<p>in-the-sky.org planetarium mode is your friend. It will pick up your location and time zone from your ip address, but you can manually set it to anywhere else.</p>

<p>I've added screenshots at both <code>20:52</code> where the Moon is about 26 degrees above the horizon, and at the stated time of<code>22:52</code> where the Moon is more like 53 degrees.</p>

<h2>20:52:</h2>

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

<h2>22:52:</h2>

<p><a href="https://i.sstatic.net/AZuf4.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/AZuf4.jpg" alt="enter image description here"></a></p>

<hr>

<p>Here's a plot of altitude and azimuth of the Moon on that night. Python script as well.</p>

<p><a href="https://i.sstatic.net/Z8AGp.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Z8AGp.png" alt="enter image description here"></a></p>

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

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

load = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files


data    = load('de421.bsp')
earth   = data['earth']
moon    = data['moon']
ts      = load.timescale()

print earth.at(ts.now()).position.km   # for no reason at all

local_hours = np.linspace(18, 24, 13)
UTC_hours   = [h-8 for h in local_hours]
times       = ts.utc(2018, 12, 22, UTC_hours, 0, 0)

KL     = earth + Topos(3.1333, 101.6833, elevation_m=66)
Ampang = earth + Topos(3.1255, 101.7588, elevation_m=66)

alt, az, dist = KL.at(times).observe(moon).apparent().altaz()
KLalt, KLaz   = alt.degrees, az.degrees

alt, az, dist        = Ampang.at(times).observe(moon).apparent().altaz()
Ampangalt, Ampangaz  = alt.degrees, az.degrees

if True:
    plt.figure()

    pts = [6, 8, 10]

    plt.suptitle('Moon  22-Dec-2018 KL &amp; Ampang', fontsize=16)

    plt.subplot(4, 1, 1)
    plt.plot(local_hours, KLalt)
    plt.plot(local_hours, Ampangalt)
    plt.plot(local_hours[pts], Ampangalt[pts], 'ok')
    plt.ylabel('Altitude (degs)', fontsize=16)
    plt.xlabel('hours (local)', fontsize=16)
    plt.ylim(0, 90)

    plt.subplot(4, 1, 2)
    plt.plot(local_hours, KLaz)
    plt.plot(local_hours[pts], Ampangaz[pts], 'ok')
    plt.ylabel('Azimuth (degs)', fontsize=16)
    plt.xlabel('hours (local)', fontsize=16)
    plt.ylim(0, 90)

    plt.subplot(2, 1, 2)
    r = (1. - Ampangalt/90.)
    r[r&gt;1] = np.nan
    theta = rads*Ampangaz
    x, y = r*np.sin(theta), r*np.cos(theta)
    plt.plot(x, y)
    plt.plot(x[pts], y[pts], 'ok')
    th = np.linspace(0, twopi, 361)
    xc, yc = np.sin(th), np.cos(th)
    plt.plot(xc, yc, '-k', linewidth=2)
    d = 0.05
    plt.plot([-d, d], [ 0, 0], '-k')
    plt.plot([ 0, 0], [-d, d], '-k')
    plt.xlim(-1.1, 1.1)
    plt.ylim(-1.1, 1.1)

    plt.show()
</code></pre>

CC BY-SA 4.0
2018-12-23T11:24:53.550Z
P1 was taken at 22:47 & P2 was 22:52. I've check back many times, there's no way the time was wrong.
not_Prince
CC BY-SA 4.0
2018-12-23T11:41:00.757Z
@BoostedNub okay sounds good. I've adjusted the wording accordingly.
uhoh
CC BY-SA 4.0
2018-12-23T12:09:36.830Z
I'm not sure if the location (Kuala Lumpur) as per shown in your screenshot would affects the output or not. Precisely the area I live is outside the Federal Territory.
not_Prince
CC BY-SA 4.0
2018-12-23T12:55:38.723Z
@BoostedNub Each hour is almost 1,700 km east or west near the equator, so it won't have a significant effect. I've added a plot.
uhoh
CC BY-SA 4.0
2018-12-23T09:53:03.150Z
2018-12-23T12:56:44.670Z
2018-12-23T12:56:44.670Z
uhoh
<p>You could try using an online planetarium to identify the stars. Right now, it's hard to tell because of the scale, but given <a href="https://in-the-sky.org/skymap.php?no_cookie=1&amp;latitude=3.14&amp;longitude=101.69&amp;timezone=8.00&amp;year=2018&amp;month=12&amp;day=22&amp;hour=22&amp;min=52&amp;PLlimitmag=2&amp;zoom=200&amp;ra=6.51140&amp;dec=30.90950" rel="nofollow noreferrer">this view</a>, stars 1-3 in the first picture belong to the Orion constellation (1 is Betelgeuze, 3 is Rigel, and 2 is one of the three belt stars) and 4 is Sirius, the brightest star in the night sky.</p>

<p>The second compass must be wrong; going to the East should move the Moon to the left in the picture, not to the right.</p>

