Laboratoriya ishi Triangulyatsiya tarmog'ini korrelata usulida tenglashtirishga misol


Download 52.9 Kb.
bet1/2
Sana18.06.2023
Hajmi52.9 Kb.
#1557178
  1   2
Bog'liq
12-13-laboratoriya


12-13-Laboratoriya ishi
Triangulyatsiya tarmog'ini korrelata usulida tenglashtirishga misol
Uncha katta bo`lmagan 2-sinf tarmog'ining chizmasi (12.1-rasm). Tekislikdagi Gauss-Kruger proektsiyasidagi yo'nalishlar, punktlar markazlariga keltirilgan, dastlabki ma'lumotlar 12.1-jadvalda berilgan.


12.1-rasm. Triangulyatsiya tarmog'i chizmasi
Boshlang`ich ma`lumotlar 12.1-jadval



Boshlang`ich
punktlar

Koordinatalar, m

Direktsion
burchak

Tomon
uzunligi, m

X

Y

1

5 963 124,8

8 412 617,83

7Β° 28'37,81"

14 311,32

2

5 977 314,44

8 414 480,18

Burchaklar bo`yicha tenglashtirilganda r shartli tenglamalarning umumiy
soni:

π‘Ÿ = 𝑛 βˆ’ 2π‘ž = 9 βˆ’ 4 = 5.


Yo'nalishlar bo'yicha tenglashtirilganda π‘Ÿ = 4, gorizont sharti mavjud bo`lmaydi, bu yerda uchta shakl va bitta qutb shartlari mavjud.

Punkt raqami

Yo`nalish
raqami

Tekislikdagi
yo`nalishlar

Tenglashdan
tuzatmalar

Tenglangan
yo`nalishlar

1


1
2
3

0Β° 00' 00,00"
37 11 06,71
68 08 59,63

+0,13"
+0,06
-0,19

0Β° 00' 00,00"
37 11 06,64
68 08 59,31

2


4
5
6

0Β° 00' 00,00"
25 12 57,42
58 19 54,61

-0,18
+0,41
-0,23

0Β° 00' 00,00"
25 12 58,05
58 19 54,56

3


7
8
9

0Β° 00' 00,00"
28 01 30,27
53 31 06,58

+0,06
-0,09
+0,02

0Β° 00' 00,00"
28 01 30,12
53 31 06,54

4


10
11
12

0Β° 00' 00,00"
109 41 57,13
238 59 22,59

+0,04
-0,28
+0,23

0Β° 00' 00,00"
109 41 56,81
238 59 22,78


Shakl shartli tenglamalarini tuzish 12.3-jadvalda keltirilgan.
12.3-jadval




Uchburchak
raqami

Punktlar
raqami

Yo`nalishlar
farqi

Tekislikdagi
burchaklar

Shakl shartli
tenglamalar




1

2-1

37Β° 11'06,71"

-(1)+(2)-(5)+(6)-

1

2

6-5

33 06 57,19

-(10)+(11)+




4

11-10

109 41 57,13

+ 1,03"=0







βˆ‘

180 00 01,03




πœ”1

+ 1,03




2

5-4

25 12 57,42

-(4)+(5)-(8)+(9)-

2

3

9-8

25 29 36,31

- (11)+(12) -




4

12-11

129 17 25,06

- 1,21"= 0







βˆ‘

179 59 58,79




πœ”2

-1,21




1

3-2

30 57 52,92

-(2)+(3)-(7)+(8)+

3

4

10-12

121 00 37,41

+(10)-(12)+




3

8-7

28 01 30,27

+0,60"=0







βˆ‘

180 00 00,60




πœ”3

+0,60




βˆ‘ πœ”2
= 2,88β€²β€²
; π‘š =
βˆšβˆ‘ πœ”2
3𝑛
= √2,88
9
= 0,57β€²β€²

π‘š
π‘šπ‘š = =
√2(𝑛 βˆ’ 1)
0,57β€²β€²



√4
= 0,28β€²β€²

Markaziy sistemaning qutb shartlarini tuzish:

𝑆41 βˆ™ 𝑆42 βˆ™ 𝑆43
sin(6 βˆ’ 5) sin(9 βˆ’ 8) sin(3 βˆ’ 2)

𝑆42

  • 𝑆43

  • 𝑆41

= sin(2 βˆ’ 1) sin(5 βˆ’ 4) sin(8 βˆ’ 7) = 1.

Tenglamaning chiziqli shakli
𝑐𝑑𝑔(2 βˆ’ 1)(1) βˆ’ [𝑐𝑑𝑔(2 βˆ’ 1) + 𝑐𝑑𝑔(3 βˆ’ 2)](2) + 𝑐𝑑𝑔(3 βˆ’ 2)(3) +
+𝑐𝑑𝑔(5 βˆ’ 4)(4) βˆ’ [𝑐𝑑𝑔(6 βˆ’ 5) + 𝑐𝑑𝑔(5 βˆ’ 4)](5) + 𝑐𝑑𝑔(6 βˆ’ 5)(6) +
+𝑐𝑑𝑔(8 βˆ’ 7)(7) βˆ’ [𝑐𝑑𝑔(9 βˆ’ 8) + 𝑐𝑑𝑔(8 βˆ’ 7)](8) + 𝑐𝑑𝑔(9 βˆ’ 8)(9) + πœ” = 0

Ushbu tenglamaning 𝑐𝑑𝑔𝛽 koeffitsientlari va ozod hadlari 12.4-jadvalda hisoblanadi.


