# Transform pixel coordinates (in FITS file) to equatorial

Say I have a FITS file with an observed field of stars, each one with x,y coordinates given in pixels. Assume the image is aligned so that the x axis is the right ascension, and the y axis the declination.

I'm looking for a tutorial to transform the pixel coordinates of each star to equatorial, even if the transformation is approximate.

I'v been looking for a while but found no tutorial/book/site that enumerates the steps needed to produce this transformation.

Assuming that each (square) pixel has the same angular scale (not a given if the field of view is large) of $\theta$ degrees/pixel.

Then the declination (in degrees): $\delta \simeq \delta_0 + \theta y$

The right ascension (in degrees): $\alpha \simeq \alpha_0 - \theta x/\cos \delta$

where $\alpha_0$ and $\delta_0$ are the RA and Dec at $x=0$, $y=0$. [The minus sign is there because right ascension increases towards the left of a sky image.]

This approximatimation becomes poor as the field of view gets larger.

A slightly more complex, but accurate, approach is described in http://gtn.sonoma.edu/data_reduction/astrometry.php

• Thanks Rob, this is the same formula I arrived at, but I can not find any source for this. Do you know where these steps might be mathematically demonstrated? A book, an article, anything that could be used as a source. Mar 15 '16 at 22:43
• A source? It's just from the definition of a spherical coordinate system. If a ship travels at constant speed this is how you would work out the latitude and longitude. Mar 15 '16 at 23:06
• No, this is the definition of a spherical coordinate system. I'm asking if you know of a source where this recipe is shown, as in Section X: How to transform pixel coordinates to equatorial coordinates. I'm still marking your answer as accepted, I'm just curious because I could not find an explicit presentation of such a basic set of equations for such a used procedure in Astronomy, in any book/article/site. Mar 15 '16 at 23:17
• Thank you for the source Rob! I'll take a look at it right now. Mar 16 '16 at 2:01