<p>I think there might have been a problem with your compass readings, or "in the north" really means <em>just a little north of east</em>.</p>

<p>Cuba is in the northern hemisphere at around 20 to 23°N (tropic of Cancer) and so the Sun can be directly over head and the Moon with an additional 5 degrees inclination with respect to the ecliptic can go several degrees North. (see <a href="https://astronomy.stackexchange.com/a/27750/7982">this answer</a> for a bit about the Moon's motion).</p>

<blockquote>
  <ul>
  <li>it started in the north position around 4 pm. </li>
  <li>By 7pm, it was in the east position, </li>
  <li>and by 11 pm, it was in the south position. </li>
  </ul>
</blockquote>

<p>I think there might have been something affecting your compass. The <a href="https://en.wikipedia.org/wiki/Magnetic_declination" rel="nofollow noreferrer">Magnetic declination</a> there is about 6-8 degrees, that might have had a tiny effect, but you might look for problems like </p>

<ol>
<li>iron or other ferromagnetic material, </li>
<li>mechanical compass from the <a href="https://outdoors.stackexchange.com/q/21360/12619">wrong hemisphere</a> sticking due to dip angle</li>
<li>or if it was an electronic compass, incorrect <a href="https://electronics.stackexchange.com/q/249086/102305">degaussing/calibration</a>.</li>
</ol>

<p>The other possibility is that by "in the north" you really mean only "slightly north of east" or "east-north-east".</p>

<hr>

<p>I wrote a short script in Python for your trip, and the three lines in the plot represent January 15 (lowest, blue), 17, and 19 from Santiago de Cuba, the southernmost area of Cuba. The moon does get to the zenith and perhaps a tiny bit past it (towards North) but I don't think that would be visually noticeable.</p>

<p>The Large black dots represent 7PM local time, and each dot to the west is one more hour.</p>

<p><a href="https://i.sstatic.net/RSVlk.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/RSVlk.png" alt="enter image description here"></a></p>

<p>Some Python:</p>

<pre><code>import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Loader, Topos

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads        = 180/pi, pi/180

load  = Loader('~/Documents/fishing/SkyData')  # single instance for big files
ts    = load.timescale()
de421 = load('de421.bsp')
earth = de421['earth']
moon  = de421['moon']

Santiago_de_Cuba = earth + Topos(latitude_degrees  = 20.019833,
                                 longitude_degrees = -75.813917,
                                 elevation_m       = 10.)

hours = np.arange(16, 23.1, 0.5) + 5

days   = (15, 17, 19, 21)

altazs, lines, linez= [], [], []
for day in days:
    times = ts.utc(2019, 1, day, hours)
    alt, az, d = (Santiago_de_Cuba).at(times).observe(moon).apparent().altaz()
    alt, az    = [thing.degrees for thing in (alt, az)]
    alt[alt&lt;0] = np.nan
    altazs.append((alt, az))
    r = (1 - alt/90.)
    theta = rads * az
    lines.append((r, theta))
    x, y = [r*f(theta) for f in (np.sin, np.cos)]
    linez.append((x, y))

if True:
    plt.figure()
    plt.subplot(2, 1, 1)
    for alt, az in altazs:
        plt.plot(hours, alt)
        plt.plot(hours[::2], alt[::2], 'ok')
    plt.subplot(2, 1, 2)
    for alt, az in altazs:
        plt.plot(hours, az)
        plt.plot(hours[::2], az[::2], 'ok')
    plt.show()

th = np.linspace(0, twopi, 201)
cth, sth = [f(th) for f in (np.cos, np.sin)]
if True:
    plt.figure()
    plt.plot(cth, sth, '-k', linewidth=1.5)
    plt.plot([0], [0], 'or', markersize=8)
    plt.plot([-0.1, 0.1], [ 0,   0  ], '-k')
    plt.plot([ 0,   0  ], [-0.1, 0.1], '-k')

    plt.text(-0.05,  0.85, 'N', fontsize=16)
    plt.text( 0.9,  -0.05, 'E', fontsize=16)
    plt.text(-0.95, -0.05, 'W', fontsize=16)
    plt.text(-0.05, -0.90, 'S', fontsize=16)

    for x, y in linez:
        linewidth= 1
        plt.plot(x, y)
        plt.plot(x[:-1:2], y[:-1:2], '.k')
        plt.plot(x[:1], y[:1], 'ok')
    plt.show()
</code></pre>

