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Coefficient of drag question

  1. Oct 23, 2008 #1
    It is said that Lance Armstrong can maintain a average power output of 500 watts for a time trial. The additional power (above that needed to fight gravity) is primarily used to fight aerodynamic drag. On a flat time trial Lance is able to maintain an average speed of about 54 km/hr. Almost all of the power he generates on a flat time trial is used to fight aerodynamic drag. The force of aerodynamic drag is given by
    Fdrag=1/2CDPairAv^2 (if any of that is confusing i posted the link to the pic

    The parameter CD is the coefficient of drag. It depends on the shape of the object moving through the air. A value of CD=0.90 is a reasonable estimate for a person on a bike. The parameter A is the cross-sectional area facing the airstream. It depends on how the rider is positioned on the bike. The parameter rair = 1.29 kg/m3 is the mass density of air, and v is the speed through the air. First show that the power needed to fight aerodynamic drag is proportional to v3. Estimate a value for Lance's cross-sectional area A during a flat time trial. Other losses (rolling friction, drive train friction, etc. are relatively small effects.)
    so i am just unsure of why they are wanting to know anything about v^3.
    some stuff that was given to us before this problem is this.
    In 2004, for the first time ever the Tour de France bicycle race organizers set a time trial on the famous climb from Bourg-D'Oisans to the top of L'Alpe d'Huez. The stage was won by Lance Armstrong with a winning time of 39 minutes and 42 seconds. The time trial consisted of 1.5 km of relatively flat road followed by 14.0 km of climbing. The Tour organizers classify the difficulty of climbs using Categories 1 (hardest) through 5 (easiest). The L'Alpe d'Huez climb is rated 'H.C' (Haute Categorie or beyond category) due to its length and steepness. The bottom of the mountain, the 'Pied de cote', is at 740 m above sea level and the top is 1850 m above sea level. Lance covered the first 1.5 km in just over two minutes (2:05).
    that should help some i think

    2. Relevant equations
    well i am pretty sure that power can by found by using Fd/t and since d/t = v. so just cube everything?

    3. The attempt at a solution
    well i am not really sure where to start. i know that for finding A we would just plug everything into the given equation. What i also dont know is if you go to the link and look then it looks like part of the equation says P(air) and we are given r(air), so i dont know if that is a typo or if we need to do something else also.
  2. jcsd
  3. Oct 23, 2008 #2


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    Re: Drag

    Fdrag = 1/2*CD*Pair*A*v2

    But what is power? Force * distance / time? And Distance / time is ...?

    As for the rest just calculate it based on his estimated 500 watts right?
  4. Oct 23, 2008 #3
    Re: Drag

    yeah thanks for the help man.

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