## 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.