12.4-jadval



Sur`at

Maxraj

Yo`n
alish

𝛽𝑖

𝑠𝑖𝑛𝛽𝑖

𝑐𝑑𝑔𝛽𝑖

Yo`n
alish

𝛽𝑖

𝑠𝑖𝑛𝛽𝑖

𝑐𝑑𝑔𝛽𝑖

6-5

33Β°06'57,19"

0,5463342

1,533

2-1

37Β°11'06,71"

0,6043933

1,318

9-8

25 29 36,3

0,4304074

2,097

5-4

25 12 57,42

0,4260312

2,124

3-2

30 57 52,92

0,5145098

1,667

8-7

28 01 30,27

0,4698579

1,879




П1

= 0,1209851







П2

= 0,1209839




П1 =0,5463342*0,4304074*0,5145098= 0,1209851


П2 =0,6043933*0,4260312*0,4698579=0,1209839


πœ” = П1 βˆ’ П2 βˆ™ πœŒβ€²β€² = 0,0000012 βˆ™ 206265β€²β€² = 2,05β€²β€²; βˆ‘ 𝑐𝑑𝑔2𝛽 = 19,306 ;
П1 0,1209851




πœ”π‘β„Žπ‘’π‘˜π‘™π‘– = 2.5 βˆ™ π‘š βˆ™ βˆšβˆ‘ 𝑐𝑑𝑔2𝛽 = 2,5 βˆ™ 1β€²β€² βˆ™ √19,306 = 10,98β€²β€².

Hisoblangan koeffitsientlarni inobatga olgan holda qutb sharti quyidagi shaklga ega bo`ladi.
1,318(1) βˆ’ 2,985(2) + 1,667(3) + 2,124(4) βˆ’ 3,657(5) + 1,533(6)
+ 1,879(7) βˆ’ 3,976(8) + 2,097(9) + 2,05 = 0


Vaznli funktsiyalarini tuzish

𝑆34 βˆ’ tomonning direktsion burchagini aniqlashning aniqligini baholash uchun (tarmoqning zaif nuqtasida) bizda 𝛼34 = 𝑓𝛼 = βˆ’(1) + (3) βˆ’ (7) + (8) mavjud.


Eng uzoq tomon (𝑆34) ning o'rtacha kvadratik xatosini aniqlash uchun uning
teskari vaznini aniqlash kerak, buning uchun 𝑆34 ni boshlang`ich tomon (𝑆12) dan
124 va 234 uchburchaklar orqali eng qisqa yo'l bo'ylab tenglashtirilgan yo'nalishlarning funktsiyasi sifatida ifodalanishi kerak (12.1-rasmga qarang).
sin(6 βˆ’ 5) sin(3 βˆ’ 2)
𝐹 = 𝑆34 = 𝑆12 sin(11 βˆ’ 10) sin(8 βˆ’ 7). (12.1)
Ushbu funktsiyaning teskari vaznini aniqlash uchun uning orttirmasi topiladi
βˆ†πΉ = 𝑓𝑠 . Bazis shartni olishda qilingan xulosaga o'xshash xulosa chiqarish natijasida biz ega bo`lamiz.



𝑓 = βˆ†π‘†


= βˆ’ 𝑆34 𝑐𝑑𝑔(3 βˆ’ 2)(2) + 𝑆34 𝑐𝑑𝑔(3 βˆ’ 2)(3) βˆ’ 𝑆34 𝑐𝑑𝑔(6 βˆ’ 5)(5)


𝑠 34
πœŒβ€²β€²
πœŒβ€²β€²
πœŒβ€²β€²

+ 𝑆34 𝑐𝑑𝑔(6 βˆ’ 5)(6) + 𝑆34 (8 βˆ’ 7)(7) βˆ’ 𝑆34 (8 βˆ’ 7)(8)

πœŒβ€²β€²
πœŒβ€²β€²
πœŒβ€²β€²

+ 𝑆34 𝑐𝑑𝑔(11 βˆ’ 10)(10) βˆ’ 𝑆34 𝑐𝑑𝑔(11 βˆ’ 10)(11)
πœŒβ€²β€² πœŒβ€²β€²

Qulaylik uchun 𝑆34 tomoni odatda detsimetrda ifodalanadi. (12.1) formulaga yo'nalishlar farqi va (𝑆12) ni qo`yib, topamiz



𝑠𝑖𝑛33Β°06β€²57′′𝑠𝑖𝑛30Β°57β€²53β€²β€²
𝑆34 = 14311,32 𝑠𝑖𝑛109Β°41β€²57′′𝑠𝑖𝑛28Β°01β€²30β€²β€² = 90940π‘‘π‘š,
𝑆34 = 0,4409.
πœŒβ€²β€²

12.3-jadvaldagi yo'nalish farqlarini hisobga olgan holda koeffitsientlarning qiymatlarini hisoblash orqali ega bo`lamiz.



