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Speed of an arrow from a bow

  1. Jun 17, 2006 #1

    npu

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    I am trying to determine the function for the velocity of an arrow out of a bow (leaf spring).
    I am talking without considering factors such as air resistance and friction. I came up with:

    [tex]
    V=\sqrt{ \frac {2F_{av}S_f10}{m_d+ \frac {1}{2}m_fR^{-1}}}
    [/tex]

    [tex]S_f[/tex]= Spring tip movement in meters;
    [tex]S_d[/tex]= Arrow's run in meters;
    [tex]m_f[/tex]= Spring weight in Kg;
    [tex]m_d[/tex]= Arrow weight in Kg;
    [tex]F_{av}[/tex]= Average force exerted by the spring along [tex]S_f[/tex] in Newtons;
    [tex]
    R=\frac {S_d}{S_f}
    [/tex]

    Using this formula I got results which appear right:

    Overall [tex]V[/tex] appears to be directly proportional to arrow to spring ratios in mass and distance.

    Still I am unsure about it. Is it right?
    If so, is there a more elegant formulation?
     
    Last edited: Jun 17, 2006
  2. jcsd
  3. Jun 17, 2006 #2

    Andrew Mason

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    It would help if you explained your reasoning behind that equation!

    AM
     
  4. Jun 17, 2006 #3

    npu

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    [tex]V=\sqrt{2aS}[/tex] and [tex]a=\frac {F}{m}[/tex] so

    [tex]V_d=\sqrt{ \frac {2FS_d}{m_d} }[/tex];

    The above is certainly correct for the arrow by itself and a force acting on it but it says nothing of the system bow+arrow. In the bow system the spring's must be taken into account, the energy goes into moving the limbs as well as the arrow.
    I assimilated the bow system to a lever system with spring on the left
    and arrow on the right, the spring's mass must then be brought to the
    same side as the arrow,

    [tex]m_f\frac {S_f}{S_d}[/tex],

    now this can be moved to the right, it needs to be divided by two because only one end of the limb is moving as indicated while the other end is fixed.
    The force, average between initial and maximum, exerted by the spring
    also has to be similarly taken to the right side

    [tex] F\frac {S_f}{S_d} [/tex]

    The 10 in the numerator I have no explanation for and reasoning around
    Newtons and Kilograms relationship also doesn't seem to justify it.
    However it appears to work and who am I to question it.:biggrin:

    Gives:
    [tex]
    V=\sqrt{ \frac {2F_{av}S_f10}{m_d+ \frac {1}{2}m_fR^{-1}}}
    [/tex]

    May be unorthodox but it gives coherent results.
     
  5. Jun 17, 2006 #4

    Andrew Mason

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    Why does it give coherent results? What are you testing it with?

    The bow would not exert the same force as you pull the arrow back. The force would increase as you pull it back. If you assume the force increases linearly with pullback distance, the stored energy in the bow would be [itex].5kx^2[/itex] where k is the spring constant F/x and x is the displacement.

    Then assume that the potential energy of the drawn bow is converted entirely into the arrow's kinetic energy:

    [tex]\frac{1}{2}m_dv^2 = \frac{1}{2}kx_{max}^2 = F_{av}x_{max}[/tex] where [itex]F_{av} = .5kx_{max}[/tex]

    This works out to:
    [tex]
    v = \sqrt{ \frac {2F_{av}x_{max}}{m_d}}
    [/tex]

    AM
     
  6. Jun 17, 2006 #5

    npu

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    The formula I indicated shows results in line with manufacturers specs and it also shows decreasing efficiencies for decreasing arrow masses and shorter acceleration runs. Over dozens of tries with widely varying input data it never showed clearly impossible values.

    The [tex] F_{av} [/tex] is the average force along the limb's tip movement. In this sense There's no need to consider increase in force as the spring is pulled. I just take the average and the distance moved by the tip.


    In your formula there isn't the mass of the spring, surely that influences the way energy is allocated between limbs and arrow.
    Also what's the point of assuming the spring's potential energy is converted entirely into kinetic? The whole point is to determine how much energy goes into the arrow not to assume 100% does.

    Or rather why should I calculate speed by assuming 100% energy conversion if I can calculate the actual value.
     
  7. Jun 17, 2006 #6

    Andrew Mason

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    It depends on how you determine the average force. How do you determine that? It would go from 0 to maximum force at the displacement used to launch the arrow. You could just measure the force at maximum displacement and use that in determining the potential energy of the bow: [itex]U = .5F_{max}x_{max}[/itex]

    It would account for some amount of the energy of the bow not transferring to the arrow but I think the mass of the string would be more of a factor since it would vibrate after the arrow has been launched. The physical analysis of the bow movement is fairly complicated I would expect. To determine just how much energy is retained in the bow and string is not an easy thing to analyse.

    In order to determine the efficiency of your bow, you could measure the potential energy of the bow and the kinetic energy of the arrow immediately after launch.

