The hydrostatic shock theory?

can you now explain to me what the big difference is here?

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Sure. It centers on whether the only way to get incapacitation is to pucture vessels and destroy neural tissue (in which "permanent wound channel" is the key), or whether incapacitation can also be caused by a pressure wave (in which case "energy dump" becomes important).

That would mean the difference between selecting heavy, tough bullets (for maximum penetration) or lighter bullets that upset dramatically (for more energy dump) for a given caliber. So the question at issue is not point of aim, but bullet choice (and perhaps caliber choice).

I guess I'm still not seeing what a big deal this is.

If I wanted to puncture vessels and destroy neural tissue, I'd choose an expanding bullet, since the expanded bullet creates a larger permanent wound channel, and also because of the increased likelihood that the bullet will stay inside the target. Not because of energy dump, but because of the chance that if it goes completely through, it can hit something or someone that I was not intending to cause harm.

If I wanted to cause an energy dump in the target, I'd still want the bullet to get far enough inside that it can dump the energy deep in all the important internal organs, so I'd choose (again) an expanding bullet that still has enough weight to penetrate inside where the energy dump will be the most effective.

The best explanation of this is as follows

Yes it DOES exist, there is a measurable HYDROLIC shock, just as you will get one from CPR or taking a punch (just about anywhere) and it varies, so as a wounding dynamic, it becomes controversial.

Now, the next question is what does this mean to the meat, mainly CAVITATION- or the temporary cavity as TISSUE MOVES OUT OF THE BULLETS WAY.

Shoot elastic tissue and it stretches, exceed its ability to stretch with the amount of energy dumped, and you have a nasty wound. Don't, and you have SOME tissue damage from the little bits that have a hard time stretching compared to surrounding tissue, like capillaries etc. BUT

At handgun velocities, the effect is NOT a consistent wounding dynamic.

So yes it exists
but in handgun

SO WHAT...
once again, penetration (ability to TOUCH)
then like real estate...
LOCATION..LOCATION..LOCATION
(placement placement placement)

Actually, three separate issues:

1) At certain energies (velocities), a passing bullet disrupts the tissue cells and small (capillary) vessels, producing "bloodshot" meat.

2) The temporary cavity, when large enough, can tear muscle. At lower velocities, the temporary cavity can be "small enough" that it doesn't produce the "bloodshot" effect or muscle tears, but still fractures vulnerable tissues that do not have the elasticity of muscle. Such tissues include liver, spleen, brain and perhaps the largest blood vessels (which are often fixed in place by fascia, and so are subject to sheer, rather than being able to move freely out of the way and then return as the temporary cavity forms and collapses).

3) And the question at hand: whether a "pressure wave" produced by a bullet to the chest or abdomen can be transmitted (presumably via the blood vessels) to the brain and cause immediate brain dysfunction.

Stated that way, I'd have to say NO
as I have personally seen a guy who was shot 3x by a .50BMG, and if that didn't do it, don't know what will.

, it can hit something or someone that I was not intending to cause harm.

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Know what is behind your target. Even when hunting.

Sorry folks, I'm going with the Wiki quote here...

I see lots of folks quoting a lot of theory.. I have read a few of the same books.. But I know, what I know from seeing the results up close and personal.. I worked in a VERY busy Trauma Center for 2 to 3 nights a week for about 6 years.. I also have made more than one trip to the Morgue, as well as a few pretty good crime scenes.. I also have spent my whole life shooting, and running my own terminal ballistic test..For many of which my employer footed the bill...

Fact, any projectile that enters the body at speed, any speed, does so with enough force to set up some hydraulic shock. The question is how much..

There are two ways to get this shock to increase.. One is through bullet weight and size, the other through speed..

One way to increase the hydraulic shock value it through efficient bullet design.

Most handgun bullets will not effectively open, or mushroom unless they are traveling at or above 1000 FPS,, some designs, but few, will open at 870 fps, but not reliably. The old WW 180 grain Silvertip in 45ACP comes to mind here.. and the old 146 gr FBI load for 38 Spl..

If the bullet is moving fast enough to open, this expansion takes place in milliseconds.. It is a very rapid and very violent action. It places body fluids in motion, it disrupts the central nervous system. It does destroy more flesh, but only by about 10 to 15%.. Fatalities are only about 20% higher... This hydrostatic shock, HOWEVER MINOR IT MAY BE.. is what we are looking for in a defense situation.. It causes the entire nervous system to blank out.. kind of like putting your finger in a light socket.. It will cause the attacker to loosen his grip, drop what he/she is holding, to become disoriented, much like an epileptic after a seizure.. These are the minor effects, they work to our advantage.. More on that later..

The faster a bullet moves, the more violent the expansion, the more the hydraulic shock is amplified.. The greater effect upon the body..

The draw back is shot placement, abdominal shots place the most fluid in motion, however they are the least terminal, as their are fewer vital organs in the abdomen, save the liver, but even the liver is capable of sustaining major trauma, and still be repairable..

