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Need help with motor sizing/torque calculation - Building a Ski Rope Tow.

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Nov30-12, 01:53 PM
P: 5
I am building a homemade ropetow in my backyard. I have done a lot of research and unfortunately I have been away from physics for about 15 years, so I am having trouble calculating my motor size and transmission options. I will lie out what I am trying to accomplish below.

Here are the specs:

-The hill is 300 feet long at a 20% grade
-The rope is approximately 650 feet long and 200 lbs.
-I can use anywhere between a 1 to 2 hp motor operating at 1750 RPMs (would prefer to stay on 110 power and definitely can't go to 3 phase)
-The gear reducer can be 10:1 15:1 or 20:1
-The pulleys to attach to the output shaft range from 6" to 14" mostly in even numbers.
-I would like to be able to pull three 200 lb people up said rope at once
-I would like the rope speed to be approximately 7-10 fps.

I am open to ideas. I have nothing purchased at this point.

My plan was to use a 15:1 reducer with a 14" pulley for speed that works out to

1750rpm/15 = 116.667 RPM output

circumference of pulley is 3.663'

speed is 427.39011 feet/minute or 7.1 fps

This would be completely fine for speed if it can handle the load. My biggest concern is not overloading the motor.

Any help or suggestions would be appreciated.

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Nov30-12, 04:06 PM
P: 159
I gently suggest that your are approaching this problem backwards. You must size the motor & gear ratio for the load. Not try to fit the load to the motor & gearing available.

Focus your analysis on the TORQUE required of the output driveshaft. That's the torque required to move rope, people, gear, bota bags of wine, and the neighbor's dog. Up the hill, against gravity. There are physics calculations involved for linear & rotational motion, F=ma & T=Jα. Your target quantity is PEAK TORQUE that is steady state torque load but also the torque required to un-stick everybody/everything from the snow & accelerate them to your desired speed. In the required time, that is.

So then you'd have various quantities of torque, force, & linear & rotational speed. Use arithmetic to get the peak torque and driveshaft RPM. I don't have my formulas in front of me, but there is a relationship between horsepower and torque and RPM.

I'd suggest you track down catalogs from any of the major gearmotor manufacturers and seek out the Engineering sections of those catalogs. There you are very likely to find all the calculations you need. SEW Eurodrive, Dodge, Falk, Emerson come to mind. Websearch for the "Smart Motion Cheat Sheet" for a quick summary of motion & torque calculations.

Your little motor is probably too small to do the task, unless you want to move very slowly.
Nov30-12, 04:35 PM
P: 5
I completely understand what you are saying and that is what I am trying to do.

I know that 1750 RPM is a standard speed for electric motors.

I also know that I have several options for gear reduction, but my issue is if I use a 2hp motor going into a 15:1 reducer I do not know what the torque output on the shaft is. When I put a 12", or 14" pulley on said shaft will it have enough torque to move several hundred pounds up the slope.

I understand that there is a direct relationship between the speed I want to go and the HP of the motor used. I need help figuring out what size motor will get the rope moving at a realistic speed.(hopefully 7-10 feet/second)

I am by no means a physics expert. That is why I am here.

Nov30-12, 07:01 PM
Sci Advisor
HW Helper
PF Gold
P: 6,753
Need help with motor sizing/torque calculation - Building a Ski Rope Tow.

Power = T*N/5252, P in horsepower, T in ft-lbs, N in RPM
So P = 2, N = 1750, therefore T = 2 * 5252 / 1750 = 6 ft-lbs torque from motor.

Using 15:1 reduction, torque from motor is multiplied by 15.

Therefore, output torque from motor w/speed reduction = 6*15 = 90 ft-lbs
Nov30-12, 07:16 PM
P: 5
Thanks. That answers 1/2 of the question. If the output is 90ft-lbs how do I now determine whether or not that is enough to move the rope and 500 lbs of skier/snowboarder up the 20% grade with a 14" pulley on?

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