<p>Meeus gives the equation for the angle between the ecliptic and the horizon in Astronomical Algorithms, eq 14.3.
<span class="math-container">$$
\cos I = \cos \epsilon \sin \phi - \sin \epsilon \cos \phi \sin \theta
$$</span></p>
<p><span class="math-container">$ \epsilon $</span> is the obliquity of the ecliptic (e.g. 23.4°), <span class="math-container">$\phi$</span> is latitude, <span class="math-container">$\theta$</span> is the Local Sidereal Time.  And <span class="math-container">$I$</span> is the angle between the ecliptic and horizon.</p>

CC BY-SA 4.0
2024-05-11T10:33:27.120Z
could you please show the derivation of the same, can we make a spherical triangle with the specified coordinates, if so could you please show that triangle as well?
Celestia Novalis
CC BY-SA 4.0
2022-11-25T15:15:00.633Z
2022-11-25T15:15:00.633Z
Greg Miller
<p>For an observer in a city with latitude 35°, at what sidereal time is the angle that the ecliptic makes with the horizon equal to the angle that the equator makes with the horizon? What is the angular distance of these two contact points at this time?</p>
<p>I could calculate the max and min angles between horizon and ecliptic. I get that the horizon-ecliptic angle is going to be (90-35) degrees here. But I just don't understand how the angles between horizon and ecliptic change. Is there a formula for that?</p>

<observational-astronomy><celestial-mechanics>
What is the horizon-ecliptic angle formula?
Astro enthusiast
# What is the horizon-ecliptic angle formula?

(Asked by: username_0 on 2022-11-25)

For an observer in a city with latitude 35°, at what sidereal time is the angle that the ecliptic makes with the horizon equal to the angle that the equator makes with the horizon? What is the angular distance of these two contact points at this time?

I could calculate the max and min angles between horizon and ecliptic. I get that the horizon-ecliptic angle is going to be (90-35) degrees here. But I just don't understand how the angles between horizon and ecliptic change. Is there a formula for that?

**Tags:** observational-astronomy, celestial-mechanics

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## 1 answers

### Answer by username_1 on 2022-11-25. Score: 2

Meeus gives the equation for the angle between the ecliptic and the horizon in Astronomical Algorithms, eq 14.3.
$$
\cos I = \cos \epsilon \sin \phi - \sin \epsilon \cos \phi \sin \theta
$$

$ \epsilon $ is the obliquity of the ecliptic (e.g. 23.4°), $\phi$ is latitude, $\theta$ is the Local Sidereal Time. And $I$ is the angle between the ecliptic and horizon.

> **Comments:**
> **username_2**: could you please show the derivation of the same, can we make a spherical triangle with the specified coordinates, if so could you please show that triangle as well?  

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