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Date: 24 May 2004 22:35 (UTC)At a latitude of 30, say, stuff at a declination of 30 passes right overhead. And it looks like an RA of 18 will be overhead in Arizona in a couple hours...ie, around midnight whereever you are. So you can probably see the galactic center up to an latitude of around +30 (Tucson: +32, so my story checks out), maybe up to +40 or +45 if you have really clear horizons.
Second bit is trickier...there's a 23 degree angle between dec=0 and the ecliptic, and about 30 degrees between dec=0 and galactic plane. Now, for a bit of basic astronomy...during the summer, the earth's north axis points away from the sun, and around that time, the galactic center is about 3 hours off from midnight, or 45 degrees off the line between the sun and the earth, in the plane of the ecliptic... call that angle (phi), and that's a tricky problem, so I'm going to take the easy way out. If phi was 0, we'd just add up the two angles and get 53. If phi was 180, the two would subtract, and you'd get 7 degrees. The very wrong way to solve the problem, then, is to take the value at phi=0 and subtract 1/4 the difference between these two values, which is precisely what I"ll do.
53 - .25*(53-7) = 41.5 So about 40 degrees. I'll check a textbook or something tomorrow.
So that's significantly off from makovette's answer.
What do you want from me? I'm tired and a pretty lousy astronomer.
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Date: 24 May 2004 22:53 (UTC)