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Which is more important in ballistics?

Discussion in 'Handguns: General Discussion' started by IMTHDUKE, Sep 22, 2012.

  1. easyg

    easyg Senior Member

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    No one here is making this argument.

    Everyone here agrees that shot placement is of the utmost importance.

    But that does not mean that caliber, velocity, energy, and bullet design doesn't matter.
    All of that matters a great deal.

    As for the original question....which is more important, velocity or energy?
    Energy is more important.

    Whether one prefers a heavy and slower high energy round, or a light and faster high energy round, it does not matter so long as adequate energy is reached.

    Now exactly what is the adequate amount of energy?
    That is the real debate.

    I think that any handgun load that offers 400+ ft. lbs. of energy will work just fine for stopping aggressive human threats, so long as it's not too powerful for the shooter to quickly and accurately put lead on target.
     
  2. 481

    481 Senior Member

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    I agree.

    Most people do not realize that the T/Cs shown in those videos are nowhere near large enough to exceed the elastic strength of the vast majority of the tissues that a bullet will travel through as it passes through a human body.

    The T/Cs in the videos appear to be impressive, but as they say- "Appearances can be deceiving."
     
  3. BADUNAME37

    BADUNAME37 Senior Member

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    This is one of those arguments that has no right or no wrong answers.

    It all depends on exactly what is being shot -- at exactly what distance -- and, whether or not overpenetration can (or cannot) be an issue.:rolleyes:
     
  4. Kleanbore

    Kleanbore Moderator

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    MacPherson is correct when he observes the following:

    But: the fact that momentum is conserved in either an elastic (think billiard balls) or inelastic (punching bag) collision does not have a thing to do with penetration. It simply enables one to determine the velocity of the target with the bullet in it if one knows the mass of the target, the mass of the bullet, and the velocity of the bullet upon impact, and if the bullet does not exit the target. McPherson would agree with that, were he alive today. But that's something one does not need to know.

    One can also solve for the bullet's velocity upon impact by working backwards from the mass of the bullet, the mass of the target , and the velocity if the target plus the bullet after impact. Again, the equations used for that purpose involve momentum. That's how a ballistic pendulum works. At least some of us have done that in engineering school. I did it in high school, too. One wold not measure velocity that way today, but it's and excellent teaching device.

    The conversion of kinetic energy into thermal energy is part of what stops the bullet--i.e., what determines penetration--just as the heating of your brakes is what stops your car. The kinetic energy prior to impact is therefore a major determinant of penetration, (how far the bullet travels into the target after impact), and your car travels after the brakes are put on.

    What it takes to put your car or a bullet into motion is work, which is by definition energy, thermal and kinetic. It's the same thing when it comes to stopping either one, or anything that is moving, for that matter. That's indisputable. And no, I wouldn't attempt to model it. But one can measure the behavior of the items. Many of us have done that. too.

    Penetration in a given substance will be determined by a number of things, including the shape of the bullet, the mass, how it deforms, and how fast it is going. But it is the square of the velocity with which penetration will vary, all other things being equal.

    Back to the OP's question: what determines a bullet's effectiveness is what it damages. And adequate penetration is a big part of the answer to that question, along with wound channel and where the bullet strikes. That's consistent with McPherson's conclusions.

    By the way, at handgun velocities, the only important thing that energy affects is penetration. Forget "energy dump", "shock", whatever. That's consistent with McPherson's conclusions, too.

    Just an aside: the term "hydrostatic shock" is a misnomer. I'm a little embarrassed to say that I ate that stuff up while reading Jack O'Connor's books about hunting rifles at the same time I was studying engineering. I should have known better. But we all know what people mean when they use the term.
     
  5. coalman

    coalman Member

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    Momentum. I favor mass. I want heavier moving as fast as possible opposed to lighter moving as fast as possible.
     
  6. 2zulu1

    2zulu1 Senior Member

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    I'm not sure how billiard balls or punching bags relate to fluid dynamics, but I do use MacPherson's book as a reference source. I needed tutoring to grasp some of the formulas he used to validate the momentum model as it related to bullet penetration. An engineer should not have any issues understanding MacPherson's empirical research.

    Here's a link to Double Tap's gel data for a number of handgun bullet designs and calibers.

    http://www.glocktalk.com/forums/showthread.php?t=336612

    As we can see, there is a bullet penetration overlap between calibers independent of kinetic energy. Of particular interest for our discussion is the comparison between the much higher KE 10mm/180gr Gold Dot and the 45auto/230gr Gold Dot. In spite of the fact that the 10mm had a significant KE advantage, both bullets expanded to the same diameter and penetrated the same distance in ballistic gel.

