Help with an Argument on "knock down" power

The exaggeration response is a very real phenomenon. I've known my father-in-law for over 40 years, and he has always responded in a very exaggerated way to any physical contact, loud noises, or input. Damnedest thing I have seen. I quite sure he would be knocked down by a .22 lr round fired into a vest.:rolleyes:
 
tangolima said:
First we need to properly definite "knock down". Without a definition, we just get hand wavings of different kinds.
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Here is what I would use. The force required to knock an average sized person standing normally while conscious and sober off their feet to the ground by only the force applied (not physio or psychological reasons) of the bullet striking them. Is that ok?

Hand waving? Do I know you?:D
 
JohnKSa said:
The dummy was rigged with an armor plate in the torso. The bullet did not pass through the dummy in the test--it was stopped in the torso.
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So the dummy received the full force and it was not dissipated by penetration?
 
I quite sure he would be knocked down by a .22 lr round fired into a vest
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I'm not so sure I wouldn't be knocked down by a .22LR round fired into a vest! ;)

I've never been shot by a .22lr fired into a vest, so I don't know, for SURE, but I think it might be likely, if I didn't expect it. (and since I'm a LONG way from 25 and the peak of health and balance, its quite possible!)

One of the funnier things I've ever seen was a news clip that was in a "bloopers" show I saw ages ago.

An reporter, standing in front of the camera, holding his mic and wearing a bulletproof vest over his suit jacket. The reporter tells us about the new vests the local cops are getting, and how they "really work" And then he says, "and to prove it, I'm going to let them shoot me!!!"

At this point, a hand holding a small shiny revolver (nickle or stainless) appears out of the right side of the frame, distance a couple of feet away.
There is a POP sound (all gunshots become "pops" on video tape) and no apparent recoil of the revolver.

The reporter staggers back several steps, stumbles and falls, dropping the mic. He slowly gets to his hands and knees, picks up the mic and stands up, and we hear a slightly muffled "Oh F(bleep), that hurt!!"

:D

I do not think the reporter had been adequately briefed on what was going to happen. He wasn't incapacitated, but he was KNOCKED DOWN by the little round hitting the vest.

The bullet didn't physically knock him over, but he was knocked down by his own reaction to being shot.

Plug that into various formulas and see how it computes! :p

I've talked with people who have been shot. Some describe it as a "hammer blow", others said "a fierce burning sensation", one guy (a cop) said he didn't even know he had been shot, until after the gunfight was over....

Everyone is different, and some people are very different. I don't think there is any math that can take that into account, accurately.

Formulas and statistics may be useful for comparing different rounds to each other, but their accuracy in predicting real world results is 50/50, each and every shot. The real world either behaves in line the formula, or it doesn't.
Flip a coin, its just as accurate a predictor...
 
Tennessee Gentleman said:
Here is what I would use. The force required to knock an average sized person standing normally while conscious and sober off their feet to the ground by only the force applied (not physio or psychological reasons) of the bullet striking them. Is that ok?

Hand waving? Do I know you?:D
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Stick one hand out, palm facing down. Rotate the forearm back and forth in a oscillating manner, so that the sticking out hand moves in mid air like ocean wave. While doing that, you are making statements filled with "something like that", "like", "stuff" and lol. That's hand waving, not the kind you do when seeing your neighbor.

Your definition is good. I would add the person's feet are tied together side-by-side, so that he can't be in a "horse stance". I think energy of 1000 ft-lb should suffice. I haven't tried it myself and I am not going to suggest any experiments. America is no longer land of the free, but land of the blamed.

-TL

Sent from my SM-G930T using Tapatalk
 
Hey, no, I know more than you do!

Oh yeah, well I've got three doctorates
in physics!

Your doctorates don't amount to a hill
of beans compared to my experience!

Heck, none of you know as much as I do!

Wait, I have the answer, it's, it's, it's
time for dinner. Afterward, let's try
for a three-page thread or maybe four.
 
your friend is wrong. He doesn't understand physics or guns, and you are wasting your time.

If he can't watch a guy shooting steel plates that just bounce a little and understand the facts, he's never going to get it and it's not worth the effort to try and break through that thick head.

I watched a guy shoot a 100 yard partridge target with a .44 magnum, I guess the target weighed five or more pounds, the thing stopped about ten to fifteen feet from impact after falling and rolling.

