The hydrostatic shock theory?

[Pyro]

Seems that much of that energy is transferred into the stretch/permanent cavity as seen in this image from BrassFetcher.

Every fragmenting round I have tested always have quite large cavities before total fragmentation compared to JHP or other rounds.

Boy I should have listened more in physics class.

 

[KodiakBeer]

Any shotgunner knows the consequences of not mounting the gun properly, so that it is allowed some free travel before your shoulder and cheek begin to decelerate it.

True enough, but does that destroy your shoulder? You can also find threads here about various rifle calibers and their "punishing recoil". The Mosin Nagant is the classic example where one person shoots it hunkered down on a bench, while another stands on his hind legs allowing his upper body to flex. The "hunkerer" says the rifle has savage recoil, the "stander" says it shoots like a pussy cat.
It's the same with a handgun, while it's true the flex of your wrist and arm slow down and "brake" the recoil, you can also shoot that handgun off a bench with your arms braced and feel far more recoil. Still, there isn't enough to damage anything.

The bigger point isn't arguable really, since so many people hunt with handguns. I've shot deer with .44 mag, .45 Colt and .45 acp. It kills them but it isn't the same dramatic kill you often get from a high velocity rifle. And when you dress a handgun killed deer, you don't get the same "lung soup" you often find in a rifle killed deer.

Really, while gelatin and mathematical formulas are all well and good, the question should really be addressed in the hunting forum. Over there you'll get 1st person accounts of both the kills and observations of what the organs of the animal looked like when dressed.

 

[Prosser]

Try this:
Go to a recoil calculator, and put in a 7000 grain, that's a one bore, one pound bullet, at 1000 fps. Enter the gun weight as 7000 grains, or one pound. Recoil Energy of 16,474 foot pounds, and Recoil Velocity of 1,030 fps. You'll notice that if you put in a one pound gun, you get a recoil velocity nearly identical to the velocity of the projectile.

The ft-lbs of energy also is very close to the recoil energy above:
Energy of 15,547 foot-pounds for a 7000 grain bullet at 1000 fps. The variance I think, is due to the large powder amount I put in for the above load.

So yes, equal and opposite reaction does appear to be the case.

Gunweight REALLY slows down recoil when using bullets that are lighter then the gun. Going up a pound on a handgun usually cuts the recoil in half.

As far as letting a heavy recoiling rifle get a running start on your shoulder: We do the reverse. We pull the gun hard into our shoulder with one arm, so it doesn't get a chance to
hit us at it's recoil velocity. That .510 Van Horn recoils at Recoil Energy of 98 foot pounds, and Recoil Velocity of 26 fps. I'm making it look like I can take that much, but, without a PAST recoil
pad spreading the impact, I've got a real good chance of having my shoulder separate from that energy. That's like getting hit by a heavyweight boxer, just not as fast.

At a certain point, and gun design, that which works with lower calibers no longer works. I have a Ruger Maximum with Bisley grips. The Bisley grip was originally designed for
light recoiling target guns, to take the recoil straight back. Some have found it allows the gun to roll with the shot. I have not. When shooting maximum loads, 525's at 1350 fps,
it feels like the tissue in my right palm has been hit with a baseball at high velocity. Feels like the tissue is actually strained and separated. This recoil level is:
Recoil Energy of 53 foot pounds, and Recoil Velocity of 31 fps.
At this point you have the gun moving so hard it doesn't get a chance to roll up, and drives straight back into your arm. It's not fun.

Kodiakbeer:
If you take your shotgun load, say 525 grains at 1350 fps, and use a .5 pound shotgun, the recoil would be Recoil Energy of 381 foot pounds, and Recoil Velocity of 221 fps. I would suspect
this would destroy your shoulder.

 

[Eb1]

I don't think anyone on this thread has answered the question. Nor do they have a scientific answer that provides a true backup to their answer.

In real world experience for me. A Hogue mono grip makes the recoil from my 5.5" 44 Magnum way less than with the cowboy grips.
A .223 55 grain SP out of a 20" barrel AR-15 will blood shot meat like nothing I have ever shot a deer with inside of 50 yards.
A .25-06 will most likely drop a deer. Where a 30-30 will not with an equal located hit. These are my real world experiences, and I can't tell you why. I just know that it is.

I have seen a .22-250 totally destroy ever vital organ in a deer @ 200 yards with a 55 grain SP. I have see a CT BT totally destroy ever organ on a quartering shot on a deer @ 300 yards. I have seen a 170 grain 30-30 round nose plow through a deer @ 50 yards, and it run for 150 yards more than once. I have see a 125 grain HPFN from 30-30 drop deer with the same shot placement as the 170 grain RN. I don't know why. It just is from point of view.

