Trying to calculate motor kW/reduction drive for lift of 100kg at 0.8m /sec

  • #1
surfguy
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0
TL;DR Summary
need to calculate motor kw/reduction drive for lift of 100kg at .8mtrs /sec
Hi
need to calculate 240v/315v motor size attached to reduction gearbox (reduction unknown) to lift a weight of 100kg on cable drum .8mtr/sec.
Drum dia not determined yet.
motor size not determined yet.
any help is greatly appreciated as this seems to be out of depth of many motor drum winch suppliers and myself.
thanks
 
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  • #2
Welcome to PF.

What is the application? How far do you have to lift it?
 
  • #3
Single or 3 phase?

Step 1: calculate the power you need, then double it.

Step 2: pick a common motor rpm and "reasonable" drum size, then calculate the reduction. Adjust as seems appropriate.
 
  • #4
Ignoring acceleration.
The change in PE per second = m·g·h = 100 * 9.8 * 0.8 = 784 watt ≈ 1 HP.
 
  • #5
Finally, the gear reduction ##GR##, drum diameter ##d## (in meters), and motor ##RPM## are linked in the following way:
$$v = \frac{\pi}{60} \times d \times \frac{RPM}{GR}$$
Or
$$d = \frac{19.1 \times GR \times v}{RPM}$$
Where ##v## is the cable speed in meters per second. For example, given a motor of 1750 RPM and a gear reduction of 10:1, the drum diameter would be 0.087 meters or 8.7 cm to get a cable velocity of 0.8 m/s.
 
  • #6
I was hoping OP would calculate them...
 
  • Like
Likes Tom.G and berkeman

FAQ: Trying to calculate motor kW/reduction drive for lift of 100kg at 0.8m /sec

How do I calculate the power needed to lift a 100kg load at 0.8 m/s?

To calculate the power needed, use the formula: Power (P) = Force (F) x Velocity (v). The force required to lift the load is equal to the weight of the load (mass x gravity). So, P = 100 kg x 9.81 m/s² x 0.8 m/s = 784.8 Watts or approximately 0.785 kW.

What factors should I consider when selecting a motor for this lift application?

When selecting a motor, consider the required power (kW), the operating speed (RPM), the torque requirements, the type of motor (e.g., AC or DC), the duty cycle, and the environmental conditions (e.g., temperature, humidity). Additionally, consider the efficiency of the motor and any potential losses in the drive system.

How do I determine the appropriate reduction ratio for the drive system?

The reduction ratio is determined by the desired output speed and the motor's base speed. If your motor operates at 1500 RPM and you need the output speed to be 60 RPM, the reduction ratio would be 1500 / 60 = 25. This means you need a gearbox with a 25:1 reduction ratio.

What is the relationship between torque and power in this lifting application?

Torque (T) and power (P) are related through the rotational speed (ω) by the formula: P = T x ω. For lifting applications, torque is crucial because it determines the force available to lift the load. Ensure that the motor and gearbox can provide sufficient torque to overcome the load's weight and any additional resistance.

How do I account for efficiency losses in the motor and drive system?

Efficiency losses can be accounted for by considering the efficiency ratings of the motor and gearbox. If the motor has an efficiency of 90% and the gearbox has an efficiency of 85%, the overall system efficiency is 0.9 x 0.85 = 0.765 or 76.5%. To find the required input power, divide the calculated power by the system efficiency: 0.785 kW / 0.765 ≈ 1.03 kW. This is the power that needs to be supplied to achieve the desired lifting performance.

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