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

1. How do you calculate the motor power required to lift 100 kg at a speed of 0.8 m/s?

To calculate the motor power required, you first need to determine the force needed to lift the load, which is equal to the weight of the load (mass x gravity). For a 100 kg mass, the force (F) = 100 kg x 9.81 m/s² = 981 N. Next, calculate the power (P) needed using the formula P = F x v, where v is the velocity (0.8 m/s). Therefore, P = 981 N x 0.8 m/s = 784.8 Watts. Since 1 kW = 1000 Watts, the motor power required is approximately 0.785 kW.

2. What is a reduction drive, and why is it used in lifting mechanisms?

A reduction drive, or gearbox, is a mechanical device used to reduce the speed provided by the motor while simultaneously increasing the torque. This is crucial in lifting mechanisms because lifting heavy loads often requires high torque at lower speeds. By using a reduction drive, you can use a motor with higher speed and lower torque, which are generally more available and cost-effective, to achieve the necessary torque for lifting.

3. How do you select the appropriate reduction ratio for a lift system?

The appropriate reduction ratio can be selected based on the motor speed and the desired output speed of the lifting mechanism. If the motor's speed is higher than needed, a reduction drive can reduce it to the desired speed. The reduction ratio is calculated by dividing the motor speed by the required output speed. For instance, if a motor runs at 2400 RPM and you need the output speed to be 300 RPM to achieve a lift speed of 0.8 m/s, the required reduction ratio would be 2400/300 = 8:1.

4. How do you ensure the motor and reduction drive can handle the load of 100 kg?

To ensure that the motor and reduction drive can handle the load, you must check the rated torque of the motor and the gearbox's maximum permissible torque. Calculate the torque required to lift the load using the formula Torque (T) = Force x radius, where the radius is the distance from the center of the motor shaft to the point where the force is applied. Compare this required torque with the motor's and gearbox's specifications to ensure they exceed the calculated value, thereby providing a safety margin.

5. What are the safety considerations when designing a lift system for 100 kg at 0.8 m/s?

Safety considerations include ensuring that all components such as the motor, reduction drive, and lifting mechanism are rated for at least the maximum load and speed. Additionally, incorporating safety features such as brakes, limit switches, and overload protection is crucial. It's also important to consider the stability of the entire system, including the structure that supports the lift, and to adhere to relevant safety standards and regulations to prevent accidents or failures during operation.

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