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9mm vs .45 ACP

Discussion in 'Band of Glockers' started by isuzu, Nov 5, 2006.

  1. isuzu


    Jul 3, 2005
    North America
    Found this at the Caliber Corner. The article is from vs 45.htm

    Taken from the article:

    A bullet's momentum can be thought of as how much "push" or "shove" it has. Though it has been known for decades (at least since the century old Thompson-Legard tests), neither the .45 or any other round tested at that time has enough "shove" to knock a human being down. Indeed, when bullets were shot into corpses suspended by the neck using rope, movement was documented as minimal at best. In other words, the .45 auto will not physically "knock" a man down…and neither will a 9mm. If you see claims otherwise, the writer may actually be reported a physical event that he truly thought he observed, when in fact it was caused by the target's nervous system's response to the bullet impact, but it was not from momentum. None of the commonly used defensive handgun calibers pack enough momentum to "knock" any human down.
  2. OB1


    Feb 6, 2005
    This is a subject that has of course been beat to death over the last century......however, I think of this "knock-down business" being like getting an electric shock. You may find yourself thrown aganst the opposite wall, but was it the electricity that threw you there, or your own muscles responding to the shock? It's your own muscles obviously, but folks will still say something like; "when I grabbed that wire, that electricty threw mw clear accross the room!!"
    Oh well, people still call magazines "clips".

  3. isuzu


    Jul 3, 2005
    North America
    Yep! Lots of people still think that the bullet knocked the target down. It's actually a reaction of the muscles/nervous system.
  4. PMMA97

    PMMA97 TagaBundok

    Nov 25, 2003
    I saw an episode in mythbusters where they used a .50 Barret rifle into a ballistics gel human that was suspended and it still did not knock the dummy down.
  5. 3/325

    3/325 Infidel Artist

    Sep 28, 2006
    Kingston, WA
    This is always a fun debate, particularly because of how little clinical evidence people provide and how seldom the debaters stay on topic. You can usually expect the following:

    1. People chiming in with other calibers
    (Apparently they missed the title of the topic)

    2. "I'll shoot you with both and you tell me which is worse"
    (Not very helpful to an intelligent discussion)

    3. Shot placement!
    (Yeah, no kidding, thanks for the tip.)

    4. "There is no magic one-shot-stop bullet!"
    (Even though the debate is a comparison between two rounds and no one actually mentioned one-shot-stops)

    The bottom line in the debate between 45 and 9mm? When looking at the best commercially available modern ammunition, 45 is a marginally better defensive round if ALL other factors are clinically equal.

    That's it. That's all there is to the debate between these two rounds.

    Now, if you want to talk about which one is right for YOU.... then all bets are off. 9mm rounds are cheaper and therefore allow you to put more practice rounds downrange for the same money. Do you need/want the extra practice? Are you a good enough shot under pressure that you can go with the slightly better (and more expensive) 45 ammo?

    I'm still trying to decide for myself, but for budgetary reasons and the desire to maximize my practice, I'm leaning toward the 9mm. (My second purchase will most likely be a 45!)

    Anyway, if you happen to come across any hard data that clearly indicates a specific round as superior PLEASE share! :hugs:
  6. zorkd


    Sep 2, 2004
    action always has an equal and opposite reaction, if you have a load that LITERALLY knocks down (meaning by sheer force and not by any inury to neuromusculoskeletar system) a 200lbs target or so (a big man for examople), ergo firing this load will also knock YOU down.

    i don't know if you ever came across it, but some guys were playing around with a wildcat .577 T-rex round they called it, they let some of their friends try the rifle, not telling them that the compensator was not attached, now THAT round really flung the shooter back and left him sitting on the floor.
  7. HEAVY

    HEAVY "Verify!"

    May 31, 2006
    could you post the link on the t-rex bit?

    should be fun to watch. :)
  8. Poodle


    Jul 4, 2006
    This is an interesting topic.

    Now the Physics of Ballistics would include: [1] the law of conservation of momentum; [2] the law of conservation of energy; and [3] the coefficient of restitution in collisions.

