Designing a Small Ski Lift: Facts and Assumptions

[web4deb]

I'm planning on building a small ski lift for my kids. It's been 20 years since I've taken a physics class so I'm lucky to remember what gravity is.
Here's a poor image of the hill. I would like to know the following:

A) how much lbs of tension is against the lift wire?
B) how heavy should the counter weights (D&E) be so they don't rip out of the ground?

Facts and Assumptions
C) distance between towers A&B and B&C are 200' each
D) Elevation between towers A&B is 60', B&C is 40'
E) Would like to be able to lift a total of 4 people at a time...assume 175 lbs/person
F) Not as important...lift wire weight is 300 lbs total

Feel free to assume anything else that's missing. I'm basically looking for some formulas I can use to play with the numbers.

http://www.walchem.com/email/movies/lift.jpg

 

[berkeman]

Hmmmm. Looks like something I'd like to get a PE (licensed professional engineer) design for me. Even if you way overbuild it, you'll want to think about the liability ramifications. Not to discourage you. I just think that a project like this deserves a professional design. Plus, there are probably already industry-standard guidelines for things like ski lifts, and an appropriate PE will be familiar with them. Now a tow rope, that you could do on your own...

 

[web4deb]

I am well aware of my legal liabilities. This is not a lift where people would get lifted off the ground, but more of a rope toe with a pull rope hanging from the main line. (The hump in the middle of the hill makes it impossible to put in a rope toe) If you could point me in the right direction for some colculations, I would appreciate it. Thanks.

 

[uwmengineer]

Friction around a pulley: add rope and skier
V-shaped pulley: T[2]/T[1] = exp(mu * beta / sin (alpha/2) ); mu = coeff of friction, beta = angular meas of belt around pulley in radians, alpha = angular meas of the v in the pulley
rope and pulley: T[2]/T[1] = exp(mu * beta ); mu = coeff of friction, beta = angular meas of tow rope around pulley in radians, T[2]= the tension in the rope going uphill, T[1]= the tension in the rope going downhill, T[0]= tension at the middle of the catenary

Tow Rope catenary this is the rope part only
T[2] = [T[2][0] + w[2] * s[2] ]^[0.5];s = length of rope (lowest point to highest point), w is the unit weight ie lbs/ft so W=ws where W is the total weight of the rope segment
c = T[0] / w; c is a constant so T[0] = wc
s = c * sinh (x/c); x = distance horizontally from the lowest point in the catenary to the highest
y = c * cosh (x/c); y = sag in cable + c [note: solve for c using these two eq.]
y^2 - s^2 = c^2
T[2] = w*y [note: max]
T[0] = w * c [note: min]Skier skier part only
initially the skier is not moving, draw a force diagram, x is positive going uphill, y is positive above the ski slope and resolve the forces:
Friction = mu * N; mu= coeff of friction (.04? not much, you could assume frictionless), N=normal force
x direction forces=friction + weight * cos(theta); theta = angle between the ski slope and the weight vector
y direction forces = normal force N = weight * sin(theta)
after the skier is moving use F = ma, but consider highest forces and add a factor of safety

horse power of motor = rpm*torque*K; K=constant to make the units agree, torque will depend on the radius of the pulley and the tension in the rope: tau = r X F ;this is, as you recall, a vector cross product where tau is torque

you might use energy as a check: kinetic energy = translational energy (1/2 mv^2) + rotational (1/2 Iw^2)

 

[rdl]

I would first like to commend you on your efforts to build a small ski lift for your kids. It's a great way to introduce them to the world of physics and engineering.

In terms of the questions you have asked, let me provide some information and formulas that can help you with your design.

A) To determine the amount of tension on the lift wire, we need to consider the weight of the lift, the weight of the riders, and the angle of the wire. Assuming an angle of 30 degrees between the lift wire and the horizontal ground, we can use the following formula:

Tension = (Weight of lift + (Weight of riders x Number of riders))/sin(30)

Since you mentioned that you would like to lift 4 people at a time, the total weight would be 4 x 175 lbs = 700 lbs. Assuming the lift itself weighs 300 lbs, the total weight would be 1000 lbs. Plugging these values in the formula, we get:

Tension = (1000 lbs)/sin(30) = 2000 lbs

Therefore, the tension on the lift wire would be 2000 lbs.

B) To determine the weight of the counterweights, we need to consider the distance between the towers and the elevation difference between them. Assuming a pulley system, the amount of counterweight required can be calculated using the following formula:

Counterweight = (Weight of lift x Distance between towers)/(Elevation difference between towers)

Using the values from the given facts and assumptions, we get:

Counterweight = (300 lbs x 200 ft)/(60 ft) = 1000 lbs

Therefore, the counterweights should weigh 1000 lbs each to prevent them from ripping out of the ground.

C) The given distances between towers A&B and B&C are 200 ft each.

D) The elevation difference between towers A&B is 60 ft and B&C is 40 ft.

E) To lift a total of 4 people at a time, we need to consider the weight of each person, which is assumed to be 175 lbs.

F) The weight of the lift wire is provided as 300 lbs.

Some other factors that you may want to consider in your design are the maximum wind speed and slope of the hill. These can affect the tension on the lift wire and the stability of the lift.

I hope this information helps you in designing your small ski lift. Good