# Calculate the minimum motor power required to lift a human?

I'm working on a Home-Made Automatic Grappling Hook of sorts, as a project for university. I've designed a Blueprint accounting for everything but (embarrassingly) the motor, as I'm unsure how to calculate the minimum motor power required to lift me. Is there an equation/simple solution to work this out?

• Start with high school Newtonian mechanics, I'm sure you know the equation for potential energy. If there's no upper limit on the time available, there's no lower limit to the power required. – user_1818839 Nov 7 '16 at 14:07
• Pulleys make a big difference. – Carl Witthoft Nov 7 '16 at 16:21
• Welcome to Engineering Stack Exchange! You'll probably need to make some adjustments once the motor is sorted out - you probably don't have enough energy storage, However, you have a great question here! – Mark Nov 7 '16 at 22:15

Power is work divided by the time interval required to complete the work,

$$P = \frac{W}{\Delta t} \\$$

Work is force applied for a particular distance,

$$W = F\Delta x \\$$

Force is mass times acceleration:

$$F = ma \\$$

And acceleration is change in speed divided by the time interval required to complete the speed change:

$$a = \frac{\Delta v}{\Delta t} \\$$

Now, if you're moving up, you need to counteract gravity in addition to the acceleration required to change speeds, so you would find the total force required to be:

$$F_{\mbox{peak}} = m\left(\frac{\Delta v}{\Delta t} + g\right) \\$$

The peak force you need will occur during this acceleration segment, but the peak power you will need will occur during the constant-speed hoist. It's important to check both portions of the drive cycle for numbers - the acceleration requires a large force (torque) capability, the hoist requires slightly less force, but requires that force to be delivered at a higher speed.

The constant-speed portion of the drive cycle (assuming accelerate-constant speed-decelerate) requires a force of:

$$F = mg \\$$

At the end of the motion you will decelerate, so the power required will be even less. I'll point out that these delta values - the definition given for work, acceleration, and power - all depend on constant application of even force applied along a linear path. If this isn't true then you need to integrate those values instead of averaging them.

Example time! Say you weigh 165lb (75kg). You want to go from a stop to moving up at a speed of 2.2mph (1 m/s), and you want to reach that speed in 2 seconds.

So, first find the peak force:

$$F_{\mbox{peak}} = 75\mbox{kg}\left( \frac{1\mbox{m/s}}{2\mbox{s}} + 9.81 \mbox{m/s}^2 \right) \\ \boxed{F_{\mbox{peak}} = 773\mbox{N}} \\$$

Work is force applied for a particular distance. Because I'm assuming constant acceleration, the average speed is the average of the start and end speed values.

$$v_{\mbox{average}} = \frac{0 + 1}{2} \mbox{m/s} \\ \boxed{v_{\mbox{average}} = 0.5 \mbox{m/s}} \\$$

Distance is then speed times time:

$$x = v_{\mbox{average}} \Delta t \\ x = 0.5 \mbox{m/s} 2 \mbox{s} \\ \boxed{x = 1 \mbox{m}} \\$$

So, the work applied is:

$$W = F\Delta x \\ W = 773 \mbox{N} 1 \mbox{m} \\ \boxed{W = 773 \mbox{Nm}} \\$$

Finally, the power you need for this section is:

$$P = \frac{W}{\Delta t} \\ P = \frac{773 \mbox{Nm}}{2 \mbox{s}} \\ \boxed{P = 387 \mbox{W}} \\$$

Compare this to the power required to go straight up 10 meters, after the initial stopped-to-speed transient.

$$F = 75\mbox{kg}(9.81 \mbox{m/s}^2) \\ F = 736 \mbox{N} \\$$

$$x = 10 \mbox{m} \mbox{(From the problem statement)} \\$$

$$W = F\Delta x \\ W = 736 \mbox{N} (10 \mbox{m}) \\ W = 7360 \mbox{Nm} \\$$

$$x = v\Delta t \\ \Delta t = \frac{x}{v} \\ \Delta t= \frac{10 \mbox{m}}{1 \mbox{m/s}} \\ \Delta t = 10 \mbox{s} \\$$

$$P = \frac{W}{\Delta t} \\ P = \frac{7360 \mbox{Nm}}{10 \mbox{s}} \\ \boxed{P = 736 \mbox{W}} \\$$

So here you can see that, for this example, acceleration needs 5% more force (torque), but the constant-speed lift requires 90% more power. You can tweak those numbers to see how the various parameters affect the results; in particular you should notice the impact that a shorter acceleration time makes on the acceleration power requirement.

I did all this math with linear numbers because you didn't provide any drawings of what your device looks like or how anything's attached (read: gearing, drive radii, etc.), so I can't give you anything specific on your motor needs. I will say, though, that there is a very straightforward way to convert linear motion to rotational motion: $s = r\theta$. Arc length is drum distance times the total rotational angle.

You can gear the motor however you want, but power is power - you either provide lower torque at a higher speed, or higher torque at a lower speed, but either way you're going to provide the power required as given above or you're not going to get the performance you want.

For the example I gave, I'll point out that a 750W motor is about 8 pounds, which happens to be the same weight as a gallon of milk. This doesn't include the frame to hold the motor, the battery pack, the winch drum, 10 meters of cable, the gearbox, the grappling hook, or the grappling hook ejector. One gallon of milk, for just the motor.

Also, as a final comment, none of the math here took friction into account. The power I gave was the output mechanical power required to do the work. I'd probably estimate the efficiency of the mechanical drive system at 80 percent, so you could very likely wind up needing a motor 25% larger to achieve the desired performance - over 900 Watts. You can reduce weight or lower the top speed to enable use of a smaller motor, but then you're looking more at speeds of a winch like Chris Johns referenced, which operates at 8 meters per minute, meaning it would elevate about one story every thirty seconds. Not exactly breathtaking.

The minimum force required to lift an object vertically > it's weight (mass.g) however some excess force will be required to produce a useful acceleration according to F = ma

What you consider an appropriate acceleration is up to you to specify but there is obviously an upper limit which is reasonably safe.

Depending on how your design works you will need to convert this force to a torque.

For the sake of argument you decide that you want to be able to just support a mass of 100kg (ie this is the load at which the motor stalls.

F=mg gives us 100kg x 9.8m/s^2 = 9800N

If you connect the motor directly to a pulley of radius (r) 0.01m then the torque requirement is Fr = 98 Nm in practice this will require a fairly high reduction ratio gearbox but that's fine as it will keep the speed of the cable feed in a sensible range.

With appropriate gearing this is probably equivalent to about a 250-500 watt motor.

As an extra reality check this electric hoist has a capacity of 125kg with a 500W 240V motor.

The power can be as small as you want. It is just a matter of how slow you find to be acceptable. Disconsidering friction: $$P[W] = m[kg]\cdot g[m/s^2]\cdot v[m/s]$$where
P = required power