservation of momentum. I think you are confusing momentum and energy.

      • ostiariusalpha

        Yeah, that’s not the way it works, Tom.

        • Giolli Joker

          Indeed while floating in a tin can a few laws of physics might seem to work differently. 😆

      • Major Tom

        Let’s define momentum shall we?

        Momentum in Newtonian physics is the retention of kinetic energy while in motion. It is defined as the product of an object’s mass by its velocity.

        Momentum works hand in hand with inertia aka the resistance to change in velocity, direction, momentum, etc.

        In the concept of firearms, the cartridge creates an equal force but does not impart equal momentum upon the bullet or the gun. The force accelerates the bullet to a high velocity which despite its small mass gives a high momentum. The same force applied to the gun accelerates the gun only to a tiny fraction of the velocity which gives it a relatively low momentum.

        Momentum as mentioned works hand in hand with inertia. Upon firing, the bullet has very low inertia but high momentum. The gun on the other hand has very high inertia relative to the force involved but low momentum. Like inertia, how much force it takes to slow an object to rest will vary according to its momentum. It takes a lot less force to slow a gun to rest than it does a bullet.

        Were the inertia and momentum between bullet and gun an equal, then in the absence of gravity you the shooter would fly back accelerated at the same velocity and momentum as the bullet in accordance with Newton’s Third Law.

        • Oh I am just dying to hear how a higher mass object can have the same velocity and momentum at the same time as a lower mass object.

  • 2hotel9

    Little brother? You ain’t helping your cause with these posts. Beating this dead horse is simply covering you with bloody bits of dead horse. Let it go.

  • Edeco

    I kind of believe in impulse… force over (very little) time. Not that it’s a major concern of mine.

  • m-cameron

    Your “test” proved nothing…..

    This is how the bullet reacts on a person as a whole……..the body is acting like a dampener , softening the blow to the vital organs……but this isn’t how bullets work…..bullets penetrating into the body.

    Wrap a hear in soft body armor and shoot it with a .22……..the bullet won’t penetrate, and since the energy is so low, there may be some minor bruising

    Repeat that same test with buckshot…….the shot won’t penetrate…..but the heart will be destroyed from the energy transfer.

    You have to account t for the energy transfer to the soft squishy organs…….

    Anyone who says momentum or energy transfer has no bearing on stopping power is a moron

    • Then didn’t the test in fact prove that momentum doesn’t correlate to “stopping power”, but energy does?

      • MarkVShaney

        I think what you need to define is stopping power. Let’s say “stop” means “immediate incapacitatation to further advance toward the opponent.

        SP = E + F + B +/- T(v)

        Stopping Power = Kinetic Energy (momentum) + friction + biological factors (shot placement) +/- Target velocity at POI.

        Let’s say 22 rimfire produces a stopping power quotient of 1.

        A squirrel has a mass + biology + friction (fur) + Velocity < 1 then he will be stopped.

        A Elephant.. may have a quotient of 500. The 22 rimfire lacks the "Stopping Power" to "stop" an elephant.

        If a freak shot hits the elephant in the Medulla the B factor could go much higher- in which case the Elephant would be "stopped".

        If you shot a rimfire at 1000 fps at a target moving toward, or away from you at 500 fps that would impact the equation as well.

        In any case momentum ad a function of energy absolutely contributes to stopping power… perhaps moreso than all other factors. but to say that it alone is the 'only' factor is as erroneous as saying it is "no" factor. Your basic premise to that effect- that the rifle should kill the shooter and the shootee is plainly obviously false for nearly the same reason. Velocity + mass without intertia does not describe the mechanics at work.

        • What you just wrote is completely spurious nonsense.

          • MarkVShaney

            Nice. I think this entire article is laughably ignorant nonsensical clickbait. So there.

          • Okeedokee.

        • ostiariusalpha

          I hope you were trying to be funny, because this is usually the point where some pulls out that Billy Madison quote about “Everyone in this room is now dumber for having listened to it.”

          • MarkVShaney

            I think it’s Adam Sandler who is “trying to be funny”… because he hasn’t been funny in a decade at least. I suppose anyone who would pull out a Billy Madison quote… well that pretty much speaks for itself.

          • ostiariusalpha

            Aw gee, only 6 days late to the party there, Mr. Shaney. It might surprise you to know that I’ve never seen any of Mr. Sandler’s movies, including from when he was “funny” (when was that, exactly?). In fact, I noticed when I looked up the quote afterwards that the caption on the pic I appended is actually wrong; it should read “A simple ‘Wrong’ would have done just fine.” But, considering the cultural value of the source material, I can’t summon up much remorse.

  • MPWS

    This is an excellent class in understanding intermediate ballistics (aka what happens immediately when bullet is out of barrel). I like all what you say and do, Nathaniel. And yes, your appearance is confidence building. And also… I will recommend you for Medal of Honour you deserve!

    You probably know as well why you did not get hurt. Yes, the “momentum” was there, but the speed with which it was applied to you body was anemic enough so you body was able to successfully resist it. Repel it, literally.

    One thing seemingly non-elated, but rather important I have on mind is this: our bodies (both good and bad guy’s) are immensely more complex that gun mechanics. Therefore, we should be more judicious and separate them, although ultimately, they are related.

  • TDog

    Stopping power is partially dependent upon the target. A nine-pound cat is going to be stopped by a lot less than a half-ton bear.

    In my opinion, stopping power is a lot like those dating sites: which ammunition is the best match for your target? Things such as energy, mass, velocity, surface area, etc. all matter and it’s all but impossible to say bullet A has a uniform stopping power (such as it is commonly referred to) given every situation.

  • Some Rabbit

    Would not the mass of the rifle absorb some energy? Try switching the rifle around and put the muzzle against your shoulder. Then average out the results.

    • You are confusing energy and momentum. They are two different quantities.

      • Some Rabbit

        I was just trying to be snarky, I agree, momentum ain’t the cause of trauma. I believe Mythbusters debunked that in an episode by shooting a variety of weapons at a hanging hog carcass. None moved the pig, except the 12 ga. slug and that, likely owing to it’s low velocity and broad surface area. Even then, not by much. The rest zipped clean through.

  • Don Ward

    Nathaniel F. I think you need a weapon of higher caliber and more stopping power to penetrate the skulls of some of your commenters next time you do a physics lesson.

    • georgesteele

      This explains why USA students are not doing well in comparison with those of other countries – if this discussion exemplifies the state of Physics understanding imparted by our schools. Ouch.

  • Jay

    Sunday morning dodgy physics from Nathaniel. Hahaha.

    Sorry man. I don’ t think you paid attention in school.

    The reason you didn’t hurt yourself is because the rifle is about 1000 times heavier than the bullet.
    You have equal FORCE pushing both, the rifle and the bullet, for a fraction of a second, but because your rifle is 800-1000 times heavier than your bullet,, the rifle won’t be able to achieve enough velocity to penetrate your skin.
    F=m×a. The heavier the object, the lower acceleration and lower velocity will achieve.
    You need to stop posting about this things, until you have a better understanding of the basic physics.

    • So what you’re saying is that I need to attend whatever school told you there was a law of conservation of force?

  • Joe B

    If momentum is mass times velocity, and energy is 1/2 times mass times velocity squared (or 1/2 times momentum times velocity), the the greater the momentum of a projectile, the greater it’s kinetic energy. If you believe kinetic energy is directly related to “stopping power,” then you have to admit that more kinetic energy equals more stopping power. If more momentum equates to greater energy, then more momentum equals greater stopping power. Admittedly, the effect of momentum is not a huge factor when dealing with the relatively small mass and velocity of a shoulder fired projectile. But it does contribute. Fire a large cannon from your shoulder and see what happens. Finally momentum is conserved, but it is conserved in all directions. Your test only demonstrates the effect of momentum on the y axis (the plane of the barrel bore). Momentum is also being exerted on the X and Z axis as well. So, the effect on your shoulder is negligible. Not like there is much momentum from a .22lr to begin with.

    • Let’s do a simple mathematical test here:

      10 grams * 900 m/s = 9 kg-m/s momentum

      20 grams * 500 m/s = 10 kg-m/s momentum

      0.5 * 10 grams * (900 m/s)^2 = 4,050 J energy

      0.5 * 20 grams * (500 m/s)^2 = 2,500 J energy

      So, no, greater momentum does not imply greater energy.

      Also, momentum cannot be “conserved” such that it dissipates outward in all directions. That’s runs directly counter to how conservation works in physics.

    • Zapp Brannigan

      “If momentum is mass times velocity, and energy is 1/2 times mass times velocity squared (or 1/2 times momentum times velocity), the the greater the momentum of a projectile, the greater it’s kinetic energy.”

      Not necessarily. If mass is tripled and velocity halved. Momentum 50% greater but energy has decreased to 75% of starting.

