air friction

(I’ve taken the unicycling people off the followups, I don’t think they’re
into speed?)

Ramsn Fuentes (ramon.survey@arconet.es) wrote:
i: I’m studing the influence of the wind in the air friction while cycling.[/i]

So far so good…

i: When air is in calm the power you must do is: 0.0041Pb/TSa*V^3[/i]

In what units?

i: where Pb is barometric pression, T is temperature, Sa is the body area and V[/i]
i: is velocity.[/i]

T is doubtless the absolute temperature, add 273 to the temperature in celcius,
Sa is the frontal area of the object (cycle and cyclist), there doesn’t appear
to be any coefficient of drag included unless it is in the constant, 0.0041 and
V is…

i: My problem is how can I know this power when there is air motion.[/i]

The velocity of the object (cyclist etc) through the air… So if you do 10m/s
into a 5m/s headwind then V=15m/s. 10m/s with 5m/s tailwind, V=5m/s, and 10m/s
with a 5m/s side wind, all bets are off, the equation is unlikely to be
meaningful.

btw, very few people can produce more than a kilowatt for a short burst, or
about 200w for a prolonged period, and 40w is a more realistic amount to expect
to get from someone for more than a few minutes.


Graham. zebedee@union4.su.swin.edu.au

B’Doing!

I’m studing the influence of the wind in the air friction while cycling.

When air is in calm the power you must do is: 0.0041Pb/TSa*V^3

where Pb is barometric pression, T is temperature, Sa is the body area and V
is velocity.

My problem is how can I know this power when there is air motion.

thanks.

Re: air friction

> .0041Pb/TSa*(V+Vw)^3
>
> If you’re luck enough to get a tail wind, then subtract Vw from V.

It’s not that simple in practice. An orthogonal crosswind slows me down nearly
as much as a pure headwind. Turbulence/local wind effects are very important.
Time for experimentation!

Paul Smith