    Momentum is the mass of a particle multiplied by its velocity. It is a vector quantity.

    so P = Momentum = mass x velocity
    P = mv

    Kinetic energy is also a quantity possessed by a moving body and it's one half of the mass multiplied by the square of its velocity. It is a scalar quantity.

    so K = 1/2 x mass x velocity squared
    K = (1/2)mvv
    Both momentum and energy are conserved in systems where the velocity is not appreciable with respect to the velocity of light, in which case the beta correction factor will have to be factored in.

    From Newton's 2nd Law of Motion, Force = mass x acceleration

    Fnet = ma; but a = dv/dt

    Fnet = m(dv/dt), but m(dv) = dP or the change in momentum.

    So Fnet = dP/dt.

    Fnet(dt) is the IMPULSE of collision (let's call it J) and is equal to dP or the change in momentum.

    In the collision of bodies, the time involved Dt is usually very short and the force F exerted by the bodies on one another varies. Thus the shorter the time interval of the collision, the greater is the average force.

    Conservation of Linear Momentum
    In the equation Fnet(dt) = dP, if the net force Fnet acting on a system is zero, then dp is zero or Psystem is constant. Therefore, when there is no net external force on a sytem, the total linear momentum is constant. This is the principle or law of conservation of linear momentum

    Conservation of Mechanical Energy

    The Law for the Conservation of Mechanical Energy is thus stated
    ½ m(vi)(vi) +Ui = ½ m(vf)(vf) +Uf

    where m = mass; vi = initial velocity; vf = final velocity and U is potential energy. In our case, we can neglect potential energy.

    Elastic and Inelastic Collisions

    While the total momentum of an isolated system is conserved in all collisions, the total kinetic energy of the system may or may not change. There are two extreme cases of collision with respect to changes in kinetic energy. One is the case in which there is no change in the kinetic energy of the system. Such a collision is called perfectly elastic. In the other extreme case, the two bodies stick together and move together as one body. This is a perfectly inelastic collision. The kinetic energy after collision is less than the total kinetic energy before collision. The kinetic energy lost during collision is lost to other forms of energy, mainly heat. In between the two extreme cases, the collision is partially elastic.

    For a perfectly elastic collision, the relative velocity of approach is equal to the negative of the relative velocity of separation.
    So therefore, (v1 – v2) = – (u1 – u2) where (v1 – v2) is the relative velocity of approach and (u1 – u2) is the relative velocity of separation.

    Coefficient of Restitution:
    For collisions which are not perfectly elastic, the equation (v1 – v2) = – (u1 – u2) does not hold. They are related by the elasticity of the collision which is given in terms of a certain coefficient. The ratio of the negative of the relative velocity of separation to the relative velocity of approach is e, the coefficient of restitution.
    e = – (u1 – u2)/ (v1 – v2)
    For a perfectly elastic collision, e = 1; for a perfectly inelastic collision, the two bodies stick together and do not separate, hence there is no relative velocity of separation and e = 0. For collisions in between the two types, e is between 0 and 1, the value of the coefficient depending on the nature of the bodies colliding.

    Thus said, these three physical principles enter into consideration when we speak of ballistics. A lot would depend on whether the collision is perfectly elastic, perfectly inelastic or partially inelastic. A perfectly elastic collision between a bullet and a human body is highly improbable. A perfectly inelastic collision would be the case where the bullet does not exit. A partially inelastic collision would be the case where the bullet exits through and through. The coefficient of restitution in this third case, would have to be established with extreme difficulty through experimentation with preferably living organisms (something that is evil and immoral). Without the coefficient of restitution, it would be difficult to establish the loss of kinetic energy to heat.

    Example of a Perfectly Inelastic Collision:

    The ballistic pendulum is a good example of a perfectly inelastic collision . A bullet is fired into a block which is initially at rest and is suspended by a thread from the ceiling. The bullet gets imbedded in the block but does not pas through and through. The ballistic pendulum is used to measure the speed of the bullet just before it hits the block. The block with the bullet swings as a pendulum and reaches a maximum height of h which can be determined by the angle q and the length of the pendulum.