  • Zapp Brannigan

    Stopping power of a bullet is more about the internal damage done by the bullet. A 45 ACP round right through a person’s head won’t impart that much energy or momentum, but it will stop that person right then and there.

    Look at the videos of ballistic gel being shot. It’s a relatively lightweight blob just resting on top of a bench. A bullet hitting that gel doesn’t make the gel fly away, even when the bullet is stopped within the gel.

    There’s more energy and momentum in a ninety mile per hour fastball than a 45 ACP round. People get hit by baseballs thrown by major league pitchers and they don’t drop immediately to the ground. The baseball will hurt, but it won’t do damage to vital internal bits as a bullet can do.

    Am I the only one not in daily gun battles where stopping power is a primary concern? I really do not worry about this.

    • Pazuzu’s_Proletariat_Pal

      A MLB fastball has more than twice the momentum of the 45 ACP, but the 45 ACP has more than 2.5x the KE that the baseball. Energy != momentum

  • georgesteele

    I linked here from your comment on the other article. I think you might have confused my comments with those of others. First, my comment: “the force in recoil is distributed over a very large area, so that the unit area force against your hand is insufficient to cause material damage. Were you to distribute the same force with a pencil point against your hand, however, it would be a lot less comfortable”, referred – in context – to firing a .45 from a handgun.

    And it is correct, and not disproven by firing a 40 grain bullet from a 6 pound rifle when the comment related to firing a 230 grain bullet from a 1 1/2 – 2 pound handgun. I don’t think you want to reproduce the above experiment with a .45 with a sharpened pencil duct-taped to the handle against your bare skin. But I also disavowed any statement that such recoil force per unit area would be equal in *lethality* to that of the bullet being fired.

    I made the point, in fact, that momentum is a poor metric to use in *absolute* terms, when I said that a .45 has the same momentum as a 10 pound bag of sugar thrown at walking speed and a .220 swift at 4400 f/s (my handloads). But it is less bad a metric, in *relative* terms, and within similar context, when comparing a 40 grain .22 and a 230 grain .45 at about the same velocity. No one would argue seriously that the .22 is more lethal, unless other factors were included in the analysis.

    I think the other way to look at this is to equate the effect of a .45 pistol pushing back on your hand and the impact of the bullet on a 2 lb. steel plate held in your hand, or taped in front of your chest. In such a case, the net momentum exchange would be the prevalent measure, and the impact on your hand or chest would be similar to the impact of the gun on your hand – a minor push backwards, overcoming the inertia of the plate and your hand.

    In any case, I appreciate your efforts to advance the analytical basis on which we judge stopping power. It is a far more complex issue than simple Newtonian physical measures can evaluate. But anecdotally, and all other things being equal, more people would take a chance with being hit by a .22 than a .45, and they’d be right.

  • georgesteele

    After reading the comments below, I think the net is: if momentum were stopping power, the gun would have been invented, and then immediately discarded as a useless tool.

  • ozzallos .

    Iraqveteran8888’s videos infamous videos on how far will ___ caliber kill seem extremely relevant here. Most rounds easily go out past 400 yards, but 9mm ball will go through ballistic gelatin at 400 yards. They didn’t use the gel on the 45acp tests, but it too passes through the same reinforced plywood target. Clearly these rounds are no longer accelerating, which indicates momentum (simplistically energy+mass) does appear to be component factor in “killing power”.

    That said, I’m not sure how “momentum” is somehow analogous to guns being “just as deadly to the shooter as they are to the target”. In fact, it’s a horribly flawed analogy to even postulate, frankly. Since the mass of the gun and the body holding it is greater than the projectile being fired, momentum has next to nothing to do with the gun killing its user in any way, shape or form. I mean, it that really a thing for people?

    • Out of the Blue

      That does play into Nathaniel’s argument about energy being the best indicator. Momentum is conserved, meaning that the gun moves at a far lower velocity, so that it gets less energy than the bullet, due to the whole v^2 aspect of the energy value.

      • ozzallos .

        Sure, energy can substitute for mass to a certain point (and yes, more energy will punch a toothpick through your shoulder), but mass also conserves energy via newton’s 1st Law; the outside force being atmosphere or bodily mediums. Still, I’ll contend the combined definition of momentum does contribute to the nebulous concept of “killing power”.

        I’m not saying energy doesn’t matter, but there’s a reason why .40sw was seen as the sweet spot in terms of mass *and* energy.

  • bull

    ke = 0.5 mass times velocity squared.

  • georgesteele

    Just a thought on the Momentum issue: If we were to compare, say, a .22 with a .45 – and assuming all other things were equal, which they are not, but let’s assume – then the .22 has about 1/4 the frontal area of the .45. But a 230 grain bullet has about 5 3/4 times the momentum of the 40 grain .22.

    So if we ignore bullet deformation (i.e. use an FMJ .22), and assume identical CDs (CD=coefficient of drag; times frontal area equals drag – in “identical” frictional environments), then the .45 will be less frictionally retarded than the .22 at the same velocity. There are other issues, of course – energy spill from contact with bone, etc. – but the .45 would penetrate more deeply than the .22. And the wound channel, neglecting other factors, would be 4 times larger, leading to faster bleed-out. The likelihood is higher that the heavier bullet would penetrate more deeply as a result, which should increase lethality, and possibly incapacitation (stopping power).

    I’ve held steady for velocity in the above. Note that the .22 would have to travel 5.75 times the speed of the .45 – say 5750 f/s – to equal its momentum, which is off the rails – especially in a handgun. But the hydrodynamic drag also increases as the square of the speed (not true, but let’s use it), so the retarding (drag) force acting on the equal momentum but radically faster .22 would start off 8.2 (33/4) times higher than for the .45, lessening penetration despite that equal momentum.

    The mass increase effect (volume of a “sphere” increases as the cube of the radius – thus, so does the mass) advantages larger bullets in a momentum (likely penetration) comparison. I think this starts to get at the issue of why heavier, larger caliber bullets tend to be more effective – although there are lots of caveats in the above, purely theoretical, analysis. Empirical tests on the battlefield, however, seem to substantiate the general idea.

  • gunsandrockets

    Even though momentum does not equal lethality, I can see why such a crude metric was used as a rule of thumb for judging firearm projectile lethality on large game.

    I would compare such a measurement to another crude and false measurement, the notion that heavier objects fall faster than lighter objects.

    • MarkVShaney

      They do generally…although I wonder how many people on the internet live in a vacuum…or more aptly, possess one between their ears. Just as this “demonstration” proves… cherry picking pieces of Newtonian mechanics and ignoring key effects of friction, inertia etc. leads to, at the very least… clickbait.

      • It’s possible to cherry pick physics? Please, tell NASA, that will make spaceflight so much easier.

        • MarkVShaney

          You mean my former employer?

      • gunsandrockets

        Depends on what you mean by generally, even when close to the surface of the Earth.

        In the case of firearm bullets which depart a barrel parallel to and five feet above the ground surface and fired at the same 1000 fps velocity and with the same bullet density and the same shape, the lighter bullet will “fall” faster and hit the ground before the heavier bullet.

        • georgesteele

          Whaaat? Galileo just threw up a little in his coffin. Nononono. The bullets will hit the ground at the same time – if, in fact, you find a wizard to make you a bullet that has the same density and the same shape as another, but that weighs more. Let’s assume you really meant two bullets with the same frontal area, similar form factors (but different lengths) *or* the same frontal area and identical shape, but different densities.

          The bullets will hit the ground at the same time. The heavier bullet will have traveled a longer distance, as it will have retained its velocity better in response to aerodynamic drag. In fact, both will hit the ground at the same moment that a duplicate of either bullet dropped from the height of the gun at the moment either cleared the muzzle.

          The above is true to a very high level of accuracy, modified only by such things as crosswind lift effects on the spinning, rifling-engraved bullet.

          OK, Galileo – you can rest now. You, too, David Scott.

          • gunsandrockets

            A) Wizard? Sigh. Apparently the commonly defined terms of density and shape left some room for confusion. Let me make my examples a bit more explicit and perhaps more understandable. The two cartridges are the .44 Henry rimfire, and the .22 short rimfire. Both bullets are the same shape, just different in scale, and both are made of lead, therefore having the same density, and one bullet weighs much more than the other.

            B) “In fact, both will hit the ground at the same moment that a duplicate of either bullet dropped from the height of the gun at the moment either cleared the muzzle.” Uh, no they won’t. You failed to account for the small yet significant factor of the projectiles horizontal inertia. Remember, even 1,000 fps is 1/25 of full orbital velocity.

            So yes the smaller and lighter fired bullet will hit the ground before the heavier projectile because it slows down faster through the air than the heavier projectile. Think of it like the orbital decay of satellites in LEO.

