Handgun stopping power

[Guv]

We had an explosion of a test stand at work and had to build a 3/4" Lexan box around it for future testing. Just to prove a point a couple of engineers shot a spare piece with various cartridges. All failed to penetrate the Lexan, 223 55gr FMJ rifle, 45ACP 230HydraShock, 9mm 124 +p, 380. I put one in it on a different day with a 180gr Speer out of a 10" T/C 357 Max and blew a nice hole all the way through. I won some wings on that deal!

 

[Brian Pfleuger]

I still say stopping power is knockdown power when hunting. Like a deer as said before. Shooting the animal through a vital area such as the heart will stop the animal and it will fall down because of the shot hence knocking it down.

So long as you specify that you are defining words in a way that is exclusive to you so that others know that the words have a different meaning than they normally would, you are entirely correct. You have to be correct, actually, since you are defining the words.

However, the way you are using those words is completely different than they are typically used.

It's also inconsistent with your own usage of the words in the first post of this thread. Originally, you equated "stopping power" with penetration.

Stopping power is not knock down power, knock down power is not stopping power, penetration is not knock down power, penetration is not stopping power.

 

[Nanuk]

I love when people say there is no such thing as stopping power. Sure there is you just need to define it and keep your responses in context. For example if you have to shoot to defend your self and shoot said BG and he flee's immediately after being struck is that not a stop? In all the shootings I worked or saw very few people were instantly stopped and the vast majority of those were decisive CNS hits.

The other one I really like is rifle velocity VS handgun velocity.

http://www.ar15.com/ammo/project/Fackler_Articles/Theodore_Kocher.pdf

I was shooting round bales one year with my bow, bad move the arrows were lost in the bale and the horses chewed them up.

 

[Brian Pfleuger]

I love when people say there is no such thing as stopping power.

A "stop" is not the same thing as "stopping power". All guns and all bullets have some measure of stopping power.

I could poke your finger with a needle and if we let it bleed long enough you might pass out. Does that means a needle has "stopping power"?

Stopping power is simply the ability to force something or someone to stop doing what they're doing. The "power" side of it is some sort of measure as to how quickly the cartridge/firearm in question is likely to effect that stop. If Cartridge A makes someone stop in an average of 3 seconds and Cartridge B makes someone stop in an average of 5 seconds, we can say that A has more "stopping power". (If only it were that simple.)

I say again, Stopping Power is not the same as Knockdown Power. "Knockdown Power" is something that no (civilian) man-potable weapon has... no rifle, no handgun, no shotgun. It is physically impossible. Physically Impossible. Physically, as in the laws of physics. Impossible, as in violates those laws of physics.

And again, as relates to the OP of this thread, stopping power does not equal penetration.

 

[Ruger480]

I say again, Stopping Power is not the same as Knockdown Power. "Knockdown Power" is something that no (civilian) man-potable weapon has... no rifle, no handgun, no shotgun. It is physically impossible. Physically Impossible. Physically, as in the laws of physics. Impossible, as in violates those laws of physics.

I will agree with 100% except first I need to know: will those riot control beanbags fired from shotguns knockdown a person if say...shot in the face or ear?

 

[Brian Pfleuger]

They may cause a physical reaction such as flinching/jumping away or a loss of leg control from the response but they do not "knock down" the person.

If they did, the gun would also knock down the shooter. The laws of physics require it. Momentum must be conserved. The gun has the same momentum as the beanbag. If the beanbag has enough momentum to knockdown the person it hits, the gun has enough momentum to knockdown the person who pulled the trigger.

 

[buckhorn_cortez]

If they did, the gun would also knock down the shooter. The laws of physics require it. Momentum must be conserved.

Partially correct.

That's not how physics works. You are conflating momentum created through acceleration with kinetic energy.

The bullet is accelerated by the gun powder. The gun is accelerated in the opposite direction with the same amount of force. That has nothing to do with the bullet's kinetic energy.

That equation is: F (momentum) = M x A

Where F is the force created from the Mass x the Acceleration.

