{"id":263,"date":"2015-08-15T17:22:13","date_gmt":"2015-08-15T14:22:13","guid":{"rendered":"http:\/\/denizcisin.com\/?p=263"},"modified":"2015-08-15T17:22:40","modified_gmt":"2015-08-15T14:22:40","slug":"263","status":"publish","type":"post","link":"https:\/\/denizcisin.com\/index.php\/2015\/08\/15\/263\/","title":{"rendered":"Goss Seyir Problem \u00c7\u00f6z\u00fcmleri"},"content":{"rendered":"

G\u00fcne\u015fin do\u011fu\u015f zaman\u0131n\u0131n hesaplanmas\u0131:<\/strong><\/span><\/h4>\n

4 May\u0131s 1999 g\u00fcn\u00fc 38 derece N \u2013 026 derece E mevkiinde bulunan bir g\u00f6zlemci i\u00e7in G\u00fcne\u015f\u2019 in do\u011fu\u015f zaman\u0131 UT ( GMT ) nedir? ( Ek cetveller 07 \u2013 08 \u2013 38 \u2013 39 )<\/span><\/h4>\n

\u00c7\u00f6z\u00fcm: ek cetvel 38\u2019 e gir N 35 \u2013 40 derece enlemler aras\u0131ndaki g\u00fcne\u015fin do\u011fu\u015fu \u00e7izelgesinden de\u011ferleri al;<\/span><\/h4>\n

40 \u00e0 04.56 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 35\u00e0 05.06 Aradaki fark\u0131 bul; dlat= 5 derece =05.06 \u2013 04.56= 10 dakika<\/span><\/h4>\n

38 \u2013 35 = 3 derece fark var. 5 derece 10\u2019 ise 3 derecede dlat 6\u2019 eder<\/span><\/h4>\n

Ek cetvel 8 de enlem \u00e7izelgesine gir 5 derece s\u00fctunundan 3 derece yi bul ve onun s\u0131ras\u0131ndan 10 dk, s\u00fctunundaki de\u011feri al. ( 6 \u2018 ) Zaman fark\u0131 6 dakika.\u00a0\u00a0\u00a0\u00a0\u00a0 10\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 15\u2019<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3\u00e0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 6\u2019<\/span><\/h4>\n

Enlem de\u011feri b\u00fcy\u00fcd\u00fck\u00e7e zaman de\u011feri eksildi\u011fi i\u00e7in 6 dk, 35 derece kar\u015f\u0131l\u0131\u011f\u0131 zaman de\u011feri olan 05.06 dan \u00e7\u0131kar\u0131lacak. Lat 35 N 05.06\u2019 \u2013 Lat 38 N \u2013 06\u2019= 05.00 LMT<\/span><\/h4>\n

26 derece E boylam\u0131 i\u00e7in ek cetvel 7\u2019 ye gir ( 0 \u2013 59 dereceler ) aras\u0131 26 derece kar\u015f\u0131l\u0131\u011f\u0131n\u0131 bul; 1h44m. Do\u011fuda bulunuldu\u011fu i\u00e7in bulunan de\u011fer 05.00 LMT den \u00e7\u0131kar\u0131lacak.Cevap; 05.00 \u2013 1.44= 3h.16m<\/span><\/h4>\n

<\/h4>\n

G\u00fcne\u015fin bat\u0131\u015f zaman\u0131n\u0131n hesaplanmas\u0131:<\/strong><\/span><\/h4>\n

04 May\u0131s 1999 g\u00fcn\u00fc 42 derece N \u2013 039 derece E mevkiinde bulunan bir g\u00f6zlemci i\u00e7in G\u00fcne\u015fin bat\u0131\u015f zaman\u0131 UT ( GMT ) nedir? ( ek cetvel 07 \u2013 08 \u2013 38 \u2013 39 )<\/span><\/h4>\n

Ek cetvel 38 e gir G\u00fcne\u015fin bat\u0131\u015f\u0131 b\u00f6l\u00fcm\u00fcnden 40 \u2013 35 derece enlemleri aras\u0131ndaki zaman de\u011ferlerini al. 45\u00e0 19.10\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 40\u00e0 18.58\u00a0 aradaki fark\u0131 bul; 19.10 \u2013 18.58= 12 dk, dlat= 5 derece<\/span><\/h4>\n

45-42= 3 derece.<\/span><\/h4>\n

Ek cetvel 8 e gir 5 derece s\u00fctunundan 3 dereceyi bul ve 3 derece s\u00fctunu kar\u015f\u0131l\u0131\u011f\u0131 dk, b\u00f6l\u00fcm\u00fcnden 12 dk i\u00e7in 10\u2019 ve 15\u2019 de\u011ferlerini al \u00a0\u00a0\u00a0 10\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 15\u2019<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3\u00e0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 6\u2019\u00e0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 9\u2019<\/span><\/h4>\n

10\u2019 i\u00e7in 6\u2019 fark varsa 12\u2019 i\u00e7in nedir? 12×6\/10=7.2 7.2= -7×60=12\u201d saniye.=7\u2019.12\u201d<\/span><\/h4>\n

Lat 45N\u00e0 19.10\u2019 \u2013 Lat 42N\u00e0 07\u201912\u201d= 19.02\u2019.48\u201d<\/span><\/h4>\n

Ek cetvel 7 ye gir derecenin zamana \u00e7evrilmesi \u00e7izelgesinden ( 0 \u2013 59 ) s\u00fctun undan 39 derece boylam\u0131n\u0131n kar\u015f\u0131l\u0131\u011f\u0131n\u0131 bul. ( 2.36\u2019 ) E yani do\u011fuda bulunuldu\u011fu i\u00e7in LMT den bulunan de\u011ferleri \u00e7\u0131kar.<\/span><\/h4>\n

Cevap; 19.02\u201948\u201d \u2013 2.36\u2019= 16h26m48s<\/strong><\/span><\/h4>\n

<\/h4>\n

<\/h4>\n

Sabah Sivil alaca karanl\u0131k hesaplanmas\u0131:<\/strong><\/span><\/h4>\n

05 May\u0131s 1999 g\u00fcn\u00fc 39 derece N \u2013 007 derece E mevkiinde bulunan bir g\u00f6zlemci i\u00e7in sabah sivil alaca karanl\u0131k zaman\u0131 UT ( GMT ) olarak nedir? ( ek cetvel 07 \u2013 08 \u2013 38 \u00a0)<\/span><\/h4>\n

Ek cetvel 38 e G\u00fcne\u015fin do\u011fu\u015fu b\u00f6l\u00fcm\u00fcnden gir 35 ve 40 derece enlem s\u00fctununun kar\u015f\u0131s\u0131ndaki sivil alacak karanl\u0131k de\u011ferlerini al.<\/span><\/h4>\n

Lat 35\u00e0 04.38\u2019\u00a0 Lat 40\u00e0 04.26\u2019 dlat= 5 derece. 40 \u2013 39 = 1 derece fark. 04.38\u2019 \u2013 04.26= 12\u2019 fark<\/span><\/h4>\n

Ek cetvel 8 e gir 5 derece i\u00e7in 1 derece kar\u015f\u0131l\u0131\u011f\u0131n\u0131 bul onun s\u0131ras\u0131ndaki dakika s\u00fctunundan 10\u2019 ve 15\u2019 de\u011ferlerini al.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 15\u2019<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\u00e0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2\u2019\u00e0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3\u2019<\/span><\/h4>\n

10\u2019 i\u00e7in 2\u2019 ise 12\u2019 i\u00e7in nedir? 12×2\/10= 2\u201924\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2.\u201924= -2×60= 24\u201d = 2\u2019.24\u201d<\/span><\/h4>\n

Enlem de\u011ferleri azald\u0131k\u00e7a zaman de\u011feri b\u00fcy\u00fcd\u00fc\u011f\u00fc i\u00e7in \u00e7\u0131kan de\u011feri Lat 40 zaman de\u011feri ile toplar\u0131z.<\/span><\/h4>\n

04.26\u2019 + 2\u201924\u201d= 04.28\u2019 24\u201d<\/span><\/h4>\n

007 E derecesini zamana \u00e7evirmek i\u00e7in Ek cetvel 7 ye gir ( 0 \u2013 59 ) s\u00fctunundan 007 derece kar\u015f\u0131l\u0131\u011f\u0131n\u0131 bul. ( 0.28\u2019 ) boylam E oldu\u011fu i\u00e7in LMT den \u00e7\u0131kar\u0131l\u0131r. Cevap; 04.28\u201924\u201d \u2013 0.28\u2019= 04.00\u2019 24\u201d<\/span><\/h4>\n

Ak\u015fam sivil alaca karanl\u0131k hesaplanmas\u0131:<\/strong><\/span><\/h4>\n

04 May\u0131s 1999 g\u00fcn\u00fc 24 S \u2013 017 W mevkiinde bulunan bir g\u00f6zlemci i\u00e7in ak\u015fam sivil alaca karanl\u0131k zaman\u0131 UT ( GMT ) nedir? ( Ek cetvel, 07 \u2013 08 \u2013 38 \u2013 39 )<\/span><\/h4>\n

Ek cetvel 38 de g\u00fcne\u015fin bat\u0131\u015f\u0131 b\u00f6l\u00fcm\u00fcnden 20S ve 30S derece enlem s\u00fctunu kar\u015f\u0131s\u0131ndaki sivil alacak karanl\u0131k de\u011ferlerini bul.<\/span><\/h4>\n

Lat 20S\u00e0 17.59\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Lat 30 S\u00e0 17.47\u2019 dlat= 10 derece zaman fark\u0131 = 17.59 \u2013 17.47= 12\u2019 dk.<\/span><\/h4>\n

24S \u2013 20S= 4 derece fark.<\/span><\/h4>\n

Ek cetvel 8 e gir 10 derece s\u00fctunundan 4 dereceyi bul\u00e0 onun kar\u015f\u0131l\u0131\u011f\u0131 dakika s\u00fctunundan 10\u2019 ve 15\u2019 de\u011ferlerini bul.\u00a0\u00a0\u00a0 10\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 15\u2019<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4\u00e0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4\u2019\u00e0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 6\u2019<\/span><\/h4>\n

10\u2019 4 \u2018 ise 12\u2019 nedir? 4×12\/10=4.8\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4.8= -4 x60=48\u201d = 4\u201948\u201d<\/span><\/h4>\n

G\u00fcneyde enlem de\u011feri artt\u0131k\u00e7a zaman de\u011feri azald\u0131\u011f\u0131 i\u00e7in bulunan de\u011fer Lat 20 S zaman de\u011ferinden \u00e7\u0131kar\u0131lacak. Lat 20S\u00e0 17.59\u201900\u201d \u2013 4\u201948\u201d= 17.54\u201912\u201d LMT<\/span><\/h4>\n

17W boylam\u0131 i\u00e7in Ek 7 \u00e7izelgesine gir ( 0 \u2013 59 ) boylam s\u00fctunundan 17 derece i\u00e7in zaman\u0131 bul.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 17= 1h08dk. Bulunan de\u011feri boylam W oldu\u011fu i\u00e7in LMT ile topla. 17.54\u201912\u201d + 1.08\u201900\u201d= 19.02\u201912\u201d<\/span><\/h4>\n

Cevap: GMT: 19.02\u201912\u201d<\/span><\/h4>\n

<\/h4>\n

Ay\u0131n GHA ve Dec de\u011ferleri hesaplanmas\u0131:<\/strong><\/span><\/h4>\n

04 May\u0131s 1999 g\u00fcn\u00fc 02h24m00s GMT de ay\u0131n GHA ve Dec de\u011ferleri nedir?( ek cetvel 13 \u2013 38 )<\/span><\/h4>\n

Ek cetvel 38 e gir 04 May\u0131s g\u00fcn\u00fc 02 GMT de Ay s\u00fctunundan Ay\u0131n GHA \u2013 v –\u00a0 Dec\u00a0 ve d de\u011ferlerini bul.<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 v\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d<\/span><\/h4>\n

02 00 00\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 351.02\u20198\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 12.2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 S18.56\u20199\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3.3<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 24 00\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 5.43.6<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 V \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5.0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1.3<\/span><\/h4>\n

\u00a0\u00a0 GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 356.51.4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 S18.58.2<\/span><\/h4>\n

<\/h4>\n

Ek 13 e gir ( artmalar ve d\u00fczeltmeler \u00e7izelgesi ) 24 dakika \u00e7izelgesinden 00 saniye i\u00e7in kar\u015f\u0131l\u0131\u011f\u0131 Ay i\u00e7in olan de\u011feri al. 5.43\u20196. gene 24 dk \u00e7izelgesinden v ( 12.2 ) d\u00fczeltmesi i\u00e7in de\u011feri al. 12.2\u00e05.0 d ( 3.3 ) i\u00e7in g\u00f6r\u00fclen de\u011ferleri al. 3.3\u00e0 1.3 bulunan de\u011ferleri GHA ve Dec de\u011ferlerine ekle.<\/span><\/h4>\n

Cevap: GHA: 356.51.4\u00a0\u00a0\u00a0 Dec: S18.58.2<\/span><\/h4>\n

Ay\u0131n GHA ve Dec de\u011ferlerinin hesaplanmas\u0131<\/strong><\/span><\/h4>\n

03 May\u0131s 1999 g\u00fcn\u00fc 04h 25m15s GMT de Ay \u0131n GHA ve Dec de\u011ferleri nedir. ( ek cetvel 13 \u2013 39 )<\/span><\/h4>\n

Ek cetvel 39 a gir 03 May\u0131s saat 04 i\u00e7in Ay s\u00fctunundan Ay\u0131n GHA \u2013 v \u2013 Dec ve d de\u011ferlerini bul.<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 v\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d<\/span><\/h4>\n

04 00 00\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 031.27\u20193\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 12.9\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 S17.22.6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5.1<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 25 15\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 6.01\u20195\u00a0<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 V \u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a05.5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2.2<\/span><\/h4>\n

GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 037.34\u2019.3 Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 S17.24\u20198<\/span><\/h4>\n

<\/h4>\n

<\/h4>\n

Ek cetvel 13 e gir 25 dakika b\u00f6l\u00fcm\u00fcnden 15s kar\u015f\u0131l\u0131\u011f\u0131 de\u011feri bul. 15\u00e06.01.5 v (12.9 ) kar\u015f\u0131l\u0131\u011f\u0131n\u0131 bul 12.9\u00e05.5. d( 5.1 ) d\u00fczeltemesi i\u00e7in kar\u015f\u0131l\u0131\u011f\u0131n\u0131 bul 5.1\u00e02.2 Bulunan de\u011ferleri GHA ve Dec de\u011ferlerine ekle.<\/span><\/h4>\n

Cevap: GHA: 037.34.3 Dec: S17.24\u20198<\/span><\/h4>\n

<\/h4>\n

<\/h4>\n

Ay\u0131n GHA ve Dec de\u011ferlerinin hesaplanmas\u0131<\/strong><\/span><\/h4>\n

03 May\u0131s 1999 g\u00fcn\u00fc saat 20h 24m 48s ZT da 48.00 N 020.30 E mevkiinde ay\u0131n GHA ve Dec de\u011ferleri nedir? ( ek cetvel 13 ve 39 )<\/span><\/h4>\n

\u00c7\u00f6z\u00fcm: ZD= -1 GMT = ZT \u2013 ZD\u00e0 GMT= 20.24\u2019.48\u201d- 1 = 19h 24m.48s bulunur. ( boylam 20 E oldu\u011fu i\u00e7in de\u011fer ( – )<\/span><\/h4>\n

Ek 39 a gir 03 May\u0131s i\u00e7in Ay s\u00fctunundan saat 19.00 i\u00e7in Ay \u0131n GHA \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 v \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec ve d de\u011ferlerini bul.<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 v\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d<\/span><\/h4>\n

19 00 00\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 249.23\u20190\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 12.5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 S18.31\u20192\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3.9<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 24\u201948\u201d\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 5.55\u20191<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 V\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5\u20191 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\u20196<\/span><\/h4>\n

GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 255.23\u20192\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 S18.32\u20198<\/span><\/h4>\n

Ek cetvel 13 e gir 24 m s\u00fctununda 48 s i\u00e7in de\u011feri bul 48\u00e0 5.55\u20191. Gene ayn\u0131 s\u00fctundan v ( 12.5 ) de\u011feri d\u00fczeltme kar\u015f\u0131l\u0131\u011f\u0131n\u0131 bul.12.5\u00e0 5.1 d ( 3.9 ) kar\u015f\u0131l\u0131\u011f\u0131n\u0131 bul. 3.9\u00e0 1.6 bulunan de\u011ferleri GHA ve Dec de\u011ferlerine ekle.<\/span><\/h4>\n

Cevap: GHA: 255.23\u20192\u00a0\u00a0\u00a0 Dec: S18.32.8<\/span><\/h4>\n

G\u00fcne\u015fin GHA ve Dec de\u011ferlerinin hesaplanmas\u0131<\/strong><\/span><\/h4>\n

03 May\u0131s 1999 g\u00fcn\u00fc saat 16h 25m 38s ZT\u2019 da 40.45\u2019N \u2013 029. 35\u2019E mevkiinde G\u00fcne\u015fin GHA ve Dec de\u011ferleri nedir?<\/span><\/h4>\n

\u00c7\u00f6z\u00fcm: ZD \u2013 2. GMT= ZT -\/+ ZD \u00e0 GMT= 16h25m 38s \u2013 2h= 14h 25m 38s GMT: 14h 25m 38s<\/span><\/h4>\n