          • You are absolutely incorrect, and in fact NASA spent quite a lot of money to prove it:

            😉

            Accounting for atmosphere, the smaller projectile will hit the ground after the heavier one, because the effect of wind resistance on it is greater. The horizontal velocity decay has naught to do with it, because the Earth is too large for the shape to be anything but negligible given a 1,000 ft/s muzzle velocity.

          • gunsandrockets

            I find it really insulting that you responded with so little understanding of what I wrote and assuming I have so little understanding of Gravity.

            BTW, I didn’t need NASA to prove anything to me, I find the inclined planes experiments of Galileo quite adequate for proof.

          • I think I understood you perfectly well, and I’ve read all your posts so far all the way through. So far as I can tell, what you’ve said is incorrect, and I’m not shy about saying so.

            If you want to take another whack at explaining it, be my guest.

          • georgesteele

            My comment 9 hours ago, seen above: “OK, Galileo – you can rest now. You, too, David Scott” refers exactly to this experiment he (David Scott) performed on Apollo 15. Again, in a vacuum, they hit the ground at exactly the same time.

            In air, they also hit the ground at – within measurement error – the same time, because the maximum downward velocity that they hit in 6/10th of a second is so low that aerodynamic retardation of the fall is negligible for such a dense object.

            And it would most likely be less than the vortex lift caused by a bullet spinning about 1000 revolutions per second, generating a differential pressure (greater for the .44, because the linear surface velocity is higher than for the .22) above and below the bullet in even the minutest crosswind.

            On the other hand, the truth will set you free; if you can provide some sort of link to a test that measures this “horizontal inertia” you refer to – whatever that means – then I’d be very interested to read it and learn something.

          • georgesteele

            Now you have been clear about what you mean when you say shape – you mean approximately the same profile.

            Horizontal inertia? It’s just this simple; the gravitational attraction of the earth will accelerate all bodies at identical rates, independent of mass, downward. In a vacuum, no atmospheric or aerodynamic forces will affect that outcome. In air, the effect of a bullet spinning will cause a vortex that will interact slightly with crosswinds. That gravitational force doesn’t start when the bullet is fired; it is affecting the bullet while it is sitting in the mouth of the cartridge in the chamber. Firing doesn’t change that.

            The lighter bullet will usually slow down faster, because the aerodynamic forces on it will overcome its momentum – in your case, the larger bullet has about 4 times the frontal area of the smaller, but may have more than 4 times the mass – you didn’t state the grain weight – but let’s assume it’s a typical .44 Henry 200 grain bullet, which is 5 times the mass of a 40 grain .22.

            Accordingly, the momentum of the .44 will provide it with greater ability to resist wind drag, so it will travel a greater *distance*, in the unit of time over which it travels before hitting the ground, than the .22, which will be more retarded by air resistance because of its lower momentum. Again, we’re assuming equal muzzle velocity here, which is a reasonable approximation of the truth.

            But that unit of *time* will be identical to that of the .22, because each bullet will be acted upon by gravity, and accelerated downward at the same rate. The gravity vector is normal to the surface of the earth, which is a plane over the distance we are talking – at less than 6/10ths of a second, both bullets will have hit the ground from 5 feet up. Drop anything from 5 feet and you’ll see. The .44 will hit at a greater distance from the muzzle, and at a higher velocity, but it will hit at the same time as the .22. That’s what Galileo established regarding falling bodies – they fall at the same rate irrespective of mass (at least until they reach terminal velocity, if dropped in air).

            Satellites in low earth orbit are moving at a velocity that allows the centrifugal force acting on them to balance the centripetal force that is gravitational attraction. As the thin atmosphere drops their velocity, their centrifugal force decreases, and eventually they fall to earth once the equilibrium is lost. That’s not the case for a bullet traveling 150 to 180 yards.

  • gunsandrockets

    Duct tape the 10-22 action shut? Huh?

    What kind of gun owner are you Nathan? You don’t own any manually operated .22 rifles?
    LOL!

  • marine6680

    Whatever your point… This test was fundamentally flawed to prove much if anything.

    Your assertion that a firearm would need be as deadly to the user as the bullet to the target is also flawed.

    On the face correct only on the shallowest of levels.

    A car rolling a few miles an hour toward a cement block wall has more energy and momentum than a 10lb sledge swung by an average person… Yet the car is unlikely to damage the wall or even itself much if at all… While the sledge will damage it.

    Simply calculating the momentum and energy is not enough… Be they the same or different in the comparison… It’s how the momentum and energy is applied that matters.

    A firearm does not injure the user because how that energy is applied vs how the energy of a bullet is applied.

    Taping a rod to the rifle is basically nonsense to prove a point, as that rod effectively weighed 4lb or 28,000gr (just a rough estimate of the rifle’s weight) vs the 40gr projectile… Momentum and energy may have been the same, but everything else that matters was vastly different.

    Stopping power is a non-starter, as it is not really something that can be defined scientifically, as it’s meaning is not even defined well.

    As far as a projectiles effects on a living target… Well that is a complicated science that can not simply be boiled down to momentum and energy equations.

    For handgun rounds, ballistic experts agree that momentum is important, in that it is the force that directly aposes the forces that are acting to slow the bullet after it hits the target… The body causes drag, and has barriers like bone, momentum is what counteracts the force. Higher momentum equates to deeper penetration… All other factors being equal. Meaning same type of bullet and same caliber. Comparing momentum between calibers is meaningless.

    Hollowpoint designs complicate direct comparison within a caliber and even brand, as the different load types behave differently, different expansion profiles…

    The energy in handgun calibers is too low for energy to be much of a determining factor. Statistics bear that out as all pistol calibers perform vastly similarly.

    Rifle rounds start reaching into energy levels that do make the energy of a round relevant, but it’s bullet design that plays a huge role, as effective energy transfer is not obtained in rounds that pass through the body with little disruption.

    Pump velocity up too much and even all the energy in the world is no good if it zips straight through.

    Though not likely with our chemical propellant firearms, magnetically driven rounds can reach speeds that make even light weight bullets zip through without disruption. A human is unlikely to be able to weld a weapon capable of using pure energy to destroy the target. The body offers up relatively little resistance, and the energy required to get sci-fi destruction of a target is huge.

    While it’s tempting to try to use easy to understand and calculate, concepts like momentum and energy… Especially for ammo marketers… It’s just not anywhere near enough to describe the situation accurately.

    • You have written an awful lot here, but it appears that you’re not clear on how momentum, energy, velocity, and mass are related…

      • marine6680

        I am very much aware… The math is not difficult.

        The test was a stunt, it did not prove anything.

        But I do understand that many can not wrap their heads around the math, so dramatics must be applied.

        • Alright, marine, what about how I applied the momentum to my shoulder made the test invalid, then?

          • marine6680

            I explained it in the first post.

            First, ignoring any loses that holding and/or resting the rifle imparted. (The rifle was not in any form of isolation, allowed free movement, or rather, as free as possible)

            Total momentum and energy within the system were indeed equal. Newton’s laws and all that.

            But as I said… Simply tapeing a rod to the rifle is not showing that momentum applied to the same diameter does not cause harm to the user.

            The official specs say the 10/22 is 5lb.

            A 40gr projectile is .006lb

            Let’s assume a high velocity 22lr, and 1200ft/s… Momentum is 7.2lb ft/s and energy is 134.27ft-lb (I hate doing these calculations in imperial units, it’s a mess)

            Since momentum is the main focus in this article, let’s see what that breaks down to.

            If the momentum and energy are conserved and equal between the bullet and rifle.

            A 5lb rifle with 7.2lb ft/s of momentum will have a velocity of 1.44ft/s. Even if that is applied to a .22in diameter rod, it is not enough to pierce the skin.

            Sure, momentum and energy are the same, and even the diameter of the object that force is applied upon is the same… But changes in the variables mean that affects are widely different.

            So, while momentum does not equal stopping power of a bullet. (It is better thought of as penetration potential)

            But energy is likewise not a measure of stopping power.

            Stopping power is a useless concept, as it is ill defined at best… And even if it was, it can not be deduced from simple mass and velocity figures.

            Stopping power is a useless talking point made up by ammo marketing teams to sell ammo, and gun enthusiasts as something to argue over.

            Massive energy dumped shallow in the body does little to stop the target… Momentum may give a rough idea how a bullet will penetrate the target when compared to similar bullets of different momentum, it does not tell you how effective the round will be at transferring energy. And due to the fact that many rounds are designed to expand, and the fact that expansion can vary even when it is the same style and brand of bullet, but different weight… Momentum is not cut and dry as a simple indication of penetration either.

            So I may agree that momentum is not stopping power, I do not think your impromptu experiment was a good way to show that.

            If it helped convince those unfamiliar with physics and proper experimentation methodology of your point… Then rock on.

            Just don’t expect the rest of us to not see through it.

          • You’re under the impression that momentum and energy are both conserved. They obviously are not (in fact, it’s a mathematical impossibility). That sort of colors your whole analysis, and it’s something you should have noticed.

          • marine6680

            In a perfect system both are conserved quantities… The fact that the system is not perfect, simply proves the fact that the test was not proof of anything.

            A quote from the article…

            “This should be self-evident: Since momentum is conserved, if it were directly correlated to “stopping power”, then guns should be just as deadly to the shooter as they are to the target. Guns are not deadly to their shooters, Q.E.D., momentum is not correlated with stopping power.”

            This is faulty logic for a couple reasons… The forst is in the math I outlined before.

            The second lies directly in the premise of your argument. As the same argument made in that quote, can be used to refute the claim that energy is related to stopping power. Just simply substitute the word momentum with “energy”.

            It is not a good argument to base a conclusion off of.

            For the most part, I agree with the overall claim of the article.

            I’m simply arguing that the experimentation methodology is flawed, and the arguments/logic used to support the claim, as not being sound.

            And as I stated before, if it convinces those less knowledgeable in physics to your point, then fine…

          • We’re generally concerned with the kinetic energy only, and usually not thermal energy or chemical or potential energy (those do come up, but in more complex problems than this one). So kinetic energy, the energy of motion, is not at all conserved the way momentum is. Doing a simple word substitution trick here doesn’t work the way you say it does, therefore.

          • marine6680

            You may be confusing total energy in a system and the kinetic energy… Or I am reading you wrong here.

            While all the chemical energy of the system is not completly converted to kinetic energy… Once the kinetic energy is produced, it will be conserved.

            The system isn’t perfect, but roughly equivalent enough that simplifying is not counterproductive in regards to kinetic energy.

            So the kinetic energy imparted to the rifle will be conserved. It’s mass and rearward velocity will result in an energy figure that will be roughly the same as that of the bullet.

            Yes the energy is scalar, and some of the energy used to propel the bullet in a forward direction will in fact be imparted in the circumference of the barrel, not simply in opposition to the direction of travel. Much of that energy will be used to drive the rifle rearward. (If you are up for it, do the math of how much of the energy is absorbed by the barrel rather than going into any rearward motion… I could be wrong and the amount is greater than I think, and this particular argument would be invalid, it would be a Delta calculation, s the surface area increases as the bullet moves forward)

            So no, it’s not a direct correlation, I will grant you that, and the counter argument of word substitution may not be a perfect fit.

            I still contend that the experimentation methodology was flawed… And much of the arguments presented to be flawed.

            I also accept that imperfect arguments are needed to explain to the layman on the audience.

            So I’m not calling you out completely, nor as being ignorant in the subject. Just that I don’t agree with the logic presented, and I accept that it may be a necessity to help explain the subject.

          • Only about a third of the chemical energy is converted into kinetic energy, the rest is turned into waste heat.

            How can kinetic energy be conserved? We know that the momentum is conserved, so we can look at the figures I produced in another comment:

            “34 grain projectile at approximately 1,100 ft/s = 37,400 grn-ft/s

            2 grain propellant at approximately 2,000 ft/s = 4,000 grn-ft/s

            Total momentum of the ejecta: 41,400 grn-ft/s

            Weight of rifle, approximately 5.5lbs, 38,500 grains

            41,400 grn-ft/s divided by 38,500 grains = approximately 1.1 ft/s recoil velocity.”

            So we can calculate the kinetic energy of the projectile, about 124 J

            the propellant, about 24 J,

            and the rifle, about 0.151 J.

            So therefore, kinetic energy is far from conserved!

          • marine6680

            OK… Yeah… Oversight on my part. Energy is a conserved force, but I did not apply it in the proper sense.

            So I got off the rails with that side of the argument.

            But, my argument that the experimentation method is flawed as a method of showing momentum as not being related to stopping power… Still stands.

            It oversimplified the setup, as how momentum is applied/imparted matters when it comes to destructive ability. A larger, heavier, faster object and a small, lighter, faster object can have the same momentum, but how that momentum is transferred makes a big difference.

            While you may have reduced the area of the surface imparting the force, to be the same as the bullet diameter… All the other factors were very different.

            Like I said, correct conclusion, but less than clear methodology.

            I get the whole idea and point, velocity is a key factor in this. Frontal surface area means little, if the force behind it can not cause that area to pierce the target.

            I am more inclined to not dislike the setup upon closer examination and our discussion. So while I don’t think you are completely right, I do think you are much less wrong than before. Understanding your thoughts and going over it a couple times in my head…

            Momentum alone can not do it… But neither can energy.

            I think we both agree that stopping power is a crappy concept.

          • So, momentum is a quantity. You’re talking about how the momentum is “applied”, but I’ve already controlled for that. It sounds like what you’re really saying is “momentum is indicative of terminal effect when it’s coupled with an additional quantity of velocity”, well that’s (essentially) true, except that we call that kinetic energy, not momentum.

            Now, kinetic energy is not completely indicative of terminal effectiveness either, as you’ve observed in your posts. However, it is highly relevant, unlike momentum, but that is a subject for another day, as that gets into energy, its relationship with work, and how work and the forces involved affect the bullet and allow it to affect the tissues of a target.

          • marine6680

            I’m East coast, and should have been in bed 4.5 hours ago.

            Math and logic plus no sleep equals no bueno.

            Heck I just called energy a force… So yeah, need sleep.

            I will see your reply on the Morrow.

          • No problem with that, I’ve definitely screwed math up before because of sleep deprivation. Come back to it, and then we’ll carry on.

          • marine6680

            The momentum thing is a bit of a pet peeve, as so many people try to claim it as a useless figure, when it does play a key role… Especially in handgun rounds.

            So I may have jumped in without a full evaluation.

          • I think the “key role” momentum plays in handgun effectiveness is the result of over-use of short hands to estimate ballistic capability. We can discuss this later.

          • marine6680

            From my research and reading, momentum is important to penetration.

            Those who hunt using the archery disciplines try to balance out speed and momentum for good effect on game.

            Light and fast is good for flatter trajectory and less time of flight, which aids in good accurate hits.

            But without adequate momentum, the arrow/bolt will not penetrate well.

            The same basic principles apply in firearms, but more so handguns, as rifle rounds tend to be contending with more energy and force overall, and that changes things.

            Momentum is why the heavy for caliber rounds are recommended more in handguns.

            Boiled down to the basics, a built traveling through a target is experiencing drag.

            Drag tries to slow the projectile, and momentum is the opposition to that.

          • As I mentioned in the video, momentum is certainly not irrelevant to ballistics, but it is not what causes lethality, either. Inertia is very important for how bullets retain their velocity (although aerodynamics is a much, much more complex subject, of which inertia is just one part), because velocity is the major component in energy. However, the momentum of a projectile itself does not correlate with the projectile’s ability to do work to injure the target once it gets there.

          • marine6680

            Yeah it doesn’t give potential work, but as you said, it helps retain velocity.

            For a handgun round, where the primary focus is adequate penetration, momentum is a key figure.

            The energy levels of handgun rounds are generally held to be too low for the differences in energy between calibers to actually be meaningful.

            If that is indeed the case, then discussing energy as a measure of effectiveness between… 9mm, 45acp, etc is pointless. If the rounds all perform equally. (poorly as many would claim)

            So being left with only adequate penetration as a measure of performance… As I said, momentum is key.

            But it isn’t a measure of stopping power or wounding ability, which is the distinction.

          • I am actually going to write a post on how to work this problem for pistol JHPs… Someday. Hahah.

          • marine6680

            We can continue our talks further then. Should be fun.

          • I look forward to it.

          • georgesteele

            Minor nit: inertia is not the same as momentum. Inertia is just a function of mass – resistance of a body to move when acted upon by a force; momentum is mass times velocity.

          • Did what I wrote give you the impression I thought it was?

          • georgesteele

            Yes and no. You wrote: “Inertia is very important for how bullets retain their velocity”; in physics, inertia is the attribute of mass that resists forces on that mass directed at moving it – as opposed to momentum, which is the attribute of a moving mass that resists forces directed at stopping it. As I said – a minor nit – but the sentence should read “Mass is very . . .” So what you *wrote* gives that impression.

            But based on the careful reasoning and exactitude of your other writings, no – I don’t think you *thought* it was the same. That’s why I said minor nit.

          • Fair enough. 🙂

  • El Duderino

    Physics phail. The speed and violence of the recoil is attenuated by the weight of the rifle and the working of the action.

    • Um, yes, that’s my point. The momentum of an object does not correlate to it’s “violence”, to use your term.