The most effective shot placement is in the chest, or close to the spine and the central nervous system (CNS)... or the "fatal T" this is the best area for placement.. It provides the most shock transfer to the the CNS.. it also provides the most fatal results.. Why, because that is where the heart, lungs, and brain live..

Does a bullet have to move at 1000, or 2000 fps to deliver this shock... no.. a good punch in the gut from a 6 year old can take your breath away... Why? because it produces hydraulic shock... It can knock the breath out of you.. while you are trying to catch your breath, is that moment of incapacitation.. you can't do anything else except try and catch your breath..

When the body is hit by a bullet, it, to varying degrees. is doing the same thing.. and the effects can range anywhere from, a grimace and a "dang that hurt".. to immediate unconsciousness..or in a good brain stem shot, drop like a sack of potatoes..

My point being, that all bullets, handgun, and rifle will produce shockwaves within the body.. the question is how much, AND how much is enough?

Well, as stated, and known by all here that, bigger is better, AND faster is better, what is best, a debate for all time.. one that will not be settled here.. Personally I think it is a combination of both..

But for someone to say that it takes a 2000 fps projectile to create hydrostatic shock... sorry, not completely so... now a 2000 fps handgun, is a bomb, and yes it most certainly creates MORE shock.. but to say that it has to move that fast to be effective... Sorry I can't go with it..

I have seen with my own eyes, a robbery suspect shredded with Black Talons, 115 gr 9mm, probably somewhere in the 12-1300 fps range.. was the damage prolific, oh yes it was nasty. was the wound channel evident? yes, through the edema, or busing of the adjacent tissue along the wound track.. was there massive damage, no not really, no torn tissues, ruptured organs, just a lot of really red, bloody meat.. any deer hunter has seen it.. I have also seen folk hit with a 32 ACP FMJ that were just as dead.. just not near as torn up... the question is, how much fight was in them after the 1st hit... or hits..

I have seen a great many others, all bear out the same thing to one degree or another, a previous poster says the same thing, and that too, is from hospital, real world observations..

There are too many studies, to many real world experiences to show that hydraulic shock is a real deal, it happens to varying degrees with all gunshot wounds.. The degree and the effect depends upon, bullet size, bullet design, shot placement, speed, how much clothing, bone encountered.. (Secondary projectiles is whole new chapter here folks).Size of the target, Drug use, excitement or adrenaline levels.. all of it is a factor as to the immediate effectiveness of a round.

Above all, the one thing that I have learned, to an ABSOLUTE CERTAINTY... that No Two Gunshot Wounds are the same, and a bullet will never do EXACTLY the same thing twice.. But it will repeat to a degree where it's outcome can be fairly closely predicted...

But even a 22 short creates a certain amount of shock... how much effect it has depends upon it's size in relation to the target..

The whole reason that we look for ammunition that will give us the edge of this Hydraulic Shock Theory, is because we seek to cause an immediate stop the the attack, of threat.. The most effective way to do that is through good shot placement, with an effective round. One that will reliably expand.. regardless of caliber..

Good ammo, good shot placement, Caliber selection of sufficient size, speed and weight to reliably do the job.. And there are a LOT of good ones out there,, Just not that many of them move at 2000 FPS in a handgun...

3) And the question at hand: whether a "pressure wave" produced by a bullet to the chest or abdomen can be transmitted (presumably via the blood vessels) to the brain and cause immediate brain dysfunction.

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And the answer is "no". Not at handgun velocities and only rarely at rifle velocities. There are millions of big game animals taken every year from high velocity hunting rifles. Occasionally, you get a dramatic knock down that "might" be due to a neurological shut-down, who can really say? You simply don't see that with slower bullets.

If we assume that the ideal gas law holds (which I am not sure is valid at the combustion temperature/pressure...but for the sake of argument), then shouldn't pressure decrease inversely with volume? Since the barrel is (for our intents and purposes) approximately a right circular cylinder, the volume is Base x Height (pi r^2 times bullet position in barrel as a function of time). Perhaps as a function of time, the graph of chamber pressure might appear to be decreasing somewhat 'exponentially'. I think that as a function of volume or bullet position, it would appear inversely linear (ignoring any gas or structural non-linearities). Thoughts?

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PV=nrt describes how the pressure reduces. This is somewhat trivial, but it lends itself to identifying the lose of pressure due to bullet displacement. Not much to say here, as the bullet moves down the barrel, the gases expand to fill the volume created by bullet displacement. The frame of the gun meanwhile is absorbing the gases thermal energy, in effect creating a one-two punch on pressure lose. Pressure is reduced by the bullet being moved down the barrel as well as the gun absorbing the thermal energy of the gases. I thought more on it, and inverse is the correct term for pressure lose based on bullet displacement. However, point is that there is little time for pressure to reduce and the pressure at the muzzle is significantly reduced from what it was in the cartridge.

