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Help Design a Human-Powered Helicopter |
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| Dec13-09, 11:35 PM | #120 |
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Help Design a Human-Powered Helicopter |
| Dec13-09, 11:37 PM | #121 |
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| Dec14-09, 12:22 AM | #122 |
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| Dec14-09, 03:59 AM | #123 |
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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. |
| Dec14-09, 04:14 AM | #124 |
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![]() 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, dont even think about it). |
| Dec14-09, 07:20 AM | #125 |
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Recognitions:
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| Dec14-09, 12:57 PM | #126 |
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Let's just forget this thread ever happened.
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| Dec14-09, 01:03 PM | #127 |
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a badger? did I hear a badger?
dinsdale? dr |
| Dec14-09, 07:10 PM | #128 |
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How is the standard day air density obtained? |
| Dec14-09, 07:49 PM | #129 |
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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?) |
| Dec14-09, 10:37 PM | #130 |
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Blog Entries: 1
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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 accomodate the pilot's additional weight gain. Thus, the new total weight of airframe and powerplant would be 3,050.7 lbs. |
| Dec14-09, 10:52 PM | #131 |
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| Dec14-09, 11:01 PM | #132 |
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| Dec14-09, 11:31 PM | #133 |
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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. |
| Dec14-09, 11:49 PM | #134 |
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| Dec15-09, 12:00 AM | #135 |
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| Dec15-09, 08:58 PM | #136 |
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However, aerodynamic forces, Lift and Drag will increase by the factor 22, from [tex]L = k V^2 L^2 [/tex] [tex]D = k V^2 L^2 [/tex] 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. [tex]M = k V^2 L^3[/tex] |
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