Ek cetvel 39 a gir G\u00fcne\u015f s\u00fctunu 3 May\u0131s saat 14 00 i\u00e7in GHA, Dec ve d de\u011ferlerini al.<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d<\/span><\/h4>\n

14 00 00\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 30.46\u20196\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 N15.38.9\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 0.7<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 25 38\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 6.24\u20195\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 0.3<\/span><\/h4>\n

GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 37.11.1 Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 N15.39.2<\/span><\/h4>\n

Not cevapta. GHA 37.11.1 Dec N15.38.6 g\u00f6z\u00fck\u00fcyor. Normalde zaman de\u011feri artt\u0131k\u00e7a GHA ve Dec de\u011ferleri art\u0131\u011f\u0131 i\u00e7in ( + ) olmas\u0131 gerekir. Zaman de\u011feri artt\u0131k\u00e7a de\u011ferler d\u00fc\u015f\u00fcyorsa o zaman ( – ) olur.<\/span><\/h4>\n

Ek cetvel 13 e gir 25m s\u00fctunundan 38 s i\u00e7in de\u011ferleri al. 38s\u00e0 6.24\u20195. d ( 0.7 ) d\u00fczeltme de\u011ferini al\u00e0 0.3 bulunan de\u011ferleri GHA ve Dec de\u011ferine ekle<\/span><\/h4>\n

<\/h4>\n

<\/h4>\n

Ay \u0131n \u00fcst meridyenden ge\u00e7i\u015f hesaplanmas\u0131<\/strong><\/span><\/h4>\n

05 May\u0131s 1999 g\u00fcn\u00fc 42 N\u00a0 – 028 W mevkiinde bulunan bir g\u00f6zlemci i\u00e7in Ay\u0131n \u00fcst meridyenden ge\u00e7i\u015f zaman\u0131 nedir? ( ek cetvel 7 – 8 ve 38 )<\/span><\/h4>\n

\u00c7\u00f6z\u00fcm: Ek cetvel 38 e gir sa\u011f alt k\u00f6\u015feden Ay \u0131n meridyen ge\u00e7i\u015f b\u00f6l\u00fcm\u00fcnden 5 May\u0131s ve 6 May\u0131s de\u011ferlerini bul<\/span><\/h4>\n

5 May\u0131s\u00e0 03.25\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 6 May\u0131s\u00e0 04.14 aradaki fark\u0131 bul. 04.14 \u2013 03.25= 49 dakika.<\/span><\/h4>\n

Ek cetvel 7 ye gir 28 w boylam\u0131n\u0131n saat de\u011ferini\u00a0 bul. 28W\u00e0 1h 52m. Boylam i\u015fareti W oldu\u011fu i\u00e7in 03.25 zaman de\u011feri ile + ( 03.25 + 01.52= 05.17 )<\/span><\/h4>\n

Ek cetvel 8 e gir II \u00e7izelgeden 28 W boylam\u0131 i\u00e7in 49 dakika de\u011ferini bul. ( 28 W ve 49m i\u00e7in yakla\u015f\u0131k de\u011fer 20 derece ve 50m oldu\u011fu i\u00e7in onlar\u0131n de\u011ferleri al\u0131nacak )28\u00e0 50\u00e0 4 m bulunan de\u011feri boylam i\u015fareti W oldu\u011fu i\u00e7in 05.17 ye ekle. 05h17m + 4m= 05h21m. Cevap 05h21m<\/span><\/h4>\n

Ay\u0131n \u00fcst meridyenden ge\u00e7i\u015f hesaplanmas\u0131<\/strong><\/span><\/h4>\n

05 May\u0131s 1999 g\u00fcn\u00fc 37 N 025 E mevkiinde bulunan bir g\u00f6zlemci i\u00e7in Ay \u0131n \u00fcst meridyenden ge\u00e7i\u015f zaman\u0131 UT ( GMT ) nedir? ( ek cetvel 7\u00a0 \u2013 38 )<\/span><\/h4>\n

Ek cetvel 38 e gir sa\u011f alt k\u00f6\u015feden Ay \u0131n meridyen ge\u00e7i\u015f b\u00f6l\u00fcm\u00fcnden 4 ve 5 May\u0131s i\u00e7in de\u011ferleri bul.<\/span><\/h4>\n

Ek cetvel 7 ye gir 025 derec E boylam\u0131n\u0131n zaman de\u011ferini bul. 025\u00e0 1h 40m<\/span><\/h4>\n

Bulunan de\u011feri 03.25 den \u00e7\u0131kar. 03.25 \u2013 01.40= 01.45<\/span><\/h4>\n

Not: burada normalde ek cetvel 8 de verilmesi gerekir ve do\u011fu boylam\u0131 oldu\u011fu i\u00e7in bir \u00f6nceki tarih yani 04 may\u0131s de\u011feri de al\u0131nmas\u0131 gerekir. 05 may\u0131s de\u011feri aras\u0131ndaki zaman fark\u0131 bulunup ek cetvel 8e giri\u015f yap\u0131p bulunan de\u011fer 01.45 ten boylam do\u011fu oldu\u011fu i\u00e7in \u00e7\u0131kar\u0131lmas\u0131 gerekir.<\/span><\/h4>\n

04 May\u0131s 02.37 03.25- 02.37= 48 dakika fark. Ek cetvel 8 den 25 E\u00e0 48 kar\u015f\u0131l\u0131\u011f\u0131 bulunur (3\u2019,5 = 4\u2019) bulunan de\u011fer 01.45 ten boylam E oldu\u011fu i\u00e7in \u00e7\u0131kar\u0131l\u0131r. 01h45m \u2013 4m= 01.41 do\u011fru cevap. \u00a0Fakat \u015f\u0131klarda yok.<\/span><\/h4>\n

G\u00fcne\u015fin meridyenden ge\u00e7i\u015fi hesaplanmas\u0131<\/strong><\/span><\/h4>\n

05 May\u0131s 1999 g\u00fcn\u00fc 36.20\u2019N \u2013 032.42\u2019 E mevkiinde bulunan bir g\u00f6zlemci i\u00e7in G\u00fcne\u015fin meridyenden ge\u00e7i\u015f zaman\u0131 UT ( GMT ) nedir? ( ek cetvel 7 \u2013 38)<\/span><\/h4>\n

Ek cetvel 38 sa\u011f alt k\u00f6\u015feden G\u00fcne\u015f in \u00fcst meridyen ge\u00e7i\u015fini bul. \u00e011.57<\/span><\/h4>\n

Ek cetvel 7 ye gir ( 0 \u2013 59 ) \u00e7izelgesinden 32 E boylam\u0131n\u0131n zaman de\u011ferini bul 32\u00e02h 08m<\/span><\/h4>\n

Ek cetvel 7 de 42m i\u00e7in de\u011feri bul 42\u00e0 2\u201948\u201d bulunan de\u011ferleri topla. 2h08m+2m48s= 2h10m48s<\/span><\/h4>\n

Bulunan de\u011ferleri boylam E oldu\u011fu i\u00e7in Meridyen ge\u00e7i\u015finden \u00e7\u0131kar ( – )\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 11h57m00s \u2013 2h10m48s=09h 46m12s Cevap: 09h46m12s<\/span><\/h4>\n

<\/h4>\n

G\u00fcne\u015fin meridyenden ge\u00e7i\u015f hesaplanmas\u0131.<\/strong><\/span><\/h4>\n

02 May\u0131s 1999 g\u00fcn\u00fc 22.10\u2019S 046.18\u2019W mevkiinde bulunan bir g\u00f6zlemci i\u00e7in G\u00fcne\u015fin meridyenden ge\u00e7i\u015f zaman\u0131 UT ( GMT ) nedir? ( ek ctvel 7 ve 39 )<\/span><\/h4>\n

Ek cetvel 39 a gir sa\u011f alt k\u00f6\u015feden G\u00fcne\u015fin meridyen ge\u00e7i\u015f zaman\u0131n\u0131 bul\u00e011.57<\/span><\/h4>\n

46 W boylam\u0131 i\u00e7in ek cetvel 7 ye gir ( 0 \u2013 59 ) s\u00fctunundan 46 W i\u00e7in zaman de\u011ferini bul\u00e03h04m ek cetvel 7 de 18m i\u00e7in de\u011feri bul\u00e01m12s bulunan de\u011ferleri topla 3h04m+1m12s= 3h05m12s<\/span><\/h4>\n

W boylam\u0131nda bulunuldu\u011fu i\u00e7in bulunan de\u011feri G\u00fcne\u015fin meridyenden ge\u00e7i\u015f zaman\u0131na ekle ( + )<\/span><\/h4>\n

11H57m + 3h05m12s=15h02m12s Cevap: 15h02m12s<\/strong><\/span><\/h4>\n

Pollux y\u0131ld\u0131z\u0131 meridyen ge\u00e7i\u015fi hesaplanmas\u0131.<\/strong><\/span><\/h4>\n

05 May\u0131s 1999 g\u00fcn\u00fc 43 N\u00a0 – 036 E mevkiinde bulunan bir g\u00f6zlemci i\u00e7in Pollux y\u0131ld\u0131z\u0131n\u0131n meridyen ge\u00e7i\u015f zaman\u0131 UT ( GMT ) olarak nedir? ( ek cetvel 7 ve 42\u00a0 )<\/span><\/h4>\n

\u00c7\u00f6z\u00fcm:Ek cetvel 42 ye gir; Pollux SHA= 243 41\u20193 bulunan de\u011feri 036 E boylam\u0131 ile topla<\/span><\/h4>\n

243.41\u20193+036=279.41\u20193\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 GHA ( Aries )= 360 \u2013 279.41\u20193= 80.18\u20197= 80.18\u201942\u201d ( 7×6=42\u201d)<\/span><\/h4>\n

Ek cetvel 42 de Aries in 5 May\u0131s s\u00fctununda 80.18\u201942\u201d ile gir. Zaman olarak 1400 ve 1500 aras\u0131na denk gelir. Bu zamanlar\u0131n GHA de\u011ferlerini al ve aradaki fark\u0131 bul.<\/span><\/h4>\n

1400\u00a0\u00a0\u00a0\u00a0\u00a0 72.59\u20199<\/u><\/span><\/h4>\n

1500\u00a0\u00a0\u00a0\u00a0\u00a0 88.02\u20194\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 88.02\u20194 \u2013 72.59\u20199= 15.02\u20195=15.02\u201930\u201d ( 5×60=30\u201d) aradaki fark: 15.02\u201930\u201d<\/u><\/span><\/h4>\n

80.\u201918\u201942\u201d ( 42\u201d \/ 6 = 7 ) aradaki fark\u0131 bulmak i\u00e7in saniye yi ondal\u0131\u011fa \u00e7evirmek gerekir. 80.18\u20197<\/u><\/span><\/h4>\n

72.59\u20199\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 aradaki fark \u00a080.18\u20197 \u2013 72.59.9= 07.18\u20198 ( ondal\u0131\u011f\u0131 saniye ye \u00e7evir 8×6 48\u201d= 07.18\u201948\u201d<\/u><\/span><\/h4>\n

Ek cetvel 7 ye gir 7 derece kar\u015f\u0131l\u0131\u011f\u0131n\u0131 bul\u00e0 0.28\u2019 ek cetvel 7 de 18\u2019 kar\u015f\u0131l\u0131\u011f\u0131n\u0131 bul\u00e0 1\u201912\u201d bulunan de\u011ferleri topla 28\u2019+1\u201912\u201d=29\u201912\u201d bulunan bu de\u011feri saat 1400 e ekle 1400\u2019 + 29\u201912\u201d=14.29\u201912\u201d<\/span><\/h4>\n

Cevap: 14h29m12s Soru cevab\u0131 14.29.10<\/span><\/h4>\n

Rigel y\u0131ld\u0131z\u0131n\u0131n meridyenden ge\u00e7i\u015f zaman\u0131 hesaplanmas\u0131<\/strong><\/span><\/h4>\n

06 May\u0131s 1999 g\u00fcn\u00fc 35 S \u2013 052 W mevkiinde bulunan bir g\u00f6zlemci i\u00e7in Rigel y\u0131ld\u0131z\u0131n\u0131n meridyenden ge\u00e7i\u015f zaman\u0131 nedir? ( ek cetvel 7 ve 42 )<\/span><\/h4>\n

Ek cetvel 42 ye gir rigel y\u0131ld\u0131z\u0131n\u0131n SHA de\u011ferini bul\u00e0 281.22\u20199 boylam W oldu\u011fu i\u00e7in bulunan Sha de\u011ferinden \u00e7\u0131kar. 281.22\u20199 \u2013 52 = 229.22\u20199<\/span><\/h4>\n

GHA Aries = 360 \u2013 229.22\u20199 = 130.37\u20191 bulunan de\u011fer ile 6 may\u0131s s\u00fctunundan Aries GHA de\u011ferine gir. Girilecek b\u00f6l\u00fcm zaman olarak 1700 ve 1800 zaman aral\u0131klar\u0131.<\/span><\/h4>\n

1700 GHA= 119.06\u20194<\/u><\/span><\/h4>\n

1800 GHA= 134.18\u20199 aradaki fark\u0131 bul; fark = 15.12\u20195<\/u><\/span><\/h4>\n

130.37\u20191 ile 1700 Aries GHA aras\u0131ndaki fark\u0131 bul; 130.37\u20191 \u2013 119.06\u20194= fark. 11.30\u20197<\/u><\/span><\/h4>\n

Ek cetvel 7 ye gir 11 derece i\u00e7in ( 0 \u2013 59 ) s\u00fctunundan zaman de\u011ferini bul\u00e0 0.44\u2019. 30\u2019 i\u00e7in de\u011feri bul\u00e02\u201900\u201d bulunan de\u011ferleri topla 0.44\u2019+2\u2019.00= 0.46\u2019<\/span><\/h4>\n

Bulunan de\u011feri saat 1700 aries zaman\u0131na ekle = 17.00 + 0.46= 17.46\u201900\u201d<\/span><\/h4>\n

Not soru cevab\u0131nda 17 45 55 g\u00f6z\u00fck\u00fcyor.<\/span><\/h4>\n

Alphard y\u0131ld\u0131z\u0131 GHA ve Dec de\u011ferlerinin bulunmas\u0131.<\/strong><\/span><\/h4>\n

02 May\u0131s 1999 g\u00fcn\u00fc 17h 25\u2019 26\u201d ZT da 30.09\u2019 N \u2013 078.45\u2019 W mevkiinde Alphard y\u0131ld\u0131z\u0131n\u0131n GHA ve Dec de\u011ferleri nedir? ( ek cetvel 13 \u2013 44 )<\/span><\/h4>\n

ZD\u00e0 78\u00e0 78\/15=5h12m ZD ( W ) +5 oldu\u011fu i\u00e7in<\/span><\/h4>\n

GMT= ZT +\/-ZD\u00e0 17h25m26s + 5h= 22h25m26s GMT olur.<\/span><\/h4>\n

Alphard SHA 218.06\u20198\u00a0\u00a0\u00a0 Dec S8.39\u20195<\/span><\/h4>\n

2 May\u0131s 1999 i\u00e7in saat 22 00 Aries GHA 190.22\u20192<\/span><\/h4>\n

<\/h4>\n

Ek cetvel 13 e gir 25m s\u00fctunundan 26 saniye i\u00e7in de\u011feri bul 25\u00e0 26\u00e0 aries de\u011feri 6.22\u20195 bulunan de\u011feri Aries GHA ya ekle 190.22\u20192 + 6.22\u20195 = 196.44\u20197<\/span><\/h4>\n

Bulunan de\u011feri Alphard SHA ile topla. 196.44\u20197+ 218.06\u20198= 414.51\u20195 de\u011fer 360 dereceden b\u00fcy\u00fck oldu\u011fu i\u00e7in 360 dereceden \u00e7\u0131kar 414.51\u20195 \u2013 360= 054.51.5<\/span><\/h4>\n

Cevap: GHA 54.51\u20195 Dec.S8.39\u20195<\/strong> Soru cevab\u0131; 24.46\u20196 \u2013 S8.39\u20195 g\u00f6z\u00fck\u00fcyor.<\/span><\/h4>\n

\u00a0Capella y\u0131ld\u0131z\u0131n\u0131n GHA ve Dec de\u011ferleri bulunmas\u0131<\/strong><\/span><\/h4>\n

06 May\u0131s 1999 g\u00fcn\u00fc 06 24 37 UT ( GMT ) de Capella y\u0131ld\u0131z\u0131n\u0131n GHA ve Dec de\u011ferleri nedir.( ek cetvel 13 ve 42 )<\/span><\/h4>\n

\u00c7\u00f6z\u00fcm: GHA ( y\u0131ld\u0131z ) = GHA ( Aries ) + SHA ( y\u0131ld\u0131z )<\/span><\/h4>\n

Ek cetvel 42 ye gir 6 may\u0131s 0600 i\u00e7in Aries s\u00fctunundan aries in 0600 GHA de\u011ferini al\u00e0 Aries GHA =313.39\u20193. Ek cetvel 42 de Capella y\u0131ld\u0131z\u0131n\u0131n SHA ve Dec de\u011ferlerini al. SHA 280.51\u20191 Dec.N45.59\u20198<\/span><\/h4>\n

