What weights are required for this progressive braking system?

In summary, Jack explains how to calculate the kinetic energy of a person traveling down a cable at a certain speed. The person would need 8 times the weight to compensate for the lever effect. If the plane goes up, it will slow the person down.
  • #1
Puzzler24
2
0
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!
 

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  • #2
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]
 
  • #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
 
  • #4
Response included in quote:

Puzzler24 said:
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:

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


Thanks again
 
  • #5


I would recommend first determining the necessary braking force to slow a 250lb person from 30mph to zero in 10 feet. This can be calculated using the formula F=ma, where F is the force, m is the mass, and a is the acceleration. In this case, the acceleration would be the change in velocity (30mph) divided by the time (10 feet).

Once the necessary braking force is determined, it can be divided among the four weights (A, B, C, and D) based on their placement and distance from the person. The weights should be positioned in a way that creates a gradual increase in braking force as the person moves along the zipline.

To accommodate a weight range of 70lbs to 275lbs, the braking system should be designed with adjustable weights or a range of weights that can be added or removed as needed. The calculation of the necessary braking force can be repeated for different weight ranges to ensure the system will work effectively for all users.

It is also important to consider the materials and construction of the braking system to ensure it can withstand the necessary forces and maintain safety for the users. Consulting with a professional engineer or experienced zipline builder may also be helpful in designing and implementing a safe and effective braking system. Good luck with your zipline project!
 

1. How do I determine the required weights for this progressive braking system?

The required weights for a progressive braking system can be determined by first determining the total weight of the vehicle and the desired braking force. Then, using the formula F=ma (force = mass x acceleration), the required weight can be calculated based on the desired deceleration rate and the weight distribution between the front and rear axles.

2. What factors affect the required weights for a progressive braking system?

The required weights for a progressive braking system are affected by several factors, including the weight and distribution of the vehicle, the desired deceleration rate, and the type of braking system being used (e.g. disc brakes vs. drum brakes).

3. Can I use different weights for the front and rear axles in a progressive braking system?

Yes, it is possible to use different weights for the front and rear axles in a progressive braking system. However, it is important to consider the weight distribution and balance of the vehicle to ensure safe and effective braking.

4. How can I test and adjust the weights in a progressive braking system?

The weights in a progressive braking system can be tested and adjusted through trial and error. Start with a lower weight and gradually increase it until the desired braking force is achieved. It may also be helpful to consult with a professional or refer to the manufacturer's guidelines for weight adjustments.

5. Are there any safety considerations when determining the required weights for a progressive braking system?

Yes, there are safety considerations when determining the required weights for a progressive braking system. It is important to ensure that the braking force is not too high, as this can cause instability and potentially lead to accidents. It is also important to regularly check and maintain the braking system to ensure optimal performance.

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