AA Institute wrote:

-60° 50' declination.
Is there an easy calculation to work out how many degrees that
direction is off the ecliptic plane of our solar system?

Not quite as easy as the transverse motion calculation. What you'd be
doing is converting from equatorial coordinates to ecliptic coordinates.
The conversion involves a single rotation, but in three dimensions. The
axis of the rotation is the intersection of the equatorial and ecliptic
planes, and the amount of the rotation is the obliquity of the ecliptic
(the tilt of the Earth). You'd start with both RA and Dec (you need
both, since it's a 3D transformation) and end up with lamda and beta,
ecliptic longitude and latitude.

For more precise calculations, or calculations over long time frames,
you'd have to account for various effects that cause the two planes to
move with respect to each other. Nutation is a slight wobble of the
equatorial system caused by the Moon. The Earth's axis also precesses
slowly.

A search would probably turn up online calculators to do the conversion
for you, along with any number of pages on celestial coordinates, e.g.

http://www.seds.org/~spider/spider/S....html#ecliptic

Most star charting software is capable of displaying grid lines for
several coordinate systems, and some can provide the location of
specified objects in your choice of coordinate systems.

Approximate current ecliptic coordinates for Alpha Centauri are

beta = -42.6°
lamda = 239.5°

But don't take my word for it. I just did this on a calculator and may
have gotten it wrong, and I haven't told you the epoch or whether I
accounted for nutation. You'll get more satisfaction from understanding
and doing the calculation yourself.

- Ernie http://home.comcast.net/~erniew