Posted on July 10, 2026
The Tour de Fiddler is back. This time, the model for a cyclist's speed is
\[ v=\frac{P}{m\sin\theta+10}, \]
where \(P\) is pedaling power, \(m\) is mass, and \(\theta\) is the road's angle of inclination.
Cycling roads are marked by their gradient \(g\), which is slope expressed as rise over run. Thus a 45-degree incline has \(g=1\), or a 100 percent gradient.
Consider two riders:
At what gradient will the climber and sprinter cycle at the same speed?
Let \(s=\sin\theta\). Equal speed means
\[ \frac{300}{60s+10}=\frac{325}{80s+10}. \]
Cross-multiplying gives
\[ 300(80s+10)=325(60s+10). \]
So
\[ 24000s+3000=19500s+3250, \]
and therefore
\[ 4500s=250,\qquad s=\frac1{18}. \]
The question asks for the gradient \(g\), not \(\sin\theta\). Since
\[ g=\tan\theta=\frac{\sin\theta}{\sqrt{1-\sin^2\theta}}, \]
we get
\[ g=\frac{1/18}{\sqrt{1-1/18^2}} =\frac{1}{\sqrt{323}}. \]
Answer: \(\displaystyle\boxed{\frac1{\sqrt{323}}}\), about \(0.05564\), or \(5.56\%\).
The climber and the sprinter are racing up a perfectly sinusoidal hill. They go from the base, where the gradient is 0 percent, to the peak, where the gradient is again 0 percent.
For them to reach the top at the same time, what should the maximum gradient of the hill be?
The formula for \(v\) is for speed along the ground, so when the ground is inclined, the same road speed covers less horizontal distance per unit time.
Let \(x\) be horizontal distance and let the hill's gradient be
\[ g(x)=G\sin\left(\frac{\pi x}{L}\right),\qquad 0\le x\le L, \]
where \(G\) is the maximum gradient. This has zero gradient at the base and the peak, and maximum gradient \(G\) halfway up.
The angle satisfies
\[ \sin\theta=\frac{g}{\sqrt{1+g^2}}, \qquad \cos\theta=\frac{1}{\sqrt{1+g^2}}. \]
The cyclist's speed along the ground is \(v\), so their horizontal speed is \(v\cos\theta\). Thus the time to cover a horizontal bit \(dx\) is
\[ dt=\frac{dx}{v\cos\theta} =\frac{\sqrt{1+g^2}}{v}\,dx. \]
Substituting the speed formula, a rider with mass \(m\) and power \(P\) takes total time
\[ T=\frac1P\int_0^L \left(mg(x)+10\sqrt{1+g(x)^2}\right)\,dx. \]
Define
\[ A=\int_0^L g(x)\,dx,\qquad B=\int_0^L \sqrt{1+g(x)^2}\,dx. \]
Then
\[ T_{\text{climber}}=\frac{60A+10B}{300}, \qquad T_{\text{sprinter}}=\frac{80A+10B}{325}. \]
Set these equal:
\[ 325(60A+10B)=300(80A+10B). \]
This simplifies to
\[ 250B=4500A,\qquad \frac BA=18. \]
For the sinusoidal gradient,
\[ A=\int_0^L G\sin\left(\frac{\pi x}{L}\right)\,dx =\frac{2GL}{\pi}. \]
Also, using \(u=\pi x/L\),
\[ B=\frac{L}{\pi}\int_0^\pi \sqrt{1+G^2\sin^2 u}\,du. \]
So the maximum gradient \(G\) must solve
\[ \frac{\int_0^\pi \sqrt{1+G^2\sin^2 u}\,du}{2G}=18. \]
This equation does not simplify to an elementary closed form, but it is a one-variable equation and is easy to solve numerically. Its solution is
\[ G\approx 0.0874330021. \]
Extra credit answer: approximately \(\boxed{0.08743}\), or \(\boxed{8.743\%}\).
Adjust the road gradient or the sinusoidal hill's maximum gradient to compare the riders.