<p>The full formula for when an object rises or sets is as follows:</p>
<p><em>Zenith distance = 90 degrees + refraction + semi-diameter - horizontal parallax.</em></p>
<ul>
<li>For stars, the semi-diameter and horizontal parallax are insignificant. Only the atmospheric refraction (34 arcminutes is typical) is important.</li>
<li>For the Sun, the semi-diameter (the apparent angular radius) is important. 16 arcminutes is generally chosen for the apparent radius of the sun giving a total zenith distance of 90 degrees 50 arcminutes, or the center of the Sun is 0.8333 degrees below the horizon when it rises or sets.</li>
<li>For the Moon, all three corrections are important. The diameter is approximately the same as the Sun (16 arcminutes), and the horizontal parallax is on the order of 1 degree. (The horizontal parallax is the shift in the position of the Moon as seen from the Earth's surface compared to a theoretical position seen from the center of the Earth. It is important for the Moon because its distance is close in comparison to the Earth's radius.) The net is the center of the Moon is <em>above</em> the horizon when it rises or sets; that is, the zenith distance is less than 90 degrees.</li>
</ul>
<p>The U.S. Naval Observatory website <a href="https://aa.usno.navy.mil/faq/RST_defs" rel="nofollow noreferrer">Rise, Set, and Twilight Definitions</a> gives the following details for moonrise  moonset:</p>
<blockquote>
<p>Here again, a constant of 34 arcminutes (0.5666 degree) is used to account for atmospheric refraction. The Moon's apparent radius varies from 15 to 17 arcminutes and its horizontal parallax varies from 54 to 61 arcminutes. Adding all the terms above together, the center of the Moon at rise or set is geometrically 5 to 10 arcminutes above the observer's &quot;geocentric horizon&quot; - the horizontal plane that passes through the Earth's center, orthogonal to the observer's local vertical.</p>
</blockquote>
<p>The average of 5 and 10 is 7.5 arcminutes = 0.125 degrees.</p>

CC BY-SA 4.0
2024-12-24T04:32:19.133Z
2024-12-24T04:32:19.133Z
JohnHoltz
<p>In formulas for calculating rise &amp; set times for astronomical objects, you often see a list of conventional angles used for various rise/set events.  For example, <a href="https://astronomy.stackexchange.com/users/25729/greg-miller">Greg Miller</a> <a href="https://astronomy.stackexchange.com/questions/51658/math-tool-to-compute-time-of-sunrise-and-sunset?newreg=12d95ed779444b7189924d4255d0b457">wrote in answer to a previous question</a>:</p>
<blockquote>
<p>h0: Apparent rise or set angle, -0.8333 for the Sun, +0.125 for the Moon, and -0.5667 for most other objects.</p>
</blockquote>
<p>That question also describes the rise/set equations as formulated by Meeus - and <a href="https://astronomy.stackexchange.com/users/25729/greg-miller">implemented here by Greg on CelestialProgramming.com</a>, which is the context I encountered this question.</p>
<p>The same convention for lunar h0 is mentioned in <a href="https://astronomy.stackexchange.com/questions/33849/jean-meeus-calculating-moon-rise-transit-and-set-times-help-isolating-error">this question (also implementing Astronomical Algorithms by Meeus)</a>.</p>
<p>My question is this:  The angles for sun and star seems to allow for atmospheric refraction, and thus an object is visible a fraction of a degree below the horizon as calculated without taking the atmosphere into account.</p>
<p>But the moon's angle does not!  Why?</p>
<p>So for example, sunrise starts at -0.83333 degrees, hits the mid-point of the sun at -0.56667 degrees, and then sunrise ends (bottom of sun clearing the visual horizon) at -0.3 degrees.</p>
<p>(-0.3 degrees is another 'conventional angle' you will see listed for end-of-sunrise or beginning-of-sunset.)</p>
<p>The position of stars &amp; planets agrees with this mid-point of the sun angle, at -0.56667 degrees.</p>
<p>However, for the moon the angle is then listed as +0.125 degrees. It seems like, since the moon has almost exactly the same apparent diameter as the sun, the number for h0 would be the same.  And that by the time +0.125 degrees is reached, the moon would not only be peaking above the horizon, or halfway up at the midpoint, or just clear of the horizon, but actually up and above the horizon by nearly a full moon diameter (0.125 deg - -0.3 deg = 0.425 deg, whereas moon apparent diameter is ~0.5 deg).</p>
<p>So I realize a lot of this is down to convention.  These angles are chosen given some idealized conditions for not only a perfectly spherical earth, a certain standardized altitude, idealized atmospheric conditions and so on.</p>
<p>So I am not interesting in knowing how to calculate these numbers more precisely (there is software available), but rather: Why are or were these particular numbers chosen as some kind of a standard, and why is the figure used for the moon so different from that used for other objects?</p>
<p>Is there a reason behind choosing such a different figure for moon rise and set than for the other astronomical objects?  Or is it just a convention that someone, maybe only Meeus, settled on at some point?</p>
<p>Is this the same convention used universally - when you see published moon rise &amp; set times by anyone from the Naval Observatory to the Farmer's Almanac, are they using this some convention for h0, or a different one?</p>

