# How to calculate hour angle of a moon?

I am trying to construct a standing stone calendar for my world Jasmi, located at latitude 53.8 degrees South. To do this, I need the hour angle of moonrise and moonset for my world. The problem is that I have no idea how to calculate the hour angle of the Earth's moon, much less the hour angle of an exoplanet's moon from that planet.

Given planetary obliquity 14.92 degrees, local solar time of 24.02 Earth hours, and a lunar orbital period of 32.002 Earth days, how can I find the hour angle h0, or the angle of moonrise and moonset, for each of the four phases from a latitude of 53.8 degrees South? Is this possible or do I need more information?

ETA: To make it easier on all of us, I've made the descending node of my moon occur at new moon and the ascending node occur at full moon.

• I think your missing axial tilt. You're also missing the orbit of the moon and whether it moves retro-grade or pro-grade relative to the planets spin. All the characteristics of the moons orbit matter here. Mar 3 '18 at 1:26
• the axial tilt of the planet is specified at 14.92 degrees, but I will rustle up my moon's orbital parameters! Mar 3 '18 at 1:28
• I just don't know astro-lingo that well. Mar 3 '18 at 1:48
• Have you considered asking this question on Astronomy.SE? You might get a better and more detailed answer there. Mar 3 '18 at 2:18
• I second the suggestion of trying Astronomy.SE first.
– rek
Mar 3 '18 at 2:55

## I'm a geographer, not an astronomer, so there's your grain of salt.

If all you need is the hour angle, then you could just substitute the moon for the sun in this equation you posted here:

How do I calculate solar altitude and hour angle from the surface of another planet?

...But as you've posted here:

https://astronomy.stackexchange.com/questions/25352/how-do-i-calculate-the-hour-angle-of-a-planets-moon-from-a-point-on-that-planet

what you're really after is

a standing stone calendar that'll accurately denote where to look for my planet's moon at its four major phases

And that's not something the hour angle can tell you. The tl;dr is that

1. your full moons will not be in the same places every month;

2. the hour angle seems to only give you you degrees longitude;

3. and you'll also need a predictor of latitude to tell where they are.

The long version is that this is due to a combination of two reasons.

## First of all...

You have exactly 20 lunar months for one solar year (640.04 days around the sun / 32.002 days around the planet). You're going to have exactly twenty new moons, twenty full moons, etc throughout your year. And that's fine.

You also have 768.68804 solar hours per lunar month (24.02 times 32.002).

This is also fine, but that .68804 is important. I'll get back to it near the end.

## Secondly...

More importantly, your moon's orbital inclination is 9.92 degrees off the ecliptic in relation to a planet whose axial tilt is 14.92 degrees (for reference, our moon has an inclination of 5.145 degrees in relation to Earth's 23.5 degree axial tilt). And that's also fine.

So, assuming your planet and moon have the same dimensions as Earth and our Moon, here's a not-to-scale diagram of what your system will look like if we assume your moon reaches its highest latitude in the northern sky on the same day as the northern summer solstice.

And here's a closeup of your planet on that hypothetical day. The upper red line denotes the highest northern latitude that will, from their perspective, see the moon as directly overhead.

Now, let's both flip the planet and reflect the moon's orbital inclination 9.92 degrees below the pink line so that your moon reaches its highest latitude in the southern sky during the southern summer solstice. In this case, it's the lower red line that is the highest southern latitude that will see the moon as directly overhead.

## By these powers combined...

The result of these two things is that your moon is going to create a sinusoidal path, like the ones described here: