Speed reducer, shaft, and pulley lifting problem 2

[mechengineer18]

I need to calculate the torque required from a speed reducer to lift a load. The lifting system involves a motor attached to a dual speed reducer which will attach via a coupler on either side to 1” shafts. On the end of each shaft will be an 8” V-grooved drum. To each drum I will fasten two cords opposite of each other in such a way that when the shafts turn the drums, the drums will wrap all four cords and wind them in. The cords are redirected by pulleys and then attached to the four corners of the load to be lifted which is rectangular with an open middle so that it surrounds the surface where the motor and other equipment is mounted. If that is too confusing I will gladly email you a sketch of my design. I want to make sure that I am working through the problem correctly but I am not accounting for friction yet.

For the required Torque out of the speed reducer I am using:
Torque = (diameter/2)* Load
Divide by two to get the Torque per shaft.

The total load in my problem is 2,000 Lbs. so:
Torque = (8”/2)*2,000 Lbs = 8,000 In.-Lbs
or 4,000 In.-Lbs out of each side of the speed reducer

I think I have this lined up correctly but it has been a while and I would love corrections, confirmation, or factor of safety suggestions.

Thanks,
Mike

 

[Ross0684]

I feel you are confusing yourself by referring to two shafts,
should you not treat the calculation as simply :
total load (2000lbs)x distance at which force is applied (4")=8000in.lbs.(as if it was a single cord taking the load)
The design capacity of the speed reducer is simply this, irrespective whether it has one or two ouput shafts.
The selection of the speed reducer should also take into account speed of lift, duty cycle etc( and friction!)

 

[mechengineer18]

Ross,

I see. I was under the (apparently erroneous) impression that the speed reducer output shafts were independent of each other. That makes it clear that I must use a smaller drum which will be useful anyway as I was coming up with a required output speed of only 2.87 - 4.30 RPM to lift the load 9' in 60-90 seconds.

Thank you very much for clearing that up for me. I think I can figure out the friction and duty cycle but I may bother you some more.

Mike

 

[rb1957]

you're lifting the load 9' in approx. 1 minute, by coiling rope on a drum. approx. 3 rev/min implies a drum circumference on 3'.

think too about the speed reduction ... work done is constant, T*w; so you want 8,000 in.lbs of torque at 3rpm, so shouldn't your motor (as an example) output 8 in.lbs at 3,000 rpm ?