How much extra power for one person to ride a chairlift?

milk_manBeen wondering for awhile..

How much power (theoretically) would an empty chairlift use? How much power would that same chairlift with only one 150 lb skier on it use? I want to know how much extra energy that is, not sure how to figure it out or if it's possible.

LonelyIt's definetly possible and it can be estimated quite simply. The issue arises when you have to decide what variables you will have to use or take into account. Then it's just a matter of plugging in the numbers.

You would need to know the change in height and the overall weight of the chairlift to calculate the work done. Then you'd have to factor in any mechanical "loss" of energy. Then you'd probably have to worry about tension and some friction because it is not a perfect system.

Once you have those variables it's just a matter of finding the amount of work done, and the amount of energy inefficieny of the chairlift. Then you'll have an answer.

daannnnieelwouldn't you also need to know the angle of the line between the towers...? I'm super rusty on this but if the towers are all steep it's not going to be as easy as if they weren't. I don't think it's something you can just factor out or estimate really.

finding the tension you would need to know all the angles for the entire system more or less

in math, everything depends on variables. this question just has a lot of shitty useless ones we can leave to the magicians at poma or whatever the fuck that chairlift company is.

skiermanIt doesn't fucking matter. The initial power output from the lift is enough to keep the lift moving at the same pace whether no one is riding the lift or 50 fat fucks from Texas with guts filled with rodeo burgers that can barely zip up their Cowboys jackets.

God you people are stupid.

DaphActually the motor load is significantly more with a full line than all empty chairs. Every lift should have a gauge that shows percent load on the motor and drive system. With a full line of chairs a lift typically only sees a 80% load on the motor, any more than that and you start to see drive faults and motor failure especially when the lift is trying to ramp up after a slow or stop.

Like many people have already said, there are lots of variables to consider and they differ from lift to lift. Say you have a high speed quad with a bottom drive, each chair weighs about 500 lbs, there are 30 towers, 5 of which are depressions. That will take much more power to run than a fixed grip 2 seater with a top drive and only 14 towers.

VinnieFsort of related: are there advantages/disadvantage to have a bottom or top drive on a lift? Or is it usually wherever is closest to the power supply.

DaphActually the motor load is significantly more with a full line than all empty chairs. Every lift should have a gauge that shows percent load on the motor and drive system. With a full line of chairs a lift typically only sees a 80% load on the motor, any more than that and you start to see drive faults and motor failure especially when the lift is trying to ramp up after a slow or stop.

Like many people have already said, there are lots of variables to consider and they differ from lift to lift. Say you have a high speed quad with a bottom drive, each chair weighs about 500 lbs, there are 30 towers, 5 of which are depressions. That will take much more power to run than a fixed grip 2 seater with a top drive and only 14 towers.

skiermanYou're wrong and I'm right. I know chairlifts, more than anyone. I know things you don't know about chairlifts because I'm an expert on chairlifts. I can build the best chairlifts so I know more about chairlifts than you.

LonelyIt's definetly possible and it can be estimated quite simply. The issue arises when you have to decide what variables you will have to use or take into account. Then it's just a matter of plugging in the numbers.

You would need to know the change in height and the overall weight of the chairlift to calculate the work done. Then you'd have to factor in any mechanical "loss" of energy. Then you'd probably have to worry about tension and some friction because it is not a perfect system.

Once you have those variables it's just a matter of finding the amount of work done, and the amount of energy inefficieny of the chairlift. Then you'll have an answer.

daannnnieelwouldn't you also need to know the angle of the line between the towers...? I'm super rusty on this but if the towers are all steep it's not going to be as easy as if they weren't. I don't think it's something you can just factor out or estimate really.

finding the tension you would need to know all the angles for the entire system more or less

in math, everything depends on variables. this question just has a lot of shitty useless ones we can leave to the magicians at poma or whatever the fuck that chairlift company is.

intifadaThese wouldn't be variables they'd be constants. The biggest variable would be the size of your electric motor. On almost all lifts the difference would be negligible.

intifadaThese wouldn't be variables they'd be constants. The biggest variable would be the size of your electric motor. On almost all lifts the difference would be negligible.

LonelyWhat? Something like gravity is a constant. They are variables. They change the equation depending on what number you put in. If anything the electric motor would be a constant. It is the one of the only things constant between the lift having a person or not.

Lol

intifadaThe difference between a150 pound rider vs an empty line would be negligible.

skiermanYou're wrong and I'm right. I know chairlifts, more than anyone. I know things you don't know about chairlifts because I'm an expert on chairlifts. I can build the best chairlifts so I know more about chairlifts than you.

intifadaI guess I was thinking about this in regards to a single lift.
Length, weight, and steepness of a lift compared to the size of its electric motor will definately give you different results, but again these differences are nothing you need to calculate as they will show up in the information the electric motor gives you.
Then again I believe electric motors get more mechanically efficient at a higher load so maybe that 150 pound person is saving you power.

eheathYoure really losing your touch.