When you say 'exponential', I believe you mean inversely proportional to a cubic or quartic. That is, x^-n is not the same as exp(-x). Your example was 2^-4 = 1/16. An exponential function grows (or shrinks) much faster than a polynomial. Again, you might be right about the exponential decay as a function of time, I just haven't given the equations much real thought.

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You're correct. My vocabulary at 11pm isn't as good as it normally is

Later you describe force as a pressure times an area times a time. I believe you mean an impulse (similar to momentum). I think you called it that as well. Is a 'symmetric pressure recoil force' another term for impulse or momentum? Also, I believe you mean microseconds which is only really fast, rather than picoseconds (10^-12 s), which is crazy fast.

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Yes, symmetrical cartridge pressure is a specific impulse. Its the force that results from the length of time that passes when the powder (for our intents & purposes) has turned to a gas, but the cartridge hasn't deformed enough to release the bullet. Good catch on the time as well. I was trying to think of what 10^-9 seconds was, but it didn't quite come to me. Now that you mention it, that's probably too fast. Its probably nearer to 10^-6 seconds or even 10^-3 seconds.

When you describe kinetic energy, I believe you mean KE = 1/2 m v^2. This will actually produce substantially larger numerical quantities (in Imperial or SI units) than the mass times the velocity (which is momentum, a useful value for the discussion of recoil).

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I was trying to point out that KE for the bullet will always be larger than the momentum of recoil. Forgive me, late night physics is always fun but I tend to miss a few things (I get it in my head, but it doesn't always get typed correctly).

Anyway, this is a good discussion. I enjoyed your analysis about rigid structures in addition to the mass relationships between momentum transfer. You know you are having fun when you start discussing material non-linearities and structural dynamics on a gun forum!

I'm interested to hear your thoughts about my pressure vs time and pressure vs bullet position hypothesis.

Cheers!

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I always enjoy a good structural dynamics talk. Keeps the grey matter flowing

but when all is said and done the same energy is directed backwards as is driving the projectile forwards

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Issac Newton's Second Law still applies, but not in a linear fashion. Once the bullet is released from the cartridge, the pressure becomes asymmetrical with a majority of the pressure creating force on the bullet base (remember, pressure follows the path of least resistance). The remaining force of the pressure is converted to either tension or compression forces that are reacted by the grain structure of the metals in the gun with very little of that force being applied to the cartridge base.

Lets use the 357 Mag as an example. For simplicities sake, and the fact that calculus would take forever to do (let alone create a formula for), the pressure is constant and only observed at the point in time when its at its maximum: symmetrical case pressure at the exact moment the cartridge releases the bullet.

First off, the knowns/assumptions:

357 max case pressure: 35000psi
357 cartridge base area: .1127"^2
Length of time for symmetrical case pressue: .005s
Pistol weight: 2lbs
Recoil Energy of a 158gr round at 1250fps: approx. 9ft/lbs

Second, equations to solve problem:

Recoil: F=ma
Specific Impulse: F=PAλ

Start by finding a for recoil:

9ftlbs/2lbs=4.5fps^2 acceleration

Next, find F for specific impulse:

35000psi*.1127"^2*.005s= 19.72lbs

Beyond this, recoil is a net force and I for the life of me can't come up with the formula to explain how force on the cartridge base gets turned into ftlbs.

Ole Humpback said:

First off, you need to understand that the bullet kinetic energy is not reacted into the gun, rather it is the force of rapid pressure change of the burning powder in the cartridge. When the primer is struck, it ignites the powder, which rapidly burns and changes into a gas. Using the combined gas law of PV=nrt to determine cartridge pressure, we can then set about figuring out the forces being translated into the frame of the gun and bullet..

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The powder gasses provide the energy that acts on the bullet and the gun. PV=nrt is making it more complicated than it needs to be. All we need is the equation for momentum.

p=mv

In this case the momentum of the gun is equal to the momentum of the bullet plus the momentum of the powder gasses.

Ole Humpback said:

Since pressure is defined as P=F/A, we look at the internal surface area of the cartridge and the surface area of the base of the bullet. Once these two are known, then we can determine forces acting on the cartridge and bullet. The forces inside the cartridge prior bullet release are symmetrical, equal amounts of force all around the cartridge. However, when the bullet is released the forces are asymmetrical, most of the force acts on the bullet and not on the gun..

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Pressure doesn't tell you much without the frictional drag between the bullet and the barrel. The part I put in bold is simply wrong. The force acts equally on the gun and bullet until the bullet leaves the barrel. Once the bullet leaves the barrel the momentum of the powder gasses acts primarily on the gun.

