Designing a 600 kg Lifter: Calculating Power Needs

• engineeringstudent01
In summary, the conversation discusses a platform that can lift 600 kg at 0.75 m/s up to a height of 37 m with a maximum angle of inclination of 87 degrees. The issue at hand is determining the power required for this application, and the calculations done so far have resulted in a power rating of 4.65 kW without the factor of safety. The conversation also mentions the use of a pulley and the possibility of incorporating it into the calculations. Further discussion includes the overall efficiency factor for the system and the inclusion of a factor of safety. The conversation ends with a calculation of the potential and kinetic energy involved in this device.f

engineeringstudent01

TL;DR Summary
Having some issues calculating the power required to lift a specific mass on the platform of a furniture lifter.
I have a platform which can lift 600 kg (the platform itself weighs 95 kg so their combined mass is actually 695 kg) at 0.75 m/s up to around 37 m in height at a maximum angle of inclination of 87 degrees. The issue I'm coming across is determining the power required to lift said mass as the results I'm getting at the moment seem too small for such an application. Does the pulley need to be considered too at this point or is it negligible? I've tried to incorporate it too but it still seems as if I'm missing something.
What I've manage to calculate so far is the acceleration of the platform when it's accelerating or decelerating, the maximum tension in the cable (of diameter 8 mm) and the force required to lift the platform with its combined maximum payload.
Please do message me on this if you have any ideas on how this problem can be tackled.

How does this device look like (can you share any pictures or sketches) ? What kind of drive is used (just rotary motor with single pulley) ?

How does this device look like (can you share any pictures or sketches) ? What kind of drive is used (just rotary motor with single pulley) ?
I looks similar to the picture attached. It has a simple electric winch at the base, with a single pulley at the very top of the boom.

Attachments

• lifter.jpg
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The issue I'm coming across is determining the power required to lift said mass as the results I'm getting at the moment seem too small for such an application.
What answer did you get and how did you calculate it?
Does the pulley need to be considered too at this point or is it negligible? I've tried to incorporate it too but it still seems as if I'm missing something.
It will add inefficiency, but it is otherwise an energy conserving device, so you don't need to account for it in the first pass -- maybe use an overall efficiency factor for the system.

What answer did you get and how did you calculate it?
I'm getting a power rating of 4.65 kW without the factor of safety.
It will add inefficiency, but it is otherwise an energy conserving device, so you don't need to account for it in the first pass -- maybe use an overall efficiency factor for the system.
I'm using an overall factor of safety of 1.5 (apart form the cable which is higher). The calculations I've done so far are specifically without the factor or safety included so as to obtain maximum values. The factor of safety is going to be included later on.

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I'm getting a power rating of 4.65 kW without the factor of safety.
I get 5.1 kW. I multiplied the total mass by 9.81 N/kG and 0.75 m/s and then by the sine of an 87 degree angle (which is still pretty close to 1.0). We're in the ballpark though.

Acceleration is a different animal, and the force is f=ma, where a is 9.81 (gravity) + your acceleration. But 0.75 m/s is so slow I suspect you can ignore the acceleration.

Baluncore
I have a platform which can lift 600 kg (the platform itself weighs 95 kg so their combined mass is actually 695 kg) at 0.75 m/s up to around 37 m in height at a maximum angle of inclination of 87 degrees.
Assume the ramp is vertical.
In one second the potential energy will increase by a maximum of;
PE = 695 kg * 9.8 m/s2 * 0.75 m = 5108.25 joule (per second) = 5.1 kW.

The kinetic energy is only; KE = ½·m·v² = 0.5 * 695 * 0.75² = 195.5 joule.