How is ft/lbs energy of a bullet defined

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Thread: How is ft/lbs energy of a bullet defined

  1. #1
    Member Array defensive007's Avatar
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    Thumbs up How is ft/lbs energy of a bullet defined

    Been wondering how this is determined?
    I have yet to be attacked by a block of ballistic gelatin but.........

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  3. #2
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    Kinetic energy is determined by the formula E = 1/2m x v(2), that is, one half mass times velocity squared.

    Mass, m, is usually given in grains and the speed, v, in feet per second but kinetic energy, Ek, is typically given in foot-pounds. If m is specified in grains and v in feet per second, the following equation can be used, which gives the energy in foot-pounds:

    E = 1/2m x v(2) x (1 ft lb / 7000gr x 32.1739 ft(2)/sec(2))

    Clear as mud, right?


    An easier to remember formula that gives a number very close to the right answer is:

    fps x fps x grains x 222 and then divide by 100,000,000 (or just move the decimal point 8 places to the LEFT). It's easy to remember the 222 and you can remember the 8 decimal places because 8 is 2 x 2 x 2.

    That, or use m (in grains) times velocity (in fps) squared, and divide by the constant 450,400.
    A man fires a rifle for many years, and he goes to war. And afterward he turns the rifle in at the armory, and he believes he's finished with the rifle. But no matter what else he might do with his hands - love a woman, build a house, change his son's diaper - his hands remember the rifle.

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    Member Array natimage's Avatar
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    Arrow Kinetic Energy Calculator:Dewclaw Archery Supplies

    This is a calculator I use for my archery equipment but it'll work the same for bullets... It's Kenetic Energy= 1/2 (mass * velocity) squared
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    Member Array defensive007's Avatar
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    This is clear as mudd
    I have yet to be attacked by a block of ballistic gelatin but.........

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    It's not that hard, really. If you are asking broader questions about "energy" in the scientific definition (the ability of one physical system to do work on another physical system), some real research might be in order. If you're just asking how to figure out muzzle energy for your pet load, there you have it.
    A man fires a rifle for many years, and he goes to war. And afterward he turns the rifle in at the armory, and he believes he's finished with the rifle. But no matter what else he might do with his hands - love a woman, build a house, change his son's diaper - his hands remember the rifle.

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    Member Array natimage's Avatar
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    Kenetic Energy doesn't tell the whole story though...What's more important with guns is how well the round holds together (maintains its mass) and how much of that energy it actually dumps into the target (based on how well the round itself expands)
    Psalm 23
    In God I trust, it's the rest of you I'm concerned about

    Certified Smith & Wesson M&P pistol and MP15 rifle armorer

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    Quote Originally Posted by defensive007 View Post
    This is clear as mudd
    OPFOR gave you the straight story. But look at an ammo catalog listing from any of the major manufacturers. They list the cartridge and the bullet weight, then velocity and energy numbers at the muzzle, 100 yds, 200 yds, etc.

    The bottom line is that energy is calculated from the mass (weight) of the bullet and its velocity, using various conversion factors to make the units come out right. The 'trick' is that energy is calculated using the square of the velocity, it's not just bullet mass times velocity.

    Any clearer?


    One more thing, terminology is important. Energy is given (in the English system of units) in foot-pounds, not "ft/lbs". Using the slash implies "per", so when you write "ft/lbs" that is interpreted as "foot per pounds," which does not equate to an energy term at all. Don't feel bad, even car ads get that one wrong. I even heard a radio ad describing a truck engine's torque as oh-so-many "feet per pound of torque"... (should be foot-pounds). We geeky engineer types grind our teeth when we hear stuff like that.
    Smitty
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    Senior Member Array boscobeans's Avatar
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    Just as an example of how kinetic energy can be delivered:

    A baseball bat swung by an average professional baseball player generates about the same kinetic energy as a .38special 110grain Hornady Critical Defense round. ( 240 +/- ft/lbs).

    bosco

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    VIP Member Array glockman10mm's Avatar
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    Quote Originally Posted by boscobeans View Post
    Just as an example of how kinetic energy can be delivered:

    A baseball bat swung by an average professional baseball player generates about the same kinetic energy as a .38special 110grain Hornady Critical Defense round. ( 240 +/- ft/lbs).

    bosco
    I think I'd rather be hit by the bullet.
    Ignorance is a long way from stupid, but left unchecked, can get there real fast.

  11. #10
    VIP Member Array TedBeau's Avatar
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    OK, this is interesting: I followed the link that natimage gave above for calculating energy in archery equipment. At the bottom of the calculator is a paragraph explaining another way to put the numbers into perspective.

    If an arrow generates 58 ft-lb of energy that would mean that if you shot an arrow vertically into a one pound weight the arrow would be able to lift it 58 feet! Whats more impressive I guess is that that means my 38 special could lift the same one pound weight 240 feet, or lift a 240 pound man 1 foot. I know this assumes a soild target and no energy loss in bullet or target deformation but man wouldn't that look cool to have a target go flying up 240 feet when you shoot it!

    Unfortunatly the Mythbusters tried a simular test trying to knock a pig carcuss off a peg. Even using a shotgun they couldn't get a hundred pound carcuss to move sideways even a fratction of an inch. I think they would have had better luck if they had put a steel plate or ballistic vest on the carcuss to give the bullet something to push against.

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