Now if a .44 magnum could barely knock down a steel block, doesn't it also follow that a .44 magnum would have a really hard time "knocking down" a nice water filled jelly bag that weighs about forty times as much? If a .44 magnum can't knock down that bag of gunk and bones, could a .45 acp do it?

Measuring "knock down power" is the simplest thing that you could ever test. Start knocking things over. If you do some very extensive testing, you will probably find that anything much heavier than a log will "fall" down, and knock down is the absolutely worst way to describe that.
 
So the dummy received the full force and it was not dissipated by penetration?
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The .50 BMG bullet did not penetrate through and through, but it did penetrate into the dummy considerably, even going through the armor plate. After passing through the armor plate, it was stopped in the steel "spine" of the dummy and did not exit.
 
Imagine a bucket of water. If you smack the water with an open hand, it will make a big splash and displace a lot of water. This is because your hand is striking the water with considerable force, but it is spread over a wide area, so it doesn't penetrate very deep. If I plunge my fist into that same bucket of water, I will likely hit the bottom with ease. This is because, even though I am moving my arm at the same speed and with the same force, my fist has less surface area than my splayed hand. The force is more "concentrated".

It's the same way with bullets. Most bullets simply aren't wide enough to reliably knock someone off their feet from impact alone.
 
I noticed that some posts here compared the foot pound measurements of bullets to actual feet/pound measurements. This isn't a proper use of those measurements. The kinetic energy of a bullet is not comparable to the low velocity heavy weight. The bullet expends large amounts of energy just blowing through the tissues. Hitting a steel plate and expending almost all of that energy moving the plate is one thing, hitting a squishy pile of gunk and spreading that out in damaging the squishy stuff is another. Energy is lost in the splatter.

this is a super duper .44 load. it did a good job on the block, maybe a 30 pound block?

This round will not 'knock down' a man, but being hit that brutally will almost certainly make him fall over. A guy standing on two feet has a pretty tall center of gravity and he's standing on only a narrow platform with weak ankles. Would you expect a guy's legs to go pretty limp when he takes a hit like that?

It all depends on what "knock down" means, and what all of the other language means. Something to think about, were you ever hit by a baseball? 1/3 pound, 60 fps. All I ever did was swear very loudly.

https://www.youtube.com/watch?v=NU6ZKQrfr9M

This second video uses my personal carry load. That certainly failed to knock down the gel, but it really scared it.

https://www.youtube.com/watch?v=PjxSr-TYFVA
 
This could kinda be fun to think about but it becomes almost a catch 22 situation.

Newton's 2nd law: basically F=ma (Force = mass * acceleration)
3rd law: all actions have an equal and opposite reaction

With a constant acceleration while in the barrel to propel a bullet from rest to a certain velocity, there will be a steady, not-too-unpleasant force applied to the bullet, and a (equal and opposite) force applied to the firearm/shooter.

If we assume point blank range, and that the bullet does not pass all the way through the target, that same velocity that was reached must be slowed down from its peak back to zero fps by the target.

If the target is soft tissue, the bullet could take even longer to decelerate (negative acceleration) than it took to leave the gun barrel since it is passing through tissue over time. That means it would have lower acceleration, and therefore would apply lower force to the target than it applied to the gun/shooter system.

However, theoretically if the bullet impacted a target that did not give at all, such as some armor steel anchored such that it doesn't move at all from the impact, the negative acceleration of the bullet would be nearly instant. That means a huge number for acceleration which means a huge number for the force applied to the target, which could theoretically knock someone down. Of course the large force was only possible because the bullet impacted something immovable.

TLDR: A bullet could theoretically apply enough force to knock someone over, but only if they cannot physically be knocked over. Catch 22!
 
How the force is applied, and the properties of the target medium are critical to determining how much and whether the force will actually cause motion.

Here is a quick thought experiment.

Go out to your driveway, take the parking brake off, put your car in neutral and apply 100lbs of force to the car by pushing on the trunk with your hands. The car will move, and if you continue to apply force it will gradually pick up speed.

Now, put the car back where it was, brake off and in neutral and go get a sledgehammer and swing it at the back of the car so that it applies 100lbs of force. Same amount of force applied to the same target but now the force will be used up putting a large dent in the trunk and won't result in any noticeable movement.
 
He is an industry expert and knows what he is talking about. If he uses the term, then it is valid:
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https://www.logicallyfallacious.com/tools/lp/Bo/LogicalFallacies/231/Argument-from-False-Authority

This is a classic appeal to false authority fallacy.