    How does your formula factor in the efficiency of the bow?

    AM
     
  8. Jun 17, 2006 #7

    npu

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    Are you saying that average force is an unnecessary assumption?
     
  9. Jun 17, 2006 #8

    Andrew Mason

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    Well, it is not an assumption. Energy is the integral [itex]\int Fdx[/itex]. If F is a linear function of x then it works out to [itex]F_{avg}(x_f - x_i)[/itex]. F is not a perfectly linear function of x as the angle of the string increases, but it is probably not a bad approximation.

    AM
     
  10. Jun 18, 2006 #9

    npu

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    What I propose, with my formula, which I brought to your attention for evaluation is to factor in mass and distance ratios and their impact on overall efficiency.

    It also unveils the "mysterious" efficiency factor that one finds in archery literature where they would have you calculate the arrow's speed by factoring in arbitrary factor or as you propose derive efficiency experimentally by measuring speed. Which is the same thing. These approaches I find unacceptable as they deny the possibility of predicting speed.
    Well, I contend that the mysterious efficiency factor is none other than the masses-distances ratios of the apparatus and that my formula correctly predicts arrow speed for a broad range of machines.

    Efficiency can be predicted without experiment because the formula accounts for the losses which are in my opinion essentially due to the limb's mass with this being more so for low projectiles mass to limb's mass designs.

    I concede that I make assumptions about the distribution of the energy over the draw (and that this is also altered by the change in wire to limb angle as the draw progresses) but I mantain that there is less error involved there than in ignoring altogether arrow to spring mass and distance ratios.

    [tex]
    V=\sqrt{ \frac {2F_{av}S_f10}{m_d+ \frac {1}{2}m_fR^{-1}}}
    [/tex]

    with [tex]
    R=\frac {S_d}{S_f}
    [/tex]

    is a more complete and powerful tool for exploring an Elastic to Kinetic conversion apparatus than

    [tex]
    v = \sqrt{ \frac {2F_{av}x_{max}}{m_d}}
    [/tex]
     
    Last edited: Jun 18, 2006
  11. Jun 18, 2006 #10

    Andrew Mason

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    What is the essential difference between your formula in which you arbitrarily insert a factor of 10 in the numerator in order to fit experimental results, and inserting a factor e in:
    [tex]
    v = e\sqrt{ \frac {2F_{av}x_{max}}{m_d}}
    [/tex]

    to account for the experimentally determined efficiency of the energy conversion from bow to arrow?

    The latter should accurately predict the speed of the launched arrow.

    Besides, I can't follow the analysis behind your formula. You don't explain why the mass of the bow should be factored in like that. Shouldn't the distribution of that mass along the bow be a factor in how much energy is retained in the bow? If most of the mass is in the handle and the bow tips are made of lighter or thinner material, I would think that it would retain less energy than a bow in which the mass distribution is uniform.

    AM
     
  12. Jun 18, 2006 #11

    npu

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    [tex]V=\sqrt{ \frac {2FS_d}{m_d} }[/tex]

    If we want to obtain the arrow's speed starting from the spring then the 'F' we want to use is

    [tex]F_{av}\frac {S_f}{S_d}[/tex],

    with [tex]F_{av}[/tex]= Average force exerted by the spring along [tex]S_f[/tex] in Newtons;

    giving:

    [tex]V_d=\sqrt{ \frac {2 F_{av}\frac {S_f}{S_d} S_d }{m_d} }[/tex]

    then the limbs mass is similarly carried over

    [tex]V_d=\sqrt{ \frac {2 F_{av}\frac {S_f}{S_d} S_d }{m_d + \frac {1}{2}m_f\frac {S_f}{S_d} } }[/tex]

    [tex]m_f[/tex] is total flexing elements masses, no handle. It needs to be divided by two because half the limb is moving.

    [tex]
    V=\sqrt{ \frac {2F_{av}S_f10}{m_d+ \frac {1}{2}m_fR^{-1}}}
    [/tex]
    with [tex]
    R=\frac {S_d}{S_f}
    [/tex]

    I am not concerned how the spring's energy is transmitted to the arrow, just how a high force low displacement movement speeds a mass along an acceleration run.

    A factor of 10 in the numerator could be a units mixup but saying the speed should be calculated with a factor of unexplained origin defeats the purpose.
     
    Last edited: Jun 18, 2006
  13. Jun 18, 2006 #12

    Andrew Mason

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    I don't follow you there. Doesn't the displacement of the bow string play a part here? How is it factored into your equation?

    From what part of the bow do you measure the flexing elements? Doesn't the whole bow flex? Don't you have to take into account the fact that the bow does not have a uniform distribution of mass along the bow? Where is that taken into account?

    What you have to do is try to analyse how much of the stored energy of the drawn bow is left behind in the motion of the bow after the arrow leaves. Presumably, you want to minimize that energy. To do that, you would have to make the bow as light as possible in the parts that move fastest when the arrow is let go (ie. the tips and the string).

