WCS questions

[Mark Calabretta]

On Tue 1996/12/10 22:45:08 GMT, William Thompson wrote
in a message to: fitsbits at fits.cv.nrao.edu

>The trouble is that the sun is not a simple sphere like a planetary body.  It's
>a three dimensional object.  Chromospheric structures such as prominences
>extend several tens of thousands of kilometers above the surface, and
>chromospheric loops extend well beyond that.  The corona can be observed out to
>30 solar radii.  One has to be able to express data in a system which is valid
>both for pixels which are on the disk and for pixels which are above the limb.

Such a mapping from 2-D pixel coordinates (i,j) to 3-D spherical coordinates
(r,phi,theta) is under-determined for a single image, although a sequence of
such images taken at different viewing angles can be assembled via tomographic
techniques into a data-cube representation of a 3-D object.  See for example
http://www.atnf.csiro.au/people/toosterl/jupiter/.  However, such 3-D images
do not require WCS since the transformation from (i,j,k) -> (x,y,z) ->
(r,phi,theta) does not involve a spherical projection.

In any case, there's still nothing to prevent using WCS for heliographic
coordinates of the solar surface in a 2-D map even if it extends past the
solar limb.  The heliographic lng/lat of a feature beyond the limb would be
undefined, which I believe makes sense.  However, such features would still
have meaningful (x',y') coordinates.

>It's clear from Stephen Walton's original message that it's this kind of data
>which he was considering.

It wasn't clear to me.

>Another way to express the data would be in units of physical distance, rather
>than in arcseconds.  The solar radius is 6.96x10^5 km, or 696 Mm.  Expressing
>image pixels in units relative to the solar radius is really expressing it as a
>distance.

How do you interpret this "distance" if you don't know the angle between the
corresponding position vector and the plane of projection?  The best you can
say about the projected distance is that it is less than the true distance
from the solar centre so I think it could be misleading to assign it units of
metres.

Also, since the sun's rotational axis is not perpendicular to the ecliptic the
solar poles do not usually lie on the solar limb which means that not even
they will have fixed coordinates in this Cartesian system. 

Mark Calabretta
ATNF