<p>In 2017 International Astronomy Olympics, there is a question to calculate moon's shadow speed during solar eclipse:</p>
<p><a href="https://astroarena.github.io/astroarena/olympiads/iao/2017/theory/TH_solution_2017.pdf" rel="nofollow noreferrer">https://astroarena.github.io/astroarena/olympiads/iao/2017/theory/TH_solution_2017.pdf</a></p>
<p>The core part is:</p>
<hr />
<p>Assuming here the pass of the eclipse was going along the geographical parallel (parellel to East-West latitude line)</p>
<p>we know the speed of the Moon in its synodic movement around the Earth is:</p>
<p><span class="math-container">$$
v_1 = 2\pi \times 384400\,\mathrm{km} / 29.53\,\mathrm{d}
$$</span></p>
<p>and we know the azimuth (south to east) <span class="math-container">$\chi$</span> and altitude <span class="math-container">$h$</span> of the Sun (which is also the Moon)</p>
<p><span class="math-container">$v_1$</span> is the speed of the Moon's shadow on perpendicular to the Sun direction unmoving surface. To take into account the inclination during the eclipse to the perpendicular unmoving surface, the speed is</p>
<p><span class="math-container">$$
v_2 = \frac{v_1}{\sin(\arctan(\tan{h}/\sin{\chi}))}
$$</span></p>
<p>Last, minus the speed due to sidereal rotation of the Earth at latitude <span class="math-container">$\phi$</span>:</p>
<p><span class="math-container">$$
v_3 = v_e\cos\phi = (2\pi \times 6378\,\mathrm{km} / 24\,\mathrm{h}) \times \cos\phi
$$</span></p>
<p>so the moon's shadow speed during solar eclipse is:</p>
<p><span class="math-container">$$
v = v_2 - v_3 = \frac{v_1}{\sin(\arctan(\tan{h}/\sin{\chi}))} - v_e\cos\phi
$$</span></p>
<h3>MY QUESTION</h3>
<p>CAN ANYONE EXPLAIN THE ANGLE:</p>
<p><span class="math-container">$$\theta = \arctan(\tan{h}/\sin{\chi})$$</span></p>
<p>I have struggled for 2 days and I cannot understand how to get this angle.</p>

CC BY-SA 4.0
2024-10-28T13:19:56.190Z
That looks like it comes from a spherical right triangle. See https://en.wikipedia.org/wiki/Spherical_trigonometry#Napier's_rules_for_right_spherical_triangles "sine of the middle part = the product of the tangents of the adjacent parts"
PM 2Ring
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2024-10-28T13:20:44.753Z
OK I will check R4 and R5
Firestar-Reimu
CC BY-SA 4.0
2024-10-28T13:28:58.590Z
if it is on a spherical right triangle, $h$ and $\chi$ are both right arcs (not angles)
Firestar-Reimu
CC BY-SA 4.0
2024-10-30T03:00:33.427Z
Perhaps it is some sort of projection of lunar velocity?
Polaris5744
<solar-system><solar-eclipse>
How to calculate moon's shadow speed during solar eclipse
Firestar-Reimu
# How to calculate moon's shadow speed during solar eclipse

(Asked by: username_0 on 2024-10-28)

In 2017 International Astronomy Olympics, there is a question to calculate moon's shadow speed during solar eclipse:

<https://astroarena.github.io/astroarena/olympiads/iao/2017/theory/TH_solution_2017.pdf>

The core part is:

---

Assuming here the pass of the eclipse was going along the geographical parallel (parellel to East-West latitude line)

we know the speed of the Moon in its synodic movement around the Earth is:

$$
v\_1 = 2\pi \times 384400\,\mathrm{km} / 29.53\,\mathrm{d}
$$

and we know the azimuth (south to east) $\chi$ and altitude $h$ of the Sun (which is also the Moon)

$v\_1$ is the speed of the Moon's shadow on perpendicular to the Sun direction unmoving surface. To take into account the inclination during the eclipse to the perpendicular unmoving surface, the speed is

$$
v\_2 = \frac{v\_1}{\sin(\arctan(\tan{h}/\sin{\chi}))}
$$

Last, minus the speed due to sidereal rotation of the Earth at latitude $\phi$:

$$
v\_3 = v\_e\cos\phi = (2\pi \times 6378\,\mathrm{km} / 24\,\mathrm{h}) \times \cos\phi
$$

so the moon's shadow speed during solar eclipse is:

$$
v = v\_2 - v\_3 = \frac{v\_1}{\sin(\arctan(\tan{h}/\sin{\chi}))} - v\_e\cos\phi
$$

### MY QUESTION

CAN ANYONE EXPLAIN THE ANGLE:

$$\theta = \arctan(\tan{h}/\sin{\chi})$$

I have struggled for 2 days and I cannot understand how to get this angle.

**Tags:** solar-system, solar-eclipse
> **Comments:**
> **username_1**: That looks like it comes from a spherical right triangle. See https://en.wikipedia.org/wiki/Spherical_trigonometry#Napier's_rules_for_right_spherical_triangles "sine of the middle part = the product of the tangents of the adjacent parts"  
> **username_0**: OK I will check R4 and R5  
> **username_0**: if it is on a spherical right triangle, $h$ and $\chi$ are both right arcs (not angles)  
> **username_2**: Perhaps it is some sort of projection of lunar velocity?  

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## 0 answers

