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GET /hyperkitty/api/list/[email protected]/email/JW7YUIYR52KSRJHFXUEDWQ4RQQ5EUIM6/?format=api
{ "url": "https://mailman.amsat.org/hyperkitty/api/list/[email protected]/email/JW7YUIYR52KSRJHFXUEDWQ4RQQ5EUIM6/?format=api", "mailinglist": "https://mailman.amsat.org/hyperkitty/api/list/[email protected]/?format=api", "message_id": "[email protected]", "message_id_hash": "JW7YUIYR52KSRJHFXUEDWQ4RQQ5EUIM6", "thread": "https://mailman.amsat.org/hyperkitty/api/list/[email protected]/thread/P65A2TURZC3YLUHXCZ5QACGFK5QSXUG5/?format=api", "sender": { "address": "8p6sm (a) anjo.com", "mailman_id": null, "emails": null }, "sender_name": "Gus 8P6SM", "subject": "[amsat-bb] Re: maximum AO-7 distance", "date": "2012-10-02T01:43:27Z", "parent": "https://mailman.amsat.org/hyperkitty/api/list/[email protected]/email/P65A2TURZC3YLUHXCZ5QACGFK5QSXUG5/?format=api", "children": [], "votes": { "likes": 0, "dislikes": 0, "status": "neutral" }, "content": "On 10/01/2012 02:47 AM, Bob- W7LRD wrote:\n> to my AO-7 afictionados...what is the maximum distance one can\n> work edge to edge, and how did you figure that out?\n> 73 Bob W7LRD\n\nMaximum communications distance:\n\n MCD = 2 R arccos [ R / (R + h) ]\t\n\nwhere\n R = Radius of earth (Spherical earth: 6371 Km)\n h = Height at apogee\n\n (Eq 12.4, Sat. Experimenters Handbook 2nd Ed; Davidoff)\n\t\t\t\nAll you need is the apogee height for the satellite in question, and a \ncalculator. For AO-7 the apogee height USED to be 1460 Km, and probably \nhasn't altered much over the years. Therefore:\n\n MCD = 2 * 6371 * arccos [ 6371 / (6371 + 1460) ]\n = 12756 * arccos [ 6371 / 7831 ]\n = 12756 * arccos [ 0.8135614864 ]\n = 12756 * 0.620545318575\n\t= 7915.7 Km (4918.59 miles)\n\nWhat is the \"official\" distance?\n\nWith a good station, the non-spherical earth working for you at your QTH \nand hoping for some sub-horizon love, who knows?\n\n-- \n73, de Gus 8P6SM\nThe Easternmost Isle\n", "attachments": [] }