<p>I'd recommend shooting somewhere else than the Moon; your phone will probably increase exposure time, making more stars visible than right now in the pictures. Phones aren't really suitable for astronomical photos because of their small sensors and lenses, though.</p>

CC BY-SA 4.0
2018-12-22T22:28:33.380Z
Two is Alnilam.
Alphecca
CC BY-SA 4.0
2018-12-23T02:46:52.510Z
The compass of the right hand side is for P1 in which the moon in P1 is positioned at left hand side. I'm confused though, which one supposedly wrong? Aside that, I can't get a dark sky at moment; I'm sure I would shoot better pictures if I could. Besides, high-end smartphones are actually capable of shooting stars if meet the environment requirements (dark sky etc.) with the help of a tripod and picture editor software (of course not to compare to professional high-end camera).
not_Prince
CC BY-SA 4.0
2018-12-23T09:42:35.673Z
Assuming the very bright object is the moon, and not a street light, then 1, 2, and 3 can't be orion's belt, because the moon never gets that far south.
Dr Chuck
CC BY-SA 4.0
2018-12-23T09:45:41.630Z
@DrChuck I thought so, too, but the planetarium told me otherwise.
Glorfindel
CC BY-SA 4.0
2018-12-23T11:18:02.087Z
@Dr Chuck Yes it is the moon. It was full moon, that's why I've shortened the exposure time.
not_Prince
CC BY-SA 4.0
2018-12-22T20:57:49.740Z
2018-12-22T20:57:49.740Z
Glorfindel
<p>The weather tonight here is great and I decided to shoot some hand-held (too bad I don't have any tripod) night mode pictures shot with P10. I have no experience in stars observation therefore I can't tell what was I looking at. But I'd definitely love to know what are those. Forgive me if the quality of pictures are bad; I've tried my best. At some point I found some of the stars look clear on my phone but fainter over my laptop monitor.</p>

<p><sup>Picture 1 - Taken under hand-held night mode: ISO/400 S/0.5s F/2.2</sup><a href="https://i.sstatic.net/dn4Zx.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/dn4Zx.jpg" alt="enter image description here"></a></p>

<p><sup>Picture 2 - Taken under hand-held night mode: ISO/800 S/0.5s F/2.2</sup>
<a href="https://i.sstatic.net/zd9QJ.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/zd9QJ.jpg" alt="enter image description here"></a></p>

<p><sup><strong>Latitude, Longitude &amp; Compass details:</strong> Right is for P1; left belongs to P2. Keep in mind that it may not 100% accurate.</sup></p>

<p><sup><strong>Time:</strong> 22 Dec 2018 / 22:52 P.M. UTC+8</sup>
<sup><strong>Location:</strong> Ampang, Malaysia.</sup>
<a href="https://i.sstatic.net/mlShS.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/mlShS.jpg" alt="enter image description here"></a></p>

<p>Star labelled 4 is the only brightest among the others. It shines and flickers so bright that visible very well with naked eyes. The rest are barely to be seen if not focus well.</p>

CC BY-SA 4.0
2018-12-27T03:40:15.463Z
@barrycarter Aware of that too! I'd probably take better quality pictures once I got the instruments I need in hand.
not_Prince
<amateur-observing><photography><deep-sky-observing>
I've seen some stars tonight. What could I possibly have been looking at with these pictures?
not_Prince
# I've seen some stars tonight. What could I possibly have been looking at with these pictures?

(Asked by: username_0 on 2018-12-22)

The weather tonight here is great and I decided to shoot some hand-held (too bad I don't have any tripod) night mode pictures shot with P10. I have no experience in stars observation therefore I can't tell what was I looking at. But I'd definitely love to know what are those. Forgive me if the quality of pictures are bad; I've tried my best. At some point I found some of the stars look clear on my phone but fainter over my laptop monitor.

Picture 1 - Taken under hand-held night mode: ISO/400 S/0.5s F/2.2[![enter image description here](https://i.sstatic.net/dn4Zx.jpg)](https://i.sstatic.net/dn4Zx.jpg)

Picture 2 - Taken under hand-held night mode: ISO/800 S/0.5s F/2.2
[![enter image description here](https://i.sstatic.net/zd9QJ.jpg)](https://i.sstatic.net/zd9QJ.jpg)

**Latitude, Longitude & Compass details:** Right is for P1; left belongs to P2. Keep in mind that it may not 100% accurate.

**Time:** 22 Dec 2018 / 22:52 P.M. UTC+8
**Location:** Ampang, Malaysia.
[![enter image description here](https://i.sstatic.net/mlShS.jpg)](https://i.sstatic.net/mlShS.jpg)

Star labelled 4 is the only brightest among the others. It shines and flickers so bright that visible very well with naked eyes. The rest are barely to be seen if not focus well.

**Tags:** amateur-observing, photography, deep-sky-observing
> **Comments:**
> **username_0**: @barrycarter Aware of that too! I'd probably take better quality pictures once I got the instruments I need in hand.  