GHA ( Aries ) + SHA ( CApella ) 313.39\u20193 + 280.51\u20191= 590.30\u20194 bulunan de\u011fer 360 dereceden b\u00fcy\u00fck oldu\u011fu i\u00e7in 360 dereceden \u00e7\u0131kar. 590.30\u20194 \u2013 360= 234.30\u20194<\/span><\/h4>\n

Ek cetvel 13 e gir 24m s\u00fctunundan 37s de\u011ferinin kar\u015f\u0131l\u0131\u011f\u0131n\u0131 Aries s\u00fctunundan derece olarak bul\u00e0<\/span><\/h4>\n

6.10\u20193 bulunan de\u011feri GHA ya ekle. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 06 00 00\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 234.30\u20194\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 N45.59\u20198<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 24\u201937\u201d\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 6.10\u20193\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Cevap: GHA: 240.40\u20197\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec:N45.59\u20198<\/span><\/h4>\n

Acrux y\u0131ld\u0131z\u0131n\u0131n GHA ve Dec de\u011ferlerinin hesaplanmas\u0131.<\/strong><\/span><\/h4>\n

04 May\u0131s 1999 01h 25m 10s UT ( GMT ) de, Acrux y\u0131ld\u0131z\u0131n\u0131n GHA ve Dec de\u011ferleri nedir? ( ek cetvel 13 \u2013 38 ve 42 )<\/span><\/h4>\n

Ek cetvel 42 ye gir 4 May\u0131s saat 01.00 i\u00e7in aries y\u0131ld\u0131z\u0131n\u0131n GHA de\u011ferlerini al. Aries GHA\u00e0 236.28\u20197<\/span><\/h4>\n

Ek cetvel 42 de Acrux y\u0131ld\u0131z\u0131n\u0131n SHA ve Dec de\u011ferlerini al. Acrux SHA\u00e0 173.21\u20199 \u2013 Dec\u00e0 S63.05\u20199<\/span><\/h4>\n

GHA ( Aries ) + SHA ( y\u0131ld\u0131z )= 236.28\u20197 + 173.21\u20190= 409.49\u20197 \u2013 360 = 049.49\u20197 ( de\u011fer 360 dereceden\u00a0 b\u00fcy\u00fck oldu\u011fu i\u00e7in 360 dereceden \u00e7\u0131kar\u0131l\u0131r.<\/span><\/h4>\n

Ek cetvel 13 e gir Aries b\u00f6l\u00fcm\u00fcnden 25m s\u00fctunundan 10s de\u011ferini bul. 25m\u00e06.18\u20195 bulunan de\u011feri GHA ya ekle.<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec<\/span><\/h4>\n

01 00 00\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 049.49\u20197\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 S63.05\u20199<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 25\u201910\u201d\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0 6.18\u20195<\/span><\/h4>\n

GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 056.08\u20192\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 S63.05\u20199 Dec<\/span><\/h4>\n

\u00a0Ven\u00fcs gezegeni GHA ve Dec de\u011ferleri hesaplanmas\u0131<\/strong><\/span><\/h4>\n

04 May\u0131s 1999 g\u00fcn\u00fc 16h 24m 35s GMT de Ven\u00fcs \u00fcn GHA ve Dec de\u011ferleri nedir? ( ekcetvel 13 \u2013 38 \u2013 42 )<\/span><\/h4>\n

Ek cetvel 42 ye gir Ven\u00fcs s\u00fctunundaki 4 May\u0131s saat 16.00 kar\u015f\u0131l\u0131\u0131\u011f\u0131 de\u011ferleri bul. 16 00\u00e0GHA 17.05\u20198 Dec.N25. 43\u20198, v -0.7, d0.2<\/span><\/h4>\n

Ven\u00fcs\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 v\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d<\/span><\/h4>\n

16 00 00\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 17.05\u20198\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -0,7\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 N25.43\u20198\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 0.2<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 24m35s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 6.08\u20198\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 V\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 -0\u20193\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 0.1<\/span><\/h4>\n

GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 23.14\u20193\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 N25.43\u20199<\/span><\/h4>\n

<\/h4>\n

<\/h4>\n

Ek cetvel 13 e gir ( g\u00fcne\u015f gezegen )24m s\u00fctunundan 35 saniye de\u011ferini bul.24.35s\u00e06.08\u20198 v -0.7 de\u011ferini bul\u00e00.3, d 0.2 de\u011ferini bul\u00e00.1 bulunan de\u011ferleri GHA ve Dec de\u011ferlerine ekle. V de\u011feri (-) eksi oldu\u011fu i\u00e7in GHa de\u011ferinden \u00e7\u0131kar\u0131lacak. Cevap: GHA 23.14\u20193\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec.N25.43\u20199<\/span><\/h4>\n

<\/h4>\n

<\/h4>\n

Ven\u00fcs GHA ve Dec de\u011ferlerinin hesaplanmas\u0131.<\/strong><\/span><\/h4>\n

04 May\u0131s 1999 g\u00fcn\u00fc 46.14\u201900 N \u2013 006.15\u201900 W DR mevkiinde 06h 24m 22s ZT da Ven\u00fcs gezegeninden rasat yap\u0131ld\u0131\u011f\u0131nda bu gezegenin GHA ve Dec de\u011ferleri nedir.( ekcetvel 13 ve 42 )<\/span><\/h4>\n

\u00c7\u00f6z\u00fcm: ZD= 0\u00e0 GMT= 06h 24m 22s olur.<\/span><\/h4>\n

Ek cetvel 42 ye gir Ven\u00fcs gezegeninin 4 May\u0131s saat 06.00 i\u00e7in GHA, v, Dec, d de\u011ferlerini\u00a0 al.<\/span><\/h4>\n

Ven\u00fcs\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 v\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d<\/span><\/h4>\n

06 00 00\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 227.12\u20194\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -0\u20197\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 N25.42\u20190\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 0.2<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 24m22s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0 6.05\u20195<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 V\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 0\u20193\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 0\u20191<\/span><\/h4>\n

GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 233.17\u20196\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 N25.42\u20191<\/span><\/h4>\n

Ek cetvel 13 e gir G\u00fcne\u015f, gezegen 24m s\u00fctunundan 22s de\u011ferini bul\u00e06.05\u20195, ayn\u0131 s\u00fctundan v -0.7 de\u011ferini bul\u00e00.3, d 0.2 de\u011ferini bul\u00e00.1. bulunan de\u011ferleri GHA ve Dec de\u011ferlerine ekle. V nin de\u011feri ( – ) oldu\u011fu i\u00e7in GHA dan \u00e7\u0131kar\u0131lacak. Cevap: GHA:233.17\u20196\u00a0\u00a0\u00a0 Dec:N25.42\u20191<\/span><\/h4>\n

Mars gezegeninin GHA ve Dec de\u011ferleri hesaplanmas\u0131.<\/strong><\/span><\/h4>\n

06 May\u0131s 1999 g\u00fcn\u00fc 19h 24m 42s GMT de Mars gezegeninin GHA ve Dec de\u011ferleri nedir? ( ek cetvel 13 \u2013 38 \u2013 42 )<\/span><\/h4>\n

Ek cetvel 42 ye gir Mars gezegeninin 6 May\u0131s s\u00fctunundan saat 1900 i\u00e7in GHA, v, Dec ve d de\u011ferlerini al.<\/span><\/h4>\n

Mars\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 v\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d<\/span><\/h4>\n

19 00 00\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 301.16\u20193\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3.3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 S10.34\u20197\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 0.2<\/span><\/h4>\n

\u00a0\u00a0\u00a0 24m42s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 6.10\u20195<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 V\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\u20193\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d\u00a0\u00a0\u00a0\u00a0 – 0\u20191<\/span><\/h4>\n

GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 307.28\u20191\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 S10.34\u20196<\/span><\/h4>\n

<\/h4>\n

Ek cetvel 13 e gir g\u00fcne\u015f ve gezegenler 24m s\u00fctunundan 42s de\u011fer kar\u015f\u0131l\u0131\u011f\u0131n\u0131 bul\u00e0 6.10\u20195, v 3.3 de\u011ferini bul\u00e01.3, d 0.2 de\u011ferini bul\u00e0 0.1. bulunan de\u011ferleri GHA ve Dec de\u011ferlerine ekle. Zaman artt\u0131k\u00e7a de\u011fer d\u00fc\u015ft\u00fc\u011f\u00fc i\u00e7in d de\u011feri ( – ) olarak uygulan\u0131r. Cevap: GHA 307.28\u20191 Dec: S10.34\u20196<\/span><\/h4>\n

<\/h4>\n

<\/h4>\n

<\/h4>\n

G\u00fcne\u015fin Meridyen ge\u00e7i\u015fi hesaplanmas\u0131.<\/strong><\/span><\/h4>\n

04 May\u0131s 1999 g\u00fcn\u00fc 44 derece N \u2013 033 derece E mevkiinde bulunan bir g\u00f6zlemci i\u00e7in G\u00fcne\u015f in meridyen den ge\u00e7i\u015f zaman\u0131 UT ( GMT ) olarak ka\u00e7t\u0131r? ( ek cetvel 07 \u2013 38 )<\/span><\/h4>\n

Ek cetvel 38 e gir G\u00fcne\u015fin meridyen ge\u00e7i\u015f zaman\u0131n\u0131 sa\u011f alt k\u00f6\u015feden GMT olarak bul\u00e0 11.57 ek cetvel 7 ye gir( 0 \u2013 59 s\u00fctunundan ) 33 derece E boylam\u0131 i\u00e7in zaman de\u011ferini bul\u00e02h 12m bulunan de\u011feri GMT de\u011ferinden boylam i\u015fareti E oldu\u011fu i\u00e7in \u00e7\u0131kar. 11.57 \u2013 2.12=09.45 Cevap: 09.45<\/span><\/h4>\n

G\u00fcne\u015fin meridyen ge\u00e7i\u015fi hesaplanmas\u0131.<\/strong><\/span><\/h4>\n

06 May\u0131s 1999 g\u00fcn\u00fc\u00a0 38.00 N –\u00a0 030.00 W mevkiinde bir g\u00f6zlemci i\u00e7in G\u00fcne\u015f in meridyenden ge\u00e7i\u015f zaman\u0131 UT ( GMT ) ka\u00e7t\u0131r? ( ek cetvel 7 \u2013 38 )<\/span><\/h4>\n

Ek cetvel 38 e gir G\u00fcne\u015fin meridyenden ge\u00e7i\u015f zaman\u0131n\u0131 sa\u011f alt k\u00f6\u015feden GMT olarak bul\u00e0 11.57 ek cetvel 7 ye gir ( 0 \u2013 59 s\u00fctunundan ) 030 W boylam\u0131 i\u00e7in zaman de\u011ferini bul\u00e02h00m bulunan de\u011feri GMT de\u011feri ile boylam i\u015fareti W oldu\u011fu i\u00e7in topla. 11.57 + 2= 13.57. Cevap: 13.57<\/span><\/h4>\n

Ay\u0131n \u00fcst meridyenden ge\u00e7i\u015fi hesaplanmas\u0131. <\/strong><\/span><\/h4>\n

05 May\u0131s 1999 g\u00fcn\u00fc 37.00 N \u2013 025.00 E mevkiinde bulunan bir g\u00f6zlemci i\u00e7in Ay\u2019\u0131n \u00fcst meridyenden ge\u00e7i\u015f zaman\u0131 UT ( GMT ) nedir? ( ek cetvel 7 \u2013 38 )<\/span><\/h4>\n

Ek cetvel 38 e gir sa\u011f alt k\u00f6\u015feden4 ve 5 may\u0131s i\u00e7in meridyen ge\u00e7i\u015fi de\u011ferini al\u00e04 may\u0131s\u00e002.37<\/span><\/h4>\n

5 may\u0131s\u00e003.25 aras\u0131ndaki zaman fark\u0131n\u0131 bul. 03.25 \u2013 02.37= 00.48\u2019<\/span><\/h4>\n

Ek cetvel 7 ye gir ( 0 \u2013 59 s\u00fctunundan ) 25 E boylam\u0131 i\u00e7in zaman de\u011feri bul\u00e01h40m<\/span><\/h4>\n

Bulunan de\u011feri boylam i\u015fareti E oldu\u011fu i\u00e7in 03.25 de\u011ferinden \u00e7\u0131kar. 03.25 \u2013 01.40= 01.45<\/span><\/h4>\n

Not: cevapta do\u011fru cevap 01.45 g\u00f6z\u00fck\u00fcyor. Normalde ek cetvel 8 e girip 2. \u00c7izelgeden boylam de\u011feri ve 48\u2019 de\u011ferleri aras\u0131ndaki d\u00fczeltmeyi bulup bulunan de\u011ferden boylam i\u015fareti E oldu\u011fu i\u00e7in \u00e7\u0131karmak gerekir fakat cevap de\u011ferlerinde g\u00f6z\u00fckm\u00fcyor. 25\u00e048\u2019\u00e04\u2019<\/span><\/h4>\n

01.45\u2019 \u2013 4\u2019= 01.41\u2019 Do\u011fru cevap bu olmas\u0131 gerekir.<\/span><\/h4>\n

Ay\u0131n \u00fcst meridyenden ge\u00e7i\u015fi hesaplanmas\u0131.<\/strong><\/span><\/h4>\n

05 May\u0131s 1999 g\u00fcn\u00fc 42.00 N \u2013 028 W mevkiinde bulunan bir g\u00f6zlemci i\u00e7in Ay\u2019 \u0131n \u00fcst meridyenden ge\u00e7i\u015f zaman\u0131 UT ( GMT ) nedir?( ek cetvel 7 \u2013 38 )<\/span><\/h4>\n

Ek cetvel 38 e gir sa\u011f alt k\u00f6\u015feden 5 ve 6 may\u0131s i\u00e7in meridyen ge\u00e7i\u015fi de\u011ferini al\u00e05 may\u0131s\u00e003.25<\/span><\/h4>\n

6 may\u0131s\u00e004.14 aras\u0131ndaki zaman fark\u0131n\u0131 bul. 04.14 \u2013 03.25= 00.49\u2019<\/span><\/h4>\n

Ek cetvel 7 ye gir ( 0 \u2013 59 s\u00fctunundan ) 28 W boylam\u0131 i\u00e7in zaman de\u011feri bul\u00e01h52m<\/span><\/h4>\n

Bulunan de\u011ferleri boylam i\u015fareti W oldu\u011fu i\u00e7in topla\u00e003.25 + 1.52=05.17<\/span><\/h4>\n

Normalde ekcetvel 8 e girip 2. \u00e7izelgeden 28 W\u00e0 ve 49\u2019\u00e0 de\u011ferleri aras\u0131ndaki d\u00fczeltmeyi bulup bulunan de\u011fere boylam i\u015fareti W oldu\u011fu i\u00e7in eklemek gerekir.4\u2019 05.17\u2019 + 4\u2019=05.21\u2019 do\u011fru cevap.<\/span><\/h4>\n

Mars gezegeninin meridyen ge\u00e7i\u015f zaman\u0131 hesaplanmas\u0131.<\/strong><\/span><\/h4>\n

06 May\u0131s 1999 g\u00fcn\u00fc 46.00 N \u2013 052.00 W mevkiinde bulunan bir g\u00f6zlemci i\u00e7in Mars gezegeninin meridyenden ge\u00e7i\u015f zaman\u0131 UT ( GMT ) nedir? ( ek cetvel 7 \u2013 42 )<\/span><\/h4>\n

Ek cetvel 42 ye gir sa\u011f alt k\u00f6\u015feden Mars gezegeninin meridyen ge\u00e7i\u015f zaman\u0131n\u0131n de\u011ferini al\u00e022.59 ekcetvel 7 ye gir ( 0 \u2013 59 s\u00fctunundan ) 52 W boylam\u0131 i\u00e7in zaman de\u011ferini al\u00e03.28 boylam i\u015fareti W oldu\u011fu i\u00e7in bulunan de\u011ferleri topla. 22.59 + 3.28= 26.27\u00e0 02h27m 07 May\u0131s g\u00fcn\u00fc.<\/span><\/h4>\n

Markab y\u0131ld\u0131z\u0131ndan <\/strong>al\u0131nan, Sextant alt 20.14\u20193 IC +2.5 ve g\u00f6z y\u00fcksekli\u011fi 9 m oldu\u011funa g\u00f6re. D\u00fczeltilmi\u015f y\u00fckseklik ( Ho ) nedir? ( ek cetvel 9 )<\/span><\/h4>\n

20.14\u20193 Sex. Alt<\/span><\/h4>\n

\u00a0\u00a0\u00a0 + 2.5 IC<\/span><\/h4>\n

20.16\u20198<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0 –\u00a0 5.3 \u00e7evren al\u00e7al\u0131m\u0131 g\u00f6z y\u00fcksekli\u011fi d\u00fczeltmesi<\/u><\/span><\/h4>\n

20.11\u20193<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0 –\u00a0 2.6 y\u0131ld\u0131zlar\u0131n g\u00f6z y\u00fcksekli\u011fi d\u00fczeltmesi<\/u><\/span><\/h4>\n

20.08\u20199Ho<\/u><\/strong><\/span><\/h4>\n

Ek cetvel 9 a gir 9m i\u00e7in d\u00fczeltme de\u011ferini bul\u00e0-5.3 bulunan de\u011feri \u00e7\u0131kar ( – )<\/span><\/h4>\n

20.11.3 ile ek cetvel 9 da y\u0131ld\u0131zlar ve gezegenler s\u00fctununda g\u00f6z y\u00fcksekli\u011fi d\u00fczeltmesine gir\u00e0 – 2.6<\/span><\/h4>\n

