<p>That's because it isn't simple</p>
<blockquote>
<p>A really precise determination of the Moon's position requires the calculation of hundreds of periodic terms to account for all the myriad gravitational forces acting on the Moon. Meeus gives a simplified set of calculations in chapter 45 [ch 47 in the second edition] of Astronomical Algorithms, with 59 periodic terms in longitude and 30 in latitude. (<a href="https://airmass.org/notes" rel="nofollow noreferrer">source</a>)  <em>Meeus, is the author of Astronomical Algorithms, the go-to book for simple calculations of the position of astronomical bodies)</em></p>
</blockquote>
<p>This gives about 10 arcseconds of error. Meeus considers this to be a simplified algorithm, and notes that &quot;hundreds of periodic terms&quot; would be necessary for a really accurate calculation. Now, you might not need this level of precision, and some of the terms in this algorithm are pretty small and probably can be ignored for many purposes.</p>
<p>But you can't have an algorithm that is both accurate and simple, because the motion of moon, as it is flung around by the gravity of the Earth, Sun, Jupiter, Venus, tides etc, is far from simple.</p>
<p>The practical solution, if you just want to know where the moon will be, is to use a software package or website that implements the calculations for you. This won't, of course, help you to learn about the calculation process.</p>
<p>The formula then is quite simple:</p>
<blockquote>
<p>True longitude of Moon = mean longitude + major inequality + evection + variation + annual inequality + reduction to ecliptic + parallactic inequality + more terms  (<a href="http://www.geoastro.de/moonmotion/index.html" rel="nofollow noreferrer">source</a>)</p>
</blockquote>
<p>And the mean longitude increases linearly <span class="math-container">$L_0 = 218.31617 + 481267.88088T + \ldots$</span> Where T is time in centuries, and derived from the Julian Date:  <span class="math-container">$T = (JD - 2451545)/36525$</span>.  That gives you a simple, but inaccurate (by several degrees) answer to &quot;what is the longitude of the moon?&quot;</p>
<p>If you want a more accurate model, you can see any of the implementations of Meeus for example <a href="https://www.celestialprogramming.com/meeus-elp82.html" rel="nofollow noreferrer">https://www.celestialprogramming.com/meeus-elp82.html</a> That gives the formula for longitude in the form of a javascript class which implements &quot;Mean Longitude + ....&quot;</p>

CC BY-SA 4.0
2024-07-21T22:40:44.677Z
According to https://en.wikipedia.org/wiki/Lunar_theory The lunar theory of E. W. Brown, developed around the turn of the twentieth century, used over 1400 terms.
PM 2Ring
CC BY-SA 4.0
2024-07-21T23:34:51.000Z
I need a calculation for it I believe it has to do with trigonometric expression.
Carla G.
CC BY-SA 4.0
2024-07-22T00:00:17.893Z
My answer says you can have simple but not accurate.  Or accurate but not simple. Or just use a package.
James K
CC BY-SA 4.0
2024-07-22T00:05:26.370Z
Either way both works for me I just want to know the formula and I edit my question I updated it.
Carla G.
CC BY-SA 4.0
2024-07-22T03:44:40.150Z
@CarlaG. We cannot give the simplified equation (from Meeus which has only 59 terms) because it is protected by copywrite. Purchase the book Astronomical Algorithms and you will have the equation. Or use the 1000s of terms that is given in the link from Greg Miller since that material (from ELP82-2000) is open source.
JohnHoltz
CC BY-SA 4.0
2024-07-22T05:00:52.113Z
I've added a "simple but not accurate" formula.  I'm not sure I agree with John. The formula that Meeus uses is "Mean longitude + terms", and the coefficients of the terms are just numbers. You can't copyright a number, and there are plenty of implementations of Meeus around from which you could get the numbers.  I'm not including it here because of length:  But you can seen an implementation https://www.celestialprogramming.com/meeus-elp82.html
James K
CC BY-SA 4.0
2024-07-21T21:14:56.343Z
2024-07-22T05:07:12.873Z
2024-07-22T05:07:12.873Z
James K
<p>I keep searching on it on Internet but yet they can help yet and the way they reach is not straight to the point at all.</p>
<p><strong>I really need to learn this Lunar Longitude</strong> for example the Solar Longitude is 119° and the <strong>Lunar Longitude is 310°</strong>.</p>
<p>I know how to determine the Solar Longitude like this:
<a href="https://i.sstatic.net/jtWLbTdF.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/jtWLbTdF.jpg" alt="enter image description here" /></a></p>
<p>but I don't know and I don't have enough knowledge on Lunar Longitude for example like this<a href="https://i.sstatic.net/Dc7vP74E.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Dc7vP74E.jpg" alt="enter image description here" /></a></p>
<p>But the another one which is easy but not accurate enough for example if the Sagittarius enters the moon there's a possibility the moon is in 258° Lunar Longitude</p>

CC BY-SA 4.0
2024-07-21T23:43:04.883Z
One of the most common algorithms for accurate Moon position is ELP82-2000.  It's output is ecliptic cartesian coordinates, which can easily be converted to polar coordinates.  One implementation is at the link below, viewing the source code, you'll find it is far from simple though. https://celestialprogramming.com/snippets/elp2000-82b/elp2000-82.html
Greg Miller
CC BY-SA 4.0
2024-07-22T00:02:05.153Z
I just want to learn the Lunar Longitude or Moon's Ecliptic Topocentirc so that I can calculate the moon phase base on Indian Lunisolar Calendar and Chinese Lunisolar Calendar for example (Lunar Longitude) - (Solar Longitude) = N (1st Result) | N (1st Result) ÷ 12= N (Moon Phase) another 1 (Lunar Longitude) - (Solar Longitude) = N (Negative Number/s) +360* = N (1st Result) | N (1st Result) ÷ 12 = N (Moon Phase) *360 is needed to add, if the answer is negative.
Carla G.
CC BY-SA 4.0
2024-07-22T15:50:10.413Z
Here is a low precision algorithm for the moon position, similar to the low precision one you're using for the Sun. https://celestialprogramming.com/lowprecisionmoonposition.html
Greg Miller
<the-moon>
How to calculate the Moon's Longitude or Moon's Ecliptic Topocentirc?
Carla G.
# How to calculate the Moon's Longitude or Moon's Ecliptic Topocentirc?

