Patrick is describing polar coordinates, where the actual area subtended by a square minute-of-arc gets smaller as declination increases (gets closer to the poles), because the circle of right-ascension gets smaller. The alt-az example is polar coordinates used in a non-equatorial mode. Now, if describing the true FOV of an eyepiece, for example, or say, the apparent diameter of the moon, the definition of M.O.A. is typically the equatorial, or greatest, value. On Sun, Feb 8, 2009 at 10:58 PM, Patrick Wiggins <paw@wirelessbeehive.com>wrote:
On a small scale I think that would be ok. But keep in mind the further away from the horizon (in an Alt-Azi scale) the smaller each of those "squares" will get since while the vertical dimension will remain the same all the way to the zenith, the horizontal will shrink to nothing at the zenith.
patrick
On 08 Feb 2009, at 22:47, Joe Bauman wrote:
Many thanks, Patrick. Does it make sense to multiply the two to arrive at the total number of square arc-minutes? Best wishes, Joe
--- On Sun, 2/8/09, Patrick Wiggins <paw@wirelessbeehive.com> wrote: From: Patrick Wiggins <paw@wirelessbeehive.com> Subject: Re: [Utah-astronomy] quick question To: "Utah Astronomy" <utah-astronomy@mailman.xmission.com> Date: Sunday, February 8, 2009, 10:41 PM
That would depend on which way you are measuring.
Going from the horizon (assuming a flat horizon) to the zenith and back down to the opposite horizon would be 180 degrees. Multiply times 60 (60 arc minutes per degree) = 10,800 arc minutes.
But measuring around the horizon is 360 degree or 21,600 arc minutes.
patrick
On 08 Feb 2009, at 22:29, Joe Bauman wrote:
Hi, Can anybody tell me right off-hand how many arc-minutes the night sky is? Thanks, Joe
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