So, there is an opposite and equal reaction. The bullet is extremely light in comparison to the gun so the gun is accelerated in the opposite direction with a very small relative force at a much slower rate because of the mass & inertia of the gun.

Once the bullet is accelerated to it's maximum rate of speed the kinetic energy formula becomes:

KE = 1/2M x V^2.

Notice the difference - it is the velocity squared and that's where all of the foot pounds of energy come from. This has NOTHING TO DO WITH THE FORCE OF ACCELERATION THAT CREATES RECOIL.

Also notice that the amount of energy is directly linked to the velocity so that as the bullet slows down, the amount of energy is decreased proportionally - exactly how it works in real life.

Think of it this way. A 4,000 pound car can be accelerated to 60 MPH with very little energy. However, the kinetic energy released from the 4,000 pound car when it hits a brick wall is thousands of times more than the original amount of energy required to get it to 60 MPH.

Or, look at it this way. If, in fact, recoil was related to the energy imparted by a bullet at impact, then recoil would have to be infinitely variable depending upon the distance from the gun at which the bullet hit an object all the way to zero (bullet hits nothing, travels until it hits the ground).

Does that even sound plausible?

 

[Nanuk]

I could poke your finger with a needle and if we let it bleed long enough you might pass out. Does that means a needle has "stopping power"?

Stopping power is simply the ability to force something or someone to stop doing what they're doing. The "power" side of it is some sort of measure as to how quickly the cartridge/firearm in question is likely to effect that stop. If Cartridge A makes someone stop in an average of 3 seconds and Cartridge B makes someone stop in an average of 5 seconds, we can say that A has more "stopping power". (If only it were that simple.)

Which is why I said you need to definitively define the term used in the context.

If hostilities stop at the instant the person is struck and they are now more concerned with saving themselves or are unable to continue the attack, I would say that the attack was successfully stopped.

Penetration, momentum, muzzle energy, crush cavity, and pain compliance all play some role. In my LE career I have fought people on PCP, LSD, Heroin, adrenaline or any combination. Not to forget emotionally disturbed people, some people are just plain mean. Some people literally feel no pain, they are very, very hard to stop with anything.

So to put in context I carry the most powerful gun I can conceal and shoot well. Right now it is a S&W 44 magnum.

 

[Theohazard]

And by no means I'm I trying to argue. It's just my opinion/thought on knockdown and stopping power. I understand a handgun round won't physically knock some one down or stop them physically even though it may.

When you say "knockdown power", people will assume you mean the bullet is physically knocking someone down. And many of those people will therefore assume that you have no idea what you're talking about. But I think the problem here is that you have appear to have an entirely different definition of "knockdown power" than everyone else.

When people say "knockdown power", they're referring to a bullet's supposed ability to inflict enough force to actually knock someone off their feet. But no conventional bullet has that ability. Not even a shotgun slug. All the hunting stories of deer being knocked down or flipped over by a bullet or slug; those are stories told by hunters who simply don't understand what they saw. Deer often react strangely to being shot; they might lurch to one side or jump straight in the air. But that's simply their reaction to being shot, that's not because of the kinetic energy of the bullet itself. Think of it this way: If I poked you with a pin, you'd probably jump. But did force of the pin itself cause you to move? No, it was your reaction to being poked with a pin.

Here's the classic video of a man wearing body armor being shot in the chest with a .308 rifle while standing on one foot. The first time he's shot he slightly loses his balance and has to put one foot down. The second time he's shot he doesn't even lose his balance (fast-forward to 5:22 in the video to see the second shot). And this is a .308 rifle with about 2,600 foot-pounds of muzzle energy; your handguns only have about 300 - 500 foot-pounds of energy.

 

[buckhorn_cortez]

"Stopping / knockdown power" is a very complex subject.

Power is the amount of work over time causing displacement.

Work = force (F) x displacement (d) x Cosine (Theta)

The most complicated part of the work equation and work calculations is the meaning of the angle theta in the above equation. The angle is not just any stated angle in the problem; it is the angle between the F and the d vectors.

