INSTYTUT GEODEZJI OLSZTYN 18.03.1999r.

SPRAWOZDANIE NR 3

TEMAT: ZADANIE ODWROTNE - METODA GAUSSA.

WYKONAŁ:

BARTŁOMIEJ

OLSZEWSKI

GR. 1 III ROK

NR 10

1. B_S = 49°51'22,1800''

L_S = 37°52'01,3600''

∆B = B₂ - B₁ = - 0°09'34,3500''

∆L = L2 - L1 = - 0°18'02,7900''

M(B_S) = (a*(1 - e²)) / (1 - e² * sin²B_S)^3/2 = 6 372 908,468 m

N(B_S) = a/√( 1 - e² * sin²B_S) = 6 390 748,929 m

e² = 0,00669

e'² = 0,00674

2. η² = e'² * cos²B_S = 0,00280

t = tg²B = 1,40587

V = √(1 + η²) = 1,00140

a_B = 1 + [(∆L²* cos²B_S)/24]*[2 + 3t² + 2 η²] + [(∆B* η²)/8V⁴]*[t² - 1 - 4η²*t²]

a_B = 1,000011668

3. a_L = 1 + [(∆L²* sin²B_S)/24] - [(∆B²)/24V⁴]*[1 - 9η²*t²]

a_L = 1,000001926

4. ∆B = (s*cosA*a_B)/(M) ⇒ cosA = (∆B*M)/ (s*a_B)

∆L = (s*sinA*a_L)/(N*cosB) ⇒ sinA = (∆L*N*cosB)/ (s*a_L)

sin²A + cos²A = 1 ⇒ s = 27976,951 m

5. A = arccos[(∆B*M)/ (s*a_B)] = 50°37'58,5700''

6. a_A = 1 + [(∆L²* cos²B_S*V²)/12] + [(∆B)/24V⁴]*[3 + 8η² +8η⁴]

a_A = 0,9996522428

7. ∆A = ∆L*sinB_S* a_A = - 0°13'47,4300''

8. A₁ = A - (∆A /2) = 230°44'52,2800''

A₂ = A + (∆A /2) = 50°31'04,8500''

A₁ = 230°44'52,2800'' A₂ = 50°31'04,8500'' s = 27976,951 m

Dane :

B₁=49°56'09,3536''

L₁=38°01'02,7546''

B₂=49°46'35,0000''

L₁=37°42'59,9600''

a=6 378 245,0 m

b=6 356 863,0 m

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