C[omp]ute

From: andrew cooke <andrew@...>

Date: Thu, 21 Jan 2021 22:52:02 -0300

Cycling is generally more efficient, per km travelled, than walking.  However,
someone just pointed out to me that if you go fast enough this is not the case
(since losses to air resistance increase rapidly with velocity).

So what velocities are we talking about?

A typical walking speed is 3.5 mph.  That's roughly 1.5 m/s.  Looking at
http://sprott.physics.wisc.edu/technote/walkrun.htm that correponds to roughly
300 W.

So walking burns 300 / 1.5 J/m = 200 J/m.

Now let's consider cycling.  A typical CdA on a road bike is 0.3.  Force is
CdA x p x v^2 where p is air density (1.2kg/m^3).  So F = 0.36 v^2 N.  The
energy required to travel 1m is then 0.36 v^2 J (force x distance).  Setting
this equal to 200 (above) we can find the break-even velocity: 200 = 0.36 v^2
so v = 23 m/s (50 mph).

I may have made a mistake, but that ties in with my intuition, so I'll go with
that for now.

Andrew