Ole Humpback said:

As the bullet moves down the barrel, pressure is being reduced due to an increase in volume (as the bullet is displaced, gases take its place). The farther down the barrel the bullet goes, the less pressure there is acting on the gun. As the pressure is decreasing, the area for the pressure to act on is increasing which means the force is decreasing. Since both pressure & area are square units, pressure decreases by a square root for each unit increase in area (a 1 unit increase in area results in a 4 unit decrease in pressure, a 2 unit increase in area means a 16 unit decrease in pressure, ect.). The progression of pressure reduction is not linear, its exponential. This results in an exponential reduction of force the farther down the barrel the bullet goes..

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I'm not sure why you are using area here. Pressure and volume have a direct but inverse relationship. If you double the volume, you cut the pressure in half. You can see this from the PV=nrt. If you hold nrt constant, then the the value of PV is also a constant. The increase in volume leads to a linear decrease in pressure. There is another very serious problem with your analysis. nrt is not a constant. More gas is being produced as the powder burns. Therefore the the pressure decrease, or increase, is related to the increase in volume and the burning rate of the powder. A powder that is optimal for a certain cartridge will usually show a slower decrease in the pressure verses time curve. This gives you a higher average pressure and more velocity.

Ole Humpback said:

Once the bullet leaves the barrel, all the remaining pressure exits the muzzle following the path of least resistance since the primer was struck. The path of least resistance in the case of firing a gun for the cartridge gases is to push the bullet out of the gun..

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No, the force acting on the bullet and gun is the same until the bullet exits the barrel. Powder gasses for a service pistol will be around 5-15% of the total recoil. So the gun will have this amount of momentum more than the bullet.

Ole Humpback said:

Now onto recoil. Recall how the pressure goes from symmetrical to asymmetrical once the bullet is released, this defines the recoil you feel. To define the symmetrical pressure recoil force, you need to add another variable: time. The symmetrical pressure recoil force is F=PAλ. λ is the time that is experienced at that specific PA. Since λ in the case of symmetrical cartridge pressure is defined in pico-seconds (millionths of a second), the max recoil force attainable is extremely low. This is called a specific impulse. The forces during a specific impulse can be tremendous, but short time frames drastically reduce the actual force of the impulse...

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This section is just about hopeless. I'll admit I don't know whether F=PAλ is an appropriate equation. If it is the only way it can be used is with an integral and λ would be infinitesimally small as pressure is constantly changing. Pressure does not define felt recoil or total recoil. Total recoil is defined by momentum. The momentum of the gun will equal the combined momentum of the bullet and the powder gasses. The felt recoil will be determined by the mass of the gun and time frame over which the momentum is transfered to the gun and your hand. For recoil specific impulse, symbolized by I, is better defined as the change in momentum. Since initial momentum is 0, the change in momentum of the gun is equal to the momentum of the bullet and gasses and is equal to the total recoil. For a revolver the gun has about 90% of the total recoil when the bullet exits the barrel. So you hand will also experience the recoil in this same time. For a semi-auto the gun also has about 90% of the recoil when the bullet exits the barrel. But it initially goes to the slide. Then the slide uses the recoil spring to transfer the momentum to the rest of the gun and your hand. This spreads the recoil over a longer time and will make the felt recoil less.

Edit to add - F=PAλ is completely inappropriate as you're using it. You seem to bee using it to represent recoil that occurs between ignition and when the bullet starts to move. There is no recoil generated during this time. Recoil is not generated until the bullet starts to move. You want an obvious example, look at a propane cylinder. Lots of pressure and time, but it just sits there.

Ole Humpback said:

Also, the cartridge base that pushes against the firearm itself is exponentially bigger than the base of the bullet, which leads to lower force imparted on the frame of the firearm than the base of the bullet during the specific impulse. Once the cartridge pressure becomes asymmetrical, the force acting against the gun drops exponentially..

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This is actually back wards. Since F = PA, or force = pressure x area, the force on the breach face is is higher than the force on the bullet base. I know it's counter intuitive, but pressure is not the cause of recoil. The movement of the bullet and powder gasses determine recoil.

Edit to add - most service pistol cartridges have a straight or slightly tapered case, .357 sig excepted. So there is little difference between force on the breach face and the bullet base.

Ole Humpback said:

Now, to sort out kinetic energy vs recoil. Kinetic energy is F=ma. In the case of a firearm, a is acceleration with a given direction: velocity. Convert grains to pounds and multiply by muzzle velocioty to get ft/lbs of kinetic energy. As for recoil, the specific impulse that acted upon the cartridge base is greater than the weight of the gun. Subtract out the weight of the gun in pounds and you get the actual recoil force in pounds. Since the body isn't a rigid structure, your hand & arm displace to absorb the small amount of force involved..

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kinetic energy = 1/2 m v^2
momentum = mv

Acceleration is not velocity. Acceleration is the rate change of velocity. Pounds is not a unit of mass, it's a unit of force. You have to change to slugs and use the right equation and diners ready so I'm not going there.

Ole Humpback said:

As for hydraulic shock, apply the same math plus the average speed of sound in flesh to determine the rate at which the bullet kinetic energy will be imparted, determine the λ for bullet/tissue interaction time and you get a good idea of how much hydraulic shock will be imparted by any given round.

