How many watts per pound does a multirotor require?

[Wildflyer]

The title pretty well says it.

I would like to know how many watts per pound does it require to hover a multirotor copter.

Then how many W/lb does it take for good flight characteristics ?

I would like to build a lift for a large camera.

[kevinbuckley70]

Doesn't the energy equation look like this:

Hovering a mass 'm' is the same as accelerating a mass 'm' at 9.81m/s/s ('g')

So - after 1 sec - you have increased the speed of 'm' by 9.81 m/s

So - as kinetic energy is E = 1/2 mv^2 - the increase of energy is (9.81^2)m/2 = 96.2m/2 = 48.1m joules

And, as you are hovering, you are doing this every second so the energy consumption to hover (assuming perfect ESC/motor efficiency & a prop which converts ALL the energy into 'lift') is: 48.1m joules/sec

But as we know (!) 1 j/s = 1 Watt so, to hover a 1Kg multirotor with perfect 100% efficiency in the power chain would burn 48.1 Watts.

If you assume 90% motor efficiency, 90% ESC efficiency, 90% from everywhere else & about 40% prop/air efficiency (I'm making this up now!) then that increases the power consumed to hover by 1/0.29 = 3.4 so the power to hover something with an AUW of 1Kg would be 164W.

With a 3s LiPo at 11v, that would equate to a hovering current of 14.9 amps.

Anyone care to confirm or deny that is the correct energy calculation?

[Wildflyer]

Your figures come out at ~75 watts per pound.
That is almost exactly what I was told on another forum.
Thank you much

[Beemerider]

Originally Posted by kevinbuckley70

Doesn't the energy equation look like this:

Hovering a mass 'm' is the same as accelerating a mass 'm' at 9.81m/s/s ('g')

So - after 1 sec - you have increased the speed of 'm' by 9.81 m/s

So - as kinetic energy is E = 1/2 mv^2 - the increase of energy is (9.81^2)m/2 = 96.2m/2 = 48.1m joules

And, as you are hovering, you are doing this every second so the energy consumption to hover (assuming perfect ESC/motor efficiency & a prop which converts ALL the energy into 'lift') is: 48.1m joules/sec

But as we know (!) 1 j/s = 1 Watt so, to hover a 1Kg multirotor with perfect 100% efficiency in the power chain would burn 48.1 Watts.

If you assume 90% motor efficiency, 90% ESC efficiency, 90% from everywhere else & about 40% prop/air efficiency (I'm making this up now!) then that increases the power consumed to hover by 1/0.29 = 3.4 so the power to hover something with an AUW of 1Kg would be 164W.

With a 3s LiPo at 11v, that would equate to a hovering current of 14.9 amps.

Anyone care to confirm or deny that is the correct energy calculation?

You lost me after the first 3 words.

[MustangMan]

Originally Posted by kevinbuckley70

Doesn't the energy equation look like this:

Hovering a mass 'm' is the same as accelerating a mass 'm' at 9.81m/s/s ('g')

So - after 1 sec - you have increased the speed of 'm' by 9.81 m/s

So - as kinetic energy is E = 1/2 mv^2 - the increase of energy is (9.81^2)m/2 = 96.2m/2 = 48.1m joules

And, as you are hovering, you are doing this every second so the energy consumption to hover (assuming perfect ESC/motor efficiency & a prop which converts ALL the energy into 'lift') is: 48.1m joules/sec

But as we know (!) 1 j/s = 1 Watt so, to hover a 1Kg multirotor with perfect 100% efficiency in the power chain would burn 48.1 Watts.

If you assume 90% motor efficiency, 90% ESC efficiency, 90% from everywhere else & about 40% prop/air efficiency (I'm making this up now!) then that increases the power consumed to hover by 1/0.29 = 3.4 so the power to hover something with an AUW of 1Kg would be 164W.

With a 3s LiPo at 11v, that would equate to a hovering current of 14.9 amps.

Anyone care to confirm or deny that is the correct energy calculation?

The physics make perfect sense. Your "made up" efficiency values are pretty reasonable in the real world. It would seem that most 3D airplanes are considerably less efficient than your figures, however, since it generally takes well over 100 w/lb. to hover. That's probably to be expected though since the area of the propeller disk to weight ratio is probably considerably lower than the typical quad.

[kevinbuckley70]

Originally Posted by MustangMan

The physics make perfect sense. Your "made up" efficiency values are pretty reasonable in the real world. It would seem that most 3D airplanes are considerably less efficient than your figures, however, since it generally takes well over 100 w/lb. to hover. That's probably to be expected though since the area of the propeller disk to weight ratio is probably considerably lower than the typical quad.

Are there more accurate ways to calculate the efficiency with which a propeller converts power-in to thrust? I have to admit it is something I have never really understood (I like the concept that: the propeller changes the vector momentum of the air which passes over the blades & thrust is the reaction to that. But how to turn that into a formula!).

Interestingly, I remembered that I had a youtube video of my F450 hovering with the OSD showing: http://www.youtube.com/watch?v=zavtJIt31NA

I happen to know that the AuW of that flight was 1.5Kg. That was close to the limit for that quad & I tore it down shortly afterwards & fitted everything onto a bigger, F550 hex.

So I think the reported 18.1A at 11.4v during the early hovering (& 1.5Kg) equates to 62.5W/lb