方差分量估計(jì)算例

1、【例104】如圖101邊角網(wǎng)，點(diǎn)為已知點(diǎn)，為待定點(diǎn)，同精度獨(dú)立觀測(cè)了12CBA、、ED、個(gè)角度和6條邊長(zhǎng)，據(jù)分別列于表101和表102。先驗(yàn)測(cè)角中誤差，先???5.1??驗(yàn)邊長(zhǎng)測(cè)量中誤差為。試按cmS0.2???間接平差法進(jìn)行赫爾默特方差分量估計(jì)，并求出：（1）觀測(cè)值的方差估值；（2）待定點(diǎn)坐標(biāo)平差值及其方差估值。表101基準(zhǔn)數(shù)據(jù)表坐標(biāo)（m）點(diǎn)號(hào)XY至點(diǎn)坐標(biāo)方位角′″邊長(zhǎng)（m）ABC4899.8468751.9454548.795130

2、.8121099.4437572.622BC140035.771231057.974001.1177734.443表102觀測(cè)值數(shù)據(jù)表編號(hào)角度觀測(cè)值′″編號(hào)角度觀測(cè)值′″編號(hào)邊長(zhǎng)觀測(cè)值m123456840738.2374634.9580544.1330303.21260155.7205502.3789101112741816.8772759.1281343.2552109.9722225.8521620.5131415161718246

3、3.943414.615216.166042.945085.085014.99設(shè)置本例置本例題的目的：的目的：理解、熟悉赫爾默特方差分量估計(jì)方法的方差估計(jì)過(guò)程。解：解：分析：分析：此題為邊角網(wǎng)，因此，將角度、邊長(zhǎng)作為兩類(lèi)觀測(cè)值，按照赫爾默特方差分量估計(jì)模型進(jìn)行估計(jì)即可。1第一次平差（預(yù)平差）第一次平差（預(yù)平差）（1）第一次定權(quán)）第一次定權(quán)設(shè)，則，????5.10???（無(wú)量綱）1220??????P圖101181716151413121

4、110987654321ABCP1P2設(shè)編號(hào)為的角度，測(cè)站點(diǎn)點(diǎn)號(hào)為，第一照準(zhǔn)點(diǎn)點(diǎn)號(hào)為，第二照準(zhǔn)點(diǎn)點(diǎn)號(hào)為，則ijhk角度誤差方程按下式組成ikkjkkjhhjhhjjhjkjjhjkjihjkjilybxaybxaybbxaalv?????????????????)(?)(????其中).(00ihikiiLl?????組成結(jié)果列于表75邊長(zhǎng)誤差方程：邊長(zhǎng)誤差方程：設(shè)編號(hào)為的觀測(cè)邊長(zhǎng)，兩端點(diǎn)點(diǎn)號(hào)為和，則角度誤差方程按下式組成ijk（系數(shù)無(wú)

5、量綱）iiSjkjjkjjkjjkjSlyxyxV???????sin?cos?sin?cos0000????表105誤差方程組成表.0iiiSSl??表105誤差方程組成表序號(hào)Dxa?Dyb?Exc?Eyd?iliP1234567891011121314151617180.550.250.800.250.640.390.800.840.040.040.350.390.310.910.21001.000.820.560.260.560.

6、640.080.260.150.410.410.330.080.950.410.98000.090000000.220.040.180.300.040.260000.640.831.000000000.340.410.070.190.410.220000.770.550.092.660.170.310.130.880.190.931.012.841.940.431.430.61.82.413.410.310.41111111111110