The area of a band around a latitude is R*R*width * 2 * pi * cos(lat), not considering oblateness. So between two latitudes, it is the integral, or 2*pi*(sin(lat1)-sin(lat2)), or for your example 4*pi*0.3987. For the whole sphere it is 4*pi, so 40% is your answer. The uninhabited space rarely comes into play - even if it decays W-E across Sahara, it would usually be visible from one of the coasts or central states, and just two or three scattered nomads would be sufficient to cover the area - but they would probably not post a message on SeeSat-L. If a decay is visible over a rectangle (2*500) * (2000+2*500) sq.km., that is 0.6% of the Earth's surface. Seen on a time scale, if EUVE has a mean motion of 16.55, it is above 10 degrees from Honolulu on 9 passes/day, totalling 12 minutes/day. Adding 5 minutes/pass of decay duration, increases its chances to about 4%, not considering daylight. This is because about 50% of Earth never saw EUVE, and Hawaii can see up to northern apex, ie a lot more passes than areas on lower latitudes. ----- Original Message ----- > > > http://wwwvms.utexas.edu/~ecannon/oddsofseeingareentry.htm > > Two of the obvious factors are 50% daylight and 70% > ocean (and nn% of land uninhabited), and there's the > Earth's equatorial bulge (and maybe its oblateness), > and of course the percentage of cloud cover (which I > don't know). What is the percentage of the Earth's > surface within the tropics (+/- 23.5 degrees)? I > don't know how to do that math. So I would like to > learn more of the technicalities on this topic. > ----------------------------------------------------------------- Unsubscribe from SeeSat-L by sending a message with 'unsubscribe' in the SUBJECT to SeeSat-L-request@lists.satellite.eu.org http://www.satellite.eu.org/seesat/seesatindex.html
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