Engineering Lifts
CAski
07 May 2005
When a manufacturer engineers a lift, what assumptions are they making? For example, I have heard that friction forces are easiest to determine experimentally. Does the manufacturer actually go through the time to calculate the friction involved in a lift, does the manufacturer determine the force experimentally in the first test run, or does the manufacturer just ignore the friction entirely and let the lift compensate with a computer and a tachometer? Similarly, does the manufacturer determine the inertia of every last sheave, or is there some estimation involved? I have some specific questions about estimations in the tension force needed in a lift, but I need to determine how to calculate that myself first.
Aussierob
07 May 2005
When manufacturers were first designing lifts there would have been a lot of calculation and probably some assumptions which would have been tested. That has all been distilled into a computer program that essentially does all the calculations for you. I've seen the output from Poma and Doppelmayrs software and it gives you every relevant detail you need to build the thing. I would guess that it is checked by an engineer to see if it looks OK, then it gets built.
liftmech
08 May 2005
Friction is calculated into the original tension and power requirements. After many years of using the same sheaves, the manufacturers pretty much know how much they will add to the final calculation and it is around 3% of the calculated tension.
Calculating tension on a top drive lift is relatively simple, and once you've got that you can derive the horsepower needed to drive the lift. Factors involved are rope weight, carrier weight, carrier capacity, design load (per ANSI, at least, it is 170 pounds per person), and vertical rise. Horizontal and slope lengths don't matter.
An example: I'll start with the profile of the lift, say an 1100' x 4500' span. I'm building a fixed triple, so I already know the weight of my carriers (say, 325 pounds each). My experience tells me that the rope I'll need is probably going to be a 1 1/2" rope, which weighs 3.27 pounds/foot. The lift will run at 500 FPM, so the carrier spacing neded for six-second loading is 50'. I divide up the carrier weight by the carrier spacing to give me an easier number to work with, which ends up being 6.5 pounds/foot. I add that to my bare rope weight, and I get a total unit weight of 9.77 pounds/foot. Multiply that by 1100 (remember our total vertical from the beginning) and I get 10,747 pounds. this is the total change in tension from bottom to top on an empty line. For a ful line, take three passengers by 170 pounds each (510 pounds), divide by 50 again (10.2 pounds/foot), and multiply by 1100 (11,220 pounds). Add this to your empty line weight for the total change on the heavy side, which ends up being 21,967 pounds. Since this is a top drive, there is going to be more load on the drive bullwheel than the return, so our total tension (weight of counterweight or pressure of hydraulic cylinder) doesn't need to be extreme. Let's say that it's 10,000 pounds for the sake of math. Tomorrow's task is to figure out if this is enough and calculate our horsepower requirements. I need to go t work now.
Calculating tension on a top drive lift is relatively simple, and once you've got that you can derive the horsepower needed to drive the lift. Factors involved are rope weight, carrier weight, carrier capacity, design load (per ANSI, at least, it is 170 pounds per person), and vertical rise. Horizontal and slope lengths don't matter.
An example: I'll start with the profile of the lift, say an 1100' x 4500' span. I'm building a fixed triple, so I already know the weight of my carriers (say, 325 pounds each). My experience tells me that the rope I'll need is probably going to be a 1 1/2" rope, which weighs 3.27 pounds/foot. The lift will run at 500 FPM, so the carrier spacing neded for six-second loading is 50'. I divide up the carrier weight by the carrier spacing to give me an easier number to work with, which ends up being 6.5 pounds/foot. I add that to my bare rope weight, and I get a total unit weight of 9.77 pounds/foot. Multiply that by 1100 (remember our total vertical from the beginning) and I get 10,747 pounds. this is the total change in tension from bottom to top on an empty line. For a ful line, take three passengers by 170 pounds each (510 pounds), divide by 50 again (10.2 pounds/foot), and multiply by 1100 (11,220 pounds). Add this to your empty line weight for the total change on the heavy side, which ends up being 21,967 pounds. Since this is a top drive, there is going to be more load on the drive bullwheel than the return, so our total tension (weight of counterweight or pressure of hydraulic cylinder) doesn't need to be extreme. Let's say that it's 10,000 pounds for the sake of math. Tomorrow's task is to figure out if this is enough and calculate our horsepower requirements. I need to go t work now.