Bulunan de\u011feri i\u015fareti ( – ) oldu\u011fu i\u00e7in \u00e7\u0131kar. Cevap: Ho= 20.08\u20199<\/strong><\/span><\/h4>\n

Spica y\u0131ld\u0131z\u0131ndan<\/strong> al\u0131nan Sextant alt 42.38\u20196 IC: – 1.5 ve g\u00f6z y\u00fcksekli\u011fi 15 feet oldu\u011funa g\u00f6re, d\u00fczeltilmi\u015f y\u00fcksekli Ho nedir? ( ek cetvel 9 )<\/span><\/h4>\n

42.38\u20196 Sex.Alt<\/span><\/h4>\n

\u00a0\u00a0\u00a0 – 1\u20195 IC<\/span><\/h4>\n

42.37\u20191<\/u><\/span><\/h4>\n

\u00a0\u00a0 – 3\u20198 \u00e7evren al\u00e7al\u0131m\u0131 g\u00f6z y\u00fcksekli\u011fi<\/span><\/h4>\n

42.33\u20193<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 – 1\u20191 y\u0131ld\u0131zlar\u0131n g\u00f6z y\u00fcksekli\u011fi d\u00fczeltmesi.<\/span><\/h4>\n

42.32\u20192 Ho<\/u><\/strong><\/span><\/h4>\n

Ek cetvel 9 a gir 15 feet i\u00e7in g\u00f6z y\u00fcksekli\u011fi d\u00fczeltmesini bul. Bulunan de\u011fer\u00a0 ( – ) oldu\u011fu i\u00e7in \u00e7\u0131kar ve bulunan de\u011fer ile ek cetvel 9 da y\u0131ld\u0131z g\u00f6z y\u00fcksekli\u011fi d\u00fczeltmesine gir. Bulunan de\u011fer ( – ) oldu\u011fu i\u00e7in \u00e7\u0131kar. Cevap: Ho 42.32\u20192<\/span><\/h4>\n

20 a\u011fustos 1999 tarihinde Ven\u00fcs gezegeninden<\/strong> al\u0131nan Sextant Alt 32.10\u2019, IC: -1 ve g\u00f6z y\u00fcksekli\u011fi 6.0m oldu\u011funa g\u00f6re, d\u00fczeltilmi\u015f y\u00fckseklik Ho nedir? ( ek cetvel 9 )<\/span><\/h4>\n

32.10\u2019 Sex.Alt<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0 – 1 IC<\/span><\/h4>\n

32.09\u2019<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0 -4\u20193 g\u00f6z y\u00fcksekli\u011fi d\u00fczeltmesi<\/span><\/h4>\n

32.04\u20197<\/u><\/span><\/h4>\n

\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0-1.5 y\u0131ld\u0131zlar\u0131n g\u00f6z y\u00fcksekli\u011fi d\u00fczeltmesi<\/span><\/h4>\n

32.03.2 Ho<\/u><\/strong><\/span><\/h4>\n

Ek cetvel 9 a gir 6.0 m i\u00e7in g\u00f6z y\u00fcksekli\u011fi d\u00fczeltmesini bul \u2013 4.3 bulunan de\u011fer i\u015fareti ( – )oldu\u011fu i\u00e7in \u00e7\u0131kar. IC de\u011feri ( – ) oldu\u011fu i\u00e7in \u00e7\u0131kar ve elde edilen de\u011fer ile y\u0131ld\u0131zlar\u0131n g\u00f6z y\u00fcksekli\u011fi d\u00fczeltme cetveline gir. Bulunan de\u011fer i\u015fareti ( – ) oldu\u011fu i\u00e7in \u00e7\u0131kar. Cevap: 32.03\u20192 Ho<\/span><\/h4>\n

18 Haziran 1999 tarihinde Sat\u00fcrn gezegeninden <\/strong>al\u0131nan Sex.Alt 28.40 IC: + 2 ve g\u00f6z y\u00fcksekli\u011fi 5.1 m oldu\u011funa g\u00f6re d\u00fczeltilmi\u015f y\u00fckseklik Ho nedir? ( ek cetvel 9 )<\/span><\/h4>\n

28.40 Sex.Alt<\/span><\/h4>\n

\u00a0\u00a0 +\u00a0 2 IC<\/span><\/h4>\n

28.42<\/u><\/span><\/h4>\n

\u00a0\u00a0 –\u00a0\u00a0 4 g\u00f6z y\u00fcksekli\u011fi d\u00fczeltmesi<\/span><\/h4>\n

28.38<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 – 1.8 y\u0131ld\u0131zlar\u0131 g\u00f6z y\u00fcksekli\u011fi d\u00fczeltmesi<\/span><\/h4>\n

28.36.2 Ho<\/u><\/strong><\/span><\/h4>\n

Sex.Alt de\u011feri ile IC de\u011ferini topla ( + )Ek cetvel 9 a gir g\u00f6z y\u00fcksekli\u011fi 5.1 kar\u015f\u0131l\u0131\u011f\u0131 d\u00fczeltmeyi bul\u00e0 – 4 bulunan de\u011fer ile y\u0131ld\u0131zlar\u0131n g\u00f6z y\u00fcksekli\u011fi d\u00fczeltmesine gir ( ek cetvel 9 ) bulunan de\u011fer ( – ) oldu\u011fu i\u00e7in \u00e7\u0131kar. Cevap: Ho 28.36\u20192<\/span><\/h4>\n

<\/h4>\n

<\/h4>\n

<\/h4>\n

<\/h4>\n

<\/h4>\n

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\u00a0Ay\u0131n Alt kenar\u0131ndan<\/strong> al\u0131nan Sex.alt 32.20 IC: -2 ve g\u00f6z y\u00fcksekli\u011fi 8m oldu\u011funa g\u00f6re, d\u00fczeltilmi\u015f y\u00fckseklik Ho nedir? ( HP 54.4 al\u0131nacakt\u0131r.) ( ek cetvel 9 \u2013 43 )<\/span><\/h4>\n

32.20 Sex.Alt<\/span><\/h4>\n

\u00a0\u00a0\u00a0 –\u00a0 2 IC<\/span><\/h4>\n

32.18<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0 -5 g\u00f6z y\u00fcksekli\u011fi d\u00fczeltmesi ( 8 m )( ek cetvel 43 de bulunan de\u011fer )<\/span><\/h4>\n

32.13<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0 57.9 ek cetvel 43 de 32.13 de\u011fer kar\u015f\u0131l\u0131\u011f\u0131 bulunan de\u011fer<\/span><\/h4>\n

33.10.9<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1.2 ek cetvel 43 de HP 54.4 kar\u015f\u0131l\u0131\u011f\u0131 2. \u00c7izelge L s\u00fctununda bulunan de\u011fer \u00a0( +)<\/span><\/h4>\n

33.12.1 Ho de\u011feri<\/u><\/strong><\/span><\/h4>\n

\u00a0———————————————————————————————————————————-<\/span><\/h4>\n

04 May\u0131s 1999 g\u00fcn\u00fc 22 00 00 UT ( GMT ) de Ay\u0131n \u00fcst kenar\u0131ndan <\/strong>al\u0131nan Sex.Alt 34.25\u20190, IC. +1 ve g\u00f6z y\u00fcksekli\u011fi 7m oldu\u011funa g\u00f6re, d\u00fczeltilmi\u015f y\u00fckseklik Ho nedir? ( HP 54.6 al\u0131nacakt\u0131r.) ( ek cetvel 7 \u2013 43 )<\/span><\/h4>\n

34.25\u20190 Sex.Alt<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 +1\u00a0\u00a0\u00a0 IC<\/span><\/h4>\n

\u00a034.26<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0 -4.7 g\u00f6z y\u00fcksekli\u011fi d\u00fczeltmesi ( 7 m ) ( ek cetvel 43 de bulunan de\u011fer )<\/span><\/h4>\n

34.21.3<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0 56.9 ek cetvel 43 de 34.21\u20193 de\u011fer kar\u015f\u0131l\u0131\u011f\u0131 bulunan de\u011fer.<\/span><\/h4>\n

35.18\u20192<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 30\u20190 Ay\u0131n \u00fcst kenar\u0131ndan i\u015flem yap\u0131ld\u0131\u011f\u0131 i\u00e7in 30\u2019 d\u00fczeltme de\u011feri ( – )<\/span><\/h4>\n

34.48\u20192<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1.8 ek cetvel de HP 54.6 kar\u015f\u0131l\u0131\u011f\u0131 2. \u00c7izelge U s\u00fctununda bulunan de\u011fer. ( + )<\/span><\/h4>\n

34.50\u20190 Ho de\u011feri<\/u><\/strong><\/span><\/h4>\n

<\/h4>\n

<\/h4>\n

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06 May\u0131s 1999 g\u00fcn\u00fc G\u00fcne\u015f in \u00fcst kenar\u0131ndan<\/strong> al\u0131nan Sex.Alt 37.28\u2019, IC: +1 ve g\u00f6z y\u00fcksekli\u011fi 9m oldu\u011funa g\u00f6re, d\u00fczeltilmi\u015f y\u00fckseklik Ho nedir?( ek cetvel 9 )<\/span><\/h4>\n

37.28\u2019 Sex.Alt<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 +1 IC<\/span><\/h4>\n

37.29<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 – 5.3 ek cetvel 9 da bulunan 9m g\u00f6z y\u00fcksekli\u011fi d\u00fczeltmesi<\/span><\/h4>\n

37.23\u20197<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0 -17.1ek cetvel 9 da bulunan 37.23\u20197 i\u00e7in may\u0131s ay\u0131 \u00fcst kenar d\u00fczeltmesi ( – )<\/span><\/h4>\n

37.06\u20196 Ho<\/u><\/strong><\/span><\/h4>\n

Not: Cevapta 37.06\u20194 bulunmakta ek cetvel 9 da \u00fcst kenar d\u00fczeltmesi i\u00e7in ekim \u2013 mart d\u00fczeltmesinde 37.23\u20197 kar\u015f\u0131l\u0131\u011f\u0131 -17\u20193 bulunmakta. Onu referans olarak kabul etme durumunda cevap 37.06\u20194 g\u00f6z\u00fckmektedir. Fakat i\u015flem may\u0131s ay\u0131nda olmaktad\u0131r.<\/span><\/h4>\n

06 May\u0131s 1999 g\u00fcn\u00fc G\u00fcne\u015fin alt kenar\u0131ndan <\/strong>al\u0131nan sextant de\u011feri 46.12\u2019 IC:-2 ve g\u00f6z y\u00fcksekli\u011fi 6m oldu\u011funa g\u00f6re, d\u00fczeltilmi\u015f y\u00fckseklik Ho nedir? ( ek cetvel 9 )<\/span><\/h4>\n

46.12\u2019 Sex.Alt<\/span><\/h4>\n

\u00a0\u00a0 –\u00a0\u00a0 2 IC<\/span><\/h4>\n

46.10<\/u><\/span><\/h4>\n

\u00a0 \u00a0\u00a0–\u00a0 4.3 ek cetvel 9 da bulunan 6m g\u00f6z y\u00fcksekli\u011fi d\u00fczeltmesi ( – )<\/span><\/h4>\n

46.05\u20197<\/u><\/span><\/h4>\n

\u00a0\u00a0 +15\u20191 ek cetvel 9 da may\u0131s ay\u0131 i\u00e7in 46.05\u20197 kar\u015f\u0131l\u0131\u011f\u0131 bulunan de\u011fer ( + )<\/span><\/h4>\n

46.20\u20198 Ho <\/u><\/strong><\/span><\/h4>\n

08 A\u011fustos 1999 g\u00fcn\u00fc G\u00fcne\u015fin alt kenar\u0131ndan<\/strong> al\u0131nan Sex.Alt 12.20\u2019 IC: -2 ve g\u00f6z y\u00fcksekli\u011fi 9 m hava s\u0131cakl\u0131\u011f\u0131 40 derece ve hava bas\u0131nc\u0131 999 milibar oldu\u011funa g\u00f6re, d\u00fczeltilmi\u015f Ho nedir? (ek cetvel 9 \u2013 10)<\/span><\/h4>\n

12.20 Sex.Alt<\/span><\/h4>\n

\u00a0\u00a0 –\u00a0 2 IC<\/span><\/h4>\n

12.18<\/u><\/span><\/h4>\n

\u00a0\u00a0 –\u00a0\u00a0 5.3 ek cetvel 9 da bulunan 9 m g\u00f6z y\u00fcksekli\u011fi kar\u015f\u0131l\u0131\u011f\u0131 d\u00fczeltme ( – )<\/span><\/h4>\n

12.12\u20197<\/u><\/span><\/h4>\n

\u00a0 + 11\u20197 ek cetvel 9 da A\u011fustos ay\u0131\u0131nda 12.12\u20197 i\u00e7in bulunan alt kenar d\u00fczeltmesi ( + )<\/span><\/h4>\n

12.24\u20194<\/u><\/span><\/h4>\n

\u00a0+0\u20196 ek cetvel 10 da 40 derece ve 999 mb kar\u015f\u0131l\u0131\u011f\u0131 M s\u00fctunundan bulunan d\u00fczeltme (+0.6 ) G\u00f6z a\u00e7\u0131sal y\u00fckseklik ile girilecek ( 12\u201924\u20194 elimizdeki de\u011fer.)<\/span><\/h4>\n

12.24\u20194 + 0\u20196= 12.25\u20190 Ho<\/strong><\/span><\/h4>\n

02 Kas\u0131m 1999 g\u00fcn\u00fc G\u00fcne\u015fin alt kenar\u0131ndan <\/strong>al\u0131nan Sex.Alt 8,42\u2019 IC \u2013 2 g\u00f6z y\u00fcksekli\u011fi 4.2 m hava s\u0131cakl\u0131\u011f\u0131 -20 derece ve hava bas\u0131nc\u0131 980 mb oldu\u011funa g\u00f6re, d\u00fczeltilmi\u015f y\u00fckseklik Ho nedir?( ek cetvel 9 ve 10 )<\/span><\/h4>\n

8.42\u2019 SexAlt<\/span><\/h4>\n

\u00a0 –\u00a0 2\u2019 IC<\/span><\/h4>\n

8.40<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0 -3.6 ek cetvel 9 da 4.2m i\u00e7in bulunan g\u00f6z y\u00fcksekli\u011fi de\u011feri ( – )<\/span><\/h4>\n

8.36\u20194<\/u><\/span><\/h4>\n

\u00a0+10.8 ek cetvel 9 da bulunan alt kenar i\u00e7in d\u00fczeltme 8.36\u20194 kar\u015f\u0131l\u0131\u011f\u0131 yok. 9.34\u2019 kar\u015f\u0131l\u0131\u011f\u0131 al\u0131nd\u0131.<\/span><\/h4>\n

8.47\u20192<\/u><\/span><\/h4>\n

\u00a0\u00a0 – 0.8 ek cetvel 10 da bulunan -20 C 980 mb ve 8.47\u20192 g\u00f6z a\u00e7\u0131sal y\u00fcksekli\u011fi de\u011feri.<\/span><\/h4>\n

8.46\u20194 Ho<\/u><\/strong><\/span><\/h4>\n

Not cevap 8.45\u20199<\/strong> olmas\u0131 gerekir. Soru cevab\u0131 yanl\u0131\u015f \u00e7\u00fcnk\u00fc ek cetvel 9 da 8.36\u20194 de\u011ferinin kar\u015f\u0131l\u0131\u011f\u0131 yok. Soru hatal\u0131. \u00d6rnek olmas\u0131 a\u00e7\u0131s\u0131ndan yap\u0131ld\u0131.<\/span><\/h4>\n

02 May\u0131s 1999 g\u00fcn\u00fc g\u00f6z y\u00fcksekli\u011fi 18 feet olan bir seyirci 10 00 00 GMT de Ay \u0131n \u00fcst kenar y\u00fcksekli\u011fini 32.46\u20197 olarak \u00f6l\u00e7m\u00fc\u015ft\u00fcr. G\u00f6zlem an\u0131nda hava s\u0131cakl\u0131\u011f\u0131 20 C ve hava bas\u0131nc\u0131 990 mb d\u0131r. D\u00fczeltilmi\u015f sextant y\u00fcksekli\u011fi Ho nedir? ( ek cetvel 10 \u2013 39 \u2013 43 )<\/span><\/h4>\n

32.46\u20197 Ay\u0131n \u00fcst kenar\u0131<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0 -4\u20191 18 feet g\u00f6z y\u00fcksekli\u011fi kar\u015f\u0131l\u0131\u011f\u0131 bulunan d\u00fczeltme ( – )<\/span><\/h4>\n

32.42\u20196<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 +0.1 hava s\u0131cakl\u0131\u011f\u0131 20 C ve 990 mb J b\u00f6l\u00fcm\u00fcnden bulunan d\u00fczeltme ( + )<\/span><\/h4>\n

32.42\u20197<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 57\u20197 Ay\u0131n \u00fcst kenar\u0131 i\u00e7in d\u00fczeltme ek 43 32.42\u20197 kar\u015f\u0131l\u0131\u011f\u0131 ( + ) ek cetvel 43 da bulunacak.<\/span><\/h4>\n

33.40\u20194<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0 \u00a0\u00a030\u20190<\/u> Ay\u0131n \u00fcst kenar\u0131ndan yap\u0131lan i\u015flem sonucu indirilmesi gereken dakika de\u011feri<\/span><\/h4>\n

