I doubt it works
所有跟贴·加跟贴·新语丝读书论坛
送交者: HunHunSheng 于 2009-03-24, 13:30:10:
回答: 如果没有大气折射,用日出日落时间差(小于12小时)可估算。但大气的存在使 由 chouqilozi 于 2009-03-24, 13:19:35:
所有跟贴:
- Would work on Mars. (无内容) - chouqilozi (0 bytes) 2009-03-24, 13:42:23 (308796)
- Neither. - HunHunSheng (111 bytes) 2009-03-24, 13:46:01 (308797)
- Without the atmosphere, time between the first glance of sun's center - chouqilozi (112 bytes) 2009-03-24, 13:56:02 (308803)
- 那只是火星的自转造成的,和太阳和火星的距离毫无关系 (无内容) - HunHunSheng (0 bytes) 2009-03-24, 14:03:19 (308808)
- see diagram, - chouqilozi (281 bytes) 2009-03-24, 14:15:11 (308810)
- 我已经说过了火星根本没有大到能让你测量出角度 - HunHunSheng (49 bytes) 2009-03-24, 14:17:44 (308811)
- any deviation of time between sunrise and sunset from half day is evidence - chouqilozi (34 bytes) 2009-03-24, 14:20:42 (308814)
- 太阳这么大你看到的那个center可能都比地球火星还大。18世纪根本区分不出来 (无内容) - HunHunSheng (0 bytes) 2009-03-24, 14:39:27 (308825)
- 太阳这么大你看到的那个可能都比地球火星还大。18世纪根本区分不出来 (无内容) - HunHunSheng (0 bytes) 2009-03-24, 14:39:00 (308824)
- that you can just stare at the center of the sun, not the whole volume. (无内容) - chouqilozi (0 bytes) 2009-03-24, 14:37:06 (308822)
- you can just stare at the center of the sun, ignore the volume. (无内容) - chouqilozi (0 bytes) 2009-03-24, 14:36:19 (308821)
- just stare at the center of the sun, ignore the volume. (无内容) - chouqilozi (0 bytes) 2009-03-24, 14:35:44 (308820)
- 凭着18世纪以前的技术估计也检测不出来这个deviation (无内容) - HunHunSheng (0 bytes) 2009-03-24, 14:30:46 (308817)
- 太阳比火星大的多而且你不知道太阳有多大。可是你是假设太阳是一个点。 - HunHunSheng (139 bytes) 2009-03-24, 14:26:15 (308816)
加跟贴