# How to find the Andromeda Galaxy

I would like to find the Andromeda Galaxy ($M31$) with the unaided eye. One way is to use the following image and graphic

but I can’t find it just by looking in a dark sky. It is hard to detect these stars as I have no idea which way to look. Can I find it by using Wolfram Matematica (GeoPosition, location-based Entity objects, etc.. )

and if the answer is yes, how?

Thank you in advance to any one who may be able to give me some ideas.

• As an aside, you can use the Google's Sky Map app on your phone. It makes use of your phone's GPS, compass, and accelerometer to point you directly at what you want to see. – Edmund Feb 29 '16 at 17:27
• I would recommend Stellarium over Google Sky Map; the desktop version is free, however the mobile phone app is not. – shrx Feb 29 '16 at 18:27
• total aside, but unless you are somewhere with very little light pollution you are unlikely to see it at all. – george2079 Feb 29 '16 at 18:45
• Not true, the core is visible on clear nights from the edge of a small city. Averted vision helps. – shrx Mar 1 '16 at 15:01

There is some verry handy stuff you can use with our favorite tool:

So, Andromeda is Constellation, let's see,

And M31 has to be with it some where:

So the red Dot is the spot where M31 stays at the time of calulation for the current position of the CPU of mine :)). Not neat to spot ...

If you like to learn, read more see

Andromeda galaxy is Milky Way’s next-door neighbor

Messier Monday: Andromeda, the Object that Opened Up the Universe, M31

Top 10 Night Sky Objects for Astronomy Beginners

Have a look at:

http://www.sky-map.org/

http://www.stellarium.org/

http://www.shatters.net/celestia/

http://www.astrosurf.com/c2a/english/index.htm

Have fun

maybe something to get you started: (warning this is very slow)

p=Entity["Constellation","Andromeda"]["BrightStars"];
data=((Print[#["Name"]];{
QuantityMagnitude@{
UnitConvert[#["GalacticLongitude"],"Degrees"],
UnitConvert[#["GalacticLatitude"],"Degrees"]},
#["ApparentMagnitude"]})&)/@p;

Graphics[{PointSize[ #[[2]]/200 ],Point[#[[1]]]}&/@data]


To make it look right you need to use an appropriate projection, and I'm afraid I don't know what that should be.