    Given equal expansion, the 230gr Gold Dot retained more of its sectional density than the 180gr. Both bullets are in the same sectional density group and both loadings basically share the same momentum.

    Given McNett's ballistic gel data, there still isn't any way to predict wound trauma Incapacitation times.
     
  7. 1911Tuner

    1911Tuner Moderator Emeritus

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    I'm also in the mass/momentum corner, but with higher and higher velocities, you can reach the point of diminishing returns. Newton 3 will spoil the show. The harder the bullet hits the target, the harder the target hits the bullet. This is why a car being pushed for absolute top speed eventually runs into a stalemate with the wind...which is essentially "penetrating" a fluid medium.

    Studying long-range ballistics tables will reveal much on this issue.

    Two .30 caliber bullets with similar shapes and BCs...150 and 165 grains at 2800 fps and 2600 fps initial velocities...the lighter, faster bullet loses a greater percentage of its initial velocity at 100 yards than the heavier, slower bullet. At 400 yards, the speed gap is rapidly closing. At about 500 yards, the 165 catches the 150 and passes it, beating it in both energy and momentum.

    Certainly the ballistic coefficient factors in, but there's more to it than simply that. Sectional density and how efficiently the bullet conserves its momentum are equally important. Here, the advantage goes to the more massive/heavier bullet. The old-timers observed this, and they said that the heavy bullet "carried" better than the light one. The 19th century buffalo hunters favored heavy bullets for good reason, even with lower velocities and rainbow trajectories.
     
  8. bikerdoc

    bikerdoc Moderator Staff Member

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    This
     
  9. calaverasslim

    calaverasslim Senior Member

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    Your all way above me. Energy, Muzzle velocity etc etc etc.

    I want a larger caliber bullet, with enough ooomph to put the bad guy down. 9 or 10 or 41 or 44 or 45, whatever I am proficient with and can get a good solid hit.
     
  10. Kleanbore

    Kleanbore Moderator

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    This review of Duncan McPherson's book ssems to get right to the net upshot, as an engineering professor of mine used to say.

    Excerpt:

    The same things are concluded in the FBI Handgun Wounding Factors and Effectiveness report.

    Knock down power and shock are notably missing from both.

    While McPherson (1) showed how it is possible to predict penetration in gelatin from momentum data and from the expansion behavior of the bullet in water and (2) pointed out that kinetic energy (the effects of which many people incorrectly refer to as to as "hydrostatic shock") does not determine projectile effectiveness alone, I really do not think he ever believed that momentum really determines penetration in an inelastic collision; that would go against the laws of physics. He was able to correlate penetration with momentum, given other information that takes into account the effects of, among other things, energy.

    Personally, I have no interest in trying to model penetration myself. There are far too many variables. One can learn all one needs to know about penetration in gelatin, through glass, metal, and fabric, from other people's work, and even then the penetration into a 'real' target will be what it turns out to be.

    Slow heavy bullet or lighter faster one? Well, in practice, proper "placement" in a three dimensional fast-moving target is to a large extent really a matter of luck, so more rounds, and less recoil, are also key advantages. They help with Frank's contribution that more holes are better than fewer holes and holes in the right places are better than holes in the wrong places.

    By the way, lower recoil means less momentum or a heavier firearm.
     
  11. 481

    481 Senior Member

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    Well, it seems that he does, and with good reason-

    From pages 7 & 8 of Bullet Penetration by Duncan MacPherson:
    Momentum is conserved even in inelastic collisions. MacPherson explains this later in the chapter that I've quoted from (above). If you haven't read the book, I'd recommend it highly. The Schwartz bullet penetration model is also directly derived from F=ma, Newton's second law of motion, which describes momentum transfer (aka "impulse"). I find it hard to believe that both Schwartz and MacPherson are incorrect in their position.

    If there were too many variables, you wouldn't have the various soft-tissue penetration models set forth by MacPherson (n=400+), Schwartz, (n=700+) and CE Peters (n= ???).

    All of them take into account all of the necessary variables, are highly correlated, and quite accurate.
     
  12. MistWolf

    MistWolf Senior Member

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    Neither one. What's most important is terminal performance. Whether using a light fast bullet or slow heavy bullet, what really counts is what that bullet does once it hits the target.

    The bullet has to carry enough velocity and weight to get the job done but both are really only a part of the whole picture. Bullet construction has to match velocity and weight to optimize terminal performance. What difference does it make if a bullet is light & fast or heavy & slow if it leaves a long narrow wound channel, or a shallow one because of poor bullet construction and shape?

