Help Design a Human-Powered Helicopter

[Cyrus]

My mistake. Drag increases as omega squared, but not proportional to the mass.

What do you think of wing tips on the rotors? I didn't seen any on the attempted craft. Would they contribute to adverse to the individual rotors around their axiis? Or result in flutter?

To be formally correct, it increases with the tangential velocity, (r*omega)^2. I thought about wingtips but there was a reason why they were not justified. I can't remember right now, but I'll look up why and post later. Aerodynamically, there is only so much you can do here. In my mind, the key to getting this to work is a very clever structural design that is extremely light weight while meeting the stress requirements. This is much easier said that done. Quad anything means huge weight penalties, but inherent stability. A tip driven rotor, or coaxial means *significant* weight savings, but an unstable monster. Pick your poison.

 

[Phrak]

To be formally correct, it increases with the tangential velocity, r*omega^2. I thought about wingtips but there was a reason why they were not justified. I can't remember right now, but I'll look up why and post later. Aerodynamically, there is only so much you can do here. In my mind, the key to getting this to work is a very clever structural design that is extremely light weight while meeting the stress requirements. This is much easier said that done. Quad anything means huge weight penalties, but inherent stability. A tip driven rotor, or coaxial means *significant* weight savings, but an unstable monster. Pick your poison.

I have concern over the upward bending of each rotor as a result of lift. This is forth order, isn't it? Do you number for this?

 

[Cyrus]

I have concern over the upward bending of each rotor as a result of lift. This is forth order, isn't it? Do you number for this?

It's called coning. All rotors, including regular helicopters, do this.

 

[Phrak]

Sorry to be criptic. All I know is that torsional rigidity of a tube is to the forth power of the radius.

 

[Phrak]

We've gotten out of sync. But now that I know you are talking about a hinged rotor rather than rigid, I think you will have too much coning, won't you? Is there a way to suppress it?

 

[Phrak]

You are under the false premise that they were not made of carbon fiber - they were.]

carbon-carbon

 

[Cyrus]

We've gotten out of sync. But now that I know you are talking about a hinged rotor rather than rigid, I think you will have too much coning, won't you? Is there a way to suppress it?

I didn't say anything specific to a hinged rotor. The rotor will cone no matter what the hub attachment. The only way to minimize this is to increase stiffness, which will inevitably come from a heavier blade - unless you can find a material that is stiff in the direction you need for the same weight (good luck).

 

[Phrak]

I didn't say anything specific to a hinged rotor. The rotor will cone no matter what the hub attachment. The only way to minimize this is to increase stiffness, which will inevitably come from a heavier blade - unless you can find a material that is stiff in the direction you need for the same weight (good luck).

Then we are talking about the same thing, where the limiting factor upon weight considerations in rididity over strength.

Materials to compare are carbon fiber/carbon from mesophase pitch that has higher Youngs modulus compared to carbon fiber derived from the more common polyacrylonitrile--popular for it's strength.

However the carbon fiber/carbon may be prohibitively expensive. It is carbon fiber composite that undergoes a second and sometimes third process of reheating then reintroduction of matrix material.

 

[dr dodge]

"4.1.4 No devices for storing energy either for takeoff or for use in flight shall be permitted. Rotating aerodynamic components, such as rotor blades, used for lift and/or control are exempt from consideration as energy storing devices"

its not against the rules
I can see how the added mass and complexity would start you down the road of diminishing returns
thanks for the explain

dr

 

[rplatter]

Interesting problem.

I would suggest starting with something that already exists, like a gyrocopter.
And yes, I know they require forward momentum to get moving, but they are light and run on low power.
If you could get the blade moving fast enough, you could achieve lift off.
Stability is another factor. Some Gyrocopters use gravity similar to the way a hanglider does. Shifting the weight of the pilot angles the collective.
Height would be controlled by speed of the blade.
Rotation becomes the difficult part here. Both of the blades and of the craft.
Possibly a counter rotating blade unit, or an angled fin projecting into the down draft.
attaching power to the blade unit becomes touchy if you are using a free hanging pilot compartment. maybe a belt mechanism or a universal joint. I would offset the drive shaft from the blade hub so you could use some type of gearing at that point and to minimize the difficulty of construction.