33.10\u20194 <\/u><\/span><\/h4>\n

2 May\u0131s saat 10 00 00 da Ay\u0131n HP si 54.0; HP 54.0 kar\u015f\u0131l\u0131\u011f\u0131 d\u00fczeltme 54.0\u00e0 U s\u00fctunu\u00e0 1.5<\/span><\/h4>\n

33.10\u20194<\/span><\/h4>\n

\u00a0\u00a0\u00a0 + 1.5 HP 54.0 kar\u015f\u0131l\u0131\u011f\u0131 ek cetvel 43 de U s\u00fctununda bulunan d\u00fczeltme.<\/span><\/h4>\n

33.11\u20199 Ho Cevap<\/u><\/strong><\/span><\/h4>\n

03 May\u0131s 1999 g\u00fcn\u00fc g\u00f6z y\u00fcksekli\u011fi 48 feet olan bir seyirci IC de\u011feri +2\u20197, fabrikasyon hata de\u011feri +0\u20191 olan bir sextant ile ak\u015fam rasat zaman\u0131 Ven\u00fcs y\u00fcksekli\u011fini<\/strong> 50.19\u20199 olarak \u00f6l\u00e7m\u00fc\u015ft\u00fcr. Ven\u00fcs \u00fcn d\u00fczeltilmi\u015f Ho de\u011feri nedir? ( ek cetvel 9 )<\/span><\/h4>\n

50.19\u20199 Ven\u00fcs y\u00fcksekli\u011fi<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 +2\u20197 IC de\u011feri<\/span><\/h4>\n

50.22\u20196<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 +0\u20191 fabrikasyon hata de\u011feri<\/span><\/h4>\n

50.22\u20197<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 – 6\u20197 ek cetvel 9 da 48 feet y\u00fcksekli\u011fi bulunan d\u00fczeltme de\u011feri. ( – )<\/span><\/h4>\n

50.16\u20190<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 – 0\u20198 ek cetvel 9 dan bulunan 50.16\u20190 kar\u015f\u0131l\u0131\u011f\u0131 g\u00f6z y\u00fcksekli\u011fi de\u011feri ( – )<\/span><\/h4>\n

50.15\u20192 Ho<\/u><\/strong><\/span><\/h4>\n

04 May\u0131s 1999 g\u00fcn\u00fc g\u00f6z y\u00fcksekli\u011fi 18m olan bir seyirci, G\u00fcne\u015fin \u00fcst kenar\u0131ndan<\/strong> yapt\u0131\u011f\u0131 \u00f6l\u00e7\u00fcm sonucu y\u00fcksekli\u011fi 63.41\u20194 olarak tespit etmi\u015ftir. IC de\u011feri +1\u20192, \u00f6l\u00e7\u00fcm an\u0131nda hava bas\u0131nc\u0131 970 mb ve hava s\u0131cakl\u0131\u011f\u0131 20 F<\/strong> oldu\u011funa g\u00f6re, d\u00fczeltilmi\u015f sekstant y\u00fcksekli\u011fi nedir? ( ek cetvel 9 ve 10 )<\/span><\/h4>\n

63.41\u20194 G\u00fcne\u015f \u00fcst kenar\u0131<\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0 +1\u20192 IC de\u011feri<\/span><\/h4>\n

63.42\u20196<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0 –\u00a0 7\u20195 ek cetvel 9 da bulunan 18 metre g\u00f6z y\u00fcksekli\u011fi kar\u015f\u0131l\u0131\u011f\u0131 d\u00fczeltme ( – )<\/span><\/h4>\n

63.35\u20191<\/u><\/span><\/h4>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 0\u20190 ek cetvel 10 970 mb 20 F kar\u015f\u0131l\u0131\u011f\u0131 d\u00fczeltme ( E s\u00fctunundan) ( – \/ + 0)<\/span><\/h4>\n

63.35\u20191<\/u><\/span><\/h4>\n

\u00a0\u00a0 -16\u20193 ek cetvel 9 da 63.35\u20191 kar\u015f\u0131l\u0131\u011f\u0131 bulunan \u00fcst kenar d\u00fczeltmesi ( – )<\/span><\/h4>\n

63.18\u20198 Ho<\/u><\/strong><\/span><\/h4>\n

<\/h4>\n

Yatay ve d\u00fc\u015fey sextant a\u00e7\u0131s\u0131 ile mesafe hesab\u0131.<\/strong><\/span><\/h4>\n

Yatay sextant a\u00e7\u0131s\u0131 ile form\u00fcl:<\/strong> Mesafe=( 57.3 x haritadan \u00f6l\u00e7\u00fclen geni\u015flik ) \/ Derece olarak sextant.<\/span><\/h4>\n

1.D\u00fc\u015fey sextant a\u00e7\u0131s\u0131 ile form\u00fcl:<\/strong> Mesafe de\u011feri feet olarak verilmi\u015fse;<\/span><\/h4>\n

Mesafe = ( 0.565 x haritadaki madde y\u00fcksekli\u011fi ( feet ) \/ maddenin dakika olarak sextant d\u00fc\u015fey a\u00e7\u0131s\u0131.<\/span><\/h4>\n
    \n
  1. \n

    D\u00fc\u015fey sextant a\u00e7\u0131s\u0131 ile form\u00fcl:<\/strong> Mesafe de\u011feri metre olarak verilmi\u015fse;<\/span><\/h4>\n<\/li>\n<\/ol>\n

    Mesafe= (1.854 x haritadaki madde y\u00fcksekli\u011fi ( metre ) \/ maddenin dakika olarak sextant d\u00fc\u015fey a\u00e7\u0131s\u0131.<\/span><\/h4>\n

    Bir gemi seyir an\u0131nda adan\u0131n sextant la iki ucu aras\u0131ndaki a\u00e7\u0131y\u0131 7 derece \u00f6l\u00e7m\u00fc\u015ft\u00fcr, adan\u0131n haritadan \u00f6l\u00e7\u00fclen geni\u015fli\u011fi 1\u2019.7 mildir. Geminin adadan olan mesafesi nedir?<\/span><\/h4>\n

    Form\u00fcl: Mesafe= ( 57.3 x haritadan \u00f6l\u00e7\u00fclen geni\u015flik)\/ derece olarak sextant.<\/span><\/h4>\n

    Cevap: 57.3 x 1\u2019.7=97.41 \/ 7 = 13.91 mil. Mesafe: 13.91 mil.<\/span><\/h4>\n

    Bir gemi seyir an\u0131nda adan\u0131n d\u00fc\u015fey a\u00e7\u0131s\u0131n\u0131 sextant ile 5.36\u2019 derece \u00f6l\u00e7m\u00fc\u015ft\u00fcr. Haritada ayn\u0131 tepenin y\u00fcksekli\u011fi 250 feet tir. Geminin adadan olan mesafesi nedir?<\/span><\/h4>\n

    Cevap: 1 derece sektant a\u00e7\u0131s\u0131= 60\u2019 \u2013 5 x 60= 300\u2019 \u2013 300\u2019 + 36\u2019= 336\u2019 eder.<\/span><\/h4>\n

    Form\u00fcl: Mesafe=( 0.565 x haritadaki madde y\u00fcksekli\u011fi ( feet ) \/ maddenin dk. Olarak sextant d\u00fc\u015fey a\u00e7\u0131s\u0131.<\/span><\/h4>\n

    ( 0.565 x 250 ) = 141.25 \u2013 141.25 \/ 336\u2019= 0\u2019.42 Cevap: 0\u201942 mil.<\/span><\/h4>\n

    Bir gemi seyir an\u0131nda adan\u0131n d\u00fc\u015fey a\u00e7\u0131s\u0131n\u0131 sextant ile 3.36\u2019 derece olarak \u00f6l\u00e7m\u00fc\u015ft\u00fcr. Haritada ayn\u0131 tepenin y\u00fcksekli\u011fi 300 feet tir. Geminin adadan olan mesafesi nedir?<\/span><\/h4>\n

    Cevap:\u00a0 1 derece sektant a\u00e7\u0131s\u0131= 60\u2019 \u2013 3 x 60= 180\u2019 \u2013 180\u2019 + 36\u2019= 216\u2019 eder.<\/span><\/h4>\n

    Form\u00fcl: Mesafe=( 0.565 x haritadaki madde y\u00fcksekli\u011fi ( feet ) \/ maddenin dk. Olarak sextant d\u00fc\u015fey a\u00e7\u0131s\u0131.<\/span><\/h4>\n

    ( 0.565 x 300 ) = 169.5 \u2013 169.5 \/ 216\u2019 = 0\u201978 mil.<\/span><\/h4>\n

    Bir gemi seyir an\u0131nda bir adan\u0131n iki ucu aras\u0131ndaki mesafeyi sextant ile 10 derece olarak \u00f6l\u00e7m\u00fc\u015ft\u00fcr. Adan\u0131n haritadan \u00f6l\u00e7\u00fclen geni\u015fli\u011fi 2 mildir. Geminin adadan olan mesafesi nedir?<\/span><\/h4>\n

    Form\u00fcl: Mesafe= ( 57.3 x haritadan \u00f6l\u00e7\u00fclen geni\u015flik ) \/ derece olarak sextant<\/span><\/h4>\n

    Mesafe: ( 57.3 x 2 )= 114.6 \u2013 114.6 \/ 10 = 11\u201946 mil. Cevap: 11\u201946 mil<\/span><\/h4>\n

    Bir gemi seyir an\u0131nda bir adan\u0131n iki ucu aras\u0131ndaki mesafeyi sextant ile 5 derece olarak \u00f6l\u00e7m\u00fc\u015ft\u00fcr. Adan\u0131n haritadan \u00f6l\u00e7\u00fclen geni\u015fli\u011fi 1 mildir. Geminin adadan olan mesafesi nedir?<\/span><\/h4>\n

    Form\u00fcl: Mesafe= ( 57.3 x haritadan \u00f6l\u00e7\u00fclen geni\u015flik ) \/ derece olarak sextant<\/span><\/h4>\n

    Mesafe: ( 57.3 x 1 )= 57.3 \u2013 57.3 \/ 5 = 11\u201946. Cevap: 11\u201946 mil<\/span><\/h4>\n

    Bir gemi seyir an\u0131nda adan\u0131n d\u00fc\u015fey a\u00e7\u0131s\u0131n\u0131 sextant ile 2.54\u2019 derece \u00f6l\u00e7m\u00fc\u015ft\u00fcr. Haritada ayn\u0131 tepenin y\u00fcksekli\u011fi 300 m dir. Geminin adadan olan mesafesi nedir?<\/span><\/h4>\n

    Maddenin d\u00fc\u015fey y\u00fcksekli\u011fi metre olarak verilmi\u015fse form\u00fcl \u015f\u00f6yledir.<\/span><\/h4>\n

    Mesafe= ( 1.854 x haritadaki madde y\u00fcksekli\u011fi ( metre ) \/ maddenin dakika olarak sextant d\u00fc\u015fey a\u00e7\u0131s\u0131.<\/span><\/h4>\n

    2.54\u2019= 2 x 60\u2019 = 120\u2019 \u2013 120\u2019 + 54\u2019 = 174\u2019<\/span><\/h4>\n

    Mesafe= ( 1.854 x 300 )556.2 \u2013 556.2 \/ 174 = 3\u2019196 = 3\u20192 mil eder. Cevap: 3\u20192 mil<\/span><\/h4>\n

    Seyir halinde bir geminin boyu 300 feet. Pruvadan b\u0131rak\u0131lan y\u00fczer bir cismin 10 saniye sonra pupadan ge\u00e7ti\u011fi saptanm\u0131\u015ft\u0131r. Geminin s\u00fcrati nedir?<\/span><\/h4>\n

    1 feet= 30.48 cm 300 x 30.48 = 9144 cm \/ 100 = 91.44 m.<\/span><\/h4>\n

    1 saat = 60 dakika. 60 dakika= 3600 saniye.<\/span><\/h4>\n

    1 mil= 1852.3m<\/span><\/h4>\n

    10 saniyede 91.44 metre gitti\u011fine\u00a0 g\u00f6re 3600 saniyede ne kadar yol al\u0131r.<\/span><\/h4>\n

    3600 x 91.44\/ 10= 32918.4\/ 1852.3= 17.77 Cevap: 17.77<\/span><\/h4>\n

    Bir geminin pruvas\u0131nda pupas\u0131na kadar olan mesafe 367.5 feet oldu\u011funa g\u00f6re, pruvadan b\u0131rak\u0131lan y\u00fczer bir cismin 14 saniye sonra pupadan ge\u00e7ti\u011fi saptanm\u0131\u015ft\u0131r. Geminin s\u00fcrati a\u015fa\u011f\u0131dakilerden hangisidir?<\/span><\/h4>\n

    1 feet= 30.48 cm\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1 saat = 60 dakika. 60 dakika= 3600 saniye. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1 mil= 1852.3m<\/span><\/h4>\n

    367.5 x 0.3048= 112.014 m<\/span><\/h4>\n

    14 saniyede 112.014 m ise 3600 saniye de ne kadar yol al\u0131r.<\/span><\/h4>\n

    3600 x 112.014 \/ 14 =28803.6 m. Gider. 28803.6\/ 1852.3= 15\u201955 mil Cevap: 15\u201955 mil.<\/span><\/h4>\n

    Bir filikan\u0131n s\u00fcratini belirlemek i\u00e7in filikan\u0131n ba\u015f taraf\u0131ndan y\u00fczer bir cisim denize at\u0131l\u0131yor, cisim filikan\u0131n ba\u015f ve k\u0131\u00e7 bodoslamalar\u0131ndan ge\u00e7erken zaman tespiti yap\u0131l\u0131yor ve s\u00fcre 18 saniye olarak bulunuyor. Filikan\u0131n boyu 28 kadem oldu\u011funa g\u00f6re filikan\u0131n saatteki h\u0131z\u0131 nedir?<\/span><\/h4>\n

    1 kadem=\u00a0 30.48 cm. \u00a0\u00a0\u00a0\u00a0 1 saat = 60 dakika. 60 dakika= 3600 saniye. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1 mil= 1852.3m<\/span><\/h4>\n

    28 x 0.308=8.5344<\/span><\/h4>\n

    18 saniyede 8.5344 m giderse 3600 saniyede ne kadar yol al\u0131r.<\/span><\/h4>\n

    8.5344 x 3600 \/ 18 = 1706.88 m\u00a0 gider.<\/span><\/h4>\n

    1706.88 \/ 1852.3 = 0.92 mil Cevap: 1 saatteki h\u0131z\u0131 0\u201992 mil<\/span><\/h4>\n

    <\/h4>\n

    <\/h4>\n

    10 knot ile ilerleyen bir tekne \u015famand\u0131ralara yak\u0131n seyrederken, bir \u015famand\u0131ray\u0131 g\u00f6rmesi ile kaybetmesi aras\u0131ndaki s\u00fcreyi 10 dakika olarak belirlemi\u015fse g\u00f6r\u00fc\u015f mesafesi ka\u00e7 mildir?<\/span><\/h4>\n

    Cevap: 60\u2019 da 10 mil giderse 10\u2019 da ne kadar mil g\u00f6rm\u00fc\u015ft\u00fcr.<\/span><\/h4>\n

    10 x 10 = 100 \u2013 100 \/ 60 = 1\u2019666 mil g\u00f6rm\u00fc\u015ft\u00fcr.<\/span><\/h4>\n

    1.666 \/ 2 =0\u201983. Cevap: G\u00f6r\u00fc\u015f mesafesi 0\u201983 mildir.<\/span><\/h4>\n

    Zaman hesaplar\u0131:<\/strong><\/span><\/h4>\n

    26 Ekim g\u00fcn\u00fc 40.00 N \u2013 117.19\u20194 W boylam\u0131nda, Notik almanaktan g\u00fcne\u015fin do\u011fu\u015fu 06.58 olarak bulunmu\u015ftur. An\u0131lan boylamda g\u00fcne\u015f ZT olarak saat ka\u00e7ta do\u011facakt\u0131r.<\/span><\/h4>\n

    Cevap: 120 \u2013 117.19\u20194= 2.40\u20196 = ( 6×6=36\u201d ) 2 derece 8 dakika eder.<\/span><\/h4>\n

    1 derece60mil\u00a0 ve 1 derece\u00a0 4 dakika etti\u011fine g\u00f6re 40 x 4 \/60=2.666 \u2013 2 x 60 =40 = 2\u201940\u201d<\/span><\/h4>\n

    8\u2019 + 2\u201940\u201d=10\u201940\u201d\u00a0 – 06.58\u201900\u201d \u2013 10\u201940\u201d= 06.47\u201920\u201d Cevap: 06.47\u201920\u201d<\/span><\/h4>\n

    088 W boylam\u0131nda ZT 16.55 iken LMT de\u011feri nedir?<\/span><\/h4>\n

    090 \u2013 088 = 2 x 4 = 8 dakika. 16.55 + 00.08= 17.03<\/span><\/h4>\n

    10.42\u201936\u201d E boylam\u0131ndaki bir mevki ile 36.16\u201924\u201d E boylam\u0131ndaki bir mevki aras\u0131ndaki zaman fark\u0131 nedir?<\/span><\/h4>\n

    Cevap: 36.16\u201924\u201d \u2013 10.42\u201936\u201d = 25.33\u201948\u201d ( do\u011fuya gidildi\u011fi i\u00e7in koordinat de\u011ferleri ( – )<\/span><\/h4>\n

