From: Ted Viens <firstname.lastname@example.org> Date: Mon, 13 Oct 1997 19:29:56 -0700 Fwd Date: Wed, 15 Oct 1997 01:43:35 -0400 Subject: Re: The sky over Roswell and estimates of speed > To: UFO UpDates - Toronto <email@example.com> > From: Mark Cashman <firstname.lastname@example.org> > Subject: re: UFO UpDate: Re: The sky over Roswell and estimates of speed > Date: Mon, 13 Oct 1997 12:11:16 -0700 ---big snip--- > Since the size of a degree does not vary from horizon to zenith, > the speed should be the same at any point on the arc. > Since your observation was of an object travelling approx. 1.5 > degrees per second across the sky, I believe that my estimates > are correct, but I look forward to any further comments. Howdy Mark... The problem is in your choice of pivot point. Your calculations are for the sweep about the observer. This is where a degree at the zenith is equal to a degree at the horizon. The bogey is sweeping around the center of the earth. Now, back to our bogey passing at a constant 100 mile altitude. A ninety degree sweep for the observer centered at the zenith equals only 2.8 degrees of arc for the bogey. A ninety degree sweep for the observer from the zenith to the horizon equals some 12.7 degrees of arc for the bogey. This is about 4.5 times as far. A degree is not always a degree. Bye... Ted..
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