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Old 21-12-2008, 07:16   #31
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Quote:
Originally Posted by GordMay View Post
Disclaimer: I don’t “do” celestial navigation.
Since one nautical mile corresponds to an arc of one minute on the surface of earth, and the earth goes one complete revolution (360 deg) in 24 hours, each hour corresponds to 15 degrees̊ (15 x 60 = 900 minutes) of longitude, or 900 Nautical miles.
Hence, each minute corresponds to 15 nm, and each second to 0.25 nm.
Hmm....
Disclaimer: I don't "do" celestial navigation except very occasionally
IRRC, one nautical mile corresponds to an arc of one minute (of degrees) of latitude ~ 6080 feet; however one minute of time = 15 minutes of longitude. Therefore one second of time = ~0.25 nm at the equator but b*gger all the poles.
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Old 21-12-2008, 15:35   #32
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The error in longitude for an error in seconds of time is approximately equal to time x .25 x cos(latitude) expressed in nautical miles. Thus, for one second of error in time at the equator is ~.25 nm (cos(0)=1), at 30 degrees north or south the error would be ~.22 nm (cos(30)=.866), and at 45 degrees north or south the error would be ~.18 nm (cos(45)=.707).

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Old 21-12-2008, 21:13   #33
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Thanks VS, I had forgotten the formula of x cos(lat)
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