Home made body armor, amazing

[TMF]

Why does an MLB pitcher not knock the catcher over with a 98 mph fast ball? Same concept.

A baseball weighs 149 grams and travels at 98 mph. Kinetic energy = 0.5 x mass x velocity squared = 1135 ft-lbs.

Muzzle energy from a typical .44 Mag = 1100 ft-lbs. And Harry Callahan taught us all that a .44 mag sends people flying backwards.

I was awake during that part of physics.

[Chucktshoes]

Why does an MLB pitcher not knock the catcher over with a 98 mph fast ball?  Same concept.

 

A baseball weighs 149 grams and travels at 98 mph.  Kinetic energy  = 0.5 x mass x velocity squared = 1135 ft-lbs. 

 

Muzzle energy from a typical .44 Mag = 1100 ft-lbs.  And Harry Callahan taught us all that a .44 mag sends people flying backwards. 

Harry Callahan used .44 specials, so the energy would be even less.

[peejman]

Why does an MLB pitcher not knock the catcher over with a 98 mph fast ball?  Same concept.

 

A baseball weighs 149 grams and travels at 98 mph.  Kinetic energy  = 0.5 x mass x velocity squared = 1135 ft-lbs. 

 

Muzzle energy from a typical .44 Mag = 1100 ft-lbs.  And Harry Callahan taught us all that a .44 mag sends people flying backwards. 

 

 

I'll quote myself and say this is really an incomplete answer...   Conservation of momentum must also apply here.  The baseball at 98 mph actually has more momentum than the bullet.  Hence the baseball actually hits you harder....

 

200 grain bullet @ 1500 ft/s = 300,000 units of momentum. 

 

149 gram baseball = 2300 grains.  98 mph = 144 ft/s ... => 331,200 units of momentum. 

 

(ok purists... that's not typical units, but it works for this illustration...  :P )

 

Assuming you catch both the bullet and the ball in the center of your chest, your body (legs) must exert a larger force against the ground to maintain your balance and not fall over when the baseball hits you compared to the bullet.

 

Ain't math fun? 

 

 

ok.  I'll hush now.  :)

[Jonnin]

(ok purists... that's not typical units, but it works for this illustration...  :P )

Its ok.  Its a new thing, never been done before: I saw we call it the "iphailedphysix factor" instead of momentum.......    :devil:

 

 

let me add additional complexity.... vector math, remember busting the 2-d trajectory of a lobbed rock into X and Y parts and adding gravity to Y and so on? 

 

Well, take mr baseball catcher:  He is going to move his arm in the same direction as the ball, reducing the relative velocity of the ball.  Its like hitting the slowpoke in front of you on the interstate, he is doing 75, you are doing 85, so its a 10 mph fender bender rather than an 85 mph smashup.  The other way, head on, is quite another story!   The baseball guy is like the 10 mph example, he moves WITH the ball.  Also, moving with the ball slowing it down over time, is a deceleration rather than an impact.   You could try to do that with the bullet in a kevlar glove, but its moveing a LOT faster, weighs a LOT less --- you cant whip your body and arm enough to produce the same results.  FOR THIS REASON the bullet catcher will have a harder impact than the baseball catcher unless both held their arms/bodies perfectly still to neutralize this effect.  OW!

Edited March 13, 2013 by Jonnin

[peejman]

Its ok.  Its a new thing, never been done before: I saw we call it the "iphailedphysix factor" instead of momentum.......    :devil:

 

 

let me add additional complexity.... vector math, remember busting the 2-d trajectory of a lobbed rock into X and Y parts and adding gravity to Y and so on? 

 

Well, take mr baseball catcher:  He is going to move his arm in the same direction as the ball, reducing the relative velocity of the ball.  Its like hitting the slowpoke in front of you on the interstate, he is doing 75, you are doing 85, so its a 10 mph fender bender rather than an 85 mph smashup.  The other way, head on, is quite another story!   The baseball guy is like the 10 mph example, he moves WITH the ball.  Also, moving with the ball slowing it down over time, is a deceleration rather than an impact.   You could try to do that with the bullet in a kevlar glove, but its moveing a LOT faster, weighs a LOT less --- you cant whip your body and arm enough to produce the same results.  FOR THIS REASON the bullet catcher will have a harder impact than the baseball catcher unless both held their arms/bodies perfectly still to neutralize this effect.  OW!

 

If the distance is far enough that the drop due to gravity is worth considering, it's probably also worth considering the lift/drag caused by the seams on the ball and how it's rotating. 

 

 

What I meant by "catch the ball in the center of the chest" was to specifically exclude using the arms as a spring to absorb some of the energy. 

 

 

oh, and... :taunt:

[Guest Lester Weevils]

Well the rifle or pistol recoil would be greater than the momentum expended on the target, because of the "rocket effect" of gasses leaving the barrel behind the bullet. That contributes to recoil but doesn't transfer to the other end of the trajectory.

 

In addition to surface area discrepancies (small bullet cross section versus large stock or grip cross section), and mass discrepancies, light bullet versus heavy rifle-- Also possibly significant is the time period of the energy transfer? Like the big pulsed lasers used for fusion research-- Charge the capacitor banks with a "small" energy feed over a longish time, then discharge all the stored energy at once, and you get forty eleven kajillion watts for a couple of nanoseconds. :) The high energy short discharge can do things the same amount of energy can't do if if is more "spread out" in time.

 

The recoil of the gun is a "fairly long" pulse, but the bullet hitting a body (or body armor), that same energy is released in a much shorter time frame? Possibly the shortest time frame with hard body armor, slightly longer time frame with soft body armor that "gives", and longest time frame if the bullet penetrates flesh?

 

The bullet from a 16" barrel gun acquires its energy over a 16" path, but when it hits hard armor it dumps its energy in maybe an inch path?

[KaNuckles]

Here is what I have been thinking. keep in mind, I am not a physicist so I may not be using the correct terms but hopefully you'll get the idea. The action/reaction is not the bullet being launched/the affect on the target. The action is the explosion in the chamber. The reaction is the launching of the bullet + the recoil. So the recoil and bullet propulsion can receive unequal amounts of energy as long as their sum is equal to that of the explosion in the chamber. Most of the energy is directed down the barrel because that is the path of least resistance...by design. Whatever portion of that energy is imparted into the gun is transferred to the shooter. The rest (most) of the energy is imparted into the bullet and then transferred to the target. So the action/reaction there is the bullet striking the target/whatever effect it has on the target. So what ever happens to the shooter is irrelevant to what happens to the target. It is 2 different actions/reactions altogether. Since the explosion can't move the shooter as much it can't release as much of its energy that direction so it releases it down the barrel (path of least resistance) and in turn transfers that energy to the bullet.

 

I think I'm saying while the force of the explosion pushes equally in all directions, the energy is not released equally in all directions. It is it is released in the path of least resistance which is down the barrel. 

Edited March 13, 2013 by KaNuckles

[Guest Lester Weevils]

http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion

 

Newton's third law is the "equal and opposite reaction" one, though various parts of the problem would involve the other two laws as well.

 

The first law, inertia. Objects at rest want to stay at rest and objects in motion want to stay in motion. Before pulling the trigger, neither the gun or bullet want to do anything except remain at their current direction and velocity. So either setting on a range bench, or flying in an airplane, the rifle and bullet are happy to "stay together" not moving in relation to each other.

 

When the powder explodes, the chemical energy in the powder turns into heat energy and a pressure differential. Some of the energy is lost, and only part of the energy pushes the bullet away from the gun. The bullet thinks the gun is pushing it away, and the gun thinks the bullet is pushing it away. Third law.