CC BY-SA 4.0
2019-02-21T08:09:23.870Z
2019-02-21T08:50:07.093Z
2019-02-21T08:50:07.093Z
uhoh
<p>To avoid confusion with compass directions on the ground, let's express the orientation of the Moon by analogy to an imaginary clock dial.</p>

<p>Relative to the celestial equator and pole, the bright limb of a first quarter moon is always oriented between 2 and 4 o'clock depending on the time of year.
In astronomical terms, its <a href="https://en.wikipedia.org/wiki/Position_angle" rel="nofollow noreferrer">position angle</a> is between 240&deg; and 300&deg; counterclockwise from celestial north.
On the evening of 2019-01-15, JPL HORIZONS says the Moon's sub-Sun position angle was about 252&deg;, or 3:36 on an equatorially mounted clock dial.</p>

<p>Relative to the horizon and zenith, an observer in the Arctic would see the Moon in a similar orientation, near 3 o'clock all evening.
In general, however, the celestial equator is inclined to the horizon at an angle of 90&deg; minus the observer's geographic latitude.
As celestial objects cross the sky parallel to the equator, they roll through a range of <a href="https://en.wikipedia.org/wiki/Parallactic_angle" rel="nofollow noreferrer">parallactic angles</a>.
<a href="https://astronomy.stackexchange.com/a/20625">This answer</a> illustrates it with the Sun, and <a href="https://astronomy.stackexchange.com/a/26215">this answer</a> gives a standard formula.</p>

<p><a href="https://i.sstatic.net/YGAER.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YGAER.png" alt="Moon at 13:40"></a>
<a href="https://i.sstatic.net/Q4jdJ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Q4jdJ.png" alt="Moon at 19:50"></a>
<a href="https://i.sstatic.net/hVBFa.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/hVBFa.png" alt="Moon at 02:00"></a>
<br/><em>Moon and <a href="https://en.wikipedia.org/wiki/Equatorial_coordinate_system" rel="nofollow noreferrer">equatorial grid</a> rendered by Stellarium</em></p>

<p>Observing from Havana (23&deg;N) on January 15-16, the Moon's parallactic angle went from -67&deg;  at 1:40pm, to 0&deg; while crossing the meridian at 7:50pm, to +67&deg; at 2:00am.
Subtracting this from the bright limb's position angle, we get angles of 319&deg;, 252&deg;, and 185&deg; counterclockwise from the zenith, or 1:22, 3:36, and 5:50 on a vertical clock dial.
From a more northerly latitude, the rolling effect would be smaller; from the southern hemisphere, it would be reversed.</p>

CC BY-SA 4.0
2019-02-21T21:35:28.727Z
2019-02-25T03:00:28.900Z
2019-02-25T03:00:28.900Z
Mike G
<p>One picture tells more than thousands of words.<br><br>
Check this out:</p>

<ul>
<li>Moon rising (near East):</li>
</ul>

<p><a href="https://i.sstatic.net/nG1sf.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/nG1sf.jpg" alt="Moon in the east"></a></p>

<ul>
<li>Moon setting (near West):</li>
</ul>

<p><a href="https://i.sstatic.net/eMNoG.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/eMNoG.jpg" alt="Moon in the west"></a></p>

<p>That picture can you show, that Moon rises in the East and sets in the West (in fact just little more to the North - 25°).<br>
It's probably just a technical problem with a compass because Moon cannot rise in the North. (The earth always revolves around its axis, which runs along the north-south direction.)<br><br>
<em>(This picture is from the site "theskylive.org" for location Cuba and date 16/17th January 2019.)</em></p>

CC BY-SA 4.0
2019-02-21T15:01:42.110Z
2019-02-22T05:45:38.113Z
2019-02-22T05:45:38.113Z
User123
<p>I was in Cuba from January 15th to the 22nd this year and there was a quarter moon. If I use a compass to explain the orientation of the quarter moon, it started in the north position around 4 pm. By 7 pm, it was in the east position, and by 11 pm, it was in the south position. Normally the quarter moon would start in the east and end in the east as it travels across the sky.</p>

