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OWAO 2022, Практически тур — Задача 1. Stone Ecliptic

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Всички материали от OWAO 2022

2022 ◇ Practical Round ◇ Problems

Условие

You are given the map of Calakmul, a Maya archaeological site in the Mexican state of Campeche. On the main square (Gran Plaza) there are several buildings; in particular, building VI (Estructura VI) and building IV (Estructura IV). The latter is the complex of three temple towers.

It is known that if one looks from building VI to building IV, on the days of equinoxes and solstices, the rising Sun touches the tops of the respective towers (marked). The latitude of Calakmul is $18^{\circ}$ North.

Map of the Calakmul archaeological site (INAH, Calakmul) on a green background with a north arrow at the top right; labelled structures include Gran Acrópolis, Estructura VI, Estructura IV, Gran Plaza, Estructura II, Estructura I, Pequeña Acrópolis and others; dots mark the observation point on building VI and the towers of building IV; a human-scale figure (Escala humana en la Estela 43) at the bottom left and a scale bar at the bottom right.

Estimate the height of the towers relative to the observation point. Consider the Sun as a point source.

Please find calakmul.jpg to zoom in.

Отговори

Решение

The same map of Calakmul as in the statement, reprinted in the solutions: structures of the Gran Plaza with Estructura VI and Estructura IV, marked points, north arrow and scale bar.

The middle tower corresponds to the position of the Sun on the equinoxes. However, it can be seen that it is not located exactly east of the observation point, but approximately $14^{\circ}$ south. This is because the Sun is observed not at the moment of crossing the horizon, but at a certain height corresponding to the height of the tower.

Let us find the height of the Sun at the moment of “rise” above the tower. From the parallactic triangle (taking into account $\delta = 0^{\circ}$ ):

\sin\delta = \sin\varphi\sin h - \cos\varphi\cos h\cos A,

\tan h = \frac{\cos A}{\tan\varphi} \quad\Rightarrow\quad h = 36.5^{\circ}.

The distance from the observation point to the tower is $L = 80\ \mathrm{m}$ . The height of the tower

H = L\tan h = 80\ \mathrm{m}\cdot\tan 36.5^{\circ} = 59\ \mathrm{m}.

The difference in azimuths between the middle and side towers is approximately $23^{\circ}$ – $24^{\circ}$ , which corresponds to the difference in the azimuths of sunrise at the equinoxes and solstices at the latitude of Calakmul. We can consider the towers to be the same height.

Note. “Flat” approximation gives

h = 14^{\circ}\cdot\tan(90^{\circ} - \varphi) = 43^{\circ},

H = L\tan h = 75\ \mathrm{m}.

The answer seems to be somewhat exaggerated: the height of building I, the main building in Calakmul, is estimated at 55 m. However, the calculation results are quite sensitive to the determination of the observation point and azimuth.

Marking Scheme:

Distance $L$ between buildings — 2 pt.
Azimuth $\Delta A$ of the tower — 4 pt.
Altitude $h$ of the Sun — 8 pt. In particular, a flat approximation — 4 pt.
Height $h$ of the tower — 2 pt.
Heights of other towers (explanation or calculation) — 4 pt.

Оригинал в Архива: OWAO_2022_Practical_Problems.pdf · официални решения: OWAO_2022_Practical_Solutions.pdf