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On his website http://www.humanbirdwings.net/ the dutch engineer Jarno Smeets claims to have successfully build a set of 17 m^2 bird-like wings from material of a kite. It is claimed that it uses sensors taken from Wii controllers and a smart phone as well as two motors on the back of the "pilot" which amplify the flapping from the "pilots" arms which are connected with robes to the wings.

Apparently this has since been debunked/fessed as a hoax, but I wanted to try a back-of-envelope calculation to see if it was even plausible or not.

The closest thing I had to hand was this formula (pinched from here)

$$P_{total} = P_{drag} + P_{lift} = \frac 1 2 c_d \rho A_p v^3 + \frac 1 2 \frac { (mg)^2 } {(\rho v^2 A_s ) }$$

Where

  • $C_d$: drag coefficient (1.15)
  • $\rho$: density of air (1.3 kg/m3)
  • $A_p$: frontal area of human (1 m2, adding a bit for wing)
  • $v$: speed (5 m/s - optimum from $P_{drag} = P_{lift}$)
  • $m$: mass of man (80kg)
  • $g$: gravitational acceleration (9.8 m/s2)
  • $A_s$: square of wingspan (100 m2)

Plugging all that in gives me 188W, which is about 5 times more than what an average human can produce with their arms (accordingly to this source, only thing I could find).

However, a 1kg lithium-ion battery could apparently (? not sure of my interpretation) contain 150Wh, which could make up the difference.

This makes the claim seem far more feasible than I feel it ought to. Am I missing something?

UPDATE

As @zephyr points out below, I made a mistake at some point when transcribing the formula, the correct one is:

$$P_{total} = P_{drag} + P_{lift} = \frac 1 2 c_d \rho A_p v^3 + \frac 1 2 \frac { (mg)^2 } {(\rho v A_s ) }$$

Plugging the numbers in to that, and optimising $v$ gives me a $P_{total} \approx 630W$, which leaves birdman needing 4kg of batteries... (or, as Jim points out, settling for a shorter flight).

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Where did you get 100m^2 for the wing area? That is enormous. –  user2963 Mar 23 '12 at 15:04
    
also, you have an error in the second term of your formula, and when I apply your numbers I get 566 W, not 188W –  user2963 Mar 23 '12 at 15:17
    
@zephyr, sorry, I should have written 'the square of the wingspan' (see here). Well spotted for the error. Back to the drawing board –  Benjol Mar 24 '12 at 10:42
    
By using Watt-hours and Watts you are in effect assuming you want to fly for an hour. If you are willing to fly for a shorter time, you could in theory use lighter batteries. –  Jim Graber Mar 26 '12 at 6:32
    
@Jim, true, I'll update to suit. –  Benjol Mar 26 '12 at 8:12
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2 Answers 2

Man powered flight is certainly possible, and in fact a man powered aeroplane has crossed the English Channel. See http://en.wikipedia.org/wiki/Gossamer_Albatross for details.

If you're asking about the more conventional "flapping" type of flight, then if it were as efficient as the propellor of the Gossamer Albatross it should be possible; obviously so since the Gossamer Albatross managed the feat. However the musculature of the human chest isn't designed for flapping wings so it's probably not possible to produce the forces required.

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Helicopters are harder than airplanes (can only be kept in the air for 30 seconds before exhaustion), and flapping would be even harder than helicopters. –  endolith Mar 23 '12 at 18:41
    
Wikipedia says 'in still air, required 300W', so same ballpark... –  Benjol Mar 26 '12 at 5:11
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http://articles.latimes.com/1986-05-18/news/mn-20955_1_pterodactyl-flight
A motorized "life-sized" pterodactyl model did fly by flapping a few times before it crashed.
A little bit bigger wing span, a little more powerful motor... who knows? Maybe you also need a much smarter computerized controller. But your conclusion that physics does not prevent this is probably correct. Of course the motor will be a big part of the power story.

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