Hi All, In reply to my earlier statement: > The 90-degree inclination case simply reduced to the global > land-to-water ratio. Aaron Brown wrote: > I don't think this is true. The surface near the poles would > be weighted much more heavily. Imagine if the only land was > between 80 and 90 degrees latitude. The global land/water > ratio would be far less than 10/180, yet the satellite would > be over land 10/180 of the time. Aaron is absolutely correct -- my apologies. I completely neglected the cosine effect with latitude. This actually makes the land/water ratio vs. orbital inclination function an even more interesting calculation. The only hard part of solving this problem is creating a function which when given a specific latitude and longitude returns "water" or "land". Or taking it one step further, returns a geographic label (e.g. country name, ocean, sea, island, etc.) This function, in conjuction with a specific decaying orbit, could then be used to assess impact probabilities in an automated fashion. With a little more complexity (taking orbital height, day/night, and minimum elevation angle into consideration), the function could be expanded to assess the chances of observing the reentry from a given location. Best, Rob ----------------------------------------------------------------- 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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