    The FBI specifies pistol ammo must meet a minimum and maximum penetration in ballistic gel (among other criteria). To meet the specifications, manufacturers must balance velocity, mass and construction to give the best terminal performance. Complicating things further is other mission needs such as penetration of barriers, shooting through heavy clothing, price and so on. It's also known that pistol calibers suitable for a duty handgun are unreliable fight stoppers meaning more than one round will probably be needed to stop the altercation.

    When selecting self defense ammo, the only thing that's important about velocity and mass is whether or not the load offers enough of both to get the job done. Even so, neither will do any good if the bullet constructed improperly, resulting in poor terminal performance
     
  13. Kleanbore

    Kleanbore Moderator

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    True, and so I have said. But that does not mean that momentum is the determinant of the distance that an object will travel after being captured in an inelastic collision.

    To stop a moving object it is, as you certainly understand, necessary to apply a force to it. And in an inelastic collision, that force involves converting kinetic energy into other forms of energy and/or transferring it to other objects. McPherson said as much.

    In stopping a car or truck or a train, friction brakes convert KE to heat, and regenerative brakes convert it to electrical energy.

    And guess what: given a constant braking force, the stopping distance is proportional to the square of the initial velocity--to kinetic energy. When I was young, every college freshman studying engineering performed calculations and experiments showing that.

    And penetration is nothing but stopping distance.

    Things get a little more complex when it comes to the stopping an airplane on a carrier, where the brakes convert some of the kinetic energy to heat, and some of it is transferred to the arresting equipment; and in addressing the the slowing a vehicle reentering the atmosphere from space, where kinetic anergy is converted into heat, until aerodynamic drag takes over, and perhaps unless a parachute is deployed to shed much of the remaining kinetic energy.

    If two billiard balls collide (elastic collision), kinetic energy is conserved as kinetic anergy. But if a car smashes into a deformable building, its kinetic energy is shed in several ways--into heat, into material that is put in motion, and so on. The same thing is true in bullets.

    As McPherson said, "kinetic energy is not only not conserved in real collisions, but is transferred into thermal energy in a way that usually cannot be practically modeled." So he decided to come up with correlative models based on momentum. But it would be a mistake to conclude that he ever believed that the stopping distance of a car under a braking force (a far easier thing to model--I've done it) was proportional to momentum. Because it is not, and never has been.

    I've thought about it. I have long ago accepted his conclusions about what causes wounding and what does not, but I have no interest in reading about his correlative modeling of penetration, however brilliant it may be.

    I'm not one of those fires bullets in to gelatin or pays a lot of attention to test results. I've chosen my carry loads, and I do not continue to meditate about how far a single bullet would go into gelatin.

    I do stay up with those fine people here who opine about big and slow vs light and fast, but I do not concern myself much about what a single bullet properly placed would do under ideal circumstances.

    So, when it comes to momentum vs energy, I will likely choose low momentum over high, to keep recoil down. And even though a bigger bullet is more effective than a smaller one, there's the issue of round count to keep in mind.

    I am not disputing McPherson's assertion that many people have placed too much emphasis on kinetic energy in assessing wounding effectiveness. Full disclosure: until some years ago, I was among them.
     
  14. Kleanbore

    Kleanbore Moderator

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    Say again? Really?

    Well, yeah, and that's why I can't get too interested in choosing a firearm on the basis of its performance with a single hit under ideal circumstances.

    Experts can certainly make informed judgments on what hits would be most effective. They'll say CNS, but that does not mean that anyone would recommend aiming for the brain of a moving target in a violent encounter.

    I's pretty well accepted that one should aim for center mass and shoot more than once. A hit to the upper spinal column, or to the ulna 0f the arm holding the gun, or to a tendon holding the knife or maybe a few more....
     
  15. 481

    481 Senior Member

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    Actually, it is. It is called transfer of momentum. This occurs through the action of a force exacted upon a bullet by contact with a medium which causes it to decelerate. It is F = ma or more specifically F = m ∆v/∆t, which indicates a transfer of momentum. Your statement here says just that:

    It is also a momentum transaction.


    That's a good example of F = m (-a) and it is still a mometum transaction. -think "impulse"-func. (F/∆t)--- [ m; ∆v/∆t; ∆v/∆x
     
    Last edited: Sep 26, 2012
  16. Kleanbore

    Kleanbore Moderator

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    Posted by 481, in response to "but that [that momentum is conserved in an inelastic collision] does not mean that momentum is the determinant of the distance that an object will travel after being captured in an inelastic collision":
    Do you really and seriously believe that the distance required to stop a moving object by applying a constant force is proportional to the velocity, rather than to the square of the velocity? That is, to momentum, rather than to energy?