    • ostiariusalpha

      Speed and violence (I’m supposing you mean energy) are indeed attenuated by the weight of the rifle, momentum is not. Also, if you’ll notice, Nate has the action taped down, so it doesn’t effect the outcome.

  • A Fascist Corgi

    Well said. You simply can’t beat the magical 5.56x45mm NATO round (especially the M855A1). Not only is it light weight, but it has superior stopping power than a .50 BMG ball round (an actual argument made by Nathaniel in a debate with me). When combined with the perfect rifle (the AR-15), you’ve created the finest weapon known to mankind. No other gun or round should be featured on this blog. I mean, what’s the point when the American gun community has already achieved perfection?

    • I don’t remember running over anyone’s dog, so did your wife cheat on you with me, or something? Is that why you comment on every article I write, waddling in with your massive hate-boner?

  • Soless

    I’ve never really heard momentum equals stopping power. I’ve only entertained the correlation between momentum, mass, and penetration.

  • TheSmellofNapalm

    What is the fragmentation velocity threshold for M855A1?

    • I have endeavored to get that piece of information, but I don’t know. I think it is no higher than 2,100 ft/s and no lower than 1,600 ft/s, but I cannot be more specific than that.

  • AlanHan

    While it doesn’t do to redefine the laws of Newtonian mechanics, the importance of momentum versus energy can arise for the hunter from biological/target characteristics. Velocity obviously is very important as a contributor to total energy and so varies with the square. Momentum does not vary with the square: This means that it does not increase as much with increased velocity, but also that it does not decrease as drastically with a slowing of velocity. The energy of a bullet is dissipated in varying ways. Much of it is converted to heat. The more and denser tissue a bullet needs to penetrate, the less important the energy number is, compared to the momentum and sectional density.

    It isn’t whimsical to choose a very heavy high SD bullet when shooting a cape buffalo or elephant: The heavy high SD bullet proves more successful at penetrating very thick hide and a foot or two of flesh before it gets to the vitals. Such a bullet in a typical loading may well have less muzzle energy than a super-magnum small boare, but will have more momentum that the lighter lower-mo bullet after penetrating elephant hide and a foot of meat and bone. For a six inch penetration of gray matter, a fast .275 has proven sufficient. For a heart/lung shot it has proven again and again unreliable. Readers here seem to think of the issue just in terms of human targets, which are fragile and shallow in depth.

  • Byron Cole

    So I guess the Navy wasted it’s money on the rail gun that shoots one pound aluminum projectiles that destroy ships, after all the velocity has nothing to do with impact damage or so you say.

  • Martin

    You have not taken into account the inertia of the rifle. This experiment is not scientific at all.

    To do it properly: take a barrel with no breech (of whatever you call it, sorry I don’t know a lot about firearms) the barrel must have no “back” where the cartridge is placed in the barrel, like a loose shotgun barrel, place the barrel in a vice so it cannot move. Place a shotgun slug in the barrel. Behind that put a rod the same diameter as the shotgun round with a flat end cap to hold the back of the bullet that also has a small pin for firing. Cause the firing pin to go off (not sure how to do this one) with the rod against your shoulder. When you have recovered the use of your arm (hopefully) write to me and tell me I was right.

    I also find it strange that you used a .22LR even though I doubt that caliber is EVER used in a discussion of stopping power despite your strange long winded excuse for doing so. It only makes your odd experiment even less credible.

    Even though I agree with your stance that momentum is not directly related to stopping power I think your “experiment” is not only completely unscientific but does more damage than good because it does not actually create any clarity around the issue and as a result only further muddies the waters that you were trying to clear.

    Good luck.

    Thanks.

    Regards, Martin.

  • Martin

    Wow, after thinking for some time that you’re serious about this issue you state this in the comments: “”Stopping power” doesn’t really exist as such” yet your headline states: ”

    Does Momentum Equal Stopping Power? Let’s Find Out!” and most of your article is about stopping power, a quantity you now admit doesn’t exist.

    I am actually angry about how much time I wasted on this issue. Sorry mate I’m outta here, will be avoiding NathanielF as an author in the future. Good luck with your bogus articles in the future.

    • OK.

    • georgesteele

      C’mon, man – the essence of the article is to differentiate between momentum, which is an absolutely measurable entity in physics, and stopping power, which is a combination of physics, human physiology and anatomy, psychology, etc. – and not absolutely measurable.

      The level of dialog that is present on this forum, and the issues being discussed, have dissected the issue very well, and have been – I am sure – an eye-opener for many, and a useful exercise in thinking out loud.

      There are real conclusions that can be taken away from the article, its precursor, and the contributions of the other commenters – one of which is that momentum is a lousy predictor, as a metric, of the ability of a projectile to stop a person in his or her tracks, because it is not correlated with the mechanisms that do so. The video embodies that fact very well, in a simplified manner.

  • Mark

    Actually you experienced the combined masses of the rifle, spent case and taped-on cleaning rod acting against your shoulder. Because the mass of this system is so much larger than that of a bullet, its velocity is much, much lower. Hence the lack of injury!

    If you had an idealised set-up that weighed NOTHING (apart from the cleaning rod tip) then you might well have been injured, but there would be very little time for the rod tip to freely accelerate as it is in contact with your shoulder already.

    • Um, yes, that’s how conservation of momentum WORKS.

  • GhostofBrowningNagle

    momentum might equal stopping power, the issue is a bullet doesn’t have nearly the momentum to stop a person.

    a 180 lb grown man moving at 5 miles per hour has momentum of about 183 N*s.

    a 147 gr 9mm round traveling at say.1000fps has momentum of 2.9 N*s

    If i have momentum of 183 N*s and something with a measly 2.9 N*s hits me along a vector completely opposite to mine, i still hardly notice.

    • GhostofBrowningNagle

      the same reason if i run full speed into a truck, the truck doesn’t even notice

  • Cal S.

    I guess the only question left is, why don’t we hunt bear or moose with a .22lr?

    There has to be a reason we use larger caliber for bigger game. Is it ballistics alone? That we only use larger bullets with heavier weights because anything lighter or smaller would destabilize? Yes, velocity is paramount, the faster something is going, the longer and deeper it will cut. But which one wins in a fight if two projectiles going the same speed and one is heavier? For instance, Liberty Ammunition offers both 9mm and .40 SW rated at 2000fps out the barrel. Now that they’re booking it at the same speed, is it my understanding that they will both do the same damage? But isn’t it a criticism of the .40 that it over penetrates even with normal loads? Let’s not get into a caliber war, folks, it’s a legitimate question.

    I’m not necessarily disagreeing with you, but you realize you fired a ~40 and ~150 grain object (.0057lb and .02142lb, respectively) out of a 4lb and 8lb gun in an effort to prove the conservation of momentum. The weight of the firearm had no effect on the bullet, but did have a clear effect on your shoulder. Just sayin’. Also, try firing a 12 gauge with a pointy dowel rod next time, the stock bruises bad enough with poor form…

    • GhostofBrowningNagle

      my guess is a bigger projectile is better for making sure energy is transferred into the target

      so like you said, smaller bullet too prone to over penetrating. a teeny tiny bullet with all the kinetic energy in the world is useless if it dumps next to nothing into the target as it passes through

      • Cal S.

        But allegedly, ‘energy dumping’ is also a myth. I don’t think anybody really knows what’s going on. Either that or they’re too scared to say anything about it.

        ‘They’ also say surface area doesn’t matter…

        • GhostofBrowningNagle

          energy dumping isn’t a myth. if a bullet has say, 500j upon entering a target and the bullet doesn’t pass through, all that energy had to go somewhere.

          some of it is deforming the projectile itself, but the rest is spent deforming the target (meat explosion!)

        • I actually feel I have a pretty good grasp on what’s going on. It’s more complex than a lot of simulations assume, but it’s not that conceptually difficult to grasp as it’s entirely mechanical in nature (at least, disregarding the psychological and physiological components.

          • Cal S.

            Actually, I did have a bone to pick with your testing methodology. I know you’ve taken a lot of heat, so take this with a grain of salt. I just put my finger on why my previous point on the bullet weight and firearm bothered me so much.

            Harkening back to my ME classes, the conservation of momentum, simplified, is m1v1=m2v2. Well, your shoulder never had to worry, since the mass of the .22lr bullet was offset by the mass of the firearm (including the now empty casing). V2=m1v1/m2. It’s academic. The bigger the m2, the slower the v2 and the less felt recoil.

            Not so much an argument of the results or theory as it is the testing method. Plug the stats for a .500 nitro express into that equation, and you’ll see where your methodology falls out. I’d do that one with pressure sensors if I were you.

          • I could get injured if the cartridge were powerful enough or the rifle light enough, true, but the fact that I didn’t get injured here versus the very obvious fact that a speeding .22 Long Rifle bullet will surely injure you if it hits is the most illustrative scenario, I think.