    The two Conservation Laws may be applied, the Law of Conservation of Momentum for the collision and the Law of Conservation of Mechanical Energy for the swing of the block.
    For the Law of Conservation of Momentum, we have
    mV = (m + M)u [Equation 1]
    where m = mass of the bullet, V is the velocity of the bullet before it hit the block, M is the mass of the block and u is the common velocity of the block and the bullet after the collision.
    The Law of Conservation of Mechanical Energy states that the kinetic energy of the block and the bullet at the start of the swing is equal to the potential energy of the system at the top of the swing. Hence,
    ½ (m + M) u2 = (m + M) gh [Equation 2]
    u = √¬(2gh)
    From the geometry of the figure,
    h = L – L cos q = L (1-cos q)
    From Equation 1, we get
    V = [(m + M) u]/m
    V = [(m + M) √¬(2gh)]/m
    V = [(m + M)/m] √¬(2gL(1-cos q))
    In a ballistic pendulum, the coefficient of restitution = 0 and the kinetic energy of the bullet before impact is not equal to the kinetic energy of the block and bullet together after impact because as the bullet penetrates the block, heat is generated at the expense of kinetic energy.

    Now a human body (or a good facsimile of a human body) suspended from a rope is not at all analogous to a block of wood in a ballistic pendulum. The block of wood is more or less homogenous as compared to a human body and more solid. A gel like humanoid may be reasonably similar to the human frame but it lacks bones and sinews. A human cadaver would be desirable in any case but as mentioned previously, it is not solid as the block of wood and there are many variables in this case unlike in the block of wood which is more predictable. If the bullet passes through soft tissue and managed to avoid hitting bone, the energy absorbed on impact may be considerably less than if it were to hit bony material. The shape of the bullet also enters the equation because the shape would determine the increment of time that the collision will take or whether the bullet would pass through and through. A more streamlined bullet would more likely pass through and through. All of these would determine the velocity and the energy that would be imparted to the target by the bullet. Velocity imparted and energy absorbed play a great part on whether or not the target is felled down by a bullet. Of course we are still not considering the motor reflexes of a live target or shot placement.

    These all said, we can PROBABLY assume that determining the relative desirability of the stopping power of a .45 cal round as compared to a 9 mm parabellum would be subject to extreme difficulty because of the many variables included in the equation.
  9. OB1


    Feb 6, 2005
    All that math reminded me of the old equation that "the
    angle of the dangle increases with the heat of the meat in direct proportion to the sag of the bag provided that the mass of the a-- remained constant". (sorry, just couldn't help it!)

    :rofl: :banana:
  10. 1-2man

    1-2man Part Time

    Sep 28, 2006
    :rofl: :rofl: That's a little easier to understand.:rofl: :rofl:
  11. 3/325

    3/325 Infidel Artist

    Sep 28, 2006
    Kingston, WA
    I checked your math. Your forgot to carry the "1".
  12. ok that's it... no more tequila for this man. ;) ;) :supergrin:
  13. PMMA97

    PMMA97 TagaBundok

    Nov 25, 2003
    Father may picture pa ba ni Ruffa? :supergrin:
  14. ang layo ng connection bro. :supergrin:
    pero mas madali nga namang isipin si Ruffa kaysa e calculate ang na post sa itaas.:laughabove:
  15. zorkd


    Sep 2, 2004
    i tried the link at, seems like their host for the videos is down, perhaps though youtube has it, very very funny.
  16. toxic


    Jan 15, 2005
    doin' time
    :supergrin: I guess ..this is Father Poodles way of saying you go, read and learn and dont talk nonsense:supergrin:
  17. vega


    Sep 29, 2001
    Guys, it's really simple.
    Can't you see the relationship of equation 1 & 2?
    45ACP makes bigger hole than 9mm.

    And to top it all:
    It can also be Force / mass = acceleration.
    where F/m = a

    In layman's term:

    Father over mother = anak
  18. 3/325

    3/325 Infidel Artist

    Sep 28, 2006
    Kingston, WA
    That can't be right. Whenever MY father got over my mother there was NEVER any "anak", you can bet on that!