Re: Drag force
- From: Ken Freeman <kenfreeman096@xxxxxxxxx>
- Date: Wed, 29 Apr 2009 14:34:59 -0700 (PDT)
On Apr 29, 7:46 am, Stephen Bauman <sbau...@xxxxxxx> wrote:
On Wed, 29 Apr 2009 11:28:19 +0200, Dieter Britz wrote:
It seems to be assumed that the drag force on a cyclist due to the speed
is proportional to velocity squared. Where does this come from? In fluid
dynamics texts (assuming a spherical cyclist {:] ) this is far from
clear.
Force is the time rate of change of momentum. The only mass being moved
is a volume of air, on a level course. So, it's simply calculating the
mass of that air, multiplying it by velocity and dividing it by time to
calculate the force.
Assume the air has a constant density per unit volume: r.
Assume there is no air velocity.
Assume the rider is traveling at a constant velocity: v.
Assume the rider profile presents a cross-sectional area perpendicular to
the direction of motion: A.
Then in time T, the rider will travel a distance vT. This is also the
depth of the volume of air that must be displaced.
So, the volume of the air is A vT. Its mass is r AvT. Is momentum is
rAvTv or rATv^2.
Finally, this is the momentum per unit time:T so the force exerted is:
rAv^2.
One result of this is that the power required to overcome air drag, in
watts of course, is proportional to the cube of the airspeed. This is
the basis of the fact that the resistance to rider propulsion, the
limit of speed, is aerodynamic, not rolling, resistance.
Ken Freeman
.
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