Stopping Power - Knock Down Power

[fredj338]

From another angle: When staying at a hotel, I am concerned about innocents that may be staying in rooms across the hall or around me. I think I would rather be killed myself than to kill a child.

What say you about that?

If you don't like this line of thinking, just remember; I am a frog. What do we know?

Practice enough so you dont miss a man sized target at 21ft. Practice at high speed, you should be moving. I worry more about misses than over penetration. 9mm JHP & up, dont miss, gtg. Unless you stay in some pretty rank motels, I feel pretty safe in my room with bolted door. If you are really afraid, travel with a door wedge. Anyone trying to breech the door is in a fatal funnel, impossible to miss.

 

[Frogwalking]

Fred; you sound like you have a burr under your saddle.

 

[JayFramer]

Be that as it may, there is a correlation that a .22 short has less effectiveness on a human adversary than a 12 gauge rifle slug.

So does the 12 gauge have more “stopping power” than the .22 short? As in does it have more of a power to reliably “stop” an aggressor, statistically, than .22 short?

Well I think the answer is yes. Not sure why so many dislike the term. I think few understand what they’re talking about and just want to jump on the “stopping power is a myth” bandwagon in an attempt to sound more educated than they really are.

 

[shouldazagged]

I'm not a physicist or ballistics expert, nor am I much impressed by ballistic gel testing, covered or otherwise. If something has worked well for a lot of people over a long time, and I can shoot it, it suits me well enough. The key words are "if I can shoot it".

Being a traditionalist, I've always gone with handgun ammunition, revolver or automatic, that had a proven history of effective threat-stopping. It might not be the red hot newest load, but it has been shown to work many times over.

Count me in the Keep It Simple camp.

 

[Ziggy2525]

Some YouTube star posts a video talking about no such thing as knock down power and everyone believes him,jeeze luis.
The logic used is that they try to say that the handgun would knock you down when you shoot it , if it had enough force to knock down what you are shooting.
And that is totally an incorrect statement! Percieved recoil is about 1/8 ( depending on the firearm) . So a .45 acp delivering 850 ft lbs to your target is only projecting less than 50 ft lbs to the shooter.
Now go to your garage and grab a torque wrench and torque a bolt to 850 ft lbs. Yep, you will need your neighbors to come help, even with a 4ft cheater bar. When your done, come back and post how 850 ft lbs is not enough to knock you down.
I'll be waiting!

Aren’t you confusing momentum and kinetic energy? Momentum (mass*velocity) is what controls felt recoil, bullet penetration, and knockdown force. The 850 ft lbs you mention is kinetic energy. Kinetic energy (1/2*m*v^^2) is how much energy is available to damage tissue.

 

[Ribwizzard]

Aren’t you confusing momentum and kinetic energy? Momentum (mass*velocity) is what controls felt recoil, bullet penetration, and knockdown force. The 850 ft lbs you mention is kinetic energy. Kinetic energy (1/2*m*v^^2) is how much energy is available to damage tissue.

Please explain your point better, are you arguing that they are equal?

 

[Old TexMex]

Will any of this actually be the deciding factor in what anyone chooses to use as a defensive handgun? And how disappointed will they be when they realize they’ve spent more time on youtube or talking about what they’ve seen or heard there, instead of getting properly trained.....

 

[Ziggy2525]

Please explain your point better, are you arguing that they are equal?

No, not equal. I’m saying what’s in your post is wrong three ways.

Way 1 - When you say something has “850 ft lb” that’s kinetic energy, not momentum. Kinetic energy is about destroying tissue. Momentum is about knocking things down.

Way 2 - More of a side point, a 230gr .45 acp +p round has about 450 ft lb of KE, not 850 ft lb. To get 850 ft lb of KE from a 230gr bullet, you’d need a velocity of about 1,300 f/s. Which .45 acp round has that velocity from a pistol?

Way 3 - You said

The logic used is that they try to say that the handgun would knock you down when you shoot it , if it had enough force to knock down what you are shooting.
And that is totally an incorrect statement!