Matt
08 May 2005
I think I can finish the horsepower part off for you. The tension on the cable due to the cable itself, and the chairs attached to it will be the same on both sides of the drive bullwheel. This would cancel out.
The force we need to calculate the horsepower will then be just the tension caused by the riders. Useing the same numbers as liftmech (3 x170lb passengers every 50 feet, with 1100 feet of vertical), we get 11220lb force.
Power is force times distance divided by time. The line speed was arbitrarily chosen to be 500fpm.
11220lb * 500 ft/min = 5610000 lb ft / min
Since one horsepower is equal to 33000 lb ft / min, 5610000/33000 = 170HP. So the power requirements for the lift, neglecting friction, will be 170HP.
If you take into account the 3% friction losses, the horsepower is determined to be 175.1HP. You could add the 3% part to the tension, but it propagates throught the calculations to give you the same result.
This post has been edited by Matt: 08 May 2005 - 08:38 AM
The force we need to calculate the horsepower will then be just the tension caused by the riders. Useing the same numbers as liftmech (3 x170lb passengers every 50 feet, with 1100 feet of vertical), we get 11220lb force.
Power is force times distance divided by time. The line speed was arbitrarily chosen to be 500fpm.
11220lb * 500 ft/min = 5610000 lb ft / min
Since one horsepower is equal to 33000 lb ft / min, 5610000/33000 = 170HP. So the power requirements for the lift, neglecting friction, will be 170HP.
If you take into account the 3% friction losses, the horsepower is determined to be 175.1HP. You could add the 3% part to the tension, but it propagates throught the calculations to give you the same result.
This post has been edited by Matt: 08 May 2005 - 08:38 AM
CAski
08 May 2005
Wait...
You might be making some assumptions that are not true. For instance, the tension force is not necessarily equal on both sides of a pulley system due to inertia.
Also, what about the sag in a lift; doesn't that need to be factored in? And as far as friction, it seems as though it is just a huge estimate. Perhaps there is a correlation between the number of sheaves in the lift and the coefficient of friction for the entire lift.
Liftmech, it seems as though you have not calculated the tension force in a lift, but the difference in potential energy between the top and the bottom terminals for a one foot long section of the haul rope. I don't exactly see how that can be useful in determining the tension force needed in the lift.
This post has been edited by CAski: 08 May 2005 - 11:13 AM
You might be making some assumptions that are not true. For instance, the tension force is not necessarily equal on both sides of a pulley system due to inertia.
Also, what about the sag in a lift; doesn't that need to be factored in? And as far as friction, it seems as though it is just a huge estimate. Perhaps there is a correlation between the number of sheaves in the lift and the coefficient of friction for the entire lift.
Liftmech, it seems as though you have not calculated the tension force in a lift, but the difference in potential energy between the top and the bottom terminals for a one foot long section of the haul rope. I don't exactly see how that can be useful in determining the tension force needed in the lift.
This post has been edited by CAski: 08 May 2005 - 11:13 AM
Allan
08 May 2005
Does this answer any questions, or just create more?
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CAski
08 May 2005
That does, for the most part, answer my question. There does appear to be a crapload of calculations involved. Now, I just wish I knew what some of the stuff (P, chord, etc.) meant.
Allan
08 May 2005
Heh, I was hoping you would know, and explain them! 
I can tell you the P/PX/PY are only negative on hold-down assemblys, and are near 0 on the combination asemblies.
I can tell you the P/PX/PY are only negative on hold-down assemblys, and are near 0 on the combination asemblies.
Kelly
08 May 2005
Some of this topic was discussed back on Dec 03.
http://www.skilifts....p?showtopic=216
http://www.skilifts....p?showtopic=189
Be sure to scroll down to the bottom of each page.
Ryan B
http://www.skilifts....p?showtopic=216
http://www.skilifts....p?showtopic=189
Be sure to scroll down to the bottom of each page.