    25 x 4 =100 \/ 60\u2019=1.666 \u2013 1.666 \u2013 1 x 60 =40\u2019 = 1 h 40m.<\/span><\/h4>\n

    1 derece 60 mil. 1 derece 4\u2019 zaman fark\u0131 oldu\u011funa g\u00f6re 33\u2019 x 4\u2019 \/ 60\u2019=2.2 –\u00a0 2.2 \u2013 2 x 60=12\u201d = 2\u201912\u201d<\/span><\/h4>\n

    1.40\u2019+2\u201912\u201d= 1.42\u201912\u201d\u00a0\u00a0\u00a0 1.42\u201912\u201d + 0.00\u201902\u201d= 1.42\u201914\u201d\u00a0\u00a0\u00a0\u00a0 Cevap: 1.42\u201914\u201d<\/strong><\/span><\/h4>\n

    60\u2019= 4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 30\u2019 = 2\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 15\u2019 = 1\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 00\u2019 = 0\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 60\u201d = 0.4\u201d<\/span><\/h4>\n

    48\u201d x 0.4\u201d \/ 60\u201d=0.32 \u2013 ( 0.32 \u2013 0 x 60\u201d= 19\u20192 = 2\u201d ) 2 saniye.<\/span><\/h4>\n

    20.15\u201936\u201d boylam\u0131nda olan bir mevki ile 42.28\u201918\u201d W boylam\u0131nda olan bir mevki aras\u0131ndaki zaman fark\u0131 nedir?<\/span><\/h4>\n

    Cevap: 42.28\u201918\u201d + 20.15\u201936\u201d=62.43\u201954\u201d ( do\u011fudan bat\u0131ya gidildi\u011fi i\u00e7in koordinat de\u011ferleri + )<\/span><\/h4>\n

    \u00a062 x 4\u2019 = 4. 133 \u2013 ( – 4 x 60=8 ) = 4h 08m.<\/span><\/h4>\n

    1 derece 60 mil. 1 derece 4\u2019 zaman fark\u0131 oldu\u011funa g\u00f6re 43\u2019 x 4\u2019 \/ 60=2.866 \u2013 ( -2.866 x 60 = 52\u201d) =2\u201952\u201d 4h 08m + 2m 52s= 4h 10m 52s +0.00.02 =\u00a0 4h.10m54s Cevap: 4h 10\u201956\u201d<\/strong><\/span><\/h4>\n

    60\u2019= 4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 30\u2019 = 2\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 15\u2019 = 1\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 00\u2019 = 0\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 60\u201d = 0.4\u201d<\/span><\/h4>\n

    54\u201d x 0.4\u201d\/ 60\u201d=0.36\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ( 0.36 -0 x60 =2\u201d) veya 1\u2019 x 2\u2019 \/ 30\u2019= 0.066 \u2013 0 x 60= 4\u201d<\/strong><\/span><\/h4>\n

    05 May\u0131s 1999 tarihinde Greenwich hakiki zaman\u0131 GAT 19.34\u201920 ise UT (GMT) nedir?(ek cetvel 38)<\/span><\/h4>\n

    Ek cetvel 38 gir sa\u011f alt k\u00f6\u015fede sol s\u00fctunda zaman denklemi b\u00f6l\u00fcm\u00fcnden 5 may\u0131s i\u00e7in de\u011feri bul\u00e003.15<\/span><\/h4>\n

    Bulunan de\u011feri eldeki GAT zaman\u0131 ile topla ( + ) 19.34\u201920 + 3\u201915= 19.37\u201935 Cevap: 19.37\u201935<\/strong><\/span><\/h4>\n

    05 May\u0131s 1999 g\u00fcn\u00fc saat 10.24\u2019 UT ( GMT ) de, Greenwich hakiki zaman\u0131 nedir?( ek cetvel 38 )<\/span><\/h4>\n

    Ek cetvel 38 e gir sa\u011f alt k\u00f6\u015fede zaman denklemi b\u00f6l\u00fcm\u00fcnden sa\u011f s\u00fctunda 05 may\u0131s i\u00e7in de\u011feri bul\u00e003.17<\/span><\/h4>\n

    Bulunan de\u011feri eldeki Greenwich zaman\u0131ndan \u00e7\u0131kar. ( – ) 10.24\u2019 \u2013 03\u201917\u201d= 10.20\u201943\u201d Cevap:10.20\u201943\u201d<\/strong><\/span><\/h4>\n

    115 W mevkiinde g\u00fcne\u015fin bat\u0131\u015f\u0131 LMT olarak 16.48\u201900\u201d bulunmu\u015ftur. G\u00fcne\u015f bat\u0131\u015f zaman\u0131 Zt olarak ka\u00e7t\u0131r?<\/span><\/h4>\n

    Cevap: 120 \u2013 115 = 005\u00a0 5 x 4=20\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 16.48\u201900\u201d \u2013 28\u2019 = 16.28\u201900\u201d Cevap:16.28\u201900\u201d<\/strong><\/span><\/h4>\n

    Boylam\u0131 060.12\u2019 W olan bir mevkide LMT 08h21m37s iken boylam\u0131 018.03\u2019 olan bir noktada LMT ka\u00e7t\u0131r?<\/span><\/h4>\n

    Cevap: 060.12\u2019 + 018.03\u2019= 78.15\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 78 x 4 \/ 60 =5.2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -5 x 60= 12\u2019 = 5h 12m<\/span><\/h4>\n

    15\u2019 x 4\u2019\/ 60\u2019= 1\u2019 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5h 12m + 1m= 5h 13m Cevap: 08h 21m 37s + 5h 13m= 13h 34m 37s <\/strong><\/span><\/h4>\n

    15 Aral\u0131k g\u00fcn\u00fc g\u00fcne\u015f 36.27\u2019 N \u2013 072.30\u2019 E mevkiinde GMT olarak 20.30 da batm\u0131\u015ft\u0131r. LMT olarak g\u00fcne\u015fin bat\u0131\u015f saati nedir?<\/span><\/h4>\n

    \u00c7\u00f6z\u00fcm: 72 x 4 =288\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 288\/60=4.8\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -4 x 60=48\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4h 48m<\/span><\/h4>\n

    30\u2019 x \u20184 \/ 60\u2019=2\u2019 4h 48m + 2m= 4h 50m\u00a0\u00a0 20h 30m + 4h 50m= 01h 20m ( 16 aral\u0131k )<\/strong><\/span><\/h4>\n

    Seyir halindeki bir gemiden 26 may\u0131s g\u00fcn\u00fc ZT 16h34m14s de g\u00fcne\u015ften rasat yap\u0131lm\u0131\u015ft\u0131r. DR boylam\u0131m\u0131z 156.19\u20195 E dir. Rasat yap\u0131ld\u0131\u011f\u0131 andaki GMT nedir?<\/span><\/h4>\n

    156 x 4 \/ 60=10.4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 veya 156 \/ 15=10.4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 16h 34m 14s \u2013 10h= 06h 34m 14s<\/strong><\/span><\/h4>\n\n\n\n\n
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    \u00a0<\/span><\/h4>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    \u00a0<\/strong><\/span><\/h4>\n

    \u00a0<\/strong><\/span><\/h4>\n

    7,5\u00a0\u00a0\u00a0\u00a0 0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 7,5\u00a0\u00a0\u00a0\u00a0\u00a0 22.5\u00a0\u00a0\u00a0 37.5\u00a0\u00a0\u00a0 52.5\u00a0\u00a0\u00a0 67.5\u00a0\u00a0\u00a0 82.5\u00a0\u00a0\u00a0\u00a0 97.5\u00a0\u00a0 112.5\u00a0\u00a0 127.5\u00a0 142.5\u00a0\u00a0 157.5\u00a0 172.5<\/strong><\/span><\/h4>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0– 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 – 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 – 3\u00a0\u00a0\u00a0\u00a0\u00a0 – 4\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0– 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 – 6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -7\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 – 8\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0– 9\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0-10\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0-11<\/strong><\/span><\/h4>\n

    <\/h4>\n

    Seyir durumundaki bir gemi 26 Nisan g\u00fcn\u00fc Zone time olarak 16h 36m 14s de g\u00fcne\u015ften rasat yapm\u0131\u015ft\u0131r. Rasat an\u0131ndaki DR mevkiin boylam\u0131 159.19\u20195 E dir. Rasat yap\u0131ld\u0131\u011f\u0131 anda GMT zaman\u0131 nedir.<\/span><\/h4>\n

    Cevap: 16h 34m 14s \u2013 11h= 05h 36m 14s<\/strong><\/span><\/h4>\n

    38.41\u2019 N 125.20\u2019 W olan bir A mevkiinde 21 A\u011fustos 2003 tarihinde vardiya saati 13.15 ise 55.17\u2019N 082.45\u2019 E mevkiinde bulunan bir g\u00f6zlemci i\u00e7in vardiya saati ka\u00e7t\u0131r?<\/span><\/h4>\n

    <\/h4>\n

    <\/h4>\n

    +127.5+112,5 +97.5\u00a0 +82.5\u00a0 +67.5\u00a0 +52.5\u00a0\u00a0\u00a0 +37.5\u00a0 +22.5\u00a0\u00a0 7.5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 0\u00a0\u00a0\u00a0 -7.5\u00a0\u00a0\u00a0 \u00a0-22.5\u00a0\u00a0 \u00a0-37.5\u00a0\u00a0 \u00a0-52.5\u00a0 \u00a0-67.5<\/span><\/h4>\n

    \u00a0\u00a0\u00a0\u00a0 +8\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 +7\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 +6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 +5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 +4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 +3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 +2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 +1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/h4>\n

    \u00a0<\/u><\/span><\/h4>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -82.5\u00a0\u00a0\u00a0\u00a0 -97.5<\/span><\/h4>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -6<\/span><\/h4>\n

    Aradaki zaman fark\u0131 14 saat oldu\u011funa g\u00f6re 13.114=27.15 27.15 \u2013 24.00= 03.15 22 A\u011fustos.<\/span><\/h4>\n

    \"12321\"<\/a><\/span><\/h4>\n

    <\/h4>\n

    DO\u011eU YARIM K\u00dcRES\u0130<\/span><\/h4>\n

    (ZD) Saat Fark\u0131:-l -2 -3 -4 -5 -6 -7 -8 -9 -10 -11 -12<\/span><\/p>\n

    B\u00f6lge Harfi :ABCDEFGHIKL M<\/span><\/p>\n

    Standart Zaman<\/span><\/p>\n

    Boylam\u0131 : 15\u00b0E 30\u00b0E 45\u00b0E 60\u00b0E 75\u00b0E 90\u00b0E 105\u00b0E 120\u00b0E 135\u00b0E 150\u00b0E 165\u00b0E 172\u00b0,5 E -180\u00b0<\/span><\/p>\n

    V<\/span><\/p>\n

    ts)<\/span><\/p>\n

    oo<\/span><\/p>\n

    BATI YARIM K\u00dcRES\u0130<\/span><\/p>\n

    Saat Fark\u0131 :+1 +2 +3+4+5 +6 +7 +8 +9 +10 +11 +12<\/span><\/p>\n

    B\u00f6lge Fark\u0131 :NOPQRSTUVWX Y<\/span><\/p>\n

    Standart Zaman<\/span><\/p>\n

    Boylam\u0131 : 15\u00b0W 30\u00b0W 45\u00b0W 60\u00b0W 75\u00b0W 90\u00b0W 105\u00b0W 120\u00b0W 135\u00b0W 150\u00b0W 165\u00b0W 172\u00b0,5 W-180\u00b0<\/span><\/p>\n

    Greenwich S\u0131f\u0131r B\u00f6lgesi Z<\/span><\/p>\n

    B\u00f6lgesel saatler belirtilirken b\u00f6lge (ZONE) harfi vaktin sonuna konmal\u0131da-. \u00d6rne\u011fin 63\u00b0 35′ E boylam\u0131ndaki b\u00f6lge saatine g\u00f6re saat 15h<\/sup> 30m<\/sup> 20S <\/sup>belirtmek i\u00e7in 15h<\/sup> 3GP <\/em>20S<\/sup> D olarak yaz\u0131lmal\u0131d\u0131r. Burada kullan\u0131lan D harfi, belirtilen vaktin hangi b\u00f6lge saatine g\u00f6re oldu\u011funu g\u00f6sterdi\u011fi gibi, GMT bulmak i\u00e7in saat fark\u0131n\u0131n (Zone Discription’\u0131n) ZD-4 oldu\u011funuda belirtir.<\/span><\/p>\n

    (ZONE TIME) ZT 18h<\/sup>35m<\/sup>42s<\/sup>D ZT 15h<\/sup> 42m<\/sup> 30S<\/sup> Q<\/span><\/p>\n

    B\u00f6lgesel Saatin ZD -_4 ZD +4 <\/span><\/p>\n

    GMT D\u00f6n\u00fc\u015ft\u00fcr\u00fclmesi: GMT 14h<\/sup> 35m<\/sup> 42S<\/sup> GMT19h<\/sup>42m<\/sup>30s<\/sup><\/span><\/p>\n

    x<\/span><\/p>\n

    \"54545\"<\/a><\/span><\/h4>\n

    Yay par\u00e7as\u0131n\u0131 veya boylam\u0131 zamana de\u011fi\u015ftirmek i\u00e7in kural;<\/strong> derece, dakika. ( x ) ve saniyeleri 15 e b\u00f6l \u00e7\u0131kan de\u011feri ( – ) yani saat, dakika veya saniyeyi \u00e7\u0131kar kalan k\u00fcsurat\u0131 60 ile \u00e7arp.<\/span><\/h4>\n

    \u00d6rnek: 68 derece 52\u2019 35\u201d<\/span><\/h4>\n

    68\/15=4.533 saat\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -4 x 60=32\u2019<\/span><\/h4>\n

    52\u2019\/15=3.466 dakika\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -3 x 60=28\u201d<\/span><\/h4>\n

    35\u201d\/15=2.333 saniye\u00a0\u00a0\u00a0\u00a0\u00a0 -2 x 60=20 salise\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 bulunan de\u011ferleri topla.<\/span><\/h4>\n

    4h 32m<\/span><\/h4>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3m 28s<\/span><\/h4>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2s20salise<\/span><\/h4>\n

    +———————–<\/span><\/h4>\n

    4h35m30s2salise eder.<\/span><\/h4>\n

    Zaman\u0131 boylama veya a\u00e7\u0131ya de\u011fi\u015ftirmek i\u00e7in kural;<\/strong> Saatleri 15 ile \u00e7arp ( x ), dakika ve saniyeleri 4 e b\u00f6l ( \/ ) \u00e7\u0131kan de\u011feri\u00a0 ( – ) yani dakika ve saniyeyi \u00e7\u0131kar kalan k\u00fcsurat\u0131 60 ile \u00e7arp.<\/span><\/h4>\n

    \u00d6rnek:<\/span><\/h4>\n

    8h58m27s<\/span><\/h4>\n

    8×15=120 derece<\/span><\/h4>\n

    58\/4=14.5 derece\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -14×60=30\u2019 dakika<\/span><\/h4>\n

    27\/4=6.75 dakika\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -6×60=45\u201d saniye. Bulunan de\u011ferleri topla<\/span><\/h4>\n

    120 derece<\/span><\/h4>\n

    \u00a0\u00a0 14.30\u2019<\/span><\/h4>\n

    \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 6\u201945\u201d<\/span><\/h4>\n

    +—————-<\/span><\/h4>\n

    134.36\u201945\u201d bulunur.<\/span><\/h4>\n

    GMT=ZD \u2013 ZT d\u0131r.\u00a0\u00a0 MERKEZ BOYLAM = ZD x 15\u00b0<\/span><\/h4>\n

    ZD rakam\u0131n\u0131 bilmek, seyirciye; bulundu\u011fu b\u00f6lgenin, b\u00f6lge zaman\u0131n\u0131 (ZT) bulmas\u0131nda<\/span><\/h4>\n

    yard\u0131mc\u0131 olur.<\/span><\/h4>\n

    \u00d6RNEK : <\/strong>Boylam\u0131: 127\u00b0 13\u20192 W<\/span><\/h4>\n

    Boylam\u0131: 127\u00b0 42\u20194 W olan mevkilerin b\u00f6lge tan\u0131tma (ZD) rakamlar\u0131 nedir ?<\/span><\/h4>\n

    \u00c7\u00f6z\u00fcm: 127\u00b013\u20192\/15 = 8 kalan 7\u00b0,5 den k\u00fc\u00e7\u00fck (7\u00b013\u20192) bu durumda ZD = +8 dir.<\/span><\/h4>\n

    127\u00b042\u20192\/15 = 8 kalan 7\u00b0,5 den b\u00fcy\u00fck (7\u00b042\u20194) bu durumda ZD = +9 olur.<\/span><\/h4>\n

    GMT ile ZT aras\u0131ndaki ili\u015fki<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/span><\/h4>\n

    Astronomik Seyirde g\u00f6k cisimlerinin koordinatlar\u0131n\u0131 bulurken gereksinme duydu\u011fumuz<\/span><\/h4>\n

    zaman ZT den \u00e7ok GMT dir. Zira almanakta verilen bilgiler GMT zaman\u0131na g\u00f6redir.<\/span><\/h4>\n