Pretty sure you’re incorrect. Recoil is about momentum. Basic physics. There’s conservation of momentum. m(gun)*v(gun)=m(bullet)*v(bullet)=m(person shot)*v(person shot)

 

[hkcavalier]

Guys, who cares about formulas?

The number of North American big game animals taken with cartridges like the .44-40 and even .32-20 is quite large. Headshots to whitetails and the like with .22LR, albeit at pretty close range. Last time I checked, the largest moose in Idaho history was taken with a .30-30.

While not the top choice, 7mm Mausers have taken bull elephant with a single shot.

Natives to the Arctic successfully hunt polar bear with .223s and .243s all the time.

What's the common theme here? BULLET PLACEMENT.

 

[Ribwizzard]

No, not equal. I’m saying what’s in your post is wrong three ways.

Way 1 - When you say something has “850 ft lb” that’s kinetic energy, not momentum. Kinetic energy is about destroying tissue. Momentum is about knocking things down.

Way 2 - More of a side point, a 230gr .45 acp +p round has about 450 ft lb of KE, not 850 ft lb. To get 850 ft lb of KE from a 230gr bullet, you’d need a velocity of about 1,300 f/s. Which .45 acp round has that velocity from a pistol?

Way 3 - You said
Pretty sure you’re incorrect. Recoil is about momentum. Basic physics. There’s conservation of momentum. m(gun)*v(gun)=m(bullet)*v(bullet)=m(person shot)*v(person shot)

No, it has nothing to do with destroying anything, anything, we are talking about the energy that a projectile has, not what was delivered to a target. It has not hit anything yet .

And recoil is not momentum, it is how much energy is applied over what period of time. The energy is applied over a longer period of time to the handgun vs the projectile

How about doing some research before you start stating people are wrong. Simple Google search will show the actual formulas. Even wikipedia goes into detail about it.

 

[Ribwizzard]

Its simple, we know a .44 mag will knock down a bowling pin , whereas a .22 wont becouse the .44 mag carries more kinetic energy
It is measurable and real

 

[Ziggy2525]

No, it has nothing to do with destroying anything, anything, we are talking about the energy that a projectile has, not what was delivered to a target. It has not hit anything yet .

And recoil is not momentum, it is how much energy is applied over what period of time. The energy is applied over a longer period of time to the handgun vs the projectile

How about doing some research before you start stating people are wrong. Simple Google search will show the actual formulas. Even wikipedia goes into detail about it.

Energy applied over time is Work. Work is only indirectly related to impulse recoil. Since you seem fond of internet explanations, here is an animated Youtube video explaining momomentum and recoil.

YouTube

 

[Ziggy2525]

Its simple, we know a .44 mag will knock down a bowling pin , whereas a .22 wont becouse the .44 mag carries more kinetic energy
It is measurable and real

A .22 bullet fired from a rifle has a lot more kinetic energy than a bowling ball rolling down a bowling alley. The bowling ball has a lot more momentum than the .22 bullet.

The bowling ball knocks down bowling pins. The .22 bullet doesn’t. The .22 damages the bowling pins when it hits them, even though it doesn’t knock them down. The bowling ball doesn’t damage the pins when it knocks them down.

How do you explain that given your idea of KE?

 

[Ribwizzard]

A .22 bullet fired from a rifle has a lot more kinetic energy than a bowling ball rolling down a bowling alley. The bowling ball has a lot more momentum than the .22 bullet.

The bowling ball knocks down bowling pins. The .22 bullet doesn’t. The .22 damages the bowling pins when it hits them, even though it doesn’t knock them down. The bowling ball doesn’t damage the pins when it knocks them down.

How do you explain that given your idea of KE?

I believe you are way wrong when you say bowling ball has less kinetic energy. That's how.
Why would you think it would have less?

 

[Muss Muggins]

Given E=MC(squared), with the only variable being mass, the both the bowling ball and the .44 bullet have more energy than the .22 bullet, as they have more mass . . .

