测量学复习总结 - 下载本文

测量学复习总结

例 ① 在相同的观测条件下,对某段距离测量了五次,各次长度分别为:121.314m, 121.330m, 121.320m, 121.327m, 121.335m。试求:(1)该距离算术平均值;(2)距离观测值的中误差;(3)算术平均值的中误差;(4)距离的相对误差。 解:(1)算术平均值 L = 121.325m

(2)观测值的中误差 m = ±[ [vv]/(n-1) ]1/2 = ± 0.0083m (3)算术平均值的中误差 mL= ±[ [vv]/n*(n-1) ] 1/2= ±0.0037m (4)距离的相对误差为:mL /L = 1:32685

例 ② 今用钢尺丈量得两段距离:S1 = 60.25 ± 6 cm, S2 =80.30 ± 7 cm,S3 =102.50 ± 8 cm,距离S4 = (S1 + S2 + S3 )/3,分别计算 S4的距离值、中误差和相对误差。

解:S4 = 81.017m m42 = (m12 + m22 + m32) / 9 = 16.56

m4 = ±4.07cm

相对误差为:0.0407 / 81.017 = 1/1993

例 ③对某角度进行了6个测回,测量角值分别为42°20′26″、42°20′30″、42°20′28″、42°20′24″、42°20′23″、42°20′25″,试计算:(1)该角的算术平均值;(2)观测值的中误差;(3)算术平均值的中误差。

解:(1)算术平均值为:42°20′26″(2分) (2)观测值的中误差为:±2.6″(6分)

(3)算术平均值的中误差为:±1.16″(10分)

例 ④下今用钢尺丈量得两段距离:S1 = 120.63 ± 6.1cm, S2 =114.49 ± 7.3cm,试求距离S3 = S1 + S2 和S4 = S1 - S2 的中误差和它们的相对中误差。

解:S3 = S1 + S2 = 235.12m (1分)

m3 = (m1* m1 + m2* m2 ) 1/2 = 9.5 cm(3分) ρ3 = m3 / S3 = 1/2475(5分) S4 = S1 - S2 = 6.14m(6分)

m4 = (m1* m1 + m2* m2 ) 1/2 = 9.5 cm(8分) ρ4 = m4 / S4 = 1/65(10分)

例⑤ 用钢尺丈量某一段距离,6次测量的距离值分别为(单位m):20.290, 20.295, 20.298, 20.291, 20.289, 20.296,试计算:(1)距离最或是值;(2)距离观测值中误差;(3)最或是值的中误差;(4)相对误差。

解:

(1) 算术平均值 L = 20.293m

1/2

(2) 观测值的中误差 m = ±[ [vv]/(n-1) ] = ± 0.0037m

1/2

(3) 算术平均值的中误差 mL= ±[ [vv]/n*(n-1) ] = ±0.0015m 距离的相对中误差为:mL /L = 1:13434

4.闭合水准路线(计算高程):

例 ① 根据下图所示水准路线中的数据,计算P、Q点的高程。

测量学复习总结

BM-3.0011.4kP -4.740HBM1 = 163.751m HBM2 = 157.732m BM1.7193.5kQ 6.3k 解:(1)计算高差闭合差: △h = HBM2 - HBM1 = 157.732 – 163.751 = -6.019 m

∑h = -3.001 – 4.740 + 1.719 = = - 6.022m

fh = ∑h - △h = -6.022 – (-6.019) = -0.003m = -3mm

(2)分配闭合差,计算改正数

∑L = 1.4 + 6.3 + 3.5 = 11.2 km v1 = - (L1/∑L) * fh = 0mm v2 = - (L2/∑L) * fh = 2mm

v3 =- (L3/∑L) * fh =1mm (3)计算改正后的高差的高程

HP = HBM1 + h1 + v1 = 163.751 – 3.001 + 0 = 160.750m

HQ = HP + h2 + v2 = 160.750 – 4.740 + (0.002) = 160.750 – 4.738 = 156.012m 或HQ = HBM2+ (h3 + v3) = 157.732 – 1.719 –0.001 = 160.750 – 4.738 = 156.012m

例 ② 闭合水准路线高差观测如图,已知A点高程HA = 41.20m,观测数据如图所示(环内单位为m的为两点高差,环外单位为km为两点距离),计算B、C、D、E点的高程。(此题图不知哪里去了!)

解:(1)计算高差闭合差:fh = ∑h = -0.024m = -24 mm (2)分配闭合差,计算改正数 ∑L = 12km

v1 = - (L1/∑L) * fh = 2mm v2 = - (L2/∑L) * fh = 6mm v3 =- (L3/∑L) * fh = 4 mm v4 =- (L4/∑L) * fh = 7mm

v5 =- (L5/∑L) * fh = 5mm (3)计算改正后的高差的高程 HB = HA+ h1 + v1 = 39.784m HC = HB + h2 + v2 = 37.338m

测量学复习总结

HD = HC + h3 + v3 = 39.399m HE= HD + h4 + v4 = 40.184m

例 ③根据下图所示水准路线中的数据,计算P、Q、R点的高程。

M N

+1.326

2.1k-3.517

3.4k

+3.718+2.674

P R

HM = 26.201m

2 4 Q HN = 30.425m

解:(1)计算高差闭合差:fh = ∑h - △h = -0.023m = -23 mm

(2)分配闭合差,计算改正数 ∑L = 11.5km

v1 = - (L1/∑L) * fh = 4.2mm v2 = - (L2/∑L) * fh = 8mm v3 =- (L3/∑L) * fh = 4 mm

v4 =- (L4/∑L) * fh = 6.8mm (3)计算改正后的高差的高程 HP = HM+ h1 + v1 = 27.531m HQ = HP + h2 + v2 = 30.213m

HR = HQ + h3 + v3 = 33.935m

李翔 于成都信息工程学院

2010年11月27日