# Hovering Drone....W/Kg Calculation.....

Discussion in 'General Electronics Discussion' started by Fish4Fun, Aug 13, 2015.

1. ### Fish4FunSo long, and Thanks for all the Fish!

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Aug 27, 2013
While not strictly electronics, the physics of a "hovering quad-copter" is a modern manifestation of a problem that has plagued me since high school physics.....the definition of work is Force x Distance and since "hovering" by definition is not moving.....there is a tad of a SNAFU calculating the energy requisite to keep it hovering .....Obviously it requires some type of force to cancel the effect of gravity....in a uniform gravitational field we can calculate the Potential Energy of any given mass with PE= 1/2 MGH....but "H" (the distance) is not particularly germain to calculating the power required to hover any particular mass at any given height.....There is certainly a solution because helicopters and quad/hex copters can in fact hover.....and fall like rocks if the power suddenly terminates!.....The solution obviously involves the vector sum of the mass and velocity of the particles acted on by the rotors (which is likely summed up as a normalized density constant with an average acceleration/velocity matching the mass of the copter multiplied by the acceleration due to gravity)....but that doesn't answer the underlying question....how much energy does it require to "hover"....more precisely, ignoring efficiency, what is the theoretical minimum energy per unit mass required to offset a uniform gravitational field? Am I over-thinking this?.......

So...on again, off-again....I have wrestled this conundrum for 3 decades and never really found a satisfactory answer in physics....but with the advent of Drone-Mania a new approach to the problem presented itself.....The DJi3 Drones have specifications that include: Battery Capacity (Voltage x Current Capacity)....average Flight Time and Flight Weight.....some quick and easy arithmetic and I finally found an empirical solution: 140W/Kg.....Obviously this number is only valid for one particular drone variant, and obviously this number includes losses associate with converting chemically stored energy to moving air particles....but it is a "real number"....a starting place if you will.....And one should assume that the actual energy required to hover must be less than the empirically derived figure...but how much less? Is 140W/Kg scalable from 1.2Kg to 1200Kg? I certainly don't know any way of proving that it is or isn't mathematically....

An approach to an empirical solution is to carefully test the efficiency of the motors/drivers/peripherals and then remove the losses from the 140W/Kg.....this would still leave the efficiency of the rotors and the importance of (air) density in the figure...but it should be "closer" to the "pure number"....

But I am hoping someone can just point me to some simple formula (that I am obviously too dumb to find) relating the energy required to balance a mass in a uniform gravitational field.....

**I am aware that there is a "drone flight calculator" on the web...I haven't checked it out...suppose to be really cool....but my inquiry is not drone specific, and I am more interested in how the energy required is calculated than any particular real-world solution....(even though my best/only answer to date is, in fact, derived from a real-world solution, lol)**

Fish

2. ### BobK

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Jan 5, 2010
It takes no energy to maintain position in a gravitational field. For example, when we are standing on the ground, the earth is pushing up on us just enough to balance the gravitational force, but since there is no movement against the force, no energy is expanded.

The question you need to deal with is how can we supply that force in at atmosphere of a specific density. We need to move air around to do this. Theoretically, there is no minimum energy required to do this, it is all a matter of efficiency. So the calculations are not straightforward.

Bob

3. ### Alec_t

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Jul 7, 2015
The background equations in this link give propellor thrust. Air exit velocity from the propellor would be calculated from the propellor diameter and blade pitch and the motor power.

4. ### AnalogKid

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Jun 10, 2015
F = m x a Force equals mass times acceleration. Force is the dead weight of the drone. mass is the mass of air being displaced by the rotors. Acceleration is the acceleration of the air being displaced.
- or -
F = 1/2 m x v^2 Force equals one-half times mass times velocity squared.

In this scenario, another term for force is thrust.

This is what makes an airplane fly. From an old NOVA program on the 50th anniversary of the DC3, we learned that a DC10 "throws down" 5 tons of air per second to maintain level flight. That's the mass part. As the air stream leaves the trailing edge of the wing at a downward angle, the vertical component of that vector is the velocity.

ak

5. ### duke37

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Jan 9, 2011
The energy required will depend on the rotor design. Aircraft produce thrust by throwing air backwards, this is matched by the wings pulling air forwards.
The thrust as @AnalogKid says can be done by throwing a lot of air back at low velocity or a little air back at high velocity however the energy required in the latter case is higher. Passenger jets use fans as large as possible whereas military jets use smaller lightweight fans and a lot of fuel.

Wind turbines have an efficiency of about 25% maximum if I remember rightly.

