Cycling Equipment · Public discussion

Friction vs speed

Started by Dieter Britz · · Last activity · 6 posts · 1,009 views

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Cycling Equipment
Published
4 May 2004
Last activity
4 May 2004
Original author
Dieter Britz
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  1. 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.

  2. "Dieter Britz" <[email hidden]> wrote in message
    news:[email hidden]...

    Quoted message said:

    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..

  3. "Dieter Britz" <[email hidden]> wrote in message
    news:[email hidden]...

    Quoted message said:

    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

  4. Dieter Britz <[email hidden]> wrote in message news:<[email hidden]>...

    Quoted message said:

    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%

  5. In article <[email hidden]>, [email hidden] says...

    Quoted message said:

    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).

    --
    Remove the ns_ from if replying by e-mail (but keep posts in the
    newsgroups if possible).

  6. Dieter Britz said:

    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?


    No, it's wrong ;-) The force required to overcome aerodynamic drag
    varies with the SQUARE of the speed and the power required varies with
    the cube.

    Friction is roughly proportional to speed and is negligible compared to
    aerodynamic drag at race speeds.

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