Hello Everybody:
I am trying to figure out how to calculate this for a investigation for a math/physics class in college:
My investigation question is how much potential energy of riders pushing downward on a fully loaded chair (6 person chair, 169 chairs) would it take to overcome the friction placed on the bull wheel, causing a roll back. Generally, I would need to calculate the potential energy of all the riders on the lift (multiplying average weight of rider X chair capacity X number of chairs) and then transfer that into lateral motion, and then subtraction the friction of the sheaves...
Any guidance without how to conduct this calculation would be greatly appreciated. Also, if anyone has any other ideas of physics related topics that I could calculate regarding lifts, that would also be great.
Thank you!
Best,
Owen
1
Lift Calculation Question!
Started by SkiForChairlifts, Aug 17 2017 12:32 PM
4 replies to this topic
#3
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Posted 20 August 2017 - 10:40 AM
You would also need the lift's vertical rise over its horizontal distance, as a flat lift (obviously) won't roll back no matter how loaded it is.
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#4
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Posted 22 August 2017 - 07:50 AM
Bull wheel liner friction isn't an issue, but the friction of its bearings is. Though I think the friction from the sheave bearings will be much higher, and there's also the flexion of the haul rope to think about. You could start with the equations for a ball rolling down a ramp and work from there.
#5
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Posted 03 September 2017 - 06:42 PM
Friction on each sheave (including bearings and rope-liner) is generally accepted as 2 to 3% of load on each sheave. Bullwheel friction is less, from 1% to 0.3%.
These are dependent on whom you ask.
Since you are asking about a 6-place chair, you may be talking about a detachable lift, and if the rope is used to drive the station mechanism, it will also require force to run that, adding drag to your hypothetical situation.
These are dependent on whom you ask.
Since you are asking about a 6-place chair, you may be talking about a detachable lift, and if the rope is used to drive the station mechanism, it will also require force to run that, adding drag to your hypothetical situation.
This post has been edited by Tramway Guy: 03 September 2017 - 06:45 PM
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