I believe that speed with the right projectile is a better killer. This is until you get into .45 calibers with weight of 200+ grains. Then the slow and big argument comes into play.
Either way. Fast and small or Big and slow are the key. Medium speed with Medium size bullets to me are a toss up of DRT kills, and I am learning to stay away from the med./med. rifles for hunting. Yes, they kill, but the chase may be on with a good shot placement where a .243, .25-06, 6mm, etc will most likely give you the results you want over the medium calibers with mediocre velocities.

 

[KodiakBeer]

Kodiakbeer:
If you take your shotgun load, say 525 grains at 1350 fps, and use a .5 pound shotgun, the recoil would be Recoil Energy of 381 foot pounds, and Recoil Velocity of 221 fps. I would suspect
this would destroy your shoulder.

I don't think so, though no doubt you wouldn't want to repeat the experiment. However, concentrate that force into a buttstock .68 caliber in size and you'd certainly sustain some permanent damage.
But really, it is that dispersion of force that is the heart of the matter here. When you shoot a firearm, the recoil force is dissipated by a number of factors, all mentioned in earlier posts.
At the other end when the projectile strikes, similar factors come into play since we are talking about the secondary damage around the actual projectile. That force is also being dissipated (absorbed) by the cone of tissue and fluid around the actual projectile.

Again, I'll point to handgun hunters. My experience with that is (relatively) limited, but the prevailing wisdom is usually "you can eat right up to the hole". For example, a deer hit through the liver with a rifle means no liver and onions for you. A deer hit through the liver with a handgun means a liver with a hole through it.

 

[Loosedhorse]

But really, it is that dispersion of force that is the heart of the matter here.

It's a big part of it, but...

When a gun fires, the ratio of kinetic energy of the bullet to kinetic energy of the rifle is the same as the ratio of the mass of the rifle to the mass of the bullet. To the extent that the rifle and shooter form a solid connection (that is, the rifle is not allowed to free-recoil), the effective mass of the rifle is made even greater, and even less energy is delivered to the rifle-shooter system.

That, coupled with the diffusion of impulse over the surface of the pistol grip or butt, plus the elastic properties of a butt pad/rubber grip and of the shooter's joints and flesh, result in a less destructive result at the butt end than at the muzzle end.

 

[Prosser]

I will say again:
There ARE handgun rounds, factory rounds, that equal rifle ballistics, and create Hydrostatic/dynamic shock. A 200 grain, .452" caliber bullet, at 2300 fps, in .460 S&W is a combination that is in the rifle category.

Also keep in mind that from game observations, there are combinations that flat out work, regardless of ballistics.

Something happens when a .475 LFN, 420 grains, hits a deer in the ham, and the bullet travels the length of the deer, hits nothing vital, and exits. Yet the deer drops dead on the spot.

Likewise the guys that hunt with these things swear by heavy, LFN bullets, at 1350 fps, some even slower, like 525's at 1100 fps, and that they kill like the Thor's Hammer.

I think that the bullet meplat has a huge impact, if you'll excuse a pun, on impact with the target. The larger the meplat, the geometrically more energy
is put on target in the initial impact. This may create hydrodynamic shock. Also, I think that maintaining the velocity through the target, due to the extremely heavy bullets creates havock, using both temporary and permanent wound channels.

The other alternative is also there. With these huge cases we can out Lee Jurras' Lee Jurras, by loading 180-220 grain bullets at rifle velocities, and
create hydrodynamic shock.

So yes, you can create loads that create hydrostatic shock in a handgun.
It may have to be a LARGE handgun, but it is possible, and, some factory loads do it right now.

 

[KodiakBeer]

There ARE handgun rounds, factory rounds, that equal rifle ballistics, and create Hydrostatic/dynamic shock. A 200 grain, .452" caliber bullet, at 2300 fps, in .460 S&W is a combination that is in the rifle category.

Yeah, but we aren't talking about those. We are talking about common defense cartridges like the .38 special posed by the OP.

 

[KodiakBeer]

That, coupled with the diffusion of impulse over the surface of the pistol grip or butt, plus the elastic properties of a butt pad/rubber grip and of the shooter's joints and flesh, result in a less destructive result at the butt end than at the muzzle end.