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The speed of sound is not really that important. You really need the energy and the penetration depth to get an average for energy transfer. To get specific about the change, it's best to study the size and shape of the temporary stretch cavity.

Ole Humpback, I don't mean any disrespect but I've got a few bones to pick with some of your statements.

Yes, symmetrical cartridge pressure is a specific impulse.

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Pressure is not an impulse. Impulse is a change in momentum with units of force*time. Specific impulse is a term from rocketry and has little relevance here (it's equal to force/(mass flow rate*g)).

I was trying to point out that KE for the bullet will always be larger than the momentum of recoil.

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KE and momentum are different physical quantities and their comparative numerical values have little meaning.

Issac Newton's Second Law still applies, but not in a linear fashion. Once the bullet is released from the cartridge, the pressure becomes asymmetrical with a majority of the pressure creating force on the bullet base (remember, pressure follows the path of least resistance). The remaining force of the pressure is converted to either tension or compression forces that are reacted by the grain structure of the metals in the gun with very little of that force being applied to the cartridge base.

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Regardless of whether its application is linear over time, Newton's 3rd law still applies at any given instant. Superman cannot fly because he cannot generate a net force against anything. Likewise, if the chemical explosion inside the chamber generates a force on the bullet, it generates the same force on the gun. While the bullet is stationary in the case, the chemical energy generated by the powder is transferred into heating the cartridge/chamber and slight case deformation. However, the instant the bullet starts moving, the gun must start moving in the opposite direction, though slower on account of its greater mass. In other words, there is a net force on both the gun and the bullet. If there weren't, there would be no recoil.

Recoil: F=ma
Specific Impulse: F=PAλ

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Again, impulse is in units of force*time. Impulse includes, but is not, a force.

Edited to add: I didn't see THplane's excellent post while composing mine!

Ole Humpback said:

the pressure becomes asymmetrical with a majority of the pressure creating force on the bullet base (remember, pressure follows the path of least resistance).

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However, when the bullet is released the forces are asymmetrical, most of the force acts on the bullet and not on the gun

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No.

As the bullet leaves the barrel, the previously "closed" pressure system develops a "leak" at the bullet-muzzle gap. Pressure will therefore be lost first there, with that loss being transmitted quickly (at the speed of sound) to the breech face.

So, the bullet will "feel" the loss of pressure first.

KodiakBeer said:

And the answer is "no". Not at handgun velocities and only rarely at rifle velocities.

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Well, that is currently in dispute. The authors answer that point (about deer not often experiencing "shock" drop) by supposing that deer's dependence on head- and antler-butting dominance games has made them less susceptible to such shock than humans.

I'm not saying that's true, but it is their theory.

the capability of a round to inflict hydrostatic shock has nothing to do with the speed alone and everything with the amount of energy released in the medium (in this case the body of the target) in function of penetration and time (deceleration).
Ideally there would be massive expansion and not over penetration (loss of energy).
A 300 grain 50 caliber hp 'soft' slug at 900fps will produce a killer wave vs a .22 55gr fmj at 3000fps that might cut clean if it doesn't tumble.
Normally more speed means more energy but if there is over penetration all potential is lost.
I have seen both type of wounds and trauma and even before I read the Cornell University report it didn't take too much to understand the simple effects of physics. I see if I can find it and post it here.

A couple different ways to create shock.

One that doesn't really get mentioned much is:

http://www.kettering.edu/physics/drussell/Demos/doppler/doppler.html

Depending on bullet design, weight and velocity a similar shock wave would occur in fluid mediums as well.

I suspect as the bullet maintains velocity through the liquid medium it can create a similar pressure wave inside the body.

http://www.grosswildjagd.de/penetrat.htm

The above link has a bunch of references to bullets designed to maximize
that shock wave.

Kind of fills in a missing part of the discussion.

To elaborate: as velocity through the target increases, so does the wound channel. I am pretty sure this is due to the pressure wave created by the bullet as it goes through the medium.
The design of the nose of the bullet is vital for stability through the target, and really effects how the pressure was is created, and how much of one is created.

The problem I see with most service caliber hollow points is the bullet expands suddenly: It also slows down VERY quickly, since the bullet hasn't enough mass to maintain it's velocity through the target.
You get an increase wound cavity usually right after penetration, but, sacrifice wound channel after the bullet expands and deaccelerates. One of the great things about 45 caliber or larger pistols is that
they can be loaded with heavier hollow points, that have a much better chance of moving through the medium, expanding, and continuing to create a pressure wave as they go through the target.

