Need help figuring required horsepower of electric motor

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Discussion Overview

The discussion revolves around determining the required horsepower for an electric motor to power a conveyor-like trolley system designed to move materials up an incline. Participants explore calculations related to weight, speed, and energy requirements, while also considering the effects of gear reduction on torque and horsepower.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant outlines the specifications of their trolley system, including weight capacity, incline angle, and desired speed, and expresses a preference for oversizing the motor.
  • Another participant calculates the potential energy change and derives a minimum horsepower requirement of 1 HP based on the system's parameters.
  • A third participant provides an English units conversion of the horsepower calculation, confirming the earlier result of approximately 0.972 HP.
  • A subsequent question is raised about the impact of using a 40:1 speed reducer on the motor's torque and horsepower.
  • Another participant explains that while the gearbox changes the torque and RPM ratio, the overall power remains largely unchanged, aside from minor losses in the gearbox.

Areas of Agreement / Disagreement

Participants generally agree on the calculations leading to a minimum horsepower requirement of around 1 HP. However, there is a question regarding the effects of the speed reducer on horsepower and torque, which remains somewhat contested.

Contextual Notes

Participants express uncertainty regarding friction losses and the exact speed of the motor, indicating that adjustments may be necessary once the system is operational. The discussion does not resolve the implications of using a speed reducer on horsepower definitively.

Who May Find This Useful

Individuals interested in mechanical design, motor specifications, or those working on similar conveyor or lifting systems may find this discussion relevant.

HandymanExpres
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I have built a conveyor system but still need to power it. It is not a conveyor belt, but rather a trolley of sorts for moving materials up to the second floor. It runs parallel to stairs for about 17 or 18 feet in my garage at an angle of apx 38°. It needs to be able to move at least 500 lbs. but I would rather calculate it to accommodate more like 1000. It needs to move at a rate of about 105' per minute. It rolls on hard rubber wheels but I may switch to steel wheels. The motor will have a pulley on it which winds a high tensile strength material just like used for seatbelt material. I will be using a reversible motor with a speed of 1750 RPM and a reducer. I saw a 40:1 reducer available, so I made calculations based on it for my pulley. I came up with a pulley size of 9.14" in diameter and a circumference of 28.7". I know that with every wind the pulley will get larger, thereby increasing the speed, but it will be negligible for what I am doing. I know that there is friction loss here and there, and the exact speed of the motor may not be 1750, so once this is operational I may remake the pulley size up or down to my liking. I just need to know the HP requirements needed. My motto is "Overkill is underrated" so I would oversize the motor in a heartbeat, but I do not want to undersize it.
 
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Welcome.
I assume 105' is 105 feet.
The time needed will be (18 feet / 105' ) = 0.171 minutes = 10.28 seconds
Up 18 feet * Sin(38°) = 11 feet vertical = 3.3528 metres.
m = 500 pounds = 226.8 kg
Potential Energy change = m*g*h = 226.8 * 9.8 * 3.3528 = 7452. joule
That is 7452. joule / 10.28 seconds = 724.9 joule/sec = 724.9 W = 0.972 HP

The absolute minimum motor would be 1 HP.
 
In English Units: ( to avoid metric conversions if you wish )

1 HP = 550 foot-pounds/second

500 pounds x 11 feet / 10.28 seconds = 535 ft-lbs/sec

535 ft-lbs/sec / ( 550 ft-lbs/sec )/HP = 0.972 HP
 
OK. Next question. If I use a 1 HP motor, and put a 40/1 speed reducer on it, does that just change the torque or the HP also?
 
Power = torque * RPM
The gearbox is a transformer, it transforms the ratio of torque to RPM.
The power does not change, apart from a couple of percent loss in the gearbox.
 

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