Penentuan Posisi
1. Gambar
A = 40° 20’ S - 180° 00’
B = 56° 10’ S - 067° 10’ B
∆Lt = 15° 50’ ∆Bu = 112° 50’ = 950’ = 6770’
- cos AC = sin Lt1 x cosec Lt2
= sin 40° 20’ x cosec 56° 10’
= 0.779175
AC = 2328.9 mil
-cos λ1 = tg Lt1 x cotg Lt2
= 40° 20’ x 1 tg 56° 10’
= 0.56911
λ1 = 55° 18.7’
- λ2 = ∆Bu - λ1
= 112° 50’ - 55° 18.7’
= 57° 31.3 = 3451.3 mil
- Jauh (CB) = λ2 x cos Lt2
= 3451.3 x 0.55678
= 1921.6 mil
a. Jauh menurut composite track :
= AC + CB
= 2328.9 + 1921.6
= 4250.5 mil
b. Letak Vertex
- cos a = cos (Lt1+Lt2) - cosLt1 x cosLt2 x sinv ∆Bu
=cos(15° 50’)–cos 40° 20’x56° 10’x(1-cos112° 50’)
= 0.37293
a = 68° 06.2’
- cos LtV = cos Lt1 x cos Lt2 x sin ∆Bu x cosec a
= cos 40° 20’xcos 56° 10’xsin 112° 50’x 1 sin68° 06.2’
= 0.42158
LtV = 65° 3.9’ S
- Bujur Vertex = Bujur tolak + λ1
= 180° + 55° 18.7’
= 124° 41.3 B
Jadi letak Vertex = 65° 3.9’ S - 124° 41.3 B
2. Sebuah kapal berlayar mengikuti lingkaran besar dari H ke Y.Letak H = 22°25’ U - 159°25’B dan letak Y = 34°45’U - 146°46,5’T
Hitunglah : a. Titik potong lingkaran besar tsb dgn
Katulistiwa
b. Titik potong lingkaran besar dgn
derajah 180°
c. Titik potong lingkaran besar dgn jajar
30° U
d. Jarak lingkaran besar dgn suatu titik L
yg posisinya 28° 15’ U - 177° 25’ B
Jawab
a.Titik potong lbs dgn katulistiwa
cos ∆bu = tg l x cotg LV
= tg 0° x 1
tg 34° 56.3’ = 0 x 1.43142
= 0
Bujur Vertex + 90°
= 146° 46.5’ + 90°
= 236° 46.5’ - 180° = 56° 46.5’ B
Jadi titik potong = 00°00’ - 56° 46.5 B
b.Titik potong lbs dgn derajah 180°
tg lx = cos ∆bu’ x tg lv
∆bu’= 180°- bujur vertex
= 180°- 146°46.5’
= 33°13.5’
tg lx = 33°13.5’ x tg 34° 56.3’
= 0.83653 x 0.69860
= 0.58440
lx = 30° 18.1 U
jadi titik potong 30° 18.1 U - 180°
c. Titik potong lbs dgn jajar 30° U
cos ∆bu’ = tg ly x cotg lv
= tg 30° x 1
tg 34° 56.3’ = 0.57735 x 1.43142
= 0.82643
∆bu’ = 34° 16’
= 181° 02.5’
= 360° - 181° 02.5’
= 178° 57.5’ B
jadi titik potong 30° U - 178° 57.5’ B
d. Jarak lbs dgn titik L
∆bu < S = bu L + bu V
= 177° 25’ + 146°46.5’
= 324° 11.5’
maka 360° - 324° 11.5’ = 35° 48.5’ B
cos < S = sin lv x sin ∆bu < S
= sin 34° 56.3’ x sin 35° 48.5’
= 0.33507
< S = 70° 25.4’
tg LS = tg lv x cos ∆bu < S
= tg 34° 56.3’ x cos 35° 48.5’
= 0.56655
ls = 29° 32’ U
LS = ls – l L
= 29° 32’ - 28° 15’
= 01° 17’
sin LN = sin LS x sin< S
= sin 01° 17’ x sin 70° 25.4’
= 0.02111
LN = 01° 12.6’ = 72.6 mil
3. Manado
SF = 37° 48’ U - 122° 40’ B
∆Lt = 35° 48’∆bu=112° 20’ = 2148’ = 6740’
A.cos a=cos(Lt1-Lt2)-cosLt1 x cosLt2 x sinv ∆Bu
cos(35°48’)–cos02°00’xcos37°48’x(1-cos112°20’)
= - 0.27868
a = 106° 10.9’ = 6370.886 mil
cos LtV = cos Lt1 x cos Lt2 x sin ∆Bu x cosec a
= cos02°00’xcos 37° 48’xsin112°20’x 1 cos6°10.9’ = 0.760514
LtV = 40° 29.4’
- cos λ1 = cotg Ltv x tg Lt1
= cotg 40° 29.4’ x tg 02° 00’
= 0.0409
λ1 = 87° 39.3’
Bujur Vertex = Bujur Tolak + λ1
= 125° 00’T + 87° 39.3’
= 147° 20.6 B
B. Letak titik vertex diutara :
= 40° 29.4’ U - 147° 20.6 B
Letak titik vertex diselatan :
= 40° 29.4’ S - 032° 39.4 T
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