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Thread: Arrow Flight Problem

  1. #1 Arrow Flight Problem 
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    Hello, brand new here and in need of some expertise. An arrow being shot is a tricky thing.

    The archery industry uses the term Kenetic Energy to describe the energy available to an arrow upon impact. Energy to perform a penetration. I suggested that KE is not the term that should be used to describe this, but rather that KE best describes the energy required to do the work of the launch phase of the arrow. The amount of work required to impart a certain speed to the arrow.

    KE= 1/2 mass times velocity squared
    Where mass is in kilograms, velocity in meters per second, and the result in joules.

    Another problem is that the industry, and my friends on the crossbow forum, use grains to measure the 'weight' of their arrows, feet per second to measure the 'speed', and of course 'KE' to measure the power available upon impact. So many issues here all at once.

    They are a distinguished bunch, these fellow archers, and I wish to 'change' some things while stepping lightly on old toes, and help point them, and myself, to a more accurate manner with which to determine this.

    Can you manage a formula that would allow us to use the weight of an arrow in grains, traveling at speeds measured in fps by our chronographs, and compute the amount of 'Impact Energy" a particular arrow would potentially have. This is important to us because an arrow loses speed over distance and at our maximum range of fifty yards, we would like to know if our arrows retain enough energy to penetrate our targets.

    Oh....and it would be most helpful for it to be simple and in layman's terms, otherwise my friends will simply curl up into a little ball and ignore the strange doodles on their screens.


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    I'm not sure what the problem is (with using KE), apart from using non-metric units; you just need to convert the units to metric.

    weight in kg = 0.065 * weight in grains
    speed in m/s = 0.3048 * speed in f/s

    KE = 0.5 * 0.065 * (weight in grains) * (0.3048 * (speed in f/s))^2

    or rather:

    KE = 0.00302 * (weight in grains) * (speed in f/s) squared

    Is that what you were looking for?

    The kinetic energy at impact will be smaller than at launch (because of heat lost to air friction), the formula gives the energy of the point at which you measured speed. So the closer to impact you measure it, the better it will be.

    (Note that as the arrow falls it will accelerate downwards, which will tend to increase it's absolute speed; I suspect this effect will be smaller than the speed loss due to air friction under normal conditions, but may compensate for some of it)


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  4. #3  
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    Thank you Doctor


    Once launched, the arrow encounters both gravity and air resistance ( mostly the feathers that guide the tail end of the arrow) and it slows. We archers will allow for this effect by aiming over the intended target, knowing beforehand the trajectory. Indeed the arrow may increase speed slightly as it travels down the backside of the arc, and just as you suggest, that is probably very minimal when compared to the forces trying to slow it.

    But..... isn't kinetic energy different than impact energy ?

    I understood kinetic energy to be the amount of 'work' required to move an object up to a certain speed.

    Maybe I am confusing myself. But the measurement of how much energy is transferred from the arrow to the target is what I am seeking, and probably should be expressed in foot pounds of force.

    For example....an arrow that weighs 450 grains, engages the target at a speed of 250 feet per second, and imparts X foot pounds of force.

    Already you have helped me very much by simplifying the formulas and allowing for a proper use of the KE formula. Thank you.
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    The energy is not expressed in foot pounds of force. It can be expressed in foot-pounds of energy, and is related to the velocity by KE=1/2mv^2. Knowing the mass in grains and the velocity in feet per second, you can use the appropriate conversion factors to get foot-pounds of energy. Google calculator is very handy for such conversions. Example if you type "0.5*450 grains*250*250*ft^2/sec^2 in foot-pounds" into the Google search box you get (0.5 * 450 grains * 250 * 250 * (ft^2)) / (sec^2) = 62.4394088 foot-pounds, which tells you that a projectile weighing 450 grains and traveling 250 feet per second has 62.4 foot-pounds of kinetic energy.

    If you use a chronograph, that will give you the speed at a few feet from the bow, and as you know, it will slow down from there. To find the velocity at the target, you could probably use some sort of ballistic calculator. There is probably some kind of application on the web that will do that for you. Certainly, there is for rifle bullets. I don't know about archery.
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  6. #5  
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    Thank you very much. I do appreciate your responses. This is a wonderful forum.
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  7. #6 Re: Arrow Flight Problem 
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    Quote Originally Posted by FishinMortician
    Hello, brand new here and in need of some expertise. An arrow being shot is a tricky thing.