<p>Could someone please explain this phenomenon?</p>

<p>Thanks!</p>

CC BY-SA 4.0
2019-02-20T21:08:46.160Z
Are you talking about the angle between the Moon's bright limb and the observer's zenith?
Mike G
CC BY-SA 4.0
2019-02-21T02:48:31.713Z
Could you please clarify? I edited it so it a bit more organized and clear, but still...
Max0815
CC BY-SA 4.0
2019-02-21T16:47:21.707Z
OP said "explain," not "measure" - crucial distinction.
Mike G
CC BY-SA 4.0
2019-02-21T17:55:30.373Z
Possibly related: [Changing appearance of the moon during a single night](https://astronomy.stackexchange.com/q/20620)
Mike G
<the-moon>
The movement of the moon
James MacIntosh
# The movement of the moon

(Asked by: username_0 on 2019-02-20)

I was in Cuba from January 15th to the 22nd this year and there was a quarter moon. If I use a compass to explain the orientation of the quarter moon, it started in the north position around 4 pm. By 7 pm, it was in the east position, and by 11 pm, it was in the south position. Normally the quarter moon would start in the east and end in the east as it travels across the sky.

Could someone please explain this phenomenon?

Thanks!

**Tags:** the-moon

> **Comments:**
> **username_1**: Are you talking about the angle between the Moon's bright limb and the observer's zenith?  
> **username_2**: Could you please clarify? I edited it so it a bit more organized and clear, but still...  
> **username_1**: OP said "explain," not "measure" - crucial distinction.  
> **username_1**: Possibly related: [Changing appearance of the moon during a single night](https://astronomy.stackexchange.com/q/20620)  

---

## 3 answers

### Answer by username_3 on 2019-02-21. Score: 5

I think there might have been a problem with your compass readings, or "in the north" really means *just a little north of east*.

Cuba is in the northern hemisphere at around 20 to 23°N (tropic of Cancer) and so the Sun can be directly over head and the Moon with an additional 5 degrees inclination with respect to the ecliptic can go several degrees North. (see [this answer](https://astronomy.stackexchange.com/a/27750/7982) for a bit about the Moon's motion).

> * it started in the north position around 4 pm.
> * By 7pm, it was in the east position,
> * and by 11 pm, it was in the south position.

I think there might have been something affecting your compass. The [Magnetic declination](https://en.wikipedia.org/wiki/Magnetic_declination) there is about 6-8 degrees, that might have had a tiny effect, but you might look for problems like

1. iron or other ferromagnetic material,
2. mechanical compass from the [wrong hemisphere](https://outdoors.stackexchange.com/q/21360/12619) sticking due to dip angle
3. or if it was an electronic compass, incorrect [degaussing/calibration](https://electronics.stackexchange.com/q/249086/102305).

The other possibility is that by "in the north" you really mean only "slightly north of east" or "east-north-east".

---

I wrote a short script in Python for your trip, and the three lines in the plot represent January 15 (lowest, blue), 17, and 19 from Santiago de Cuba, the southernmost area of Cuba. The moon does get to the zenith and perhaps a tiny bit past it (towards North) but I don't think that would be visually noticeable.

The Large black dots represent 7PM local time, and each dot to the west is one more hour.

[![enter image description here](https://i.sstatic.net/RSVlk.png)](https://i.sstatic.net/RSVlk.png)