6. ### AnalogKid

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Jun 10, 2015
EDIT: Twice in post #4 I said thrust instead of lift:

"In this scenario, another term for force is lift."
"As the air stream leaves the trailing edge of the wing at a downward angle, the vertical component of that vector is the lift."

ak

7. ### BGB

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Nov 30, 2014
the way I had imagined it before (very informally) was that you have to provide enough power to accelerate the vehicle upwards at the same speed gravity tries to pull it downwards.

so, to hover, you need about 9.8N for every kg. naively, this would imply about 9.8W per kg (since 1W and 1N are roughly equivalent). if your vehicle weighs 10kg, in this ideal world, it could fly with about 100W.

it doesn't take long to realize though that this is a bit overly optimistic.

for a propeller, the force is split: lift vs drag, naively, if you have an L/D ratio of 4, it becomes 125W. L/D ratio depends a fair bit on the propeller design and airspeed. for example, thin propellers with a shallow pitch are good for high-speed with light loads (high L/D ratio), whereas thicker propellers and steeper pitch are better for increasing thrust but have worse L/D.

in my case, in tests, I had used some fairly thick and steep propellers (for sake of increasing maximum weight), so had an L/D ratio IIRC closer to 2. also I was aiming for a somewhat higher weight limit (around 40kg), which would require closer to 600W (at full load).

now, say, you have a motors that reaches peak efficiency of 70% at about 10A and 100W. assuming this, you could get by with about 858W of input power. but, in the above situation, 600W (output) is a little over 100W for the efficiency point. at 150W output per motor, this drops efficiency to about 55%. actual input power required then becomes around 1.1kW (and about 23A per motor at 12v).

also, the original battery I was going to use (an LiFePO4 starter battery, approx 9Ah) would be able to supply this much power for all of 4 minutes (had idly considered the possibility of putting a small generator on it).

hat tried building a prototype gearbox (with a 14 inch propeller), but it was quickly revealed in tests (via catastrophic failure), and confirmed by looking it up, that the forces I was dealing with are outside what is readily doable with ABS plastic parts (would need something a little more durable). (though, before it failed, it was able to use air-flow to blow a shoe off the floor and back around into the propeller from the top and send it flying across the room).

so, project status: effectively derailed for the time being.

also pretty much out of money for now... (haven't been keeping track, but in general materials for this project and a few other sub-projects is probably somewhere between \$1.5k or \$2k over the past year).

8. ### Fish4FunSo long, and Thanks for all the Fish!

464
105
Aug 27, 2013
Thanks For All The Responses!

I at least feel a little better about having failed to find a "simple formula" .... I guess intuitively I "knew" the answer was in F = 1/2 mV^2.....I just couldn't think of a good way to work it out....If one thinks of a drone in water rather than air it is easy to grasp the importance of "media density" wrt lift/drag/thrust....and it also brings home the importance of the density of the hovering mass wrt the density of the media in which it is to hover....for instance a boat "hovers" on the surface of water with zero energy....(well, unless it is raining and it requires a bilge pump ;-) ).....In a vacuum the mass of the hovering craft itself would have to be reduced over time to cancel the gravitational forces....the length of time any particular mass could hover would be a function of the exit velocity per particle unit mass...(this is assuming the exit vector was kept directly opposing the gravitational field) ....ie, each particle would provide 1/2 (particle mass) * V^2 (exit velocity) in "thrust"...and as long as the sum of the Forces created by accelerating the particles equaled the force of gravity on the remaining mass, the mass would continue to "hover".... hrmmmm....I think I got it...

Note to self: If you happen to find yourself near the event-horizon of a black hole....make certain to have plenty of disposable mass (and some way to accelerate it to near light-speed in the length of the ship ;-) )

In a non-vacuum using Newtonian velocities the entire problem is then reduced to the efficiency of accelerating air molecules....or in accelerating some portion of the total mass (ie like the case of "solid rockets" where the fuel itself provides the mass that is being accelerated)....While there is a "real minimum energy required to hover", it is a relativistic function of mass loss and is not practical for any non-relativistic calculations....that is Relativistic theoretical minimum energy required to hover involve C^2 which is generally half a dozen orders of magnitude greater than practical Newtonian Energies.......Since there are fairly fixed upper limits to the efficiency of propellers in air, once we select the most efficient propellers, the problem is reduced to the Conversion Efficiency of "fuel" to mechanical energy and the energy density of the fuel in terms of Joules/Kg....and, of course, our ability to control the conversion of fuel to thrust in a manner appropriate for hovering....(That is, solid rocket engines may have a much higher energy capacity in terms of Joules/Kg, but they are extremely difficult to control and typically have a relatively brief period of usefulness...)

While a bit off topic, assuming we use F = 1/2mV^2 to balance F = mG to calculate the energy required to hover and we assume the speed of light as the upper limit of V (ie Vmax = C) and we assume I am correct in assuming we should be able to calculate the minimum energy required to "hover" as a function of mass converted to energy (E = mC^2)?......Would this imply that magnetic levitation has an associate loss in mass (perhaps in both the "fixed magnets" and the "Hovering Magnet(s)") ? Or would the case of magnetic levitation be more like a "string" suspending the mass? Does the Third Law apply to magnetic levitation?....or is it somehow circumvented? Things that make you go Hrmmmm....

