# What weights are required for this progressive braking system?

1. Jun 20, 2010

### 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!

#### Attached Files:

• ###### Pulley brake.JPG
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2. Jun 23, 2010

### 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:

$$h = \frac{d}{8}$$

3. Jun 23, 2010

### 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

4. Jun 23, 2010

### jack action

Response included in quote: