# Is it possible to fly like a bird using semi-motorized wings?

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.

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

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).

• 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