    \u00d6RNEK<\/strong><\/span><\/h4>\n
      \n
    1. \n

      B\u00d6LGE ZAMANI (ZT) VE YEREL ZAMAN (LMT) ARASINDAK\u0130 \u0130L\u0130\u015eK\u0130 :<\/strong><\/span><\/h4>\n<\/li>\n<\/ol>\n

      LMT; b\u00f6lge zaman\u0131ndan, b\u00f6lgenin merkez boylam\u0131ndan itibaren, do\u011fuya veya bat\u0131ya<\/span><\/h4>\n

      do\u011fru olan boylam fark\u0131n\u0131n (dlo) zaman cinsinden de\u011feri kadar bir farkl\u0131l\u0131k g\u00f6sterir. Notik<\/span><\/h4>\n

      almanaklarda b\u00fct\u00fcn g\u00f6ksel olaylar (alaca karanl\u0131k, do\u011fu\u015f ve bat\u0131\u015f ve boylam ge\u00e7i\u015f zamanlar\u0131)<\/span><\/h4>\n

      orta boylama g\u00f6re ve LMT zaman\u0131 olarak verilmi\u015ftir. E\u011fer seyirci merkez boylam\u0131n do\u011fusunda<\/span><\/h4>\n

      ise, g\u00f6ksel olaylar merkez boylam\u0131ndan \u00f6nce olu\u015fur. Bu hususun hi\u00e7bir zaman unutulmamas\u0131<\/span><\/h4>\n

      gerekir. G\u00f6ksel olay\u0131n ZT olarak hesaplanabilmesi i\u00e7in almanaktan bulunan LMT de\u011ferine<\/span><\/h4>\n

      boylam fark\u0131n\u0131n zaman de\u011feri ilave edilir veya \u00e7\u0131kart\u0131l\u0131r. (Merkez boylam\u0131n do\u011fusunda ise<\/span><\/h4>\n

      \u00e7\u0131kart\u0131l\u0131r, Bat\u0131s\u0131nda ise toplan\u0131r). A\u015fa\u011f\u0131daki \u00f6rnekte G\u00fcne\u015f\u2019in do\u011fu\u015f zaman\u0131n\u0131n ZT olarak<\/span><\/h4>\n

      bulunu\u015fu g\u00f6sterilmi\u015ftir.<\/span><\/h4>\n

      \u00a0<\/strong><\/span><\/h4>\n

      B\u00fcy\u00fck daire seyri;<\/strong><\/span><\/h4>\n

      Ba\u015flang\u0131\u00e7 rota a\u00e7\u0131s\u0131 38 derece ba\u015flang\u0131\u00e7 noktas\u0131 koordinatlar\u0131 47 derece N \u2013 050 derece W olan bir b\u00fcy\u00fck daire seyri i\u00e7in vertex noktas\u0131na mesafe nedir? ( ek cetvel 26 )<\/span><\/h4>\n

      Cevap: LHA 38 olur. Ek cetvel 26 ya gir. 47 derece s\u00fctunundan en b\u00fcy\u00fck Hc de\u011feri al\u00e0 s\u00fctundan al\u0131nmas\u0131 gereken bilgiler. Dec \u2013 Hc \u2013 d \u2013 Z<\/span><\/h4>\n

      Dec\u00e054\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 en b\u00fcy\u00fck Hc\u00e0 65.10\u20193\u00a0 d -1\u20199\u00a0\u00a0\u00a0 Z 59.1\u2019<\/span><\/h4>\n

      Mesafeyi bulmak i\u00e7in; 90 \u2013 Dec<\/span><\/h4>\n

      Vertex ba\u015flang\u0131\u00e7 noktas\u0131na olan mesafeyi bulmak i\u00e7in 90 \u2013 54=36 derece bulunan dereceyi dakikaya \u00e7evir. 36 x 60\u2019=2160 mil eder. Cevap.2160 mil<\/strong><\/span><\/h4>\n

      Kalk\u0131\u015f noktas\u0131 51.N \u2013 065 W var\u0131\u015f noktas\u0131 40.37\u2019 N \u2013 010.00\u2019W olan bir b\u00fcy\u00fck daire seyri i\u00e7in kalk\u0131\u015f rota a\u00e7\u0131s\u0131n\u0131 ve mesafeyi bulmak i\u00e7in NP 401 in ilgi sayfas\u0131na girilerek a\u015fa\u011f\u0131daki de\u011ferler elde edilmi\u015ftir.<\/span><\/h4>\n

      ( ek cetvel 28 )<\/span><\/h4>\n

      Hc\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Z<\/u><\/strong><\/span><\/h4>\n

      48.32\u20196\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 +28.6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 81.2<\/span><\/h4>\n

      Ekte bulunan interpolation cetvelini kullanarak mesafeyi bulun.<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm; Enterpolasyon cetveline d de\u011feri olan 28.6 de\u011feri ile girilir.<\/span><\/h4>\n

      Dec.Inc\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 20\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Decimals\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u2026\u20268\u2019<\/u><\/strong><\/span><\/h4>\n

      28.6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 9.5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 .6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4.1<\/span><\/h4>\n

      Sonra double second corr. S\u00fctunundan 28.6 i\u00e7in d\u00fczeltme bulunur. ( 28.0 \u2013 29.6 aras\u0131nda ) 1.8<\/span><\/h4>\n

      Sonra enterplasyon tablosundan bulunan b\u00fct\u00fcn de\u011ferleri topla.<\/span><\/h4>\n

      TTL= 9.5 + 4.1 + 1.8=15.4 ve mesafeyi hesaplamaya ge\u00e7.<\/span><\/h4>\n

      Distance: 90 \u2013 Hc + TTL\u00a0 Distance= 90 \u2013 48.32\u20196 = 41.27\u20194 + 15.4= 41.42\u20198=2502.8 mil<\/strong> bulunur.<\/span><\/h4>\n

      \u00a0Kalk\u0131\u015f noktas\u0131 51.N \u2013 065 W, var\u0131\u015f noktas\u0131 61 N \u2013 005 W olan bir b\u00fcy\u00fck daire seyri i\u00e7in kalk\u0131\u015f rota a\u00e7\u0131s\u0131 ve mesafe a\u015fa\u011f\u0131dakilerden hangisidir? ( problem ekte bulunan NP 401 sayfas\u0131 kullan\u0131larak \u00e7\u00f6z\u00fclecek.)<\/span><\/h4>\n

      ( ek cetvel 27 )<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: LHA= dlong\u00e065W -5W = 60 ek cetvel 51 derece s\u00fctunundan Lat2= dec 61 dereceye gir.<\/span><\/h4>\n

      51 derece<\/u><\/strong><\/span><\/h4>\n

      Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hc\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Z<\/u><\/strong><\/span><\/h4>\n

      61\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 56.19\u20199\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 10\u20192\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 49.2<\/span><\/h4>\n

      Kalk\u0131\u015f rota a\u00e7\u0131s\u0131= N 49.2 E= 49.12\u2019<\/strong><\/span><\/h4>\n

      Mesafeyi bulmak i\u00e7in; Distance= 90 \u2013 Hc x 60\u00a0\u00a0\u00a0\u00a0 90 \u2013 56.19\u20199= 33.40\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 33 x 60=1980 + 40=2020 mil<\/strong><\/span><\/h4>\n

      Kalk\u0131\u015f noktas\u0131 46 derece N \u2013 065 W var\u0131\u015f noktas\u0131 50 derece N \u2013 027 W olan bir b\u00fcy\u00fck daire seyri i\u00e7in kalk\u0131\u015f rota a\u00e7\u0131s\u0131 ve mesafeyi bulun. ( problem ekte bulunan NP 401 sayfas\u0131 kullan\u0131larak \u00e7\u00f6z\u00fclecek. Ek cetvel 26 )<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: LHA y\u0131 bulmak i\u00e7in dlong bulunur. \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dlong= 65W \u2013 27W= 38 derece bulunur.<\/span><\/h4>\n

      Ek cetvel LHA 38 derece ile girilir ve Dec s\u00fctunundan 50 derece nin kar\u015f\u0131l\u0131\u011f\u0131 ve46 derece s\u00fctunundaki de\u011ferler \u00a0bulunur. Lat 2 = Dec.<\/span><\/h4>\n

      46 derece s\u00fctunu<\/u><\/strong><\/span><\/h4>\n

      Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hc\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Z\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/u><\/strong><\/span><\/h4>\n

      50\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 64.32\u20196\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 +4\u20199\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 67.0\u2019<\/span><\/h4>\n

      Kalk\u0131\u015f rotas\u0131= Z de\u011feri 67.0\u2019 i\u015fareti ise N\u00a0 ve E dir. Kalk\u0131\u015f rotas\u0131 N 67 E<\/strong><\/span><\/h4>\n

      Mesafe\u00e0 Dist= 90 \u2013 Hc \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 90 \u2013 64.32\u20196= 25.27\u20194\u00a0\u00a0\u00a0\u00a0 25 x 60=1500 + 27=1527 mil bulunur.<\/strong><\/span><\/h4>\n

      <\/h4>\n

      <\/h4>\n

      <\/h4>\n

      <\/h4>\n

      <\/h4>\n

      Kalk\u0131\u015f noktas\u0131 48 N \u2013 050 W ve kalk\u0131\u015f rota a\u00e7\u0131s\u0131 N60 E olan bir b\u00fcy\u00fck daire seyri i\u00e7in vertex noktas\u0131n\u0131n koordinatlar\u0131 nedir? ( problem ekte verilen NP 401 sayfas\u0131 kullan\u0131larak \u00e7\u00f6z\u00fclecek.) ( ek cetvel 27 )<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: LHA 60 olur. Ek cetvel 27 de 48 derece s\u00fctunundan gir ve en b\u00fcy\u00fck Hc yi bul.<\/span><\/h4>\n

      48 derece<\/u><\/strong><\/span><\/h4>\n

      Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hc\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Z<\/u><\/strong><\/span><\/h4>\n

      66\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 54.35\u20191\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -1\u20191\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 37.4\u2019 bulunur.<\/span><\/h4>\n

      Vertex noktas\u0131n\u0131n koordinatlar\u0131 Lat= Hc ve Long = Long1 \u2013 Z de\u011feridir. Bu durumda \u00e7\u00f6z\u00fcm.<\/span><\/h4>\n

      Lat. 54.35\u20191N ( yani Hc de\u011feri )\u00a0 Long = 050W \u2013 37\u20194Z= 12.36\u2019 W<\/span><\/h4>\n

      Cevap: Lat: 54.35\u20191N\u00a0\u00a0\u00a0\u00a0\u00a0 Long:12.36\u2019W<\/span><\/h4>\n

      Kalk\u0131\u015f noktas\u0131 koordinatlar\u0131 45 derece N \u2013 060 W, vertex noktas\u0131 koordinatlar\u0131 52.12\u20197 N \u2013 023 18 W, kalk\u0131\u015f rota a\u00e7\u0131s\u0131 60 derece olan bir b\u00fcy\u00fck daire seyrinde, vertex noktas\u0131ndan sonraki ilk ilk ara noktan\u0131n koordinatlar\u0131 nedir? ( Ara noktalar yakla\u015f\u0131k 5 derece dlong fark\u0131 ile al\u0131nacak ve ekteki NP 401 sayfas\u0131 kullan\u0131larak problem \u00e7\u00f6z\u00fclecektir.)( ek cetvel 27 )<\/span><\/h4>\n

      LHA= 60 ek cetvel ile 45 derece s\u00fctununa gir. Vertex noktas\u0131 kar\u015f\u0131l\u0131\u011f\u0131 Dec de\u011ferinden 5 derece \u00f6nceki Hc de\u011ferini bul.<\/span><\/h4>\n

      45 derece<\/u><\/strong><\/span><\/h4>\n

      Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hc\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Z<\/u><\/strong><\/span><\/h4>\n

      60\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 52.06\u20194\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -3\u20199\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 44.8<\/u><\/span><\/h4>\n

      65\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 52.12\u20197\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -2\u20198\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 36.7<\/u><\/span><\/h4>\n

      Vertex noktas\u0131ndan sonraki ilk ara nokta koordinatlar\u0131 Dec 60 de\u011ferinde g\u00f6r\u00fclen Hc de\u011feridir.<\/span><\/h4>\n

      Lat= Hc long= Long1 \u2013 Z. Long= 60 \u2013 44\u20198= 15.12\u2019W dir\u00a0\u00a0 Cevap: 52.06\u20194 N \u2013 15\u201912 W<\/strong><\/span><\/h4>\n

      Enlem \u2013 Boylam ve mesafe hesaplamalar\u0131<\/strong><\/span><\/h4>\n

      \u00a0<\/strong>40\u201900\u2019 N \u2013 026 W mevkii ile 35\u201940 N \u2013 023\u201920 W mevkii aras\u0131ndaki deprature ka\u00e7 mildir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: Dist= dlong x Cos MLat.<\/span><\/h4>\n

      Dlong= 026.30\u2019 W \u2013 023.20\u2019 W =\u00a0 03.10\u2019 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3 x 60\u2019=180\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 180\u2019 + 10\u2019 = 190\u2019<\/span><\/h4>\n

      MLat= 40.00\u2019 N + 35.40\u2019 N = 75.40\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 75.40\u2019 \/ 2 = 37.50\u2019<\/span><\/h4>\n

      Dist= 190 x Cos37.50\u2019= 150\u201973 mil<\/strong> eder.<\/span><\/h4>\n

      <\/h4>\n

      <\/h4>\n

      \u00a042.10\u2019 N \u2013 036.20\u2019 E mevkii ile 46.20\u2019 N \u2013 033.00\u2019 E mevkii aras\u0131ndaki departure ka\u00e7 mildir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: Dist= dlong x Cos MLat.<\/span><\/h4>\n

      Dlong= 036.20\u2019 E \u2013 033.00\u2019 E = 03.20\u2019\u00a0\u00a0\u00a0\u00a0\u00a0 03 x 60\u2019= 180\u2019 + 20 = 200\u2019<\/span><\/h4>\n

      MLat= 42.10\u2019 N + 46.20\u2019 N = 88.30\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 88.30\u2019 \/ 2 = 44.15\u2019<\/span><\/h4>\n

      Dist= 200 x Cos44.15 = 143.5 mil<\/strong><\/span><\/h4>\n

      36.20\u2019 N \u2013 040.20 E mevkii ile 36.20\u2019 N \u2013 044.40\u2019 E mevkii aras\u0131ndaki mesafe yakla\u015f\u0131k ka\u00e7 mildir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: Dist= dlong x cosMlat<\/span><\/h4>\n

      Dlong= 044.40\u2019 \u2013 040.20\u2019 = 4.20\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 4 x 60\u2019= 240\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 240\u2019 + 20\u2019 = 260\u2019<\/span><\/h4>\n

      Mlat= 36.20\u2019 + 36.20\u2019 = 72.40\u2019\u00a0\u00a0\u00a0 72.40\u2019 \/ 2 = 36.20\u2019<\/span><\/h4>\n

      Dist= 260\u2019 x Cos36.20\u2019= 209\u20198 mil<\/strong><\/span><\/h4>\n

      120 rotas\u0131na 100 mil seyrederek 40.0 N \u2013 020.0\u2019 mevkiine demirleyen bir geminin seyre kalkt\u0131\u011f\u0131 mevkiinin enlemi nedir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: 180 -120 = 60<\/span><\/h4>\n

      Dlat= Dist x Cos60\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 100\u2019 x Cos60 = 50\u2019<\/span><\/h4>\n

      Kalk\u0131\u015f enlemimiz Lat A kuzeyden g\u00fcney istikametine oldu\u011fu i\u00e7in 40.50\u2019 N olur. Cevap: 40.50\u2019 N<\/strong><\/span><\/h4>\n

      \u00a0G\u00f6ksel seyir<\/strong><\/span><\/h4>\n

      ZD +4 ( W )\u00a0 olan bir saat diliminde ve 32 enlemindeki bir mevkide ZT 04 00 00 iken LMT 04 30 00 ise be mevkiin boylam\u0131 nedir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: GMT= ZT – \/ + ZD = 04.00.00 + 4= 08 00 00 GMT:08.00.00<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 LMT= GMT \u2013 Long W 04.30.00= 08.00.00 \u2013 Long W\u00e0 Long W= 03.30.00\u00a0 <\/strong><\/span><\/h4>\n

      03.30= ( 3 x 15 ) + ( 0,5 x 55)<\/span><\/h4>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 45\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 +\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 7.5 = 52.5 = 52.30\u2019<\/strong> bulunur.<\/span><\/h4>\n

      5 derece E boylam\u0131 \u00fczerinde bir mevkide ZT 08.15 iken GMT nedir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: GMT= ZT \u2013 ZD = 08.15 \u2013 0 = 08.15 GMT<\/strong><\/span><\/h4>\n

      ZD \u2013 4 olan bir saat diliminde ve 32 N enlemindeki bir mevkide ZT 14. 00. 00 iken LMT 14.15.00 ise, bu mevkiin boylam\u0131 nedir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: GMT= 14.00.00 \u2013 4 = 10.00.00 GMT<\/strong> bulunur.<\/span><\/h4>\n

      LMT= GMT \u2013 Long E 14.15.00 = 10.00.00 \u2013 Long E \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Long E = – 04.15 saat.<\/span><\/h4>\n

      04.15= ( 4 x 15 ) + ( 0,5 x 15 )\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 60 + 7.5= 67.5 = 67.30\u2019 Cevap: 67.30\u2019<\/strong><\/span><\/h4>\n