Ryan B
CAski
08 May 2005
Ah, then P must be the force on the tower by the cable. PX and PY are the vectors of the force.
Thanks for the links to the explainations, Ryan. It cleared up some things but still leaves me fuzzy on others.
Thanks for the links to the explainations, Ryan. It cleared up some things but still leaves me fuzzy on others.
Matt
08 May 2005
P is the vector of the force, PX and PY are the X and Y components of the vector. If you use the pythagorean theorem you will find that PX^2 + PY^2 = P^2 for the values in the table.
The tension in the cable due to the mass of the cable and chairs will be nearly the same on both sides of the bullwheel. There may be a little more of the cable on the loaded side due to sag. 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. The calculations could not be performed using a calculator.
The tension in the cable could theoretically be zero, but tension is required to minimize cable sag, and to ensure traction of the drive bullwheel. One of the things favouring top drive terminals is the tension created by the cable's mass pulling down the slope. This reduces the tension required in the cable at the bottom bullwheel.
Calculation of friction losses is probably not done thoroughly. There are many losses that are difficult, and others are simply not possible to calculate. An example is the wind resistance of a loaded chair will be different from that of an unloaded chair. There's also a loss inherent in the flexing of the cable itself. Another loss is the flexing of the rubber sheeves. It's possible to estimate, but most of the data on losses in a system as complicated as a chairlift is obtained empirically.
The tension in the cable due to the mass of the cable and chairs will be nearly the same on both sides of the bullwheel. There may be a little more of the cable on the loaded side due to sag. 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. The calculations could not be performed using a calculator.
The tension in the cable could theoretically be zero, but tension is required to minimize cable sag, and to ensure traction of the drive bullwheel. One of the things favouring top drive terminals is the tension created by the cable's mass pulling down the slope. This reduces the tension required in the cable at the bottom bullwheel.
Calculation of friction losses is probably not done thoroughly. There are many losses that are difficult, and others are simply not possible to calculate. An example is the wind resistance of a loaded chair will be different from that of an unloaded chair. There's also a loss inherent in the flexing of the cable itself. Another loss is the flexing of the rubber sheeves. It's possible to estimate, but most of the data on losses in a system as complicated as a chairlift is obtained empirically.
liftmech
09 May 2005
Thanks for finishing up my calculations, Matt.
CAski- I hadn't gotten to the actual tension force required because I hadn't finished the rest of the equation. One starts out with a basic setup like this, and then goes on to calculate sag (which, as Matt mentions, is possible only on a computer). Once you've gotten the sag figured, you'll know whether your original tension that you arbitrarily picked is enough or too much. One also has to factor in the rope breaking strength; this must be divided by 5 (safety factor) to give you your maximum tension. If you cannot get your sag where you want it with the rope you've chosen, you must go up either a size or a grade (from IPS to XIPS, for example). One final factor is the adherence of a steel rope to a rubber drive sheave liner. Various manufacturers can give you the coefficient of friction for their proprietary material, and it is possible to calculate how much tension is needed to go over that number.
There are, as you say, a crapload of numbers involved in designing a lift; most of these can be just plugged into a computer programme and it will spit out results. By the way- in Allan's sheet for Motherlode (Silverlode?) 'chord' is the angle between towers as measured from tower cap to tower cap. This number is used to determine sag.
CAski- I hadn't gotten to the actual tension force required because I hadn't finished the rest of the equation. One starts out with a basic setup like this, and then goes on to calculate sag (which, as Matt mentions, is possible only on a computer). Once you've gotten the sag figured, you'll know whether your original tension that you arbitrarily picked is enough or too much. One also has to factor in the rope breaking strength; this must be divided by 5 (safety factor) to give you your maximum tension. If you cannot get your sag where you want it with the rope you've chosen, you must go up either a size or a grade (from IPS to XIPS, for example). One final factor is the adherence of a steel rope to a rubber drive sheave liner. Various manufacturers can give you the coefficient of friction for their proprietary material, and it is possible to calculate how much tension is needed to go over that number.
