Nomadic tribe route

The position on the surface of the earth is determined by latitude $\phi$ and longitude $\lambda$ . While the circles of longtitude go from pole to pole and have a radius $R$ equal to earth's radius, the radius $R' = R \cos \phi$ of the circle of latitude depends on the latitude $\phi$ and becomes smaller towards the pole. If the nomadic tribe travels along a circle about the angle $\Delta \phi$ or $\Delta \lambda$ , the distance traveled results to $\Delta s = R \Delta \phi$ and $\Delta s = R' \Delta \lambda$ , respectively.

Therefore, for individual sections of the route follows $\begin{aligned} 100 \,\text{km} &= R \cos (\phi_A) \Delta \lambda \\ 100 \,\text{km} &= R \Delta \phi \\ 98 \,\text{km} &= R \cos (\phi_A + \Delta \phi) \Delta \lambda \end{aligned}$ where $\phi_A$ is the latitude of the winter quarter and $\Delta \phi = 100\,\text{km}/6371 \,\text{km}\approx 0.0157 \approx 0.9^\circ$ is the covered latitude. Therefore, $\begin{aligned} \frac{98 \,\text{km}}{100\,\text{km}} = 0.98 &= \frac{\cos(\phi_A + \Delta \phi) }{\cos(\phi_A)} \\ &= \frac{\cos(\phi_A )\cos(\Delta \phi) - \sin(\phi_A )\sin(\Delta \phi)}{\cos(\phi_A)} \\ &\approx \frac{\cos(\phi_A) - \Delta \phi \sin(\phi_A)}{\cos(\phi_A)} \\ &=1 - \Delta \phi \tan(\phi_A) \\ \Rightarrow \qquad \phi_A &\approx \arctan \left(\frac{0.02}{\Delta \phi}\right) \\ &\approx 0.905 \approx 51.9^\circ \end{aligned}$ Here we used the addition theorem for the cosines and the small angle approximations $\cos \Delta \phi \approx 1$ and $\sin \Delta \phi \approx \Delta \phi$ . A more accurate evaluation results $\Delta \phi_A = 51.7^\circ$ .

As the travel at constant longitude is along a geodesic, both cover an angle $\Delta\phi$ such that $\Delta\phi·R=100km$ , so $\Delta\phi\approx 0.015696$

If $\lambda$ is the angle traveled at constant latitude, we know that the radius of the circle of that latitude is $r_{\phi}=R·cos(\phi)$ , and $\lambda·r_{\phi}=100km$ , so we have $\lambda=\frac{100}{R·cos(\phi)}$ .

We also know that $\lambda·r_{\phi+\Delta\phi}=98km$ , which simplifies into $\frac{cos(\phi+\Delta\phi)}{cos(\phi)}=\frac{98}{100}$

7 steps of a dichotomic search for $\phi$ in the range $[0, \pi/2]$ yields an approximate value of $51.7$ .

The latitude, $\phi$ , is the angle above the equator from Earth's center. Thus, we know that an arc traversed at constant longitude (along a meridian) is always of length $R\Delta\phi$ .

The longitude, $\theta$ , is the angle from the prime meridian to other meridians, as measured from Earth's axis. The distance from any point on Earth's surface to its axis is $R \cos \phi$ , and hence the length of any arc traversed at constant latitude (parallel to the equator) is always of length $R\cos\phi \Delta \theta$ .

Let $a = 100\,{\rm km}$ , $b = 98\,{\rm km}$ , and $\phi_0$ be the latitude at $A$ . Then we may immediately write $\Delta\phi = \frac{a}{R}$ , and we may determine $\phi_0$ as follows: $\begin{aligned} \Delta\theta = \frac{a}{\not{R} \cos\phi_0} &= \frac{b}{\not{R} \cos \left( \phi_0 + \tfrac{a}{R} \right)} \\ \cos \left( \phi_0 + \tfrac{a}{R} \right) &= \tfrac{b}{a}\cos\phi_0 \\ \cos\phi_0 \cos \tfrac{a}{R} - \sin\phi_0\sin\tfrac{a}{R} &= \tfrac{b}{a}\cos\phi_0 \\ \cos\phi_0\left( \cos\tfrac{a}{R} - \tfrac{b}{a} \right) &= \sin\phi_0 \sin \tfrac{a}{R} \\ \tan\phi_0 &= \cot\tfrac{a}{R} - \tfrac{b}{a}\csc\tfrac{a}{R} \\ \phi_0 & = \arctan \left( \cot\tfrac{a}{R} - \tfrac{b}{a}\csc\tfrac{a}{R} \right) \\ &\approx 0.9024\,{\rm rad} \approx 51.7041^\circ \end{aligned}$

As the arcs of the circles of latitude trailed in spring and in autumn differ by 2%, their radii will do so as well.

This difference $0.02R \cos \phi$ is one leg of a right-angled triangle, the hypotenuse is the summer trail=100, and the angle is $\phi$ ,

yielding: $\sin \phi = 0.02 R \cos \phi/100,\\ \tan \phi = 0.02*6371/100 = 1.2742\\ \phi = 51.875°$

The metric on the sphere of radius $R$ can be written as $dS^2=R^2d\theta^2 +R^2\sin^2\theta d\phi^2, \quad \theta \in [0,\pi/2], \quad \phi \in [0,2\pi]$ We have to note that we start counting the angle $\theta$ from the north pole up to the equator. Angle $\phi$ doesn't have a particular point, and we can start counting from wherever we please. The nomads start from $(\theta_0,0)$ in our choice of coordinates. After the first journey they end up at $(\theta_0, \phi_0)$ . The proper distance is given by $100=\int_0^{\phi_0} R\sin\theta_0 d\phi$