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My original goal was to investigate the relationship between voltage and lift force in toy helicopters, however I've been advised that it will be easier to find help with the relationship between power or current with lift force rather than voltage. Really anything will help me.

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All the maths is in this reply, but I'll try to explain the theory:

https://aviation.stackexchange.com/questions/93859/could-ducted-propellers-improve-efficiency-in-small-drones/93866#93866

Lift is calculated as the same as the force required to accelerate air downwards. Every second the motor produces a certain amount of energy, which is converted into kinetic energy in the air moving downwards through the rotor disk. You don't know the speed of the air, but if it's 10m/s, then a 10 meter cylinder of air is accelerated from zero to 10m/s every second.

You know the disk area, and the density of air, so you can work out the mass of that cylinder of air. The mass and speed give you the kinetic energy, and thus the power.

Then you can use F=MA to work out the force required to accelerate it that much - or you can flip it around to work out what the speed would be for the amount of lift you require.

If you play around with the equations, the speed terms cancel out and you get the conversion from power to lift in the linked question.

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