The fact that Mr. Potterfield is obviously very good at and clearly very knowledgeable when it comes to making money by selling firearms products does not imply that he is an expert when it comes to terminal ballistics, that he has any special insight into the topic, nor that his use of a term in that field provides any proof that it is valid.

Here's the link that explains what those numbers mean. Apparently, they are someone's idea at Midway USA of how various cartridges should be ranked. In other words, even if one chooses to accept them as valid, they aren't a measure of effectiveness, they are merely a ranking.

https://www.midwayusa.com/content/HowToGuides/CCH-Rating-Chart.htm

For some arbitrary reason, they topped out their ranking at 400 which means that nothing can rank higher on the list than a .45ACP 185gr JHP +P load. Not even a 12ga shotgun or a .50BMG rifle.
 
Here are two points.

Red, you applied newton's law, that is correct, but misapplied. Newton's law applies to pool much better. The cue imparts motion to cue ball, to next ball, then all balls are impacted during the break. There is very little energy lost in the process. A pool cue is very heavy, and imparts high velocity movement to the ball, and a enormous amount of other impacts give all of those other balls a share of that motion, the motion is expended on friction and deformation of the bumpers.

The energy in a bullet is conserved, following the other law, conservation of energy. It is simply converted to lots of different forms. Some of that energy is converted to friction that releases energy as heat. Then we have the energy lost when the bullet deforms, as much energy as it would take to hammer in a nail. (Look at blank powered nail setters). The next thing that absorbs energy is deforming anything that it touches, but if you are talking about a lead bullet on steel, obviously, the only deformation of the steel that occurs is on essentially microscopic and extremely low levels.

Blah blah blah, more still, the last I have to point out is that motion is imparted two ways, but only a part is usually transferred as movement. What energy isn't used up in all of those other processes is expended by pushing against the target. In meat, most of that energy seems to be not used in pushing, shoot a gallon of water, it will barely move. Steel applies a lot of energy as movement, but still, a lot of that movement is nothing but vibration, transferred to sound through the air.

So, most energy is expended in other ways while the rest is transferred as push.

You can't 'knock a man over' by hitting him with a bullet, the vast majority of that 'push' is converted into other forms. We like that. We want that energy converted into destruction.

Model twelve, as was said,lets not start up with an argument with one 'expert' against another.

Factor isn't a science term, really. They used some formula to calculate how much energy is available for push, that is obvious and simple, and it would work just fine if nobody used anything but fmj ball.

But once you change a variable like adding a hollow point, The force that would be used as 'push' or 'knock down' is being converted into heat and destruction. No 'knockdown' factors exists that can be applied to real world shootings on anything anything that deforms, or by any item that deforms.

You can transfer motion, per Newton, or convert it. The best science is still inadequate to create a genuinely effective calculator for the effects of a gun on a person. There are too many possible variations in the target or projectile to factually declare a scientific law based on only factors of velocity, diameter, and weight.
 
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Energy is not conserved. Momentum is ALWAYS conserved.

Even in a perfectly inelastic collision, momentum is always conserved.

Barring overpenetration, it doesn't actually make any difference whether a FMJ or JHP is fired into a ballistic pendulum. Deformation of the bullet and target, noise, heat, etc, is all irrelevant as that corresponds to wasted energy instead.

Midway's "powerfactor" is an ordinal unitless madeup number that doesn't correspond to any physical value.

There isn't enough momentum in a bullet to knock down a person.

And that's fine, as "knockdown power" is a physiological phenomena instead. Consider the "knockdown power" of a groin tap, for example.
 
Kozak, have you ever heard of the law of conservation of energy?

https://en.m.wikipedia.org/wiki/Conservation_of_energy

There you have it.

You can convert, divert, whatever sort of energy you want into another, maybe undesirable form of energy or work but it is still there, you can't 'waste' energy unless you mean that you aren't getting a desired result.

Your assessment of whatever you are referring to as 'wasted energy' is wrong. Momentum cannot ever possibly be entirely transferred from one physical thing to another, elastic or inelastic, because in EVERY collision there is energy converted from momentum into some other form suck as heat.

But, whatever. Believe whatever you have to believe.
 
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Yeah, lotsa foolishness out there. My two buddies came home from Paris Island Boot Camp in 1968, advising that the reason the 5.56 was so effective was that it starting tumbling out the barrel.......... As far as knockdown, the Taylor Knock Out formula comes the closest to field performance for me.
 
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