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Force, initial velocity, kinetic energy...

Discussion in 'Mathematics and Physics' started by Externet, Jan 3, 2017.

  1. Externet

    Externet Member

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    Not exactly in the electronics domain, but am working on a mechanical analogy of impedance matching. Not homework either. Did that too many years ago and remember too little. Please help with the calculation :

    A speargun takes 80 lbs_force to arm its elastic sling (is that 36 Kg_force or 356 Newtons ?)
    When triggered, its 1 Kg_weight_in_water / 1 metre long spearshaft with 10mm diameter pointy arrowhead shoots with what
    -Initial velocity,
    -Kinetic energy
    The media is water, density 1 g/ml.

    [ The shape constant for the spearhead is ? (cannot find a table)
    The initial impulse at full force gradually decreases to zero impulse in 0.1 seconds at firing when the spearshaft exits the gun and slings have collapsed ]

    Formulas I believe are pertinent:

    Vinitial = square root of 2 x kinetic energy / mass

    Deceleration due to water resistance = density of media x area x shape constant x Vinitial^2 / 2 mass

    Can the calculation steps and rationale be explained, please ?
     
  2. alec_t

    alec_t Well-Known Member Most Helpful Member

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    It's gonna be a bit more complicated than that. One important factor missing is water viscosity. Resistance to motion (drag) of the spear depends on that, also on cross-section of the spear head and spear velocity. Any turbulence around the spearhead would also create drag.
    Note the initial velocity is zero, not as per your formula. The spear accelerates to some maximum velocity while the sling is still forcing it forward, then decelerates due to viscous and turbulent drag.
    If you have simulation software (e.g. LTspice) you could run a sim with behavioral voltage sources representing sling force, acceleration, velocity, drag.
     
  3. Externet

    Externet Member

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    Thanks, alec_t

    Viscosity of water is 1 and density is 1 at 20°C; Cross section of spear is 1cm diam = 0.0000785 m^2 cross section.
    There is a constant factor for 'shape' for sharp conical point I cannot find. It can be assumed as optimal 1, for minimum drag, I believe.

    Before jumping to electrical, would like to know the initial velocity and kinetic energy, and if the formulas posted are not enough, would like to learn which are proper.

    I have no simulator, and probably no brain to apply it if I had such software.
     
  4. dave

    Dave New Member

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  5. Les Jones

    Les Jones Well-Known Member

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    To know the initial kinetic energy storded in the elastic you need to know how far you have to pull it. I assume that 80 lb force is the maximum force you have to apply. I can't remember the formula for the stored energy but I would make a guess at it being half the maximum force times the distance of the pull.

    Edit. Correction. It is potential energy that is stored in the elastic. (Not kinetic.) It becomes kinetic energy when it is transfered to the moving mass of the projectile. The kinetic energy of the projectile will be Half M x V^2

    Les.
     
    Last edited: Jan 4, 2017
  6. Externet

    Externet Member

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    Thanks. To extend o.5 m the elastic band to the cocked position takes 80 lbs_force = 36 Kgf. That is the force that is applied to impel the spearshaft.
    When firing, contracts 0. 5 metres in 0.1 second. That should calculate acceleration for a 1 Kg_weight_in_water spearshaft.

    That calculated acceleration applied to 1 Kg during 0.1 seconds along 0.5 metre travel should bring the kinetic energy; right ?

    ----> https://shouldersofgiants.com/wp-co...eys_Spearfishing_CJE-15913-e1455203988540.jpg

    ----> http://prime.primescubainc.netdna-c...8d6e5fb8d27136e95/i/m/image_33760_1_17207.jpg

    ----> http://www.spearfishing.de/images/product_images/popup_images/771_0.jpg

    ----> http://www.spearfishingproducts.com.au/image/cache/data/shafts/P1060077 (640x480)-800x600.jpg
     
  7. Externet

    Externet Member

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    Check and do corrections to my rationale please :

    Initial velocity Vi is 0
    Impulse force F applied is 356 Newtons
    Time the force is applied is 0.1 seconds = Δt
    Final velocity Vf exiting muzzle is unknown = Vf = Vfinal - Vinitial = Δv

    Acceleration is a = Δv/Δt = F/m
    Δv = Δt F /m = 0.1s x 356N / 1Kg = 35.6 m/s

    The kinetic energy Ek is mv^2 / 2
    1 x 35.6 x 35.6 /2 = 633.68 Joules.

    Edited :

    I do not see any need to know the distance traveled while the force is applied to the spearshaft. ¿? but the force is applied in full 365 Newtons at time zero and decreases probably linearly until becomes zero Newtons 0.1 seconds later. That means the force applied is not constant, but decreases in time. Can you guide me for doing the integration ?
     
    Last edited: Jan 3, 2017
  8. alec_t

    alec_t Well-Known Member Most Helpful Member

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    The geometry of the speargun defines the distance over which the spear is accelerating while still being propelled by the sling. You cannot assume a 0.1 sec time, since that would imply a known acceleration.
    You need to take drag into account. This background re the drag equation may be useful.

    Edit:
    Here's a sim, where k1 is an empirical proportionality constant to cover the viscosity and geometry parameters in the drag equation. The sling force is assumed to reduce linearly to zero over a 1 metre distance. Note that the formulae used to define the voltage source outputs are interdependent, chicken and egg style, e.g. the one for acceleration depends indirectly on distance and the one for distance depends on acceleration.
    Speargun.PNG
     
    Last edited: Jan 4, 2017

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