(edit: That's all I got . . .)

A .22 bullet fired from a rifle has a lot more kinetic energy than a bowling ball rolling down a bowling alley. The bowling ball has a lot more momentum than the .22 bullet.

The bowling ball knocks down bowling pins. The .22 bullet doesn’t. The .22 damages the bowling pins when it hits them, even though it doesn’t knock them down. The bowling ball doesn’t damage the pins when it knocks them down.

How do you explain that given your idea of KE?

 

[Ziggy2525]

I believe you are way wrong when you say bowling ball has less kinetic energy. That's how.
Why would you think it would have less?

Because you can easily calculate both. KE increases by velocity^^2. Momentum increases linearly with velocity. The v^^2 of the .22 way exceeds the mass difference between the ball and bullet.

ETA - I'm done. Bye.

 

[Ribwizzard]

Please show your calculations

 

[Ziggy2525]

Please show your calculations

AhhhUgggg. I said I'll stop, but I'll show you the numbers. I did these off the cuff.

Assumptions.
40 grain .22 lr traveling at 1,000 f/s.

A child rolling a 6 lb bowling ball at 10 mph (14.6 f/s) when it hits the pin - It will probably still knock down a pin even at 5 mph, but lets use 10 mph. The ball has a radius of 4.25" or .375 feet.

For the bullet the momentum is m*v where m is in slugs and v is in f/s. To convert 40 grain to slugs m=40/7000/32.2 = .000177. The momentum = .000177*1100 = 0.195 slug-ft/s.

The KE of the bullet is .5*m*v^^2 = .5*.195*(1100^^2) = 107 ft-lb.

The total momentum of the bowling ball = the transitional momentum + angular momentum. Transitional momentum = m*v. Angular momentum = Inertia (I) * Angular Velocity (w)

The mass of the bowling ball is 6/32.2 = .186 slugs. The Transitional Momentum is .186*14.6 = 2.72 slug-ft/s.

I = 2/5*m*r^^2 = 2/5*.186*(.354^^2) = .009 slug-ft^^2

w = v/r = 14.6/.354 = 41.2 rad/sec

Transitional momentum = 2.72 + .385 = 3.11 slug-ft/s

The bowling ball kinetic energy = Transitional KE + Rotational KE = .5*m*v^^2 + .5*I*w^^2

Trans KE = .5*.186*(14.6^^2) = 19.9 ft-lb
Rot KE = .5*.009*41.2 = 7.94 ft-lb
Total KE = 27.8 ft-lb

Bottom line...
A .22 with 107 ft-lb of KE and .195 slug-ft/s of momentum can't knock down a bowling pin, but can damage it.

A 6 pound bowling ball rolled by a child at 10 mph with 27.8 ft-lb of KE and 3.11 slug-ft/s of momentum can knock down a bowling pin and won't damage it.


I didn't double check the numbers, but I think they close. Like I said, I'm done.

 

[JayFramer]

Whether you chose “knockdown power”, “stopping power”, “dropping factor”, it all boils down to what is the most effective cartridge for self defense. For revolver, it is the .357 magnum 125 grain hollowpoint. For handguns, it is the .45 ACP 230 grain FMJ. For shottys, the 1 oz slug OR the 9 pellet 00.

It makes sense to me: use what has shown on the streets and in war if you want the most stopping power, no matter if you like the term, or otherwise.

 

[Ribwizzard]

First of all, who uses a 10lb ball???
And why are you calculating momentum?
Use formula; Energy= bullet weight x Velocity ^2/450,437.

But either way, that is only going to give you the energy of the projectile in flight. The knock down potential. And all that information is already posted on the box of ammo, so why your arguing this point I do not understand.
Please be more clear about what it is you do not understand so as I can focus on that area.
Is it perceived recoil?
Kinetic energy of the projectile?
Or how the projectile transfer that energy to the target?
It's a fairly busy day, but I will do my best to get back to you and help you out.