With bullets, stopping power is interactive with the object itself and its makeup as it governs how the bullet's kinetic energy is converted to other types of energy and "power."

The problem with quantifying "knockdown power" is that the bullet itself reacts to the energy it contains. For example, if you shoot a bullet at a target made of A500 steel, the target moves only a relatively small amount when hit compared to the theoretical kinetic energy contained in the bullet.

Why? Because the bullet expends its energy on itself as it shatters into fragments as it is stopped when hitting the steel target.

If the bullet could transfer all of its energy to the target itself - then you would see the total amount of Work (or power) contained in the bullet's kinetic energy.

But, it can't do that. As the bullet interacts with an organic object containing water, some of the energy is expended making the bullet travel forward, some of it is expended as the bullet expands, some of it is turned into heat, some of it is turned into hydrostatic pressure waves, etc.

So, the bullet's calculated, theoretical foot/pounds of kinetic energy; and the Work (power) that the bullet is capable of doing is turned into a number of different types of energy so it doesn't all go into displacing (moving) the object. All of the theoretical kinetic energy IS expended (conservation of energy) - but, it is not all turned into Work that has displacement as part of the power calculation.

It's a fairly complicated interaction as it happens over time, and has a number of factors like the diameter of the bullet, bullet meplat (frontal surface area), velocity, and bullet construction.

 

[Theohazard]

Thank you for that explanation. I've never heard it explained that way before, but now that I think about it, it makes sense.

 

[JD0x0]

A firearm can't knock someone over purely from the force of the bullet, unless the gun being fired recoils hard enough to knock down the shooter.

Mythbusters did an episode on this. IIRC no cartridge even moved their human target, with the exception of 12 GA at point blank, which moved the target about 3 inches backwards.

I've seen videos of people being shot with 9mm and you wouldn't even know, when they were hit, because most of the time they don't even flinch, unless they are hit in the CNS.

The only way to 'knock down' a person with a pistol is to make them choose to lay down out of fear for their life, or when they die and drop dead from the wounds, and in that case, it's not the bullet knocking you down, it's more likely fainting from blood loss or shock.

 

[Brian Pfleuger]

Partially correct.

That's not how physics works. You are conflating momentum created through acceleration with kinetic energy.

The bullet is accelerated by the gun powder. The gun is accelerated in the opposite direction with the same amount of force. That has nothing to do with the bullet's kinetic energy.

That equation is: F (momentum) = M x A

Where F is the force created from the Mass x the Acceleration.

So, there is an opposite and equal reaction. The bullet is extremely light in comparison to the gun so the gun is accelerated in the opposite direction with a very small relative force at a much slower rate because of the mass & inertia of the gun.

Once the bullet is accelerated to it's maximum rate of speed the kinetic energy formula becomes:

KE = 1/2M x V^2.

Notice the difference - it is the velocity squared and that's where all of the foot pounds of energy come from. This has NOTHING TO DO WITH THE FORCE OF ACCELERATION THAT CREATES RECOIL.

Also notice that the amount of energy is directly linked to the velocity so that as the bullet slows down, the amount of energy is decreased proportionally - exactly how it works in real life.

Think of it this way. A 4,000 pound car can be accelerated to 60 MPH with very little energy. However, the kinetic energy released from the 4,000 pound car when it hits a brick wall is thousands of times more than the original amount of energy required to get it to 60 MPH.

Or, look at it this way. If, in fact, recoil was related to the energy imparted by a bullet at impact, then recoil would have to be infinitely variable depending upon the distance from the gun at which the bullet hit an object all the way to zero (bullet hits nothing, travels until it hits the ground).

Does that even sound plausible?

No, it doesn't sound plausible and I am not confusing kinetic energy and momentum. (I have an AS degree in physics, BTW)

Momentum is conserved, kinetic energy is irrelevant. Kinetic energy does no create recoil. Momentum is responsible for recoil.