Another approach would be to use laminar air flow and instead of a bunch of blades just have a saucer.
If you spin a smooth plate it will force air out from the center. (Tesla turbine)
shape it in a dome shape and the air going out will be directed down also.
The dome/saucer will also provide structural support to the entire structure.
I would imagine that a fairly soft but strong material could be used, a large mylar sheet or something. With a belt around the outside edge to provice rigidity. Just stretch it tight and smooth.
You still have to deal with the craft rotation, but that may be simple.

 

[Cyrus]

Interesting problem.

I would suggest starting with something that already exists, like a gyrocopter.
And yes, I know they require forward momentum to get moving, but they are light and run on low power.

How is a gyrocopter going to hover?

If you could get the blade moving fast enough, you could achieve lift off.

That's fairly obvious...

Stability is another factor. Some Gyrocopters use gravity similar to the way a hanglider does. Shifting the weight of the pilot angles the collective.
Height would be controlled by speed of the blade. Rotation becomes the difficult part here. Both of the blades and of the craft. Possibly a counter rotating blade unit, or an angled fin projecting into the down draft. attaching power to the blade unit becomes touchy if you are using a free hanging pilot compartment. maybe a belt mechanism or a universal joint. I would offset the drive shaft from the blade hub so you could use some type of gearing at that point and to minimize the difficulty of construction.

Err...okay. Try getting a person to pedal around 1 HP and see if they are also able to shift their body weight around (This isn't going to happen).

Another approach would be to use laminar air flow and instead of a bunch of blades just have a saucer. If you spin a smooth plate it will force air out from the center. (Tesla turbine) shape it in a dome shape and the air going out will be directed down also.
The dome/saucer will also provide structural support to the entire structure.
I would imagine that a fairly soft but strong material could be used, a large mylar sheet or something. With a belt around the outside edge to provice rigidity. Just stretch it tight and smooth. You still have to deal with the craft rotation, but that may be simple.

Errr...okay. I'd like to see some calculations as to why you think this would work.

 

[FredGarvin]

Interesting problem.

I would suggest starting with something that already exists, like a gyrocopter.
And yes, I know they require forward momentum to get moving, but they are light and run on low power.
If you could get the blade moving fast enough, you could achieve lift off.
Stability is another factor. Some Gyrocopters use gravity similar to the way a hanglider does. Shifting the weight of the pilot angles the collective.
Height would be controlled by speed of the blade.
Rotation becomes the difficult part here. Both of the blades and of the craft.
Possibly a counter rotating blade unit, or an angled fin projecting into the down draft.
attaching power to the blade unit becomes touchy if you are using a free hanging pilot compartment. maybe a belt mechanism or a universal joint. I would offset the drive shaft from the blade hub so you could use some type of gearing at that point and to minimize the difficulty of construction.

Another approach would be to use laminar air flow and instead of a bunch of blades just have a saucer.
If you spin a smooth plate it will force air out from the center. (Tesla turbine)
shape it in a dome shape and the air going out will be directed down also.
The dome/saucer will also provide structural support to the entire structure.
I would imagine that a fairly soft but strong material could be used, a large mylar sheet or something. With a belt around the outside edge to provice rigidity. Just stretch it tight and smooth.
You still have to deal with the craft rotation, but that may be simple.

It's not an interesting problem because it is fundamentally flawed. It is a dead issue. Period.

You can throw terms like "laminar flow" or Tesla turbine but I don't think you really have any clue as to what you are talking about. A gyrocopter, while using less power, still requires ORDERS OF MAGNITUDE more power than a single human can provide (and those are olympian athletes). On top of it, they can't hover.

Let's stick to actual engineering discussions and not turn this into a thread that belongs in Skepticism & Debunking.

 

[Phrak]

Interesting problem.

I appreciate your input rplatter. You're thinking far afield, and outside what at first blush appears unworkable. Nothing wrong with that, that I can see. It has a chance of lending inspirational direction.

 

[dr dodge]

I would like to "refine" my previous input.
I am still sticking with the dual rotor assembly. if the inside "stability" rotor has more mass than the outside one it should work as an inertial gyro that should give some added stability to the system. you would drive the outer rotor off the inside one.
as far as the angle of attack problem, how about this. the outer blade is completely flat to start and then, when "critical rotation" is reached you inflate a series of tubes, one at a time. the "tube bladder" takes the flat rotor blade, and gently makes it an air foil. no blade angle changes. Using pressurized internal structural members would add rigidity making the material used potentially thinner than if just stand alone. the hardest part I can see is evenly inflating both rotors evenly to avoid instability. then use compressed air or air water "retro-rockets" for the lateral corrections, drive the compressor off of the inner disc. and use 2 people, that way one person is the "engine" and the second is "second engine" until run up is completed, then they assist for power and pilot the craft. I can see an advantage in 2 persons because it would be very hard to concentrate on control while pedaling your butt off.

and a side note, if the rotors were on the ground you'd get an "extension" of ground effect

you may fire at will...lol

dr

 

[Phrak]

Another approach would be to use laminar air flow and instead of a bunch of blades just have a saucer.
If you spin a smooth plate it will force air out from the center. (Tesla turbine) shape it in a dome shape and the air going out will be directed down also. The dome/saucer will also provide structural support to the entire structure. I would imagine that a fairly soft but strong material could be used, a large mylar sheet or something. With a belt around the outside edge to provice rigidity. Just stretch it tight and smooth. You still have to deal with the craft rotation, but that may be simple.

This may have some traction if it can stay within the bounds of mutable rules.

Though these so-called laminar flow lifting things don't appear to be very efficient, it may be possible to produce an efficient version powered by a central propeller.

The more air one can grab per unit load, the greater efficiency to maintain altitude--less dv/dt is required for each parcel of air. This is accomplished on fixed wing aircraft by increasing the wing span.

If the skirt can be made large and self supporting, or nearly so, it would have a small overhead in weight penalty, and support the total load upon a larger volume of air.

 

[Cyrus]

If the skirt can be made large and self supporting, or nearly so, it would have a small overhead in weight penalty, and support the total load upon a larger volume of air.

With no calculations to show the power requirements, or the estimated weight, this is really unsupported speculation. I would avoid such statements with nothing to back it up. His idea sounds absolutely terrible.

 

[Phrak]

With no calculations to show the power requirements, or the estimated weight, this is really unsupported speculation. I would avoid such statements with nothing to back it up. His idea sounds absolutely terrible.

Yeah, I know, but so is human powered flight.

 

[Brian_C]

Just for giggles, I made a spreadsheet which calculates the minimum required rotor diameter as a function of rotor thrust based on elementary momentum theory (see attached). For a 180 lb man who can output 300W (0.4 hp) of power, the minimum rotor diameter is 2656 ft. The rotor disc area is an incredible 127 acres. Note that the disc loading is measured in lb/acre. Also note that we're ignoring the weight of the vehicle, figure of merit, transmission losses, and a thousand other variables.

Not a very practical idea.

 

[Cyrus]

Just for giggles, I made a spreadsheet which calculates the minimum required rotor diameter as a function of rotor thrust based on elementary momentum theory (see attached). For a 180 lb man who can output 300W (0.4 hp) of power, the minimum rotor diameter is 2656 ft. The rotor disc area is an incredible 127 acres. Note that the disc loading is measured in lb/acre. Also note that we're ignoring the weight of the vehicle, figure of merit, transmission losses, and a thousand other variables.

Not a very practical idea.

This is completely wrong. Go back and check your work. (I've done all the calculations, in much, much more depth than you can possibly imagine -and no, I'm not going to share those results). You are orders of magnitude wrong. To give you an idea: I know the weight down to each rib in the past HPHs. When I see people say "such and such might work and the weight is possible" I simply shake my head at the lack of depth in their study of this topic. It's quite apparent too many people are making bogus claims, and/or doing bogus calculations. This is the second wrong calculation I've seen so far, especially considering I did a calculation for you that showed it was just over 1HP for a 100ft rotor... does a 2656 foot rotor seem reasonable to you? Or a disc area in acres...?? Such a statement would/should get you fired in the real world! If you presented this to my helicopter professor in graduate school he would promptly chew you out.

