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Format of RA/DEC that I am working with

Hello, I am collecting some data and need to convert some units, so I built a small script to convert RA/DEC to galactic coordinates. The data has specifically formatted the coordinates as shown in the photo. I have an example in the code, but I don't really have anything to compare my answer to so I don't even know if it is correct. I don't really have familiarity with this so please let me know if I am overlooking something! Here is the code:

from astropy import units as u
from astropy.coordinates import SkyCoord

ra="21:33:56.568"
dec="+04:29:07.84"


#The following function takes str arguments for ra and dec measured as formatted in the TNS #data, and outputs galactic coordinates l and b (l,b)

def ICRStoGal(ra,dec):
    
    coord = ra + dec

    c_icrs = SkyCoord((coord), unit=(u.hourangle, u.deg))

    c_gal =c_icrs.galactic
    l,b=c_gal.l.degree, c_gal.b.degree
    return l,b

ICRStoGal(ra,dec)[0]

As an output to this, I got (58.720664132924085, -32.823605289663924). Or, if anyone has any practice problems with this conversion where I can look at the solutions that would also be amazing. Thank you to anyone who takes a look!

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Applying spherical trigonometry, it follows that the conversion from equatorial coordinates $\alpha$ and $\delta$ to galactic coordinates $l$ and $b$ is:

$$\left. \begin{aligned} \sin b &=\sin \delta_{NGP} \sin \delta + \cos \delta_{NGP} \cos \delta \cos ( \alpha - \alpha_{NGP}) \\ \sin ( l_{NCP}-l) &=\dfrac{\cos \delta \sin ( \alpha - \alpha_{NGP})}{\cos b} \\ \cos ( l_{NCP}-l) &=\dfrac{\cos \delta_{NGP}\sin \delta - \sin \delta_{NGP} \cos \delta \cos ( \alpha - \alpha_{NGP})}{\cos b} \end{aligned} \right \}$$

The equatorial coordinates (J2000) of the North Galactic Pole are:

$\alpha_{NGP}=12^h 51^m 26.282^s=192.8595º=3.36603 \ rad$

$\delta_{NGP}=+27º 07’ 42.01’’=27.1283º=0.473479 \ rad$

And the galactic longitude of the North Celestial Pole is:

$l_{NCP}=122.932º=2.14557 \ rad$

You can check if your code gives the same results as these equations.

Best regards.

PD. For your example:

$\alpha = 21^h \ 33^m \ 56.568^s$

$\delta = +04^{\circ} \ 29' \ 07.84''$

The equations give as a result:

$l = 58.720766^{\circ}$

$b = -32.823544^{\circ}$

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    $\begingroup$ Thanks, that is a much simpler way to show the work than what I've been finding online $\endgroup$ Feb 19 at 16:27

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