The initial momentum of the system (gun and bullet and powder/etc) is zero. After the bullet leaves the barrel, the total momentum of the system must still be zero. The bullet has a momentum vector in whatever direction, the gun has a momentum vector that is exactly, perfectly opposite. (Ignoring the powder and other trivialities)

This video proves the point(s) nicely:
http://www.liveleak.com/view?i=310_1212367354

 

[jimbob86]

But after seeing jhp's go through 4' of wet compressed hay and plywood it made me think the jhp would also go through and out a human. It just made me think other wise of things I've read that's all.

Hay is a pretty poor testing medium, in my amatuer opinion.

I do not doubt that a 158gr XTP launched from a .357 Magnum revolver at around 1250f/sec from a range of 3' would pentrate a round bale of hay. It certainly could happen.

OTH, I know that 4 out of 4 XTP's launched from a .357 Magnum revolver at around 1250f/sec from a range of 3' did not completely penetate a 2 1/2 year old whitetail buck on one occasion ....... I had dropped him where he stood (knockdown? no, he pretty much dropped like a sack of potatoes) with a .270WIN to the chest .... I set the rifle down and walked over to the animal, knife in hand and it jumped to it's feet...... just a few yards from a very large unpicked cornfield. I panicked and drew my belt gun and shot until it fell again. 5 (the .270 SGK and 4 XTP's) entrance wounds, no exits..... deer are tough.

 

[buckhorn_cortez]

(I have an AS degree in physics, BTW)

I'm thrilled for you. I work with PhD's from Sandia National Laboratories and Los Alamos National Laboratories as part of my job.

Now that we've gotten the bona fides out of the way...

Momentum is conserved but momentum IS NOT KINETIC ENERGY. Momentum is the Force (energy) being expended to accelerate an object.

Momentum - as you've stated - creates recoil. Recoil is the opposite and equal reaction of the bullet being accelerated and that force is momentum. It is IRRELEVANT TO THE AMOUNT OF FORCE APPLIED BY THE OBJECT WHEN IT HITS A TARGET.

The energy contained in a moving object is KINETIC ENERGY and NOT momentum. Those are two separate and different parts of energy.

So the statement that if something was knocked down by a bullet would cause the shooter to be knocked down IS NOT TRUE.

The kinetic energy (power) contained in a projectile is calculated as stated:

KE = 1/2M x V^2.

When something is hit by a moving object the kinetic energy contained in the moving object governs the amount of force (power) - this is independent of the amount of energy used to accelerate the object to it's velocity - or the velocity of the object when it hits something.

Kinetic energy is the amount of energy contained in a moving object. That is NOT momentum.

 

[buckhorn_cortez]

A firearm can't knock someone over purely from the force of the bullet, unless the gun being fired recoils hard enough to knock down the shooter.

NOT TRUE. The force of acceleration of the bullet is momentum. The opposite force is recoil - or the momentum of the gun.

Once the bullet (or any object) attains its maximum speed, then the amount of energy contained in the object is described as kinetic energy.

That HAS NOTHING TO DO WITH THE MOMENTUM CREATED AS THE OBJECT WAS ACCELERATED.

See the discussion on "power" and the fact that all of the theoretical kinetic energy cannot be converted to power as that is 100% as Work.

All of the Kinetic Energy is conserved, but it is converted to other types of energy and not Work.

 

[Brian Pfleuger]

Well, sir, working with PhDs is not exactly an actual qualification.

Kinetic energy is conserved, but it is converted to things like heat that have nothing to do with motion transfer.

What is transferred between a moving body and a still body is momentum. Purely, if the collision is inelastic.

Kinetic energy does not dictate "impact" force, the ability of the bullet to move (knock down) a target.

There is no way to know what part of kinetic energy will remain kinetic energy and what part will convert to heat energy, deformation, etc.

Momentum is always, always conserved and it is a VECTOR quantity. It is conserved in both quantity and direction.

 

[buckhorn_cortez]

You're right Brian I only have a Master's degree...but, I manage to hang with the propeller heads from the national laboratories....

