INSTYTUT GEODEZJI OLSZTYN 18.03.1999r.

SPRAWOZDANIE NR 2

TEMAT: ZADANIE WPROST - METODA CLARKE'A.

WYKONAŁ:

BARTŁOMIEJ

OLSZEWSKI

GR. 1 III ROK

NR 10

1. ε = ( s² * sin A₁ * cos A₁ ) / 2 * Q₁² = 0°00'00,9700''

e²= 0,00669

Q₁ = √(M(B₁)*N(B₁)) = 6 381 881,271 m

M(B₁) = (a*(1 - e²)) / (1 - e² * sin²B₁)^3/2 = 6 372 996,555 m

N(B₁) = a/√( 1 - e² * sin²B₁) = 6 390 778,373 m

2. P₁ - 1/3 * ε = A₁ - 1/3 * ε ⇒ P₁ = 230°44'50,9600''

P₂ - 1/3 * ε = 90°- A₁ - 2/3 * ε ⇒ P₂ = 219°15'08,3900''

P₂' - 1/3 * ε = 90° - 1/3 * ε ⇒ P₂' = 90°

3. u = s * cos (A₁ - 2/3 * ε) = - 17 702,214 m

v = s * sin (A₁ - 2/3 * ε) = - 21 664,352 m

4. ΔB = u/M₁(B₁) = - 0°09'32,9400''

B_S = B₁ +1/2 * ΔB = 49°51'22,8800''

M_S = M(B_S) = 6 372 908,683 m

B₂' = B₁+ u/M_S = 49°46'36,4100''

5. B₂ = B₂' - v²/2Q₂'² * tg B₂'= 49°46'35,0000''

M(B₂') = (a*(1 - e²)) / (1 - e² * sin²B₁)^3/2 = 6 372 820,772 m

N(B₂') = a/√( 1 - e² * sin²B₂') = 6 390 719,615 m

Q₂' = √(M(B₂')*N(B₂')) = 6 381 763,918 m

6. ΔL = (v/N(B₂')) /cos ( B₂+1/3 * ε₁) = - 0°18'02,7900''

ε₁ = B₂' - B₂ = 0°00'01,4100''

γ = ΔL*sin(B₂+2/3 * ε₁) = - 0°13'46,7500''

A₂ = A₁ ± 180°+γ - ε = 50°31'02,8000''

L₂ = 37°42'59,9600'' A₂ = 50°31'02,8000'' B₂= 49°46'35,0000''

Dane :

A₁=230°44'50,9600''

B₁=49°56'09,3536''

L₁=38°01'02,7546''

a=6 378 245,0 m

b=6 356 863,0 m

s=27 977,0 m

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