第一章 习题参考答案 1题.略
2题.解 (1)4?1?9?2
（2）(L1?3L2)222?12?9L12?22
22?cosL1cos(L1?L2)?sinL1sin(L1?L2)?2?sinL1sin(L1?L2)?2??(3) ??1???2 22?????cos(L1?L2)?\??cos(L1?L2)?\?3题. 解DAD?ADTTTXL?LL,DYL?BADLL,DXYLLAB
4题.
解 设路线总长S公里，按照测量学上的附合路线计算步骤，则路线闭合差
fh?HA?h1?h2?HB
由于是路线中点，故v1?v2?12f1h??2?HA?h1?h2?HB? 则线路中点高程
H?中点?HA?h1?v1?HA?h11?2?HA?h1?h2?HB??12h?12h1112?2HA?2HB ???1?2?1?2????h?1??h?12???2?HA?HB?设每公里高差观测中误差为?0，则?h1??h2?(s/2)?0按误差传播定律
2?20?1?H?中点???1?2?1???h1??2????2????0?2?h12???????2???14?21211h1?4?h1?4?(s/2?0)2?4?(s/2?0)2
?1?(s/2?5)2?1?(s/2?5)2?25S?102444,S?16(km)5.解 设每个测回的中误差为?0，需要再增加n个测回，则
?020??0.42,??0.28,?0??0.4220?0??0.28n?20(1)
?0n?20(2)由上式可解出n.即
20?0.422?3?n??20?20????20?25 20.28?2?再增加25个测回
6题.解
2?x??pL??p???p?(p1L11?p2L2?...?pnLn)?p??1??P?p2?P??L1???pn??L2?...,????P??????Ln???pn???...??P??????1p10?001?p1?0?????p??p2?0?????p??,0?????1??pn??pn????p??
Qxx?1px?p??1??P?p2?P??p2??0
?p1??p?p1?1p1??p1p2?p??p2?p??1p2?p2
?p??....?pn?p??1pn?pn?p?
?p?2?p?2???pn?p?2?1?p??px??p?7题。解 PDd? D8题. 解 由题意可求出Qxx,Qyy,Qxy即：
Qxx???1?2?????n??????1p10?0n01p20??0?00????1??n2?????2?????i 0????i?1pi??1???n?pn??同理可得：Qyy??pi?1n?i2i,Qxy??i?1?i??ipi?Qyx
QFF?f211???f1pF?Qxxf2???Qyxni?1?pi?1n?i2i?2f1f2??i??ipiQxy??f1?22???f1Qxx?2f1f2Qxy?f2Qyy?Qyy??f2? ?f22?pi?1n?i2i
9题。解：(1)L 的协因数阵为单位阵E.
Qxx?(BTB)?1BTE((BTB)?1BT)T?(BTB)?1BTB((BTB)?1)T
?(BTB)?1??L?V?L?BX??L?BX? LTT?1TQL?L??BQLLB?B(BB)B
(2).因为 V=BX-L,
?QVX?BQxx?QLX?B(BTB)?1?B((BTB)T)?1?0 ??L?V,L?BX?V 因为LQL?V?QLV?QVV?BQXV?QVV?QVV?BQXV?0
?0?10题。解（1）每千米观测高差的中误差的估值为?2?pdd??9.2,???2n0?3mm
?2?（2）第二段观测高差的中误差??0?5mm s2???2?2?4mm
?2中?（3）第二段观测高差的平均值的中误差?（4）全长一次观测高差的中误差?全?（5）全长高差平均值的中误差?平??0?12mm S???8mm
?全2