𝑓𝑠 = βˆ†π‘†34 = βˆ’0,735(2) + 0,735(3) βˆ’ 0,676(5) + 0,676(6) + 0,828(7)
βˆ’ 0,828(8) βˆ’ 0,158(10) + 0,158(11).

12.5-jadvalda shartli tenglamalar va 𝑓𝛼 π‘£π‘Ž 𝑓𝑠 vaznli funktsiyalarining koeffitsientlari keltirilgan.


12.5-jadval



Tuzat
malar

a

b

c

d

π‘“π‘Ž

𝑓𝑠

𝑠′

𝖯

(1)

βˆ’1







+1,318

βˆ’1




βˆ’0,682

+0,13

(2)

+1




βˆ’1

βˆ’2,985




βˆ’0,735

βˆ’3,720

+0,06

(3)







+1

+1,667

+1

+0,735

+4,402

βˆ’0,19

(4)




βˆ’1




+2,124







+1,124

βˆ’0,18

(5)

βˆ’1

+1




βˆ’3,657




βˆ’0,676

βˆ’4,333

+0,41

(6)

+1







+1,533




+0,676

+3,209

βˆ’0,23

(7)







βˆ’1

+1,879

βˆ’1

+0,828

+0,707

+0,06

(8)




βˆ’1

+1

βˆ’3,976

+1

βˆ’0,828

βˆ’3,804

βˆ’0,09

(9)




+1




+2,097







+3,097

+0,02

(10)

βˆ’1













βˆ’0,158

βˆ’0,158

+0,04

(11)

+1

βˆ’1










+0,158

+0,158

βˆ’0,28

(12)




+1

βˆ’1










0,000

+0,23

πœ”

+1,03

βˆ’1,21

+0,60

+2,05







[𝖯2] = 0,455

Nazorat

βˆ‘ π’‚π’Š = βˆ‘ π’ƒπ’Š = βˆ‘ π’„π’Š = βˆ‘ π’…π’Š = βˆ‘ π’‡π’Š = βˆ‘ 𝒔′ = βˆ‘ 𝝑 = 𝟎
π’Š

βˆ’[π‘˜πœ”] = 0,458




0

0

0

0,000

0

0,000

0,000

βˆ’0,02

Normal tenglamalarning koeffitsientlari 12.6-jadvaldan olinadi.


12.6-jadval






a]
π‘˜1

b]
π‘˜2

c]
π‘˜3

d]
π‘˜4

π‘“π‘Ž

𝑓𝑠

πœ”

𝑠

Nazorat
𝑠° = [π‘Žπ‘ β€²] + πœ”

[a

6,000

-2,000

-2,000

0,887

+1,000

+0,933

+1,03

+5,850

+5,850

[b




6,000

-2,000

0,292

-1,000

-0,006

-1,21

+0,076

+0,076

[c







6,000

-1,203

+3,000

-0,344

+0,60

+4,053

+4,053

[d










57,398

-5,506

+11,776

+2,05

+65,694

+65,694

[π‘“π‘Ž













4,000

-0,921




+0,573

+0,573

[𝑓𝑠
















+3,415




+14,853

+14,853

Normal tenglamalarni tuzish


[π‘Žπ‘Ž]π‘˜1 + [π‘Žπ‘]π‘˜1 + [π‘Žπ‘]π‘˜3 + β‹― + πœ”1 = 0,
{[π‘Žπ‘]π‘˜1 + [𝑏𝑏]π‘˜2 + [𝑏𝑐]π‘˜3 + β‹― + πœ”2 = 0,
… … … … … … … … … … … … … … … … … …
Korrelatlarni hisoblash nazorati:
([π‘Žπ‘Ž] + [π‘Žπ‘] + β‹― )π‘˜1 + ([π‘Žπ‘] + [𝑏𝑏] + β‹― )π‘˜2 + ([π‘Žπ‘] + [𝑏𝑐] + β‹― )π‘˜3 +
+ β‹― (πœ”1 + πœ”2 + β‹― ) = 0

Normal tenglamalarni yechish 12.7-jadvalda keltirilgan


Korrelatlarni hisoblash nazorati:
(6,000 βˆ’ 2,000 βˆ’ 2,000 + 0,887)(βˆ’0,1774) + (βˆ’2,000 + 6,000 βˆ’ 2,000 +
0,292)(0,0999) + (βˆ’2,000 βˆ’ 2,000 + 6,000 βˆ’ 1,203)(βˆ’0,1331) + (0,887 + 0,292 βˆ’
1,203 + 57,398)(βˆ’0,0363) + (1,03 βˆ’ 1,21 + 0,60 + 2,05) = 0,002



Download 52.9 Kb.

Do'stlaringiz bilan baham:
  1   2



Ma'lumotlar bazasi mualliflik huquqi bilan himoyalangan ©fayllar.org 2024
ma'muriyatiga murojaat qiling