Some Python:

```
import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Loader, Topos

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads        = 180/pi, pi/180

load  = Loader('~/Documents/fishing/SkyData')  # single instance for big files
ts    = load.timescale()
de421 = load('de421.bsp')
earth = de421['earth']
moon  = de421['moon']

Santiago_de_Cuba = earth + Topos(latitude_degrees  = 20.019833,
                                 longitude_degrees = -75.813917,
                                 elevation_m       = 10.)

hours = np.arange(16, 23.1, 0.5) + 5

days   = (15, 17, 19, 21)

altazs, lines, linez= [], [], []
for day in days:
    times = ts.utc(2019, 1, day, hours)
    alt, az, d = (Santiago_de_Cuba).at(times).observe(moon).apparent().altaz()
    alt, az    = [thing.degrees for thing in (alt, az)]
    alt[alt<0] = np.nan
    altazs.append((alt, az))
    r = (1 - alt/90.)
    theta = rads * az
    lines.append((r, theta))
    x, y = [r*f(theta) for f in (np.sin, np.cos)]
    linez.append((x, y))

if True:
    plt.figure()
    plt.subplot(2, 1, 1)
    for alt, az in altazs:
        plt.plot(hours, alt)
        plt.plot(hours[::2], alt[::2], 'ok')
    plt.subplot(2, 1, 2)
    for alt, az in altazs:
        plt.plot(hours, az)
        plt.plot(hours[::2], az[::2], 'ok')
    plt.show()

th = np.linspace(0, twopi, 201)
cth, sth = [f(th) for f in (np.cos, np.sin)]
if True:
    plt.figure()
    plt.plot(cth, sth, '-k', linewidth=1.5)
    plt.plot([0], [0], 'or', markersize=8)
    plt.plot([-0.1, 0.1], [ 0,   0  ], '-k')
    plt.plot([ 0,   0  ], [-0.1, 0.1], '-k')

    plt.text(-0.05,  0.85, 'N', fontsize=16)
    plt.text( 0.9,  -0.05, 'E', fontsize=16)
    plt.text(-0.95, -0.05, 'W', fontsize=16)
    plt.text(-0.05, -0.90, 'S', fontsize=16)

    for x, y in linez:
        linewidth= 1
        plt.plot(x, y)
        plt.plot(x[:-1:2], y[:-1:2], '.k')
        plt.plot(x[:1], y[:1], 'ok')
    plt.show()

```


---

### Answer by username_1 on 2019-02-21. Score: 2

To avoid confusion with compass directions on the ground, let's express the orientation of the Moon by analogy to an imaginary clock dial.

Relative to the celestial equator and pole, the bright limb of a first quarter moon is always oriented between 2 and 4 o'clock depending on the time of year.
In astronomical terms, its [position angle](https://en.wikipedia.org/wiki/Position_angle) is between 240° and 300° counterclockwise from celestial north.
On the evening of 2019-01-15, JPL HORIZONS says the Moon's sub-Sun position angle was about 252°, or 3:36 on an equatorially mounted clock dial.

Relative to the horizon and zenith, an observer in the Arctic would see the Moon in a similar orientation, near 3 o'clock all evening.
In general, however, the celestial equator is inclined to the horizon at an angle of 90° minus the observer's geographic latitude.
As celestial objects cross the sky parallel to the equator, they roll through a range of [parallactic angles](https://en.wikipedia.org/wiki/Parallactic_angle).
[This answer](https://astronomy.stackexchange.com/a/20625) illustrates it with the Sun, and [this answer](https://astronomy.stackexchange.com/a/26215) gives a standard formula.

[![Moon at 13:40](https://i.sstatic.net/YGAER.png)](https://i.sstatic.net/YGAER.png)
[![Moon at 19:50](https://i.sstatic.net/Q4jdJ.png)](https://i.sstatic.net/Q4jdJ.png)
[![Moon at 02:00](https://i.sstatic.net/hVBFa.png)](https://i.sstatic.net/hVBFa.png)
  
*Moon and [equatorial grid](https://en.wikipedia.org/wiki/Equatorial_coordinate_system) rendered by Stellarium*

Observing from Havana (23°N) on January 15-16, the Moon's parallactic angle went from -67° at 1:40pm, to 0° while crossing the meridian at 7:50pm, to +67° at 2:00am.
Subtracting this from the bright limb's position angle, we get angles of 319°, 252°, and 185° counterclockwise from the zenith, or 1:22, 3:36, and 5:50 on a vertical clock dial.
From a more northerly latitude, the rolling effect would be smaller; from the southern hemisphere, it would be reversed.


---

### Answer by username_4 on 2019-02-21. Score: 2

One picture tells more than thousands of words.  
  
Check this out:

* Moon rising (near East):

[![Moon in the east](https://i.sstatic.net/nG1sf.jpg)](https://i.sstatic.net/nG1sf.jpg)

* Moon setting (near West):

[![Moon in the west](https://i.sstatic.net/eMNoG.jpg)](https://i.sstatic.net/eMNoG.jpg)

That picture can you show, that Moon rises in the East and sets in the West (in fact just little more to the North - 25°).  
It's probably just a technical problem with a compass because Moon cannot rise in the North. (The earth always revolves around its axis, which runs along the north-south direction.)  
  
*(This picture is from the site "theskylive.org" for location Cuba and date 16/17th January 2019.)*


---