            That’s not a fault of the methodology, but a perk. You and I both know how momentum works, this test was designed to illustrate that to people who probably don’t have a good grasp.

          • Cal S.

            Forgive me if I’m not understanding your premise or stepping out of the scope of this test with this, but…

            All it really illustrated was the effects of felt recoil over a smaller area (dowel rod versus butt pad). Myth Busters showed that the momentum of a projectile does not impart enough force on the target to knock it down. So, yes, a projectile will only impart the same force on the target that it imparts on the shooter (plus weapon), but that hasn’t been my question or the question of others (that have realized the myth of the “Knock-down power” of a cartridge).

            The real question for me is, what effect does momentum have in terminal ballistics (penetration/damage/wound cavity)? I think a more accurate test, and maybe you have this in the works, is to have two identical (in size) projectiles traveling at identical velocities with the only difference being the weights. I think we both know that the heavier projectile will penetrate deeper and cause more damage, no?

          • A 34 grain .22 Long Rifle bullet striking a target at 1,100 ft/s will produce a serious wound, and it may be enough to kill. However, that projectile has less momentum and the same frontal area as the cleaning rod attached to the rifle when I fired it, yet I was completely unharmed by the cleaning rod.

            This demonstrates that it’s not momentum that is inducing the lethality, but something else (that something else being energy).

            Repeating this test with larger caliber rifles would prove nothing more, although I suspect they wouldn’t be nearly as injurious as some people here think they would be. Of course, you could conceive of a variation on this test where the rifle was light enough to have enough energy to injure you, but that isn’t very illustrative of what the momentum is doing.

          • Cal S.

            Ok, I think I’ve finally wrapped my head around it. I’m an audio-visual learner, mostly audio, so I have to talk it out a little first (yes, I talk to myself as I’m typing, don’t judge).

            Thank you for your patience in explaining it. I was about to type out a reply about a 68gr .22lr @ 1100fps having twice the momentum and trying to equate that to penetration when I realized that would also double the KE. Also, the ‘target’s’ own mass would cancel out the momentum more so than the gun’s mass had canceled out the momentum of the projectile in my previous example.

            My bad, I’ve got it now. Forgive me, my engineering principles are a tad rusty. They’ve been wasting away for 4 years under my current job as a computer programmer/technical writer and haven’t been oiled well lately.

          • That’s a bingo. Thank you for being patient and listening!

          • georgesteele

            Not exactly. The momentum of the firearm was essentially that of the bullet. The rearward movement of the firearm imparted that momentum to his shoulder. It’s a transitive force, acting through a transfer function. All this proves is that momentum is a lousy metric to use for “stopping power” effectiveness, since it is a product of mass and velocity, and at very low velocity, heavy objects are ineffective at stopping people. Kinetic energy is more reflective of stopping power, but it is not a good metric without taking other factors into account.

          • Cal S.

            Yes, momentum is conserved, but all he did was prove why “Felt recoil” is called what it is. You know, why firing a .357 magnum out of a pocket-sized revolver and a 16″ lever-action feels so different even though the cartridge didn’t change.

            I guess it does dispel the myth that momentum can actually “Knock down” anything because the momentum produced by a .0057lb-.021lb bullet isn’t going to be enough to do that. However, and maybe I’m misunderstanding the scope of this test, I’m more interested in the effect that momentum has on terminal ballistics and wound cavities. Does momentum help a projectile penetrate further and cause larger permanent/temporary wound cavities?

  • GhostofBrowningNagle

    Consider 3 objects:

    First, traveling due east, an 80 kg man moving at 2.5 m/s. He as momentum of 200 Ns and kinetic energy of 250j
    Second and third, traveling due west. a .01kg bullet moving at 305 m/s. It has momentum of 3 Ns and KE of 465j and a 500kg blocking moving at 1 m/s, with momentum of 500 Ns and KE of 250j.

    if the man and the bullet collide, the man keeps moving forward, barely impeded
    if the man and the block collide, the man is now moving west.

    momentum = stopping power.

    • GhostofBrowningNagle

      i mean, obviously the man notices because he now has a hole in him, but the effect on his forward momentum from the bullet is negligible

      • ostiariusalpha

        Great! So now all you need is a weapon that shoots 500kg blocks.

        • GhostofBrowningNagle

          it would be the gold standard for knockdown power, thats for sure

        • GhostofBrowningNagle

          if you wanted to just stop the man, as opposed to sending him in the other direction, roughly a 10,000 gr projectile at 1000 ft/s should do it provided it hit him square and they stayed together…..in reality it’d probably just tear off a chunk and keep going

    • georgesteele

      Misdirection. What your argument proves is that an 1100 pound block will move a man backwards. That’s not stopping power – that’s arresting of forward motion, in an example entirely unrelated to bullet impact effect. Unless, of course, your pistol fires 1100 pound blocks. Big pistol.

      The man in question is not “stopped” – in fact he’s still moving, but backwards – but that wouldn’t stop him from firing a gun at you.

      Stopping power in the context of firearm effectiveness is a shorthand term for immediate incapacitation; that is, once hit, does an assailant ~immediately no longer have the capacity to inflict harm on the victim.

      Put another way, if the man and the bullet collide, and the impact point is the man’s forehead, then the inverse is true: the lower momentum object has greater stopping power.

      • GhostofBrowningNagle

        well, as i said in an earlier post, i equate “stopping power” with “arresting of forward” motion. but i readily concede there’s no scientific definition

        if you want to say in the context of firearm effectiveness, my definition is wrong, i can’t argue. .

        • georgesteele

          Ah – I didn’t see your earlier post, which would have given an alternative context. In that case, we’re both right!

          • GhostofBrowningNagle

            as long as we agree: regardless of earthly calculations of momentum or kinetic energy, the 45 acp wholly removes from the physical realm the entire spiritual essence of an individual, as if his consciousness never existed.

            perhaps THAT is the true definition of ‘stopping power’ ? :p

          • georgesteele

            Perhaps the .45 ACP should be called the Stalin Pogrom round, then. But wouldn’t that better be called “erasing power?”

          • GhostofBrowningNagle

            heheh.

            though if we named if after a Rooskie, we’d probably need to use the metric designation as well. 11.43×23 doesn’t roll off the tongue the same as “forty five aaaaay ceeeee peeee”

  • Oh yeah, I remember when thy brought that sketch artist in, he was a really nice guy, although he kept spitting sunflower seeds all over my floor.

    Still, I think his representation of me was very true to life, and I’m pretty happy with the final work.

  • Well, guess what, this post has nothing to do with any of that, so even in the weird alternate universe where you’re a voice of sanity and reason, your comment is gratuitous and misplaced.

    Did she wear a purple cardigan? I remember someone with a purple cardigan…

    • A Fascist Corgi

      “Well, guess what, this post has nothing to do with any of that, so
      even in the weird alternate universe where you’re a voice of sanity and reason, your comment is gratuitous and misplaced.”

      Here’s what you wrote in the first part of your “Top 4 Ballistics Myths Most People Believe” series of posts:

      “If you have some handy, take a .45 ACP and compare it to a 9mm, or a .308 Winchester against a .223; any two cartridges that are
      significantly different in size and weight will do. It’s so obvious, in a
      way that is maybe hard to explain, that the bigger round is better,
      that it will do so much more damage.”

      It’s extremely obvious that the underlying reason why you made these posts was to justify your embarrassingly stupid comments about the 5.56x45mm NATO round having superior terminal performance than .308 and .50 BMG ball ammo.

      “Did she wear a purple cardigan? I remember someone with a purple cardigan…”

      Did you remember to lick the Cheetos dust off of your fingers before you wrote that? And is it even physically possible for you to have sex?

      • You’re obsessed with me. It’s creepy. Go away.

  • Darren Hruska

    Let’s, say, compare a 750-grain .50 BMG bullet with a muzzle velocity of 2,900 ft/s (which would be a pretty damn hot loading) to a 150 pound (1,050,000 grain) man charging at you at 10 mph (14.666… ft/s). Clearly, we all know which one we’d like to be hit by the very least. However, what about their momentum, their “knockdown” power? The bullet will have a momentum of about 9.656 ft-lbs/s while the man will have a momentum of about 68.372 ft-lbs/s (about seven times the momentum the bullet has). Furthermore, this is a time unit, meaning that the slower object will impart its momentum on its target for a longer period of time (and a person is MUCH slower than a supersonic projectile). So, if a bullet was able to throw somebody backwards like in the movies (especially with one as powerful at the example), having somebody charge into you would probably be a semi-legitimate way of air travel. However, this isn’t the case in the slightest. So, momentum is pretty much irrelevant in the world of terminal ballistics, even with very powerful cartridges.

