Mountain road

Using cylindrical coordinates we can parametrize the road by the angle $\phi$ : $\vec r(\phi) = \left( \begin{array}{c} \rho(\phi) \cos \phi \\ \rho(\phi) \sin \phi \\ z(\phi) \end{array} \right)$ The height $z(\phi)= z(\rho(\phi))$ is determined by the radial distance $\rho$ and has a constant slope on the cone: $\frac{dz}{d\rho} = - \tan \beta$ The road has a pitch angle of $\alpha = \arctan \frac{1}{10} = 5.71^\circ$ , so that $\frac{dz}{ds} = \tan \alpha$ with the distance $ds$ of the road inside the xy-plane $ds = \sqrt{dx^2 + dy^2} = \sqrt{d\rho^2 + \rho^2 d\phi^2 } = -\sqrt{1 + \left(\frac{\rho(\phi)}{\rho'(\phi)}\right)^2 } d\rho$ Therefore, $\begin{aligned} & & \tan \alpha &= \frac{dz}{ds} = \frac{dz}{d\rho} \frac{d\rho}{ds} = \frac{\tan \beta}{\sqrt{1 + \left(\frac{\rho(\phi)}{\rho'(\phi)}\right)^2 }} \\ \Rightarrow & & 1 + \left(\frac{\rho(\phi)}{\rho'(\phi)}\right)^2 &= \left( \frac{\tan \beta}{\tan \alpha} \right)^2 \\ \Rightarrow & & \rho'(\phi) &= - \gamma \rho(\phi)\,,\quad \gamma = \left[ \left( \frac{\tan \beta}{\tan \alpha} \right)^2 - 1 \right]^{-1/2} \approx 0.1759 \end{aligned}$ The solution for the differential equation is the exponential function $\rho(\phi) = r_0 e^{- \gamma \phi}$ with the radius $r_0 = r_1 + \cot (\beta) h = 973\,\text{m}$ of the mountain. The road reaches the summit at an angle $\phi_0$ , so that $\begin{aligned} & & \rho(\phi_0) &= r_0 e^{- \gamma \phi_0} = r_1 \\ \Rightarrow & & \phi_0 &= \frac{1}{\gamma} \ln \frac{r_0}{r_1} \approx 12.553 \approx 4 \pi \end{aligned}$ Therefore, the road makes a total of 2 turns around the mountain.