# I've seen some stars tonight. What could I possibly have been looking at with these pictures?

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

Picture 2 - Taken under hand-held night mode: ISO/800 S/0.5s F/2.2

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.

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.

• I'd normally recommend nova.astrometry.net but this image might be bright enough to stump it – user21 Dec 26 '18 at 20:38
• @barrycarter Aware of that too! I'd probably take better quality pictures once I got the instruments I need in hand. – Boosted Nub Dec 27 '18 at 3:40
• If you remove the numbering/lettering, you can at least try running it through that. The image may have a lot more information that is invisible (or very difficult to see) to the human eye without graphical enhancement. – user21 Dec 28 '18 at 20:00

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.

## 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

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

earth   = data['earth']
moon    = data['moon']

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
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()

• P1 was taken at 22:47 & P2 was 22:52. I've check back many times, there's no way the time was wrong. – Boosted Nub Dec 23 '18 at 11:24
• @BoostedNub okay sounds good. I've adjusted the wording accordingly. – uhoh Dec 23 '18 at 11:41
• 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. – Boosted Nub Dec 23 '18 at 12:09
• @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 Dec 23 '18 at 12:55

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, 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.

• Two is Alnilam. – Alphecca Dec 22 '18 at 22:28
• 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). – Boosted Nub Dec 23 '18 at 2:46
• 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 Dec 23 '18 at 9:42
• @DrChuck I thought so, too, but the planetarium told me otherwise. – Glorfindel Dec 23 '18 at 9:45
• @Dr Chuck Yes it is the moon. It was full moon, that's why I've shortened the exposure time. – Boosted Nub Dec 23 '18 at 11:18