An alternative approach for service calibers is to use a truncated nose, or some sort of flat point that still feeds, in a non-expanding bullet. At over 1350 fps, LFN's tend to widen, and create a larger then caliber
flat point bullet that still penetrates straight, and maintains it's velocity through the target. Lee Jurras used to do this with the 44 magnum. He used silouhette flat points at 1900 fps, 185 grains, and the speed and
shock wave of the bullet did the damage. Using this method he gets a large, long wound channel, without the various problems associated with HP's. I've often considered, and do with 9MM, use 147 Grain Flat points
as an alternative, or in conjunction with 147 grain HST. I'm not convinced that the standard party line on penetration of 12-14" is enough, since after the bullet expands, about the last 6 inches produces minimal wound
channel, because the HP is first expanded, and second unstable, and third moving very slowly.

Buffalobore makes a 45 ACP offering similar to what I'm thinking of, though it's a bit heavy:
http://www.buffalobore.com/index.php?l=product_detail&p=214
Still, it makes the .45 ACP=45 Colt, the Holy Grail standard for service calibers.

Strykervet said:

If it doesn't penetrate the target completely, then the round dumped all of its energy. That is the physics fact for the day.

If you can get something going fast enough, even air, it will kill. Bombs kill via shockwave as an example. That is 100% pure hydrostatic shock since there is no other way for the wave to hurt you. Also, if you are in a swimming pool (or water) when explosives go off, you'll feel it much more than if not. It doesn't take much to kill in water.

The FBI Miami shootout. One of the cops got hit in the neck with a 5.56. It passed through, didn't hit anything major, but it dropped him and he spent the rest of the gunfight paralyzed. Apparently the shock wave separated his spinal column causing it to pinch a nerve. Very lucky.

I've seen ballistics gel tests, and if they are similar to soft tissue, I can say that having my stomach shoved into my chest looks pretty uncomfortable to me, even if it is temporary. I've had bad gas before, it was temporary too, but I almost ended up in the ER.

Too many variables to say for certain how damage occurs and what does what, but I can say that it is very plausible.

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Lots of outstanding points to consider here.

Shear_stress and Loosedhorse already covered most of the problems in this post so I'm going to tackle this one.

Ole Humpback said:

Lets use the 357 Mag as an example. For simplicities sake, and the fact that calculus would take forever to do (let alone create a formula for), the pressure is constant and only observed at the point in time when its at its maximum: symmetrical case pressure at the exact moment the cartridge releases the bullet..

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It doesn't take 35000 psi to seat a bullet on a loading press. It does not take 35000 psi to get the bullet moving. As soon as the pressure causes enough force to
1) expand the brass or
2) force on the base of the bullet exceeds the bullet pull force or
3) a combination of these happens
the bullet starts moving. The bullet doesn't wait for max psi to start moving. Then it hits the lands in the barrel. At this point the acceleration of the bullet slows because of the engraving force of pushing the bullet into the lands. Then the bullet accelerates again. The attempt to use the 35000 psi as some magical number just doesn't work.

Ole Humpback said:

First off, the knowns/assumptions:

357 max case pressure: 35000psi
357 cartridge base area: .1127"^2.

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You've got some explaining to do here. You come up with .012701 sq-in.
It should be

a = pi r^2= (3.14)(.1785)^2 = .1 sq-in.

Ole Humpback said:

Length of time for symmetrical case pressue: .005s.

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Even if this number correct it is basically meaningless. Recoil doesn't start until the bullet moves.

Ole Humpback said:

Pistol weight: 2lbs
Recoil Energy of a 158gr round at 1250fps: approx. 9ft/lbs

Second, equations to solve problem:

Recoil: F=ma
Specific Impulse: F=PAλ.

Click to expand...

Specific Impulse is a measure of the efficiency of a rocket motor. While it could be applied to the powder, it has little if anything to do with recoil.

Ole Humpback said:

Start by finding a for recoil:

9ftlbs/2lbs=4.5fps^2 acceleration.

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9ft/lbs is a measure of kinetic energy. It's not a measure of force so it's use here is wrong. The 9ft/lbs comes from

KE = 1/2mv^2.

KE and force are not the same thing.

Ole Humpback said:

Next, find F for specific impulse:

35000psi*.1127"^2*.005s= 19.72lbs.

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Again, this is simply meaningless

Ole Humpback said:

Beyond this, recoil is a net force and I for the life of me can't come up with the formula to explain how force on the cartridge base gets turned into ftlbs.

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That's good because it doesn't. You might want to do some reading on a few formulas

F = ma
KE = 1/2mv^2
Momentum = mv
Impulse = change in momentum.

Since the initial velocities of the all components is 0 fps,
Impulse = momentum = momentum of pistol = momentum of bullet + momentum of powder gasses

I think Impulse or free recoil energy or free recoil velocity are all useful measures of total recoil. These can all be calculated from the above equations. You need to know the gun mass, bullet mass, powder mass, bullet velocity and powder gasses average velocity. For a typical service pistol you can SWAG the powder gasses at about 2000 fps.

Clean-One-Shot Kill.....

KodiakBeer said:

And the answer is "no". Not at handgun velocities and only rarely at rifle velocities. There are millions of big game animals taken every year from high velocity hunting rifles. Occasionally, you get a dramatic knock down that "might" be due to a neurological shut-down, who can really say? You simply don't see that with slower bullets.

