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IOAA 2025, Наблюдения, звездни карти (OM) — Задача 3. Know your Time

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Всички материали от IOAA 2025

OM02: Solution and Marking Scheme · 18 т.

Условие

The given sky map (in Mercator projection) shows the constellation of Orion as seen from a certain location X (longitude $\lambda_{\mathrm{X}} = 70^{\circ}\,\mathrm{E}$ ) on 21 March 2025 at 22:00 local time. The point “W” marks the cardinal West point. The altitude and azimuth values are marked on the grid.

Mercator altitude–azimuth grid, altitude 10° to 50° on the left axis and azimuth 220° to 320° along the bottom with 270° marked W; the constellation Orion is drawn with points A, B, C, D, E, F; the horizon is a dark band at the bottom.

(OM03.1) What is the approximate latitude ( $\phi_{\mathrm{X}}$ ) of the location X? [4 т.]

(OM03.2) The grid provided on the Summary Answersheet has the same projection (Mercator), and the angular scale in both azimuth and altitude are identical to that of the grid provided in the question. On this grid draw to scale the constellation of Orion as it will be at another location Y (with latitude $\phi_{\mathrm{Y}} = 40^{\circ}\,\mathrm{S}$ and longitude $\lambda_{\mathrm{Y}} = 50^{\circ}\,\mathrm{W}$ ) on 21 January 2026 at 18:00 local time. An approximate outline of the constellation is enough, with the points A–F being marked clearly. Identify the cardinal point “P” shown on the grid (tick (✓) the appropriate box in the Summary Answersheet). You may make suitable approximations to arrive at your answer.

You may use the following relations between the Hour Angle ( $H$ ), Declination ( $\delta$ ), Altitude ( $a$ ), Azimuth ( $A$ ) and Latitude ( $\phi$ ):

\cos H = \frac{\sin a - \sin\delta \sin\phi}{\cos\delta\cos\phi}

\sin\delta = \sin\phi\sin a + \cos\phi\cos a\cos A

[14 т.]

Отговори

Решение

The question's altitude–azimuth grid repeated at the head of the OM03 solution: altitude 10° to 50°, azimuth 220° to 320° with 270° marked W, Orion drawn with points A–F.

(OM03.1)

Draw the celestial equator (line passing through W and Mintaka ( $\delta$ Ori).
Measure the angle $\phi_{\mathrm{X}}$ between the Celestial Equator and the vertical at $270^{\circ}$ azimuth.

$\boxed{\phi_{\mathrm{X}} = 21^{\circ}\,\mathrm{N}}$ 1.0

Half credit lower limit	Full credit range	Half credit upper limit
$19^{\circ}$	$20^{\circ}$ to $22^{\circ}$	$23^{\circ}$

Drawing the equator	0.5
Equator passing through Mintaka (point C)	1.0
Indication of hemisphere either through “+”, or “N”, or “North”	1.5

(OM03.2)

The star Mintaka (which lies very close to the Celestial Equator, i.e., $\delta \approx 0^{\circ}$ ) has an approximate altitude at X at 22:00 hrs on 21 March $a_{\mathrm{X,21Mar,22:00}} = 25^{\circ}$ from the West horizon.

Therefore its hour angle at 22:00 hrs local time on 21 March is given by

H_{\mathrm{21Mar,22:00}} = \cos^{-1}\left[\frac{\sin 25^{\circ} - \sin 0^{\circ}\sin 20^{\circ}}{\cos 0^{\circ}\cos 20^{\circ}}\right]

= 63.27^{\circ}

Therefore, the hour angle of Mintaka at 22:00 hrs local time on 21 Jan 2026 (306 days later) will be

H_{\mathrm{21Jan,22:00}} = H_{\mathrm{21Mar,22:00}} + 306 \times \frac{360^{\circ}}{365.2564}

= 63.27^{\circ} + 301.60^{\circ} = 364.87^{\circ} \equiv 4.87^{\circ}

The hour angle of Mintaka at 18:00 hrs local time on 21 Jan 2026 will be

H_{\mathrm{21Jan,18:00}} = H_{\mathrm{21Jan,22:00}} - 4 \times \frac{360^{\circ}}{24}

= 4.87^{\circ} - 60^{\circ} = -55.13^{\circ}

The hour angle of an object at a certain local time on a given day will be the same at all locations.

Since the hour angle is negative, it indicates that Mintaka is to the east of the meridian at 18:00 hrs local time.

Therefore, the point P represents the cardinal point East.

The altitude of Mintaka at 18:00 hrs local time on 21 Jan 2026 at the location Y will be

a_{\mathrm{Y,21Jan,18:00}} = \sin^{-1}\left[\cos H_{\mathrm{21Jan,18:00}}\cos\phi_{Y}\right] \qquad \text{since } \delta \approx 0

= \sin^{-1}\left[\cos(-55.13^{\circ})\cos(-40^{\circ})\right]

\approx 26^{\circ}

Now, the location Y lies in the Southern hemisphere. The Celestial Equator lies towards the North (i.e., tilted towards the left of the cardinal point E in the given projection) near the Eastern horizon in the Southern hemisphere.
Further, the latitude of Y is $40^{\circ}\,\mathrm{S}$ , which gives the inclination of the Celestial Equator w.r.t. the vertical at $90^{\circ}$ azimuth.
The position of Mintaka will be nearly on the Celestial Equator at an altitude of $\approx 26^{\circ}$ .

The orientation of Orion can be inferred by realizing that the shield of Orion is ahead of the belt as the sky moves.

Solution for (OM03.2): the answer-sheet grid with altitude 10° to 50°; Orion drawn inverted with points A–F, C at about 26° altitude, and a grey celestial-equator line through C down to the horizon point P, the angle to the vertical marked ϕ = 40°.

“P” marked as “E”	2.0
Angle of line joining Mintaka (C) and cardinal point (P) w.r.t. vertical, between $37^{\circ}$ to $43^{\circ}$	2.0
Angle of line joining Mintaka (C) and cardinal point (P) w.r.t. vertical, between $35^{\circ}$ to $37^{\circ}$ or between $43^{\circ}$ to $45^{\circ}$	1.0
All 6 of A–F points lie to the left of the vertical at $90^{\circ}$ azimuth	2.0
Only 5 among A–F points lie to the left of vertical at $90^{\circ}$ azimuth	1.0
Only 4 among A–F points lie to the left of vertical at $90^{\circ}$ azimuth	0.5
Altitude of Mintaka (point C) between $24^{\circ}$ to $28^{\circ}$	2.0
Altitude of Mintaka (point C) between $22^{\circ}$ to $24^{\circ}$ or between $28^{\circ}$ to $30^{\circ}$	1.0
Shield (point D) at lesser azimuth than belt (point C)	2.0
Legs (points A and B) are south of Celestial Equator	2.0
Scale: Distance of B–F is between 1.8 cm and 2.8 cm	2.0
Scale: Distance of B–F is between 1.3 cm and 1.8 cm or between 2.8 cm and 3.3 cm	1.0

Оригинал в Архива: Observation-SkyMap-Questions.pdf · официални решения: Observation-SkyMap-Solutions.pdf