.45 Super vs. 10mm?

Discussion in 'Reloading' started by SIG_Fiend, Apr 6, 2008.

  1. SIG_Fiend

    SIG_Fiend Administrator TGT Supporter Admin

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    Here's a question to all of you more knowledgeable guys out there. I'm wondering what caliber provides more ultimately superior ballistic capabilities, either 45 Super or 10mm? I've done a little research into both calibers, however I still have a lot to learn. Judging by what I've seen, I was under the impression 10mm brass from certain manufacturers is capable of handling significantly higher pressure levels, and because of this I also thought you could load them really hot to far surpass the capabilities of 45 Super? I remember reading something about certain 10mm brass being capable of withstanding up to ~50,000cup and 45 Super only being able to withstand ~28,000-30,000cup. The reason I'm asking is, in the not too distant future, I'd really like to experiment with one of these calibers. I may just experiment with both, however I would like to learn more about both regardless. Any input is much appreciated.
     


  2. phatcyclist

    phatcyclist Active Member

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    I'd say, go with 10mm. The fact that .45 Super is hard (or impossible) to find on shelves.
     
  3. Texas1911

    Texas1911 TGT Addict

    May 29, 2017
    Austin, TX
    It isn't peak pressure that wins the game, it's pressure over time. That's why the .45 Super develops so much power. If you go by pressure peak alone the .45 Super should be falling short of 10mm factory ammo, and empirical says otherwise.

    If I was actually going to own something, I'd go 10mm and save the ammo hassle. Then buy a .500 S&W and laugh at your puny .45 Super as I knock out T72's at 500 yards.
     
  4. SIG_Fiend

    SIG_Fiend Administrator TGT Supporter Admin

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    While .45 Super may have more energy at lower velocities (sub 1500fps range), from what I've seen 10mm can be loaded to considerably higher velocities than .45 Super, which seems to be one benefit. There is of course always something to be said for ease of obtaining brass and the other components for reloading, or even just pre-loaded ammo. I'm giving serious consideration to going with an EAA Witness Match in 10mm.
     
  5. Texas1911

    Texas1911 TGT Addict

    May 29, 2017
    Austin, TX
    Velocity is the main ticket since it is a squared value.
     
  6. comanche

    comanche New Member

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    I would have to disagree. It is more about caliber and bullet weight.
    SIZE and penetration is everything. ask any woman.
    A 240gr bullet at 1400fps will not penetrate as deep as a 300gr at 1200fps. Muzzle energy is not as important as the energy deposited in the target.
     
  7. Texas1911

    Texas1911 TGT Addict

    May 29, 2017
    Austin, TX
    [​IMG]

    You can see that in order to derive kinetic force you are better suited increasing velocity. The 240 gr. bullet at 1400 FPS has 8.9% more energy that it will deliver when it impacts the target.

    The reason for the difference in penetration is that the bullet is contacting a target primarily made of water. Since water is held to be incompressible at relatively low pressure, the bullet at the higher velocity is met with more resistance when it impacts the body. It's a duration of load, and relative hardness factorization.

    Caliber aids in preventing wound contraction which helps bleed the target, but it reduces the ability of the round to penetrate ambient objects since it has a reduced nose pressure. A 1" slug would take 20 times the force to punch through an object compared to a .22 bullet. It's further hindered by the reduction in speed, which lowers the duration of load, which makes a material a hair more resilient.

    Essentially, we all know if you slip into the water it doesn't hurt, but if you do a 20' cannon ball into the water it stings. That's because the water molecules don't compress and have significantly less time to displace. If you were to jump off a bridge and hit the water, it'd be like jumping off a building at 30' and landing on concrete.

    Just my take on it, I could be crazy.
     
  8. comanche

    comanche New Member

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    No, I don't think your crazy. But, your analogies are a little off. I'd much rather fall 30' into water than concrete.:rofl:

    The mathematical formula for the mass of a bullet is:
    M=W/7000g
    where: W=bullet weight in grains
    and g=32.174 ft/second (squared) which is acceleration due to gravity.
    The kinetic energy of a bullet can then be found from the formula:
    KE= m(VxV)/2
    where m=mass of bullet
    and V=velocity