However, let's get back to what you've said

If the beanbag has enough momentum to knockdown the person it hits, the gun has enough momentum to knockdown the person who pulled the trigger

You have obviously conflated the momentum of the object and recoil momentum of the gun with kinetic energy.

You have described the bean bag as being unable to knock someone down because its momentum is the same as the gun recoil and the shooter would have to be knocked down.

You have described only 1/2 of the force / power equation regarding a projectile.

Momentum does not equal kinetic energy = you will agree?

Momentum is the amount of energy or force used to accelerate the object.

Once at is maximum velocity the energy inherent in the object is now described as KINETIC ENERGY = agree?

The kinetic energy equation is:

KE = 1/2M x V^2 ----agree?

Then what ever theoretical energy is contained in the object when it strikes another object is described as kinetic energy. Agree?

It is NOT MOMENTUM so your original statement about the bean bags cannot be correct.

Brian - it's okay, everybody gets confused on the subject of recoil versus "knockdown power."

As I have already stated - all of the kinetic energy of a bullet cannot be transferred to power as that is work and it involves movement (displacement) of an object over time. I'm not going into it again - you can re-read my post on "knockdown power" versus work.

Just stop conflating momentum and recoil with energy contained in a moving object.

There is no way to know what part of kinetic energy will remain kinetic energy and what part will convert to heat energy, deformation, etc.

MMmmm sort of...the bullet's energy "remains kinetic" until the bullet velocity is = 0. Once the bullet starts impacting a target, the energy in converted to a number of different types of energy and forces - so the bullet is "losing kinetic energy" as it converts to other types of energy or forces including, heat, shock wave, thermal energy, etc. Meaning - the bullet is slowing down faster as the kinetic energy is converted to other types of energy.

But, all of the energy is conserved - including passing through the object and traveling down range if that's what happens until the bullet's forward velocity is equal to zero (it's resting on the ground), and the bullet has zero kinetic energy.

Momentum is always, always conserved and it is a VECTOR quantity. It is conserved in both quantity and direction.

During bullet acceleration down the barrel and resultant gun recoil - 100% correct.

After the object is at maximum speed after leaving the barrel - the energy contained in the object is described as kinetic energy - not momentum.

F(momentum) = M x V

KE = 1/2M x V^2

You do see the difference?

Beanbag not knocking someone down is not connected to not knocking down the shooter. Not knocking down the target is because the kinetic energy has been converted to other types of energy other than "power" - or "work" that includes displacement (movement) of the object.

 

[Brian Pfleuger]

You have obviously conflated the momentum of the object and recoil momentum of the gun with kinetic energy.

No, I have not.

Momentum does not equal kinetic energy = you will agree?

They are not equal and are not even related except that they are both qualities of motion. One can be very high while the other is very low.

Once at is maximum velocity the energy inherent in the object is now described as KINETIC ENERGY = agree?

A moving object has kinetic energy and momentum. They exist together but are not directly related except that they both describe motion.

KE = 1/2M x V^2 ----agree?

Then what ever theoretical energy is contained in the object when it strikes another object is described as kinetic energy. Agree?

That is the KE formula, yes.

The energy contained in the bullet is, indeed, described by that formula.

It is NOT MOMENTUM so your original statement about the bean bags cannot be correct.

Brian - it's okay, everybody gets confused on the subject of recoil versus "knockdown power.

No, you are wrong. Just because the object has kinetic energy does not mean that the object does not also have momentum. In fact, having kinetic energy means that by the very laws of physics the object ALSO has momentum.

The same laws of physics tell us that momentum is purely conserved (in an inelastic collision) and kinetic energy is not. Kinetic energy can be transferred to other forms of energy, such as heat, while momentum must remain, always and forever, as momentum.

Momentum has a formula too... p=mv (momentum equals mass times velocity)

That equation is: F (momentum) = M x A

That equation absolutely IS F=MA, where F is force, M is mass and A is the acceleration rate but Force is absolutely NOT equivalent to momentum. In fact, the equation can also be written as f=∆p/∆t, where f is still force, p is momentum and t is time. So, f is the CHANGE in momentum over time but it is NOT "momentum". In other words, force is the rate at which momentum changes.