There are, as you say, a crapload of numbers involved in designing a lift; most of these can be just plugged into a computer programme and it will spit out results. By the way- in Allan's sheet for Motherlode (Silverlode?) 'chord' is the angle between towers as measured from tower cap to tower cap. This number is used to determine sag.
Powdr
09 May 2005
Remember too, that all the calculations I've seen done above reflect a constant load, not peak load. I see that someone calculated about 175 HP needed for the 1100x4500 triple, but that only accounts for a constant, running load. You have to factor in starts & stops, hot & cold to get your actual HP requirment. Add a safety factor and you're there.
Powdr
Powdr
Matt
09 May 2005
Calculating startups is a little more tricky because then you have to take into account the moment of intertia of all the sheaves and such, and overcoming static friction. Fortunately the type of DC motor used in lifts can typically develop five or more times their full speed torque on startup (without current limiting). Normally it's not neccessary to oversize a DC motor to deal with that sort of thing unless it's frequent start and stop service, which I would personally classify a chairlift as.
Can anybody provide numbers on maximum startup current -vs- running current? Since torque is directly proportional the armature current (assuming constant field current), we could get an idea of startup torque -vs- running torque.
The 3% loss to friction did sound a little small to me, but I went with what I was given.
Can anybody provide numbers on maximum startup current -vs- running current? Since torque is directly proportional the armature current (assuming constant field current), we could get an idea of startup torque -vs- running torque.
The 3% loss to friction did sound a little small to me, but I went with what I was given.
CAski
09 May 2005
liftmech, on May 9 2005, 03:46 AM, said:
Thanks for finishing up my calculations, Matt.
CAski- I hadn't gotten to the actual tension force required because I hadn't finished the rest of the equation. One starts out with a basic setup like this, and then goes on to calculate sag (which, as Matt mentions, is possible only on a computer). Once you've gotten the sag figured, you'll know whether your original tension that you arbitrarily picked is enough or too much. One also has to factor in the rope breaking strength; this must be divided by 5 (safety factor) to give you your maximum tension. If you cannot get your sag where you want it with the rope you've chosen, you must go up either a size or a grade (from IPS to XIPS, for example). One final factor is the adherence of a steel rope to a rubber drive sheave liner. Various manufacturers can give you the coefficient of friction for their proprietary material, and it is possible to calculate how much tension is needed to go over that number.
There are, as you say, a crapload of numbers involved in designing a lift; most of these can be just plugged into a computer programme and it will spit out results. By the way- in Allan's sheet for Motherlode (Silverlode?) 'chord' is the angle between towers as measured from tower cap to tower cap. This number is used to determine sag.
<{POST_SNAPBACK}>
CAski- I hadn't gotten to the actual tension force required because I hadn't finished the rest of the equation. One starts out with a basic setup like this, and then goes on to calculate sag (which, as Matt mentions, is possible only on a computer). Once you've gotten the sag figured, you'll know whether your original tension that you arbitrarily picked is enough or too much. One also has to factor in the rope breaking strength; this must be divided by 5 (safety factor) to give you your maximum tension. If you cannot get your sag where you want it with the rope you've chosen, you must go up either a size or a grade (from IPS to XIPS, for example). One final factor is the adherence of a steel rope to a rubber drive sheave liner. Various manufacturers can give you the coefficient of friction for their proprietary material, and it is possible to calculate how much tension is needed to go over that number.
There are, as you say, a crapload of numbers involved in designing a lift; most of these can be just plugged into a computer programme and it will spit out results. By the way- in Allan's sheet for Motherlode (Silverlode?) 'chord' is the angle between towers as measured from tower cap to tower cap. This number is used to determine sag.
<{POST_SNAPBACK}>
My issue is that you have labeled your units incorrectly. The number you came up with is not the change in tension, but the change in potential energy. Matt's calculations do not work because he treats it as the total tension.
Why would you arbitrarily choose the tension needed? If you can calculate it out, why bother estimating it?
This post has been edited by CAski: 09 May 2005 - 07:47 PM
CAski
09 May 2005
Interesting...