    The archery industry uses the term Kenetic Energy to describe the energy available to an arrow upon impact. Energy to perform a penetration. I suggested that KE is not the term that should be used to describe this, but rather that KE best describes the energy required to do the work of the launch phase of the arrow. The amount of work required to impart a certain speed to the arrow.

    KE= 1/2 mass times velocity squared
    Where mass is in kilograms, velocity in meters per second, and the result in joules.

    Another problem is that the industry, and my friends on the crossbow forum, use grains to measure the 'weight' of their arrows, feet per second to measure the 'speed', and of course 'KE' to measure the power available upon impact. So many issues here all at once.

    They are a distinguished bunch, these fellow archers, and I wish to 'change' some things while stepping lightly on old toes, and help point them, and myself, to a more accurate manner with which to determine this.

    Can you manage a formula that would allow us to use the weight of an arrow in grains, traveling at speeds measured in fps by our chronographs, and compute the amount of 'Impact Energy" a particular arrow would potentially have. This is important to us because an arrow loses speed over distance and at our maximum range of fifty yards, we would like to know if our arrows retain enough energy to penetrate our targets.

    Oh....and it would be most helpful for it to be simple and in layman's terms, otherwise my friends will simply curl up into a little ball and ignore the strange doodles on their screens.
    Kiineteic energy is the correct term and the right physical quantity.

    You have two clear elements of energy involved. The first is the energy associated iwth the work performed in drawing the bow. That energy is the integral of the force distance curve between the brace height and the ultimate draw length. It is the work done by the archer in drawing the bow or in cocking a cross bow.

    Next is the kinetic energy of the arrow. That is dependent of the work done in drawing the bow, less the losses incurred when the arrow is released. Generally the heavier the arrow the more efficient is the energy transfer, but the slower is the arrow.

    The kinetid energy of the arrow is determined by 1/2 mv^2. To arrive at energy in foot-pounds the mass should be expressed in slugs and the velocity in ft/sec. A slug is 32.2 poiunds mass, and a pound mass is 7000 grains. So to get mass in slugs from mass in grains you need to take the weight in grains divide by 7000 x 32.2.

    Finally the kinetic energy at the target is given by 1/2 mv^2 where v is the striking velocity. That velocity will be somewhat less than the velocity at the bow due ot drag, but probably is not too different as drag at the low velocities of an arrow (200-300 fps) is not all that great over the distances involved.

    You can measure arrow velocity by shooting the arrow through a chronograph. This can be done at any range at which you are confident of putting the arrow across the sky screens and not hitting the screens themselves.

    One other important factor in arrow penetration is momentum, which is just mv. Heavier arrows will generally penetrate farther than lighter arrows.
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  8. #7  
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    kinetic energy is as good as impact energy cause they are totally interchangeable.
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  9. #8  
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    Quote Originally Posted by kakarot
    kinetic energy is as good as impact energy cause they are totally interchangeable.
    Not quite.

    Kinetic energy is a general term for the energy that comes with motion.

    Impact energy is the kinetic energy at impact.
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  10. #9  
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    DrRocket wrote:
    Not quite.
    Kinetic energy is a general term for the energy that comes with motion.

    Impact energy is the kinetic energy at impact.




    But in this context according to the law of conservation of energy they both result to the same and hence are interchangeable (though I might be wrong thats what i feel)
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  11. #10  
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    Quote Originally Posted by kakarot
    DrRocket wrote:
    Not quite.
    Kinetic energy is a general term for the energy that comes with motion.

    Impact energy is the kinetic energy at impact.




    But in this context according to the law of conservation of energy they both result to the same and hence are interchangeable (though I might be wrong thats what i feel)
    Your feelings are irrelevant. The arrow starts out with a certain kinetic energy and loses kinetic energy due to friction with air, or maybe gains some by gravity if the shot is downhill. The impact energy is the kinetic energy that remains at the time of impact. What is there to argue about?
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  12. #11  
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    Quote Originally Posted by FishinMortician
    Thank you Doctor
    Once launched, the arrow encounters both gravity and air resistance ( mostly the feathers that guide the tail end of the arrow) and it slows. We archers will allow for this effect by aiming over the intended target, knowing beforehand the trajectory. Indeed the arrow may increase speed slightly as it travels down the backside of the arc, and just as you suggest, that is probably very minimal when compared to the forces trying to slow it.
    If you're trying to calculate before hand the amount of energy that will be lost in flight, then you can probably leave gravity out of it. Any energy you lose by aiming your arrow upward is regained when it comes back down. So, the only really important losses come from air friction.