    • Mark

      This.

      We might need a “dummies guide” to basic physics here. There’s a lot of misconceptions about and going back to first principles might clear things up. I trained as a physicist, and gunfighter terminology drives me nuts. It takes too many liberties. I blame Jeff Cooper! 😁

  • Jim Drickamer

    I agree that momentum is not the final answer to stopping power, but trying to prove that from the momentum of recoil is flawed due to the vast difference in weight between the slug and the rifle. You might be willing to take a .22 caliber hit in the shoulder from a 3 pound rifle traveling at 3 fps, but how would you feel about a hit from a .22 caliber slug at 1300 fps? These have approximately the same momentum. I can tame the recoil of a .50 Smith & Wesson revolver by adding more weight to it. I could make it feel like a .22 revolver with enough weight. But I still would not want to get shot by it.

    • ostiariusalpha

      Right, so the greater kinetic energy of the bullet is more important than its momentum. Glad you understood what Nate was getting at.

  • African PH

    Dude your experiment kinda sux. Take a 460 W/mag, remove the barreled action from the stock, load with the heaviest bullet possible and repeat the experiment by pressing the back of the bare action against your shoulder. All you proved with this experiment is that a heavy rifle absorbs recoil nicely.

    • I think you don’t understand how conservation of momentum works…

  • Kovacs Jeno

    You’ll get the same momentum as the bullet, when there is no bolt in the gun, and you try to lock the bullet case in the chamber with your finger, when its payload is exploding. Your finger will be blown off cleanly.

  • Jwedel1231

    “1/2*m*v^2” The m is mass, NOT momentum. If you were using momentum, it would be “1/2*momentum*v”, since momentum is mass*velocity.

    I’m not trying to be pedantic, but it’s the lack of understanding of physics that got us here, and you (actually fairly successfully) tried to clear up with this article.

    • Nobody said that’s what the “m” stands for. Here’s what I wrote:

      “One of the sources of confusion here, I think, is that kinetic energy – which is clearly a very important factor in terminal effectiveness – can be described as one half times momentum times velocity,”

  • Bjørn Vermo

    You might want to look up the meaning of “correlation”. I am willing to bet that the coefficient of correlation between momentum and “stopping power” is significantly higher than zero.

    • georgesteele

      No. It is only correlated with the velocity component relative to the mass component for a given momentum. That is, for any given momentum, M, the “stopping power” (a nonsensical physical principle, but a quasi-practical one – so we’ll abbreviate it as SP, even though it should be abbreviated BS) increases as the velocity goes up and the mass goes down.

      A higher momentum resulting from a higher mass at a lower velocity correlates *inversely* with “SP”; a higher momentum resulting from a lower mass at a higher velocity correlates *directly* with “SP.” Thus the correlation coefficient of “SP” with M is zero.

  • Joshua Knott

    But physics also states that energy will find the least resistance to exploit, so i dont really see the correlation between a bullet “finding” the least resistance out of the barrel, for example if a Masuer 98 produces somewhere in the 20lbs range of “felt” recoil but the round produces over 3000 ft pounds of energy into a target by means of exploitation of path of least resistance.

    • The subject of this experiment was momentum, not energy.

      • Joshua Knott

        Relation between momentum and kinetic energy

        Sometimes it’s desirable to express the kinetic energy of a particle in terms of the momentum. That’s easy enough.

        K= 1/2M (P/M)2 =P2/2M

        I love you Nathaniel in a brotherly way but the relation is there and undeniable. Ps I hate math, makes my head hurt, I prefer history haha.

        • Um, or you could express it as KE = momentum * velocity * 0.5

          The whole point of this article is to illustrate how conservation of momentum works.

          • Joshua Knott

            wasnt trying to troll or tick you off, just to have a genuine thought on the matter. I can see where you were going with it, there’s a lot of variables and that just means more explanation no matter how much you try and simplify it.

          • I don’t think I was ticked off, but you’re OK anyway.

    • 2hotel9

      Nathaniel refuses delivery of any facts that show .22anything is inferior to any real rifle bullet, such as US .30, 8mm Mauser, 7.62 NATO or Warsaw, .303Brit etc etc. Please, spare yourself the aggravation, simply nod your head and say”cool story, bro”.

      • If that’s what you think, then you’ve grossly misunderstood what I have to say.

  • tom

    Have you considered that a simpler approach would be to look at force being the factor at play?
    I think that the criticisms are somewhat in based in reality as the area does matter, but in the equation of stress, stress=force over area. Since stress is what determines the tearing and breaking of tissue it would be important.
    (I think) the reason that you rifle won’t cause the same damage, even with the same area of action, is that it will only accelerate the tissues in front of it to its speed, assuming that it doesn’t just stop and transfer all energy. since the mass of the rifle is so much, its speed won’t be that high, therefore the force will not be as high, therefore the stress won’t be as high, therefore not the same damage.
    Still, good article.

  • Jeff Edwards

    If you’re looking to hurt yourself, get a bigger gun.

  • kcshooter

    Until you can figure out why your “test” is totally invalid, I’d suggest you save yourself the embarrassment of posting any others. Honestly, if you think this proves something, you’re an idiot. Stop trying to prove the math until you are able to understand the physics. Eat less, think more.

  • ILicence

    “…kinetic energy – which is clearly a very important factor in terminal effectiveness – can be described as one half times momentum times velocity,
    but this does not mean that momentum itself, when not multiplied by
    those other factors, directly has an effect on the “stopping power” of a
    bullet.”

    The formula for Kinetic energy is 1/2 * MASS(KG) * Velocity(M/S).
    Am I wrong?

    • ILicence

      To clarify, MASS is used in the formula for kinetic energy, not MOMENTUM.

      • georgesteele

        Well, it isn’t in *that* formula, because you haven’t squared the velocity; momentum is, in fact, used in your above formula for kinetic energy KE= .5M*V, because that is identical to the correct formula KE=.5mV^2 (where m is mass, and M is momentum), because the formula for momentum is M=mV.

        • ILicence

          You nailed it! .5*m*V^2 = KE = .5*M*V (Where m is mass, and M is momentum)
          The wording of the article was kind of murky at times and I was trying to make that distinction. Maybe what I wrote was a little murky too!

    • georgesteele

      Yes. See below.

    • Mark

      This is absolutely fine!

      The confusion is coming from people thinking the m stands for momentum. Standard notation uses p for it!

      As momentum is defined p = mv

      Then we can substitute it in

      Ek = 1/2 * mv^2

      To get…

      Ek = 1/2 * pv

      Hope this clears it up!

      • ILicence

        Agreed! The proper use of “p” helps clear up the wording used in the article a ton.

  • Wil Ferch

    The article is seriously flawed. KE = 1/2mv2…..the writer asserts that the value “m” is momentum. It is NOT MOMENTUM as claimed….but MASS ( layman’s terms..”weight”).

    So…. we have 2 measures of a fast moving bullet. Kinetic energy ( with the formula shown, but noting the correction that “m” is MASS….not momentum)….and Momentum ( as a formula, equals M x V….or mass times velocity).

    Both formulas should be uesed for making comparisons amongst various rounds for “stopping power”. Simply as one example, if you plug-in the numbers, the 45 ACP and 9mm parabellum have fairly equivalent kinetic energy numbers…..but the 45 trumps the 9mm in the momentum calcs. That is pretty obvious as the kinetic energy formula has a bias towards the object’s speed ( velocity “squared”)….whereas the momentum formula only gives directly proportional weight to the object’s speed.

    In both formulas….. “Action equals reaction”…so the mass and velocity of the bullet relates to the shooter ( in the opposite direction) using the rifle’s weight and velocity compared to the bullet’s weight and velocity in the opposite direction….using the appropriate formulas for each case ( kinetic energy or momentum).

    The article does nothing to properly relate these two formulas and the reaction as seen by the shooter. The mass and velocity of the rifle will not result in a massive “hit” to the shoulder …using either formula…compared to the similar characteristics of the fast moving but much lighter bullet in the opposite direction.

    • georgesteele

      You wrote: “but MASS (layman’s terms..’weight'”). Almost, but not quite. Mass is a physical attribute of matter independent of gravity; weight is a measure of force exerted by that matter in a gravitational field. In earth’s gravity, a kilogram of mass generates a kilogram weight force, by our convention. On the moon, a kilogram of mass does not “weigh” a kilogram – because the moon’s gravity is lower – thus the kilogram of mass does not exert a force of 1 kilogram-weight on the moon’s surface.

      Also, I don’t see where he said that the m in his formula above stands for momentum. I think you misread him. He did ask the question: ” if only there were a physical quantity like momentum, but one that emphasized velocity more, maybe by, like, having it in there twice?” He then presents the kinetic energy formula, using mass and squaring the velocity (“in there twice”). There’s no flaw in that, other than that I think you misread what he was implying.

