Help Design a Human-Powered Helicopter

[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?

 

[Cyrus]

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

No clue - and that's what worries me. If you find anything let me know because I'd be really interested to see how. These things are pretty much the 'magic crystals' and 'healing pyramids' of aerospace, IMO.

 

[Phrak]

No clue - and that's what worries me. If you find anything let me know because I'd be really interested to see how. These things are pretty much the 'magic crystals' and 'healing pyramids' of aerospace, IMO.

I've been scrolling around YouTube for one of the toy demos I once ran into, and can't find one anymore. Apparently I don't know the keywords to use.

 

[Brian_C]

I made an algebraic error. Can you tell I'm not a helicopter engineer? See attached.

I think you would need to perform an iterative calculation to come up with realistic results. For a given body weight of the operator, you have to calculate the minimum rotor diameter, then update the total weight of the operator/vehicle based on the added weight of the rotors. When you plug the updated weight (thrust) into the momentum equations, you end up with an even larger rotor diameter! The results will probably not converge unless you use a super light-weight material.

 

[Cyrus]

I made an algebraic error. Can you tell I'm not a helicopter engineer? See attached.

I can't seem to open this MATLAB file.

You should never present your results in units of acres, but otherwise the diameter is now correct. You are pardoned of your engineering sin... this time.

Edit to your edit: To go to the next level analysis, you need to write a BEMT code. Here you can account for prandtl tip losses, lift/drag for your chosen airfoil section, pitching moments, and ground effect. This is not pretty, and should not be attempted in excel (seriously, don't even think about it).

 

[FredGarvin]

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

That is the standard day air density converted to sl/ft^3. It's right.

 

[Brian_C]

Let's just forget this thread ever happened.

 

[dr dodge]

a badger? did I hear a badger?

dinsdale?

dr

 

[Phrak]

That is the standard day air density converted to sl/ft^3. It's right.

OK. I get .00237 slug/ft^2 for international standard density at sea level. Interesting that the thrust is the 1/3 power of the density, so that the variation in density is not so critical. On a crisp cold morning in Death Valley one could exptect to get about 8 to 9% better lift over standard day air density.

How is the standard day air density obtained?

 

[Phrak]

Here's one for you, Cyrus. Have you done the scaling analysis on this sort of problem?

If the weight of the pilot is doubled, how does the size of the structure increase to obtain the same material stresses. The question akin to this is to obtain the same bending radiuses based on material rigidity. I'm not sure if this one should be compared against doubling the total mass or doubling a length, or what-have-you.

The last I can think of asking is how aerodynamic forces scale with a doubling in size of the airframe. (Should fluid velocity be kept constant or also double for this?)

 

[mugaliens]

All things being equal, and discounting Reynold's numbers as the weight of the pilot is usually substantially less than than of the airframe and powerplant, whatever the overall weight increase of the pilot increases the total MGTOW, the airframe and powerplant would require a similar increase to achieve the same performance (same stall speed, time to climb, etc.)

Example: Your pilot initially weighs 150 lbs, but after feasting for two years arrives at 300 lbs. His old plane's empty weight + useable fuel was 3,000 lbs.

He has money galore, but loves his old plane, so he's commissioning the design and building of a new plane that'll match the old plane's performance characteristics exactly.

Percentage Increase: (3300-3150)/3150 = 4.8% increase in overall weight of the airframe and powerplant. Because weight increases as the cube of an single dimension, the pilot's new aircraft would have to be just 1.69% larger in any dimensional direction to accommodate the pilot's additional weight gain.

Thus, the new total weight of airframe and powerplant would be 3,050.7 lbs.

 

[Cyrus]

The last I can think of asking is how aerodynamic forces scale with a doubling in size of the airframe. (Should fluid velocity be kept constant or also double for this?)

Aerodynamic forces don't change due to airframe size, they depend on the rotor specifications. The "fluid velocity", is termed the rotor inflow, and be calculated (to first order) using the inflow equation.

 

[Cyrus]

All things being equal, and discounting Reynold's numbers as the weight of the pilot is usually substantially less than than of the airframe and powerplant, whatever the overall weight increase of the pilot increases the total MGTOW, the airframe and powerplant would require a similar increase to achieve the same performance (same stall speed, time to climb, etc.)

The Reynolds number does not change with the weight of the pilot (or weight in general), so I'm not sure where you're going with this. Also, the pilot weight here is substantial >%50 of the vehicle weight, so your analysis is not valid in this application.

Percentage Increase: (3300-3150)/3150 = 4.8% increase in overall weight of the airframe and powerplant. Because weight increases as the cube of an single dimension, the pilot's new aircraft would have to be just 1.69% larger in any dimensional direction to accommodate the pilot's additional weight gain.

Thus, the new total weight of airframe and powerplant would be 3,050.7 lbs.

This scaling rule is good for intial insights, but one can simply use the equation I provided to see exactly how the rotor radius changes in response to changes in vehicle weight.

 

[Phrak]

I'll have to reread your posts tomorrow with better consideratioin, Cyrus. But this is the reason I ask: Of the 186 tour de France entrants in one year, their weight averaged 156 pounds. We might take this as the optimum weight for best cyclists. Human flight requires some more consideration, as I'm sure you know. The mass of the pilot and how this scales the weight of the aircraft becomes a factor.

But as a baseline, after some research, the average, midline, World Class, 156 pound cyclist can deliver 449 Watts = 0.603 HP = 331 ft-lb-sec^-1 over a 5+ minute duration.

I would initially presume that HP/Mass_of_pilot is constant.

 

[Phrak]

All things being equal, and discounting Reynold's numbers as the weight of the pilot is usually substantially less than than of the airframe and powerplant, ...

In this case, such as it is, as Cyrus has said, initially consider the pilot and airframe about equal. Maybe start with an initial estimate of the pilot at 140 lb. and airframe at 30% more, and go from there.

 

[Cyrus]

I would initially presume that HP/Mass_of_pilot is constant.

It decays, but the rate of decay would have to be found experimentally for a particular person.

 

[Phrak]

All things being equal, and discounting Reynold's numbers as the weight of the pilot is usually substantially less than than of the airframe and powerplant, whatever the overall weight increase of the pilot increases the total MGTOW, the airframe and powerplant would require a similar increase to achieve the same performance (same stall speed, time to climb, etc.)

Example: Your pilot initially weighs 150 lbs, but after feasting for two years arrives at 300 lbs. His old plane's empty weight + useable fuel was 3,000 lbs.

He has money galore, but loves his old plane, so he's commissioning the design and building of a new plane that'll match the old plane's performance characteristics exactly.

Percentage Increase: (3300-3150)/3150 = 4.8% increase in overall weight of the airframe and powerplant. Because weight increases as the cube of an single dimension, the pilot's new aircraft would have to be just 1.69% larger in any dimensional direction to accommodate the pilot's additional weight gain.

Thus, the new total weight of airframe and powerplant would be 3,050.7 lbs.

OK. You motivate me to do this thing. The simplest is a rescaling of lengths. I prefer doubling. It makes things easier to consider. If materials density is constant then mass increases as 2³, as you've noted.

However, aerodynamic forces, Lift and Drag will increase by the factor 2², from

$$L = k V^2 L^2$$

$$D = k V^2 L^2$$

where L is some typical length. (This will assume, the change in typical length doesn't significantely effect Reynolds number, as you've also noted.)

Aerodynamic moments increase as 2³.

$$M = k V^2 L^3$$

I would initially presume that HP/Mass_of_pilot is constant.

It decays, but the rate of decay would have to be found experimentally for a particular person.

I'm not sure what you mean, but was saying that I would initially assume that over a population of world class cyclists that cycling power is proportional to the mass of the rider over a realistic weight range of, say 120 to 180 pounds.

 

[mugaliens]

Good catch, Phrak. It's why modern larger jets are more efficient in terms of lb-miles traveled per lb of fuel consumed than modern smaller jets.

Given my computations for a light airplane, it won't amount to much at all.

Given the fact you're desiging for a human-powered helo, however, it'll mean a great deal.

I had an idea: Have you considered using solar concentrators, built into the wings, to gather sunlight, piping it down light-tubes to the center where it's used to power a Stirling engine to sping the prop? Or, in the case of your helo, you could have engines mounted about 2/3 of the way out along the rotors, and smaller props to push the rotors around?

Just a thought. I also thought about putting the cyclists out there, as well, if you want to keep the engines fully human.

 

[Phrak]

Good catch, Phrak. It's why modern larger jets are more efficient in terms of lb-miles traveled per lb of fuel consumed than modern smaller jets.

Really, I'm not sure how this works out. Can you give details?


But there is still stress and strain to consider for human powered flight, in general.

Do you recall something called the Square-Cube rule, or square-cube law as applied to the strength of a bone or beam, or even a wing as it scales in length only? The idea is to keep material density unchanged, and the shapes of everything stay the same. It's just scaled up in size. Latently I found that the amazing Wikipedia provides it.

http://en.wikipedia.org/wiki/Square-cube_law"

For Aerodynamic Forces:

"When a physical object maintains the same density and is scaled up, its mass is increased by the cube of the multiplier while its surface area only increases by the square of said multiplier. This would mean that when the larger version of the object is accelerated at the same rate as the original, more pressure would be exerted on the surface of the larger object."

For the Stength of a Beam:

"If an animal were scaled up by a considerable amount, its muscular strength would be severely reduced since the cross section of its muscles would increase by the square of the scaling factor while their mass would increase by the cube of the scaling factor."

Unfortunately Wikipeda doesn't present this in terms of the yield strength of a cantilivered beam (or wing) (or rotor), but we can replace "muscle strength" by "yield stress".

The strength of the beam increases as L² and the mass increases as L³.

I'm still trying to find how rigidity scales.

I had an idea: Have you considered using solar concentrators, built into the wings, to gather sunlight, piping it down light-tubes to the center where it's used to power a Stirling engine to sping the prop? Or, in the case of your helo, you could have engines mounted about 2/3 of the way out along the rotors, and smaller props to push the rotors around?

The challenge in this thread is human powered flight, but check out the NASA Pathfinder.

http://www.nasa.gov/centers/dryden/news/FactSheets/FS-034-DFRC.html"

Under the hot summer midday sun, you might expect one horse power per square yard of solar insulation. The best Solar electric panels are about 15% efficient, I think...

Just a thought. I also thought about putting the cyclists out there, as well, if you want to keep the engines fully human.

Well, I did too, but no one took me seriously. It would significantly reduce the weight-per-pilot requirement of the airframe--by as much as 50%. But there's a catch. There must be at least one crew member that cannot rotate but face the same direction throughout the flight.

(BTW, very cool stuff you had on the crosswind landing thread.)

 

[Cyrus]

What do you mean by "put the pilot out there." The rules clearly state the pilot cannot be rotating.

 

[Phrak]

What do you mean by "put the pilot out there." The rules clearly state the pilot cannot be rotating.

"Cyclists", in plural, Cyrus. Say you have four blades and put the cyclists in the center of each blade. This means that the bending stress on the blades is cut in half. Belatedly, I recall (to first order, see Poison ratio) that the strain, that leads to coning, is also cut in half.

The rules state that at least one member of the crew must face in the same direction throughout the flight. So this ...complicates things.

How would you go about having at least one nonrotating crew member, where the crew, as the rules call them, are dispersed about the blades to save stuctual weight?