Just stop conflating momentum and recoil with energy contained in a moving object.

I am conflating no such thing. Kinetic energy is not relevant to the momentum transfer that is responsible for both recoil and bullet impact.

During bullet acceleration down the barrel and resultant gun recoil - 100% correct. After the object is at maximum speed after leaving the barrel - the energy contained in the object is described as kinetic energy - not momentum.

You are absolutely wrong. It is not an either/or proposition. The object has BOTH momentum AND kinetic energy.

Momentum is responsible for both recoil and the "impact" of the bullet that would knock someone down.

A bullet can have high kinetic energy and low momentum or low kinetic energy and high momentum (or any combination thereof) but if it's moving, it ALWAYS has both.

A fast, light bullet has relatively high KE and low momentum. A slow, heavy bullet has relatively low kinetic energy and high momentum.

Momentum is responsible for both recoil and "knock down" power.

Conservation of momentum:
http://www.physicsclassroom.com/class/momentum/u4l2b.cfm

Note that "collisions" is listed under conservation of momentum:
http://hyperphysics.phy-astr.gsu.edu/hbase/conser.html

Note that momentum can not be converted to other, non-motion, forms.

Note that kinetic energy can be converted to non-motion forms of energy and thereby "lost" in terms of the motion of the two objects.

 

[45_auto]

Buckhorn, you're so confused between energy, velocity, momentum and work that it actually makes me laugh!

Brian Pfleuger has been 100% correct in every statement he has made.

I only have a master's degree in mechanical engineering with a minor in physics, but I've been designing rockets for NASA since 1987 and Newton's Laws are still valid.

NOT TRUE. The force of acceleration of the bullet is momentum. The opposite force is recoil - or the momentum of the gun.

No, the force that accelerates the bullet has nothing to do with momentum. That Force is quantified as the Mass of the bullet times the Acceleration of the bullet. F = M x A.

I can accelerate a bullet of a given mass to a given velocity with a large force over a short amount of time, or small force over a large amount of time. Either one will give the bullet the exact same energy and momentum. For example, a 125 grain bullet that reaches 1000 FPS at the end of a 4" barrel in .001 seconds has the exact same energy and momentum as a 125 grain bullet that reaches 1000 FPS at the end of a 20" barrel in .004 seconds.

Hopefully it's obvious to you that the bullet that accelerated to 1000 FPS in .001 seconds in the 4" barrel had a higher rate of acceleration than the bullet that accelerated to 1000 FPS in the 20" barrel in .004 seconds.

Since F = M x A, and the bullet M (mass) is identical in both cases, that means that more F was required to generate the higher A in the short barrel.

Since both bullets have identical mass and identical velocity, their energy and momentum are also identical. If your statement I referenced above was true, then identical bullets with identical velocities would have different momentum values depending on whether they were fired from a pistol (high accelerating force) or rifle (low accelerating force). Hopefully everyone realizes that this is incorrect!

But, all of the energy is conserved - including passing through the object and traveling down range if that's what happens until the bullet's forward velocity is equal to zero (it's resting on the ground), and the bullet has zero kinetic energy.

Nope, kinetic energy is NOT conserved in an INELASTIC collision.

http://www.uta.edu/physics/labs/1401/1401Lab6.pdf

In an inelastic collision kinetic energy isn’t conserved, the ballistic pendulum is a classic example of this type of collision.

You may want to review these simple questions:

http://physics.bu.edu/~redner/211-sp06/class-energy/ballistic.html

Pay attention to the reminder at the bottom:

Reminder:
Elastic collision: both energy and momentum are conserved.
Inelastic collsion: momentum is conserved, but energy is "lost".

As Brian stated, MOMENTUM is conserved. That's why a ballistic pendulum works.

You may find these problems from my old Schaum's Engineering Mechanics enlightening.