    AM
     
  14. Jun 20, 2006 #13

    andrevdh

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    If you only need to determine the velocity of the arrow as a function of draw distance you could make a calibration curve with some experimental data:
    Load the bow with different weights and note the extension for each of these. This will give you the stored potential energy in the bow as a function of the drawing distance. Assuming that all this stored energy will be imparted to the arrow one can subsequently construct an arrow velocity versus draw distance curve for the bow.
     
  15. Jun 22, 2006 #14

    npu

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    I want to verify the correctness of the following approach in determining how a leaf spring transfers power to an arrow.
    Also keep in mind that nothing is assumed here about the actual kinematic chain from spring to arrow.

    The (corrected) formula is:

    [tex]
    v=\sqrt{ \frac {2F_{av}S_f}{m_d+ \frac {1}{2}m_fR^{-1}}}
    [/tex]
    with
    [tex]S_f[/tex]= Spring tip movement in meters;
    [tex]S_d[/tex]= Arrow's run in meters;
    [tex]m_f[/tex]= Spring weight in Kg;
    [tex]m_d[/tex]= Arrow weight in Kg;
    [tex]F_{av}[/tex]= Average force exerted by the spring along [tex]S_f[/tex] in Newtons;
    [tex]
    R=\frac {S_d}{S_f}
    [/tex]

    Derivation:
    Start from [tex]v=\sqrt{ \frac {2FS}{m}}[/tex], then, the rest is derived by assimilating the bow to a lever system with the spring on one side and the arrow on the other side and taking the spring on the other side by multiplying F and mf by the distances ratio [tex]R^{-1}[/tex]. The spring's mass needs to be divided by 2 because one end is pulled


    Verification:
    A commercial crossbow has:

    90Kg max draw weight,
    60Kg average, 600N * 0.3m ([tex]S_d[/tex]), 180Nm work in loading

    At the springs:
    [tex]m_f[/tex]=0.9Kg (measured, including cams-eccentrics)
    [tex]S_f[/tex]=0.070m (measured)
    [tex]F_{av}[/tex]= 2570N (derived to fit loading effort)

    This is the energy store and source. The motor+fuel so to speak.

    At energy release:
    [tex]S_d[/tex]=0.3m
    [tex]m_d[/tex]=1Kg, 0.03Kg, and 0.01Kg gives


    Using the formula:

    mass speed KE KE/Loading Work
    1Kg 18m/s KE 162Joules Efficiency 90%
    0.03Kg 51m/s KE 40Joules Efficiency 22%
    0.01Kg 56m/s KE 15Joules Efficiency 9%

    The above shows output for different launch masses.
    In a similar fashion the impact of distances variations can be expolored:

    [tex]S_d[/tex]=0.3m, 0.6m and 1.2m
    [tex]m_d[/tex]=0.03Kg

    0.03Kg 51m/s KE 40Joules Efficiency 22% R=4.28
    0.03Kg 66m/s KE 65Joules Efficiency 36% R=8.57
    0.03Kg 80m/s KE 96Joules Efficiency 53% R=17.1

    The efficiency of the Elastic to Kinetic conversion is directly proportional to arrow to spring ratios in mass and distance.

    The point being:

    The "efficiency factor", found in archery literature in connection with the determiination of arrow speed, is essentially due to masses-distances ratios and is superflous; the speed can be predicted with high precision without resorting to arbitrary factors or experiment.

    Or to be more blunt all the insistence on experimental data is unnecessary and with regard to the designing of a bow totally useless.

    Also the average force is derived from initial and final draw at the spring and how the energy is distributed along the launch run and is irrelevant.

    Does anybody know if the formula is correct ?
    Is the reasoning behind it valid ?
     
  16. Jun 23, 2006 #15

    andrevdh

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    Another way of measuring the exit speed of a projectile is with a ballistic pendulum. The impulse imparted to the pendulum by the arrow increases its (and the arrow stuck in it) momentum. With momentum conservation

    [tex]p_{before}=p_{after}[/tex]

    [tex]m_av=(m_a+m_p)v_b[/tex]

    [tex]m_av=m_bv_b[/tex]

    where subscripts a refer to the arrow, p to the pendulum and b to the arrow-pendulum after it sticks in the pendulum. I would think that the pendulum need to be quite long in order to absorb the impulse in a not too hard material (for the arrow to stick in).

    From the above formula [tex]v[/tex] can be calculated if [tex]v_b[/tex] is known.

    The kinetic energy of the pendulum-arrrow gets converted to potential energy as it swings up:

    [tex]1/2m_bv_b^2=m_bgh[/tex]

    [tex]v_b^2=2gh[/tex]

    from which [tex]v_b[/tex] can be calculated by measuring the height [tex]h[/tex]to which the pendulum rises.
     
    Last edited: Jun 23, 2006
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