CC BY-SA 4.0
2024-12-20T17:55:59.267Z
I believe the value for the Moon is to account for parallax, assuming the Moon position used is a geocentric position.
Greg Miller
CC BY-SA 4.0
2024-12-20T18:00:07.660Z
Aha, now that makes sense.

BTW, huge thank you for all your work on CelestrialProgramming.com & elsewhere.  I'm very busy right now implementing a bunch of astronomical stuff for Garmin watches, and all the hard parts are based directly on your code.  Really appreciate it.
Brent Hugh
CC BY-SA 4.0
2024-12-20T23:57:19.397Z
You may wish to get familiar with JPL [Horizons](https://ssd.jpl.nasa.gov/horizons/). It has a *lot* of features, so there's a bit of a learning curve, but it's a great way to verify that astronomical calculations are on-track.
PM 2Ring
<the-sun><the-moon><sunrise><algorithm><sunset>
Why is h0=0.125 deg used as the apparent rise & set angle of the moon - when the angle is negative for sun & stars?
Brent Hugh
# Why is h0=0.125 deg used as the apparent rise & set angle of the moon - when the angle is negative for sun & stars?

(Asked by: username_0 on 2024-12-20)

In formulas for calculating rise & set times for astronomical objects, you often see a list of conventional angles used for various rise/set events. For example, [Greg Miller](https://astronomy.stackexchange.com/users/25729/greg-miller) [wrote in answer to a previous question](https://astronomy.stackexchange.com/questions/51658/math-tool-to-compute-time-of-sunrise-and-sunset?newreg=12d95ed779444b7189924d4255d0b457):

> h0: Apparent rise or set angle, -0.8333 for the Sun, +0.125 for the Moon, and -0.5667 for most other objects.

That question also describes the rise/set equations as formulated by Meeus - and [implemented here by Greg on CelestialProgramming.com](https://astronomy.stackexchange.com/users/25729/greg-miller), which is the context I encountered this question.

The same convention for lunar h0 is mentioned in [this question (also implementing Astronomical Algorithms by Meeus)](https://astronomy.stackexchange.com/questions/33849/jean-meeus-calculating-moon-rise-transit-and-set-times-help-isolating-error).

My question is this: The angles for sun and star seems to allow for atmospheric refraction, and thus an object is visible a fraction of a degree below the horizon as calculated without taking the atmosphere into account.

But the moon's angle does not! Why?

So for example, sunrise starts at -0.83333 degrees, hits the mid-point of the sun at -0.56667 degrees, and then sunrise ends (bottom of sun clearing the visual horizon) at -0.3 degrees.