---

## 2 answers

### Answer by username_1 on 2018-12-23. Score: 2

in-the-sky.org planetarium mode is your friend. It will pick up your location and time zone from your ip address, but you can manually set it to anywhere else.
I've added screenshots at both 20:52 where the Moon is about 26 degrees above the horizon, and at the stated time of22:52 where the Moon is more like 53 degrees.
20:52:

22:52:


Here's a plot of altitude and azimuth of the Moon on that night. Python script as well.

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

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

load = Loader('~/Documents/fishing/SkyData')  # avoids multiple copies of large files


data    = load('de421.bsp')
earth   = data['earth']
moon    = data['moon']
ts      = load.timescale()

print earth.at(ts.now()).position.km   # for no reason at all

local_hours = np.linspace(18, 24, 13)
UTC_hours   = [h-8 for h in local_hours]
times       = ts.utc(2018, 12, 22, UTC_hours, 0, 0)

KL     = earth + Topos(3.1333, 101.6833, elevation_m=66)
Ampang = earth + Topos(3.1255, 101.7588, elevation_m=66)

alt, az, dist = KL.at(times).observe(moon).apparent().altaz()
KLalt, KLaz   = alt.degrees, az.degrees

alt, az, dist        = Ampang.at(times).observe(moon).apparent().altaz()
Ampangalt, Ampangaz  = alt.degrees, az.degrees

if True:
    plt.figure()

    pts = [6, 8, 10]

    plt.suptitle('Moon  22-Dec-2018 KL & Ampang', fontsize=16)

    plt.subplot(4, 1, 1)
    plt.plot(local_hours, KLalt)
    plt.plot(local_hours, Ampangalt)
    plt.plot(local_hours[pts], Ampangalt[pts], 'ok')
    plt.ylabel('Altitude (degs)', fontsize=16)
    plt.xlabel('hours (local)', fontsize=16)
    plt.ylim(0, 90)

    plt.subplot(4, 1, 2)
    plt.plot(local_hours, KLaz)
    plt.plot(local_hours[pts], Ampangaz[pts], 'ok')
    plt.ylabel('Azimuth (degs)', fontsize=16)
    plt.xlabel('hours (local)', fontsize=16)
    plt.ylim(0, 90)

    plt.subplot(2, 1, 2)
    r = (1. - Ampangalt/90.)
    r[r>1] = np.nan
    theta = rads*Ampangaz
    x, y = r*np.sin(theta), r*np.cos(theta)
    plt.plot(x, y)
    plt.plot(x[pts], y[pts], 'ok')
    th = np.linspace(0, twopi, 361)
    xc, yc = np.sin(th), np.cos(th)
    plt.plot(xc, yc, '-k', linewidth=2)
    d = 0.05
    plt.plot([-d, d], [ 0, 0], '-k')
    plt.plot([ 0, 0], [-d, d], '-k')
    plt.xlim(-1.1, 1.1)
    plt.ylim(-1.1, 1.1)

    plt.show()

> **Comments:**
> **username_0**: P1 was taken at 22:47 & P2 was 22:52. I've check back many times, there's no way the time was wrong.  
> **username_1**: @BoostedNub okay sounds good. I've adjusted the wording accordingly.  
> **username_0**: I'm not sure if the location (Kuala Lumpur) as per shown in your screenshot would affects the output or not. Precisely the area I live is outside the Federal Territory.  
> **username_1**: @BoostedNub Each hour is almost 1,700 km east or west near the equator, so it won't have a significant effect. I've added a plot.  

---

### Answer by username_2 on 2018-12-22. Score: 1

You could try using an online planetarium to identify the stars. Right now, it's hard to tell because of the scale, but given [this view](https://in-the-sky.org/skymap.php?no_cookie=1&latitude=3.14&longitude=101.69&timezone=8.00&year=2018&month=12&day=22&hour=22&min=52&PLlimitmag=2&zoom=200&ra=6.51140&dec=30.90950), stars 1-3 in the first picture belong to the Orion constellation (1 is Betelgeuze, 3 is Rigel, and 2 is one of the three belt stars) and 4 is Sirius, the brightest star in the night sky.

The second compass must be wrong; going to the East should move the Moon to the left in the picture, not to the right.

I'd recommend shooting somewhere else than the Moon; your phone will probably increase exposure time, making more stars visible than right now in the pictures. Phones aren't really suitable for astronomical photos because of their small sensors and lenses, though.

> **Comments:**
> **username_3**: Two is Alnilam.  
> **username_0**: The compass of the right hand side is for P1 in which the moon in P1 is positioned at left hand side. I'm confused though, which one supposedly wrong? Aside that, I can't get a dark sky at moment; I'm sure I would shoot better pictures if I could. Besides, high-end smartphones are actually capable of shooting stars if meet the environment requirements (dark sky etc.) with the help of a tripod and picture editor software (of course not to compare to professional high-end camera).  
> **username_4**: Assuming the very bright object is the moon, and not a street light, then 1, 2, and 3 can't be orion's belt, because the moon never gets that far south.  
> **username_2**: @DrChuck I thought so, too, but the planetarium told me otherwise.  
> **username_0**: @Dr Chuck Yes it is the moon. It was full moon, that's why I've shortened the exposure time.  

---