 

The force equals mass X acceleration (second law). The force pushing the bullet is equal and opposite to the force pushing the gun. So for instance if the gun weighs 1000 times as much as the bullet-- The bullet will experience 1000 times the acceleration of the gun. If you mount the gun on roller skates, then if the bullet goes forward at 1000 feet per second, then the gun will roll backwards at 1 foot per second. (If the bullet's reaction were the only factor acting on the gun)

 

But because of inertia, after the gun has been accelerated it will want to keep moving. If it rolls back then hits a back stop, it will deliver the same "push" to the backstop, the same "push" as the bullet would deliver when it hits a forward-stop, moving faster but with less weight.

 

But the gun would expend some of the energy as heat, waste energy that doesn't push the bullet and gun away from each other. And the jet of gasses exiting behind the bullet. I read that the jet can in some cases impart additional acceleration to the bullet when it has exited the barrel but is still within a few inches of the barrel. But the "waste energy" of the exiting gasses would impart extra acceleration to the gun that the bullet would not receive. The acceleration of the gun is the sum of the acceleration of the bullet and the acceleration of the gasses. That is, if there is no compensator on the gun.

 

A cunningly designed compensator, which directs nearly all the gasses "straight back toward the shooter" might get very close to a gun with no recoil at all, because the rear-directed gas jet "pulling the gun forward" might be made be nearly equal to the acceleration of the bullet "pushing the gun backward". Unfortunately, a compensator that directs all the gasses back into the shooter's face wouldn't be very practical. But various compensators can use the waste energy from the gasses, to reduce or re-direct the bullet recoil. And spread the force's impulse over a longer time period.

 

If you were an astronaut, accidentally drifted away from your craft in orbit-- If your jet pack fails, you could (with great care) steer yourself back to the ship using a slug gun as your steering rocket. Or for that matter, throwing extra tools and equipment "as hard as possible" in the opposite direction of where you want to go. Third law.

[KaNuckles]

smarty pants.... :P

[East_TN_Patriot]

This is a good way to get yourself killed if you tried this to protect yourself from a gunshot. Body armor is not just about catching the round; that's the easy part. The challenge is to absorb and distribute the energy of the round enough to minimize blunt force trauma when it strikes the body. Body armor makers do a tremendous amount of testing to minimize what they call "back face signature," which is basically the amount of energy transferred into the body from the round caught in the vest. If not enough energy is absorbed or distributed across the ballistic panel, the blunt force can destroy internal organs and kill you. This is why ceramic panels are worn over soft ballistic panels. The ceramic catches the round, the ballistic fabric panel absorbs and redistributes the energy to limit back face signature. Think about a major league fastball. Hitting the padded catcher's mitt is quite different than hitting a hard bat or the battery's helmet.

The other part of the discussion that is missing, but important is kenetic energy. The amount of energy put into the round is concentrated into a very small point. The same energy is imparted on the shooter, but the firearm is designed to absorb much of the energy. Edited March 14, 2013 by East_TN_Patriot

[NashvilleStage]

If you take a single shot pistol and fire it at another person holding that same exact pistol and the bullet hits the second pistol and does not exit the recoil will be the same. Actually a little less because the bullet has slowed down between the two.

 

Newton's law is what says any reaction will have and equal and opposite reaction.

 

Lester beat me to the punch: The above example is true, except that it doesn't take into account the high-pressure gas exiting the muzzle after the bullet leaves the barrel. So the recoil of the sending firearm will be much greater than the receiving firearm.

I'd be curious to find out how much of the recoil is generated by the exiting gasses vs. the acceleration of the bullet. My personal wet-finger-in-the-ear guess would be 50% or more of the recoil is from the gas, which would explain why compensators & suppressors are so effective at reducing recoil. Compensators redirect most of gases to the side (in two opposite streams the cancel their forces), or up (to reduce muzzle-flip), or to the rear (to cancel even the bullet acceleration force). Suppressors force the gases to exit more gradually, thus reducing their velocity. I've also heard that the *volume* of escaping gasses is reduced because the suppressor absorbs some of the heat, but I'm kinda doubtful that a whole lot of cooling can occur in the few milliseconds that the gas impulse exists. Maybe I'm wrong, though.

Switching from fun techno-babble to other fun stuff...