I agree, but the question lies in how much secondary destruction occurs around the impact point. I've shot deer with a handgun and I've shot deer with a rifle. My own lyin' eyes tell me that there simply isn't much peripheral damage generated by the rather modest velocity of a handgun round. You open the chest and the organs are still intact, with holes in them.

With a rifle the destruction is often massive - far more than can be accounted for by contact with the slug. I don't know what velocity you begin to see significant amounts of peripheral damage, but it's well beyond the 1000 - 1200 fps you get in standard defensive handguns.

 

[dcarch]

@mrmeangenes: That's got to be just about the funniest thing I've read this week!

 

[brickeyee]

So yes, equal and opposite reaction does appear to be the case.

By making the gun weight and bullet wight identical you are going to have the same momentum since it MUST be conserved.

if the recoiling gun and the bullet both have the same velocity and mass, they are going to have the same energy.

When you have a gun weight that is many hundreds of times the bullet weight (an 8 pound rifle is over373 times the weight (and mass) of a 150 grain bullet) the recoil velocity of the gun is cut by hundreds of times.

Put the velocity into the kinetic energy equation (1/2 * m * v^2)
and the energy is a very small fraction of the bullet.

You feel nothing compared to the energy the bullet has.

 

[Loosedhorse]

There ARE handgun rounds, factory rounds, that equal rifle ballistics, and create Hydrostatic/dynamic shock.

According to the Courtneys, any handgun round that produces 1000 psi of pressure wave in the chest is sufficient to cause hydrostatic shock.

Fragmenting bullets produce greater pressure magnitude than non-fragmenting bullets. Bullets penetrating 10-12" produce more pressure than bullets penetrating 14" or more. For example, a bullet which does not fragment and penetrates 14" needs over 700 ft-lbs of energy to produce 1000 PSI. In contrast, a bullet which fragments and penetrates 12" needs just under 450 ft-lbs of energy to produce 1000 PSI.

As a reference, some 124gr 9mm+P loads will go 470 ft-lbs of muzzle energy; so can some 125gr .38+Ps. It is "an emerging theory," so add salt to taste.

By making the gun weight and bullet wight identical you are going to have the same momentum since it MUST be conserved

I remember a physics quiz question in high school that ran, "If the bullet is heavier than the gun..."

The answer: "Then the shooter is in more danger than the target."

 

[MachIVshooter]

I don't know what velocity you begin to see significant amounts of peripheral damage, but it's well beyond the 1000 - 1200 fps you get in standard defensive handguns.

It starts to occur around 2,000 FPS.

Handgun bullets at normal handgun velocities do produce hydraulic shock, which creates the temporary cavity. It's just not enough to exceed the elastic limits of most human body tissues, which is why bruising is usually the extent of peripheral damage from handgun bullets (the capillaries had their elastic limits exceeded, but the tissue did not).

I also have no doubt that hydraulic shock can disrupt organ and nervous system function without actually damaging the tissue, but then the disruption is temporary, usually momentary. A sneeze or cough also disrupts these functions, as does a bump on the noggin. There are miles of separation between "disrupting function" and "damaging tissue".

The comparison to shooting a rodent has already been addressed. Using such a small body to demonstrate the effectiveness of a bullet meant to dispatch much larger creatures is flawed for obvious reasons; A squirrel is a 0.5 to 2 lb animal, and shooting one with a .45 would be comparable to hitting a human with a 203mm artillery round. It's all about scale.

 

[Ole Humpback]

What is a myth is the "dumping all its energy" mantra. When you shoot a handgun, how much energy is dumped in your palm? A similar amount of energy is dumped into the target (physics) and it isn't enough to make any difference.

You'd be wrong. As per Isaac Newton: For every action, there is an equal and opposite reaction.

Kodiak, you're playing too strict with the laws of physics. You're treating the action of firing a gun as a static situation when its actually a dynamic situation.

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.

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.

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.

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.

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. 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.

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 velocity 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.

A gun is nothing more than a pressure vessel that controls the direction and release of pressure.

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.

 

[1stmarine]

There are many studies about the subject. The most comprehensive one I found in the library of Cornell University. It describes quite in detail the effects of hydrostatic pressure in human targets. Try google it, you migh find it.

 

[aggieoutlaw]

Ole Humpback,

I follow you conceptually but have some confusion about your physics.

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?

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.

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.

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).

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!

 

[brickeyee]

Now onto recoil. Recall how the pressure goes from symmetrical to asymmetrical once the bullet is released

Recoil starts as soon as the bullet starts to move in the barrel.
F=ma and the bullet is accelerating.

 

[KodiakBeer]

A gun is nothing more than a pressure vessel that controls the direction and release of pressure.

Yeah, we've covered some of those bases, but when all is said and done the same energy is directed backwards as is driving the projectile forwards. That energy is spread out over a longer duration and the weight of the gun, the flex of your hands and/or body, etc, all serve to make that energy controllable.

One might compare it to shooting a bow. The reverse energy in that device is taken up as you retract the arrow and stored in the limbs of the bow. It's still there even if the "recoil" is prior to releasing the arrow.

 

[KodiakBeer]

It starts to occur around 2,000 FPS.

My "gut" tells me that's about right, though I don't think there's an objective way to test that. There are too many types of tissues and too many angles and variables in a gunshot wound to establish a "law".

I do know that at typical hunting rifle velocities (2600/3000 fps+), a pass-through rib shot on a deer or other critter of similar size will produce "lung soup". A handgun round even up to magnum velocities of something like 1400 fps, doesn't have that effect.

 

[Prosser]

2150 fps has been a bit of the Holy Grail for velocity for a long time.

I suggest we rethink the way we view 'magnum' handguns. The .475 Linebaugh will push a 275 grain bullet at 1560 fps, using a MINIMUM powder charge of AA 9. Max loads will run a 325 grain bullet at 1630 fps.

The .460 S&W will push 200 grain bullets over 2300 fps, and 200-260 grain bullets all over 2100 fps. That's a LOT of velocity from a handgun.

The 500 S&W will push a 275 grain bullet over 2100 fps. That's pretty much a rifle round.

 

[KodiakBeer]

Well, I don't think many will adopt a .460 or .475 for defense carry, against two-legged predators anyway...

 

[snooperman]

I found this on Wikipedia, concerning hydostatic shock from bullets...

Hydrostatic shock is real and not a myth. It describes a penetrating bullet that can produce incapacitation in living things through the hydraulic effect in liquid tissues. Neural damage can occur from hydrostatic shock. A bullet wound to the chest can cause brain hemorrhaging from fatal hits to the chest from handgun bullets. Human autopsies results have prooved that this phenomenon is real and not a myth. The journal of Neurosurgery has published data that this scientific evidence is well established. ***See Wikipedia- Hydrostatic Shock for more.

 

[Odd Job]

Hydrostatic shock is real and not a myth. It describes a penetrating bullet that can produce incapacitation in living things through the hydraulic effect in liquid tissues. Neural damage can occur from hydrostatic shock. A bullet wound to the chest can cause brain hemorrhaging from fatal hits to the chest from handgun bullets. Human autopsies results have prooved that this phenomenon is real and not a myth. The journal of Neurosurgery has published data that this scientific evidence is well established. ***See Wikipedia- Hydrostatic Shock for more.

It's all real according to Michael Courtney, yes

 

[Loosedhorse]

the same energy is directed backwards as is driving the projectile forwards.

Not strictly true.

The bullet and the gun (if unrestrained and allowed to recoil freely) will have the same momentum in opposite directions as the bullet clears the muzzle. However, the bullet will have more energy (assuming it is lighter than the firearm).

The bullet's base and the breachface (or boltface) are of course exposed to the same pressure (accelerating force) for about the same time. But during that time the bullet travels more forward than the breachface does backward. And energy is force times distance.

The reverse energy in that device is taken up as you retract the arrow and stored in the limbs of the bow. It's still there even if the "recoil" is prior to releasing the arrow.

Actually, the potential energy stored in the drawn bow is analogous to the potential energy stored in the gunpowder.

There is no "recoil" before releasing the arrow. If the arrow were released from an unrestrained bow, the bow would move backward with the same momentrum with which the arrow moves forward.

 

[RX-178]

I've said it before, and I'll say it again, I think this whole debate is two different camps writing page-long physics articles to split hairs.

Let's take hydrostatic shock theory first. In order to generate the most possible hydrostatic pressure in the target, you want to put the shot into areas of the body that have the most fluids, organs, etc, center of mass. That way you get enough of a pressure wave to travel through the bloodstream, damage the brain, and incapacitate the target.

Now for the internal bleeding/organ failure theory. You want to put your shot in the areas of the body to do enough damage to the internal organs as possible, in order to cause bleeding severe and rapid enough that the target loses the ability to function and is thusly incapacitated.

So in short, if you believe in the hydrostatic shock theory, you aim center of mass.

And if you don't believe in the hydrostatic shock theory... you aim center of mass.

Now, you guys have demonstrated you can explain Newtonian Physics and such... can you now explain to me what the big difference is here?