As it’s now July, the Tour de Fiddler is back!
This time, we’ll be looking at a model for a cyclist’s speed $v$ as a function of their pedaling power $P,$ their mass $m,$ and the ground’s angle of inclination $\theta$: $$v = \frac{P}{m \sin \theta + 10}.$$
In cycling, roads are marked with a gradient $g,$ which is a hill’s slope, typically expressed as a percentage. Thus, an incredibly steep 45-degree incline has a gradient of 1, or “100 percent.”
Consider the following two riders:
- A “climber,” who has a power of 300 and a mass of 60
- A “sprinter,” who has a power of 325 and a mass of 80
At what gradient will the climber and sprinter cycle at the same speed?
In this case, since the formula involves the ground's angle of inclination $\theta,$ but we then want an answer in terms of gradient, $g,$ we need to understand how these two quantities relate to one another. In particular, we see that $g = \tan \theta,$ or equivalently $\theta = \tan^{-1} g,$ but more on this later.
In general if $v_c$ and $v_s$ are the climber's and sprinter's velocities, then they will be equal when $$\frac{P_c}{m_c \sin \theta + 10} = \frac{P_s}{m_s \sin \theta + 10},$$ or equivalently when $$P_c(m_s \sin \theta + 10) = P_s(m_c \sin \theta + 10)$$ which in turn is equivalent to when $$(P_cm_s - P_sm_c) \sin \theta = 10 (P_s - P_c),$$ or when $$\sin \theta^* = \frac{ 10 (P_s - P_c) }{ P_cm_s - P_s m_c }.$$ In particular, when we have $P_c = 300,$ $m_c = 60,$ $P_s = 325,$ and $m_c = 80,$ then we have $$\sin \theta^* = \frac{10 \cdot ( 325 - 300 )}{ 300 \cdot 80 - 325 \cdot 60 } = \frac{250}{4500} = \frac{1}{18},$$ or equivalently $\theta^* = \sin^{-1} \frac{1}{18}.$
Since we have $$\tan (\sin^{-1} u) = \frac{u}{\sqrt{1-u^2}},$$ for any $u \in [-1,1],$ we see that the gradient at which the climber and sprinter will have the same speed is $$g^* = \tan \left( \sin^{-1} \frac{1}{18} \right) = \frac{ \frac{1}{18} }{ \sqrt{ 1 - \left( \frac{1}{18} \right)^2 } } = \frac{1}{\sqrt{18^2 -1}} = \frac{\sqrt{323}}{323} \approx 5.56414884\dots\%.$$
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