I found a nice slow-motion video of a tank firing. You can even see the shockwave if you look closely.

http://www.businessinsider.com/tank-firing-in-super-slow-motion-2011-6

...and some pics of Howitzers going BOOM.

 

Without a compensator:

 

 

 

...and with a compensator:

 

 

 

 

 

 

 

You want a suppressor? How about this one:

 

 

[Jonnin]

OO OO I know that one!

 

F=MA.

Matter is neither created nor destroyed in a simple reaction (non nuclear).  (and nuclear I am unsure of... you bust an atom into protons and such, it still is not destroyed.... *something* is coverted to energy (which in and of itself may be made up of subatomic particles??)).  I did not get to nuke physics class.

 

Therefore the weight of the gas exiting is aproximately the weight of the powder + a little more for anything it binds to during combustion - whatever residue goes in the gun as well as air that was in the barrel, but call it a wash and say its almost what the powder weighed.   The velocity is about what the muzzle velocity of the round is. 

 

So for a 9mm, for example, its like 7 grains @ 1200 fps worth of recoil, vs the bullet, which is say 115 @ 1200 fps?   So its total recoil is 122 grains at 1200, and the gas is 7/122 = not quite 6% of the recoil.   Assuming I did that right.  And I did not, I used weight instead of mass, but you get the idea.  Both of those need to be divided by gravity, and I am not doing that in my head this morn.  Because gravity is constant, the % is still correct I think.  Stupid time change makes me sleepy, I dunno lol.

Edited March 14, 2013 by Jonnin

[peejman]

OO OO I know that one!

 

F=MA.

Matter is neither created nor destroyed in a simple reaction (non nuclear).  (and nuclear I am unsure of... you bust an atom into protons and such, it still is not destroyed.... *something* is coverted to energy (which in and of itself may be made up of subatomic particles??)).  I did not get to nuke physics class.

 

 

I believe it's converted to plasma energy and radiant energy (gamma rays, etc.).

 

 

The trick for dealing with the forces involved with unburned powder exiting the barrel is to assume the barrel is long enough that all the powder is burning before the bullet exits...   Either way, my assumption is that the forces generated by unburned powder exiting the barrel are relatively small compared to everything else (as your incorrect calculation illustrates.... ;)  ).

[Jonnin]

I believe it's converted to plasma energy and radiant energy (gamma rays, etc.).

 

 

The trick for dealing with the forces involved with unburned powder exiting the barrel is to assume the barrel is long enough that all the powder is burning before the bullet exits...   Either way, my assumption is that the forces generated by unburned powder exiting the barrel are relatively small compared to everything else (as your incorrect calculation illustrates.... ;)  ).

ya but take a gamma ray --- its made up of fast moving matter, alpha or something particle?   So the mass/matter remains, but its now considered energy as its mostly velocity, and that is where I am not sure what it even IS anymore lol. 

 

 

the powder does not matter if its burnt or unburt for recoil.  Its still gonna be moving about the same velocity as the bullet, and its mass it what it is, and it adds some recoil.   I would remark that a 223 the powder weighs almost 50% of the bullet mass for a 55 grain.....   the impact of the bullet for that caliber may be far less than the recoil was? 

Edited March 14, 2013 by Jonnin

[peejman]

Good point about the relative mass of the powder.  I'm not a reloader so I just assumed the powder is much lighter than the bullet.  Obviously that's not necessarily true. 

 

I wonder if it would be easier to demonstrate with a shotgun or black powder?  Both have wadding and powder of significant mass that exits the barrel, yet it doesn't hit the target. 

[KaNuckles]

Got it figured out fellas...

 

[Krull]

There has to be more to it. Otherwise how could you push a car and wouldn't firing a gun likely break your collar bone? At least it would seem that way.

