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How dense would the atmosphere of a theoretical planet have it be, in relation to Earth's atmosphere, to fly with, say, a 10 to 15 foot wingspan? Math is acceptable and encouraged, and a set of artificial wings is assumed. I'm thinking an incredibly tough (fictional) plant fiber spread over a carbon composite frame. When not in use, it could be folded into a school backpack-size with convenient support straps so that it can actually be WORN like a backpack. It can also be taken completely apart to be stored in a large pack with other equipment. It's got a joint system for flapping, but not with all the minutiae of a real bird wing, so you'd have to angle your arms on the upflap. It will also have footholds for when in flight.

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  • $\begingroup$ ... how fast can you go? With any speed, you can make a human fly with much smaller wings, as seen here: nowiknow.com/wp-content/uploads/2012/05/… $\endgroup$
    – PipperChip
    Commented Mar 5, 2015 at 5:07
  • $\begingroup$ @PipperChip You're expected to use them coming out of a free fall $\endgroup$ Commented Mar 5, 2015 at 12:12
  • $\begingroup$ You might want to check this out: what-if.xkcd.com/30 $\endgroup$
    – KSmarts
    Commented Mar 5, 2015 at 20:36

1 Answer 1

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This is a quick answer: The Lift Equation is:

$$ L = C_L * (1/2) * \rho * V^2 * A $$

Where $L$ is Lift in Newtons, $C_L$ is the lift coefficient of the wing, $\rho$ is the air density in kg/m^3, and $A$ is wing area in m^2.

$\rho$ is ~1.2 kg/m^3 between mean sea level and 2 km MSL on earth as we know it. At 15 ft (~5 m) wingspan and a crude aspect ratio of 3 (which is low, a higher the aspect ratio of 10 would be better), A (area) is about 8 m^2. A human at 60kg is ~600N of lift required. Let C_L be about 1.0, best case, but more likely around 0.6 for this wing, for the forward-flight cruise condition. Second part of this is determining the power required for flight, which is probably 80 Watts max continuous for a typical person. $Power = Force * Velocity$, and in this, power is a constraint, the force and velocity being free variables. The force being provided is to counter drag, which we can let be about 20% of the lift force worst case, ~10% better case.

I'll assume: $$ L = 600N, C_L = 0.6, A = 8 m^2, P = 80 W, D=0.1 * L = 60 N $$ This yields $600 = 0.6 * 0.5 * \rho * V^2 * 8$, or $V = \sqrt{600/(0.6 * 0.5 * \rho * 8)}$ The second equation, $P < F * V$, can have the above spliced into it: $80 = 60 * \sqrt{600/(0.6 * 0.5 * \rho * 8)}$

Solving for rho, we have $\rho > 140 \: kg/m^3$

This is 30 times the density of the atmosphere of Titan at its surface, and 1/7th the density of water. Of course, if you tweak numbers, you can probably get this much lower, say, by putting drag to 5%, assuming C_L at 0.8 to 1.0, and increasing the wing area. You may want the wing span to be a bit higher, say, 8 to 10 meters (26 to 32 feet), which allows for more wing area and increases the aspect ratio, which makes your wing more efficient.

A few things I'm skipping are the Reynolds numbers, Oswald efficiency, and how viscosity and compressibility would work.

Bonus: Here is a website I found a while ago that shows how you could do flapping flight with not-to-distant-future technology: http://www.dcgeorge.com/

References:

http://en.wikipedia.org/wiki/Human_power

http://www.lpl.arizona.edu/~yelle/eprints/Yelle97b.pdf

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  • $\begingroup$ How does your calculated atmospheric pressure compare to earth sea level atmospheric pressure? $\endgroup$
    – Tim B
    Commented Mar 5, 2015 at 9:03
  • $\begingroup$ @TimB I just googled this, Earths air pressure at sea level and 50 C is around 1.225 kg/m cubed... $\endgroup$ Commented Mar 5, 2015 at 12:10
  • $\begingroup$ But question, how do you find the lift coefficient? $\endgroup$ Commented Mar 5, 2015 at 12:13
  • $\begingroup$ "Pressure" and "Density", make sure you get the terms correct! :) . Sorry I left out how to find the Coefficient of Lift C_L. That can be found on something called a "Lift Polar" diagram. Check this site out: airfoiltools.com/index $\endgroup$
    – Steve
    Commented Mar 5, 2015 at 20:39

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