<p>The transit altitude of an object does not only depend on your latitude ($b$), but also on the declination of the object ($\delta$):
$$h_\mathrm{tr} = \arcsin\left(\sin b \sin \delta + \cos b \cos\delta\right).$$</p>

CC BY-SA 3.0
2014-01-06T16:22:58.767Z
2014-01-06T16:22:58.767Z
AstroFloyd
<p>With a lot of difficulty if you want to do it numerically by yourself. The Moon's orbit is very complex.</p>

<p>Far and away the easiest method would be to use NASA's <a href="http://ssd.jpl.nasa.gov/?horizons" rel="nofollow">HORIZONS</a> service for calculating <a href="http://ssd.jpl.nasa.gov/?glossary&amp;term=ephemeris" rel="nofollow">ephemerides</a>, which will get you very comprehensive and accurate information on objects at your specified times.</p>

<p>I don't think that HORIZONS can directly return the time that the Moon is at its maximum altitude. However, if you can write some basic computer code then it shouldn't be difficult to query the ephemerides with a precise (say 1 minute) time interval and find the maximum altitude from there.</p>

CC BY-SA 3.0
2016-08-11T13:11:12.013Z
Can you please give references to explanations of the moon orbit, especially as it looks from earth? I'm puzzled that the moon and the sun have interchangable paths on the sky between winter and summer. Is it possible to easily explain this? Would this be a good question (on its own) to ask in astronomy.stackexchange.com? Many thanks!
George
CC BY-SA 4.0
2023-08-03T04:01:09.223Z
Horizons has an "RST flag" (rise, set, transit) in the observer table options, so it can do the computation for you.
Greg Miller
CC BY-SA 3.0
2013-12-17T17:58:41.680Z
Thanks :) But, I'm looking for a mathematical approach as it was asked in one of my astronomy homeworks.
Ken
CC BY-SA 3.0
2013-12-17T02:50:29.557Z
2013-12-17T02:58:51.857Z
2013-12-17T02:58:51.857Z
Moriarty
<p>This can be taken a bit further. The right-hand side of
<span class="math-container">$$
\sin h_{\rm tr} = \sin \phi \sin \delta + \cos \phi \cos \delta
$$</span>
is the formula for the cosine of the difference of two angles, so
<span class="math-container">$$
\sin h_{\rm tr} = \cos( \phi - \delta).
$$</span>
Noting that
<span class="math-container">$$
\sin x = \cos(90^\circ - x),
$$</span>
we have
<span class="math-container">$$
\cos(90^\circ - h_{\rm tr}) = \cos(\phi - \delta),
$$</span>
giving
<span class="math-container">$$
h_{\rm tr} = 90^\circ - \left | \phi - \delta \right | .
$$</span>
The absolute value is necessary because the cosine is an even function, i.e.,
<span class="math-container">$$
\cos(-x) = \cos x.
$$</span></p>

CC BY-SA 4.0
2023-08-03T03:20:36.440Z
2023-08-03T09:31:49.260Z
2023-08-03T09:31:49.260Z
JeffC
<p>With the known latitude coordinate of the observing position, how to find the altitude of the moon when it is high, i.e when it crosses the local meridian? </p>

CC BY-SA 3.0
2013-12-17T09:11:31.580Z
May be this is out of your scope, but I saw those programs may help you. There is a cool software [Digital planetarium application](http://www.stellarium.org/), free to download. And another one here but shareware [Starry Night](http://astronomy.starrynight.com/)
Ahmed Hamdy
CC BY-SA 3.0
2013-12-17T17:48:52.293Z
Thanks :) But, I'm looking for a mathematical approach as it was asked in one of my astronomy homeworks.
Ken
CC BY-SA 3.0
2014-02-18T04:04:20.500Z
If it is homework, shouldn't you be attempting to research the answer yourself, or figure it out for yourself? The point being to help you comprehend what is going on, not just obtain an answer?
Jeremy
CC BY-SA 3.0
2014-02-18T13:12:10.720Z
@Jeremy That is exactly what I did (sort of) :) AstroFloyd's answer did help me to comprehend the concept a bit, and later I got a clarified answer from my class.
Ken
<the-moon><amateur-observing><observational-astronomy>
How to calculate the altitude of the Moon?
Ken
# How to calculate the altitude of the Moon?

(Asked by: username_0 on 2013-12-16)

With the known latitude coordinate of the observing position, how to find the altitude of the moon when it is high, i.e when it crosses the local meridian?

**Tags:** the-moon, amateur-observing, observational-astronomy
> **Comments:**
> **username_1**: May be this is out of your scope, but I saw those programs may help you. There is a cool software [Digital planetarium application](http://www.stellarium.org/), free to download. And another one here but shareware [Starry Night](http://astronomy.starrynight.com/)  
> **username_0**: Thanks :) But, I'm looking for a mathematical approach as it was asked in one of my astronomy homeworks.  
> **username_2**: If it is homework, shouldn't you be attempting to research the answer yourself, or figure it out for yourself? The point being to help you comprehend what is going on, not just obtain an answer?  
> **username_0**: @Jeremy That is exactly what I did (sort of) :) AstroFloyd's answer did help me to comprehend the concept a bit, and later I got a clarified answer from my class.  

---

## 3 answers

### Answer by username_3 on 2014-01-06. Score: 3 | Accepted answer

The transit altitude of an object does not only depend on your latitude ($b$), but also on the declination of the object ($\delta$):
$$h\_\mathrm{tr} = \arcsin\left(\sin b \sin \delta + \cos b \cos\delta\right).$$


---

### Answer by username_4 on 2013-12-17. Score: 3

With a lot of difficulty if you want to do it numerically by yourself. The Moon's orbit is very complex.

Far and away the easiest method would be to use NASA's [HORIZONS](http://ssd.jpl.nasa.gov/?horizons) service for calculating [ephemerides](http://ssd.jpl.nasa.gov/?glossary&term=ephemeris), which will get you very comprehensive and accurate information on objects at your specified times.

I don't think that HORIZONS can directly return the time that the Moon is at its maximum altitude. However, if you can write some basic computer code then it shouldn't be difficult to query the ephemerides with a precise (say 1 minute) time interval and find the maximum altitude from there.

> **Comments:**
> **username_0**: Thanks :) But, I'm looking for a mathematical approach as it was asked in one of my astronomy homeworks.  
> **username_5**: Can you please give references to explanations of the moon orbit, especially as it looks from earth? I'm puzzled that the moon and the sun have interchangable paths on the sky between winter and summer. Is it possible to easily explain this? Would this be a good question (on its own) to ask in astronomy.stackexchange.com? Many thanks!  
> **username_6**: Horizons has an "RST flag" (rise, set, transit) in the observer table options, so it can do the computation for you.  

---

### Answer by username_7 on 2023-08-03. Score: 1

This can be taken a bit further. The right-hand side of
$$
\sin h\_{\rm tr} = \sin \phi \sin \delta + \cos \phi \cos \delta
$$
is the formula for the cosine of the difference of two angles, so
$$
\sin h\_{\rm tr} = \cos( \phi - \delta).
$$
Noting that
$$
\sin x = \cos(90^\circ - x),
$$
we have
$$
\cos(90^\circ - h\_{\rm tr}) = \cos(\phi - \delta),
$$
giving
$$
h\_{\rm tr} = 90^\circ - \left | \phi - \delta \right | .
$$
The absolute value is necessary because the cosine is an even function, i.e.,
$$
\cos(-x) = \cos x.
$$


---