    A parabolic flight path is longer than a straight line flight path, however, so it's important to factor that in.



    But..... isn't kinetic energy different than impact energy ?
    There's no reason not to call both by the name "kinetic energy" if that's an easier way for you to organize your thoughts. Just remember that the amount of kinetic energy available is different at launch than it is at impact.

    The amount of kinetic energy your arrow still has when it hits a stationary target is the amount of energy it will impart to that object.



    I understood kinetic energy to be the amount of 'work' required to move an object up to a certain speed.
    Yeah, but it's also the amount of 'work' required to stop it.


    Maybe I am confusing myself. But the measurement of how much energy is transferred from the arrow to the target is what I am seeking, and probably should be expressed in foot pounds of force.
    Energy is foot pounds of force multiplied by distance. So, you see that foot pounds are always included. It's just that there's another, additional factor that also makes a difference.

    The amount of force the arrow exerts on impact is variable depending on how suddenly it has to stop. If the stop is totally immediate, then the amount of force experience will be very strong, over a shorter time. If the stop is more gradual (like when a person falling from a tall building lands on a trampoline), then the force will be less, but last longer.

    The reason why a cheaper arrow will usually break, or have parts come loose, if you aim it at a brick wall, is because the force it experiences is actually stronger than it would be if it hits a soft target.
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  13. #12  
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    BS! the momentum trasferred to the target is the same . MV
    The rate can be different - for example a pointed arrow will penetrate a soft target and dissipate its energy more slowly than a blunt one which impacts and suddenly stops . the energy is still the same. Ke= 1/2MV^2
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  14. #13  
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    Quote Originally Posted by kojax
    Energy is foot pounds of force multiplied by distance.
    NO !

    Energy is pounds of force multiplied by distance, hence foot*pounds.

    Foot pounds is a unit of energy, not force.
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  15. #14  
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    Quote Originally Posted by fizzlooney
    BS! the momentum trasferred to the target is the same . MV
    No it is not, unless the arrow fails to penetrate the target. If the arrow passes through the target then there is residual momentum in the arrow when it exits and that momentum is quite variable, depending on the mechanics of the penetration. It is VERY common for a broadhead hunting arrow to completely penetrate the target, and in fact that is the desirable condition.

    The energy and momentum of an arrow are not all that important. It is a far different situation than that which attends a bullet. An arrow kills through hemorrhage, not through shock or tissue damage. What you want in a hunting arrow is the sharpest possible broadhead, placement in a region with many large blood vessels and complete penetration. If the arrow has enough energy to accomplish that, it has enough energy for the job, and the same goes for momentum. That one reason why, given ability to make the shot, heavy arrows are preferable to lighter arrows. It turns out that the mechanics of launching an arrow favor heavier arrow for both energy and momentum, but not velocity. It is a different trade-off from that which you find with a firearm.




    Quote Originally Posted by fizzlooney
    The rate can be different - for example a pointed arrow will penetrate a soft target and dissipate its energy more slowly than a blunt one which impacts and suddenly stops . the energy is still the same. Ke= 1/2MV^2
    Nope, not in practice. A blunt arrow will not penetrate the target and therefore will transfer all of its energy and momentum to the target. A sharp pointed arrow may and should penetrate and will have residual energy and momentum when it emerges. However, a sharp broadhead is far more effective than a dull one. The energy in an arrow is not all that great, and the arrow relies on penetration and ability to cut blood vessels for lethality.

    The exception are blunt arrow used for extremely small game (smaller birds, squirrels, etc.). In those specialized situations it is the momentum and energy that kill, and the animals in question are sufficiently small that the actual momentum and energy are not important.
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  16. #15  
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    Really great diatibe on hunting with a bow and arrow but not really dealing with the actual question, jesus this forum is for nit pickers,or is that nose pickers?
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