Tadeusz Gargula
Rachunek wyrównawczy 3U]\NáDG\RSUDFRZDQLDüZLF]H
Kraków 2005
6NU\SWRSLQLRZDá
*
SURIGUKDELQ 5RPDQ.DGDM
Autor skryptu jest pracownikiem naukowo-dydaktycznym w Katedrze Geodezji, Akademia Rolnicza w Krakowie, ul. Balicka 253A, 30-149 Kraków
Copyright © by GEODPIS
Druk, oprawa: P.W. STABIL Kraków Wydawnictwo: GEODPIS Andrzej Jagielski tel./fax (012) 411-89-43 tel. kom. 505-204-149 e-mail:
[email protected] ISBN 83-922884-1-6
6SLVWUH FL
3
6SLVWUH FL
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3UDZRSU]HQRV]HQLDVL Eá GyZ UHGQLFKREVHUZDFMLQLH]DOH*Q\FK ....................... 6
:\UyZQDQLHVSRVWU]H*HEH]SR UHGQLFK3XQNWZ ]áRZ\ZQLZHODFML ................... 8 8NáDGUyZQDOLQLRZ\FKMHGQR]QDF]QLHRNUH ORQ\FK .............................................. 10
8NáDGUyZQDOLQLRZ\FKQDGRNUH ORQ\FK.................................................................. 12
8NáDGUyZQDOLQLRZ\FKQLHGRRNUH ORQ\FK............................................................... 15
:\UyZQDQLHVWDF\MQHPHWRGSR UHGQLF]F .......................................................... 17
:\UyZQDQLHVWDF\MQHPHWRG]DZDUXQNRZDQ ....................................................... 21
:\UyZQDQLHVLHFLQLZHODF\MQHMPHWRGSR UHGQLF]F .......................................... 24
:\UyZQDQLHVLHFLQLZHODF\MQHMPHWRG]DZDUXQNRZDQ ....................................... 27 Rozwijanie funkcji nieliniowej w szereg Taylora ...................................................... 30 :\UyZQDQLHZFL FLDZVWHF] NWRZHJR PHWRGSR UHGQLF]F .......................... 32 :\UyZQDQLHZFL FLDZVWHF] NLHUXQNRZHJR PHWRG SR UHGQLF]F ................... 37
:\UyZQDQLHZFL FLDZVWHF] NLHUXQNRZHJR PHWRG SR UHGQLF]FSRSU]H]
HOLPLQDFM QLHZLDGRPHMRULHQWDF\MQHM ........................................................................ 42
%áGSRáR*HQLDSXQNWXHOLSVDEá GX UHGQLHJR ....................................................... 47 :\UyZQDQLHZFL FLDZSU]yG NLHUXQNRZHJR PHWRG SR UHGQLF]F ................. 49
:\UyZQDQLHF]ZRURERNXJHRGH]\MQHJR NWRZHJR PHWRG SR UHGQLF]F ...... 54
:\UyZQDQLHF]ZRURERNXJHRGH]\MQHJR NWRZHJR PHWRG ]DZDUXQNRZDQ ... 61 Wyrównanie czworoboku geodezyjnego (liniowego) PHWRGSR UHGQLF]F ...... 64
4
Spis trH FL
Wyrównanie czworoboku geodezyjnego (liniowego) PHWRG]DZDUXQNRZDQ ... 69
:\UyZQDQLHFLJXSROLJRQRZHJRPHWRGSR UHGQLF]F ...................................... 73
:\UyZQDQLHFLJXSROLJRQRZHJRPHWRGZDUXQNRZ ........................................... 80 3UDZRSU]HQRV]HQLDVL Eá GyZ UHGQLFKZLHONR FLVNRUHORZDQ\FK.
%áG UHGQLfunkcji. ...................................................................................................... 84
:\UyZQDQLHWUDQVIRUPDFMLZVSyáU] GQ\FK dla n>2 punktów dostosowania ......... 87
Dodatek A: Przydatne wzory matematyczne ........................................................... 92 Dodatek B: Podstawowe wzory z rachunku wyrównawczego ............................... 94 'RGDWHN& 6áRZQLF]HNZD*QLHMV]\FKWHUPinów z rachunku wyrównawczego .... 96
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czenia. •
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z-
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b-
liczania.
FLHM ZLHONR FLZ\UyZQDQHEá G\ UHGQLHLWS QDOH*\SRGNUH OLü
•
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•
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•
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*
temat GDQHJR üZLF]HQLD D QLH W\ONR
QXPHUNROHMQ\üZLF]HQLD
FLHMSRSHáQLDQHEá Gy w obliczeniach: o =E\W PDáD GRNáDGQR ü REOLF]H /LF]ED F\IU ]QDF]F\FK Z REOLF]HQLDFK SR UHGQLFK 1DMF]
nie poZLQQDE\üPQLHMV]DRGOLF]E\F\IUZZ\QLNX o
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cinku. o
1LHXZ]JO
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FLZ>P@ o 3U]\F]\Q Eá GQ\FK Z\QLNyZ V F] VWR (OH SU]HSLVDQH GDQH =DZV]H QDOH*\ MH VSUDZG]DüFRQDMPQLHMGZXNURWQLH
o
*
GRNáDGQR FL :\QLN REOLF]H SRZLQLHQ E\ü SRGDZDQ\ ] GRNáDGQR FL FR QDMZ\*HM MHGHQ U]G Z\*V] RG GRNáDdQR FLGDnych. PU]\NáDGRZR MH OLGáXJR FLSRGDQHV ]GRNáDGQR FLGR FPWR ZyQLNSRZLQLHQE\ü]aRNUJORQ\GRPP %á GHP MHVW WDN H SRGDZDQLH Z\QLNyZ ] SU]HVDGQ
6 ,PL
3UDZRSU]HQRV]HQLDVL
Eá GyZ UHGQLFKREVHUZDFMLQLH]DOH*Q\FK
1D]ZLVNR
Nr zestawu .....
û:,&=(1,(QU
Temat: 3UDZRSU]HQRV]HQLDVL Eá GyZ UHGQLFKREVHUZDFMLQLH]DOH*
nych Zadanie : ]DGDQLX ZFL FLD Z SU]yG GDQH V EH]Eá GQH ZVSyáU] GQH SXQNWyZ $ L % oraz poPLHU]RQH ZDUWR FL NWyZ α i β ZUD] ] Eá GDPL UHGQLPL W\FK SRPLDUyZ mα, mβ. 2EOLF]\ü Eá
G\ UHGQLH ZVSyáU] GQ\FK mx, my SXQNWX ZFLQDQHJR 3 RUD] EáG SRáR*HQLD
µP tego punktu. P Dane: xA = 1357,20
xB = 1393,22
yA = 1448,22
yB = 2896,53
α = 54°26′55″
mα = ± 10″
β = 63°21′08″
mβ = ± 10″
β
α
B
A 52=:,
1)
=$1,(
:]RU\QDREOLF]HQLHZVSyáU] GQ\FKSXQNWX3
xP = xA + dAP· cosAAP
AAP = AAB - α
=
G$3
Ã
VLQ
yP = yA + dAP· sinAAP 2) Obliczenie azymutu boku AB tgAAB = 3)
û\AB = 1448,31 = 40,208 û[AB 36,02
2EOLF]HQLHGáXJR
FLERNX
AB
2 2 dAB = ∆xAB + ∆y AB = 1448,76 P
:VSyáU] GQHSXQNWXZFLQDQHJR
sin • cos( AAB α ) = sin (. + sin • sin( AAB α ) = yP = yA + dAB • sin (. + xP = xA + dAB •
. + Ã
VLQ
AAB = 88°34′31″
• G
%$ 3UDZRSU]HQRV]HQLDVL
3RFKRGQHF]
VWNRZH]
Eá GyZ UHGQLFKREVHUZDFMLQLH]DOH*Q\FK
7
xP i yPZ]JO GHPα i β
sin( AAB − α ) ⋅ sin( α + β ) − cos( A AB − α ) ⋅ cos( α + β ) ∂x P = d AB ⋅ sin β ⋅ = ∂α sin 2 ( α + β ) cos( AAB + β ) = d AB ⋅ sin β ⋅ = 1460,16 P sin 2 ( α + β ) cos β ⋅ sin( α + β ) − sin β ⋅ cos( α + β ) ∂x P = d AB ⋅ cos( AAB − α ) ⋅ = ∂β sin 2 ( α + β ) sin α = d AB ⋅ cos( AAB − α ) ⋅ = 1246,98 P 2 sin ( α + β ) cos( A AB − α ) ⋅ ( −1) ⋅ sin( α + β ) − sin( A AB − α ) ⋅ cos( α + β ) ∂y P = d AB ⋅ sin β ⋅ = ∂α sin 2 ( α + β ) = d AB ⋅ sin β ⋅
sin( AAB + β ) = − 778,75 P sin 2 ( α + β )
cos β ⋅ sin( α + β ) − sin β ⋅ cos( α + β ) ∂y P = d AB ⋅ sin( AAB − α ) ⋅ = ∂β sin 2 ( α + β ) sin α = d AB ⋅ sin( A AB − α ) ⋅ = 845,12 P 2 sin ( α + β )
*
:
:]RU\SRPRFQLF]HGRREOLF]HQLDSRZ\ V]\FKSRFKRGQ\FK
sin(α + β ) = sin α ⋅ cos β + sin β ⋅ cos α cos(α + β ) = cos α ⋅ cos β + sin β ⋅ sin α
6 %á G\ UHGQLHZVSyáU] GQ\FKSXQNWX3
P
x
2
mx = ± 0,089
P
y
2
∂x m ∂x = ± P ⋅ α + P ∂α ρ" ∂β
2
P 2
2
2
2
m ⋅ β = ± ρ"
2
∂y m ∂y m = ± P ⋅ α + P ⋅ β = ± ∂α ρ" ∂β ρ"
2
2 + (1638,34m) ⋅ 10" 206265"
10" (1460,16m)2 ⋅ 206265 "
my = ± 0,055 P 7 %áGSRáR*enia punktu P mP = ± mx2 + my2 = ± ( 0,089P)2 + ( 0,055P)2 = ± 0,105 P
2
2 + (845,12m) ⋅ 10" 206265"
10" (− 778,75m)2 ⋅ 206265 "
2
2
*
8
:\UyZQDQLHVSRVWU]H H EH]SR UHGQLFK3XQNWZ
,PL
]áRZ\ZQLZHODFML
1D]ZLVNR
Nr zestawu .....
û:,&=(1,(QU
Temat: :\UyZQDQLHVSRVWU]H*HEH]SR UHGQLFK
3XQNWZ ]áRZ\ZQiwelacji
Zadanie :Z\QLNX SRPLDUX QLZHODF\MQHJR FLJyZ R GáXJR FL d1, d2, d3 X]\VNDQR Uy*Qice Z\VRNR FL ∆h1, ∆h2, ∆h3.
2EOLF]\ü Z\VRNR üZ\UyZQDQ
SXQNWX W
PHWRG REVHUZ
cji bezpo UHGQLFK Dane: R1 HR
∆h
d
R1
318,097
3.601
1.5
R2
333,078
18.596
2.0
R3
303,900
10.590
2.0
Nr punktu
52=:,
∆h1 W
∆h3
∆h2
R3
=$1,(
1) Zestawienie tabelaryczne Nr
L
d
p
l
1
314.496
1,5
0,67
14
2
314.482
2,0
0,50
0
0
3
314.490
2,0
0,50
8
4
1,67
suma
pl
pll
9,38 131,32
v
pv
-4,0 24,12 314,490
0
8,0
4,0
32,00 314,490
32
0,0
0,0
0,00 314,490
2,0
0,0
56,12
13,38 163,32
Li = HRi +(-) ∆h
Wyrazy wolne: li = Li – x0
*
:DUWR üSU]\EOL RQDQLHZLDGRPHM Z\VRNR FL
x0 = 314,482 m
Poprawki obserwacji: vi = ∆x - li
3U]\URVWV]XNDQHMZ\VRNR FL
û[ =
L+v
-6,0
:\VRNR FLSXQNWX:
pvv
[pl ] = 13,38 = 8,0 mm [p] 1.67
R2
a-
*
:\UyZQDQLHVSRVWU]H H EH]SR UHGQLFK3XQNWZ
:\UyZQDQDZ\VRNR
]áRZ\ZQLZHODFML
üSXQNWX:
xwyr = x0 + ∆x = 314,482 + 0,008 = 314,490 3) Kontrola ogólna:
2 pl ] [ [pvv ] = [pll ] − [p] 2 [ pl ] [pll ] − [p]
[pvv ] = 56,12 4)
2FHQDGRNáDGQR
13,382 = 163,32 − = 56,12 1,67
FL
%á G UHGQLREVHUZDFML
m0 = ±
[pvv] = ±
n −1
56,12 ≈ ± 5,3 mm km 2
%á G UHGQLQLHZLDGRPHM Z\UyZQDQHMZ\VRNR FL
mx = ±
m0 5,3 =± ≈ ± 4,1mm [p] 1,3
9
10
8NáDGUyZQD OLQLRZ\FKMHGQR]QDF]QLHRNUH ORQ\FK
,PL
1D]ZLVNR
Nr zestawu .....
û:,&=(1,(QU
Temat: 8NáDGUyZQDOLQLRZ\FKMHGQR]QDF]QLHRNUH ORQ\FK
Zadanie 'DQH V ZVSyáU] GQH x, y punktów A,B,C,D. 2EOLF]\ü ]D SRPRF UDFKXQNX macieU]RZHJRZVSyáU] GQHSXQNWXSU]HFL FLDVL SURVW\FKAC i BD. Dane:
x B
Nr punktu
x
y
C
.
A
102,20
102,20
B
703,30
98,90
C
698,90
896,70
D
98,90
903,30
P
D
A
y
52=:,
=$1,(
1) Równania prostych Wzory: ax + by +c = 0;
a = y1 – y2;
b = x2 – x1;
c = x1(y2 – y1) – y1(x2 – x1)
Prosta AC: –794,50.x + 596,70.y + 20215,16 = 0 Prosta BD: –804,40.x – 604,40.y + 625509,68 = 0
8NáDGUyZQD
onych
OLQLRZ\FKMHGQR]QDF]QLHRNUH O
− 794,50 ⋅ x + 596,70 ⋅ y + 20215,16 = 0 − 804,40 ⋅ x − 604,40 ⋅ y + 625509,68 = 0 3) =DSLVXNáDGXUyZQDZSRVWDFLPDFLHU]RZHM A.X=L − 794,50 596,70 A= ; − 804,40 − 604,40
x X = ; y
- 20215.16 L= - 625509,68
8NáDGUyZQD OLQLRZ\FKMHGQR]QDF]QLHRNUH ORQ\FK
4) Utworzenie macierzy blokowej B
B = [A
5)
5R]NáDGSURVWRN
− 794,50 596,70 - L] = − 804,40 − 604,40
WQHMPDFLHU]\
B na czynniki trapezowe
B = H T ⋅ G' = H T ⋅ [G − 794,50 596,70 − 804,40 − 604,40
20215,16 625509,68
- L' ]
20215,16 H11 0 1 G12 ⋅ = H 625509,68 21 H 22 0 1
0 1 − 0,75104 − 794,50 ⋅ B= 1 − 804,40 − 1208,535 0 6)
2EOLF]HQLHZDUWR
− L'1 ' − L2
− 25,444 − 500,641
FLQLHZLDGRP\FK
G . X = L’ 1 − 0,75104 x 25,444 0 ⋅ y = 500,641 1 401,444 X = 500,641 7)
:VSyáU] GQHSXQNWXSU]HFL
FLDSURVW\FK
xP = 401,444;
yP = 500,641
8) Kontrola − 794,50 ⋅ + 596,70 ⋅ + 20215,16 = 0,387 ≈ 0 − 804,40 ⋅ − 604,40 ⋅ + 625509,68 = 0,706 ≈ 0
11
12
8NáDGUyZQD OLQLRZ\FK
,PL
nadRNUH ORQ\FK
1D]ZLVNR
Nr zestawu .....
û:,&=(1,(QU
Temat: 8NáDGUyZQDOLQLRZ\FKQDGRNUH ORQ\FK
Zadanie 'DQH V ZVSyáU] GQH x, y punktów A,B,C,D,E,F. 2EOLF]\ü ]D SRPRF UDFKXnNXPDFLHU]RZHJRZVSyáU]
GQHSXQNWXQDMNRU]\VWQLHMV]HJRGODSU]HFL FLDVL WU]HFKSUo-
stych: AB, CD oraz EF.
Dane:
x
F C
Nr punktu
x
y
B
.
A
202,64
203,08
B
802,64
1403,08
C
802,86
202,64
D
202,64
1402,42
E
102,86
702,64
F
1302,42
1102,86
A
E
D
R2=:,=$1,( 1) Równania prostych Wzory: ax + by +c = 0;
a = y1 – y2;
b = x2 – x1;
c = x1(y2 – y1) – y1(x2 – x1)
Prosta AB:
–1200,00.x + 600,00.y + 121320,00 = 0
Prosta CD:
–1199,78.x – 600,22.y + 1084883,95 = 0
Prosta EF:
–400,22.x + 1199,56.y – 801692,21 = 0
8NáDGUyZQD
OLQLRZ\FK
nadRNUH ORQ\FK
- 1200,00 ⋅ x + 600,00 ⋅ y + 121320,00 = 0 - 1199,78 ⋅ [ - 600,22 ⋅ \ + 1084883,95 = 0 - 400,22 ⋅ [ + 1199,56 ⋅ \ − 801692,21 = 0
y
13
8NáDGUyZQD OLQLRZ\FKQDGRNUH ORQ\FK
3) =DSLVXNáDGXUyZQDZSRVWDFLPDFLHU]RZHM A.X=L − 1200,00 600,00 A = − 1199,78 − 600,22 ; − 400,22 1199,56
x X = ; y
− L = −
4) Macierz normalna AT.A − 1200 ,00 A ⋅A = 600 ,00 T
− 1200 ,00 − 400 ,22 ⋅ − 1199 ,78 1199 ,56 − 400 ,22 − 479955 ,952 2159208 ,242
− 1199 ,78 − 600 . 22
3039648 ,097 = − 479955 ,952
600 ,00 − 600 ,22 = 1199 ,56
wadratowej symetrycznej AT.AQDF]\QQLNLWUyMNWQH
5R]NáDGPDFLHU]\N
AT ⋅ A = R T ⋅ R 3039648,097 − 479955,952 − 479955,952 2159208,242 =
R11 0 R11 R12 R ⋅ 12 R 22 0 R 22
1743,459 − 275,290 R= 0 1443,407 2GZURWQR
üPDFLHU]\WUyMN
WQHM
R-1
R
−1
⋅ R = I
' ' R11 1743,459 − 275,290 R12 ⋅ = ' , 0 1443 407 R 0 22
R
2GZURWQR
ü
−1
5,73573 ⋅ 10 −4 = 0
macierzy normalnej:
1,09393 ⋅ 10 −4 6,92805 ⋅ 10 − 4
(AT.A)-1
5,736 ⋅ 10 −4 T ( A T ⋅ A) -1 = R −1 ⋅ (R −1 ) = 0 3,40952 ⋅ 10 −7 = −7 0,757880 ⋅ 10
1 0 0 1
1,094 ⋅ 10 −4 5,736 ⋅ 10 −4 ⋅ 6,928 ⋅ 10 − 4 1,094 ⋅ 10 − 4
0,757880 ⋅ 10 −7 4,79979 ⋅ 10 −7
0 = −4 6,928 ⋅ 10
14
8NáDGUyZQD OLQLRZ\FK
nadRNUH ORQ\FK
8) Wektor niewiadomych X X = ( A T ⋅ A ) -1 ⋅ A T ⋅ L = 3,410 ⋅ 10 −7 = −7 0,757 ⋅ 10
8)
:VSyáU]
− 121320,00 0,757 ⋅ 10 −7 − 1200,00 − 1199,78 − 400,22 ⋅ − 1084883,95 = ⋅ −7 − 600.22 1199,56 4,800 ⋅ 10 600,00 801692,21 500,750 = 824,558
GQHZ\UyZQDQHSXQNWXSU]HFL
x = 500,750;
FLDVL
WU]HFKSURVW\FK
y = 824,558
8NáDGUyZQD OLQLRZ\FKQLHGRRNUH ORQ\FK
,PL
1Dzwisko
15
Nr zestawu .....
û:,&=(1,(QU
Temat: 8NáDGUyZQDOLQLRZ\FKQLHGRRNUH ORQ\FK
Zadanie 'DQ\MHVWQLHGRRNUH ORQ\XNáDGUyZQDOLQLRZ\FK6WRVXMFUDFKXQHNPDFLe-
(
U]RZ\]QDOH üQDMOHSV]HUR]ZL
]DQLHZHGáXJPHWRG\QDMPQLHMV]\FKNZDGUDWyZ
Dane: 23 ⋅ x + 2 ⋅ y + z − 2 = 0 x + 3 ⋅ y + 23 ⋅ z − 4 = 0
52=:,
=$1,(
1) =DSLVXNáDGXUyZQDZSRVWDFLPDFLHU]RZHM A.X=L 23 2 1 A= ; 1 3 23
x X = y ; z
2 L= 4
2) Macierz normalna A .AT 23 1 23 2 1 534 52 A⋅A = ⋅ 2 3= 1 3 23 1 23 52 539 T
5R]NáDGPDFLHU]\NZDGUDWRZHMV\PHWU\F]QHM
A .AT nDF]\QQLNLWUyMNWQH
A ⋅ AT = R T ⋅ R 534 52 R11 52 539 = R 12
0 R11 R12 ⋅ R 22 0 R 22
23,108 2,250 R= 23,107 0
16
8NáDGUyZQD OLQLRZ\FK
2GZURWQR
üPDFLHU]\WUyMN
WQHM
niedoRNUH ORQ\FK
R-1 R −1 ⋅ R = I
' ' R11 23,108 2,250 1 0 R12 = ⋅ ' 23,107 0 1 0 R 22 0
4,327 ⋅ 10 −2 R −1 = 0 2GZURWQR
üPDFLHU]\QRUPDOQHM
( A .AT)-1
4,327 ⋅ 10 −2 T ( A ⋅ A T ) -1 = R −1 ⋅ (R −1 ) = 0 1,890 ⋅ 10 −3 = −3 0,1823 ⋅ 10
− 0,4214 ⋅ 10 −2 4,328 ⋅ 10 − 2
0,421⋅ 10 −2 4,327 ⋅ 10 −2 ⋅ 4,328 ⋅ 10 − 2 0,421⋅ 10 − 2
0,182 ⋅ 10 − 3 1,873 ⋅ 10 −3
6) Wektor niewiadomych X X = A T ⋅ A ⋅ AT ⋅ L =
23 1 1,890 ⋅ 10 −3 = 2 3 ⋅ 0,182 ⋅ 10 −3 1 23
5R]ZL
0,077 0,182 ⋅ 10 −3 2 ⋅ ≈ 0,027 1,873 ⋅ 10 −3 4 0,167
]DQLH
x =0,077;
y = 0,027
z = 0,167
8) Kontrola 23 ⋅ 0,077 + 2 ⋅ 0,027 + 0,167 − 2 = 0,008 ≈ 0 0,077 + 3 ⋅ 0,027 + 23 ⋅ 0,167 − 4 = 0,001 ≈ 0
0 = −2 4,328 ⋅ 10
WyrówQDQLHVWDF\MQHPHWRGSR UHGQLF]F ,PL
1D]ZLVNR
17
Nr zestawu .....
û:,&=(1,(QU
Temat: :\UyZQDQLHVWDF\MQHPHWRGSR UHGQLF]F
Zadanie: Na stanowisku st.I ]RVWDá\SRPLHU]RQH]Uy*QGRNáDGQR FLNW\ – 6. Stosu-
* WyZ
Wych jako niewiadome x, y, z, tDWDN*HSU]HSURZDG]LüDQDOL] GRNáDGQR FLSRZy-
M F PQN QDOH \ REOLF]\ü SRSUDZNL GR W\FK SRPLDUyZ RUD] Z\UyZQDQH ZDUWR FL N SU]\M
równaniu.
Dane: L1 = 31,3107g L2 = 07,1832g L3 = 18,1519g L4 = 32,1421g L5 = 88,7871g L6 = 50,2933g
52=:,
p1 = 2 p2 = 2 p3 = 2 p4 = 2 p5 = 5 p6 = 7
1 x
y
2
5
z 3
t 4
st.I
6
=$1,(
1) Równania obserwacyjne L1 + v1 = x L2 + v2 = y L3 + v3 = z L4 + v4 = t L5 + v5 = x + y + z + t L6 + v6 = z + t
5yZQDQLDEá
GyZ
v1 = ∆x v2 = ∆y v3 = ∆z v4 = ∆t v5 = ∆x + ∆y + ∆z + ∆t + 8cc v6 = ∆z + ∆t + 7cc
2):DUWR FLSU]\EOL*RQHQLHZLDGRP\FK x0 = L1 = 31,3107g y0 = L2 = 07,1832g z0 = L3 = 18,1519g t0 = L4 = 32,1421g
1DGRNUH
x = x0 + ∆x y = y0 + ∆y z = z0 + ∆z t = t0 + ∆t
ORQ\XNáDGUyZQD
OLQLRZ\FK
∆x = 0 ∆y = 0 ∆z = 0 ∆t = 0 ∆x + ∆y + ∆z + ∆t + 8 = 0 ∆z + ∆ t + 7 = 0
18
:\UyZQDQLHVWDF\MQHPHWRG
5) Zapis macierzowy
SR UHGQLF]F
6) Macierz wagowa P
A⋅ X = L 1 0 0 0 1 0
0 0 0 0 1 0 0 ∆x 0 0 1 0 ∆y 0 ⋅ = 0 0 1 ∆z 0 1 1 1 ∆t − 8 0 1 1 − 7
2 0 0 P= 0 0 0
0 0 0 0 0 0 0 5 0 0 0 0 0 7 0 2 0 0
0 0 2 0
0 0 0 2
0 0 0 0
7) Macierz normalna AT.P.A 7 5 T A ⋅P ⋅A = 5 5
5 5 5 7 5 5 5 14 12 5 12 14
8 5R]NáDGPDFLHU]\NZDGUDWRZHMV\PHWU\F]QHMAT.P.AQDF]\QQLNLWUyMNWQH
7 5 5 5
5 7 5 5
AT ⋅ P ⋅ A = R T ⋅ R 0 0 0 R11 R12 5 5 R11 R R 22 0 0 0 R 22 5 5 ⋅ = 12 0 0 0 14 12 R 13 R 23 R 33 0 12 14 R14 R 24 R 34 R 44 0 2.6458 1.8898 1.8898 0 1.8516 0.7715 R= 0 0 3.1358 0 0 0
R13 R 23 R 33 0
R14 R 24 R 34 R 44
1.8898 0.7715 2.4980 1.8956
9 2GZURWQR üPDFLHU]\WUyMNWQHM R-1 R R' 11 R' 12 0 R' 22 0 0 0 0
R' 13 R' 23 R' 33 0
−1
⋅ R = I
R' 14 2.6458 1.8898 1.8898 R' 24 0 1.8516 0.7715 ⋅ R' 34 0 0 3.1358 R' 44 0 0 0
R −1
1.8898 1 0.7715 0 = 2.4980 0 1.8956 0
0.3780 − 0.3858 − 0.1329 − 0.0447 0 0.5401 − 0.1329 − 0.0447 = 0 0 0.3189 − 0.4202 0 0 0.5275 0
0 0 0 1 0 0 0 1 0 0 0 1
WyrówQDQLHVWDF\MQHPHWRGSR UHGQLF]F
19
10) OdwrotQR üPDFLHU]\QRUPDOQHM (AT.P. A)-1
A ⋅ P ⋅ A 7
( )
= R −1 ⋅ R −1
0.3780 − 0.3858 − 0.1329 − 0.0447 0.3780 0 0.5401 − 0.1329 − 0.0447 − 0.3858 ⋅ = 0 0 0.3189 − 0.4202 − 0.3290 0 0 0.5275 − 0.0447 0 0.3113 − 0.1887 − 0.0236 − 0.1887 0.3113 − 0.0236 = − 0.0236 − 0.0236 0.2783 − 0.0236 − 0.0236 − 0.2217
T
= 0.5401 0 0 = 0 − 0.1329 0.3189 − 0.0447 − 0.4202 0.5275 − 0.0236 − 0.0259 − 0.2217 0.2783 0
0
0
11) Wektor niewiadomych X
12) Wyrównane niewiadome
X = (A T ⋅ P ⋅ A) -1 ⋅ AT ⋅ P ⋅ L =
x = x 0 + ∆x = 31.3107 g − 0 ,7 cc = 31.31063 g y = y 0 + ∆y = 7.1832 g − 0 ,7 cc = 7.18313 g z = z 0 + ∆z = 18 .1519 g − 3 ,1cc = 18.15159 g
− 0.71 − 0.71 ≅ = − 3.15 − 3.15
− 0.7 − 0.7 − 3.1 − 3.1
13) Wektor poprawek
t = t 0 + ∆t = 32 .1421g − 3 ,1cc = 32.14179 g
14) Obserwacje wyrównane
V = A⋅X − L = 1 0 0 = 0 1 0
0 0 − 0,7 cc 0 − 0.7 0 − 0,7 cc 0 − 0.7 0 − 3,1cc − = ⋅ 1 − 3.1 0 − 3,1cc 1 1 1 − 3.1 − 8 0,4 cc cc 0 1 1 − 7 0,8 0 1 0 0
0 0 1 0
15) Kontrola ogólna V T ⋅ P ⋅ V = LT ⋅ P ⋅ L − LT ⋅ P ⋅ A ⋅ X 45.5 = 663 − 617.5 45.5 = 45.5
L1 L2 L3 L4
+ v 1 = 31.3107 g − 0 ,7 cc = 31.31063 g + v 2 = 7.1832 g − 0 ,7 cc = 7.18313 g + v 3 = 18 .1519 g − 3 ,1cc = 18.15159 g + v 4 = 32 .1421g − 3 ,1cc = 32.14179 g
L 5 + v 5 = 88 .7871g + 0 ,4 cc = 88.78714 g L 6 + v 6 = 50 .2933 g + 0 ,8 cc = 50.29338 g
20
:\UyZQDQLHVWDF\MQHPHWRG
SR UHGQLF]F
16) Kontrola generalna /HZDVWURQDUyZQD
L1 L2 L3 L4
REVHUZDF\MQ\FK
3UDZDVWURQDUyZQD
REVHUZDF\MQ\Fh
x = 31.31063 g y = 7.18313 g z = 18.15159 g t = 32.14179 g
+ v 1 = 31.31063 g + v 2 = 7.18313 g + v 3 = 18.15159 g + v 4 = 32.14179 g
L 5 + v 5 = 88.78714 g
x + y + z + t = 88.78714 g
L 6 + v 6 = 50.29338 g
z + t = 50.29338 g
17 2FHQDGRNáDGQR FL a)
Eá
G
m0 = ± b)
Eá
G\
UHGQLREVHUZDFML
V T P ⋅V =± n−k
45.5 ≈ ± 4.8 CC 2
UHGQLHQLHZLDGRP\FK ZLHONR
FLZ\UyZQDQ\FK
Macierz kowariancyjna 0.3113 − 0.1887 − 0.0236 − 0.0236 − 0.1887 0.3113 − 0.0236 − 0.0259 Q = A ⋅ P ⋅ A = − 0.0236 − 0.0236 0.2783 − 0.2217 − 0.0236 − 0.0236 − 0.2217 0.2783 7
2EOLF]HQLHEá
GyZ UHGQLFK
m x = ± m 0 ⋅ Q xx = ± 2.7 cc m y = ± m 0 ⋅ Q yy = ± 2.7 cc m z = ± m 0 ⋅ Q zz = ± 2.5 cc m t = ± m 0 ⋅ Q tt = ± 2.5 cc
:\UyZQDQLHVWDF\MQHPHWRG
,PL
zawarunkowan
1D]ZLVNR
21
Nr zestawu .....
û:,&=(1,(QU
Temat: :\UyZQDQLHVWDF\MQHPHWRG]DZDUXQNRZDQ
Zadanie: Na stanowisku st.I ]RVWDá\ SRPLHU]RQH ] Uy*Q GRNáDGQR FL NW\ – 6. PrzeprowaG]LüZ\UyZQDQLHZHGáXJPHWRG\ZDUXQNRZHM
Dane: L1 = 31,3107g L2 = 07,1832g L3 = 18,1519g L4 = 32,1421g L5 = 88,7871g L6 = 50,2933g
52=:,
p1 = 2 p2 = 2 p3 = 2 p4 = 2 p5 = 5 p6 = 7
1
5
2 3 st.I
4
6
=$1,(
1) Liczba warunków w sieci niwelacyjnej r=n–k=2 n = 6 – liczba wszystkich obserwacji k = 4 –OLF]EDREVHUZDFMLQLH]E GQ\FK 2) Równania warunkowe (L1 + v1) + (L2 + v2) + (L3 + v3) + (L4 + v4) - (L5 + v5) = 0 (L3 + v3) + (L4 + v4) - (L6 + v6) = 0
5yZQDQLDEá
GyZ QLHGRRNUH
ORQ\XNáDGUyZQD
OLQLRZ\FK
v1+ v2 + v3 + v4 – v5 + ω1 = 0
ω1 = L1 + L2 + L3 + L4 – L5 = 8cc
v3 + v4 – v6 + ω2 = 0
ω1 = L3 + L4 – L6 = 7cc
22
:\UyZQDQLHVWDF\MQHPHWRG
4) Zapis macierzowy
zawarunkowan
2GZURWQR
üPDFLHU]\Z
agowej
A ⋅V = W v1 v 2 1 1 1 1 − 1 0 v 3 0 0 1 1 0 − 1 ⋅ = v 4 v 5 v 6
− 8 − 7
P −1
0 0 0 0 0.5 0 0 0.5 0 0 0 0 0 0 0.5 0 0 0 = 0 0 0.5 0 0 0 0 0 0 0 0.2 0 0 0 0 0 0.14 0
6) Macierz normalna A.P-1.AT 1 2.2 A ⋅ P −1 ⋅ AT = 1 1.14 5R]NáDGPDFLHU]\NZDGUDWRZHMV\
metrycznej A.P-1.ATQDF]\QQLNLWUyMNWQH
A ⋅ P −1 ⋅ AT = R T ⋅ R 1 R11 2.2 1 1.14 = R 12
0 R11 R12 ⋅ R 22 0 R 22
1.48 0.67 R= 0.83 0 2GZURWQR
üPDFLHU]\WUyMN
WQHM
R-1
R R 0
' 11
−1
⋅ R = I
R 1.48 0.67 1 0 = ⋅ 0.83 0 1 R 0 ' 12 ' 22
0 0.67 R −1 = − 0.55 1.21
2GZURWQR
üPDFLHU]\
normalnej:
A ⋅ P −1 ⋅ A
(A.P-1. AT)-1 7
( )
= R −1 ⋅ R −1
T
=
0 0.75 − 0.66 1.48 0.67 1.48 = ⋅ = 0.83 0.67 0.83 − 0.66 1.45 0
:\UyZQDQLHVWDF\MQHPHWRG
zawarunkowan
10) Wektor korelat K − 1.42 K = (A ⋅ P −1 ⋅ AT ) -1 ⋅ W = − 4.89 11) Wektor poprawek V − 0.7 − 0.7 − 3 . 2 V = P −1 ⋅ AT ⋅ (A ⋅ P −1 ⋅ AT ) -1 ⋅ W = P −1 ⋅ AT ⋅ K = − 3.2 0.3 0.7 12) Obserwacje wyrównane
L1 L2 L3 L4
+ v 1 = − = 31.3106 g + v 2 = 7.1832 g − 1cc = 7.1831 g + v 3 = 18 .1519 g − 3 cc = 18.1516 g + v 4 = 32 .1421g − 3 cc = 32.1418 g J
FF
13) Kontrola ogólna
V T ⋅ P ⋅V = W T ⋅ K 45 .5 = 45 .5
L5 + v 5 = 88 .7871g − 0 cc = 88.7871 g L 6 + v 6 = 50 .2933 g + 1cc = 50.2934 g 14) Kontrola generalna 31.3106g + 7.1831g + 18.1516g + 32.1418g – 88.7871g = 0.0000g 18.1516g + 32.1418g – 50.2934g = 0.0000g 1 %áG UHGQLREVHUZDFML m0 = ±
V T P ⋅V =± n−k
45.5 ≈ ± 4.8 cc 2
23
24 ,PL
:\UyZQDQLHVLHFLQLZHODF\MQHMPHWRG
SR UHGQLF]F
1D]ZLVNR
Nr zestawu .....
û:,&ZENIE nr .....
Temat: :\UyZQDQLHVLHFLQLZHODF\MQHMPHWRGSR UHGQLF]F
Zadanie :Z\QLNXSRPLDUXQLZHODF\MQHJRFLJyZRGáXJR FL d1, d2, d3, d4, d5 uzyskano Uy*QLFH Z\VRNR FL ∆h1, ∆h2, ∆h3, ∆h4, ∆h5. 6WRVXMF PHWRG SR UHGQLF]F QDOH*\ ZyUyZQDüSRPLDU\RUD]Z\VRNR FLSXQNWyZ x i y,DWDN*HSU]HSURZDG]LüDQDOL] GRNáDGQo-
FLSRZ\UyZQDQLX
Dane:
*
5y QLFH
'áXJR FL :\VRNR FL
Z\VRNR FL
FL JyZ
[m]
[km]
reperów [m]
1
2,722
1,0
107,276
2
3,274
1,0
105,223
3
0,558
1,0
106,831
4
2,277
0,5
5
1,226
0,5
Nr
R1
∆h2
∆h1 ∆h3
x
y
∆h4
∆h5 R2
R3
52=:,
=$1,(
1) Równania obserwacyjne ∆h1 + v1 = R1 – x ∆h2 + v2 = R1 – y ∆h3 + v3 = x – y ∆h4 + v4 = R3 – x ∆h5 + v5 = R2 – y 3) RównanLDEá GyZ v1 = - ∆x v2 = - ∆y v3 = ∆x – ∆y – 6mm v4 = - ∆x v5 = - ∆y – 5mm
:DUWR
*
FLSU]\EOL RQHQLHZLDGRP\FK
x0 = R1 – ∆h1 = 104,554 y0 = R1 – ∆h2 = 104,002
1DGRNUH
ORQ\XNáDGUyZQD
− ∆x = 0 − ∆y = 0 ∆x − ∆ y − 6 = 0 − ∆x = 0 − ∆y − 5 = 0
x = x0 + ∆x y = y0 + ∆y
OLQLRZ\FK
:\UyZQDQLHVLHFLQLZHODF\MQHMPHWRG
5) Zapis macierzowy
SR UHGQLF]F
25
6) Macierz wagowa P
A⋅ X = L
waga pi =
−1 0 0 0 −1 0 ∆x 1 −1 ⋅ = 6 ∆y −1 0 0 0 −1 5
1 0 P = 0 0 0
1 di
0 0 0 0 0 0 0 2 0 1 0 0
0 0 1 0
0 0 0 2
7) 8NáDGUyZQDQRUPDOQ\FK: (A T ⋅ P ⋅ A )⋅ X = A T ⋅ P ⋅ L 4 − 1 x 6 − 1 4 ⋅ y = − 16 8 5R]NáDGPDFLHU]\NZDGUDWRZHMV\PHWU\F]QHMAT.P.AQDF]\QQLNLWUyMNWQH AT ⋅ P ⋅ A = R T ⋅ R 4 −1 R11 0 R11 R12 ⋅ = −1 4 R12 R22 0 R22
→
2 −0,5 R= 0 1,94
9 2GZURWQR üPDFLHU]\WUyMNWQHM R-1 R
−1
⋅R = I
R '11 R '12 2 −0,5 ⋅ = 0 R '22 0 1,94
1 0 0 1
→
0,5 0,13 R −1 = 0 0,52
10 2GZURWQR üPDFLHU]\QRUPDOQHM (AT.P. A)-1
( )
(A T ⋅ P ⋅ A)-1 = R −1 ⋅ R −1
T
0 0,27 0,067 0,5 0,13 0,5 = ⋅ = 0 0,52 0,13 0,52 0,067 0,27
11) Wektor niewiadomych X
12) Wyrównane niewiadome
(A
x = x0 + ∆x = 104,554m + 1mm = 104,555 y = y 0 + ∆y = 104,002m − 4mm = 103,998
T
⋅ P ⋅ A )⋅ X = A T ⋅ P ⋅ L =
0,5mm = − 3,9mm
26
:\UyZQDQLHVLHFLQLZHODF\MQHMPHWRG
13) Wektor poprawek
SR UHGQLF]F
14) Obserwacje wyrównane
V = A⋅X − L = −1 0 0 −1 0,53 = 1 −1 ⋅ − −3,87 −1 0 0 −1
0 0 6 = 0 5
−0,53 3,87 −1,60 ≅ −0,53 −1,13
−1mm 4mm −2mm −1mm −1mm
∆ h1 + v1 = 2,722 − 1mm = 2, 721 ∆ h2 + v 2 = 3,274 + 4mm = 3, 278 ∆ h3 + v 3 = 0,558 − 2mm = 0, 556 ∆ h4 + v 4 = 2,277 − 1mm = 2, 276 ∆ h5 + v 5 = 1,226 − 1mm = 1, 225
15) Kontrola ogólna V T ⋅ P ⋅ V = LT ⋅ P ⋅ L − LT ⋅ P ⋅ A ⋅ X 20,9 = 86 − 65,1 20,9 = 20,9 16) Kontrola generalna /HZDVWURQDUyZQD
∆ h1 ∆ h2 ∆ h3 ∆ h4 ∆ h5
REVHUZDF\MQ\FK
3UDZDVWURQDUyZQD
R1 − x = 107,276 − 104,555 = 2, 721 R1 − y = 107,276 − 103,998 = 3, 278 x − y = 104,555 − 103,998 = 0, 557 R3 − x = 106,831 − 104,555 = 2, 276 R 2 − y = 105,223 − 103,998 = 1, 225
+ v1 = 2, 721 + v 2 = 3, 278 + v 3 = 0, 556 + v 4 = 2, 276 + v 5 = 1, 225
17 2FHQDGRNáDGQR FL a) EáG UHGQLREVHUZDFML m0 = ±
V T P ⋅V 20,9 =± ≈ ± 2,6 mm km n−k 3
b) Eá G\ UHGQLHQLHZLDGRP\FK ZLHONR FLZ\UyZQDQ\FK Macierz kowariancyjna Q = (AT ⋅ P ⋅ A )-1
2EOLF]HQLHEá
REVHUZDF\MQ\FK
0,27 0,067 = 0,067 0,27
GyZ UHGQLFK
m x = ± m 0 ⋅ Q xx = ± 1,4mm m y = ± m 0 ⋅ Q yy = ± 1.4mm
:\UyZQDQLHVLHFLQLZHODF\MQHMPHWRG
,PL
zawarunkowan
1D]ZLVNR
27
Nr zestawu .....
û:,&=(1,(QU
Temat: :\UyZQDQLHVLHFLQLZHODF\MQHMPHWRG]DZDUXQNRZDQ Zadanie :Z\QLNXSRPLDUXQLZHODF\MQHJRFLJyZRGáXJR FL d1, d2, d3, d4, d5 uzyskano * i ∆h1, ∆h2, ∆h3, ∆h4, ∆h5. 3U]HSURZDG]LüZ\UyZQDQLHVLHFLQLZHODF\MQHM
Uy Q FHZ\VRNR FL
ZHGáXJPHWRG\ZDUXQNRZHM
Dane:
*
5y QLFH
'áXJR FL :\VRNR FL
Z\VRNR FL
FL JyZ
[m]
[km]
reperów [m]
1
2,722
1,0
107,276
2
3,274
1,0
105,223
3
0,558
1,0
106,831
4
2,277
0,5
5
1,226
0,5
Nr
R1
∆h1
∆h2 ∆h3
x
∆h4
∆h5 R2
R3
52=:,
=$1,(
1) Liczba warunków w sieci niwelacyjnej a) ogólna: w = n – p + pr = 5 – 5 + 3 = 3 b) PL G]\UHSHUDPL Zr = pr -1 = 3 – 1 = 2 c) dla oczek siatki: wo = 1 2) Równania warunkowe Dla reperów: Dla oczka:
R1 – (∆h1 + v1) + ∆h4 + v4 – R3 = 0 R1 – (∆h2 + v2) + ∆h5 + v5 – R2 = 0 ∆h2 + v2 – (∆h3 + v3) – (∆h1 + v1) = 0
5yZQDQLDRGFK\áHN QLHGRRNUH ORQ\XNáDGUyZQD
– v1 + v4 + ω1 = 0 gdzie: – v2 + v5 + ω2 = 0 – v1 + v2 – v3 + ω3 = 0
y
OLQLRZ\FK
ω1 = R1 – ∆h1 + ∆h4 – R3 = 0 ω2 = R1 – ∆h2 + ∆h5 – R2 = 5 mm ω3 = – ∆h1 + ∆h2 – ∆h3 = - 6 mm
28
:\UyZQDQLHVLHFLQLZHODF\MQHMPHWRG
5) Macierz wagowa RGZURWQR ü
4) Zapis macierzowy A ⋅V = W
v1 0 1 0 v 2 − 1 0 0 − 1 0 0 1 ⋅ v = 3 1 1 1 0 0 − − v 4 v 5
8NáDGUyZQD
zawarunkowan
0 − 5 6
waga pi =
1 ; pi−1 = d i di
1 0 = 0 0 0
0 0
P −1
QRUPDOQ\FKNRUHODW
(A ⋅ P
−1
)
⋅ AT ⋅ K = W
0 1 k1 1.5 0 1.5 − 1 ⋅ k2 = 1 − 1 3 k3
5R]NáDGPDFLHU]\QRUPDOQHM
0 0 0 1 0 0 0 0 0.5 0 0 0 0 0.5 0 0 1 0
0 − 5 6
A.P-1.AT na czynniki trójkWQH
A ⋅ P −1 ⋅ AT = R T ⋅ R 0 1 R11 0 0 R11 R12 1.5 0 1.5 − 1 = R R 22 0 ⋅ 0 R 22 12 1 − 1 3 R13 R 23 R 33 0 0
R13 R 23 R 33
0 0.816 1.225 R= 0 1.225 − 0.816 0 0 1.291
2GZURWQR
üPDFLHU]\WUyMN
WQHM
R
−1
−1
⋅ R = I R' 13 1.225 0 0.816 R' 23 ⋅ 0 1.225 − 0.816 = R' 33 0 0 1.291 R
R' 11 R' 12 0 R' 22 0 0
R-1
0 − 0.516 0.816 0 0.816 0.516 = 0 0 0.775
1 0 0 0 1 0 0 0 1
:\UyZQDQLHVLHFLQLZHODF\MQHMPHWRG
2GZURWQR
üPDFLHU]\QRUPDOQHM
zawarunkowan
29
(A.P-1. AT)-1
( A ⋅ P −1 ⋅ A T ) -1 = R −1 ⋅ (R −1 ) = T
− 0.516 0.816 0 0 0 0.933 − 0.267 − 0.4 0.816 = 0 0.816 0.516 ⋅ 0 0.816 0 = − 0.267 0.933 0.4 0 0 0.775 − 0.516 0.516 0.775 − 0.4 0.4 0.6 10) Wektor korelat K − 1.07 K = (A ⋅ P ⋅ A ) ⋅ W = − 2.27 1.60 −1
T
-1
11) Wektor poprawek V − 0.53 3.87 V = P −1 ⋅ AT ⋅ (A ⋅ P −1 ⋅ AT ) -1 ⋅ W = P −1 ⋅ AT ⋅ K = − 1.60 ≅ − 0.53 − 1.13 12) Obserwacje wyrównane
13) Kontrola ogólna
∆ h1 ∆ h2 ∆ h3 ∆ h4
+ v1 = 2,722 − 1mm = 2, 721 + v 2 = 3,274 + 4mm = 3, 278 + v 3 = 0,558 − 2mm = 0, 556 + v 4 = 2,277 − 1mm = 2, 276 ∆ h5 + v 5 = 1,226 − 1mm = 1, 225
14) Kontrola generalna 107.276 – 2.721 + 2.276 – 106.831 = 0.000 107.276 – 3.278 + 1.225 – 105.223 = 0.000 3.278 – 0.556 – 2.721 = 0.001 1 %áG UHGQLREVHUZDFML m0 = ±
V T P ⋅V =± n−k
− 1 mm 4 mm − 2 mm − 1 mm − 1 mm
20.9 ≈ ± 2.6 mm km 3
V T ⋅ P ⋅V = W T ⋅ K 20.9 = 20.9
30
Rozwijanie funkcji nieliniowych w szereg Taylora
,PL
1D]ZLVNR
Nr zestawu .....
û:,&=(1,(QU Temat:
Rozwijanie funkcji nieliniowej w szereg Taylora
Zadanie 5R]ZLQüZV]HUHJSRGDQHIXQNFMHQLHOLQLRZH]SRPLQL FLHP wyrazów o pot G]HZ\*V]HMQL*2EOLF]\üZDUWR üIXQNFMLfRUD]ZDUWR FLMHMUR]ZLQL FLDOLQLRZHJRf’ dla GDQ\FKZDUWR FL]PLHQQ\FK x0, y0) i ich przyrostów (∆x, ∆y 2EOLF]\üUy*QiF f – f’. Funkcja 1. Dane: f1 = x 2 − 2y
([
5R]ZLQL
=
=
FLHIXQNFML f1' = f1 (x 0 , y 0 ) +
∆\ = )
∆[ =
\
∂f1 ∂f ⋅ ∆x + 1 ⋅ ∆y ∂x ∂y
f1' = x 02 − 2y 0 + 2 x0 ⋅ ∆x − 2 ⋅ ∆y Funkcja w postaci liniowej f1’ dla
=
=
[ \
:
f = 20 ⋅ ∆x − 2 ⋅ ∆y + 80 ' 1
f1’ dla ∆[ = ∆\ = : f = 20 ⋅ 0.5 − 2 ⋅ 0.2 + 80 = 89.60
:DUWR üIXQNFML
' 1
:DUWR üIXQN
cji f1 dla
[
= [ +∆[ = 10.5\ = \ +∆\ = 10.2
f1 = 10.5 − 2 ⋅ 10.2 = 89.85 2
*
f − f1' = 0.25
5y QLFD 1
Funkcja 2. Dane: f 2 = sin x − cos y [ = \ =
(
5R]ZLQL
J
J
∆[ =
J
∆\ =
FLHIXQNFML ∂f ∂f f 2' = f 2 (x0 , y 0 ) + 2 : ρ g ⋅ ∆x + 2 : ρ g ⋅ ∆y ∂x ∂y cos x 0 sin y 0 ⋅ ∆x + ⋅ ∆y f 2' = sin x 0 − cos y 0 + g ρ ρg
J
)
Rozwijanie funkcji nieliniowej w szereg Taylora Funkcja w postaci liniowej f2’ dla
=
=
J
[ \
J
,
(ρ
g
f = 0.015515 ⋅ ∆x + 0.002457 ⋅ ∆y − 0.831254
f2’ dla ∆[ = ∆\ = : ' f 2 = 0.015515 ⋅ 0.1 + 0.002457 ⋅ 0.2 − 0.831254 = − 0.829211 J
:DUWR üIXQNFML
f1 dla
(
J
= [ +∆[ = 10.1000 g \ = \ +∆\ = 10.2000 g
[
)
(
f 2 = sin 10.1 − cos 10.2
*
g
g
) = − 0.829206
f − f 2' = 0.000005 = 5 ⋅ 10 −6
5y QLFD 2
Funkcja 3. Dane: f3 = arctg
([ 5R]ZLQL
y x
=
=
\
∆[ =
∆\ = )
FLHIXQNFML ∂f ∂f f3' = f3 (x 0 , y 0 ) + 3 ⋅ ρ g ⋅ ∆x + 3 ⋅ ρ g ⋅ ∆y ∂x ∂y x y y f3' = arctg 0 − 2 0 2 ⋅ ρ g ⋅ ∆x + 2 0 2 ⋅ ρ g ⋅ ∆y x0 x0 + y 0 x0 + y 0
Funkcja w postaci liniowej f3’ dla
=
=
[ \
,
(ρ
f = − 3.1831 ⋅ ∆x + 3.1831 ⋅ ∆y + 50.0000 ' 3
g
g
g
g
f3’ dla ∆[ = ∆\ = : f = − 3.1831g ⋅ 0.1 + 3.1831g ⋅ 0.2 + 50.0000 g = 50.3183 g
:DUWR üIXQNFML
' 3
f3 dla [ = [ +∆[ = 10.1\ = 10.2 = 50.3136 g f3 = arctg 10.1
:DUWR üIXQNFML
*
f − f3' = − 47cc
5y QLFD 3
+ ∆\ = 10.2
\
)
= 200 g : π :
)
= 200 g : π :
' 2
:DUWR üIXQNFML
31
32
:\UyZQDQLHZFL
,PL
FLDZVWHF] NWRZHJR PHWRGSR UHdQLF]F
1D]ZLVNR
Nr zestawu .....
û:,&=(1,(QU
:\UyZQDQLHZFL FLDZVWHF] NWRZHJR PHWRG SR UHdQLF]F
Temat:
SRPLHU]RQR ] MHGQDNRZ GRNáDGQR FL NW\ – 6WRVXMF
SR UHGQLF]F QDOH*\ REOLF]\ü Z\UyZQDQH ZDUWR FL SRPLDUyZ RUD] ZVSyáU] dnych x, y punktu P,DWDN*HSU]HSURZaG]LüDQDOL] GRNáDGQR FLSRZ\UyZQaniu.
Zadanie: Na stanowisku P PHWRG
Dane: 3RPLHU]RQHN
. W
1 2 3
o
/
W\
1
//
62 39 137 48 35 430 72 28 289
:VSyáU]
Pkt 1 2 3 4
GQHSXQNWyZVWDá\FK
x 7108.75 5953.34 5282.99 5453.62
2EOLF]RQHZVSyáU]
xPo = 6307.48
52=:,
2
y -3034.50 -2142.62 -3148.85 -4528.10
1
GQHSU]\EOL*RQHpunktu P yPo = -3708.87
=$1,(
1) Równania obserwacyjne y − yP y − yP − arctg 1 β1 + v 1 = AP2 − AP1 = arctg 2 x 2 − xP x1 − x P
β 2 + v 2 = AP3 − AP2 = arctg
y3 − yP y − yP − arctg 2 x 3 − xP x 2 − xP
β 3 + v 3 = AP 4 − AP3 = arctg
y − yP y 4 − yP − arctg 3 x 4 − xP x 3 − xP
5R]ZLQL
3
2 3 P (xP=?, yP=?)
FLHZV]HUHJ7D\ORUD
∂A ∂A ∂A ∂A β1 + v 1 = AP2 o − AP1o + P2 − P1 ⋅ ∆xP + P2 − P1 ⋅ ∆y P ∂xP ∂y P ∂xP ∂y P ∂A ∂A ∂A ∂A β 2 + v 2 = AP3 o − AP2 o + P3 − P2 ⋅ ∆xP + P3 − P2 ⋅ ∆y P ∂xP ∂y P ∂xP ∂y P ∂A ∂A ∂A ∂A β 3 + v 3 = AP 4 o − AP3 o + P4 − P3 ⋅ ∆xP + P4 − P3 ⋅ ∆y P ∂xP ∂y P ∂xP ∂y P
4
:\UyZQDQLHZFL
FLDZVWHF] NWRZHJR PHWRGSR UHdQLF]F
33
3) :VSyáF]\QQLNLNLHUXQNRZH (poFKRGQHF]VWNRZH ∂AP i
aP i =
∂x P
⋅ ρ ′′ =
yi − yP ⋅ ρ ′′ ; d P2 i
bPi =
∂AP i ∂y P
xi − xP ⋅ ρ ′′ ; d P2 i
⋅ ρ ′′ = −
ρ ′′ = 20626 5′′
( i = 1, 2, 3 , 4 )
Kierunek P-i
Przyrosty x i − x P0 y i − y P0 801.27 –354.14 –1024.49 –853.86
P-1 P-2 P-3 P-4
674.37 1566.25 560.02 –819.23
.ZDGUDWGáXJR FL
d
:VSyáF]NLHUXQNRZH
2 P i0
aP i
1096808.51 2578554.20 1363202.16 1400214.69
126.822 125.288 84.736 –120.680
bP i –150.686 28.329 155.015 125.782
4) :VSyáF]\QQLNLSU]\QLHZLDGRP\FKZUyZQDQLDFKEá GyZ a1 = aP2 – aP1 = -1.534
b1 = bP2 – bP1 = 179.015
a2 = aP3 – aP2 = -40.552
b2 = bP3 – bP2 = 126.686
a3 = aP4 – aP3 = -205.416
b3 = bP4 – bP3 = -29.233
:\UD]\ZROQHUyZQD
Eá
$]\PXW\REOLF]RQH]HZVSyáU]
ϕ P1o = arctg ϕ P2o = arctg ϕ P3o = arctg ϕ P4o = arctg
y1 − y P 0
= arctg
x1 − xP0
y 2 − y P0 x 2 − xP0 y 3 − y P0 x 3 − xP0 y 4 − y P0 x 4 − xP0
GyZ
GQ\FKprzybli*RQych
674.37 = 40$05′05.5′′; 801.27
AP1o = ϕ P1o = 40$ 05′05.5′′
= arctg
1566.25 = 77$15′33.4′′; − 354.14
= arctg
560.02 = 28$ 39′44.9′′; − 1024.49
AP3o = 180$ − ϕ P3o = 151$ 20′15.1′′
= arctg
− 819.23 = 43$ 48′51.2′′; − 853.86
AP4o = 180$ + ϕ P4o = 223$ 48′51.2′′
AP2o = 180$ − ϕ P2o = 102$ 44′26.6′′
Wyrazy wolne l1 = AP 2 o − AP1o − β1 = 7.4 // l 2 = AP3 o − AP2o − β 2 = 5.5 // l 3 = AP 4 o − AP3 o − β 3 = 7.2 // 6) RyZQDQLDEá GyZ
7 1DGRNUH ORQ\XNáDGUyZQDOLQLowych
-1.534.∆xP + 179.015.∆yP + 7.4
-1.534.∆xP + 179.015.∆yP + 7.4 = 0
v2 = -40.552.∆xP + 126.686.∆yP + 5.5
-40.552.∆xP + 126.686.∆yP + 5.5 = 0
v3 = -205.416.∆xP – 29.233.∆yP + 7.2
-205.416.∆xP – 29.233.∆yP + 7.2 = 0
v1 =
34
:\UyZQDQLHZFL
FLDZVWHF] NWRZHJR PHWRGSR UHdQLF]F A⋅ X = L
8) Zapis macierzowy
− 1.534 179.015 − 40.552 126.686 ⋅ ∆x P = ∆y − 205.416 − 29.233 P
8NáDGUyZQD
− 7.4 − 5.5 − 7.2
AT.A.X = AT.L
QRUPDOQ\FK
592.946 ∆x P 1713.383 43842.551 592.946 48950.281 ⋅ ∆y = − 1811.006 P 5R]NáDGPDFLHU]\QRUPDOQHM
AT.AQDF]\QQLNLWUyMNWQH
AT ⋅ A = R T ⋅ R 592.946 43842.551 592.946 48950.281 = 2GZURWQR
üPDFLHU]\WUyMN
R R 0
' 11
R11 0 R11 R12 R ⋅ ; 12 R 22 0 R 22
−1
WQHM
2.832 R 209.386 = ⋅ 0 221.229 R
R-1
⋅ R = I
' 12 ' 22
2GZURWQR
2.832 209.386 R= 0 221.229
(AT.A)-1
üPDFLHU]\QRUPDOQHM
0.004776 − 6.113 ⋅ 10 −5 R −1 = 0 0.004520
1 0 0 1 ;
( )
A ⋅ A = R −1 ⋅ R −1 7
0.004776 − 6.113 ⋅ 10 −5 0.004776 = ⋅ −5 0 0.004520 − 6.113 ⋅ 10
T
=
0 2.281⋅ 10 −5 = 0.004520 − 2.763 ⋅ 10 −7
− 2.763 ⋅ 10 −7 2.043 ⋅ 10 −5
13) Wektor niewiadomych X 2.281⋅ 10 −5 X = (AT ⋅ A) -1 ⋅ ( AT ⋅ L ) = −7 − 2.763 ⋅ 10
− 2.763 ⋅ 10 −7 1713.383 ⋅ ≅ 2.043 ⋅ 10 −5 − 1811.006
14) Wyrównane niewiadome ZVSyáU] dne punktu P) xP = xP0 + ∆xP = 6307 .480 + .040 = 6307 .520 y P = y P0 + ∆y P = − 3708 .870 − 0.037 = −3708 .907
0.040 − 0.037
:\UyZQDQLHZFL
FLDZVWHF] NWRZHJR PHWRGSR UHdQLF]F
35
15) Wektor poprawek V = A ⋅ X − L − 1.534 179..05 0.040 V = − 40.552 126.686 ⋅ − − 0.037 − 205.416 − 29.233
− 7.4 − 5.5 = − 7.2
0.6′′ − 0.9′′ 0.2′′
16) Obserwacje wyrównane
β 1 + v 1 = 2 $ 39 ′13 .7 ′′ + .6 ′′ = 2 $ 39 ′14 .3′′ β 2 + v 2 = 48 $ 35′43 .0′′ − .9 ′′ = 48 $ 35 ′42 .1′′ β 3 + v 3 = 72 $ 28′28 .9′′ + .2′′ = 72 $ 28′29 .1′′ 17) Kontrola ogólna V T ⋅ V = LT ⋅ L − LT ⋅ A ⋅ X 1.15 = 136.85 − 135.70 1.15 = 1.15 18) Kontrola generalna
GQ\FKZyrównanych y1 − y P 674.41 = arctg = arctg = 40$ 05′16.6′′; AP1 = ϕ P1 = 40$ 05′16.6′′ x1 − xP 801.23
$]\PXW\REOLF]RQH]HZVSyáU]
ϕ P1
ϕ P2 = arctg
y 2 − yP 1566.29 = arctg = 77 $15′29.5′′; − 354.18 x 2 − xP
ϕ P3 = arctg
y3 − yP 560.06 = arctg = 28$ 39′47.7′′; − 1024.53 x3 − xP
ϕ P4 = arctg
y 4 − yP − 819.19 = 43 $ 48′41.4′′; = arctg − 853.90 x 4 − xP
/HZDVWURQDUyZQD
AP2 = 180$ − ϕ P2 = 102$ 44′30.5′′ AP3 = 180$ − ϕ P3 = 151$ 20′12.3′′ AP4 = 180$ + ϕ P4 = 223$ 48′41.4′′
3UDZDVWURQDUyZQD
REVHUZDF\MQ\FK
obserwacyjnych
β 1 + v 1 = 2 $ 39 ′14 .3 ′′ β 2 + v 2 = 48 $ 35′42 .1′′ β 3 + v 3 = 72 $ 28′29 .1′′
AP 2 − AP1 = 102$ 44′30.5′′ − 40$05′16.6′′ = 62$39′13.9′′ AP3 − AP2 = 151$ 20′12.3′′ − 102 $ 44′30.5′′ = 48 $35′41.8′′ AP 4 − AP3 = 223 $ 48′41.4′′ − 151$ 20′12.3′′ = 72$ 28′29.1′′
36
:\UyZQDQLHZFL
FLDZVWHF] NWRZHJR PHWRGSR UHdQLF]F
19 2FHQDGRNáDdQR FL a)
Eá
G
UHGQLREVHUZDFML
V T ⋅V 1.15 =± ≈ ± 1.1′′ n−k 1
m=±
b) bá G\ UHGQLHQLHZLDGRP\FK ZVSyáU] GQ\FKZ\UyZQDQ\FK Macierz kowariancyjna 2.281⋅ 10−5 Q = A ⋅ A = −7 − 2.763 ⋅ 10 7
2EOLF]HQLHEá
GyZ UHGQLFK
mx = ± m ⋅ Qxx = ± 0.005m my = ± m ⋅ Qyy = ± 0.005m
− 2.763 ⋅ 10−7 2.043 ⋅ 10−5
:\UyZQDQLHZFL
,PL
FLDZVWHF] NierunkRZHJR PHWRGSR UHdQLF]F
1D]ZLVNR
37
Nr zestawu .....
û:,&=(1,(QU
:\UyZQDQLHZFL FLDZVWHF] (kierunkowego PHWRG SR UHGQiF]F
Temat:
SRPLHU]RQR]MHGQDNRZGRNáDGQR FLNLHUXQNL – 4. Stosu SR UHGQLF]F QDOH*\ REOLF]\ü Z\UyZQDQH ZDUWR FL SRPLDUyZ RUD] ZVSyá U] GQ\FKx, y punktu P,DWDN*HSU]HSURZDG]LüDQDOL] GRNáDGQR FLSRZ\Uywnaniu.
Zadanie: Na stanowisku P
M F PHWRG
Dane: Pomierzone kierunki o / // Nr 1 94 00 132 2 263 31 082 3 325 18 492 4 26 11 117 :VSyáU]
Pkt 1 2 3 4
GQHSXQNWyZVWDá\FK
x 3301.17 397.93 858.74 2051.78
2EOLF]RQHZVSyáU]
xPo = 1803.30
52=:,
y -4408.82 -3582.04 -4936.63 -5347.76
GQHprzyblL*RQHpunktu P yPo = -4126.20
=$1,(
1) Równania obserwacyjne y1 − y P −z x1 − x P y − yP − z = arctg 2 −z x 2 − xP y − yP − z = arctg 3 −z x3 − xP y − yP − z = arctg 4 −z x 4 − xP
K 1 + v 1 = AP1 − z = arctg K 2 + v 2 = AP 2 K 3 + v 3 = AP3 K 4 + v 4 = AP 4
1 2
1 2
3
3
P (xP=?, yP=?)
4
4
38
:\UyZQDQLHZFL
5R]ZLQL
FLDZVWHF] NierunkoweJR PHWRGSR UHdQLF]F
FLHZV]HUHJ7D\ORUD
∂A ∂A K 1 + v 1 = AP1o − zo + P1 ⋅ ∆xP + P1 ⋅ ∆y P − ∆z ∂xP ∂y P ∂A ∂A K 2 + v 2 = AP 2 o − zo + P2 ⋅ ∆xP + P2 ⋅ ∆y P − ∆z ∂xP ∂y P ∂A ∂A K 3 + v 3 = AP3 o − zo + P3 ⋅ ∆xP + P3 ⋅ ∆y P − ∆z ∂xP ∂y P ∂AP4 ∂A ⋅ ∆xP + P4 ⋅ ∆y P − ∆z K 4 + v 4 = AP 4 o − zo + ∂xP ∂y P 3) :VSyáF]\QQLNLNLHUXQNRZH SRFKRGQHF]VWNRZH ai =
∂AP i y −y ⋅ ρ ′′ = i 2 P ⋅ ρ ′′ ; ∂xP dP i
Kierunek P-i P-1 P-2 P-3 P-4
bi =
∂AP i x −x ⋅ ρ ′′ = − i 2 P ⋅ ρ ′′ ; ∂y P dP i
Przyrosty x i − x P0 y i − y P0
'áXJR ü
1497.87 -1405.37 -944.56 248.48
1524.299 1507.042 1244.584 1246.576
-282.62 544.16 -810.43 -1221.56
( i = 1, 2, 3, 4 )
:VSyáF]NLHUXQNRZH
d P i0
ρ ′′ = 20626 5′′
ai
bi
-25.09 49.42 -107.92 -162.14
-132.97 127.63 125.78 -32.98
*
$]\PXW\REOLF]RQH]HZVSyáU] GQ\FKSU]\EOL RQ\FK
y1 − y P 0
ϕ P1o = arctg ϕ P2o = arctg ϕ P3o = arctg ϕ P4o = arctg
x1 − xP0
y 2 − y P0 x 2 − xP0 y 3 − y P0 x 3 − xP0 y 4 − y P0 x 4 − xP0
− 282.62 = 10 $ 41′ 06.1′′; 1497.87
AP1o = 360 − ϕ P1o = 349$18′53.9′′
= arctg
544.16 = 21$09′59.1′′; − 1405.37
AP2o = 180$ − ϕ P2o = 158$ 50′00.9′′
= arctg
− 810.43 = 40 $37′46.2′′; − 944.56
AP3o = 180$ + ϕ P3o = 220$ 37′46.2′′
= arctg
− 1221.56 = 78$ 30′08.0′′; 248.48
= arctg
AP4o = 360$ − ϕ P4o = 281$ 29′52.0′′
5) Tabela orientacji stanowiska
Stan.
Nr kier.
P
1 2 3 4
Kierunek
Azymut przybl.
K
A0
° ′ ″ 94 00 13.2 263 31 08.2 325 18 49.2 26 11 11.7
° ′ ″ 349 18 53.9 158 50 00.9 220 37 46.2 281 29 52.0 UHGQLD zo =
6WDáDRULH
ntac.
z = A0 − K ° ′ ″ 255 18 40.7 255 18 52.7 255 18 57.0 255 18 40.3 255 18 47.7 / 0
Kier. przyEOL*
K 0 = A0 − z0 ° ′ ″ 94 00 06.2 263 31 13.2 325 18 58.5 26 11 04.3
Wyraz wolny
l = K0 − K ″ -7.0 5.0 9.3 -7.4
:\UyZQDQLHZFL
FLDZVWHF] NierunkRZHJR PHWRGSR UHdQLF]F
6) RyZQDQLDEá GyZ
7 1DGRNUH ORQ\XNáDGUyZQDOLQLowych
v1 = -25.09.∆xP – 132.97.∆yP – ∆z – 7.0
-25.09.∆xP – 132.97.∆yP – ∆z – 7.0 = 0
49.42.∆xP + 127.63.∆yP – ∆z + 5.0
49.42.∆xP + 127.63.∆yP – ∆z + 5.0 = 0
v3 = -107.92.∆xP + 125.78.∆yP – ∆z + 9.3
-107.92.∆xP + 125.78.∆yP – ∆z + 9.3 = 0
v2 =
v4 = -162.14.∆xP – 32.98.∆yP – ∆z – 7.4
-162.14.∆xP – 32.98.∆yP – ∆z – 7.4 = 0
A⋅ X = L
8) Zapis macierzowy
− 25.09 − 132.97 49.42 127.63 − 107.92 125.78 − 162.14 − 32.98
8NáDGUyZQD
− 1 − 1 ∆x P ⋅ ∆y = − 1 ∆zP − 1
7 − 5 − 9.3 7.4
AT.A.X = AT.L
QRUPDOQ\FK
14.89 245.73 ∆x 41007.95 1416.89 50878.73 − 87.46 ⋅ P = ∆y P 245.73 . . 4 00 ∆z − 87 46 5R]NáDGPDFLHU]\QRUPDOQHM
− 618.91 − 2982.75 − 0.10
AT.AQDF]\QQLNLWUyMNWQH
AT ⋅ A = R T ⋅ R 14.89 245.73 41007.95 1416.89 50878.73 − 87.46 = 245.73 4.00 − 87.46
R11 0 R 12 R 22 R13 R 23
0 R11 R12 0 ⋅ 0 R 22 0 R33 0
1.213 202.5 6.997 R= 0 225.4 − 0.4256 0 0 1.532 2GZURWQR
39
üPDFLHU]\WUyMN
R −1 ⋅ R R' 11 R' 12 0 R' 22 0 0
R
−1
WQHM
R-1
= I R' 13 202.5 6.997 1.213 R' 23 ⋅ 0 225.4 − 0.4256 = R' 33 0 0 1.532
1 0 0 0 1 0 ; 0 0 1
0.004938 − 0.0001532 − 0.003954 = 0 0.004435 0.001232 0 0 0.6528
R13 R 23 ; R33
40
:\UyZQDQLHZFL
2GZURWQR
FLDZVWHF] NierunkoweJR PHWRGSR UHdQLF]F
üPDFLHU]\QRUPDOQHM
(AT.A)-1
( )
T
A ⋅ A = R −1 ⋅ R −1 = 0 0 0.004938 − 0.0001532 − 0.003954 0.004938 0 0.004435 0.001232 ⋅ − 0.0001532 0.004435 0 = = 0 0 0.6528 − 0.003954 0.001232 0.6528
4.005 ⋅ 10 −5 = − 5.553 ⋅ 10 −6 − 0.002582
− 5.553 ⋅ 10 −6 2.119 ⋅ 10 −5 0.0008045
7
− 0.002582 0.0008045 0.4261
13) Wektor niewiadomych X X = (AT ⋅ A) -1 ⋅ ( AT ⋅ L ) = 4.005 ⋅ 10 −5 − 5.553 ⋅ 10 −6 −6 2.119 ⋅ 10 −5 − 5.553 ⋅ 10 − 0.002582 0.0008045
− 0.002582 − 618.91 0.0008045 ⋅ − 2982.75 ≅ 0.4261 − 0.10
− 0.008 − 0.060 − 0.8
14) Wyrównane niewiadome xP = xP0 + ∆xP = 1803 .300 − .008 = 1803 .292 y P = y P0 + ∆y P = − 4126 .200 − 0.060 = −4126 .260 z = z0 + ∆z = 255 $18′47 .7′′ − 0.8′′ = 255 $18′46.9′′ 15) Wektor poprawek V = A ⋅ X − L − 25.09 − 132.97 49.42 127.63 V = − 107.92 125.78 − 162.14 − 32.98
− 1 − 0.008 − 1 ⋅ − 0.060 − − 1 0 . 8 − − 1
16) Obserwacje wyrównane K 1 + v 1 = 94$ 00′13.2′′ + 2.0′′ = 94 $00′15.2′′ K 2 + v 2 = 263$ 31′ 08.2′′ − 2.2′′ = 263$ 31′ 06.0′′ K 3 + v 3 = 325$18′49.2′′ + 3.5′′ = 325$18′52.7′′ K 4 + v 4 = 26$11′ 11.7′′ − 3.3′′ = 26$11′ 08.4′′ 17) Kontrola ogólna V T ⋅ V = LT ⋅ L − LT ⋅ A ⋅ X 31.71 = 215.25 − 183.54 31.71 = 31.71
7 − 5 = − 9.3 7.4
2.0′′ − 2.2′′ 3.5′′ − 3.3′′
:\UyZQDQLHZFL
FLDZVWHF] NierunkRZHJR PHWRGSR UHdQLF]F
18) Kontrola generalna $]\PXW\REOLF]RQH]HZVSyáU]
ϕ P1
GQ\FKZ\Uywnanych
y − yP = arctg 1 = 10 $ 40′57.9′′; x1 − xP
AP1 = 360 − ϕ P1 = 349 $19′02.1′′
ϕ P2 = arctg
y 2 − yP = 21$10′07.3′′; x 2 − xP
AP2 = 180 $ − ϕ P2 = 158 $ 49′52.7′′
ϕ P3 = arctg
y3 − yP = 40 $ 37′39.8′′; x 3 − xP
AP3 = 180 $ + ϕ P3 = 220 $ 37′39.8′′
ϕ P4 = arctg
y4 − yP = 78 $ 30′04.4′′; x 4 − xP
AP4 = 360 $ − ϕ P4 = 281$ 29′55.6′′
/HZDVWURQDUyZQD
Prawa stroQDUyZQDREVHUZDF\MQ\FK
obserwacyjnych K1 + v1 = 94 $00′15.2′′
AP1 − z = 349 $19′02.1′′ − 255 $18′46.9′′ = 94$00′15.2′′
K 2 + v 2 = 263 $31′ 06.0′′
AP 2 − z = 158 $ 49′52.7′′ − 255 $18′46.9′′ = 263 $31′ 05.8′′
K 3 + v 3 = 325 $18′52.7′′
AP3 − z = 220 $37′39.8′′ − 255 $18′46.9′′ = 325 $18′52.9′′
K 4 + v 4 = 26 $11′ 08.4′′
AP 4 − z = 281$ 29′55.6′′ − 255 $18′46.9′′ = 26$11′ 08.7′′
19 2FHQDGRNáDGQo FL D Eá
G
UHGQLREVHUZDFML
V T ⋅V 31.71 m=± =± ≈ ± 5.6′′′ n−k 1 E Eá
G\
UHGQLHQLHZLDGRP\FK ZVSyáU] GQ\FKZ\UyZQDQ
ych)
Macierz kowariancyjna 4.005 ⋅ 10 −5 Q = A ⋅ A = − 5.553 ⋅ 10 − 6 − 0.002582 7
2EOLF]HQLHEá
GyZ UHGQLFK
m x = ± m ⋅ Q xx = ± 0.036 m m y = ± m ⋅ Q yy = ± 0.026 m
− 5.553 ⋅ 10 −6 −5
2.119 ⋅ 10 0.0008045
− 0.002582 0.0008045 0.4261
41
Wyrównanie ... SRSU]H]HOLPLQDFM QLHZLDGomej orientacyjnej
42 ,PL
1D]ZLVNR
Nr zestawu .....
û:,&=(1,(QU
:\UyZQDQLHZFL FLDZVWHF] NLHUXQNRZHJR PHWRG SR UHGQLF]FSRSU]H]HOLPLQDFM QLHZLDGomej orientacyjnej
Temat:
GRNáDGQR FLNLHUXQNL – 4. StosuMF PHWRG SR UHGQLF]F QDOH*\ REOLF]\ü Z\UyZQDQH ZDUWR FL NLHUXQNyZ RUD] ZVSyá U] GQ\FK x, y punktu P,DWDN*HSU]HSURZDG]Lü DQDOL] GRNáDGQR FLSR Z\UyZQDQLX=aZadanie: Na stanowisku P
SRPLHU]RQR]MHGQDNRZ
VWRVRZDüVSRVyEHOLPLQDFMLVWDáHMRULHQWDFMLVW
anowiska.
Dane: Pomierzone kierunki o / // Nr 1 94 00 132 2 263 31 082 3 325 18 492 4 26 11 117 :VSyáU]
Pkt 1 2 3 4
GQHSXQNWyZVWDá\FK
x 3301.17 397.93 858.74 2051.78
y -4408.82 -3582.04 -4936.63 -5347.76
Obliczone wVSyáU] GQHSU]\EOL*RQHpunktu P xPo = 1803.30 yPo = -4126.20
52=:,
=$1,(
1) Równania obserwacyjne y1 − y P −z x1 − x P y − yP − z = arctg 2 −z x 2 − xP y − yP − z = arctg 3 −z x3 − xP y − yP − z = arctg 4 −z x 4 − xP
K 1 + v 1 = AP1 − z = arctg K 2 + v 2 = AP 2 K 3 + v 3 = AP3 K 4 + v 4 = AP 4
1 2
1 2
3
3
P (xP=?, yP=?)
4
4
Wyrównanie ... poprzez elLPLQDFM QLHZLDGomej orientacyjnej 5R]ZLQL
43
FLHZV]HUHJ7D\ORUD
∂A ∂A K 1 + v 1 = AP1o − zo + P1 ⋅ ∆xP + P1 ⋅ ∆y P − ∆z ∂xP ∂y P ∂A ∂AP2 ⋅ ∆xP + P2 ⋅ ∆y P − ∆z K 2 + v 2 = AP 2 o − zo + ∂y P ∂xP ∂A ∂A K 3 + v 3 = AP3 o − zo + P3 ⋅ ∆xP + P3 ⋅ ∆y P − ∆z ∂y P ∂xP ∂A ∂A K 4 + v 4 = AP 4 o − zo + P4 ⋅ ∆xP + P4 ⋅ ∆y P − ∆z ∂xP ∂y P 3) :VSyáF]\QQLNLNLHUXQNRZH SRFKRGQHF]VWNRZH ai =
∂AP i y −y ⋅ ρ ′′ = i 2 P ⋅ ρ ′′ ; ∂xP dP i
Kierunek P-i P-1 P-2 P-3 P-4
bi =
∂AP i x −x ⋅ ρ ′′ = − i 2 P ⋅ ρ ′′ ; ∂y P dP i
Przyrosty x i − x P0 y i − y P0
'áXJR ü
1497.87 -1405.37 -944.56 248.48
1524.299 1507.042 1244.584 1246.576
-282.62 544.16 -810.43 -1221.56
( i = 1, 2, 3, 4 )
:VSyáF]NLHUXQNRZH
d P i0
ρ ′′ = 20626 5′′
ai
bi
-25.09 49.42 -107.92 -162.14
-132.97 127.63 125.78 -32.98
*
$]\PXW\REOLF]RQH]HZVSyáU] GQ\FKSU]\EOL RQ\FK
y1 − y P 0
ϕ P1o = arctg ϕ P2o = arctg ϕ P3o = arctg ϕ P4o = arctg
x1 − xP0
y 2 − y P0 x 2 − xP0 y 3 − y P0 x 3 − xP0 y 4 − y P0 x 4 − xP0
− 282.62 = 10 $ 41′ 06.1′′; 1497.87
AP1o = 360 − ϕ P1o = 349$18′53.9′′
= arctg
544.16 = 21$09′59.1′′; − 1405.37
AP2o = 180$ − ϕ P2o = 158$ 50′00.9′′
= arctg
− 810.43 = 40 $37′46.2′′; − 944.56
AP3o = 180$ + ϕ P3o = 220$ 37′46.2′′
= arctg
− 1221.56 = 78$ 30′08.0′′; 248.48
= arctg
AP4o = 360$ − ϕ P4o = 281$ 29′52.0′′
5) Tabela orientacji stanowiska
Stan.
Nr kier.
P
1 2 3 4
Kierunek
Azymut przybl.
K
A0
° ′ ″ 94 00 13.2 263 31 08.2 325 18 49.2 26 11 11.7
° ′ ″ 349 18 53.9 158 50 00.9 220 37 46.2 281 29 52.0 UHGQLD zo =
6WDáDRULH
ntac.
z = A0 − K ° ′ ″ 255 18 40.7 255 18 52.7 255 18 57.0 255 18 40.3 255 18 47.7 i 0
Kier. przyEOL*
K 0 = A0 − z0 ° ′ ″ 94 00 06.2 263 31 13.2 325 18 58.5 26 11 04.3
Wyraz wolny
l = K0 − K ″ -7.0 5.0 9.3 -7.4
Wyrównanie ... SRSU]H]HOLPLQDFM QLHZLDGomej orientacyjnej
44
6) RóZQDQLDEá GyZ v1 = -25.09.∆xP – 132.97.∆yP – ∆z – 7.0 v2 =
49.42.∆xP + 127.63.∆yP – ∆z + 5.0
v3 = -107.92.∆xP + 125.78.∆yP – ∆z + 9.3 v4 = -162.14.∆xP – 32.98.∆yP – ∆z – 7.4
:VSyáF]\QQLNL]UHGXNRZDQHJRXNáDGXUyZQD
Ai = ai −
[a] ; n
Bi = bi −
[b] ; n
Li = l i −
Ai
Bi
n
n=4
Li
1
36.34
-154.84
-7.0
2
110.85
105.77
5.0
3
-46.48
103.61
9.3
4
-100.71
-54.85
-7.4
=UHGXNRZDQ\XNáDGUyZQD
Eá GyZ
[l ]
[a] = − 245.73; [b] = 87.46; [l ] = − 0.1; Kier.
Eá
GyZ
vi = Ai.∆xP + Bi.∆yP + Li
9 1DGRNUH ORQ\XNáDGUyZQDOLQLRZ\FK
36.34.∆xP – 154.84.∆yP – 7.0
36.34.∆xP – 154.84.∆yP – 7.0 = 0
v2 = 110.85.∆xP + 105.77.∆yP + 5.0
110.85.∆xP + 105.77.∆yP + 5.0 = 0
v3 = -46.48.∆xP + 103.61.∆yP + 9.3
-46.48.∆xP + 103.61.∆yP + 9.3 = 0
v4 = -100.71.∆xP – 54.85.∆yP – 7.4
-100.71.∆xP – 54.85.∆yP – 7.4 = 0
v1 =
A⋅ X = L
10) Zapis macierzowy
7 − 154.84 36.34 110.85 105.77 ∆x P − 5 ⋅ = − 46.48 103.91 ∆y P − 9.3 7.4 − 100.71 − 54.85 8NáDGUyZQD
QRUPDOQ\FK
AT.A.X = AT.L
25911.21 6791.93 ∆x P − 612.86 6791.93 48968.53 ⋅ ∆y = − 2984.98 P
Wyrównanie ... poprzez elLPLQDFM QLHZLDGomej orientacyjnej 2GZURWQR
45
(AT.A)-1
üPDFLHU]\QRUPDOQHM
AT ⋅ A = R T ⋅ R 25911.21 6791.93 R11 0 R11 R12 ⋅ ; 6791.93 48968.53 = R 12 R 22 0 R 22
2GZURWQR
üPDFLHU]\WUyMN
R
−1
WQHM
R-1
⋅ R = I
' ' R11 160.97 42.19 R12 ⋅ = ' 0 217.23 0 R 22
2GZURWQR
160.97 42.19 R= 0 217.23
(AT.A)-1
üPDFLHU]\QRUPDOQHM
0.006212 − 0.001207 R −1 = 0 0.004603
1 0 0 1 ;
A ⋅ A 7
( )
= R −1 ⋅ R −1
T
=
0 0.006212 − 0.001207 0.006212 4.005 ⋅ 10 −5 = ⋅ = 0 0.004603 − 0.001207 0.004603 − 5.554 ⋅ 10 −6
− 5.554 ⋅ 10 −6 2.119 ⋅ 10 −5
15) Wektor niewiadomych X 4.005 ⋅ 10 −5 X = (AT ⋅ A) -1 ⋅ ( AT ⋅ L ) = −6 − 5.554 ⋅ 10
− 5.554 ⋅ 10 −6 − 612.86 ⋅ ≅ 2.119 ⋅ 10 −5 − 29.84.98
− 0.008 − 0.060
16) Przyrost niewiadomej orientacyjnej ∆z =
[a] ⋅ ∆x n
P
+
[b] ⋅ ∆y n
P
+
[l ] = − 61.43 ⋅ (− 0.008) + 21.86 ⋅ (− 0.060) + 0.0 = − 0.8′′ n
17) Wyrównane niewiadome x P = x P0 + ∆x P = 1803 .30 − .008 = 1803 .292 y P = y P0 + ∆y P = − 4126 .20 − 0.060 = −4126 .260 z = z 0 + ∆z = 255 $18′47 .7′′ − 0.8′′ = 255 $18′46.9 ′′ 18) Wektor poprawek V = A ⋅ X − L − 154.84 36.34 110.85 105.77 − 0.008 ⋅ V = − − 46.48 103.91 − 0.060 − 100.71 − 54.85
7 − 5 = − 9.3 7.4
2.0′′ − 2.2′′ 3.5′′ − 3.3′′
Wyrównanie ... SRSU]H]HOLPLQDFM QLHZLDGomej orientacyjnej
46
19) Obserwacje wyrównane K 1 + v 1 = 94$ 00′13.2′′ + 2.0′′ = 94 $00′15.2′′ K 2 + v 2 = 263$ 31′ 08.2′′ − 2.2′′ = 263$ 31′ 06.0′′ K 3 + v 3 = 325$18′49.2′′ + 3.5′′ = 325$18′52.7′′ K 4 + v 4 = 26$11′ 11.7′′ − 3.3′′ = 26$11′ 08.4′′ V T ⋅ V = LT ⋅ L − LT ⋅ A ⋅ X 31.71 = 215.25 − 183.54 31.71 = 31.71
20) Kontrola ogólna
21) Kontrola generalna $]\PXW\REOLF]RQH]HZVSyáU]
ϕ P1
GQ\FKZ\Uywnanych
y − yP = arctg 1 = 10 $ 40′57.9′′; x1 − xP
AP1 = 360 − ϕ P1 = 349 $19′02.1′′
ϕ P2 = arctg
y 2 − yP = 21$10′07.3′′; x 2 − xP
AP2 = 180 $ − ϕ P2 = 158 $ 49′52.7′′
ϕ P3 = arctg
y3 − yP = 40 $ 37′39.8′′; x 3 − xP
AP3 = 180 $ + ϕ P3 = 220 $ 37′39.8′′
ϕ P4 = arctg
y4 − yP = 78 $ 30′04.4′′; x 4 − xP
AP4 = 360 $ − ϕ P4 = 281$ 29′55.6′′
/HZDVWURQDUyZQD
3UDZDVWURQDUyZQD
REVHUZDF\MQ\FK
obserwacyjnych AP1 − z = 349$19′02.1′′ − 255$18′46.9′′ = 94 $00′15.2′′
K1 + v1 = 94 $00′15.2′′ K 2 + v 2 = 263 $31′ 06.0′′
AP 2 − z = 158$ 49′52.7′′ − 255$18′46.9′′ = 263$ 31′ 05.8′′
K 3 + v 3 = 325 $18′52.7′′
AP3 − z = 220$ 37′39.8′′ − 255$18′46.9′′ = 325$18′52.9′′
K 4 + v 4 = 26 $11′ 08.4′′
AP 4 − z = 281$ 29′55.6′′ − 255$18′46.9′′ = 26$11′ 08.7′′
22 2FHQDGRNáDGQR FL a)
Eá
UHGQLREVHUZDFML
b)
Eá
G
G\
m=±
31.71 V T ⋅V =± ≈ ± 5.6′′′ n−k 1
UHGQLHQLHZLDGRP\FK ZVSyáU]
GQ\FKZ\UyZQDQ\FK
Macierz kowariancyjna Q = A ⋅ A 7
2EOLF]HQLHEá
4.005 ⋅ 10 −5 = −6 − 5.55 ⋅ 10
GyZ UHGQLFK
mx = ± m ⋅ Qxx = ± 0.036m my = ± m ⋅ Qyy = ± 0.026m
− 5.55 ⋅ 10 −6 2.119 ⋅ 10 −5
*
%á GSRáR HQLDSXQNWXHOLSVDEá
,PL
GX UHGQLHJR
1D]ZLVNR
47
Nr zestawu .....
û:,&=(1,(QU Temat:
%áGSRáR*HQLDSXQNWXHOLSVDEá GX UHGQLHJR
Zadanie 1DSRGVWDZLHGDQ\FK]üZLF]HQLDÄ:\UyZQDQLHNLHUXQNRZHJRZFL FLDZVWHF]”
*
QDOH \ REOLF]\ü Eá
G\ UHGQLH ZVSyáU] GQ\FK SXQNWX Z\]QDF]DQHJR EáG SRáR*HQLD
GX UHGQLHJR :\QLNL SURV] SU]HGVWDZLü UyZQLH* So-
SXQNWX RUD] SDUDPHWU\ HOLSV\ Eá
staci rysunku w odpowiedniej skali. Dane:
%á G UHGQLREVHUZDFML
V T ⋅V = ± 5.6′′′ m=± n−k
:
4.005 ⋅ 10 −5 Q = A ⋅ A = −6 − 5.55 ⋅ 10
Macierz kowariancyjna:
:\UyZQDQHZVSyáU]
7
GQH:
xp = 1803.292;
Obliczenia: 1)
%á G\
UHGQLHZVSyáU] GQ\FK
m x = ± m ⋅ Q xx = ± 0.036 m m y = ± m ⋅ Q yy = ± 0.026 m 2)
*
%á GSRáR HQLDSXQNWX
m P = ± m x2 + m y2 = ± 0.044 m 3)
. WVNU
tg (2α ) =
FHQLDHOLSV\
2Qxy Q xx − Qyy
=
− 1.1108 ⋅ 10 −5 = − 0.588971 1.8660 ⋅ 10 −5
ϕ = 30°29′49″ 2α = 360° - ϕ = 329°30′11″
α = 164°45′06″
− 5.55 ⋅ 10 −6 2.119 ⋅ 10 −5
yp = - 4126.260
48
*
%á GSRáR HQLDSXQNWXHOLSVDEá
*
'áX V]DSyáR
GX UHGQLHJR
HOLSV\
a = ± m ⋅ Q xx ⋅ cos 2 α + Qyy ⋅ sin 2 α + 2 ⋅ Q xy ⋅ sin α ⋅ cos α = ± 0.036 m .UyWV]DSyáR
HOLSV\
b = ± m ⋅ Q yy ⋅ cos 2 α + Q xx ⋅ sin 2 α − 2 ⋅ Q xy ⋅ sin α ⋅ cos α = ± 0.025 m 6]NLFHOLSV\LRNU
JXEá
GX
UHGQLHJR
Skala 1:1 x
my
mP
mx
y
P y’
b
α
a
x’
WyróZQDQLHZFL FLDZSU]yG NLHUXQNRZHJR PHWRGSR UHGQLF]F ,PL
1D]ZLVNR
49
Nr zestawu .....
û:,&=(1,(QU :\UyZQDQLHZFL FLDZSU]yG NLHUXQNRZHJR PHWRG SR UHGQLF]F
Temat:
Zadanie: Na stanowiskach 1, 2, 3 SRPLHU]RQRNLHUXQNLZFLQDMFH–6WRVXMFPHWRG F QDOH*\ REOLF]\ü Z\UyZQDQH ZDUWR FL SRPLDUyZ RUD] ZVSyáU] GQ\FK x, y punktu P,DWDN*HSU]HSURZDG]LüDQDOL] GoNáDGQR FLSRZ\UyZQDQLX SR UHGQLF]
Dane: Pomierzone kierunki o / // Nr 1 237 42 068 2 297 11 20 3 91 23 33 4 151 02 10 5 219 11 53 6 0 00 00 7 50 02 37 :VSyáU]
Pkt 1 2 3
7
x 2051.78 858.74 397.93
2EOLF]RQHZVSyáU]
5
2 P (xP=?, yP=?)
y -5347.76 -4936.63 -3582.04
GQHSU]\EOL*RQHSXQNWX3 yPo = -4126.20
=$1,(
1) Równania obserwacyjne GODNLHUXQNyZZFLQDMF\FK K 1 + v 1 = A1P − z1 = arctg
6
4
GQHSXQNWyZVWDá\FK
xPo = 1803.30
52=:,
3
y P − y1 − z1 x P − x1
K 4 + v 4 = A2P − z 2 = arctg
yP − y2 − z2 xP − x2
K 7 + v 7 = A3P − z3 = arctg
yP − y3 − z3 x P − x3
3
1
2
1
50
:\UyZQDQLHZFL
5R]ZLQL
FLDZSU]yG NLHUXQNRZHJR PHWRGSR UHGQLF]F
FLHZV]HUHJ7D\ORUD
∂A ∂A K 1 + v 1 = (A1P o − z1 ) + 1P ⋅ ∆xP + 1P ⋅ ∆y P ∂xP ∂y P ∂A ∂A K 4 + v 4 = (A2P o − z 2 ) + 2P ⋅ ∆xP + 2P ⋅ ∆y P ∂xP ∂y P ∂A ∂A K 7 + v 7 = (A3P o − z3 ) + 3P ⋅ ∆xP + 3P ⋅ ∆y P ∂xP ∂y P :VSyáF]\QQLNLNLHUXQNRZH SRFKRGQHF]
ai =
∂Ai P ∂x P
⋅ ρ ′′ = −
yP − yi ⋅ ρ ′′ ; d i2P
bi =
∂Ai P ∂y P
VWNRZH
⋅ ρ ′′ =
xP − xi ⋅ ρ ′′ ; d i2P
ρ ′′ = 20626 5 ′′
( i = 1, 2, 3 )
Kierunek i-P 1-P 2-P 3-P
xP 0
Przyrosty yP 0 − yi − xi
-248.48 944.56 1405.37
1221.56 810.43 -544.16
'áXJR ü
:VSyáF]NLHUXQNRZH
d i P0 1246.58 1244.58 1507.04
ai
bi
-162.14 -107.92 49.24
-32.98 125.77 127.63
4) Azymuty GODNLHUXQNyZZFLQDMF\FK -SU]\EOL*RQH)
ϕ 1p o = arctg ϕ 2 p o = arctg ϕ 3 p o = arctg
y P0 − y 1 x P0 − x1 y P0 − y 2 x P0 − x 2 y P0 − y 3 x P0 − x3
= arctg
1221.56 = 78 $30′08.0′′; − 248.48
= arctg
810.43 = 40$ 37′46.2′′; 944.56
= arctg
− 544.16 = 21$ 09′59.1′′; 1405.37
ϕ 12 = arctg
y 2 − y1 411.13 = arctg = 19$ 00′51.5′′; − 1193.04 x 2 − x1
ϕ 23 = arctg
y3 − y2 1354.59 = arctg = 71$12′45.0′′; − 460.81 x3 − x 2
A21 = 180$ + A12 = 340$ 59′08.5′′;
A1P o = 180$ − ϕ 1p o = 101$ 29′52.0′′ A2P o = ϕ 2P o = 40$ 37′46.2′′ A3P o = 360$ − ϕ 3 P o = 338$ 50′00.9′′ A12 = 180$ − ϕ 12 = 160$ 59′08.5′′ A23 = 180$ − ϕ 23 = 108$ 47′15.0′′
A32 = 180$ + A23 = 288$ 47′15.0′′
6WDáHRULHQWDFMLVWDQRZLVN
z1 = A12 − K 2 = 160$59′08.5′′ − 297$11′ 20′′ = 223$ 47′48.5′′ $ ⇒ z 2 = 249 35′28.7′′ $ $ $ z 2 ' ' = A23 − K 5 = 108 47′15.0′′ − 219 11′ 53′′ = 249 35′22.0′′ z3 = A32 − K 6 = 288$ 47′15.0′′ − 0 $ 00′00′′ = 288$ 47′15.0′′ z 2 ' = A21 − K 3 = 340$ 59′08.5′′ − 91$ 23′33′′ = 249$ 35′35.5′′
WyróZQDQLHZFL FLDZSU]yG NLHUXQNRZHJR PHWRGSR UHGQLF]F
51
6) Wyrazy wolne l1 = A1P o − z1 − K 1 = − 3.3′′ l 4 = A2P o − z 2 − K 4 = 7.5′′ l 7 = A3P o − z3 − K 7 = 9.0′′ 5yZQDQLDEá
GyZ
1DGRNUH
v1 = -162.14.∆xP – 32.98.∆yP – 3.3 v4 =
-107.92.∆xP + 125.78.∆yP + 7.5
v7 =
49.42.∆xP + 127.63.∆yP + 9.0
ORQ\XNáDGUyZQD
OLQLRZ\FK
-162.14.∆xP – 32.98.∆yP – 3.3 = 0 -107.92.∆xP + 125.78.∆yP + 7.5 = 0 49.42.∆xP + 127.63.∆yP + 9.0 =0
9) Zapis macierzowy
10) Macierz wagowa
A⋅ X = L − 162.15 − 32.98 − 107.92 125.78 ⋅ ∆x P = ∆y 49.42 127.63 P
8NáDGUyZQD
3.3 − 7.5 9.0
0 0 0.5 P = 0 0.67 0 0 0 0.5
QRUPDOQ\FK
AT.P.A.X = AT. P.L 52.38 22169.16 − 3267.27 ∆x P − 3267.27 19288.36 ⋅ ∆y = − 1260.80 P 5R]NáDGPDFLHU]\QRUPDOQHM
AT.P.AQDF]\QQLNLWUyMNWQH
AT ⋅ P ⋅ A = R T ⋅ R 22169.16 − 3267.27 − 3267.27 19288.36 =
148.89 − 21.94 0 137.14
R11 0 R11 R12 R ⋅ 12 R 22 0 R 22
: R =
13) Odw URWQR üPDFLHU]\WUyMNWQHM R-1 R R 0
' 11
−1
⋅ R = I
R 148.89 − 21.94 = ⋅ 0 137.14 R ' 12 ' 22
1 0 0 1
0.006716 0.001075 0 0.007292
: R −1 =
52
:\UyZQDQLHZFL
2GZURWQR
FLDZSU]yG NLHUXQNRZHJR PHWRGSR UHGQLF]F
üPDFLHU]\QRUPDOQHM
(AT.P.A)-1
( )
( AT ⋅ P ⋅ A ) -1 = R −1 ⋅ R −1
T
=
0 4.626 ⋅ 10 −5 0.006716 0.001075 0.006716 = ⋅ = 0 0.007292 0.001075 0.007292 7.835 ⋅ 10 −6
7.835 ⋅ 10 −6 5.317 ⋅ 10 −5
15) Wektor niewiadomych X 4.626 ⋅ 10 −5 X = (AT ⋅ P ⋅ A) -1 ⋅ ( AT ⋅ P ⋅ L ) = −6 7.835 ⋅ 10
52.38 − 0.007m 7.835 ⋅ 10 −6 ⋅ ≅ −5 5.317 ⋅ 10 − 1260.80 − 0.067m
16) Wyrównane niewiadome x P = x P0 + ∆x P = 1803 .30 − 0.007 = 1803 .293 y P = y P0 + ∆y P = − 4126 .20 − 0.067 = −4126 .267 17) Wektor poprawek V = A ⋅ X − L 3.4 V = − 7.6 − − 8.9
3.3 − 7.5 = − 9.0
0.1′′ − 0.1′′ 0.1′′
18) Obserwacje wyrównane K 1 + v 1 = 237$ 42′06.8′′ + 0.1′′ = 237$ 42′06.9′′ K 4 + v 4 = 151$02′10′′ − 0.1 = 151$ 02′09.9′′ K 7 + v 7 = 50 $02′37′′ + 0.1′′ = 50 $02′37.1′′
19) Kontrola ogólna V T ⋅ P ⋅ V = LT ⋅ P ⋅ L − LT ⋅ P ⋅ A ⋅ X 0.017 = 83.632 − 83.615 0.017 = 0.017
WyróZQDQLHZFL FLDZSU]yG NLHUXQNRZHJR PHWRGSR UHGQLF]F
53
20) Kontrola generalna Azymuty obliczone ze wspyáU] GQ\FKZ\UyZQDQ\FK y P − y1 1221.493 ϕ 1p = arctg = arctg = 78 $ 30′04.6′′; x P − x1 − 248.487
ϕ 2 p = arctg ϕ 3 p = arctg
yP − y2 xP − x 2 yP − y3 x P − x3
/HZDVWURQDUyZQD
= arctg
810.363 = 40 $ 37′38.6′′; 944.553
= arctg
− 544.227 = 21$10′08.0′′; 1405.363
3UDZDVWURQDUyZQD
A1P = 180 $ − ϕ 1p = 101$ 29′55.4′′ A2P = ϕ 2P = 40 $ 37′38.6′′ A3P = 360 $ − ϕ 3 P = 338 $ 49′52.0′′
REVHUZDF\MQ\FK
obserwacyjnych K 1 + v 1 = 237 $ 42′06.9′′
A1P − z1 = 101$ 29′55.4′′ − 223 $ 47′48.5′′ = 237 $ 42′06.9′′
K 4 + v 4 = 151$ 02′09.9′′
A2P − z 2 = 40 $ 37′38.6′′ − 249 $ 35′28.7′′ = 151$ 02′09.9′′
K 7 + v 7 = 50 $ 02′37.1′′
A3P − z3 = 338 $ 49′52.0′′ − 288 $ 47′15.0′′ = 50 $ 02′37.0′′
2FHQDGRNáDGQ
D Eá
o FL
G
UHGQLREVHUZDFML
V T ⋅ P ⋅V 0.017 =± ≈ ± 0.1′′ n−k 1
m0 = ±
E Eá
G\
UHGQLHZVSyáU]
GQ\F
h
Macierz kowariancyjna Q = A ⋅ P ⋅ A 7
2EOLF]HQLHEá
4.626 ⋅ 10 −5 = −6 7.835 ⋅ 10
GyZ UHGQLFK
m x = ± m 0 ⋅ Q xx = ± 0.001m m y = ± m 0 ⋅ Q yy = ± 0.001m F Eá
*
GSRáR HQLDSXQNWX3
mP = m x2 + my2 = 0.001m
7.835 ⋅ 10 −6 5.317 ⋅ 10 −5
Wyrównanie F]ZRURERNXJHRGH]\MQHJR NWRZHJR
54 ,PL
1D]ZLVNR
PHWRG SR UHGQLF]
F
Nr zestawu .....
û:,&=(1,(QU
:\UyZQDQLHF]ZRURERNXJHRGH]\MQHJR NWRZHJR PHWRG SR UHGQiF]F
Temat:
Zadanie: Na stanowiskach A, B, C, D SRPLHU]RQR]MHGQDNRZGRNáDGQR FLNW\ – 8. StosuMF PHWRG SR UHGQLF]F QDOH*\ REOLF]\ü Z\UyZQDQH ZDUWR FL SRPLDUyZ RUD] ZVSyáU] GQ\FK x, y punktów B i D. : UDPDFK RFHQ\ GRNáDGQR FL SURV] REOLF]\ü EáG
UHGQL REVHUZDFML Eá G\ UHGQLH ZVSyáU] GQ\FK Z\UyZQDQ\FK RUD] Eá G\ SRáR*HQLD punktów wyznaczanych. Dane:
3RPLHU]RQHN
W\
o
/
1 2 3 4 5 6 7 8
81 46 27 54 52 44 29 25
06 18 21 03 16 31 08 13
. W
:VSyáU]
Pkt A C 52=:,
8
//
13 12 45 30 27 21 52 37
GQHSXQNWyZVWDá\FK
x 1000.00 1000.00
D (xD=?, yD=?)
1 2 A
B (xB=?, yB=?) 2EOLF]RQHZVSyáU]
1) Równania obserwacyjne
β 3 + v 3 = ABD − ABA β 4 + v 4 = ABC − ABD β 5 + v 5 = ACA − ACB β 6 + v 6 = ACD − ACA β 7 + v 7 = ADB − ADC β 8 + v 8 = ADA − ADB
4 3
=$1,(
β 2 + v 2 = AAB − AAC
C
6 5
y 0.00 1000.00
yC − y A y − yA − arctg D xC − x A xD − x A y − yA y − yA = arctg B − arctg C xB − x A xC − x A yD − yB y A − yB = arctg − arctg x D − xB x A − xB y − yB y − yB = arctg D − arctg A x D − xB x A − xB y − yC y − yC = arctg A − arctg B x A − xC x B − xC y − yC y − yC = arctg D − arctg A x D − xC x A − xC y − yD y − yD = arctg B − arctg C xB − x D xC − x D y A − yD yB − yD = arctg − arctg x A − xD x B − xD
β1 + v1 = AAC − AAD = arctg
7
xBo = 421.65 xDo = 1852.23
GQHSU]\EOL*RQH yBo = 552.58 yDo = 133.38
Wyrównanie czworoboku geodezyjnego (NWRZHJR PHWRGSR UHGQLF]F 5R]ZLQL
FLHZV]
55
ereg Taylora
∂A ∂A β1 + v1 = AAC − AADO + − AD ⋅ ∆xD + − AD ⋅ ∆y D x ∂ D ∂y D ∂A ∂A β 2 + v 2 = AAB0 − AAC + AB ⋅ ∆xB + AB ⋅ ∆y B x ∂ B ∂y B ∂A ∂A ∂A ∂A ∂A ∂A β3 + v3 = ABD0 − ABA0 + BD − BA ⋅ ∆xB + BD − BA ⋅ ∆y B + BD ⋅ ∆xD + BD ⋅ ∆y D ∂xB ∂y B ∂xB ∂y B ∂xD ∂y D A A A A A ∂ ∂ ∂ ∂ ∂ ∂A β 4 + v 4 = ABC0 − ABD0 + BC − BD ⋅ ∆xB + BC − BD ⋅ ∆y B + − BD ⋅ ∆xD + − BD ⋅ ∆y D ∂xB ∂y B ∂xB ∂y B ∂xD ∂y D ∂ACB ∂ACB β5 + v5 = ACA − ACB0 + − ⋅ ∆y B ⋅ ∆xB + − ∂y B ∂xB ∂A ∂A β 6 + v 6 = ACD0 − ACA + CD ⋅ ∆xD + CD ⋅ ∆y D x ∂ D ∂y D ∂A ∂A ∂A ∂A ∂A ∂A β 7 + v7 = ADB0 − ADC 0 + DB ⋅ ∆xB + DB ⋅ ∆y B + DB − DC ⋅ ∆xD + DB − DC ⋅ ∆y D x y x x y ∂ ∂ ∂ ∂y D ∂ ∂ D B B D D ∂A ∂A ∂A ∂A ∂A ∂A β8 + v8 = ADA0 − ADB0 + − DB ⋅ ∆xB + − DB ⋅ ∆y B + DA − DB ⋅ ∆xD + DA − DB ⋅ ∆y D ∂xD ∂y D ∂xB ∂y B ∂xD ∂y D
3) :VSyáF]\QQLNLNLHUXQNRZH SRFKRGQHF]VWNRZH a =
y i − y St d
2 St − i
Kierunek
⋅ ρ ′′ ;
x i − x St d St2 − i
Przyrosty x i − xSt y i − y St
(St.-Cel)
852,23 -578,35 578,35 1430,58 578,35 -578,35 852,23 -852,23 -1430,58 -852,23
A-D A-B B-A B-D B-C C-B C-D D-C D-B D-A
b =−
133,38 552,58 -552,58 -419,20 447,42 -447,42 -866,62 866,62 419,20 -133,38
⋅ ρ ′′ ;
'áXJR ü
ρ ′′ = 20626 5 ′′ :VSyáF]NLHUXQNRZH
a
b
36,97 178,14 -178,14 -38,91 172,60 -172,60 -121,00 121,00 38,91 -36,97
-236,24 186,44 -186,44 -132,78 -223,11 223,11 -118,99 118,99 132,78 236,24
dSt − i 862,60 799,90 799,90 1490,73 731,21 731,21 1215,45 1215,45 1490,73 862,60
4) :VSyáF]\QQLNLSU]\QLHZLDGRP\FKZUyZQDQLDFKEá GyZ v i = aS ⋅ ∆x S + bS ⋅ ∆y S + aL ⋅ ∆x L + bL ⋅ ∆y L − aP ⋅ ∆x P − bP ⋅ ∆y P + l i aS = a P − a L , bS = b P − b L 1 2 3 4 5 6 7 8
P S S L P L
aB 0 -178,14 139,23 211,51 -172,60 0 -38,91 38,91
P S S L P L
bB 0 -186,44 53,66 -90,33 223,11 0 -132,78 132,78
L P L P S S
aD 36,97 0 38,91 -38,91 0 121,00 -82,09 -75,88
L P L P S S
bD -236,24 0 132,78 -132,78 0 118,99 13,79 103,46
56
Wyrównanie F]ZRURERNXJHRGH]\MQHJR NWRZHJR
:\UD]\ZROQHUyZQD
Eá
$]\PXW\REOLF]RQH]HZVSyáU]
ϕ ADo = arctg ϕ AC = arctg ϕ ABo = arctg
y DO − y A xD0
GQ\FKSU]\EOL*RQych
1000,00 yC − y A = arctg 0,00 xC − x A xB0 − x A
AADo = ϕ ADo = 8 $53′42,1′′
AAC = 90 $00′00′′
→
552,58 = 43 $ 41′ 40,7′′; − 578,35
= arctg
GyZ
133,38 = arctg = 8 $53′42,1′′; 852,23 − xA
y BO − y A
PHWRG SR UHGQLF]
AABo = 180$ − ϕ ABo = 136$18′19,3′′
ABAo = AABo + 180$ = 316$18′19,3′′
ϕ BDo = arctg ϕ BCo = arctg
y DO − y B0 xD0 − xB0 y C − y B0 xC − xB0
= arctg = arctg
− 419,20 = 16 $19′55,4′′; 1430,58
447,42 = 37$ 43′33,8′′; 578,35
ABDo = 360$ − ϕ BDo = 343$ 40′04,6′′ ABCo = ϕ BCo = 37$ 43′33,8′′
ACBo = ABCo + 180$ = 217$ 43′33,8′′ ACA = AAC + 180$ = 270$00′00′′
ϕ CDo = arctg
y DO − y C xD0 − xC
= arctg
− 866,62 = 45$ 28′46,8′′; 852,23
ACDo = 360$ − ϕ CDo = 314$31′ 13,2′′
ADCo = ACDo − 180$ = 134 $31′ 13,2′′ ADBo = ABDo − 180$ = 163$ 40′04,6′′ ADAo = AADo + 180$ = 188$53′42,1′′
Wyrazy wolne l1 = AAC − AADO − β1 = 4,9 // l2 = AAB0 − AAC − β2 = 7,3// l3 = ABD0 − ABA0 − β3 = 0,3// l4 = ABC0 − ABD0 − β 4 = − 0,8 // l5 = ACA − ACB0 − β5 = − 0,8 // l6 = ACD0 − ACA − β6 = − 7,8 // l7 = ADB0 − ADC0 − β7 = − 0,6 // l8 = ADA0 − ADB0 − β8 = 0,6 // 6) RyZQDQLDEá GyZ v1 = v2 = v3 = v4 = v5 = v6 = v7 = v8 =
0 -178,14 139,23 211,51 -172,60 0 -38,91 38,91
∆xB . ∆xB . ∆xB . ∆xB . ∆xB . ∆xB . ∆xB . ∆xB .
+0 -186,44 +53,66 -90,33 +223,11 +0 -132,78 +132,78
∆yB +36,97 .∆xD . ∆yB +0 .∆xD . ∆yB +38,91 .∆xD . ∆yB -38,91 .∆xD . ∆yB +0 .∆xD . ∆yB +121,00 .∆xD . ∆yB -82,09 .∆xD . ∆yB -75,88 .∆xD .
-236,24 +0 +132,78 -132,78 +0 +118,99 +13,79 +103,46
∆yD ∆yD . ∆yD . ∆yD . ∆yD . ∆yD . ∆yD . ∆yD . .
+4,9 +7,3 +0,3 -0,8 -0,8 -7,8 -0,6 +0,6
F
Wyrównanie czworoboku geodezyjnego (NWRZHJR PHWRGSR UHGQLF]F 7 1DGRNUH ORQ\XNáDGUyZQDOLQLRZ\FK 0 -178,14 139,23 211,51 -172,60 0 -38,91 38,91
.
∆xB ∆xB . ∆xB . ∆xB . ∆xB . ∆xB . ∆xB . ∆xB .
+0 -186,44 +53,66 -90,33 +223,11 +0 -132,78 +132,78
.
∆yB +36,97 .∆xD ∆yB +0 .∆xD . ∆yB +38,91 .∆xD . ∆yB -38,91 .∆xD . ∆yB +0 .∆xD . ∆yB +121,00 .∆xD . ∆yB -82,09 .∆xD . ∆yB -75,88 .∆xD .
8NáDGUyZQD
.
∆yD ∆yD . ∆yD . ∆yD . ∆yD . ∆yD . ∆yD . ∆yD .
+4,9 +7,3 +0,3 -0,8 -0,8 -7,8 -0,6 +0,6
=0 =0 =0 =0 =0 =0 =0 =0
A⋅ X = L
8) Zapis macierzowy 0 − 178 ,14 139 ,23 211,51 − 172 ,60 0 − 38 ,91 38 ,91
-236,24 +0 +132,78 -132,78 +0 +118,99 +13,79 +103,46
0 − 186 ,44 53 ,66 − 90 ,33 223 ,11 0 − 132 ,78 132 ,78
36 ,97 0 38 ,91 − 38 ,91 0 121,00 − 82 ,09 − 75 ,88
QRUPDOQ\FK
− 236 ,24 0 132 ,78 − 132 ,78 ⋅ 0 118 ,99 13 ,79 103 ,46
− 4,9 − 7,3 ∆ x B − 0,3 ∆ y B 0,8 = ∆ x D 0,8 ∆ y 7,8 D 0,6 − 0,6
AT.A.X = AT.L
128675,22 − 6600,49 − 2571,02 − 6109,22 ∆x B 1243,07 − 6600,49 130843,18 6426,85 31026,37 ∆y B 1291,83 ⋅ = − 2571,02 6426,85 31532,10 7012,16 ∆x D 716,09 7012,16 116125,50 ∆y D 1885,84 − 61,09,22 31026,37
2GZURWQR
( AT ⋅ A ) -1
üPDFLHU]\QRUPDOQHM
7,811⋅ 10 −6 2,977 ⋅ 10 −7 = 5,094 ⋅ 10 −7 −7 3,006 ⋅ 10
(AT.A)-1
2,977 ⋅ 10 −7 8,216 ⋅ 10 −6
5,094 ⋅ 10 −7 − 1,181⋅ 10 −6
− 1,181⋅ 10 −6 − 2,108 ⋅ 10 −6
3,235 ⋅ 10 −5 − 1,611⋅ 10 −6
3,006 ⋅ 10 −7 − 2,108 ⋅ 10 −6 − 1,611⋅ 10 −6 9,288 ⋅ 10 −6
11) Wektor niewiadomych X X = (AT ⋅ A)-1 ⋅ ( AT ⋅ L) = 7,811⋅ 10 −6 2,977 ⋅ 10 −7 = 5,094 ⋅ 10 −7 −7 3,006 ⋅ 10
2,977 ⋅ 10 −7
5,094 ⋅ 10 −7
8,216 ⋅ 10 −6 − 1,181⋅ 10 −6 − 2,108 ⋅ 10 −6
− 1,181⋅ 10 −6 3,235 ⋅ 10 −5 − 1,611⋅ 10 −6
3,006 ⋅ 10 −7 1243,07 0,011m − 2,108 ⋅ 10 −6 1291,83 0,006m ≅ ⋅ − 1,611⋅ 10 −6 716,09 0,019m 9,288 ⋅ 10 −6 1885,84 0,014m
57
58
Wyrównanie F]ZRURERNXJHRGH]\MQHJR NWRZHJR
12) Wyrównane niewiadome ZVSyáU] GQHSXQNWyZB i D) x B = x B0 + ∆x B = 421,65 + 0,011 = 421,661 y B = y B0 + ∆y B = 552,58 + 0,006 = 552,586 x D = x D0 + ∆x D = 1852,23 + 0,019 = 1852,249 y D = y D0 + ∆y D = 133,38 + 0,014 = 133,394 13) Wektor poprawek V = A ⋅ X − L − 2,6 − 4,9 2,3 − 3,1 − 7,3 4,2 4,5 − 0,3 4,8 − 0,8 0,8 − 1,6 = − V = − 0,6 0,8 − 1,4 4,0 7,8 − 3,8 − 2,6 0,6 − 3,2 1,2 − 0,6 1,8 14) Obserwacje wyrównane
β1 + v 1 = 81°06 /13 // + 2,3 // = 81°06 /15,3 // β 2 + v 2 = 46°18 /12 // + 4,2 // = 46°18 /16,2 // β 3 + v 3 = 27°21/ 45 // + 4,8 // = 27°21/ 49,8 // β 4 + v 4 = 54°03 / 30 // − 1,6 // = 54°03 / 28,4 // β 5 + v 5 = 52°16 / 27 // − 1,4 // = 52°16 / 25,6 // β 6 + v 6 = 44°31/ 21// − 3,8 // = 44°31/17,2 // β 7 + v 7 = 29°08 / 52 // − 3,2 // = 29°08 / 48,8 // β 8 + v 8 = 25°13 / 37 // + 1,8 // = 25°13 / 38,8 // 15) Kontrola ogólna V T ⋅ V = LT ⋅ L − LT ⋅ A ⋅ X 78,41 = 140,23 − 61,43 25°13 /37// 78,41 ≅ 78,80
PHWRG SR UHGQLF]
F
Wyrównanie czworoboku geodezyjnego (NWRZHJR PHWRGSR UHGQLF]F 16) Kontrola generalna $]\PXW\REOLF]RQH]HZVSyáU]
ϕ AD = arctg ϕ AB = arctg
yD − y A xD − x A yB − y A xB − x A
GQ\FKZ\UyZQDQ\FK
= 8 $53′44,7′′; = 43$ 41′ 43,8′′;
AADo = ϕ ADo = 8 $53′44,7′′ AABo = 180$ − ϕ ABo = 136$18′16,2′′
ABA = AAB + 180$ = 316$18′16,2′′
ϕ BD = arctg ϕ BC = arctg
yD − yB
= 16$19′54,0′′;
xD − xB
yC − y B = 37$ 43′34,4′′; xC − xB
ABDo = 360$ − ϕ BDo = 343$ 40′06,0′′ ABCo = ϕ BCo = 37 $ 43′34,4′′
ACB = ABC + 180$ = 217$ 43′34,4′′
ϕ CD = arctg
y D − yC x D − xC
= 45 $ 28′42,8′′;
ACDo = 360$ − ϕ CDo = 314$31′ 17,2′′
ADC = ACD − 180$ = 134$31′ 17,2′′ ADB = ABD − 180$ = 163$ 40′06,0′′ ADA = AAD + 180$ = 188$53′44,7′′ /HZDVWURQDUyZQD
3UDZDVWURQDUyZQD
REVHUZDF\MQ\FK
obserwacyjnych
β 1 + v 1 = 81°06 /15,3 //
AAC − AAD = 81°06 /15,3 //
β 2 + v 2 = 46°18 /16,2 //
AAB − AAC = 46°18 /16,2 //
β 3 + v 3 = 27°21/ 49,8 //
ABD − ABA = 27°21/ 49,8 //
β 4 + v 4 = 54°03 / 28,4 //
ABC − ABD = 54°03 / 28,4 //
β 5 + v 5 = 52°16 / 25,6 //
ACA − ACB = 52°16 / 25,6 //
β 6 + v 6 = 44°31/17,2 //
ACD − ACA 6 = 44°31/17,2 //
β 7 + v 7 = 29°08 / 48,8 //
ADB − ADC = 29°08 / 48,8 //
β 8 + v 8 = 25°13 / 38,8 //
ADA − ADB = 25°13 / 38,8 //
59
Wyrównanie F]ZRURERNXJHRGH]\MQHJR NWRZHJR
60
PHWRG SR UHGQLF]
17 2FHQDGRNáDGQR FL D Eá
G
UHGQLREVHUZDFML
m=±
E Eá
G\
78,41 V T ⋅V =± ≈ ± 4,4′′ n−k 4
UHGQLHQLHZLDGRP\FK ZVSyáU]
GQ\FKZ\UyZQDQ\FK
Macierz kowariancyjna
Q = ( AT ⋅ A ) -1
2EOLF]HQLHEá
7,811⋅ 10 −6 2,977 ⋅ 10 − 7 = 5,094 ⋅ 10 − 7 −7 3,006 ⋅ 10
2,977 ⋅ 10 −7 8,216 ⋅ 10 − 6 − 1,181 ⋅ 10 − 6
5,094 ⋅ 10 −7 − 1,181⋅ 10 − 6 3,235 ⋅ 10 −5
− 2,108 ⋅ 10 − 6
− 1,611⋅ 10 − 6
GyZ UHGQLFK
m xB = ± m ⋅ Q xx B = ± 0.012 m m y B = ± m ⋅ Q yy B = ± 0.013 m m xD = ± m ⋅ Q xx D = ± 0.025 m m y D = ± m ⋅ Q yy D = ± 0.013 m F Eá
*
GSRáR HQLDSXQNWyZ
BiD
m PB = m x2B + m y2B = 0.018m m PD = m x2D + m y2D = 0.028m
3,006 ⋅ 10 −7 − 2,108 ⋅ 10 − 6 − 1,611 ⋅ 10 − 6 9,288 ⋅ 10 − 6
F
Wyrównanie czworoboku geodezyjnego (NWRZHJR PHWRGzawarunkowan ,PL
1D]ZLVNR
61
Nr zestawu .....
û:,&=(1,(QU :\UyZQDQLHF]ZRURERNXJHRGH]\MQHJR NWRZHJR PHWRG zawarunkowaQ
Temat:
SRPLHU]RQR ] MHGQDNRZ GRNáDGQR FL NW\ ÷8. F PHWRG ]DZDUXQNRZDQ QDOH*\ REOLF]\ü Z\UyZQDQH ZDUWR FL SRPLarów oraz ZDUWR üEá GX UHGQLHJRREVHUZDFML
Zadanie: Na stanowiskach A, B, C, D 6WRVXM
Dane: 3RPLHU]RQHN
W\
o
/
1 2 3 4 5 6 7 8
81 46 27 54 52 44 29 25
06 18 21 03 16 31 08 13
. W
D //
13 12 45 30 27 21 52 37
8
7
C
6 5
1 2 A
4 3 B
52=:,
=$1,(
1) Liczba warunków a) liczba ogólna warunków: w = r = 4, (r – liczba obserwacji nadwymiarowych) b) OLF]EDZDUXQNyZWUyMNWyZ Ztr = 3 c) liczba warunków sinusowych: ws = 1 2) Równania warunkowe :DUXQNLWUyMN
WyZ
Warunek sinusowy:
(β1 + v1) + (β2 + v2) + (β3 + v3)+ (β8 + v8) = 180° (β4 + v4) + (β5 + v5) + (β6 + v6)+ (β7 + v7) = 180° (β2 + v2) + (β3 + v3) + (β4 + v4)+ (β5 + v5) = 180° sin(β1 + v1)⋅ sin(β3 + v3 )⋅ sin(β5 + v5 )⋅ sin(β7 + v7 ) =1 sin(β2 + v2) ⋅ sin(β4 + v 4 ) ⋅ sin(β6 + v6 )⋅ sin(β8 + v8 )
5yZQDQLDRGFK\áHN QLHGRRNUH ORQ\XNáDGUyZQD
OLQLRZ\FK
v1 + v2 + v3 + v8 = ω1 v4 + v5 + v6 + v7 = ω2 v2 + v3 + v4 + v5 = ω3 ctgβ1⋅v1 − ctgβ 2 ⋅ v 2 + ctgβ 3 ⋅ v 3 − ctgβ 4 ⋅ v 4 + ctgβ 5 ⋅ v 5 − ctgβ 6 ⋅ v 6 + ctgβ 7 ⋅ v 7 − ctgβ 8 ⋅ v 8 = ω 4
62
WyrówQDQLHF]ZRURERNXJHRGH]\MQHJR NWRZHJR PHWRG]DZDUXQNRZDQ
2GFK\áNL
ω1 = 180° – (β 1 + β2 + β3 + β8) = 13″ ω2 = 180° – (β 4 + β5 + β6 + β7) = -10″ ω3 = 180° – (β 2 + β3 + β4 + β5) = 6″ ω4 = (1 – F0).ρ″ = -0.015″ ≅ 0″, sin β1 ⋅ sin β3 ⋅ sin β5 ⋅ sin β7 = 1,00000007 gdzie: F0 = sin β2 ⋅ sin β4 ⋅ sin β6 ⋅ sin β8 5) Zapis macierzowy A ⋅V = W v1 v 2 1 1 0 0 0 0 1 v3 1 v 4 0 0 0 1 1 1 1 0 ⋅ v5 = 1 1 1 1 0 0 0 0 0.157 − 0.956 1.932 − 0.725 0.774 − 1.017 1.793 − 2.123 v 6 v7 v8
8NáDGUyZQD
4 0 2 − 0.99
2GZURWQR
QRUPDOQ\FKNRUHODW
0 4 2 0.825
(A ⋅ A )⋅ K = W T
− 0.99 k1 2 2 0.825 k2 ⋅ = 4 1.025 k3 1.025 14.552 k 4
üPDFLHU]\QRUPDOQHM
( A ⋅ AT ) -1
13 − 10 6 0
(A. AT)-1
0.128 − 0.271 0.039 0.395 0.128 0.375 − 0.253 0.0052 = − 0.271 − 0.253 0.523 − 0.0409 0.0052 − 0.0409 0.0739 0.039
8) Wektor korelat K 2.23 − 3.61 K = (A ⋅ AT )-1 ⋅ W = 2.14 0.206
13 − 10 6 0
Wyrównanie czworoboku geodezyjnego (NWRZHJR PHWRGzawarunkowan 9) Wektor poprawek V
T T -1 T V = A ⋅ (A ⋅ A ) ⋅ W = A ⋅ K =
1
0
0
1 1 0 0
0 0 1 1
1 1 1 1
0 0 1
1 1 0
0 0 0
0.157 − 0.956 1.932 2.23 − 0.725 − 3.61 ≅ ⋅ 0.774 2.14 − 1.017 0.206 1.793 − 2.123
2.3 // // 4.2 4.8 // // − 1.6 − 1.3 // // − 3.8 − 3.2 // // 1.8
10) Obserwacje wyrównane
β1 + v 1 = 81°06 /13 // + 2,3 // = 81°06 /15,3 // β 2 + v 2 = 46°18 /12 // + 4,2 // = 46°18 /16,2 // β 3 + v 3 = 27°21/ 45 // + 4,8 // = 27°21/ 49,8 // β 4 + v 4 = 54°03 / 30 // − 1,6 // = 54°03 / 28,4 // β 5 + v 5 = 52°16 / 27 // − 1,3 // = 52°16 / 25,7 // β 6 + v 6 = 44°31/ 21// − 3,8 // = 44°31/17,2 // β 7 + v 7 = 29°08 / 52 // − 3,2 // = 29°08 / 48,8 // β 8 + v 8 = 25°13 / 37 // + 1,8 // = 25°13 / 38,8 // 11) Kontrola ogólna V T ⋅V = W T ⋅ K 78.1 ≅ 77.9 12) Kontrola generalna (81°06′15,3″) + (46°18′16,2″) +(27°21′49,8″) + (25°13′38,8″) = 180°00′00,1″ (54°03′28,4″) + (52°16′25,7″) +(44°31′17,2″) + (29°08′48,8″) = 180°00′00,1″ (46°18′16,2″) +(27°21′49,8″) + (54°03′28,4″) + (52°16′25,7″) = 180°00′00,1″
sin(81°06′15,3′′) ⋅ sin(27°21′ 49,8′′) ⋅ sin(52°16′25,7′′) ⋅ sin(29°08′48,8′′) 0,174936706 = = 1,0000005 sin(46°18′16,2′′) ⋅ sin(54°03′28,4′′) ⋅ sin(44°31′ 17,2′′) ⋅ sin(25°13′38,8′′) 0,174936625
1 %áG UHGQLREVHUZDFML T m = ± V ⋅ V = ± 78,1 ≈ ± 4,4′′ r 4
63
Wyrównanie czworoboku geodezyjnego (liniRZHJR PHWRGSR UHGQLF]F
64 ,PL
1D]ZLVNR
Nr zestawu .....
û:,&=(1,(QU Wyrównanie czworoboku geodezyjnego (liniowego) PHWRGSR UHGQiF]F
Temat:
Zadanie: W czworoboku geodezyjnym SRPLHU]RQR]MHGQDNRZGRNáDGQR FLGáXJR FL – 6WRVXMF PHWRG SR UHGQLF]F QDOH*\ REOLF]\ü Z\UyZQDQH ZDUWR FL SRPLDUyZ oraz wspóáU] GQ\FK x, y punktów B i D. : UDPDFK RFHQ\ GRNáDGQR FL SURV] REOLF]\ü EáG UHGQLREVHrZDFMLEá G\ UHGQLHZVSyáU] GQ\FKZ\UyZQDQ\FKRUD]Eá G\SRáR*HQLD punktów wyznaczanych. D (xD=?, yD=?) 4
Dane:
C
[m]
5
3RPLHU]RQHGáXJR FL
Nr 1 2 3 4 5
'áXJR ü
799.82 1490.64 731.26 1215.40 862.64
2 A
3
1 B (xB=?, yB=?)
:VSyáU]
GQHSXQNWyZVWDá\FK
Pkt A C
52=:,
x 1000.00 1000.00
y 0.00 1000.00
=$1,(
1) Równania obserwacyjne d1 + v1 = dAB = d2 + v2 = dBD = d3 + v3 = dBC = d4 + v 4 = dCD = d5 + v5 = dDA =
(xB − xA)2 + (yB − y A)2 (xD − xB )2 + (yD − yB )2 (xC − xB )2 + (yC − yB )2 (xD − xC )2 + (yD − yC )2 (xA − xD )2 + (y A − yD )2
2EOLF]RQHZVSyáU]
xBo = 421.65 xDo = 1852.23
GQHSU]\EOL*RQH yBo = 552.58 yDo = 133.38
Wyrównanie czworoboku geodezyjnego (linoZHJR PHWRGSR UHGQLF]F 5R]ZLQL
FLHZV]HUHJ7D\ORUD
∂d ∂d d1 + v1 = d AB0 + AB ⋅ ∆xB + AB ⋅ ∆y B ∂y B ∂xB ∂d ∂d ∂d ∂d d2 + v 2 = dBD0 + BD ⋅ ∆xB + BD ⋅ ∆y B + BD ⋅ ∆xD + BD ⋅ ∆y D x y x ∂ ∂ ∂ ∂y D D B B ∂d ∂d d3 + v3 = dBC0 + BC ⋅ ∆xB + BC ⋅ ∆y B ∂xB ∂y B ∂d ∂d d4 + v 4 = dCD0 + CD ⋅ ∆xD + CD ⋅ ∆y D x ∂ D ∂y D ∂d ∂d d5 + v5 = dDA0 + DA ⋅ ∆xD + DA ⋅ ∆y D ∂y D ∂xD 3 :\UD]\ZROQHUyZQDEá GyZ
*
'áXJR FLSU]\EOL RQH
dAB0 = dBD0 = dBC0 = dCD0 = dDA =
(x (x (x (x (x
) ( ) − x ) + (y − y ) = 1490,734m − x ) + (y − y ) = 731,214m − x ) + (y − y ) = 1215,453m − x ) + (y − y ) = 862,604m 2
2
− xA + yB0 − y A = 799,896m
B0
2
D0
B0
2
D0
B0
2
C
B0
2
C
B0
D0
C
2
D0
C
2
2
A
D0
2
A
D0
Wyrazy wolne l1 = d AB0 − d1 = 799,896 − 799,820 = 0,076m l 2 = d BD0 − d 2 = 1490,734 − 1490,640 = 0,094m l 3 = d BC0 − d 3 = 731,214 − 731,260 = −0,046m l 4 = dCD0 − d 4 = 1215,453 − 1215,400 = 0,053m l 5 = d DA0 − d 5 = 862,604 − 862,640 = −0,036m 4) $]\PXW\SU]\EOL*RQH ϕ ABo = arctg ϕ BDo = arctg ϕ BCo = arctg ϕ CDo = arctg
y BO − y A x B0 − x A
y DO − y B0 x D0 − x B0 y C − y B0 xC − x B0 y DO − y C
ϕ DAo = arctg
x D0 − xC y A − y D0 x A − x D0
552,58 = 43 $ 41′ 40,7′′; 578 35 − ,
= arctg = arctg
− 419,20 = 16 $19 ′55,4 ′′; 1430,58
= arctg
447,42 = 37 $ 43 ′33,8 ′′; 578,35
= arctg
− 866,62 = 45 $ 28 ′46,8′′; 852,23
= arctg
− 133,38 = 8 $ 53 ′42,1′′; − 852,23
AABo = 180 $ − ϕ ABo = 136 $18′19,3′′ ABDo = 360 $ − ϕ BDo = 343 $ 40′04,6′′ ABCo = ϕ BCo = 37 $ 43 ′33,8 ′′ ACDo = 360 $ − ϕ CDo = 314 $ 31′ 13,2′′ ADAo = ϕ DAo + 180 $ = 188 $ 53′42,1′′
65
66
Wyrównanie czworoboku geodezyjnego (liniRZHJR PHWRGSR UHGQLF]F
5) RyZQDQLDEá GyZ v i = − cos A jk ⋅ ∆x j − sin A jk ⋅ ∆y j + cos A jk ⋅ ∆x k + sin A jk ⋅ ∆y k + l i v1 =
cos AAB ⋅ ∆x B + sin AAB ⋅ ∆y B
+ l1
v 2 = − cos ABD ⋅ ∆x B − sin ABD ⋅ ∆y B + cos ABD ⋅ ∆x D + sin ABD ⋅ ∆y D + l 2 v 3 = − cos ABC ⋅ ∆x B − sin ABC ⋅ ∆y B
+ l3
v4 =
cos ACD ⋅ ∆x D + sin ACD ⋅ ∆y D + l 4
v5 =
− cos ADA ⋅ ∆x D − sin ADA ⋅ ∆y D + l 5
v1 = -0,723032.∆xB +0,690815 .∆yB
+0 .∆xD
+0 .∆yD +0,076
v2 = -0,959648 .∆xB +0,281204 .∆yB +0,959648 .∆xD -0,281204 .∆yD +0,094 v3 = -0,790945 .∆xB -0,611887 .∆yB
+0 .∆xD
+0 .∆yD
-0,046
v4 =
0 .∆xB
+0 .∆yB +0,701162 .∆xD -0,713002 .∆yD +0,053
v5 =
0 .∆xB
+0 .∆yB +0,987973 .∆xD 0,154625 .∆yD
-0,036
6 1DGRNUH ORQ\XNáDGUyZQDOLQLRZ\FK -0,723032.∆xB +0,690815 .∆yB
+0 .∆xD
-0,959648 .∆xB +0,281204 .∆yB +0,959648 -0,790945 .∆xB -0,611887 .∆yB
.
+0 .∆yD +0,076 = 0
∆xD -0,281204 .∆yD +0,094 = 0
+0 .∆xD
+0 .∆yD -0,046 = 0
0 .∆xB
+0 .∆yB
+0,701162 .∆xD -0,713002 .∆yD +0,053 = 0
0 .∆xB
+0 .∆yB
+0,987973 .∆xD 0,154625 .∆yD -0,036 = 0
7) Zapis macierzowy
A⋅ X = L
0,690815 0 0 − 0,076 − 0,723032 ∆x B − 0,959648 0,281204 0,959648 − 0,281204 − 0,094 ∆y B = 0,046 − 0,790945 − 0,611887 0 0 ⋅ ∆x D 0 0 0 , 701162 0 , 713002 − ∆y D − 0,053 0,036 0 0 0,987973 0,154625
8NáDGUyZQD
QRUPDOQ\FK
AT.A.X = AT.L
0,2699 ∆x B 2,0693 − 0,2853 − 0,9209 − 0,2853 0,9307 0,2699 − 0,07908 ∆y B ⋅ = − 0,9209 0,2699 2,3886 − 0,6170 ∆x D 0,6114 ∆y D 0,2699 − 0,07918 − 0,6170
0,1088 − 0,1071 − 0,0918 0,0699
:\UyZQDQLHF]ZRURERNXJHRGH]\MQHJR OLQLRZHJR PHWRG
2GZURWQR
( AT ⋅ A ) -1
üPDF
ierzy normalnej:
SR UHGQiF]F
(AT.A)-1
0,1198 0,2063 − 0,03973 0,5968 0,1198 1,1351 0,01505 − 0,07817 = 0,2063 − 0,07817 0,6505 0,5553 0,01505 0,5553 2,2157 − 0,03973
10) Wektor niewiadomych X X = (AT ⋅ A) -1 ⋅ ( AT ⋅ L) = 0,1198 0,2063 − 0,03973 0,1088 0,5968 0,1198 − 0,07817 1,1351 0,01505 − 0,1071 ≅ ⋅ = 0,2063 − 0,07817 0,6505 0,5553 − 0,0918 0,01505 0,5553 2,2157 0,0699 − 0,03973 11) Wyrównane niewiadome ZVSyáU] GQHSXQNWyZB i D) x B = x B0 + ∆x B = 421,65 + 0,030 = 421,680 y B = y B0 + ∆y B = 552,58 − 0,100 = 552,480 x D = x D0 + ∆x D = 1852,23 + 0,010 = 1852,240 y D = y D0 + ∆y D = 133,38 + 0,098 = 133,478 12) Wektor poprawek V = A ⋅ X − L − 0,091 − 0,075 V = 0,037 − − 0,063 0,025
− 0,076 − 0,094 0,046 = − 0,053 0,036
− 0,015m 0,019m − 0,009m − 0,010m − 0,011m
13) Obserwacje wyrównane d1 + v 1 = 799,820 − 0,015 = 799,805 d 2 + v 2 = 1490,640 + 0,019 = 1490,659 d 3 + v 3 = 731,260 − 0,009 = 731,251 d 4 + v 4 = 1215,400 − 0,010 = 1215,390 d 5 + v 5 = 862,64 − 0,011 = 862,629 14) Kontrola ogólna V T ⋅ V = LT ⋅ L − LT ⋅ A ⋅ X 0,00089 = 0,02083 − 0,01991 0,00089 ≅ 0,00092
0,030m − 0,100m 0,010m 0,098m
67
Wyrównanie czworoboku geodezyjnego (liniRZHJR PHWRGSR UHGQLF]F
68
15) Kontrola generalna
'áXJR FLREOLF]RQH]HZVSyáU]
GQ\FKwyrównanych
d BC =
(xB − x A )2 + (y B − y A )2 = (− 578,32)2 + (552,480)2 = 799,805m (xD − x B )2 + (y D − y B )2 = 1430,5602 + (− 419,002)2 = 1490,659m (xC − xB )2 + (y C − y B )2 = (− 578,32)2 + (− 447,520)2 = 731,251m
dCD =
(xD
d DA =
(x A − xD )2 + (y A − y D )2
d AB = d BD =
− xC ) + (y D − y C ) = 2
/HZDVWURQDUyZQD
2
=
(852,240)2 + (− 866,522)2
= 1215,390m
(− 852,240)2 + (− 133,478)2
= 862,629m
Prawa stronDUyZQDREVHUZDF\MQ\FK
obserwacyjnych d 1 + v 1 = 799,805m
d AB =
d 2 + v 2 = 1490,659m
799,805 m
d BD = 1490,659 m
d 3 + v 3 = 731,251m
d BC =
d 4 + v 4 = 1215,390m
731,251m
d CD = 1215,390 m
d 5 + v 5 = 862,629m
d DA =
862,629 m
16 2FHQDGRNáDGQR FL a) EáG UHGQLREVHUZDFML
T m = ± V ⋅ V = ± 0.00089 ≈ ± 0,030m n−k 1
b) Eá G\ UHGQLHQLHZLDGRP\FK ZVSyáU] GQ\FKZ\UyZQDQ\FK Macierz kowariancyjna
Q = ( AT ⋅ A ) -1
2EOLF]HQLHEá
− 0,03973 0,1198 0,2063 0,5968 0,1198 − 0,07817 1,1351 0,01505 = 0,2063 − 0,07817 0,6505 0,5553 0,01505 0,5553 2,2157 − 0,03973
GyZ UHGQLFK
m xB = ± m ⋅ Q xx B = 0,030 ⋅ 0,5968 = ± 0.023 m m y B = ± m ⋅ Q yy B = 0,030 ⋅ 1,1351 = ±0.032 m m xD = ± m ⋅ Q xx D = 0,030 ⋅ 0,6505 = ± 0.024 m m y D = ± m ⋅ Q yy D = 0,030 ⋅ 2,2157 = ± 0.045 m F Eá
*
GSRáR HQLDSXQN
mPB = m
2 xB
+m
2 yB
tów B i D
= 0.039m;
mPD = mx2D + my2D = 0.051m
Wyrównanie czworoboNXJHRGH]\MQHJR OLQLRZHJR PHWRG]DZDUXQNRZDQ ,PL
1D]ZLVNR
69
Nr zestawu .....
û:,&=(1,(QU Wyrównanie czworoboku geodezyjnego (liniowego) PHWRG]DZDUXQNRZaQ
Temat:
Zadanie: W czworoboku geodezyjnym SRPLHU]RQR]MHGQDNRZGRNáDGQR FLGáXJo FL – 5. 3RPLDU\ QDOH*\ Z\UyZQDü PHWRG ]DZDUXQNRZDQ. : UDPDFK RFHQ\ GRNáDGQR FL SURV] REOiF]\üEáG UHGQLREVHUZDFML
Dane:
5
4
[m]
3RPLHU]RQHGáXJR FL
Nr 1 2 3 4 5
799.82 1490.64 731.26 1215.40 862.64
2 1
'áXJR üED]\ ]HZVSyáU]
52=:,
6
'áXJR ü
3
GQ\FK G6 = 1000.00m
=$1,(
1) Liczba warunków w = r = 1,
(r – liczba obserwacji nadwymiarowych)
2) Równanie warunkowe
γ 1w + γ 2w − γ 3w = 0 gdzie:
(5) s1
r +r −s = 2 ⋅ r1w ⋅ r2w 2 2 2 r1 + vr1 + r2 + vr2 − s1 + vs1 = arccos 2 ⋅ r1 + vr1 ⋅ r2 + vr2
γ 1w = arccos
2 1w
(
2 2w
2 1w
) ( ) ( ) ( )( )
r22w + r32w − s22w = 2 ⋅ r2w ⋅ r3w 2 2 r + v + r3 + vr3 − s2 + vs2 = arccos 2 r2 2 ⋅ r2 + vr2 ⋅ r3 + vr3
α1 β3
γ 2w = arccos
(
γ 3w
) ( ) ( ) ( )( ) (r + v ) + (r + v ) − s −s = arccos ⋅r 2 ⋅ (r + v )⋅ (r + v )
r12w + r32w = arccos 2 ⋅ r1w
2 3
3w
2
2
1
β1 α2
2
3
r1
1
r1
2 3
r3
3
r3
r1 (1)
s3
(4) s2
β2 α3
r2 (2)
γ3 γ1 γ2
r3 (3)
70
Wyrównanie czworoboku geodezyjnego (liniRZHJR PHWRG]DZDUXQNRZDQ
3) Doprowadzenie do postaci liniowej
(γ 1 + dγ 1) + (γ 2 + dγ 2) − (γ 3 + dγ 3) = 0 gdzie:
dγ 1 =
∂γ 1 ∂γ ∂γ ⋅v + 1 ⋅v + 1 ⋅v ∂r1 r1 ∂r2 r2 ∂s1 s1
dγ 2 =
∂γ 2 ∂γ ∂γ ⋅v + 2 ⋅v + 2 ⋅v ∂r2 r2 ∂r3 r3 ∂s2 s2
dγ 3 =
∂γ ∂γ 3 ⋅v + 3 ⋅v ∂r1 r1 ∂r3 r3
5yZQDQLHRGFK\áNLZSRVWDFLOLQLRZHM
:
A1 ⋅ vr1 + A2 ⋅ vr2 + A3 ⋅ vr3 + B1 ⋅ vs1 + B2 ⋅ vs2 + ω = 0 :DUWR
*
FLSU]\EOL RQHN
WyZ
α1 = arccos
2 2 2 s +r −r = arccos 862.64 + 799.82 − 1490.64 = 141.55682g 2 ⋅ s1 ⋅ r1 2 ⋅ 862.64 ⋅ 799.82
α2 = arccos
2 2 2 s22 + r22 − r32 = arccos 1215.40 + 1490.64 − 731.26 = 32.39125g 2 ⋅ s2 ⋅ r2 2 ⋅ 1215.40 ⋅ 1490.64
α3 = arccos
2 2 2 s32 + r32 − r12 = arccos 1000.00 + 731.26 − 799.82 = 58.07487g 2 ⋅ s3 ⋅ r3 2 ⋅ 1000.00 ⋅ 731.26
β1 = arccos
2 2 2 s12 + r22 − r12 = arccos 862.64 + 1490.64 − 799.82 = 28.03283g 2 ⋅ s1 ⋅ r2 2 ⋅ 862.64 ⋅ 1490.64
β2 = arccos
2 2 2 s22 + r32 − r22 = arccos 1215.40 + 731.26 − 1490.64 = 107.54115g 2 ⋅ s2 ⋅ r3 2 ⋅ 1215.40 ⋅ 731.26
β3 = arccos
2 2 2 s32 + r12 − r32 = arccos 1000.00 + 799.82 − 731.26 = 51.45503g 2 ⋅ s3 ⋅ r1 2 ⋅ 1000.00 ⋅ 799.82
2 1
2 1
2 2
γ 1 = arccos
2 2 2 r12 + r22 − s12 = arccos 799.82 + 1490.64 − 1215.40 = 30.41035g 2 ⋅ r1 ⋅ r2 2 ⋅ 799.82 ⋅ 1490.64
γ 2 = arccos
2 2 2 r22 + r32 − s22 = arccos 1490.64 + 731.26 − 1215.40 = 60.06760g 2 ⋅ r2 ⋅ r3 2 ⋅ 1490.64 ⋅ 731.26
γ 3 = arccos
2 2 2 r12 + r32 − s32 = arccos 799.82 + 731.26 − 1000.00 = 90.47011g 2 ⋅ r1 ⋅ r3 2 ⋅ 799.82 ⋅ 731.26
*
3U]\EOL RQHZDUWR
FLSyOWUyMN WyZ
r , r2 , s1 :
D WUyMN W 1
F1 =
r1 ⋅ r 2 ⋅ sin γ 1 799.82 ⋅ 1490.64 ⋅ sin 30.41035 g = = 274051,6967m 2 2 2
r , r3, s2 :
E WUyMN W 2
F2 =
r2 ⋅ r3 ⋅ sin γ 2 1490.64 ⋅ 731.26 ⋅ sin 60.06760g = = 882545.1259m2 2 2
Wyrównanie czworoboNXJHRGH]\MQHJR OLQLRZHJR PHWRG]DZDUXQNRZDQ
71
:VSyáF]\QQLNLSU]\QLHZLDGRP\FK SRSUDZNDFK
(
)
cc 1 ⋅ ctg141.55682g − ctg51.45503g ⋅ ρ cc = 1368.954 A1 = − 1 ⋅ (ctgα1 − ctgβ3 ) ⋅ ρ cc = − r1 m 799.82m
(
)
cc 1 ⋅ ctg32.39125g + ctg 28.03283g ⋅ ρ cc = −1672.042 A2 = − 1 ⋅ (ctgα2 + ctgβ1) ⋅ ρ cc = − r2 m 1490.64m
(
)
cc 1 ⋅ − ctg58.07487g + ctg107.54115g ⋅ ρ cc = 777.263 A3 = − 1 ⋅ (− ctgα3 + ctgβ 2 ) ⋅ ρ cc = − r3 m 731.26m cc s 862.64m ⋅ ρ cc = 500.976 B1 = 1 ⋅ ρ cc = m 2 ⋅ F1 2 ⋅ 548103.3934m2
B2 =
cc s2 cc 1215.40m ⋅ ρ cc = = 438.362 2 ⋅ρ m 2 ⋅ F2 2 ⋅ 882545.1259m
7) OGFK\áNDω
ω = γ 1 + γ 2 − γ 3 = 30.41035g + 60.06760g − 90.47011g = 78.4cc 2VWDWHF]QDOLQLRZDSRVWDüUyZQDQLDRGFK\áNL
1368.954 ⋅ vr1 − 1672.042 ⋅ v r2 + 777.263 ⋅ vr3 + 500.976 ⋅ vs1 + 438.362 ⋅ vs2 + 78.4cc = 0 9) Zapis macierzowy: A ⋅ V = W v1 v 2 [1368.954 − 1672.042 777.263 500.976 438.362] ⋅ v3 = [− 78.4] v 4 v 5 8NáDGUyZQD
QRUPDOQ\FKNRUHODW
(A ⋅ A )⋅ K = W T
[5717036.045]⋅ [k1] = [− 78.4] 2GZURWQR
üPDFLHU]\QRUPDOQHM
A⋅A
(A. AT)-1
7
[
= 1.749 ⋅ 10−7
]
12) Wektor korelat K
[
K = (A ⋅ AT )-1 ⋅ W = − 1.371⋅ 10−5
]
Wyrównanie czworoboku geodezyjnego (liniRZHJR PHWRG]DZDUXQNRZDQ
72
13) Wektor poprawek V 1368.954 − 0.019m − 1672.042 0.023m T T -1 T −5 V = A ⋅ (A ⋅ A ) ⋅ W = A ⋅ K = 777.263 ⋅ − 1.371⋅ 10 ≅ − 0.011m 500.976 − 0.007m − 0.006m 438.362
[
]
14) Obserwacje wyrównane
(r ) : (r ) : (r ) : (s ) : (s ) : 1w
d1 + v1 = 799.82 − 0.019 = 799.801
2w
d2 + v 2 = 1490.64 + 0.023 = 1490.663
3w
d3 + v3 = 731.26 − 0.011 = 731.249
2w
d4 + v 4 = 1215.40 − 0.007 = 1215.393
1w
d5 + v5 = 862.64 − 0.006 = 862.634
15) Kontrola ogólna V T ⋅V = WT ⋅ K 0.001075 ≅ 0.001075 16) Kontrola generalna
. W\Z\UyZQDQH
γ 1w
2 2 2 r12w + r22w − s12w = arccos 799.801 + 1490.663 − 1215.393 = 30.40653g = arccos 2 ⋅ r1w ⋅ r2w 2 ⋅ 799.801⋅ 1490.663
γ 2w = arccos γ 3w
2 2 2 r22w + r32w − s22w = arccos 1490.663 + 731.249 − 1215.393 = 60.06513g 2 ⋅ r2w ⋅ r3w 2 ⋅ 1490.663 ⋅ 731.249
2 2 2 r12w + r32w − s32 = arccos 799.801 + 731.249 − 1000.00 = 90.47225g = arccos 2 ⋅ r1w ⋅ r3w 2 ⋅ 799.801⋅ 731.249
Równanie warunkowe:
γ 1w + γ 2w − γ 3w = 30.40653g + 60.06513g − 90.47225g = −5.9cc 1 %áG UHGQLREVHUZDFML T m = ± V ⋅ V = ± 0.001075 ≈ ± 0.033m r 1
:\UyZQDQLHFL
,PL
JXSROLJRQRZHJRPHWRGSR UHGQLF]F
1D]ZLVNR
73
Nr zestawu .....
û:,&=(1,(QU :\UyZQDQLHFLJXSROLJRQRZHJRPHWRGSR UHGQLF]F
Temat:
Zadanie : FLJX SROLJRQRZ\P QDZL]DQ\P GZXVWURQQLH NWRZR L OLQLRZR QDOH*\ Z\UyZQDü PHWRG SR UHGQLF]F SRPLHU]RQH ZLHONR FL RUD] ZVSyáU] GQH SXQNWyZ Z\znaczanych. : UDPDFK RFHQ\ GRNáDGQR FL SURV] REOLF]\ü EáG UHGQL MHGQRVWNRZ\ Eá G\ UHGQLHZVSyáU] dQ\FKZ\UyZQDQ\FKRUD]Eá G\SRáR*HQLDSXQNWyZ Dane: β1
β2
β3 d2
d1
A
GQHSXQNWyZVWDá\FK
x 1027.932 1130.285 1490.281 1541.380
y 1027.521 1181.507 1541.544 1696.033
D
2
B Pkt A B C D
d3
C
1
:VSyáU]
β4
3RPLHU]RQHGáXJR FL 3RPLHU]RQHN
Nr 1 2 3
[m]
'áXJR ü
162.400 185.264 185.836
W\
o
/
//
1 2 3 4
195 142 179 218
13 10 50 05
18.8 39.4 10.0 40.0
. W
md = ± 1cm %áG UHGQLSRPLDUXNDWDmβ = ± 10″ %á G UHGQLSRPLDUXGáXJR FL
52=:,
=$1,(
1):VSyáU] GQHSU]\EOL*RQHSXQNWyZ1 i 2 Nr pkt A
.
°
Wβ ′
″
B
195-13-18.8
1
142-10-39.4
2
Azymut A ° ′ ″
d
'áXJR ü
[m]
Przyrosty ∆x ∆y
56-23-18.2
-
71-36-37.0
162.400
51.234 154.107
33-47-16.4
185.264
153.973 103.029
W VSyáU] GQH X Y 1027.932 1027.521 1130.285
Nr pkt A
181.507
B
1181.519 1335.614
1
1335.492 1438.643
2
Wyrównanie FLJXSROLJRQRZHJRPHWRGSR UHGQLF]F
74
2) Równania obserwacyjne y − yB y − yB − arctg A β1 + v1 = AB1 − ABA = arctg 1 x1 − xB xA − xB y 2 − y1 y − y1 − arctg B β2 + v 2 = A12 − A1B = arctg x2 − x1 xB − x1 yC − y 2 y − y2 − arctg 1 β3 + v3 = A2C − A21 = arctg xC − x2 x1 − x2 y D − yC y − yC − arctg 2 β4 + v 4 = ACD − AC2 = arctg xD − xC x2 − xC d1 + v5 = dB1 = d2 + v6 = d12 = d3 + v7 = d2C = 5R]ZLQL
(x1 − xB )2 + (y1 − yB )2 (x2 − x1)2 + (y 2 − y1)2 (xC − x2 )2 + (yC − y2 )2
FLHZV]HUHJ7D\ORUD
∂A ∂A β1 + v1 = AB10 − ABA + B1 ⋅ ∆x1 + B1 ⋅ ∆y1 ∂ x ∂y1 1 ∂A ∂A ∂A ∂A ∂A ∂A β2 + v2 = A120 − A1B0 + 12 − 1B ⋅ ∆x1 + 12 − 1B ⋅ ∆y1 + 12 ⋅ ∆x2 + 12 ⋅ ∆y2 ∂ ∂ ∂ ∂ ∂ x x y y x 1 1 ∂y2 2 1 1 ∂A ∂A ∂A ∂A ∂A ∂A β3 + v3 = A2C0 − A210 + − 21 ⋅ ∆x1 + − 21 ⋅ ∆y1 + 2C − 21 ⋅ ∆x2 + 2C − 21 ⋅ ∆y2 ∂y2 ∂x2 ∂y2 ∂x2 ∂y1 ∂x1 ∂AC2 ∂AC2 β 4 + v 4 = ACD − AC20 + − ⋅ ∆y2 ⋅ ∆x2 + − ∂y2 ∂x2 ∂d ∂d d1 + v5 = dB10 + B1 ⋅ ∆x1 + B1 ⋅ ∆y1 ∂ x ∂y1 1 ∂d ∂d ∂d ∂d d2 + v6 = d120 + 12 ⋅ ∆x1 + 12 ⋅ ∆y1 + 12 ⋅ ∆x2 + 12 ⋅ ∆y2 ∂y 2 ∂x2 ∂y1 ∂x1 ∂ ∂ d d d3 + v7 = d2C0 + 2C ⋅ ∆x2 + 2C ⋅ ∆y2 ∂y 2 ∂x2
3RFKRGQHF]
VWNRZH
y i − y St ⋅ ρ ′′ ; a = d St2 − i Kier. B-A B-1 1-B 1-2 2-1 2-C C-2 C-D
b =−
Przyrosty xi − xSt
-102.353 51.234 -51.234 153.973 -153.973 154.789 -154.789 51.099
x i − x St ⋅ ρ ′′ ; d St2 − i
'áXJR ü
sin A =
y i − y St ; d St − i
:VSyáF]NLHU
y i − y St
d St − i
a
-153.986 154.107 -154.107 103.029 -103.029 102.901 -102.901 154.489
184.899 162.400 162.400 185.264 185.264 185.872 185.872 162.720
-929.04 1205.24 -1205.24 619.16 -619.16 614.36 -614.36 1203.48
b 617.53 -400.69 400.69 -925.31 925.31 -924.14 924.14 -398.06
cos A =
cosA
sinA
0.315479
0.948933
0.831102 0.832773
x i − x St d St − i Kier.
B-A B-1 1-B 0.556120 1-2 2-1 0.553614 2-C C-2 C-D
:\UyZQDQLHFL
JXSROLJRQRZHJRPHWRGSR UHGQLF]F
5) :VSyáF]\QQLNLSU]\QLHZLDGRP\FKZUyZQDQLDFKEá GyZ
W:
v i = aS ⋅ ∆x S + bS ⋅ ∆y S + aL ⋅ ∆x L + bL ⋅ ∆y L − aP ⋅ ∆x P − bP ⋅ ∆y P + l i , aS = aP − aL , bS = bP − bL
D N
E GáXJR ü
v i = − cos A jk ⋅ ∆x j − sin A jk ⋅ ∆y j + cos A jk ⋅ ∆xk + sin A jk ⋅ ∆y k + l i
Nr 1 2 3 4 5 6 7
x1 P S L
y1
-1205.24
P
1824.40
S
-619.16
L
400.69
p
0.315479
k
-0.831102
p
-1326.00 925.31
S
0
L
0.948933
0
-619.16
P
925.31
1233.52
S
-1849.46
-614.36
L
924.14
0
-0.556120
k
0
p
0
y2 0
P
0 k
x2
0
0.831102
k
0.556120
-0.832773
p
-0.553614
6) Wyra]\ZROQHUyZQDEá GyZ $]\PXW\REOLF]RQH]HZVSyáU]
AAB = 56°23′18.2′′ AB10 = 71°36′37.0′′ A120 = 33°47′16.4′′ A2C0 = 33°36′55.2′′ ACD = 71°41′ 52.0′′
*
'áXJR FLSU]\EOL RQH
dB10 = d120 = d2C0 =
(x
) + (y 2
10
− xB
20
− x10 + y 20 − y10
C
− x20
(x (x
10
) ( ) + (y
− yB
2
2
C
− y20
)
2
) )
GQ\FKSU]\EOL*RQych
= 162.400m
2
= 185.264m
2
= 185.872m
Wyrazy wolne l1 = AB10 − ABA − β1 = 71°36 / 37.0 // − 236°23 / 18.2 // − 195°13 / 18.8 // = 0.0 // l2 = A120 − A1B0 − β 2 = 33°47 / 16.4 // − 251°36 / 37.0 // − 142°10 / 39.4 // = 0.0 // l3 = A2C0 − A210 − β3 = 33°36 / 55.2 // − 213°47 / 16.4 // − 179°50 / 10.0 // = −31.2 // l 4 = ACD − AC20 − β 4 = 71°41/ 52.0 // − 213°36 / 55.2 // − 218°05 / 40.0 // = − 43.2 // l5 = dB10 − d1 = 162.400 − 162.400 = 0,000m l6 = d120 − d2 = 185.264 − 185.264 = 0.000m l7 = d2C0 − d3 = 185.872 − 185.836 = 0.036m
75
Wyrównanie FLJXSROLJRQRZHJRPHWRGSR UHGQLF]F
76
7) RyZQDQLDEá GyZ . . v1 = -1205.24 .∆x1 +400.69 .∆y1 +0 ∆x2 +0 ∆y2 +0 . . v2 = 1824.40 .∆x1 -1326.00 .∆y1 -619.16 ∆x2 +925.31 ∆y2 +0 . . . . v3 = -619.16 ∆x1 +925.31 ∆y1 +1233.52 ∆x2 -1849.46 ∆y2 -31.2 . . . . v4 = 0 ∆ x1 +0 ∆y1 -614.36 ∆x2 +924.14 ∆y2 -43.2 . . v5 = 0.315479 .∆x1 +0.948933 .∆y1 +0 ∆x2 +0 ∆y2 +0 v6 = -0.831102 .∆x1 -0.556120 .∆y1 +0.831102 .∆x2 +0.556120 .∆y2 +0 . . . . v7 = 0 ∆ x1 +0 ∆y1 -0.832773 ∆x2 -0.553614 ∆y2 +0.036
8 1DGRNUH ORQ\XNáDGUyZQDOLQLRZ\FK .
-1205.24 ∆x1
.
+0 ∆x2
.
+0 ∆y2
.
+0
=0
.
.
-619.16 ∆x2
.
+0
=0
.
+400.69 ∆y1
.
-1326.00 ∆y1
-619.16 ∆x1
.
+925.31 ∆y1
+1233.52 ∆x2
. 0 ∆x1
. +0 ∆y1
. -614.36 ∆x2
1824.40 ∆x1
.
+925.31 ∆y2
. -1849.46 ∆y2 -31.2 . +924.14 ∆y2 -43.2
=0 =0
.
.
.
+0 ∆x2
.
+0 ∆y2
+0
=0
.
.
.
.
+0
=0
.
.
.
.
0.315479 ∆x1 +0.948933 ∆y1
-0.831102 ∆x1 -0.556120 ∆y1 +0.831102 ∆x2 +0.556120 ∆y2 0 ∆x1
9) Zapis macierzowy
+0 ∆y1 -0.832773 ∆x2 -0.553614 ∆y2 +0.036
=0
A⋅ X = L
400.69 0 0 − 1205.24 1824.40 − 1326.00 − 619.16 925.31 ∆x1 − 619.16 − 1849.46 925.31 1233.52 ∆y1 . . − 0 0 614 36 924 14 ⋅ ∆x2 = ∆y 0.315479 0.948933 0 0 2 . . . . − − 0 831102 0 556120 0 831102 0 556120 − 0.832773 − 0.553614 0 0 10) Macierz wagowa 0 0 0 0 0 0.01 0 0 0.01 0 0 0 0 0 0 0.01 0 0 0 0 0 P= 0 0 0 0.01 0 0 0 0 0 0 0 10000 0 0 0 0 0 0 10000 0 0 0 0 0 0 0 10000 0
0 0 31.2 43.2 0 0 − 0.036
:\UyZQDQLHFL
8NáDGUyZQD
JXSROLJRQRZHJRPHWRGSR UHGQLF]F
QRUPDOQ\FK
AT.P.A.X = AT.P.L
23710.62 ∆x 59546.80 − 27134.55 − 25840.73 1 − 27134.55 39847.88 15002.05 − 32475.65 ∆y1 ⋅ = − 25840.73 15002.05 36665.94 − 24987.73 ∆x2 57464.88 ∆y 2 23710.62 − 32475.65 − 24987.73
2GZURWQR
üPDFLHU]\QRUPDOQHM
− 193.18 288.70 417.26 20.173
(AT. P.A)-1
3.097 ⋅ 10−5 1.905 ⋅ 10−5 1.799 ⋅ 10−5 5.812 ⋅ 10−6 1.905 ⋅ 10−5 5.831⋅ 10−5 9.479 ⋅ 10−6 2.921⋅ 10−5 A ⋅ P ⋅ A = −5 9.479 ⋅ 10−6 4.927 ⋅ 10−5 1.936 ⋅ 10−5 1.799 ⋅ 10 5.812 ⋅ 10−6 2.921⋅ 10−5 1.936 ⋅ 10−5 3.993 ⋅ 10−5 7
13) Wektor niewiadomych X
3.097 ⋅ 10−5 1.905 ⋅ 10−5 = 1.799 ⋅ 10−5 −6 5.812 ⋅ 10
X = (AT ⋅ P ⋅ A)-1 ⋅ ( AT ⋅ P ⋅ L ) = 1.905 ⋅ 10−5 1.799 ⋅ 10−5 5.812 ⋅ 10−6 − 193.18 5.831⋅ 10−5 9.479 ⋅ 10−6 2.921⋅ 10−5 288.70 ≅ ⋅ 9.479 ⋅ 10−6 4.927 ⋅ 10−5 1.936 ⋅ 10−5 417.26 2.921⋅ 10−5 1.936 ⋅ 10−5 3.993 ⋅ 10−5 20.173
14) Wyrównane niewiadome ZVSyáU] GQHSXQNWyZ1 i 2) x1 = x10 + ∆x1 = 1181 .519 + 0.007 = 1181 .526 y1 = y10 + ∆y1 = 1335 .614 + 0.018 = 1335 .632 x2 = x20 + ∆x2 = 1335 .492 + 0.020 = 1335 .512 y 2 = y 20 + ∆y 2 = 1438 .643 + 0.016 = 1438 .659 15) Wektor poprawek V = A ⋅ X − L − 1.2 − 8.7 7.4 V = 2.5 − 0.019 0.010 − 0.026
0 0 31.2 43.2 = 0 0 − 0.036
− 1.2′′ − 8.7′′ − 23.8′′ − 40.7′′ 0.019m 0.010m 0.010m
0.007m 0.018m 0.020m 0.016m
77
Wyrównanie FLJXSROLJRQRZHJRPHWRGSR UHGQLF]F
78
16) Obserwacje wyrównane
β1 + v1 = 195°13 / 18.8 // − 1.2// = 195°13 / 17.6 // β2 + v 2 = 142°10 / 39.4// − 8.7 // = 142°10 / 30.7// β3 + v3 = 179°50 / 10.0 // − 23.8// = 179°49/ 46.2// β4 + v 4 = 218°05/ 40.0 // − 40.7// = 218°04/ 59.3 // d1 + v5 = 162.400 + 0.019 = 162.419m d2 + v6 = 185.264 + 0.010 = 185.274m d3 + v7 = 185.836 + 0.010 = 185.846m
17) Kontrola ogólna V T ⋅ P ⋅ V = LT ⋅ P ⋅ L − LT ⋅ P ⋅ A ⋅ X 28.61 = 41.18 − 12.51 28.61 ≅ 28.67
18) Kontrola generalna $]\PXW\REOLF]RQH]HZVSyáU]
AAB = 56°23′18.2′′ AB1 = 71°36′35.8′′
GQ\FKZ\UyZQDQ\FK
A12 = 33°47′06.5′′ A2C = 33°36′52.7′′ ACD = 71°41′ 52.0′′ /HZDVWURQDUyZQD
3UDZDVWURQDUyZQD
REVHUZDF\MQ\FK
obserwacyjnych
β 1 + v 1 = 195 ° 13 / 17 . 6 //
AB1 − ABA = 71°36 / 35.8 // − 236 °23 / 18.2 // = 195 °13 / 17.6 //
β 2 + v 2 = 42 ° 10 / 30 .7 //
A12 − A1B = 33 °47 / 06 .5 // − 251 °36 / 35.8 // = 42°10 / 30.7 //
β 3 + v 3 = 179 ° 49 / 46 .2 //
A2C − A21 = 33°36 / 52.7 // − 213 °47 / 06.5 // = 179 °49 / 46 .2 //
β 4 + v 4 = 218 ° 04 / 59 .3 //
ACD − AC2 = 71°41/ 52.0 // − 213 °36 / 52 .7 // = 218 °04 / 59.3 //
d 1 + v 5 = 162 . 419 m
dB1 =
(x1 − xB )2 + (y1 − y B )2
= 162 .419m
d 2 + v 6 = 185 . 274 m
d12 =
(x2 − x1 )2 + (y 2 − y1 )2
= 185 .274m
d 3 + v 7 = 185 . 846 m
d2C =
(xC
− x2 ) + (yC − y 2 ) = 185 .846m 2
2
:\UyZQDQLHFL
JXSROLJRQRZHJRPHWRGSR UHGQLF]F
19 2FHQDGRNáDdQR FL a)
Eá
G
UHGQLREVHUZDFML MHGQRVWNRZ\
T m0 = ± V ⋅ P ⋅ V = ± 28.61 ≈ ± 3.088 n−k 3
b)
Eá
G\
UHGQLHQLHZLDGRP\FK ZVSyáU]
GQ
ych wyrównanych)
Macierz kowariancyjna 3.097 ⋅ 10−5 1.905 ⋅ 10−5 1.905 ⋅ 10−5 5.831 ⋅ 10−5 Q = A ⋅ P ⋅ A = −5 9.479 ⋅ 10− 6 1.799 ⋅ 10 5.812 ⋅ 10− 6 2.921 ⋅ 10−5 7
2EOLF]HQLHEá
GyZ UHGQLFK
mx1 = ± m ⋅ Qxx1 = ± 0.017m my1 = ± m ⋅ Qyy 1 = ± 0.024m mx 2 = ± m ⋅ Qxx 2 = ± 0.022m my 2 = ± m ⋅ Qyy 2 = ± 0.020m
F Eá
*
GSRáR HQLDSXQNWyZ
mP1 = mx21 + my21 = 0.029m mP2 = mx22 + my22 = 0.029m
1i2
1.799 ⋅ 10−5 5.812 ⋅ 10−6 9.479 ⋅ 10− 6 2.921 ⋅ 10−5 4.927 ⋅ 10−5 1.936 ⋅ 10−5 1.936 ⋅ 10−5 3.993 ⋅ 10−5
79
Wyrównanie FLJXSROLJRQRZHJRPHWRGZDUXQNRZ
80 ,PL
1D]ZLVNR
Nr zestawu .....
û:,&=(1,(QU :\UyZQDQLHFLJXSROLJRQRZHJRPHWRGZarunkoZ
Temat:
Zadanie : FLJX SROLJRQRZ\P QDZL]aQ\P GZXVWURQQLH NWRZR L OLQLRZR QDOH*\ Z\-
ZDUXQNRZ SRPLHU]RQH NW\ L GáXJR FL
REOLF]\üEáG UHGQLMHGQRVtkowy.
UyZQDü PHWRG SURV]
: UDPDFK RFHQ\ GRNáDGQR FL
Dane: β2 β1
d2
d1
A
β3
GQHSXQNWyZVWDá\FK
Pkt A B C D
x 1027.932 1130.285 1490.281 1541.380
y 1027.521 1181.507 1541.544 1696.033
d3
C
1
2
B :VSyáU]
β4
PomieU]RQHGáXJR FL 3RPLHU]RQHNW\ o / // .W Nr 'áXJR ü [m] 1 195 13 18.8 1 162.400 2 142 10 39.4 2 185.264 3 179 50 10.0 3 185.836 4 218 05 40.0
md = ± 1cm %áG UHGQLSRPLDUXNDWDmβ = ± 10″ %á G UHGQLSRPLDUXGáXJR FL
52=:,
=$1,(
1) Liczba warunków w = r = 3,
(r – liczba obserwacji nadwymiarowych)
2) Równania warunkowe :DUXQHNVXP\N
WyZ []pr = []teor
(β1 + v1) + (β2 + v2) + (β3 + v3) + (β4 + v4) = ACD − AAB + 4 ⋅ 180
$
Warunki sumy przyrostów: [∆x]pr = [∆x]teor;
[∆y]pr = [∆y]teor
(d1 + v5 ) ⋅ cos AB1 + (d2 + v6 ) ⋅ cos A12 + (d3 + v7 ) ⋅ cos A2C = xC − xB (d1 + v5 ) ⋅ sin AB1 + (d2 + v6 ) ⋅ sin A12 + (d3 + v7 ) ⋅ sin A2C = yC − yB
gdzie:
D
AB1 = AAB + (β1 + v1) − 180$ A12 = AAB + (β1 + v1) + (β2 + v2) − 2 ⋅ 180$
A2C = AAB + (β1 + v1) + (β2 + v2) + (β3 + v3) − 3 ⋅ 180$
:\UyZQDQLHFL
JXSROLJRQRZHJRPHWRGwarunkow
5yZQDQLDRGFK\áHN SRGRSURZDG]
81
eniu do postaci liniowej)
v1 + v2 + v3 + v 4 = ω1 −
(∆yBC )0 ⋅ v
(∆y1C )0 ⋅ v − (∆y 2C )0 ⋅ v + cos(A ) ⋅ v + cos(A ) ⋅ v + cos(A ) ⋅ v = ω B1 0 1− 2 3 5 12 0 6 2C 0 7 2 ρ ′′ ρ ′′ ρ ′′ (∆xBC )0 ⋅ v + (∆x1C )0 ⋅ v + (∆x2C )0 ⋅ v + sin(A ) ⋅ v + sin(A ) ⋅ v + sin(A ) ⋅ v = ω B1 0 1 2 3 5 12 0 6 2C 0 7 2 ρ ′′ ρ ′′ ρ ′′
4) Obliczenie azymutów A]\PXW\ERNyZQDZL]DQLD y − yA = arctg 153.986 = 56$23′18.2′′; ϕ AB = arctg B xB − xA 102.353 y − yC ϕCD = arctg D = arctg 154.489 = 71$41′ 52.0′′; xD − xC 51.099
*
$]\PXW\SU]\EOL RQHSR]RVWDá\FKERNyZFL
AAB = ϕ AB = 56$23′18.2′′ ACD = ϕCD = 71$41′ 52.0′′
JX
(AB1)0 = AAB + β1 − 180 = 56 23′18.2′′ + 195 13′18.8′′ − 180 = 71 36′37.0′′ (A12 )0 = AAB + β1 + β 2 − 360 = 56 23′18.2′′ + 195 13′18.8′′ + 142 10′39.4′′ − 360 = 33 47′16.4′′ (A2C )0 = AAB + β1 + β 2 + β 3 − 540 = 56 23′18.2′′ + 195 13′18.8′′ + 142 10′39.4′′ + 179 50′10.0′′ − 540 $
$
$
$
$
$
$
$
$
$
$
$
$
$
$
$
$
=
= 33 37′26.4′′ $
*
3U]\EOL RQHSU]\URVW\ZVSyáU]
GQ\FK
(∆x2C )0 = d3 ⋅ cos(A2C )0 = 154.744 (∆x1C )0 = d2 ⋅ cos(A12)0 + d3 ⋅ cos(A2C )0 = 308.717 (∆xBC )0 = d1 ⋅ cos(AB1)0 + d2 ⋅ cos(A12)0 + d3 ⋅ cos(A2C )0 = 359.951 (∆y 2C )0 = d3 ⋅ sin(A2C )0 = 102.905 (∆y1C )0 = d2 ⋅ sin(A12)0 + d3 ⋅ sin(A2C )0 = 205.934 (∆yBC )0 = d1 ⋅ sin(AB1)0 + d2 ⋅ sin(A12)0 + d3 ⋅ sin(A2C )0 = 360.041 2GFK\áNL
&i
ω1 = ACD − AAB + 4 ⋅ 180$ − β1 − β2 − β3 − β 4 = −74.4′′ ω2 = xC − xB − (∆xBC )0 = 0.045m
ω3 = yC − y B − (∆y BC )0 = −0.004m
2VWDWHF]QDSRVWDüXNáDGXUyZQD
R
dFK\áHN
v1 + v 2 + v3 + v 4 = −74.4 − 0.001746 ⋅ v1 − 0.000998 ⋅ v 2 − 0.000499 ⋅ v3 + 0.315479 ⋅ v5 + 0.831102 ⋅ v 6 + 0.832689 ⋅ v7 = 0.045 0.001745 ⋅ v1 + 0.001497 ⋅ v 2 + 0.000750 ⋅ v 3 + 0.948933 ⋅ v5 + 0.556120 ⋅ v 6 + 0.553740 ⋅ v 7 = −0.004
Wyrównanie FLJXSROLJRQRZHJRPHWRGZDUXQNRZ
82
8) Zapis macierzowy: A ⋅ V = W
1 1 1 1 − 0.001746 − 0.000998 − 0.000499 0 0.001497 0.000750 0 0.001745
0DFLHU]ZDJRZD RGZURWQR
ü
QRUPDOQ\FKNRUHODW
(A ⋅ P
−1
)
⋅ AT ⋅ K = W
0.399201 k1 400.000000 − 0.324282 − 0.324282 0.000578 − 0.000369 ⋅ k2 = 0.000736 k3 0.399201 − 0.000369 2GZURWQR
− 74.4 0.045 − 0.004
0 0 0 0 0 100 0 0 100 0 0 0 0 0 0 0 100 0 0 0 0 P −1 = 0 0 0 100 0 0 0 0 0 0 0 0.0001 0 0 0 0 0 0 0.0001 0 0 0 0 0 0 0 0.0001 0
waga: pi = 12 ; pi−1 = mi2 mi
8NáDGUyZQD
v1 v 2 0 0 0 v3 0.315479 0.831102 0.832689 ⋅ v 4 = 0.948933 0.556120 0.553740 v5 v 6 v7
− 74.4 0.045 − 0.004
üPDFLHU]\QRUPDOQHM
−1
A⋅P ⋅A
7
2.191270 − 2.656761 0.006928 = 2.191270 3240.661252 436.821569 − 2.656761 436.821569 3016.919749
12) Wektor korelat K K = (A ⋅ P −1 ⋅ AT )-1 ⋅ W = 2.191270 − 2.656761 − 74.4 0.006928 = 2.191270 3240.661252 436.821569 ⋅ 0.045 = − 2.656761 436.821569 3016.919749 − 0.004
− 0.406727 − 17.815850 206.696055
:\UyZQDQLHFL
JXSROLJRQRZHJRPHWRGwarunkow
83
13) Wektor poprawek V − 1.5′′ − 8.0′′ − 24.3′′ V = P −1 ⋅ AT ⋅ (A ⋅ P −1 ⋅ AT )-1 ⋅ W = P −1 ⋅ AT ⋅ K = − 40.7′′ 0.019m 0.010m 0.010m 14) Obserwacje wyrównane
β1 + v1 = 195°13 / 18.8 // − 1.5 // = 195°13/ 17.3 // β2 + v 2 = 142°10/ 39.4 // − 8.0 // = 142°10/ 31.4 // β3 + v3 = 179°50/ 10.0 // − 24.3 // = 179°49 / 45.7// β4 + v 4 = 218°05 / 40.0// − 40.7 // = 218°04/ 59.3 // d1 + v5 = 162.400 + 0.019 = 162.419m d2 + v 6 = 185.264 + 0.010 = 185.274m d3 + v7 = 185.836 + 0.010 = 185.846m 15) Kontrola ogólna
V T ⋅ P ⋅V = W T ⋅ K 28.72 = 28.72
16) Kontrola generalna Azymuty wyrównane:
(AB1) = AAB + (β1 + v1) − 180° = 71°36′35.5′′ (A12 ) = AAB + (β1 + v1) + (β 2 + v 2 ) − 360 ° = 33°47′06.9′′ (A2C ) = AAB + (β1 + v1) + (β 2 + v 2 ) + (β3 + v3 ) − 540 ° = 33°36′52.6′′ Równania warunkowe: L = 195°13 / 17.3 // + 142°10 / 31.4 // + 179°49 / 45.7 // + 218°04 / 59.3 // = 735°18 / 33.7 // P = ACD − AAB + 4 ⋅ 180$ = 71°41/ 52.0 // − 56°23 / 18.2 // + 720° = 735°18 / 33.8 //
L = 162.419 ⋅ cos(71°36′35.5′′) + 185.274 ⋅ cos(33°47′06.9′′) + 185.846 ⋅ cos(33°36′52.6′′) = 359.996m P = xC − xB = 1490.281 − 1130.285 = 359.996m L = 162.419 ⋅ sin(71°36′35.5′′) + 185.274 ⋅ sin(33°47′06.9′′) + 185.846 ⋅ sin(33°36′52.6′′) = 360.037m P = yC − y B = 1541.544 − 1185.507 = 360.037m
1 %áG UHGQLREVHrwacji T m0 = ± V P ⋅ V = ± 28.72 ≈ ± 3.094 n−k 3
84 ,PL
3UDZRSU]HQRV]HQLDVL
Eá GyZ UHGQLFKZLHONR FLVNRUHORZDQ\FK
1D]ZLVNR
Nr zestawu .....
û:,&=(1,(QU Temat:
3UDZRSU]HQRV]HQLDVL Eá GyZ UHGQLFKZLHONR FLVNRUHORZanych. %áG UHGQLIXQNFML
Zadanie :\NRU]\VWXMFZ\QLNL ] ]DGDQLDÄWyrównaQLHFLJX SROLJRQRZHJR PHWRGSo UHGQLF]F´QDOH*\ REOLF]\üEá G\ UHGQLHNWyZ β 1 i β3RUD]GáXJR FL d1 i d3 po wyrównaniu. Dane: β1
β2 d2
d1
A
β3
1 B
β4 d3
C
D
2
md = ± 1cm %áG UHGQLSRPLDUXNDWDmβ = ± 10″ %á G UHGQLSRPLDUXGáXJR FL
Równania obserwacyjne dla wybranych NWyZβ1 i β3RUD]GáXJR FLd1 i d3 y − yB y − yB − arctg A β1 + v1 = AB1 − ABA = arctg 1 x1 − xB xA − xB
β3 + v3 = A2C − A21 = arctg d1 + v5 = dB1 = d3 + v7 = d2C =
yC − y 2 y − y2 − arctg 1 xC − x2 x1 − x2
(x1 − xB )2 + (y1 − yB )2 (xC − x2 )2 + (yC − y2 )2
RyZQDQLDEá GyZ NWyZβ1 i β3RUD]GáXJR FLd1 i d3 v1 = -1205.24 .∆x1 +400.69 .∆y1 +0 .∆x2 +0 . . . v3 = -619.16 ∆x1 +925.31 ∆y1 +1233.52 ∆x2 -1849.46 . v5 = 0.315479 ∆x1 +0.948933 .∆y1 +0 .∆x2 +0 . . . v7 = 0 ∆x1 +0 ∆y1 -0.832773 ∆x2 -0.553614
%á G UHGQLREVHUZDFML MHGQRVWNRZ\
T m0 = ± V ⋅ P ⋅ V = ± 28.61 ≈ ± 3.088 n−k 3
∆y2 +0 ∆y2 -31.2 . ∆y2 +0 . ∆y2 +0.036 . .
3UDZRSU]HQRV]HQLDVL
Eá GyZ UHGQLFKZLHONR FLVNRUHORZDQ\FK
Macierz kowariancyjna (teoretyczna) niewiadomych x1, y1, x2, y2 3.097 ⋅ 10−5 1.905 ⋅ 10−5 1.799 ⋅ 10−5 5.812 ⋅ 10−6 1.905 ⋅ 10−5 5.831 ⋅ 10−5 9.479 ⋅ 10− 6 2.921 ⋅ 10−5 Q = A ⋅ P ⋅ A = −5 9.479 ⋅ 10− 6 4.927 ⋅ 10−5 1.936 ⋅ 10−5 1.799 ⋅ 10 5.812 ⋅ 10− 6 2.921 ⋅ 10−5 1.936 ⋅ 10−5 3.993 ⋅ 10−5 7
52=:,
=$1,(
Macierz kowariancyjna empiryczna Qx 0.0002953 0.0001817 0.0001716 5.542 ⋅ 10−5 0.0001817 0.0005560 9.0.38 ⋅ 10−5 0.0004698 Qx = m02 ⋅ Q = −5 0.0004698 0.0001846 0.0001716 9.038 ⋅ 10 5.542 ⋅ 10−5 0.0002786 0.0001846 0.0003808
%á G\
UHGQLHIXQNFML
T
: mF2 = f ⋅ Qx ⋅ f
Funkcja 1 F1 = v1 = -1205.24 .∆x1 +400.69 .∆y1 +0 .∆x2 +0 .∆y2 T ∂F f1 = 1 ∂∆x1
∂F1 ∂∆y1
∂F1 ∂∆x2
∂F1 = [− 1205.24 400.69 0 ∂∆y 2
0]
T
mF21 = f 1 ⋅ Qx ⋅ f 1 = 0.0002953 0.0001817 0.0001716 5.542 ⋅ 10−5 − 1205.24 −5 9.0.38 ⋅ 10 0.0004698 400.69 = 342.76 ⋅ = [− 1205.24 400.69 0 0 ]⋅ 0.0001817 0.0005560 −5 0 0.0001716 9.038 ⋅ 10 0.0004698 0.0001846 5.542 ⋅ 10−5 0.0002786 0.0001846 0.0003808 0 mF = 18.5′′ 1
85
86
3UDZRSU]HQRV]HQLDVL
Eá GyZ UHGQLFKZLHONR FLVNRUHORZDQ\FK
Funkcja 3 F3 = v3 = -619.16 .∆x1 +925.31 .∆y1 +1233.52 .∆x2 -1849.46 .∆y2 -31.2 T ∂F f3 = 3 ∂∆x1
∂F3 ∂∆y1
∂F3 ∂∆x2
∂F3 = [− 619.16 925.31 1233.52 − 1849.46] ∂∆y 2 T
mF23 = f 3 ⋅ Qx ⋅ f 3 = 0.0002953 0.0001817 0.0001716 5.542 ⋅ 10−5 − 1619.16 −5 9.0.38 ⋅ 10 0.0004698 925.31 = ⋅ = [− 619.16 925.31 1233.52 − 1849.46]⋅ 0.0001817 0.0005560 −5 0.0001716 9.038 ⋅ 10 0.0004698 0.0001846 1233.52 5.542 ⋅ 10−5 0.0002786 0.0001846 0.0003808 − 1849.46 = 673.92 ′ ′ mF3 = 26.0
Funkcja 5 F5 = v5 = 0.315479 .∆x1 +0.948933 .∆y1 +0 .∆x2 +0 .∆y2 T ∂F f5 = 5 ∂∆x1
∂F5 ∂∆y1
∂F5 ∂∆x2
∂F5 = [0.315479 0.948933 0 ∂∆y 2
0]
T
mF25 = f 5 ⋅ Qx ⋅ f 5 = 0.0002953 0.0001817 0.0001716 5.542 ⋅ 10−5 0.315479 −5 9.0.38 ⋅ 10 0.0004698 0.948933 = 0.000639 ⋅ = [0.315479 0.948933 0 0 ]⋅ 0.0001817 0.0005560 −5 0.0001716 9.038 ⋅ 10 0.0004698 0.0001846 0 5.542 ⋅ 10−5 0.0002786 0.0001846 0.0003808 0 mF5 = 0.025m
Funkcja 7 F7 = v7 = 0 .∆x1 +0 .∆y1 -0.832773 .∆x2 -0.553614 .∆y2 +0.036 T ∂F f7 = 7 ∂∆x1
∂F7 ∂∆y1
∂F7 ∂∆x2
∂F7 = [0 ∂∆y 2
0
− 0.832773 − 0.553614] T
mF27 = f 7 ⋅ Qx ⋅ f 7 = 0.0002953 0.0001817 0.0001716 5.542 ⋅ 10−5 0 −5 0 9.0.38 ⋅ 10 0.0004698 = 0.000613 = [0 0 − 0.832773 − 0.553614] ⋅ 0.0001817 0.0005560 ⋅ −5 0.0001716 9.038 ⋅ 10 0.0004698 0.0001846 - 0.832773 5.542 ⋅ 10−5 0.0002786 0.0001846 0.0003808 - 0.553614 mF7 = 0.025m
:\UyZQDQLHWUDQVIRUPDFMLZVSyáU]
,PL
GQ\FK
1D]ZLVNR
87
Nr zestawu .....
û:,&=(1,(QU :\UyZQDQLHWUDQVIRUPDFMLZVSyáU] GQ\FK dla n>2 punktów dostosowania
Temat:
Zadanie 'DQH V ZVSyáU] GQH SXQNWyZ GR WUDQVIRUPDFML1÷5 Z XNáDG]LH SLHUZRWQ\P xy RUD] ZVSyáU] GQH WU]HFK SXQNWyZ GRVWRVRZDQLD 10, 20, 30 Z XNáDG]LH SLHUZRWQ\P xy i wtórnym XY1DOH*\Z\UyZQDü ZHGáXJPHWRG\QDMPQLHMV]\FKNZDGUDWyZ ZVSyáU] GQH punktów dostosoZDQLD L REOLF]\ü ZVSyáU] GQH SXQNWyZ WUDQVIRUPRZDQ\FK Z XNáDG]LH wtórnym. W ramach oceQ\ GRNáDGQR FL SURV] REOLF]\ü Eá G\ UHGQLH ZVSyáU] GQ\FK SXQNWyZGRVWRVRZDQLDRUD]Eá
G UHGQLWUDQVIRUPDFML
Dane:
[X] 8NáDGSLHUZRWQ\
Nr pkt
x
10
y
[x]
8NáDGZWyUQ\
X
XP
Y
P
400.000
600.000 1800.000 2700.000
20 1500.000
600.000 1993.444 1609.818
30
400.000 2000.000 3185.625 2945.539
1
400.000
200.000
2
700.000
500.000
3
600.000
900.000
4
200.000 1000.000
5
100.000
52=:,
xP Y0
α
YP [Y]
400.000
yP
X0
=$1,(
:VSyáU] GQHELHJXQDWUDQVIRUPDFML
x0, y0ZXNáDG]LHSLHUZRtnym
Liczba punktów dostosowania: s = 3 x0 =
1 s
s
∑ xi = 766.667; 1
y0 =
1 s
s
∑y
i
= 1066.667;
i = 1,..., s
1
2) Wzory transformacyjne X − X 0 = (x − x 0 ) ⋅ C + (y − y 0 ) ⋅ S
Y − Y0 = (y − y 0 ) ⋅ C − (x − x 0 ) ⋅ S C = k ⋅ cos α ; S = k ⋅ sin α
[y]
Wyrównanie transforPDFMLZVSyáU] GQ\FK
88
3) Równania „obserwacyjne” dla punktów dostosowania X10 + VX10 = (x10 – x0).C + (y10 – y0).S + X0 Y10 + VY10 = (y10 – y0).C – (x10 – x0).S + Y0 X20 + VX 20 = (x20 – x0).C + (y20 – y0).S + X0 Y20 + VY20 = (y20 – y0).C – (x20 – x0).S + Y0 X30 + VX 30 = (x30 – x0).C + (y30 – y0).S + X0 Y30 + VY30 = (y30 – y0).C – (x30 – x0).S + Y0 4) 8NáDGUyZQDEá GyZ z nieznanymi parametrami C, S, X0, Y0 VX10 = -366.667.C – 466.667.S + X0 – 1800.000 VY10 = -466.667.C + 366.667.S + Y0 – 2700.000 VX 20 = 733.333.C – 466.667.S + X0 – 1993.444 VY20 = -466.667.C – 733.333.S + Y0 – 1609.818 VX 30 = -366.667.C + 933.333.S + X0 – 3185.625 VY30 = 933.333.C + 366.667.S + Y0 – 2945.539 5HGXNFMDXNáDGXRSDUDPHWU\
s
Równanie sumowe „X-ów”:
∑V
Xi
X0, Y0 – ZVSyáU] GQHELHJXQDZXNáDG]Le wtórnym
= 0.000 ⋅ C + 0.000 ⋅ S + 3 ⋅ X 0 − 6979.069 = 0 :
1
X0 = 1 s s
Równanie sumowe „Y-ów”:
∑V
Yi
s
∑X 1
i
= 6979.069 = 2326.356 3
= 0.000 ⋅ C + 0.000 ⋅ S + 3 ⋅ Y0 − 7255.357 = 0 :
1
Y0 = 1 s
s
∑Y = 72553.357 = 2418.452 i
1
8NáDG]UHGXNRZDQ\
VX10 = -366.667.C – 466.667.S + 526.356 VY10 = -466.667.C + 366.667.S – 281.548
VX 20 = 733.333.C – 466.667.S + 332.912 VY20 = -466.667.C – 733.333.S + 808.634 VX 30 = -366.667.C + 933.333.S – 859.268 VY30 = 933.333.C + 366.667.S – 527.086
:\UyZQDQLHWUDQVIRUPDFMLZVSyáU]
VX10 V Y10 VX 20 = VY20 V X 30 VY30
89
V = A⋅ X − L
6) Zapis macierzowy
8NáDGUyZQD
GQ\FK
− 366.667 − 466.667 − 466.667 366.667 733.333 − 466.667 ⋅ C − − 466.667 − 733.333 S − 366.667 933.333 366.667 933.333
− 526.356 281.548 − 332.912 − 808.634 859.268 527.086
AT.A.X = AT.L
QRUPDOQ\FK
0 C 371717.638 2113333.333 = ⋅ 0 2113333.333 S 2092473.368 2GZURWQR
üPDFLHU]\QRUPDOQHM
(AT.A)-1
4.7319 ⋅ 10−7 A ⋅ A = 0
7
0 −7 4.7319 ⋅ 10
9) Wektor niewiadomych X X = (AT ⋅ A)-1 ⋅ ( AT ⋅ L) = 4.7319 ⋅ 10−7 = 0
371717.638 0 ≅ ⋅ −7 4.7319 ⋅ 10 2092473.368
0.175892 0.990129 =
C S
10) Parametry transformacji k, α k = C 2 + S 2 = 0.1758922 + 0.9901292 = 1.005631 cos α = C = 0.175892 = 0.174907; k 1.005631
sin α = S = 0.990129 = 0.98458 k 1.005631
α = 88.80752g 11) Wektor poprawek V = A ⋅ X − L − 526.554 280.965 − 333.073 V = − − 808.178 859.627 527.213
− 526.356 281.548 − 332.912 = − 808.634 859.268 527.086
− 0.198 − 0.583 − 0.161 0.457 0.359 0.127
:
90
Wyrównanie transforPDFMLZVSyáU] GQ\FK
:\UyZQDQHZVSyáU]
GQHSXQNWyZGRVWRVRZDQLDZXNáDG]LHZWyUQ\P
X 10 = X10 + VX10 = 1800.000 − 0.198 = 1799.802 Y 10 = Y10 + VY10 = 2700.000 − 0.583 = 2699.417 X 20 = X 20 + VX 20 = 1993.444 − 0.161 = 1993.283 Y 20 = Y20 + VY20 = 1609.818 + 0.457 = 1610.275 X 30 = X 30 + VX 30 = 3185.625 + 0.359 = 3185.983 Y 30 = Y30 + VY30 = 2945.539 + 0.127 = 2945.666 13) Kontrola ogólna V T ⋅ V = LT ⋅ L − LT ⋅ A ⋅ X 0.759 = 2137202.082 − 2137202.324 0.759 = 0.759 14) Kontrola generalna /HZDVWURQDUyZQD
3UDZDVWURQDUyZQD
REVHUZDF\MQ\FK
obserwacyjnych X 10 = (x10 − x 0 ) ⋅ C + (y 10 − y 0 ) ⋅ S + X 0 = 1799 .802
X10 + VX10 = 1799.802
Y 10 = (y 10 − y 0 ) ⋅ C − (x10 − x 0 ) ⋅ S + Y0 = 2699 .417
Y10 + VY10 = 2699.417
X 20 = (x 20 − x 0 ) ⋅ C + (y 20 − y 0 ) ⋅ S + X 0 = 1993 .283
X 20 + VX 20 = 1993.283
Y 20 = (y 20 − y 0 ) ⋅ C − (x 20 − x 0 ) ⋅ S + Y0 = 1610 .275
Y20 + VY20 = 1610.275
X 30 = (x 30 − x 0 ) ⋅ C + (y 30 − y 0 ) ⋅ S + X 0 = 3185 .982
X30 + VX30 = 3185 .983
Y 30 = (y 30 − y 0 ) ⋅ C − (x 30 − x 0 ) ⋅ S + Y0 = 2945 .665
Y30 + VY30 = 2945.666
15) Ocena dokáDGQR FL transformacji D Eá
G\
UHGQLHZVSyáU]
GQ\FKSXQNWyZGRVWRVRZDQLD
MX = ± MY = ±
E UHGQLEá
∑
9
;L
s
∑
9
s