(Asked by: username_0 on 2024-07-21)

I keep searching on it on Internet but yet they can help yet and the way they reach is not straight to the point at all.

**I really need to learn this Lunar Longitude** for example the Solar Longitude is 119° and the **Lunar Longitude is 310°**.

I know how to determine the Solar Longitude like this:
[![enter image description here](https://i.sstatic.net/jtWLbTdF.jpg)](https://i.sstatic.net/jtWLbTdF.jpg)

but I don't know and I don't have enough knowledge on Lunar Longitude for example like this[![enter image description here](https://i.sstatic.net/Dc7vP74E.jpg)](https://i.sstatic.net/Dc7vP74E.jpg)

But the another one which is easy but not accurate enough for example if the Sagittarius enters the moon there's a possibility the moon is in 258° Lunar Longitude

**Tags:** the-moon
> **Comments:**
> **username_1**: One of the most common algorithms for accurate Moon position is ELP82-2000.  It's output is ecliptic cartesian coordinates, which can easily be converted to polar coordinates.  One implementation is at the link below, viewing the source code, you'll find it is far from simple though. https://celestialprogramming.com/snippets/elp2000-82b/elp2000-82.html  
> **username_0**: I just want to learn the Lunar Longitude or Moon's Ecliptic Topocentirc so that I can calculate the moon phase base on Indian Lunisolar Calendar and Chinese Lunisolar Calendar for example (Lunar Longitude) - (Solar Longitude) = N (1st Result) | N (1st Result) ÷ 12= N (Moon Phase) another 1 (Lunar Longitude) - (Solar Longitude) = N (Negative Number/s) +360* = N (1st Result) | N (1st Result) ÷ 12 = N (Moon Phase) *360 is needed to add, if the answer is negative.  
> **username_1**: Here is a low precision algorithm for the moon position, similar to the low precision one you're using for the Sun. https://celestialprogramming.com/lowprecisionmoonposition.html  

---

## 1 answers

### Answer by username_2 on 2024-07-21. Score: 5

That's because it isn't simple

> A really precise determination of the Moon's position requires the calculation of hundreds of periodic terms to account for all the myriad gravitational forces acting on the Moon. Meeus gives a simplified set of calculations in chapter 45 [ch 47 in the second edition] of Astronomical Algorithms, with 59 periodic terms in longitude and 30 in latitude. ([source](https://airmass.org/notes)) *Meeus, is the author of Astronomical Algorithms, the go-to book for simple calculations of the position of astronomical bodies)*

This gives about 10 arcseconds of error. Meeus considers this to be a simplified algorithm, and notes that "hundreds of periodic terms" would be necessary for a really accurate calculation. Now, you might not need this level of precision, and some of the terms in this algorithm are pretty small and probably can be ignored for many purposes.

But you can't have an algorithm that is both accurate and simple, because the motion of moon, as it is flung around by the gravity of the Earth, Sun, Jupiter, Venus, tides etc, is far from simple.

The practical solution, if you just want to know where the moon will be, is to use a software package or website that implements the calculations for you. This won't, of course, help you to learn about the calculation process.

The formula then is quite simple:

> True longitude of Moon = mean longitude + major inequality + evection + variation + annual inequality + reduction to ecliptic + parallactic inequality + more terms ([source](http://www.geoastro.de/moonmotion/index.html))

And the mean longitude increases linearly $L\_0 = 218.31617 + 481267.88088T + \ldots$ Where T is time in centuries, and derived from the Julian Date: $T = (JD - 2451545)/36525$. That gives you a simple, but inaccurate (by several degrees) answer to "what is the longitude of the moon?"

If you want a more accurate model, you can see any of the implementations of Meeus for example <https://www.celestialprogramming.com/meeus-elp82.html> That gives the formula for longitude in the form of a javascript class which implements "Mean Longitude + ...."

> **Comments:**
> **username_3**: According to https://en.wikipedia.org/wiki/Lunar_theory The lunar theory of E. W. Brown, developed around the turn of the twentieth century, used over 1400 terms.  
> **username_0**: I need a calculation for it I believe it has to do with trigonometric expression.  
> **username_2**: My answer says you can have simple but not accurate.  Or accurate but not simple. Or just use a package.  
> **username_0**: Either way both works for me I just want to know the formula and I edit my question I updated it.  
> **username_4**: @CarlaG. We cannot give the simplified equation (from Meeus which has only 59 terms) because it is protected by copywrite. Purchase the book Astronomical Algorithms and you will have the equation. Or use the 1000s of terms that is given in the link from Greg Miller since that material (from ELP82-2000) is open source.  
> **username_2**: I've added a "simple but not accurate" formula.  I'm not sure I agree with John. The formula that Meeus uses is "Mean longitude + terms", and the coefficients of the terms are just numbers. You can't copyright a number, and there are plenty of implementations of Meeus around from which you could get the numbers.  I'm not including it here because of length:  But you can seen an implementation https://www.celestialprogramming.com/meeus-elp82.html  

---