Scotty_Bdepends on the weight of the chairlift.

energy going up or down is potential and is defined as mgZ

(Mass*Gravity*Height)

Assuming an empty 4 person chairlift weighs 500 lbs
and a loaded one with little OP weighs 650 lbs

MgZ of empty

Mgz of loaded


they are going the same distance up the hill so z's cancel

so this becomes

Mg empty = 500*32 = 16000 ftlbf/s

Mg loaded = 650*32 = 20800 ftlbf/s

divide the above by 550 (this is conversion for HP)

we arrive at...

_______________________________________________________

Empty chair uses 29.09HP

OP's Chair uses 37.81HP

________________________________________________________

OP you use around 9 HP more than the empty chair and use about 30% more energy.
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milk_manLove this! But I think those calculations are a little off. Is an empty chair 2.9 hp and a chair with a rider 3.7 hp?

Regardless, your post was very helpful! Thanks brotha. I'm definitely gonna use those calculations

eheath16000 divided by 550 is not 2.9 haha

milk_manLove this! But I think those calculations are a little off. Is an empty chair 2.9 hp and a chair with a rider 3.7 hp?

Regardless, your post was very helpful! Thanks brotha. I'm definitely gonna use those calculations

milk_manYeah I know! But it doesn't take that many hp to move 500 lbs

Scotty_BYea I fucked it up hard my bad.

I reworked the problem though so you're good.

-500lbf
-1ft vertical
-3 seconds


((500*1) / 3s ) = 166 lbf*ft/s

((650*1) / 3s ) = 216 lbf*ft/s


166/550 = 0.30 HP

216/550 = 0.39 HP

You use about 30% more energy.

I love units.
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Scotty_BYea I fucked it up hard my bad.

I reworked the problem though so you're good.

-500lbf
-1ft vertical
-3 seconds


((500*1) / 3s ) = 166 lbf*ft/s

((650*1) / 3s ) = 216 lbf*ft/s


166/550 = 0.30 HP

216/550 = 0.39 HP

You use about 30% more energy.

I love units.
_______________________________________________________

DaphI laughed, but I also agree. I miss the skierman who really rustled jimmies.

Scotty_BYea I fucked it up hard my bad.

I reworked the problem though so you're good.

-500lbf
-1ft vertical
-3 seconds


((500*1) / 3s ) = 166 lbf*ft/s

((650*1) / 3s ) = 216 lbf*ft/s


166/550 = 0.30 HP

216/550 = 0.39 HP

You use about 30% more energy.

I love units.
_______________________________________________________

VinnieFThis is flawed. You can't use the weight of the chairs in the calculation since assuming there are an equal number going up the lift as down then it should theoretically cancel. Then you need a whole bunch of other variables.

Then you need the chairlift speed. 3 m/s is about right for fixed chair lift. So a 60kg object moving vertically would require 60*(9.8+3) N m/s. Which is 0.77 kW. Then say it's at a 20 deg angle which means it would be sin20(0.77) which is 0.26 kW. This is assuming no friction. that would mean a fully loaded 100 chair quad lift would need a 140hp engine if it's a fixed grip going rather slow and being 100% efficient up a 20 degree slope.

I dunno what the efficiency is, but if its 75% and they want the lift to be running at 80% capacity when it's loaded then a slow lift that size should be about 240hp I guess.

Scotty_BTrue but don't think you needed to show all that extra work. And why discredit me when your answer fell right with both of mine? 0.26kW is 0.34hp

VinnieFjust pointing out some flaws. you did a good job at it, but it needed fine tuning. your 150 lb person came out to 0.09 hp, which is definitely in the rough area, but off by enough to redo the calculations.

milk_manIs it 0.26 kw for a 150 lb person then?

VinnieFgiven a 100% efficient system with no air resistance moving at 3 m/s up a 20 degree incline, then I think that's correct.

You could assume that if the lift is already running and you don't care to figure out how much energy it's using running empty then adding 1 person of 60 kg would amount to roughly 0.26 kw extra load on the motor.

keep in mind that's a small person (130 lb). but if you average out with children that might be correct.

So then the fun stuff is how much more money it costs to run a full lift vs empty lift.

So if it's a quad with 100 chairs (so 50 chairs full) then that would add 52 kw. the average price in the US is 12 cents per kwh, so running the chairlift full would cost $6.24 an hour more than empty.

and just for fun. If this chairlift over the season averages 50% capacity and runs 8 hours a day for 4 months, that's $3000 in electricity for the extra weight of riders.

VinnieFjust pointing out some flaws. you did a good job at it, but it needed fine tuning.