(If you haven't already noticed, I have a pet-peeve about engineers presenting results without doing sanity checks )


Side: Excel, really? Learn MATLAB.

 

[mheslep]

Such a statement would/should get you fired in the real world!

So will attitude. Maybe not quickly, but it you'll find yourself closer to the door everyday.

Side: Excel, really? Learn MATLAB.

Many very, very talented engineers manage to use spreadsheets, or octave, numerical python, etc, and avoid the basic $1950 plus endless more thousands for toolboxes.

 

[Cyrus]

So will attitude. Maybe not quickly, but it you'll find yourself closer to the door everyday.

That's why I added a , it was meant as a friendly jab.

Many very, very talented engineers manage to use spreadsheets, or octave, numerical python, etc, and avoid the basic $1950 plus endless more thousands for toolboxes.

Spreadsheets...<shudder>

 

[Cyrus]

No, but seriously, go back and rework your numbers BrianC. You should get a rotor radius of 193.2' using momentum theory with 0.4HP HOGE (I think you have a units error somewhere).

Solve for radius:R=\frac{T^{3/2}}{\sqrt{2 \rho \pi } P}

T = 300 lbs (Airplane plus pilot)
\rho = 0.002 slug/ft^3
P = 0.4 HP * 550

 

[FredGarvin]

I don't knock anyone for trying. Especially when it is in a realm where one may not have any particular feel for an expected result. That being said, there is a border where answers just aren't worth considering. If anyone thinks that practicing engineers don't make mathematical mistakes, then someone hasn't been in a real engineering job. That's why we have coworkers and others to bounce figures off of and to check behind us. There is no sane company out there that just takes an engineer's calculations and just runs with them. There should always be some kind of checker.

Spreadsheets are awesome when properly used.

 

[Cyrus]

I don't knock anyone for trying. Especially when it is in a realm where one may not have any particular feel for an expected result. That being said, there is a border where answers just aren't worth considering. If anyone thinks that practicing engineers don't make mathematical mistakes, then someone hasn't been in a real engineering job. That's why we have coworkers and others to bounce figures off of and to check behind us. There is no sane company out there that just takes an engineer's calculations and just runs with them. There should always be some kind of checker.

Spreadsheets are awesome when properly used.

If you want to blow your mind, calculate the shaft torques and see what they come out to...NASTY NASTY NASTY. The low rotor RPM makes the torque go insanely high.

 

[FredGarvin]

Well yeah! Especially for a rotor disk that is half mile in diameter...give or take a few hundred feet...

 

[Phrak]

I did my own sanity check on wikipedia's http://en.wikipedia.org/wiki/Momentum_theory" equation.

P = \sqrt{\frac{T^3}{2\rho A}}

Units[P] = \frac{md^2}{t^3}}

Assuming rho has units of mass per unit volume,

Units\left[ \frac{T^3}{2\rho A} \right]= m^2 d^4/t^6

 

[Cyrus]

I did my own sanity check on wikipedia's http://en.wikipedia.org/wiki/Momentum_theory" equation.

P = \sqrt{\frac{T^3}{2\rho A}}

Units[P] = \frac{md^2}{t^3}}

Assuming rho has units of mass per unit volume,

Units\left[ \frac{T^3}{2\rho A} \right]= m^2 d^4/t^6

How is this a sanity check? There is nothing wrong with the equation for momentum theory...

(BTW, rho is slugs/ft^3)

 

[Phrak]

How is this a sanity check? There is nothing wrong with the equation for momentum theory...

(BTW, rho is slugs/ft^3)

I don't trust anyone's equations I have not derived myself.

I think your .002 value of air density may be off, but I've only visited one web site.

 

[Cyrus]

I don't trust anyone's equations I have not derived myself.

I think your .002 value of air density may be off, but I've only visited one web site.

0.002378 slug/ft^3

BTW: Not trusting others equations is a good thing!

 

[Phrak]

With no calculations to show the power requirements, or the estimated weight, this is really unsupported speculation. I would avoid such statements with nothing to back it up. His idea sounds absolutely terrible.

Do you have any idea how to calculate the lift from one of these laminar air devices; I have no idea how to approach it?