中國(guó)礦業(yè)大學(xué)測(cè)量平差習(xí)題參考答案?jìng)浞謟sb.

1、第一章習(xí)題參考答案1 題 .略2題. 解 (1)429212（ 2）(L13L2)2129L122 222(3)cosL1 cos(L1L2 )sin L1 sin(L1 L2 )2sin L1 sin( L1L2 )2cos2 (L1L2 )"1"2cos2 ( L1 L2 )3題. 解 DXLADLL , DYLBADLL , D XYADLL T AT BT4 題 .解 設(shè)路線(xiàn)總長(zhǎng)S 公里，按照測(cè)量學(xué)上的附合路線(xiàn)計(jì)算步驟，則路線(xiàn)閉合差fhH Ah1h2H B由于是路線(xiàn)中點(diǎn)，故11v1v22 f h2 H Ah1h2H B則線(xiàn)路中點(diǎn)高程?H Ah1v1H A1H Ah1

2、h2H BH 中點(diǎn)h1211112 h12 h22 H A2 H B11h11 H AH Bh2222設(shè)每公里高差觀(guān)測(cè)中誤差為0 ，則h1h 2(s / 2 )0 按誤差傳播定律1120122 ?h1H 中點(diǎn)22021h221h121h21( s / 2 0 )21( s / 2 0 )2441441(s / 25) 21( s / 2 5) 225S10 2,S 16(km)4445.解 設(shè)每個(gè)測(cè)回的中誤差為0 ，需要再增加 n 個(gè)測(cè)回，則00.42,00.4220(1)2000.28,00.28n 20( 2)n20由上式可解出n.即200.42232n0.282202022025再增加2

3、5 個(gè)測(cè)回6 題 .解xpLp1 ( p1 L1 p2 L2 . pn Ln )pL1p1p2.pnL2PPP,Ln100p1p1p1p1p2pn010p2QxxpxPP.p20p ,P001pnpnpp11p1p21p2.pn1pnpp1ppp2pppnpp1p 2p2pnp 2p 21ppxp7 題。解PDdD8 題 . 解 由題意可求出Qxx , Qyy , Qxy 即：100p1101i02nQ xxp2212n0pi0i 1001npnn2n同理可得： Qyyi,QxyiiQyxi 1pii 1piQFF1f1f 2Q xxQxyf12Qxx2 f 1 f 2 Qxy2Q yypFQ

4、 yxQ yyf 2f1f 2n2nn22i2 f1f 2ii2if1pif 2pii 1 pii 1i 19 題。解： (1)L的協(xié)因數(shù)陣為單位陣E.Q xx(BTB) 1 BTE(BT B) 1 BT )T(BT B) 1BT B( BT B) 1)T(BTB) 1?LVL?L?LBXBXQ?BQLL BTB(BT B) 1 BTL L(2).因?yàn)?V=BX-L,QVXBQxxQLXB(BT B) 1B(BT B)T ) 10因?yàn)??LV , LBXVLQ?QLVQVVBQXVQVVQVVBQXV0L V10 題。解（ 1）每千米觀(guān)測(cè)高差的中誤差的估值為?02pdd9.2,?0 3mm2n

5、（2）第二段觀(guān)測(cè)高差的中誤差?2s2?05mm（3）第二段觀(guān)測(cè)高差的平均值的中誤差?2中?24mm2（ 4）全長(zhǎng)一次觀(guān)測(cè)高差的中誤差（ 5）全長(zhǎng)高差平均值的中誤差全S ?0 12mm全平8mm2第二章平差數(shù)學(xué)模型與最小二乘原理1. n=8,t=4,r=n-t=4H AH B0h1h2h3h4H AH B0h1h6h5h4H AH B0h1h7h40h2h8h62. n=4,t=3,r=n-t=1,u=1 固方程個(gè)數(shù)為r+u=2 個(gè)1 2(l1l2 )hs 00sx3. n=5,t=3,r=n-t=2,u=3h1Xh 2Xh 3Xh 4Xh 5X12322HHHXXAAA314. n=4,t=2

6、,r=n-t=2,u>t,s=r+u=5h1X1h2X 2h3X 2h4X 3H AH B0X1X 2X 35.n=6,t=3,r=n-t=3H AH B0h1h6h5H AH B0h1h3h40h2X 3h5h60X1h10X 2h2H AH B0X1X2X 3（考慮到 B 矩陣必須列滿(mǎn)秩，因此在一般條件式中所選的未知參數(shù)至少出現(xiàn)一次）第三章條件平差3.1.解 n=7,t=3,r=n-t=4. 平差值條件方程為：?0h1h2h5?0h5h6h7?0h3h4h6?h1h3H B 0H A改正數(shù)條件方程為：v1v2v570v5v6v770v3v4v630v1v340令 C=1 ，觀(guān)測(cè)值的權(quán)

7、倒數(shù)為：112P12112N AP 1AT,K N 1W,V P 1AT K 得：N =3-10-1-14-100-15-2-1-10-23=N0.46070.14610.12360.23600.14610.31460.11240.12360.12360.11240.32580.25840.23600.12360.25840.5843V- 0.432.8- 4.4- 0.3 - 3.8 - 1.22.0 T (mm)?hV1.359 2.0120.3590.6400.653 0.999 1.652 (m)h?h1?h2?h3下面求平差后000000 1?h7的中誤差： h7h4?h5?h6?h

8、7f0000 0 0T?0V TPV35.57301 ,r43.0(mm)Q fff T Qff T QAT N1 AQf0.7416中誤差為h7VTPVQ ff2.6mmr利用 Matlab 編寫(xiě)上題的代碼為：（%后的內(nèi)容為注釋部分）clear%清除內(nèi)存中的變量A=-1100-100;000011-1;001-10-10;10-10000, a=1 1 2 2 1 1 2;Q=diag(a),% 生成對(duì)角矩陣QP=inv(Q),%inv() 為矩陣求逆運(yùn)算W=7;-7;-3;4;N=A*Q*A',%A表示 A 的轉(zhuǎn)置K=inv(N)*W,V=Q*A'*K,f=0;0;0;0;

9、0;0;1;sigma=sqrt(V'*P*V/4),%sqrt()為開(kāi)根號(hào)運(yùn)算Qff=f'*Q*f-f'*Q*A'*inv(N)*A*Q*f,3.2.解 n=9 ,t=4.r=n-t=5. 即 5 個(gè)條件方程，選取網(wǎng)中3 個(gè)圖形條件，一個(gè)圓周條件，一個(gè)極條件。個(gè)圖形條件方程為：v1v2v7w10v3v4v8w20v5v6v9w30一個(gè)圓周條件方程為：v7v8v9w40一個(gè)極條件方程為：?sin L5sin L3sin L11 0?sin L6sin L4sin L2其改正數(shù)形式為：cot L1v1 cot L2 v2cot L3v3cot L4 v4 cot

10、L5v5 cot L6v6w5 0w5'' (1sin L6 sin L4 sin L2 )33''sin L5 sin L3 sin L1將以上改正數(shù)條件方程寫(xiě)成矩陣形式為：AV-W=01100001003.10011000101.1A000011001，W1.90000001111.81.67241.09981.50052.60862.27231.846700033P 為單位陣。 N=AQA T ,K=N -1 W,VP 1ATKN =3.0000001.00000.572603.000001.0000-1.1081003.00001.00000.42561

11、.00001.00001.00003.000000.5726-1.10810.4256021.6364-1N =0.39050.05230.0567-0.1665-0.00880.05230.39560.0531-0.16700.01780.05670.05310.3898-0.1665-0.0064-0.1665-0.1670-0.16650.5000-0.0009-0.00880.0178-0.0064-0.00090.0475?ABsin L6sin L2SCD?sin L5sin L7兩邊取全微分得：sin L6 sin L2?sin L6 sin L2?d ?ABcot L2dL2c

12、ot L5dL5ABSCDsin L5 sin L7"sin L5 sin L7"AB sin L6 sin L2?cot L6dL6sin L5 sin L7"d ?"?令 ?SCDcot L2SCDdL2AB sin L6 sin L2?cot L7dL7sin L5 sin L7"?cot L5 dL5cot L6 dL6cot L7 dL7T01.0998002.72231.84670.30280Tf0QFFf T Qff T QAT N 1 AQf 5.5072由于 ?0VTPV3.485" ,可求得該平差值函數(shù)的方差,rF

13、F ?0 QFF 8.18(cm)CDFF8.181CD 邊相對(duì)中誤差為20626525200SCD利用 Matlab 編寫(xiě)上題的代碼為：（%后的內(nèi)容為注釋部分）Clear %第二題A=110000100;001100010;000011001;000000111;1.6724-1.09981.5005 -2.6086 2.2723 -1.8467 0 0 0,Q=eye(9),P=inv(Q),W=-3.1;1.1;-1.9;-1.8;33;N=A*Q*A',K=inv(N)*W,V=Q*A'*K,f=0;1.0998;0;0;-2.7223;1.8467;0.3028;0;

14、0;sigma=sqrt(V'*P*V/5),Qff=f'*Q*f-f'*Q*A'*inv(N)*A*Q*f,sigmaf=sigma*sqrt(Qff)3.3.解 觀(guān)測(cè)值個(gè)數(shù)n=9，必要觀(guān)測(cè)數(shù)t=2*3=6, 多余觀(guān)測(cè)數(shù)r=n-t=3.可以列出一個(gè)方位角附合條件和兩個(gè)坐標(biāo)附合條件。解題過(guò)程如下。（1）近似計(jì)算各個(gè)導(dǎo)線(xiàn)邊的方位角和導(dǎo)線(xiàn)點(diǎn)的坐標(biāo)列于下表。近似坐標(biāo) (m)近似方位角2(8099.150 , 3578.571)T1 = 341?1541.23(8400.223, 4836.524)T2= 76? 3225.04(9511.116 , 5363.205)

15、T3 = 25 ?2157.35(8748.204 , 6667.676)T4= 120? 1915.5T5 = 229 ?3008.4(2)組成改正數(shù)條件方程及第3 點(diǎn)平差后坐標(biāo)函數(shù)式改正數(shù)條件方程閉合差項(xiàng)：w1(T5TBD )5.0"w2( X 5X B )4.9cmw3(Y5YB )2.9cm改正數(shù)條件方程v i 15w10cos TivSi 141(Y5Yi )v i 14w202062.65sin TivSi 141(X5X i )v i 14w302062.65即v1 + v2 + v3 + v4 + v5 5.0 = 00.9470vS1 + 0.2328 vS 2 -

16、+0.9036 vS 3 0.5053vS 4 1.2440v1 1.4976v2- 0.8878 v3 -0.5012v4 + 4.9 = 0v+0.9725vvv+1.0624vv-0.3212 S1S 2 + 0.4284S 3 +0.8632S 41+0.31472+0.1687v0.3699v4+2.9 = 03000011111A0.9470.23280.90360.50531.2441.49760.88780.501200.32120.97250.42840.86321.06240. 31470.16870.3699.0W= 5.0-4.9-2.9 T第 3 點(diǎn)平差后坐標(biāo)函數(shù)式?

17、x1x1cosT1cosT2x3x1x2s1s2?y1y1sin T1y3y1y 2s1s2 sin T2全微分得?21? 21( y3 yi )d i 1dx3cos Ti dsi?21? 2( x3 xi ) d i 1dy 3sin Ti dsi 1Tfx3= 0.94700.232800-0.3562-0.6099000 fy3= -0.32120.9725000.89360.1460000 T(2)確定邊角觀(guān)測(cè)值的權(quán)設(shè)單位權(quán)中誤差 ?0?3" ；根據(jù)提供的標(biāo)稱(chēng)精度公式?= 5 mm + 5ppm?Dkm 計(jì)算測(cè)邊中誤差D根據(jù) (3-3-26) 式，測(cè)角觀(guān)測(cè)值的權(quán)為P = 1

18、；為不使測(cè)邊觀(guān)測(cè)值的權(quán)與測(cè)角觀(guān)測(cè)值的權(quán)相差過(guò)大，在計(jì)算測(cè)邊觀(guān)測(cè)值權(quán)時(shí)，取測(cè)邊中誤差和邊長(zhǎng)改正值的單位均為厘米（ cm）。PD?02(s2 cm2)?D2則可得觀(guān)測(cè)值的權(quán)陣為5 . 26 . 87 . 25 . 7P11111(3)組成法方程，計(jì)算聯(lián)系數(shù)、改正數(shù)及觀(guān)測(cè)值平差值，得N =5.0000-4.13061.1759-4.13065.1684-1.80531.1759-1.80531.7081N-1=0.61250.54250.15170.54250.78720.45850.15170.45850.9656KN 1W (AP 1AT)W 0.03572.47444.2883TVP 1ATK

19、0.2 0.70.6 0.41.5 2.3 1.42.8 0.0 T?1628.522S1?1293.473S2?1229.415S3?1511.181S4?'291 45 26.3"1?'275 16 46.1"2?'128 49 33.7"3?'27457 21.0"4?'289 10 52.9"5進(jìn)一步計(jì)算各導(dǎo)線(xiàn)點(diǎn)的坐標(biāo)平差值，得1 (8099.144 , 3578.561) ； 2 (8400.210 , 4836.508) ；3(9511.092, 5363.198)(4)精度評(píng)定1）單位權(quán)中誤

20、差?0VTPV2.9"r2）點(diǎn)位中誤差權(quán)倒數(shù)：Qx?3f xT3 P 1 f x3f xT3 P 1 AT N 1 AP 1 f x30.1818Qy?3f yT3 P 1 f y3f yT3 P 1 AT N 1 AP 1 f y 30.1836點(diǎn)位中誤差：?Qx?31.21cm，?Q ?1.22cmx?30y30y3?3?x?23?y?321.72cm利用 Matlab 編寫(xiě)上題的代碼為：（%后的內(nèi)容為注釋部分）clear%第三章第三題A=0 0 0 0 1 1 1 1 1;0.947 0.2328 0.9036 -0.5053 -1.244 -1.4976 -0.8878 -0

21、.5012 0;-0.3212 0.9725 0.4284 0.8632 1.0624 0.3147 0.1687 -0.3699 0,a=5.2 6.8 7.2 5.7 1 1 1 1 1;P=diag(a),Q=inv(P),W=5.0;-4.9;-2.9;N=A*Q*A',K=inv(N)*W,V=Q*A'*K,fx=0.947;0.2328;0;0;-0.3562;-0.6099;0;0;0;fy=-0.3212;0.9725;0;0;0.8936;0.146;0;0;0;sigma=sqrt(V'*P*V/3),Qffx=fx'*Q*fx-fx'

22、;*Q*A'*inv(N)*A*Q*fx,Qffy=fy'*Q*fy-fy'*Q*A'*inv(N)*A*Q*fy,sigmafx=sigma*sqrt(Qffx),sigmafy=sigma*sqrt(Qffy),sigma3=sqrt(sigmafx2+sigmafy2)3.4.解（1）由條件平差的法方程可知AP 1ATW0N aa AP 1 AT102,W6246要求單位權(quán)中誤差 ?0VT PVrW 的維數(shù)即為 r,所以 r=2.V P 1ATK,VTKTAP 1V T PV K T AP 1 PP 1 AT K K T Naa KK 為可求量。 K1,V

23、 T PV K T N aa K 182VT PV183?0r2Q FFfT Q? f f T( P 1P 1 AT N aa1 AP 1 ) fL Lf T P 1 ff T P 1 AT N aa1 AP 1 f（2）由于( ff/ p)( af / p, bf/ p) * N aa1 * (af/ p, bf / p) T1011644162404444424則F0QFF3.046.03.5.解（ 1）由于 BC BD BCBDABAB而 BC?BD?sin L8,sin L1BD?ABsin(180?sin L5L1L3 )代入上式得sin(180?L1L3 ) sin L5ABBC?

24、sin L1 sin L8?sin( L1L3 ) sin L5ABBC?sin L1sin L8全微分得 AB 邊的權(quán)函數(shù)式：d ABABL3 )cot L1 )?(cot( L1dL1cot( L1 L3 ) dL3 )"?cot L8dL8 )cot L5 dL5AB (cot( L1L3 )cot L1 )0cot(L1L3 )0cot L5"T ?f dL（ 2）平差后BE?沒(méi)有參加平差，故?BCL9，由于 L9L9BEBCL9則1Q BEQL9L91P BE?dL1?0 0dL2cot L8 )?dL8L93.6.解設(shè)每公里的測(cè)量中誤差為0 , h1 的水準(zhǔn)路線(xiàn)

25、長(zhǎng)度為X ，則 h2 的水準(zhǔn)路線(xiàn)長(zhǎng)度為S-X.則?H Ah1v11 1h1H AH1v121D ? ?11h1h1h1v12H1H11v1 h1v1v1h1X0 ,h 2SX0v1Xh1 h2H B )( H ASh110h1h2XXX2020v1v1SS0(S X)20XSX 22X S 0Sh1v1 1 0 X 0200(S X)20XX 2S2X S 0S2DH?1H?11 1h1h1h1v12v1h1v1v11X 21XS上式對(duì) X 求導(dǎo)并另其等于零得： dD ? ?1 2 X0H1H1SX=S/2 時(shí) D ? ? 取 最大值。H1H13.7.解(1) 由于 AV+W=0,W=ALAP

26、 1ATKW0K(AP 1AT ) 1W法方程 VP 1ATKP 1AT (AP 1AT) 1W)P 1AT (AP 1AT ) 1 ALFf T (L V )所以f T ( L P 1 AT ( AP 1 AT ) 1f T ( E P 1 AT ( AP 1 AT ) 1AL )A) LQFFf T (E P 1 AT (AP 1 AT ) 1 A) QLL ( E AT (AP 1 AT ) 1 AP 1) ff T (QLLP 1 AT ( AP 1 AT ) 1 AQLL ) ( E AT ( AP 1 AT ) 1 AP 1 ) ff T (QLLQLL AT(AP 1AT) 1A

27、P 1P 1AT(AP 1AT ) 1 AQLLP 1 AT ( AP 1 AT ) 1 AQLL AT ( AP 1 AT ) 1 AP 1 ) ff T (QLLQLL AT ( AP 1 AT ) 1 AQLL ) f(2 由VP 1AT(AP 1AT) 1 ALFf T ( LV)f T (E P 1 AT (AP 1 AT ) 1 A) LQVFP 1 AT ( AP 1 AT ) 1 A QLL (E AT ( AP 1 AT ) 1 AP 1 ) f則( P 1AT(AP 1AT) 1A QLLP 1 AT ( AP 1 AT ) 1 A QLL AT ( AP 1 AT ) 1

28、 AP 1 ) f( P 1AT(AP 1AT) 1A QLLP 1 AT ( AP 1 AT ) 1 AP 1 ) f03.8.解 n=9,t=7,r=n-t=2.（ 1）列條件方程只需列兩個(gè)條件方程，如圖所示兩個(gè)圖形條件分別為v 1v 2v 3v 4w1 0式中w1( 1234360 )1arccos S22S52S12,2arccos S72S52S622S2 S52S7 S5S7 2S8 2S9 2,S8 2S2 2S4 23arccos2S7 S84arccos2S2S8v" ( vScosCBAvS2cosBACvS5 )11h1v 2" ( vS 6cosCA

29、EvS 5cosAECvS 7 )h2vv3" (vS 9cosCEDvS 7cosCDEvh34" ( vS4cosCBDvS 2cosCDBvh4S8S8)h1S5sinBAC , h2S5sinCAEh3S7sinCED , h4S8sinCDBv 5v 6v 7w20式中w2( 567 )S4 2S8 2S2 2, 6S8 2S9 2S7 25arccos2S2S5arccos2S7 S52227 arccos S4 S9 S3 2S7 S8v 5" (vS2cosDBCvS 4cosBCDvS8 )h5v 6" (vS 7 cosCEDvS9cosDCEvS8 )h6v 7" (vS 3cosDBEvS 4cosBDEvS 9 )h7h5S4sinD