Again, Thanks ALL for the responses! I am very happy that I **think** I finally have the hovering problem solved

Fish

9. ### AnalogKid

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Jun 10, 2015
Nope, a boat doesn't hover. It is propelled to the surface of the water by gravity, and it requires energy to keep it there.
ak

10. ### BGB

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Nov 30, 2014
though, one doesn't need to supply the energy to keep a boat floating, rather, gravity and the water do so, and the net energy is still roughly 0.

more like two magnets stuck together. a lot of force may be present holding the magnets together, but they are neither being supplied this power, nor does them sticking together use up the power in the magnets.

for some uses, a balloon or airship could be a lot more efficient than a quadrotor. nevermind any energy that goes into making H2 to keep it floating (if one doesn't have the money to buy helium). then you only need fans to fly it around, and maybe a tether line to keep the wind from blowing it away.

11. ### BobK

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Jan 5, 2010
Really? Where does that energy come from? If you believe this, then you must believe in free energy.

Bob

12. ### AnalogKid

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Jun 10, 2015
ak: Nope, a boat doesn't hover. It is propelled to the surface of the water by gravity, and it requires energy to keep it there.

BobK: Really? Where does that energy come from? If you believe this, then you must believe in free energy.

Really? *must*??? Quite a leap of nothing-resembling-semantics-or-logic. Is English your second language?

The energy that keeps the solid separate from a liquid of greater density comes from the gravity field. Without it, the boat would assume a position midway between the surface and the the ocean floor due to pressure from surface tension, assuming there are no ocean currents.

ak

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Dec 18, 2013
This is getting interesting now, carry on!

14. ### Martaine2005

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May 12, 2015
Hi Fish,
It is impossible to measure anything flying
I fly radio controlled helicopters and came 4th in the schluter cup in 94.
The formulas that have been explained or tried to explain mean nothing.
These are ideal conditions.
The real world has wind, temperature and mind set.These cannot be measured.
But you have to do do it all in a perfect environment.

Pitch, thrust or lift only works in that perfect environment
+ 4 for a hover in no wind might get you there.
ADD wind, the pitch ranges change.!!
Martin

15. ### Fish4FunSo long, and Thanks for all the Fish!

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105
Aug 27, 2013
Hey Martin.....Actually I don't even have a quad, lol, I was just using the specifications from a DJi-550 to attempt to get a "real number" for Joules/Kg requisite for hovering....that part went exceptionally well....then I went and calculated the Relativistic answer.....it turns out if I could just figure out a way to accelerate hydrogen atoms to the speed of light and direct them toward the gravitational field, roughly 100mg of them could keep a 1Kg mass hovering in a vacuum for ~29 million years before requiring more hydrogen.....The actual net energy required to "hover" is fairly trivial.....the amount of energy expended to achieve this trivial amount of energy is something all-together-different, LOL.....

Congrats on your flying skills! My experience with RC planes/copters/boats/cars is more along the lines of a proof of the laws of thermodynamics.....only a few seconds with me on the sticks and a great deal of order is replaced by enthropy....

@Boat not requiring energy to float on water.....I think it is fairly safe to say that there is a great deal of energy surrounding any given floating object, but the object itself does not require any energy expenditure to float....plenty of forces acting on it...almost certainly motion involved...but still no active energy output required to keep it floating....

Thanks All!

Fish

Martaine2005 likes this.
16. ### Martaine2005

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May 12, 2015
you are clearly in

17. ### Martaine2005

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May 12, 2015
Oops, I don't know what happened there!!
I was too busy arguing via email with my ex wife.
Let me delete that if I can.

@Fish4Fun I agree that a vacuum would be ideal conditions.
I just wanted to say that there are too many variables to be able to measure anything accurately.
In strong wind the heli can be quite stable in a nose down position and very little pitch.
In gusty wind, the oppsosite is true. It will be all over the place and pitch/throttle will need adjusting constantly.
Also, in cold conditions the engines are leaned out and more RPM required.
The opposite is true in hot weather.
It also depends on how many blades are on the head too.

So any calculations would change on a minute by minute basis.

Martin

18. ### BobK

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Jan 5, 2010

You do not understand the difference between force and energy. The gravitational field applies the force to hold up the boat. Energy is force times distance. Since the boat is not moving in opposition to the force, there is no energy expended. Think about it, if there was energy being extracted from the gravitational field, it would eventually be exhausted, and the boat would no longer stay afloat! Or do you contend that an infinite amount of energy can be extracted from a gravitational field? That, by the way, was why I said you must believe in free energy.

And no, English is my first language. Physics is my second language, as evidenced by my Bachelor of Science degree.

Bob

19. ### Externet

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Aug 24, 2009
Found this, may be of help :
Theoretical lift developed by common helicopters:
300 hp, 30' rotor develops 3,400 pounds of lift
25 hp, 12' rotor develops 347 pounds of lift
2 hp, 6' rotor develops 39 pounds of lift