What weights are required for this progressive braking system?

[Puzzler24]

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!

 

[jack action]

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

\frac{1}{2} m_{p} v^{2} = Td

Where:

m_{p} = mass of person

v = initial velocity

T = tension in cable

d = braking distance

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

2T = m_{w} g

Therefore:

m_{w} = \frac{m_{p}v^{2}}{gd}

The energy stored in the masses will be:

4m_{w} gh

Where h is the height the masses will move. So:

h = \frac{d}{8}

 

[Puzzler24]

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

 

[jack action]

Response included in quote:

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)

The appropriate unit for energy is Joule

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?

You do realize that to stop from 12.5 m/s in 3 m represents a deceleration of 26 m/s² or 2.65g, that's a lot! (a = v²/2/d). Just to scare you more, this is the mass of one counterweight; you have 4 of them (2400 kg total).

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

The appropriate unit for momentum is kg.m/s

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

Based on the deceleration you want, the time will be 0.48 s (t = v/a). This will give you 300 kg of weight required. This would be the correct answer if you had a single counterweight attach at the end of the cable (with only one pulley). But the height needed would be 3 m instead of 0.375 m.

The reason why in your proposed arrangement you have such large counterweights is that the 8 vertical cables shortens the distance traveled by 8, so the total mass must be 8 times larger to compensate (Energy stored = mgh = 2400 X g X 0.375 = 300 X g X 3). It is a lever effect: Either you go heavy and short or light and long.

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?

If the plane goes up, of course it will slow you down. You will have to subtract the energy needed to go up:

\frac{1}{2} m_{p} v^{2} - m_{p}gdsin \theta = Td

Thanks again