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Ballistics Energy Confusion

Discussion in 'General Gun Discussions' started by DMK, Sep 7, 2007.

  1. DMK

    DMK Well-Known Member

    Let's take Hornady .308 155gr TAP for example.


    It does 2785 fps and has 2669 ft/lbs of energy at the muzzle (2577fps and 2285 ft/lbs at 100 yards).

    My understanding of mechanical energy is that 1 ft/lbs of energy will move a one pound object one foot.

    So with the above TAP round, is it capable of hitting a 1 lb object sitting at 100 yards and moving it 2,285 feet?
  2. Claude Clay

    Claude Clay Well-Known Member

    the energy is not transfered all at once, but rather like a bell curve ( skewered to the right a bit ) where the total area under the curve adds up to the energy occures over time. granted , not a whole lot of time, but enough to keep the penetrated object from leaving your time zone. simplistic, yes....but i hope this helps a bit.i shot golf balls w/ a 22 & hit just right they move out fast about 60+ feet. the density /weight of the object & angle will determine what actually happens. its a complicaated question....
  3. fletcher

    fletcher Well-Known Member

    I'm not sure if that would work, since the movement will be rate dependent, and different rates consume different amounts of energy.
  4. mpmarty

    mpmarty Well-Known Member

    A foot pound of "energy" as it is referred to is equal to one pound dropped one foot onto an immovable object. It has nothing whatsoever to do with the ability to move an object weighing one pound. Thus, in theory, your 2700 ft lbs of "energy" is equal to 2700 pounds dropping one foot to an immovable object.

    Of course, this assumes a standard gravitational power such as that found at the equator.
    Last edited: Sep 7, 2007
  5. svtruth

    svtruth Well-Known Member

    Muzzle energy

    has mass in the equation. Assuming complete transfer, velocity of target will be what ever yields the same KE. One lb is 7,000 grains, so a 250 gr bullet is 1/28th of that. And, velocity is squared in the KE equation.
    Good luck.
  6. Navy_Guns

    Navy_Guns Well-Known Member

    Um... sort of!

    If an object has a kinetic energy of 1,000 foot-pounds, what it means is that if you were magically able to capture all of that energy and magically convert it to a force that lifted an object against the force of gravity, the energy COULD lift a 1,000 pound mass through a distance of 1 foot working against a standard gravity. I say MAGICALLY because no process is 100% efficient.

    There is an absolutely horrible article in the 17th edition of Handloader's Digest written by Lee Saunders called "Thoughts on Energy" where the man boldly tries to wrap his brain around the observation that when a 1,000 pound object is shot by said projectile it never does fly a foot in the air. It's one of those laws like "heat only flows from hot to cold" or "you can't push a rope" - there is no perfect process by which you can capture all the kinetic energy of a moving object and use it to do work (force x distance, lift a weight for example).

    And though the bullet does transfer energy to the target, it's a momentum transfer (more or less) that you can use to determine the resulting velocity of the struck object. Assume (1) is the bullet and (2) is the target and that the bullet sticks in the target... if the target starts at rest, M1 x V1 = (M1 + M2) x V(1+2).

    Though I'm sure there are folks in Hollywood who would try to convince you that the .44 magnum can indeed cause a 200 pound man to fly backwards at 75 mph, flip over a truck, knock down buildings, etc... :)
  7. Vern Humphrey

    Vern Humphrey Well-Known Member

    What you want is momentum, not kinetic energy.

    That same load deilvers just under 2 slug foot pounds of momentum. So no, you're not going to move much weight any significant distance.
  8. Jim Watson

    Jim Watson Well-Known Member

    I remember that article, Navy Guns. Just bloody awful, as bad as the frequent attempts to appy John "Pondoro" Taylor's TKO to pocket pistols and varmint rifles.

    As the saying goes, "Gravity, it's not a suggestion, it's the Law.'

    Energy and momentum are always conserved. Doesn't mean they have to be displayed in obvious fashion. Shoot a bullet into a big enough or hard enough target to stop it and a good deal of energy goes into deforming the bullet and the target. It ends up as material deformed into other shapes and conditions and as heat. Momentum can be transferred through the target's supporting members into the ground, thereby giving the whole world a little nudge.
  9. Navy_Guns

    Navy_Guns Well-Known Member

    that 155 grain bullet at 2785 fps, striking a 200 pound object at rest (if it's totally rigid and on a frictionless surface, no air resistance, etc...) could make the target move backwards no faster than 0.31 feet per second. If that object were living, being struck could cause nerves to fire, muscles to jerk, etc. which could give more apparent reaction but still not Hollywood special effects.
  10. DMK

    DMK Well-Known Member

    Thanks guys. Now I'm really confused. :D
  11. Vern Humphrey

    Vern Humphrey Well-Known Member

    The forumula for kinetic energy is:
    Ke = 1/2 M*V^2

    The formula for momentum is:
    m = M*V

    To get M (mass) you have to divide the weight in pounds by the Gravitational Constant, 32.2 ft/sec/sec.

    While this may look complicated, it's really simple -- Kinetic energy is what does the killing, but momentum is what does the moving. That's why a gun doesn't kill the man holding it -- because recoil is expressed as momentum, not kinetic energy.
  12. akodo

    akodo Well-Known Member

    yes, as long as you can have that 1 lb object on a frictionless surface, loose no energy to heat, deformation of the projectile, deformation of the target, etc etc.
  13. Vern Humphrey

    Vern Humphrey Well-Known Member


    If you had total transfer of momentum, the 1 lb weight would be accelerated to 57 fps. If the direction of acceleration was upward, it would rise to a height of 28.5 feet.
  14. Feanaro

    Feanaro Well-Known Member

    Vern is correct. In a closed system, with an elastic collision, the kinetic energy will not be conserved. 2,285 ft/lbs of KE would never be fully used. Momentum, on the other hand, must be conserved.

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