Tuesday, April 13, 2010

And Boy Are My Arms Tired


I’m traveling until Thursday night.   I should be able to post consistently again once I get back.  In the mean time, try calculating this: How fast would you half to flap your arms to fly?

The physics of this problem is essentially the same as the “Beat It, Mothra” problem, so I’ll use the same formulas.  A person’s wingspan and arm width are about 2.0 m (~6.6 ft) and 10 cm, respectively.  This means the “wings” have a total area of about 0.2 m2.  I’ll assume the arms/wings move 0.30 m (~1.0 ft) down with each thrust, so that each push propels 6.0×10-2 m3 of air downward.  As before, we can compute the total mass of air moved with each thrust,

air mass =  (density) · (volume)
=  (1.2 kg/m3) · (6.0×10-2 m3)
= 72 g.

For a 75 kg person, the force of gravity is

gravitational force = (mass) · (gravitational acceleration)
= (75 kg) · (9.8 m/s2)
= 740 N.

We can solve for the frequency using the equation from the Mothra problem,

frequency = {(gravitational force) / [(air mass) · (range)]}1/2
= {(740 N) / [(72 g) · (0.3 m)]}1/2

= 185 Hz

This mean you’d have to flap your arms up and down at least 185 times each second in order to just get off the ground.  I’ve simplified both this and the Mothra problem considerably.  As Rich pointed out in the Mothra comments,

“…wouldn't there be a force downward when Mothra lifts her wings up? I haven't really studied how moths wings work but I would assume that the upswing at least creates some downward force, requiring a faster wing beat in order to compensate for those downward forces.”

It’s a very good point that we neglected in both problems.  As I understand it, the down stroke (i.e. the one that pushes the air down and the animal up) is done with the wings spread out as much as possible so that the maximum amount of air is pushed down.  The up stroke occurs closer to the body so that less air is propelled.  There are reasonably good videos of these dynamics here and here.  If the two strokes were the exactly same, they would cancel each other and there would be no net up force.  In these problems, I’ve assumed the upstroke is negligible.  This assumption may be good for an order of magnitude estimate, but for something more precise you definitely need to include the effect the up stroke. 
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Posted by aaronsantosdotcom at 4:33 PM

3 comments:

  1. ZacharyApril 13, 2010 at 5:33 PM

    For an 80 kg human, w/ arms about 0.75m long, I get around 90 Hz?

    ReplyDelete
  2. aaronsantosdotcomApril 18, 2010 at 1:43 PM

    Yeah, that's about what I get. I suspect it may be appreciably bigger if we include the effect of the upstroke as Rich suggested.

    ReplyDelete
  3. hale-boppApril 19, 2010 at 2:06 PM

    I assumed the force of the arms going up and down would cancel out and the only force you could get would be from your hands (assuming they were flat on the downstroke and you turned them vertical on the upstroke to drastically reduce their area...I igonred the force on the upstroke since it would be small compared to he downward force). My hand is (close to) 10cm x 20cm, yielding only about 1/10th the area you assumed so it increased my answer about a factor of 10 (I assumed a slightly heavier person but slightly more up and down vertical movement) getting around 2,000 Hz.

    I visualized the person with the arms extended for the entire up/down stroke. If they brought the arms close to the body for the upstroke, you could substantially reduce the loss on the upstroke.

    So I guess part of this one depends on how you visualize the person flapping their arms!

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