      • Wil Ferch

        Wrong…he is clearly quoted as saying…. ” I think, is that kinetic energy – which is clearly a very important factor in terminal effectiveness – can be described as one half times momentum times velocity, “…..

        This is wrong and I stand by my previous statement.

        As to mass…yes….mass is not the same as weight and that is PRECISELY why I said “layman’s terms”…as I knew I was not speaking to a peer group of engineers.

        • georgesteele

          The equations are identities. 1/2 m V^2 = 1/2 MV. Both are kinetic energy equations by identity, because momentum, M, equals mass, m, times velocity, V – i.e. M=mV. By substitution: 1/2 MV = 1/2 (mV)*V = 1/2 mV*V = 1/2 mV^2. Kinetic Energy IS 1/2 momentum times velocity.

          • Wil Ferch

            I stand corrected…..but it was a terrible obfuscation to introduce, for a topic that is hard for normal people to follow….to make such a substitution within the body-text of the discussion.

  • Charles Newman

    The only measurement that matters is the bullet having just left the muzzle at a given velocity and weight. Forget the recoil tests as different designs produce different results. It doesn’t matter if the firearms explodes and kills the shooter, but if the bullet is flying faster and heavier given any mayhem it left behind, heavier/faster is always better. So yes, momentum equals greater stopping power. Kinetic energy wins. Math not needed, just look at 50 BMG impact results.

    • Mark

      But other measurements DO matter!

      The argument is whether or not momentum is a RELIABLE indicator of ballistic effectiveness.

      Unless we make various assumptions that everyone agrees on and knows about, it isn’t. And science is VERY fussy about communicating information accurately so that we don’t draw silly conclusions.

      If I ask an engineer to design for me a device to deliver 1000ft-lbs of momentum to a target object, they might simply hand me a sledgehammer when I wanted a rifle.

      Imagine you are standing on a sidewalk when there is a big car accident where a wheel, tyre and all, comes loose and strikes you with the same momentum and energy as a .50BMG at the muzzle.

      That’d be a BAD day, but I’m guessing you’d still pick it every time over actually being shot with a .50 BMG…

      Thus we have just learned that cross-sectional area is also an important factor.

      Etc!

  • Wil Ferch

    The whole premise presented is simply seriously flawed in a number of areas. The fact that the writer puts a rod on the buttstock to have the “rearward” momentum ( action = reaction, etc)….is also flawed. Why ? MV of the bullet = MV of the rifle ( basically…there are actually nuance problems with this too….but is close enough to make a point. Saying this so I don’t get challenged like the “Weight” vs “mass” point I made earlier…again, doing this as a simplification that doesn’t much matter in the context of the discussion).

    Ok…so the small diameter of the buttstock “rod” ….is a close approximation of the maybe the bullet going in the other direction. The problem?….. you didn’t correct for the weight of the stock/rifle. The rod becomes part of the rifle/stock “mass”.

    (MV) item 1 = (MV) item 2 … in the other direction.

    For those so inclined….do the math directly. Here’s some help. There are 7000 grains to a pound ( when looking at a bullet’s weight and compare that to the weight of the rifle. The measurement units for MV= MV…need to be the same for the 2 items moving in opposite direction.

    Ya gotta look at the mass x velocity of the bullet…..and then say it will be equal ( and opposite) to the mass x velocity of the gun/stock combo.

    The premise here to add a rod to the stock is Totally Meaningless.

  • georgesteele

    You can probably tell from the number of my posts below that I find this subject interesting. I have been trying to work up an understanding of just what does go into “stopping power” as a physical measurement (i.e. independent of physiology, anatomy, and psychology). It seems to me that the wrangling over momentum vs. kinetic energy may be the problem; it may be that putting them on opposite sides is where that problem originates. I would think that the truth is nearer something like: for a given velocity, higher momentum equals greater stopping power. That puts energy and momentum on the same side, holds velocity constant, and gives value to the mass component. Heavier bullet, more damage – or at least more potential damage because of increased penetration.

    But it’s not complete – because if we look at two different projectiles of equal mass – one an inch in diameter, and one a millimeter in diameter (think rifled slug vs. long, thin, heavy rod), intuitively we would think that the former would have a greater “stopping power” effect when impacting at the same velocity, in most cases. It would transfer its energy to the target more instantaneously (again – both objects have the same energy and the same momentum) because the force generated would operate over a larger resistive area of, say, a torso, as well as having a statistically greater chance of nicking an aorta, CNS component, or some such, post-penetration.

    The only case where that would be wrong would be one where the rifled slug was unable to penetrate the target (think bulletproof vest) because the energy was dissipated over too large an impact area, whereas the long, thin rod of the same weight did penetrate, because its energy was focused on a much smaller area (1/645th the area, in fact). So the force per unit area exerted by the long, thin projectile is almost three orders of magnitude greater, would penetrate the vest, chest, and who knows what else on its Shishkebab of Death (good name for a rock band), inflicting a wound that could well stop the fight – as opposed to just being an annoyance and causing ringing in the ears for the slug.

    So for any projectile, first it has to penetrate, then it has to reach a vital area, then it needs to damage enough of it to cause some form of disabling effect, and the effect needs to be large enough to incapacitate the target. The first item relates to form factor, velocity, and mass, the second seems to be strongly affected by momentum, the third to frontal area, and the fourth to velocity and frontal area – at least in what we think of as a traditional bullet.

    Of course, all these factors are what goes into cartridge and bullet selection – whether self-defense or hunting small, medium, or large game. But those choices seem to be empirical and anecdotal in their genesis, whereas this is a more theoretical and principles-based definitive approach.

    Just some thoughts – your article is thought-provoking to a fault, and much appreciated.

  • robert57Q

    I always thought that what matters most is the amount of energy transferred to the target, and the amount of damage to vital organs.

  • Zebra Dun

    A Feather at 250 Feet per second will tickle you.
    A BB at 250 feet per second will sting might even penetrate a little.
    A 40 mm M-79 launched solid grenade will do more damage at the same velocity.
    A semi tractor at the same velocity will kill you.
    Weight as well as speed and size have a lot to do with terminal effect.

    Recoil of the projectile is another, a feather will bounce back, a BB might bounce back, a Thump gun grenade will penetrate into your body before bouncing back a little, a five ton semi Tractor will not bounce back at all.

    All is moot, a bullet will injure or kill you a rifle recoil will hurt or not.

  • kcshooter

    Part of the reason there is so much misunderstanding when it comes to external ballistics is guys like this and the serious misinformation they spread thru formats where they are considered “experts”,

  • nicholsda

    Oh come on. Be a real man and do it with a Mosin-Nagant M-44 and the rod shaped like the bullet it fires. SAFETY WARNING: Don’t try this at home or the range. A 7.62X54R round in an M-44 ( or even a M-N 91-30 ) can make your shoulder hurt as the force is above the 10 lb-ft level. Mosin-Nagants, relocating dislocated shoulders since 1891. 🙂

  • Billca

    This is a discussion that’s been going on for a long time, probably nearly as long as guns have been around. It doesn’t matter if you call it stopping power, knock-down power, or whatever, the best we can do is describe an approximation of the expected results. Mostly because we are dealing with highly variable organic bodies which can react differently to the application of terminal ballistics.

    I’m not a physicist so I’m sure someone more educated in the subject will chime in with corrections here.

    Some will speak of the system (here referring to the gun & projectile) but we miss treating the bullet itself as a system. Not only do we have linear energy of the projectile moving at velocity, there is also rotational energy (moment of inertia) from the bullet’s spin.

    We know that bullets exit the barrel spinning to help maintain a stable direction of flight (actually a curve due to gravity and air resistance). And we know that the momentum can be expressed as energy to do work (in ft-lbs) to use as a relative measurement for comparisons.

    But we also see in various “Gel tests” and real life examples that path of a bullet often changes and some bullets rotate inside the test medium, inverting to move base-first. This means there are torque effects occurring on the bullet that are largely ignored. The energy required to change the bullet’s flight-path in soft tissue may not be large but they exist. The energies to cause a stable projectile to invert while still having forward momentum are more significant than those merely diverting it’s flight-path, I think.

    The energy to cause a bullet to “tumble” is energy from the bullet’s motions (linear and rotational) being unevenly absorbed by the target medium as it resists the passage of the bullet. This is a non-trivial event, especially if it happens rapidly in a short time/distance event — but is often ignored. The result we see is usually a fairly substantial permanent cavity as the bullet rotates and presents a larger surface area moving forward.

    It’d be an interesting avenue of investigation to determine the effects of this energy including whether things like the texture of the bullet, number of lands & grooves or left hand or right hand twist makes much difference.