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Cable rotations formulae or method

  1. May 2, 2015 #1
    Hello, and thank you for your time.

    I observed a wire vibrating is good at blocking things trying to pass through their area.
    for example a rope spinning around can stop a tennis ball thrown through that area.
    obviously the speed of the rope allows it to "protect" an larger area than its gauge.

    On this background, does anyone know how i could calculate the stopping power of
    a cable rotating.

    for example a 5mm gauge cable rotating at x speed and with enough slack to cover x amount of space could stop a projectile at x mph velocity.


    sorry if the question is poor as i have no engineering background.

    any suggestions welcome.

    thanks
     
  2. jcsd
  3. May 2, 2015 #2

    mfb

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    That does not work. For many realistic values you will have a non-zero probability that the particle gets through without a collision, and if it collides the motion afterwards depends on where exactly it hits and many other details that are hard to define in general.

    If your rope makes one revolution in the time the projectile needs to travel its own length, then you will certainly get some collision.
     
  4. May 2, 2015 #3

    davenn

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    this reminds me of back in the early days of aerial warfare where they finally figured out how to sync guns and propeller
    so as to shoot through the prop without destroying it
     
  5. May 3, 2015 #4
    thank you for your reply mfb

    davenn might be onto something there
    if they can work out when to fire than surely the opposite of when not to fire is possible.


    supposed we remove variables
    "one revolution in the time the projectile needs to travel its own length"
    also sounds interesting.
    for example if an object travels at 10mph and is 10 cm long
    how could i roughly work out the revolutions needed?


    its at times like this i wish i were as smart as an engineer.
     
  6. May 3, 2015 #5

    mfb

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    First you convert the 10mph to proper units: roughly 5m/s. It travels 10cm in 10cm/(5m/s) = 1/50s = 20ms.
    If your string makes 50 revolutions per second (quite fast), it will certainly hit it in some way.
     
  7. May 3, 2015 #6

    Baluncore

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    You must also consider the differential velocity. If the screen is like a skipping rope there will be two “walls” to pass through near the middle. But at either the top or bottom there will be a horizontal movement in the same direction as the projectile. That will allow the projectile to pass with a higher probability.
    The probability profile plotted against against height of the object's trajectory will depend on length and speed of the projectile, along with the direction and rate of rotation of the wire generated “cage” of revolution.
     
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