(-0.3 degrees is another 'conventional angle' you will see listed for end-of-sunrise or beginning-of-sunset.)

The position of stars & planets agrees with this mid-point of the sun angle, at -0.56667 degrees.

However, for the moon the angle is then listed as +0.125 degrees. It seems like, since the moon has almost exactly the same apparent diameter as the sun, the number for h0 would be the same. And that by the time +0.125 degrees is reached, the moon would not only be peaking above the horizon, or halfway up at the midpoint, or just clear of the horizon, but actually up and above the horizon by nearly a full moon diameter (0.125 deg - -0.3 deg = 0.425 deg, whereas moon apparent diameter is ~0.5 deg).

So I realize a lot of this is down to convention. These angles are chosen given some idealized conditions for not only a perfectly spherical earth, a certain standardized altitude, idealized atmospheric conditions and so on.

So I am not interesting in knowing how to calculate these numbers more precisely (there is software available), but rather: Why are or were these particular numbers chosen as some kind of a standard, and why is the figure used for the moon so different from that used for other objects?

Is there a reason behind choosing such a different figure for moon rise and set than for the other astronomical objects? Or is it just a convention that someone, maybe only Meeus, settled on at some point?

Is this the same convention used universally - when you see published moon rise & set times by anyone from the Naval Observatory to the Farmer's Almanac, are they using this some convention for h0, or a different one?

**Tags:** the-sun, the-moon, sunrise, algorithm, sunset
> **Comments:**
> **username_1**: I believe the value for the Moon is to account for parallax, assuming the Moon position used is a geocentric position.  
> **username_0**: Aha, now that makes sense.  BTW, huge thank you for all your work on CelestrialProgramming.com & elsewhere.  I'm very busy right now implementing a bunch of astronomical stuff for Garmin watches, and all the hard parts are based directly on your code.  Really appreciate it.  
> **username_2**: You may wish to get familiar with JPL [Horizons](https://ssd.jpl.nasa.gov/horizons/). It has a *lot* of features, so there's a bit of a learning curve, but it's a great way to verify that astronomical calculations are on-track.  

---

## 1 answers

### Answer by username_3 on 2024-12-24. Score: 3

The full formula for when an object rises or sets is as follows:

*Zenith distance = 90 degrees + refraction + semi-diameter - horizontal parallax.*

* For stars, the semi-diameter and horizontal parallax are insignificant. Only the atmospheric refraction (34 arcminutes is typical) is important.
* For the Sun, the semi-diameter (the apparent angular radius) is important. 16 arcminutes is generally chosen for the apparent radius of the sun giving a total zenith distance of 90 degrees 50 arcminutes, or the center of the Sun is 0.8333 degrees below the horizon when it rises or sets.
* For the Moon, all three corrections are important. The diameter is approximately the same as the Sun (16 arcminutes), and the horizontal parallax is on the order of 1 degree. (The horizontal parallax is the shift in the position of the Moon as seen from the Earth's surface compared to a theoretical position seen from the center of the Earth. It is important for the Moon because its distance is close in comparison to the Earth's radius.) The net is the center of the Moon is *above* the horizon when it rises or sets; that is, the zenith distance is less than 90 degrees.

The U.S. Naval Observatory website [Rise, Set, and Twilight Definitions](https://aa.usno.navy.mil/faq/RST_defs) gives the following details for moonrise moonset:

> Here again, a constant of 34 arcminutes (0.5666 degree) is used to account for atmospheric refraction. The Moon's apparent radius varies from 15 to 17 arcminutes and its horizontal parallax varies from 54 to 61 arcminutes. Adding all the terms above together, the center of the Moon at rise or set is geometrically 5 to 10 arcminutes above the observer's "geocentric horizon" - the horizontal plane that passes through the Earth's center, orthogonal to the observer's local vertical.

The average of 5 and 10 is 7.5 arcminutes = 0.125 degrees.


---

