48 replies to this topic

[CAski]

Matt, on May 10 2005, 04:15 PM, said:

The change in gravitational potential energy is the correct method to determine lift horsepower. All a chairlift does is change the gravitational potential energy of the passengers. I calculated the power required to give kinetic energy to the passengers, but it was like 0.17 horsepower so I never bothered mentioning it.

Liftmech's method, which I copied, works because work is force times the the component that of the displacement that is parallel to the force. You could either calculate the tension in the cable that's not cancelled out on the other side, and multiply by the line speed. You never need to know the angle. Or, you could calculate the vertical component of velocity (as CAski did), and multiply by the vertical force. Both methods yield the same result.
<{POST_SNAPBACK}>

Okay, I think I see the logic in the equations you presented. But it is a very backward, awkward, and convoluted approach to me. Your "tension in the cable that's not cancelled out on the other side" looks to be the dPE for an average foot long section of the cable. This would be in vertical foot*lbs/foot. You are saying that a foot long piece of cable will undergo x amount of work to go from the bottom to the top of the lift. The entire amount of the cable on the uphill-travelling side will make it to the top in y amount of time. By multiplying the dPE per foot by the total amount of cable, you are calculating the total change in potential energy, or the work performed. By dividing the potential energy by the y amount of time, you get the power. Since the total amount of cable divided by y equals the speed, one can cut this intermediate step and say that the dPE for an average foot long section of cable multiplied by the speed equals the power. Time saving, perhaps, but I would argue that it is error-prone unless one understands the logic behind it.

[Matt]

There's static tension in the cable from the counterweight or hydraulic tensioning system. There's a difference in tension between the cable on each side of the driving bullwheel. If there were no difference in tension, then no torque would be required to drive the lift. This is obviously not the case. Think of a tug o war, when you pull on the rope, the rope in front of you is under more tension than the rope behind you.

I think liftmech's method is far less convoluted. CAski's method uses arctan, pythagorean theorem, and sin. The method liftmech uses skips all this. You could calculate the length of the hypotenuse, and the angle from the elevation gain and horizontal distance. You could multiply the hypotenuse by the sin of the angle, but you'll wind up with 1100' again.

[CAski]

The method is simpler, but the way it works is more obscure.

[nathanvg]

I've looked at the posted math and as has been stated it is all correct.

There have been a lot of assumptions made that allow this problem to be extremely simplified. The only work the motor is doing is moving the passengers vertically due to our assumptions so:

Power = work / time = (vertical_distance * force) / ( length / speed)

Power = 1100 ft * (170 lbs/person/chair * 3 people * 90 chairs) / (4500 ft / 500 ft/sec) = 5610000 lb ft / min or 170 HP

The above logic is basically the same as lifmech but in a form that I find much more logical and simpler. Maybe this makes more sense to CAski.

[CAski]

Yes. That makes sense.

[Matt]

That doesn't look any simpler than this:

((170 lbs/person * 3 person/chair) / 50 feet/chair) * 1100 feet * 500 feet / min = 561000 lb*feet/min = 170HP

Using this method, it's not neccessary to include, or even know the horizontal distance of the lift. Also, you can substitude the bullwheel radius for the line speed, and you'll get your bullwheel torque, another important piece of information when designing a lift.

Lets say the bullwheel has a 3'6" radius. (I made that up, I have no idea what a typical value is.):

((170 lbs/person * 3 person/chair) / 50 feet/chair) * 1100 feet * 3.5 feet = 39270 lb*feet

[liftmech]

A typical fixed-grip bullwheel runs 11' in diameter, so 5'5" for a radius.

The formula I learned in class continues thus: Torque at bullwheel (this is a top drive, bottom tension lift again) is (T3-T4) x radius. T3 is the tension on the heavy side, T4 is tension on the light side. Since you are multiplying pounds by feet, your final unit will be in foot-pounds.

For horsepower required: [(T3-T4) x lift speed] / (33,000 x efficiency)
T3 and T4 are the same, lift speed must be in feet per minute, efficiency is usually 0.85 to 0.90, depending upon application. A vault drive with a vertical shaft is slightly less efficient because of the heat buildup in the u-joints on the shaft. 33,000 is a constant needed to end up with a proper number of horsepower.

[Dr Frankenstein]

But, whan you bluid a new lift, how do you calculate the position of each tower, with the # of sheaves there will be, etc.?

[liftmech]

Tower locations are decided upon using the lift's profile, which is a side view of the lift and the terrain it will be built upon. Common practice is to build shorter towers on high spots, taller ones on low ground, and try to keep abrupt changes in angle to a minimum. Towers are generally spaced close enough together that if the rope were to come off one of them it wouldn't hit the ground. This follows the discussion about rope sag we had further up in the thread. The number of sheaves is decided upon by how much load will be on that tower. You can take the change in rope angle from the profile, and figure out your assembly from that. A rule of thumb is one sheave per degree of angle, but that isn't always the case depending upon how much load the sheave can take.

[Allan]

Here is a VERY rough drawing of what a profile looks like, unfortunately I don't know how to calculate sheave loads, etc!

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[Duck]

(I've got nothing to contribute to this thread, just wanted to say I find this very interesting and I've enjoyed following along in the background )

-Iain

This post has been edited by Duck: 21 May 2005 - 03:31 AM

[iceberg210]

Quote

This can be calculated, but it involves the catenary curve equation (not pretty) between every tower. Taking this into account is only possible with a computer.

Or a slide rule if your really good.
Anyway firstly if the tension was the same on both sides and the friction was negible then on an empty lift it would just start moving if you put a sandbag on the top chair and let that chair go downhill? Assuming that there is no static friction or inertia to overcome? Is a chairlift really that effienct? Remember even simple machines aren't even close to 97% effienct so is it just that due to a chairlift's circular nature it becomes even on both sides and is no longer important enough to put in the equation?

The more I think about it the more I understand but that doesn't mean I'm making all to much headway.

Also where would one take such a class John as this is driving me crazy with this wealth of information and I would love to learn more.

[CAski]

iceberg210, on May 26 2005, 06:43 PM, said:

Or a slide rule if your really good.
Anyway firstly if the tension was the same on both sides and the friction was negible then on an empty lift it would just start moving if you put a sandbag on the top chair and let that chair go downhill?  Assuming that there is no static friction or inertia to overcome?  Is a chairlift really that effienct?  Remember even simple machines aren't even close to 97% effienct so is it just that due to a chairlift's circular nature it becomes even on both sides and is no longer important enough to put in the equation? 

The more I think about it the more I understand but that doesn't mean I'm making all to much headway. 

Also where would one take such a class John as this is driving me crazy with this wealth of information and I would love to learn more.
<{POST_SNAPBACK}>

You may not be able to humanly calculate the catenary curve. I suspect that it involves a bit of calculus, and that being the case, there is the possibility that the equation is not differentiable or able to be integrated by hand, that it must be estimated using an algorithm based off of Euler's method. Granted, I have not seen the equation.

Also, the tension is not the same all over a lift; it varies from tower to tower, as the sheet Allen has presented shows. I am fairly certain of this statement.

A sandbag would not start the lift moving because there is a force of static friction working against it. You cannot assume friction to be negligable.

Don't worry too much about these things, as nothing has really been discussed in great detail here. These are just a few important concepts to consider in the design of a lift.

[liftmech]

Here are some more random equations to play with.
Lift capacity = (3600 x carrier capacity) / loading interval
Carrier spacing = (60 x carrier capacity x lift speed) / lift capacity

Examples: a quad's capacity at six-second loading is 2400; 3600 x 4 = 14400; 14400 / 6 = 2400.

That same quad's carrier spacing is 1000 feet; 60 x 4 = 240; 240 x 1000 = 240000; 240000 / 2400 = 1000.

[iceberg210]

The catenary curve I'm sure involves calculas but remember Sir Issac Newton invented calculus before the calculator or computer (that isn't to say that it isn't humanly possible for most of us to do it however including me).

Quote

A sandbag would not start the lift moving because there is a force of static friction working against it. You cannot assume friction to be negligable.

I was saying that if the static friction was negible not that it is of course but that if the friction really took only 3% of the force out then my sandbag hypothosis might work and therefore it muct be that the friction is more then that I think.

[liftmech]

The 3% I referred to is what the line generally adds. This doesn't include the bullwheel bearings, gearbox, drive shafts, electric motors or diesel engines, and bullwheel liners. Case in point- when I participated in a load test of Baker's old C-2 (Riblet centre-pole double with 320' of vertical) we could not get the lift to roll back even with all carriers on the heavy side loaded. That's 48 chairs each loaded to 2 x 180 pounds, or 17,280 pounds total load. Of course, a steeper lift will roll back with that much load, but that goes to show you how much static friction there is in a lift.

[iceberg210]

Oh okay for the line only I could see 3% but for the entire lift that didn't make sense thanks for the clarification.

[CAski]

iceberg210, on May 27 2005, 07:18 PM, said:

The catenary curve I'm sure involves calculas but remember Sir Issac Newton invented calculus before the calculator or computer (that isn't to say that it isn't humanly possible for most of us to do it however including me).
<{POST_SNAPBACK}>

You misunderstand me. There are certain equations that simply aren't possible to integrate by hand. The reasons could be either that the equation is so complex that it would take a number of lifetimes to solve or that the equation simply cannot be solved using the laws of Calculus. The following is an example of the latter.

For example, let's say, for fun, that you have a cable hanging between two towers that are 100 m apart and that have a vertical difference of 10 m. The cable, by completely random coincidence, happens to trace out a curve that follows y=10^(-3)(x^2) from x=0 to x=100. If we want to find the length of the cable hanging between the two towers, we can use calculus to obtain an answer. Length, L =∫(√(1+(dy/dx)^2))dx over the proper interval. In the case we have here, L=∫√(1+[2*10^(-3)x]^2)dx from 0 to 100. This simplifies to L=∫√(1+[4*10^(-6)(x^2)])dx. Now, even with this simple equation derived from y=x^2, it becomes impossible to simplify any further. L=∫√(1+[4*10^(-6)(x^2)])dx is unsolvable by hand. However, using my trusty TI-83+, I know that the length of the cable is 100.663m, a mere 16.4cm more than if the cable were perfectly straight.

This post has been edited by CAski: 28 May 2005 - 11:05 PM

[Matt]

The catenary curve can be approximated by a simple parabola, which is easily differentiable, and the length of the curve can be found by integration. The results should be sufficient for designing a lift.

There is a formula for the length of the curve that is precise, but I don't recollect it at the moment. Try a quick google search.

[CAski]

See my post above.

L =∫(√(1+(dy/dx)^2))dx over the horizontal distance between the two towers in question. This is the precise length of a curve descrived by an eqation.