0
$\begingroup$

The SWARP Documentation specifies that we first set the CENTER_TYPE to MANUAL and then in the CENTER parameter we specify either one of the equatorial coordinates (RA, Dec). This is specified in the image below:

CENTER Parameters specs

In doing so, I do not understand how by giving just one of these coordinates, the SWARP tool will interpret which specific coordinate is given.

I am currently viewing the output image in matplotlib, setting its projection with the wcs header info that we get in the output fits file. It seems as though SWARP takes every coordinate specified in the CENTER parameter as its right ascension. There is also no way to specify if we want to set the declination value. How can we set a center for the output image if

  1. We can only give one coordinate (RA or Dec), and
  2. Whatever coordinate we provide is interpreted by SWARP as the right ascension?
$\endgroup$
2
  • $\begingroup$ You're correct, it doesn't make sense to provide just one of RA or Dec. What it's trying to say is: "Position of the center in RA and Dec, or longitude and latitude [e.g., for Galactic coordinates]." And then it tells you that the format of the values can be in decimal degrees, or in hh:mm:ss + dd:mm:ss for RA and Dec. $\endgroup$ Oct 28 '20 at 10:27
  • $\begingroup$ So if your desired center coordinate is RA,Dec = 12.0, +10.5 in decimal degrees, you could use that, or you could use 00:48:00.00, +10:30:00.0. (Page 20 of the manual is a bit clearer on this, and the example configuration file on page 37 shows an example of the latter usage.) $\endgroup$ Oct 28 '20 at 10:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.