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Multi-Rotors Discuss your quad copters and other multi engined choppers here.

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Old 11-11-2012, 10:49 PM   #1
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Question How many watts per pound does a multirotor require?

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.

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Old 12-04-2012, 11:14 PM   #2
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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?
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Old 12-05-2012, 05:10 PM   #3
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Your figures come out at ~75 watts per pound.
That is almost exactly what I was told on another forum.
Thank you much

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Old 12-12-2012, 03:49 AM   #4
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Originally Posted by kevinbuckley70 View Post
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.
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Old 12-12-2012, 05:33 PM   #5
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Originally Posted by kevinbuckley70 View Post
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.
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Old 12-14-2012, 01:00 AM   #6
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Originally Posted by MustangMan View Post
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
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Old 04-13-2015, 05:46 PM   #7
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Originally Posted by kevinbuckley70 View Post
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

...
I don't agree (i know it's a bit late)

You're considering the falling speed after 1 second... But if you'd have considered it after 2 sec, the result would be 192.47m joules... So, for 1 sec, it would be 192.4/2 = 96.2m joules... And 96.2 not equal to 48.1...

Your method only says how much energy is needed to stop the falling motion after a given time of fall.
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Old 04-13-2015, 06:31 PM   #8
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Originally Posted by comode View Post
I don't agree (i know it's a bit late)

You're considering the falling speed after 1 second... But if you'd have considered it after 2 sec, the result would be 192.47m joules... So, for 1 sec, it would be 192.4/2 = 96.2m joules... And 96.2 not equal to 48.1...

Your method only says how much energy is needed to stop the falling motion after a given time of fall.
But the answer is in Watts, a watt is energy per second, so acceleration is also per second... That bit of the calc is correct.

What is incorrect is that the object (the quad) is not actually accelerating. Gravitational acceleration multiplied by mass gives a downward force, otherwise known as weight. The heli only needs to produce an opposing force equal to weight to hover.
Force does not require any work to be done. Think of the quad resting on a table. The legs of the table are producing an upward force equal to the weight of the quad (plus the table top) yet the legs of the table don't need to expend any energy operate. The reason the table doesn't do any work when it supports the quad is because: work = force x distance moved. As the object doesn't move no work is done.

Fundamentally a hovering object is similar, it does not according to the most basic 'work = force x distance' formula require any energy to stay airborne. The only reason hovering does in reality expend energy is because you move air, so the movement of the air requires work to be done.

The amount of work this needs is very complex but is related to how big the props are. With bigger props the work needed to hover is less which explains why helicopters have huge 'props'.

As an illustration of what wrong with the formula in post #2 take a human power helicopter.

Post #2 says that to hover a 1Kg multirotor with perfect 100% efficiency in the power chain would burn 48.1 Watts

So if we look at a human power helicopter we can find that that with 'pilot' it weighs in at 128Kg. So if post #2 was correct to fly (even with 100% efficiency) the human power heli would need : 128Kg x 48.1W = 6157 Watts. Allowing for less than perfect efficiency that will be over closer to 10,000 Watts

Even the fittest and most powerful cyclist can only sustain around 500W... yet the human power heli can fly!... The formula is fundamentally wrong!
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Old 04-13-2015, 07:03 PM   #9
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I agree with everything jetplaneFlyer. But i don't think there's anything wrong with what i said...

The fact is that the result of the formula will vary when brought back to 1 second depending on how long the fall last. My point was that the guy who wrote that did it because he mingled "number one" and "unit of measurement", not because the speed is a unit of distance per time.

edit : got it... I said "Your method only says how much energy is needed to stop the falling motion after a given time of fall." I'm not sure it make any sense. My bad.
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