Friction vs speed

Dieter Britz

Some years ago, I read that the force required to push the bike
forward goes up as the cube of the speed. Is this correct, and
if so, why? It can't be wind resistance, as that gives a force
(for a sphere {:] ) that goes up proportionally with the speed.
Can this be friction of the bearings and tyres?

--
Dieter Britz, Kemisk Institut, Aarhus Universitet, Danmark.

S. Anderson

"Dieter Britz" <britz@chem.au.dk> wrote in message
news:c77hg7$e9c$1@news.net.uni-c.dk...
> Some years ago, I read that the force required to push the bike
> forward goes up as the cube of the speed. Is this correct, and
> if so, why? It can't be wind resistance, as that gives a force
> (for a sphere {:] ) that goes up proportionally with the speed.
> Can this be friction of the bearings and tyres?
>
> --
> Dieter Britz, Kemisk Institut, Aarhus Universitet, Danmark.
>

It is indeed aerodynamic drag that exponentially increases with speed.
Check the drag formula for automobiles out. I believe it is squared or
cubed w.r.t. speed.

Cheers,

Scott..

cheg

"Dieter Britz" <britz@chem.au.dk> wrote in message
news:c77hg7$e9c$1@news.net.uni-c.dk...
> Some years ago, I read that the force required to push the bike
> forward goes up as the cube of the speed. Is this correct, and
> if so, why? It can't be wind resistance, as that gives a force
> (for a sphere {:] ) that goes up proportionally with the speed.
> Can this be friction of the bearings and tyres?
>
> --
> Dieter Britz, Kemisk Institut, Aarhus Universitet, Danmark.
>

Fluid drag force is proportional to the square of velocity:

F=p/2*Cd*A*V^2

p= density of fluid
Cd=drag coefficient of object
A=projected area in direction of motion
V=velocity

Rik O'Shea

Dieter Britz <britz@chem.au.dk> wrote in message news:<c77hg7$e9c$1@news.net.uni-c.dk>...
> Some years ago, I read that the force required to push the bike
> forward goes up as the cube of the speed. Is this correct, and
> if so, why? It can't be wind resistance, as that gives a force
> (for a sphere {:] ) that goes up proportionally with the speed.
> Can this be friction of the bearings and tyres?

The force required is proportional to the square of velocity. The
power required is proportional to the cube of velocity (power = force
X velocity).

So the faster you cycle on level ground the more power is required to
overcome air resistance.

The next most important force required (assuming level ground) is
rolling resistance - this is related to the hub bearings, tyres and
road surface (mainly the latter two; in other words normally adjusted
hub bearing require a trivial amount of power to overcome their
resistance).

The next force required is to overcome the drivetrain - this is the
chain and derailleur.

As an example consider Chis Boardman's UCI legal world record. On a
smooth indoor velodrome using the highest qaulity tubs available on
single speed bike (without derailleur) over 92% of his power output is
required to overcome air resistance when travelling @ 49.442 km/h.

*** "Chris Boardman (GBR) Athlete's Hour Record, Manchester, 2000"
***

Input Parameters [Metric Format]
--------------------------------
Cyclist Velocity [km/h] ................ 49.442

Power Total [W] ........................ 401.2
Power Air Resistance [W] ............... 372.8 92.91%
Power Rolling Resistance [W] ........... 20.4 5.09%
Power Drive Train [W] .................. 8.0 2.00%

David Kerber

In article <c77hg7$e9c$1@news.net.uni-c.dk>, britz@chem.au.dk says...
> Some years ago, I read that the force required to push the bike
> forward goes up as the cube of the speed. Is this correct, and
> if so, why? It can't be wind resistance, as that gives a force
> (for a sphere {:] ) that goes up proportionally with the speed.
> Can this be friction of the bearings and tyres?

For fluid (air/water) resistance, the _force_ required goes up with the
square of the speed, and the _power_ requirements therefore go up with
the cube of the speed (because power is proportional to force times
speed).

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