# Wikibooksβ

Problem:

It has been estimated that a Tour-de-France champion could generate a sustained 510 Watts of power while a healthy young human male (HYHM) can generate about 310 Watts of power. The bicycling champion and HYHM are going to race (on bicycles) up a hill with a 6% upgrade, that is five miles long, and the elevation at the top of the hill is 5000 feet. Both rider/bicycle combinations weigh 170 lbs, with frontal area 0.4 ${\displaystyle m^{2}}$ and coefficient of drag 0.9 (values being typical of bicyclists in crouched racing positions). The coefficient of rolling resistance for both bicycles is 0.01. Assume ${\displaystyle \rho }$ is 1.0567 kg/cubic-m. Remember, power equals the product of force and velocity. (1) Who gets to the top first? (2) How much longer does it take the loser to make it to the top?

Solution:

The winner is obvious the Tour-de-France champion, as everything between the two is equal with the exception of power. Since the champion produces more power, he, by default, would win.

We need to find the maximum steady state speed for each racer to compute the differences in arrival time. At the steady state speed we must have:

${\displaystyle F_{t}=R_{a}+R_{rl}+R_{g}\,\!}$

Where:

• ${\displaystyle F_{t}\,\!}$ = Tractive Effort
• ${\displaystyle m\,\!}$ = Vehicle Mass
• ${\displaystyle a\,\!}$ = Acceleration
• ${\displaystyle R_{a}\,\!}$ = Aerodynamic Resistance
• ${\displaystyle R_{rl}\,\!}$ = Rolling Resistance
• ${\displaystyle R_{g}\,\!}$ = Grade Resistance

We also know:

• ${\displaystyle \rho \,\!}$ = 1.0567 kg/cubic-m
• ${\displaystyle W\,\!}$ = 756 N
• ${\displaystyle f_{rl}\,\!}$ = 0.01
• ${\displaystyle G\,\!}$ = 0.06
• ${\displaystyle A\,\!}$ = 0.4 ${\displaystyle m^{2}}$
• ${\displaystyle C_{D}\,\!}$ = 0.9

To estimate available tractive effort we can use the definition of power as time rate of work, P = FV, to get:

${\displaystyle F={\frac {P}{v}}\,\!}$

Substituting in the components of the individual formulas to the general one, we get:

${\displaystyle {\frac {P}{v}}={\frac {\rho }{2}}AC_{D}V^{2}+f_{rl}W+WG\,\!}$

If we compute rolling and grade resistance, we get:

${\displaystyle R_{rl}=f_{rl}W=0.01(756)=7.56\ N\,\!}$

${\displaystyle R_{g}=WG=756(0.06)=45.36\ N\,\!}$

Together, they add up to 52.92 N.

Aerodynamic Resistance can be found to be:

${\displaystyle R_{a}={\frac {1.0567}{2}}(0.4)(0.9)v^{2}=0.19v^{2}\,\!}$

With everything substituted into the general formula, the end result is the following formulation:

${\displaystyle v={\frac {P}{[0.19v^{2}+52.92]}}\,\!}$

This problem can be solved iteratively (setting a default value for v and then computing through iterations) or graphically. Either way, when plugging in 510 watts for the Tour-de-France champion, the resulting velocity is 7.88 meters/second. Similarly, when plugging in 310 watts for the HYHM, the resulting velocity is 5.32 meters/second.

The hill is five miles in length, which translates to 8.123 kilometers, 8,123 meters. It will take the champion 1030 seconds (or 17.1 minutes) to complete this link, whereas the HYHM will take 1527 seconds (or 25.4 minutes). The resulting difference is 8.3 minutes.