 

 

ever seen a 250lb man move a car with no wheels on it?  me either

[Guest Lester Weevils]

I googled the gas jet recoil question some months ago. Some of the answers may have been wrong, but IIRC it amounted to a decently large fraction of the total recoil. It would depend on the velocity of the rocket thrust as well as the mass of the gas+particles. Unless the gas pressure has already reduced to near atmospheric as the bullet exits the barrel, am guessing the gas exiting an un-compensated barrel will have an exit velocity much higher than the velocity of the bullet? When the bullet "gets out of the way", the gas will move much faster than the bullet if there is substantial pressure remaining. Because the mass of the gas is less than the bullet, but the same force is acting on the gas as the force acting on the bullet, after the bullet gets out of the way? Like popping a cork on a champagne bottle.

 

Dunno, just saying you would need to know the gas velocity, and the gas velocity is probably faster than the bullet, at least with such as pistols.

 

Brings to mind those fairly rare "odd looking" rifle compensators that look like rocket nozzles. Some of the old german and commie guns. They probably wouldn't reduce recoil but wonder if a properly configured nozzle compensator could boost gas velocity enough to add significant "extra push" to the bullet in a few extra inches after it exits the barrel?

 

http://en.wikipedia.org/wiki/De_Laval_nozzle

 

[Guest Lester Weevils]

smarty pants.... :P

 

Classical physics are "fairly simple" but it is easy to forget to consider all relevant factors and get all confused predicting the results, at least to me. It is so easy to forget fine nuances of the simple rules.

 

A fella who seemed to have one of the most solid "total conceptions" of the classical physics I ever met, is a guy who had a masters in physics then taught high school physics his entire life. After repeatedly pounding the concepts into skulls of mush for 40 years, the man knew that subject real solid and was excellent at immediately identifying "fuzzy thinking" and misconceptions.

[Jonnin]

you are right, probably the gas is moving faster.  I doubt its *much* faster though, and in the case of pistol, the powder is insignificant mass vs the bullet usually --- I would bet on 5-10% of total recoil is still ballpark even with fuzzie math, for a typical pistol.  When you start talking rifles, with 40 grains of powder mass....  ???  I dunno.  I better stop here.  My 55 grain 223 load uses 23 grains of powder, for that reference.  Getting fancy again, as I understand it powder has internally the oxygen to burn, so a lot of my assumptions is that the reaction of burning does not add mass from absorbing atmospheric oxygen

Edited March 14, 2013 by Jonnin

[peejman]

Based on this...

 

http://vimeo.com/48571597

 

(zip ahead about 7:30 for the super-slo-mo stuff). 

 

It would appear there's two gas clouds.  The first precedes the bullet and shows that gas leaks past the bullet while it's traveling down the barrel.   The seconds exists behind the bullet but expands nearly instantaneously to a certain size, then stops expanding almost as fast.  To me, that implies it has very little mass... very little momentum to continue expanding.  The bullet then appears to punch through the gas that's expanded around it and continue on it's path. 

Edited March 14, 2013 by peejman

[NashvilleStage]

Nice vid, peejman.

 

I found a couple of others. Man, I must be bored today.

 

These are at 1,000,000 frames/sec.

 

http://www.youtube.com/watch?v=otpFNL3yem4

 

This one is a little better, forward to 1:07

http://www.youtube.com/watch?v=G0VjdI_S_HM

 

...and on the receiving end of things. Watch it all the way through; there's a lot of variety to see.

http://www.youtube.com/watch?v=QfDoQwIAaXg

 

 

Hey peejman, how did you get your video to embed? I tried a few different things, but didn't have any luck.

[Jonnin]

Based on this...

(zip ahead about 7:30 for the super-slo-mo stuff). 

 

It would appear there's two gas clouds.  The first precedes the bullet and shows that gas leaks past the bullet while it's traveling down the barrel.   The seconds exists behind the bullet but expands nearly instantaneously to a certain size, then stops expanding almost as fast.  To me, that implies it has very little mass... very little momentum to continue expanding.  The bullet then appears to punch through the gas that's expanded around it and continue on it's path. 

 

The cloud has the mass that it has.  Each individual particle of the gas has low mass, and when it hits air, it dumps its velocity in a HURRY.  The action of each individual particle does not indicate the mass of the total particles.