Calculate Electric Motor HP Required For A Load

For static load of 320 Kg, your calculation of torque 219 N m is correct. For this load and torque we can calculate power as below:

Power = force x velocity = 2pi.d.n.F = 2pi x Fd x n = 2pi x torque x rotations per second

Where we have used the fact that torque = force x distance = Fd

Symbolically, this is usually written as P= T x w

where T is torque and w is called the angular velocity, and is equal to the

factor 2pi x rotations per second.

Your motor RPM is 1450 and speed reduction is 1/70. The speed of hoist drum will be 1450/70, Comes out 20.71 rpm or 0.345 rps

Therfore power comes out P= 219 x 2 x3.14 x 0.3452 =474.81 watts

or = 0.636 HP

This calculation is based on your static load with 100% efficiency of gear box, motor and other system without friction loss. If you assume rope weight, friction losses (efficiency) and over load considerations then accordingly torque and power requirements will change.

As you may be aware, your terminology is a little loose. "Static" implies motionless. Motionless implies no power*. The power that has been calculated is that required to move the load at a constant rpm (which translates into a constant upward speed of about 923 cm/min).

Electric motors can supply very high torques for very short times, and can supply perhaps three times rated torque for a minute or less (all this varies with the motor, but these are OK rough guides).

At the instant that the motor starts, there will no doubt be some play to be taken up, but within a few hundredths of a second, the torque on the motor will increase to (for example) 4 times its rated torque. The load will accelerate upward with 4 times the steady state load, so will accelerate at 4 G, subjecting the cable to 4 times higher load that the steady state load. You can see that if this were an engineered system, that you might want a "soft start" facility to prevent the motor from over stressing the cable, causing bounce, etc.

The electrical power during that first instant can be quite high, and the motor efficiency will be quite low. If you imagine the very first instant, the load is not moving, but the motor is starting to grunt... and heat up. Because the load is not yet moving, mechanical power is zero. On a motor this size, the input power may be 3 kW and the output is (instantaneously) essentially zero.

But the load quickly begins to accelerate and at 4G, gets to its operating speed very quickly. (If we round the 923 cm/s to 9.8m/s we can see that it should take about 1/4 second.) The motor's output power is constantly changing during this brief phase.

Efficiency of 70:1 gearboxes (especially worm gearboxes) can be low -- anywhere from about 35% to about 65%. So the motor would need to be oversized to compensate. Although you may think the motor will "just slow down" to meet its continuous hp rating, it will not. It's "slip" will increase and its hp output will increase, and it will heat up.

In practice, if you know the power required and the efficiency of the mechanism, then you look for a motor that will provide the resulting power level for as long as required (taking into account the duty cycle -- in other words how frequently and for how long must the motor operate).

If you are really building a hoist, then there are additional considerations mainly having to do with safety. But for a thought experiment, this is about how a motor would be sized.

* Mechanical power and electrical power diverge a little here. A stove burner does not "do work" in a mechanical sense but dissipates a lot of electrical power.