500 ft/min*(1/60) = 8.333 ft/sec
m = 1100/4500 = 0.244
Angle = Tan^-1(.244) = 13.736 Degrees
Sin(13.736) = 0.2375
8.333*0.2375 = 1.979 vert. feet/sec
3*170 = 510 lbs/chair
PE = Fh = 510(1.979) =1009.290 (ft*lbs)/(chair*secs)
(1100^2+4500^2)^(1/2) = 4632.494 ft
500 ft/min*(6/60) = 50 ft/ six seconds (spacing between chairs)
4632.494/50 = 92.650 chairs
1009.290*92.650 = 93510.719 ft*lbs/sec
93510.719/550 = 170 HP
My number turned out the same; I believe that I know what is going on now. The mathematical order that Matt has chosen really does not make any sense. Because of that, some of his units are labeled incorrectly. However, all of these inconsistencies cancel out in the end, yielding both the proper number and units. I am sorry, but it really does not make any sense to divide 510 pounds by 50 feet and multiply by 1100 vertical feet. The end result is in vertical foot-pounds per foot, which really is not applicable to anything. Your result is correct, but not for the reason you would believe.
This post has been edited by CAski: 09 May 2005 - 09:05 PM
500 ft/min*(1/60) = 8.333 ft/sec
m = 1100/4500 = 0.244
Angle = Tan^-1(.244) = 13.736 Degrees
Sin(13.736) = 0.2375
8.333*0.2375 = 1.979 vert. feet/sec
3*170 = 510 lbs/chair
PE = Fh = 510(1.979) =1009.290 (ft*lbs)/(chair*secs)
(1100^2+4500^2)^(1/2) = 4632.494 ft
500 ft/min*(6/60) = 50 ft/ six seconds (spacing between chairs)
4632.494/50 = 92.650 chairs
1009.290*92.650 = 93510.719 ft*lbs/sec
93510.719/550 = 170 HP
My number turned out the same; I believe that I know what is going on now. The mathematical order that Matt has chosen really does not make any sense. Because of that, some of his units are labeled incorrectly. However, all of these inconsistencies cancel out in the end, yielding both the proper number and units. I am sorry, but it really does not make any sense to divide 510 pounds by 50 feet and multiply by 1100 vertical feet. The end result is in vertical foot-pounds per foot, which really is not applicable to anything. Your result is correct, but not for the reason you would believe.
This post has been edited by CAski: 09 May 2005 - 09:05 PM
CAski
09 May 2005
Aaaaaarrrrrrhhhhhggggggg!
The mislabled units are driving me loony!
The mislabled units are driving me loony!
liftmech
10 May 2005
Can I defend myself by saying I'm not an engineer? Probably not; at any rate, I'm pretty much regurgitating a class I've taken before on lift physics. We never really went into calculating tensioning force; rather, we chose an arbitrary number based upon a similar real lift and built our imaginary one from there. You're correct that my calculations defined total change in potential energy. I think we would have gotten into calculating actual tensioning force if we hadn't been inside a warm hotel room on the last day of a conference.
Matt
10 May 2005
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.
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.
Outback
10 May 2005
Most interesting post so far.
If you look at engineered lifts in the last ten years you will notice that the friction numbers on the sheaves zero out (almost) when calculating the profile. The sag between towers and sheave friction are almost nil if the lft is designed properly to a profile......true tension and profile are determined with capacity, vertical rise, etc.
Not to say this calculation is not obseved, but once a true profile is established with rap angles, etc. the friction on the line...not turning bullwheels (or severe depression assemblies) the friction doesn't add up.
Whadda think????
If you look at engineered lifts in the last ten years you will notice that the friction numbers on the sheaves zero out (almost) when calculating the profile. The sag between towers and sheave friction are almost nil if the lft is designed properly to a profile......true tension and profile are determined with capacity, vertical rise, etc.
Not to say this calculation is not obseved, but once a true profile is established with rap angles, etc. the friction on the line...not turning bullwheels (or severe depression assemblies) the friction doesn't add up.
Whadda think????