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I think that is a rather dogmatic statement, that handguns cannot cause any hydrostatic shock sufficient enough to cause damage to organs connected via arteries, vessels, etc.

My 500 Magnum is "all-handgun" and has much more power and energy than a lot of rifles with which people use to CLEANLY-kill big game all the time!

It causes the entire nervous system to blank out.. kind of like putting your finger in a light socket.. It will cause the attacker to loosen his grip, drop what he/she is holding, to become disoriented, much like an epileptic after a seizure.. These are the minor effects, they work to our advantage.. More on that later..

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Someone sticking a red hot poker on your skin or giving you a nasty nurple will also cause you to cease whatever you're doing and most likely drop whatever is in your hands. It is a disruption to the nervous system, but not a physical one-it's a psychological one; Our nervous system cannot command any task other than reflex when sudden, excruciating pain is sensed.

There are exceptions to that shutdown rule, of course. One is adrenaline; People in extremely high-stress situations often don't even realize they've been injured until it's over. This includes being shot.

The other is simply mental preparation. This also incorporates adrenaline to some degree, and there are limits (entirely dependent on the person), but some people can experience incredible amounts of pain without loosing motor function and conscious thought if they're ready for it.

About the only type of trauma that won't cause people to let go of things is electrical shock. Not because it doesn't disrupt the nervous system, but because that electrical impulse causes every muscle to contract. Since the muscles that close the hand are far stronger than the ones that open it, the hand remains firmly clenching whatever was in it. Of course, the entire body tends to go limp noodle once the electricity ceases to flow. How long it takes the nervous system to recover depends on the amount of electricity and whether it caused only temporary disruption of neural pathways or permanent nerve damage.

Does a bullet have to move at 1000, or 2000 fps to deliver this shock... no.. a good punch in the gut from a 6 year old can take your breath away... Why? because it produces hydraulic shock... It can knock the breath out of you.. while you are trying to catch your breath, is that moment of incapacitation.. you can't do anything else except try and catch your breath..

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No, that punch produces blunt trama, and the diaphragm is very susceptible to it.

Blunt trama and hydraulic shock are not the same thing.

the capability of a round to inflict hydrostatic shock has nothing to do with the speed alone and everything with the amount of energy released in the medium

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Incorrect. A two-ton sedan moving at 2 MPH produces far more energy than a handgun bullet, yet if you are standing against a wall and that car rolls into you (Immediate "energy dump"), you're not going to experience any injury from hydraulic shock.

energy is the product of mass and velocity, but in and of itself is not a wounding mechanism. Miniscule obejcts at extreme velocities and very large objects at very low velocities can both produce energy equal to a bullet, but cause far less injury (if any). It is a balance between the mass and velocity that makes bullets so effective in damaging tissue.

If you really believe it's all about energy, try comparing a hot 10mm load and a .17 Remington on water jugs or gelatin. Both produce about 750 ft/lbs, but one has a much more violent effect on liquid or semi-liquid (tissue-like) mediums. This is the result of velocity; The hydraulic shock effect is relatively minimal at the ~1,400 FPS 10mm velocity, but very dramatic at the 4,200 FPS .17 velocity.

I hate using kills on small animals as a comparison, but in this case I think it makes the point. I've shot plenty of pesky rabbits with both of these rounds (180 gr. Golden Sabre at 1,406 FPS for 792 ft/lbs and .17 Rem. with a 20 gr. HP at 4,245 FPS for 800 ft/lbs). While the 10mm tore them up something fierce, the .17 literally blew them apart. As in, a leg over here, 15 feet to the right a head with shoulder and half a leg, 20 feet back a length of intestine......

Both rounds produce hydraulic shock that exceeded the tiny animal's tissue elasticity, but the high velocity rifle round produced so much more that it was effectively like putting a small bomb inside the animal.

Well, that is currently in dispute. The authors answer that point (about deer not often experiencing "shock" drop) by supposing that deer's dependence on head- and antler-butting dominance games has made them less susceptible to such shock than humans.

I'm not saying that's true, but it is their theory.

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Well, people shoot lots of critters other than deer. Perhaps pigs might answer that question better than deer. I have no experience with pigs, but there are plenty of people who do. I suspect their answer will be the same one that deer hunters have.

Well, there certainly is a language barrier here, to some extent.

We seem to have "hydrostatic shock", "hydrodynamic shock", and "hydraulic shock" as terms...and I'm not sure that there's general agreement on what any of them are, and whether they are different. The Courtneys use "Ballistic Pressure Wave," and wiki uses hydrostatic shock.

The ability of a 4200 fps round to disintegrate a small rodent would seem to be (as has been stated) due to the temporary cavity being bigger than the maximum elastic expansion of the small animal. That may fall under "hydrostatic" (etc.) shock, but it is not the ballistic pressure wave in the sense the Courtneys mean.

