I'm interested in the formula for how much power a handcrank motor can generate.

I dont know much about electrical engineering but I'm assuming it's related to a few factor.

  1. The speed on which you are cranking
  2. The amount of copper wrapper

But what I'm confused on is the exact formula. E.g.

What would generate more power more wrapping of the coils or thicker wire? Is it based on total mass?

  • \$\begingroup\$ How hard do you want it to be to turn? Are you charging a battery, or directly running some kind of device? AC or DC load? \$\endgroup\$ – Ron Beyer Dec 28 '19 at 21:02
  • \$\begingroup\$ Probably charging a battery so the power can be built up \$\endgroup\$ – johnny 5 Dec 28 '19 at 21:03
  • \$\begingroup\$ What do you mean how hard, are you saying the more resistance the more power right \$\endgroup\$ – johnny 5 Dec 28 '19 at 21:04
  • 1
    \$\begingroup\$ Always start with the fist law of thermodynamics: you don't get energy for nothing. This immediately falsifies "The amount of copper wrapper" as it would imply that you start with bit more copper and in lenght of days get more energy out. Which is obvious not possible. \$\endgroup\$ – Oldfart Dec 28 '19 at 21:24
  • 3
    \$\begingroup\$ more copper may reduce the losses, that is. make it more efficient, but the electrical power output is still dominated by your mechanical power input \$\endgroup\$ – Neil_UK Dec 28 '19 at 21:37

What would generate more power more wrapping of the coils or thicker wire? Is it based on total mass?

In general the more 'copper' (wire) and/or 'iron' (magnetic material) a generator has the more power it can produce. But deriving an exact formula is much more difficult than just weighing it.

The more efficient the generator is the more power it can produce without overheating, plus being more efficient makes it easier to crank. Lower efficiency means more heat, which may require a heatsink or fan which takes up room that could have been used by other parts.

Thicker wire reduces the resistance that causes loss due to current flowing through the coils, but makes the windings larger which increases the size of the generator. The more turns you have the more voltage it can produce without cranking too fast, but this also increases size unless you use thinner wire, which has higher resistance.

You can reduce the number of turns required by winding the wire around an iron core that concentrates the magnetic field. But this takes up room that could be used for copper, and introduces hysteresis and eddy current losses that increase at higher speed. Thin laminations of exotic grain-oriented silicon steel reduce core loss, but are expensive and difficult to manufacture.

You could also use larger, stronger magnets, and try to get them closer to the windings. But there's a limit to how close you can get them without touching, and core losses also increase. Large high strength magnets aren't cheap either.

So you must find a balance that produces the best result at the speed and size you want. The calculations required for accurate estimation of stator core size and shape, copper fill, magnet placement etc. are not easy because they involve 3-dimensional analysis of magnetic fields interacting with the various components. This is commonly done with a program called FEMM (Finite Element Method Magnetics).

If you can't handle that then just try to get the thickest wire and strongest magnets you can you can into it, with the number of turns needed to get the voltage you want.

  • \$\begingroup\$ Thanks, is there a specific formula like x thick copper wound y times around an iron core with dimensions spun at this velocity will produce this much power in a perfect world? \$\endgroup\$ – johnny 5 Dec 28 '19 at 23:05
  • \$\begingroup\$ Simple formulas that need parameters such as core dimensions are problematic because without finite element element analysis they are unlikely to be accurate. However if you have test results for a particular design then you can apply formulas to change some parameters. eg. voltage is proportional to speed and number of turns, so if you want to double the voltage or halve the rpm then just double the number of turns. Wire diameter can be (roughly) calculated from cross-sectional area etc. But the easiest way to determine power output might be to just build the generator and measure it. \$\endgroup\$ – Bruce Abbott Dec 29 '19 at 1:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.