    Have you ever conducted any experiments in that area?

    Consider this.

    Or if you prefer, see this.

    That is, distance varies proportionaly with energy.

    Yes, force is equal to mass times acceleration. Do not forget that distance is equal to one half the acceleration times the square of the time. We're dealing with force times distance which is work which is energy-- not momentum.

    That momentum is conserved is irrelevant--it doesn't the determine distance traveled by the moving object. Energy does. Momentum transfer does determine the resultant velocity of the combined mass of the target and the bullet--which, in the case of a 100kg target and a pistol bullet, is insignificant.

    That someone has been able to greatly simplify modeling by correlating distance and other things with momentum is a wonderful thing--but it does not mean that work is equal to momentum, or that force times distance is momentum.
     
  17. 481

    481 Senior Member

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    Yep, if the force, defined as F = m ∆v/∆t required to decelerate an object is proportional to 1/2MV^2, then it is also proportional to MV (momentum)- the first derivative of 1/2MV^2.

    Are they no longer teaching the calculus with physics? :confused:


    Just as these descriptions are descriptions of energy expenditures over a given distance and period of time, so too are they descriptions of F = m ∆v/∆t- the force you describe acts also to reduce the mass' velocity over that same ∆t changing the object's velocity and therefore, its momentum. It is called impulse.

    Yep, and distance may also be expressed as s = vt.

    Actually, the object posseses both quantities, but in the deceleration of any mass, the most direct way to analyse that motion is through the linear relatonship of s = v∆t with respect to ∆x (distance). It is as valid as any energy based analysis, but far simpler and more convenient to conduct.


    Oh, it most certainly does. Did you really just say that? :what:

    If an object has momentum, it is in motion. When a force acts to reduce or increase momentum, it does so over a period of time causing a change in velocity, ∆v/∆t, and also results in a displacement of the object.

    I've never heard Newton's third law of motion, MV = MV, referred to as being "irrelevant".

    You see those "V"s up there in the expression "MV = MV" ? That's motion (velocity) which over a period of time equates to distance unless these events occur instantaneously, that is with t = 0. Not possible. So yes, momentum is directly proportional to the distance than an object will move.


    Nope, work is proportional to momentum just as it is proportional to energy, but I digress- I covered that derivative relationship in the first sentence of this response.
     
    Last edited: Sep 26, 2012
  18. Kleanbore

    Kleanbore Moderator

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    To be proportional, the relationship must be linear.

    What is it that you are trying to say?
     
  19. Eb1

    Eb1 Senior Member

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    Simple:

    The size of the hole from an unexpanded bullet.
     
  20. 481

    481 Senior Member

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    If that is the case, then how can you say this? :confused:


    Energy vs. distance traveled (∆V w/respect to ∆x/∆t) is not a linear function-

    Are you messin' with me, KB? C'mon. I mean, you do know what a derivative is, don't you? :confused:

    Inhomogeneous first-order linear constant coefficient ordinary differential equation: cv + x^2 = dv/dx
     
    Last edited: Sep 26, 2012
  21. Kleanbore

    Kleanbore Moderator

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    Energy equals force times distance.
     
  22. 481

    481 Senior Member

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    Um, yeah...that's great, but it doesn't address the question asked.
     
  23. Certaindeaf

    Certaindeaf member

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    In a world of compromises, all men do.
     
  24. 2zulu1

    2zulu1 Senior Member

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    FWIW, consider this;

    Let's remember we are discussing fluid flow dynamics, not brake temperatures.

    An expanding bullet with 1800 ft/lbs of energy raises the temperature of three pounds of water by 1F.

    FWIW, I have both of these references, but I prefer the "Quantitative Ammunition Selection" book to MacPherson's WTI book. it's an easier read and the author does a great job of walking the reader through the mathematical equations needed if they wish to test their carry ammunition, factory or handloaded. Since I live in a very rural area, both of these books have assisted me greatly in determining bullet selection/caliber based upon potential four legged threats. Not any gel data for 38Super loaded with 357mag bullets. :)
     
  25. Certaindeaf

    Certaindeaf member

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    So what focus does this work subscribe to? Are all projectile types held constant or is it just fuzzy numbers/math? If all projectiles are not the same type, it's still just begging the question/s.
    Since you live in a very rural area, maybe just shoot prospects into some creek mud and go with that?
     

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