    It's simple Physics (math), however Ballistics is a little more complicated than kinetic energy. There are several other factors like; the various "Laws of Motion and Momentum", friction (resistance), gravity, linear acceleration, linear velocity, linear momentum, a torque, angular acceleration, angular velocity, angular momentum, ect. For our purposes, a vector is any physical quantity that has both a magnitude and a direction and it is important to realize that a vector can change in magnitude, in direction, or both. [FONT=arial,helvetica,verdana,sns-serif]Translational motions are linear motions of a body (i.e., a bullet). These motions can occur in three directions, for example, in the downrange, vertical, and crossrange directions. The translational motions of a bullet are governed by Newton’s Laws, specifically Newton’s Second Law, which states that [FONT=arial,helvetica,verdana,sns-serif]the rate of change of the linear momentum of a body with respect to time is equal to the force applied to the body. When the mass of the body is constant, not changing with time, this relationship becomes the familiar [FONT=arial,helvetica,verdana,sns-serif]mass times acceleration is equal to the force applied to the body. [/FONT][FONT=arial,helvetica,verdana,sns-serif]These are relationships among the vector quantities of force and momentum (or acceleration) of the bullet. When these vectors are resolved into components along the three translational directions of motion of the bullet, the results are called the equations of linear motion of the bullet.[/FONT] [/FONT][/FONT]
    I could go on; Rotational motions of a rigid body (a bullet) are caused by torques applied to the body and the resulting equations are called Euler’s equations of angular motion of the body.
    Then there is ballistic coefficients and aerodynamics and I don't even want to think about differential equations of motion.

    OK, enough of the mad scientist bit.;)

    It is absolutely true that, if two bullets with the same ballistic coefficient are fired with the same muzzle velocity at the same firing point and with the same weather conditions, the bullet with the greater mass will arrive first at a specific target, will have higher velocity and energy when it arrives, will suffer less wind deflection and will have less change in trajectory.

    My point is muzzle velocity or muzzle energy can be deceiving. The answer is caliber and bullet weight. These are the only CONSTANTS we have in external ballistics. Velocity is a constantly diminishing variable. Velocity and foot pounds of energy look good on paper. But, a much more reliable (simple) formula that tells the true results on game is John Taylor's Knock Out Formula : Caliber x Bullet weight x velocity divided by 7000
     
  9. Texas1911

    Texas1911 TGT Addict

    May 29, 2017
    Austin, TX
    Ah, there is a misconception involved.

    He states that if you take a heavier bullet, and fire it using the same velocity, etc. That automatically tells you that a greater amount of energy has been expelled to accelerate that heavier mass. The KE is greater, and the result is longer range, more power at the target, and a more stable flight path (reduction in migration due to the heavier body).

    However if you take a heavier bullet, and fire it using the same charge, as is normal practice in factory ammo, then you don't have these benefits since the reduction in velocity reduces the KE value of the bullet.

    The larger the caliber the greater the potential resistance. You can develop the frontal section to be more inclusive of laminar flow to reduce boundary layer separation, but you will not overcome the sectional area aspect of aerodynamic drag. As such, a lower caliber bullet will generally maintain more velocity over a longer range since drag is reduced, if fired at the same velocity. With pistol calibers, there is always the transition between super sonic and sub sonic rounds in relation to aerodynamic properties since the flows alter in response. The slower the speed, the greater the increase in cross sectional area since the airflow begins to wrap itself onto the bullet facet. Similar to a wing, where you derive cross section from the engaged surface rather than planer facet.

    In respects to drag however, it too involves a squared velocity. The faster the round, the greater the drag vector. You would have to greatly reduce the cross sectional area to make up for the difference. This is assuming the round maintains a high velocity to ensure a Reynolds number over 1000.

    Rotational spin is a method of eliminating undesirable effects of firing ball ammo out of a smooth bore. The round picks up an imparted spin along a single forward or otherwise offset axis and it has the effect of a curve ball in baseball. By purposely imparting spin on the bullet you override any undesirable vectors. Spin itself is a self-neutral force unless it's at a good range. Once the bullet is exposed to a long flight time the resistance to the torque vector causes the bullet to begin migrating. It essentially "walks on air".

    After plotting .357 Sig, .45 Auto in 145 Gr., and .45 Auto in 230 Gr. in relation to drag, velocity, and energy at 0 - 50 yards, I noticed something rather obvious, but contradictory to some arguments. The reduction in Energy of the bullet is directly proportional to the value of Aerodynamic drag. That sounds obvious right? Well ... take this into account.

    The 145 Gr. bullet exhibited GREATER loss in energy than the 100 Gr. bullet, and the 230 Gr. exhibited the least. That contradicts the notion that grain weight is the reason for maintained energy / velocity. Here's why.

    It's all based on friction. The smaller 100-grain .357 Sig round has a high velocity, but due to it's smaller cross section it's drag is less than the slightly slower, but larger 145-grain .45 Auto.

    The 230-grain .45 Auto exhibits much less dramatic degradation of performance, but it's no coincidence that it also exhibits the least drag. It still under performed both rounds in overall energy.

    The relation to energy transfer is obviously linked to frictional performance of the bullet. Grain weight is simply a fixed value of energy, and caliber depicts cross section in determining friction loss.

    Plotting rifles delivers some interesting results as well, but this is regarding pistols.
     
  10. comanche

    comanche New Member

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    Austin
    My head hurts. I'm just going to shoot 45s; because Samuel Colt didn't make a 46.
     

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