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What weights are required for this progressive braking system?

  1. Jun 20, 2010 #1
    Hi there

    I'm building a zipline and want to create a simple braking system that will progressively slow a 250lb person (me) from a top speed of about 30mph to zero in about 10'

    The braking system is as shown in the attachment.

    I'm a long time out of college, so although I recall the principles, I can't recall the math involved in calculating these forces.

    Can anyone help me figure out the weighs required for A, B, C and D? Ideally I'd like to know the formulas so I can recalculate for smaller and larger people and design a system that will accommodate a weight range of 70lbs to 275lbs.

    Thanks

    PS, Just in case you're wondering, this is NOT homework, I really am building a zipline in my backyard!
     

    Attached Files:

  2. jcsd
  3. Jun 23, 2010 #2

    jack action

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    Energy to be removed should be equal to the energy taken from braking:

    [tex]\frac{1}{2} m_{p} v^{2} = Td[/tex]

    Where:

    [tex]m_{p}[/tex] = mass of person

    [tex]v[/tex] = initial velocity

    [tex]T[/tex] = tension in cable

    [tex]d[/tex] = braking distance

    The tension in the cable is (assuming all masses A, B, C & D are equal and each is [tex]m_{w}[/tex] and that the cables are vertical):

    [tex]2T = m_{w} g[/tex]

    Therefore:

    [tex]m_{w} = \frac{m_{p}v^{2}}{gd}[/tex]

    The energy stored in the masses will be:

    [tex]4m_{w} gh[/tex]

    Where [tex]h[/tex] is the height the masses will move. So:

    [tex]h = \frac{d}{8}[/tex]
     
  4. Jun 23, 2010 #3
    Thanks Jack!

    Using your formulas I come up with the following:

    Td=0.5 * 113kg * 12.5m/s * 12.5m/s = 8828 Newtons (Kinetic Energy)

    Mw=(113 kg * 12.5m/s * 12.5m/s) / (9.8 m/s/s * 3m)=600kg

    h = 3m/8 = .375m

    Based on my experimental device 600kg seems like too much. Did I get the math right?

    The way I was trying to tackle the problems was using incline plane formula, I had the Momentum figured as Mass * Velocity = 113kg * 12.5 = 1417N

    Based on a 2 second deceleration, I therefore need 708N/s. Dividing by g, I get 72kg of weight required. Hmm, something is amiss!

    The other thing is, do I need to consider the angle of travel down the line?

    Based on inclined plane formula, if I take the Sine of the line angle (3.4 degrees), the resultant kinetic energy is reduced. Is this relevant?

    Thanks again
     
  5. Jun 23, 2010 #4

    jack action

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    Response included in quote:

     
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