Let's talk about (brain) concussions for a moment. Their mechanism is not well understood. MRIs and similar structural scans done on patients with traumatic concussions often show NO abnormal brain findings: no bleeding, bruising, or swelling. So, what's causing the dysfunction?

Given that we are unsure of the mechanism of concussion, I think it is premature to say that a ballistic pressure wave "couldn't possibly" cause a concussion.

No, that punch produces blunt trama, and the diaphragm is very susceptible to it.

Blunt trama and hydraulic shock are not the same thing.

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Perhaps we are having another language problem: what does "blunt trauma" mean in this context? Obviously, the diaphram is not directly struck by a fist to the gut, but the force of the impact is transfered to the diaphragm by over-pressure of the abdominal cavity/displacement of the abdominal contents into the diaphragm (with those contents behaving like a viscous liquid).

I think I'd need a clearer definition of both "blunt trauma" and " hydraulic shock" to decide which of those affect the diaphragm after an abdominal blow. Maybe it's both (that the blunt trauma is transmitted via a hydrostatic mechanism, for example).

I suspect their answer will be the same one that deer hunters have.

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You could be right, but how many hunters use less penetrative, fragmenting bullets on pigs? It's intersting to speculate whether or not we'd see more "shock drops" on pigs if we did. However, the ethics of hunting demand that we kill the pig as quickly as possible, not "stop him" immediately, as in the case of SD.

One of the central arguments here is that the "only dependable way" to stop a human attacker is to deilver a wound that either kills him or leaves him with permanent, devastating neurological damage. And yet it seems that a significant number of attackers are stopped by handgun wounds that are not that severe. The current theory is that such "stops" are 100% psychological.

You know, it occurs to me that veins and arteries have "valves" every few inches. Junkies regularly blow these out leading to collapsed veins. This is something easily recognized by a physician, and surely by a coroner.

If gunshot wounds were blowing enough pressure through the vascular system to create damage and neurologic shock far from the point of impact, I'm sure coroners and surgeons would have noted that long ago.

Obviously, the diaphram is not directly struck by a fist to the gut, but the force of the impact is transfered to the diaphragm by over-pressure of the abdominal cavity/displacement of the abdominal contents into the diaphram (with those contents behaving like a viscous liquid).

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There's your answer, and it is still direct physical contact causing the diaphragm to be affected.

To say that it is not the impact of the fist that causes this would be like saying that the torque applied by the hand turning the screwdriver did not loosen the screw. Transmission of force/impact through a medium does not change the mechanism. With the screwdriver, it is still torque that causes the screw to turn. With a gut punch, it is still blunt impact that causes the diaphragm to be affected. The medium can amplify or dissipate, but does not alter the mechanism. It is still torque that turns the screw, and it is still impact that "knocks the breath out of you"

KB, several things.

Coroners are asked to determine cause of death. They see a gunshot wound, they see exsanguination? Not going to put distant veins under a microscope. And since we are looking a ballistic pressure wave as a possible cause of incapacitation, not death, it is unlikely a coroner would meet a temporarily incapacitated person.

Veins are extremely elastic, arteries less so; neither would be damaged visibly by a momentary pressure wave (as distinct from a temporary cavity stretch). (BTW, arteries have no valves; and venous valves are fairly weak, as anyone with varicose veins will attest.)

And again, you are demanding evidence of damage from a mechanism that may not produce visible damage at all; just concussion-like incapacitation.

...displacement of the abdominal contents into the diaphragm (with those contents behaving like a viscous liquid).

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There's your answer, and it is still direct physical contact causing the diaphragm to be affected.

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Well, but...

As I understand it, a hydraulic system works by taking force applied to a master cylinder here and trasmitting it via displacement of hydraulic fluid from the master cylinder through lines to a slave cylinder over there.

So if your punch's force is transmitted by displacing a fluid-like material against the diaphragm...why is that not hydraulic?

And, no: there is NO direct physical contact between the fist and the diaphragm (unless your fist penetrates the abdomen and touches the diaphragm). At least that is my understanding of the term "direct contact."

Arteries do not have valves , only veins do. That said..

veins take blood away from body structures toward the heart and arteries takes blood with oxygen and nutients TO body structures. There is much medical evidence that there is hydrostatic shock in arteries from gun shot wounds to one degree or another . This can affect blood flow to the brain from pistol shots to the chest creating loss of neural blood supply, causing neural shock and death. Read " Hydrostatic shock" on wikipedia and other sites that give medical evidence for this phenomenon.

Veins are extremely elastic, arteries less so;

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Yet, if you work in an ER as I have done, you note those blown veins made just by injecting a couple of CC's of fluid without releasing the tourniquet. They are visible from the outside of a body and are diagnostic of an IV drug user. This is pretty basic ER stuff that even techs and nurses will recognize.

I'm sure it would be noted by a coroner if they were present in gun shot wounds.