      115 W mevkiinde g\u00fcne\u015fin bat\u0131\u015f\u0131 LMT olarak 16.48.00 bulunmu\u015ftur. G\u00fcne\u015fin bat\u0131\u015f zaman\u0131 ZT olarak ka\u00e7t\u0131r?<\/span><\/h4>\n

      LMT= GMT + Long W\u00e0 16.48= GMT + 115W \/ 15)\u00e0 GMT=24.28 bulunur.<\/span><\/h4>\n

      GMT= ZT – \/ + ZD\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 24.28 \u2013 8= 16.28\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ZT= 16.28<\/strong> bulunur.<\/span><\/h4>\n

      Veya 120 \u2013 115 =5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5 x 4= 20\u2019\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 16.48\u2019- 20\u2019= 16.28 ZT<\/strong><\/span><\/h4>\n

      Boylam 060.12\u2019 W olan bir mevide LMT 08.21.37 iken boylam\u0131 018.03 E olan bir noktada LMT ka\u00e7t\u0131r?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: boylamlar\u0131 topla 4\u2019 ile \u00e7arp 60 a b\u00f6l 60 78 x4 = 5.2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -5 x 60= 12 =\u00a0 5h12m<\/strong><\/span><\/h4>\n

      Boylamdaki dakikalar\u0131 topla 12\u2019 + 03\u2019 = 15\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 15 X 4 =60\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 60 \/ 60=1\u2019 dakika.<\/span><\/h4>\n

      5h 12m + 1m=5h 13m\u00a0\u00a0\u00a0 W den E e giderken toplan\u0131r\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 08 h 21m 37s + 5h 13m= 13h 34m 37s<\/strong> bulunur.<\/span><\/h4>\n

      Bir g\u00f6k cisminin almanaktan bulunan GHA de\u011feri 220 derece, Dec 15.00 N ve DR mevkiin koordinatlar\u0131 35 N \u2013 060 E oldu\u011funa g\u00f6re HA ( t ) a\u00e7\u0131s\u0131 nedir?( Boylam E GHA ile topla )<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: HA= 360 \u2013 ( GHA – \/ + Long ) 360 \u2013 ( 220 + 60 ) = 80 derece<\/strong> bulunur.\u00a0 HA( t ): 80 derece<\/strong><\/span><\/h4>\n

      Boylam\u0131m\u0131z 100 W ve GHA de\u011feri 350 ise saat a\u00e7\u0131s\u0131 ( HA veya t ) nedir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: ( boylam W GHA dan \u00e7\u0131kar ) HA= 360 \u2013 ( 350 \u2013 100 )= 110 derece olur. HA ( t ): 110 derece<\/strong><\/span><\/h4>\n

      GHAy de\u011feri 242 ve SHA H de\u011feri 103 ise GHA H de\u011feri nedir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: GHA H = GHAy + SHA H = 345 olur.<\/strong><\/span><\/h4>\n

      Kutup y\u0131ld\u0131z\u0131n\u0131n d\u00fczeltilmi\u015f a\u00e7\u0131sal y\u00fcksekli\u011fi ( Ho ) 51. 15\u2019 \u2013 a0; 14\u20198\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 a1; 0\u20196\u00a0\u00a0 a2; 0\u20196 oldu\u011funa g\u00f6re bulundu\u011funuz enlem de\u011feri nedir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: 51.15\u2019 + 14\u20198 + 0\u20196 \u2013 1 = 50.30\u20194 ( emin de\u011filim )<\/span><\/h4>\n

      Vardiya zabiti olarak g\u00fcne\u015fin bat\u0131\u015f an\u0131nda semtini 289 olarak \u00f6l\u00e7t\u00fc\u011f\u00fcn\u00fcze g\u00f6re, g\u00fcne\u015fin o andaki S\u0130A de\u011feri nedir?<\/span><\/h4>\n

      280 \u2013 270 = 019 bulunur.<\/span><\/h4>\n

      Semt a\u00e7\u0131m\u0131z g\u00fcne\u015fin bat\u0131 i\u00e7in W Semt> 270 oldu\u011funa g\u00f6re N i\u015fareti al\u0131r. Cevap: N019 W<\/strong><\/span><\/h4>\n

      Vardiya zabiti olarak g\u00fcne\u015fin do\u011fu\u015f an\u0131nda semtini 079 derece olarak \u00f6l\u00e7t\u00fcn\u00fcz. G\u00fcne\u015fin o andaki S\u0130a de\u011feri nedir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: SIA= 090 \u2013 079 = 11 derece i\u015fareti ise E ve N olur.<\/span><\/h4>\n

      Cevap: N11E<\/strong><\/span><\/h4>\n

      <\/h4>\n

      Geminizin mevkii 28.00\u2019S \u2013 42.36\u2019W dir. G\u00fcne\u015f in Dec de\u011feri N 20.30\u2019 S\u0130A de\u011feri 23.4 d\u00fcr. G\u00fcne\u015fin do\u011fu\u015funda semti nedir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: SIA = 90 \u2013 Semt\u00e0 Semt= 90 \u2013 SIA \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 SEMT= 90 \u2013 23.4= 66.6 SEMT: 66.6 olur<\/strong>.<\/span><\/h4>\n

      03 May\u0131s 1999 g\u00fcn\u00fc 09.24.22 UT ( GMT ) de G\u00fcne\u015fin GHA ve Dec de\u011ferleri nedir? ( Ek cetvel 13 \u2013 39 )<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm:<\/span><\/h4>\n

      G\u00fcne\u015f\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Dec\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d<\/span><\/h4>\n

      09 00 00 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 315.46\u20193\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 N15.35\u20192\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 0.7<\/span><\/h4>\n

      \u00a0\u00a0\u00a0\u00a0 24 22\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0 6.05.5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 d\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 0.3<\/span><\/h4>\n

      GHA\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 321.51.8 Dec\u00a0\u00a0\u00a0\u00a0\u00a0 N15.35.5 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Cevap: GHA:321.51.8 \u00a0\u00a0 Dec:N15.35.5<\/strong><\/span><\/h4>\n

      Fenerlerin co\u011frafi g\u00f6r\u00fc\u015f mesafesini hesaplamak:<\/strong><\/span><\/h4>\n

      De\u011fer feet olarak verilmi\u015fse; Form\u00fcl= ( 1.14 x harita y\u00fckselli\u011fi karesi ) + 1.14 K\u00f6pr\u00fc \u00fcst\u00fc y\u00fcksekli\u011fi karesi)<\/span><\/h4>\n

      \u00d6rnek: Gp Fl ( 2 ) 10 Sec 81 feet olan 10M \u00f6zelli\u011findeki bir feneri, 16 feet g\u00f6z y\u00fcksek olan bir zabitin feneri co\u011frafi olarak g\u00f6rme mesafesi nedir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: 81 karesi 9\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 16 karesi 4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ( 1.14 x 9 ) + 1.14 x 4 )\u00e0 ( 10.26 ) + ( 4.56 )=14.82 mil<\/strong> eder<\/span><\/h4>\n

      Harita ve K\u00f6pr\u00fc \u00fcst\u00fc mesafesi metre olarak verilmi\u015fse; Form\u00fcl=<\/span><\/h4>\n

      G\u00f6r\u00fc\u015f mesafesi= ( 2.08 x Harita y\u00fcksekli\u011fi karesi ) + ( 2.08 x K\u00f6pr\u00fc \u00fcst\u00fc y\u00fcksekli\u011fi karesi )<\/span><\/h4>\n

      \u00d6rnek: G\u00f6z y\u00fcksekli\u011fi 25 metre olan bir fener g\u00f6z y\u00fcksekli\u011fi 16 metre olan bir g\u00f6zlemci taraf\u0131ndan ka\u00e7 milden g\u00f6r\u00fcn\u00fcr?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: 25 karesi 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 16 karesi 4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ( 2.08 x 5 ) + ( 2.08 x 4 )\u00e0( 10.4 ) + ( 8.32 )=18.72 mil<\/strong> eder.<\/span><\/h4>\n

      Haritada bir fenerin y\u00fcksekli\u011fi 225 feet, 49 feet lik bir g\u00f6r\u00fc\u015f mesafesi olan bir k\u00f6pr\u00fc \u00fcst\u00fcnden co\u011frafi g\u00f6r\u00fc\u015f mesafesi nedir?<\/span><\/h4>\n

      De\u011fer feet olarak verilmi\u015fse; Form\u00fcl= ( 1.14 x harita y\u00fcksekli\u011fi karesi ) + 1.14 K\u00f6pr\u00fc \u00fcst\u00fc y\u00fcksekli\u011fi karesi)<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: 225 karesi 15\u00a0\u00a0\u00a0\u00a0 49 karesi 7\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ( 1.14 x 15 ) + ( 1.14 x 7 )\u00e0 ( 17.1 ) + ( 7.98 )=25.08<\/strong> mil eder<\/span><\/h4>\n

      Haritada bir fenerin y\u00fcksekli\u011fi 225 metre, 49 metre lik bir g\u00f6r\u00fc\u015f mesafesi olan bir k\u00f6pr\u00fc \u00fcst\u00fcnden co\u011frafi g\u00f6r\u00fc\u015f mesafesi nedir?<\/span><\/h4>\n

      G\u00f6r\u00fc\u015f mesafesi= ( 2.08 x Harita y\u00fcksekli\u011fi karesi ) + ( 2.08 x K\u00f6pr\u00fc \u00fcst\u00fc y\u00fcksekli\u011fi karesi )<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: 225 karesi 15\u00a0\u00a0\u00a0\u00a0 49 karesi 7\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ( 2.08 x 15 ) + ( 2.08 x 7)\u00e0 ( 31.2 ) + ( 14.56 ) 45.76<\/strong> mil eder<\/span><\/h4>\n

      <\/h4>\n

      G\u00f6z y\u00fcksekli\u011fi 64 feet olan bir g\u00f6zlemci i\u00e7in 144 feet y\u00fcksekli\u011findeki bir feneri ka\u00e7 milden g\u00f6rebilir?<\/span><\/h4>\n

      De\u011fer feet olarak verilmi\u015fse; Form\u00fcl= ( 1.14 x harita y\u00fcksekli\u011fi karesi ) + (1.14 K\u00f6pr\u00fc \u00fcst\u00fc y\u00fcksekli\u011fi karesi)<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: 64 karesi 8\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 144 karesi 12\u00a0\u00a0\u00a0\u00a0\u00a0 ( 1.14 x 8 ) + ( 1.14 x 12 )\u00e0 ( 9.12 ) + ( 13.68 )= 22.8 mil<\/strong> eder<\/span><\/h4>\n

      Kutup seyri:<\/strong><\/span><\/h4>\n

      80 N 040 E mevkiinde hakiki y\u00f6n 150 derece ise grid y\u00f6n\u00fc nedir?<\/span><\/h4>\n

      Grid y\u00f6n\u00fc= Hakiki y\u00f6n ( – \/ + long ) ( +W \/ – E )\u00e0 Kuzey kutbu.<\/span><\/h4>\n

      Form\u00fcle g\u00f6re boylam i\u015fareti E oldu\u011fu i\u00e7in ve Kuzey kutup ta bulunuldu\u011fu i\u00e7in boylam de\u011feri hakiki y\u00f6n de\u011ferinden \u00e7\u0131kar\u0131lacak. 150 \u2013 40 = 110 derece<\/strong><\/span><\/h4>\n

      82 S \u2013 010 W mevkiinde hakiki y\u00f6n 040 derece ise grid y\u00f6n\u00fc nedir?<\/span><\/h4>\n

      Grid y\u00f6n\u00fc= Hakiki y\u00f6n ( – \/ + long ) ( + E \/ – W )\u00e0 G\u00fcney kutbu.<\/span><\/h4>\n

      Form\u00fcle g\u00f6re boylam i\u015fareti W ve G\u00fcney kutupta bulunuldu\u011fu i\u00e7in boylam de\u011feri hakiki y\u00f6n de\u011ferinden \u00e7\u0131kar\u0131lacak. 040 \u2013 010 = 030 derece<\/strong><\/span><\/h4>\n

      79 N \u2013 130 W mevkiinde grid rotas\u0131 045 ise hakiki rota ka\u00e7t\u0131r?<\/span><\/h4>\n

      Grid y\u00f6n\u00fc= Hakiki y\u00f6n ( – \/ + long ) ( + W \/ – E )\u00e0 Kuzey kutbu.<\/span><\/h4>\n

      Form\u00fcle g\u00f6re boylam i\u015fareti W ve Kuzey kutupta bulunuldu\u011fu i\u00e7in boylam de\u011feri ve rota toplanacak<\/span><\/h4>\n

      130 + 45 = 175 derece<\/strong><\/span><\/h4>\n

      80S \u2013 050E mevkiinde grid rotas\u0131 120 ise hakiki rota ka\u00e7t\u0131r?<\/span><\/h4>\n

      Grid y\u00f6n\u00fc= Hakiki y\u00f6n ( – \/ + long ) ( +E \/ -W ) \u00e0 G\u00fcney kutbu.<\/span><\/h4>\n

      Form\u00fcle g\u00f6re boylam i\u015fareti E ve G\u00fcney kutupta bulunuldu\u011fu i\u00e7in boylam de\u011feri rota ile toplanacak.<\/span><\/h4>\n

      120 + 50 = 170 derece<\/strong><\/span><\/h4>\n

      Pusula ve Kerteriz hesaplamalar\u0131:<\/strong><\/span><\/h4>\n

      C\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 M\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 V\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 T y\u00f6ntemi:<\/u><\/strong><\/span><\/h4>\n

      C;\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/u><\/strong>Manyetik pusula rotas\u0131 ve kerterizi ( Compass &Bearing )<\/u><\/span><\/h4>\n

      D;<\/u><\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Arizi sapma veya yapay sapma ( Deviation )<\/u><\/span><\/h4>\n

      M;\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/u><\/strong>Magnetik rota veya kerteriz ( Magnetic &Bearing)<\/u><\/span><\/h4>\n

      V;<\/u><\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Do\u011fal sapma ( Variation ) ( Haritadaki pusula g\u00fcl\u00fcnde okunan de\u011fer.)<\/u><\/span><\/h4>\n

      T;\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/u><\/strong>Hakiki rota veya kerteriz ( t\u00fcre course & Bearing )<\/u><\/span><\/h4>\n

      Bu y\u00f6ntemde dikkat edilecek konu Soldan sa\u011fa do\u011fru gidilirken E de\u011ferler ( + ) W de\u011ferler ( – ) olur.<\/span><\/h4>\n

      Sa\u011fdan sola gidilirken W de\u011ferler ( + ) E de\u011ferler ( – ) olur.<\/span><\/h4>\n

      \u00d6rnek: Manyetik pusulam\u0131z ile 112 dereceye ilerlerken b\u00f6lgede do\u011fal sapma de\u011feri 2 derece E ve bu rotadaki yapay sapma 4 derece W ise hakiki rota nedir?<\/span><\/h4>\n

      C:112\u00a0\u00a0\u00a0\u00a0 D:4W\u00a0\u00a0\u00a0\u00a0 M?\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 V:2E\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 T:?<\/span><\/h4>\n

      C\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 M\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 V\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 T<\/span><\/h4>\n

      112 –\u00a0\u00a0\u00a0\u00a0\u00a0 4W\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 108\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2E+\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 110\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Cevap: 110 derece true course ( Hakiki rota ) bulunur.<\/span><\/h4>\n

      Ana cayro\u00a0 pruva de\u011feri\u00a0 078 derece iken ripiter pruva de\u011feri 080 olarak yanl\u0131\u015f devreye al\u0131nm\u0131\u015f bir ripiterle A fenerinin kerterizi 268 olarak al\u0131n\u0131yor. Ana cayro hatas\u0131 3 E oldu\u011funa g\u00f6re hakiki kerteriz ka\u00e7 derecedir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: Ripiter hatas\u0131 : 80 \u2013 78= 2 W\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 A fenerinin ger\u00e7ek kerterizi= 268 \u2013 2 W= 266<\/span><\/h4>\n

      Cayro hatas\u0131= 3 E\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hakiki kerteriz: 266 + 3= 269 derece.<\/strong><\/span><\/h4>\n

      Bu soru C D M V T\u00a0 olarak soldan sa\u011fa i\u015flemi olarak \u015f\u00f6yle yap\u0131l\u0131r.<\/span><\/h4>\n

      C : 268 ( A fenerinin kerterizi )\u00a0\u00a0 D: 2 W ( Burada ana cayro de\u011feri > ripiter de\u011feri ise W olur.<\/span><\/h4>\n

      Ana cayro de\u011feri< Ripiter de\u011feri ise i\u015faret E olur.<\/span><\/h4>\n

      M:?\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 V: 3 E \u00a0\u00a0\u00a0 T:?<\/span><\/h4>\n

      C\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 M\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 V\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 T<\/span><\/h4>\n

      268 – \u00a0\u00a0\u00a0\u00a0 2W\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 266\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3E +\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 269\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Cevap: 269 derece.<\/strong><\/span><\/h4>\n

      G\u00fcne\u015fin meridyenden ge\u00e7i\u015f an\u0131nda filika pusulas\u0131n\u0131n yapay sapmas\u0131n\u0131 belirlemek isteyen filika amiri filika pusula kart merkezine bir kalemi dik tutarak g\u00f6lgesinin 190 derece \u00e7izgisinin \u00fcst\u00fcne d\u00fc\u015ft\u00fc\u011f\u00fcn\u00fc belirlemi\u015ftir. B\u00f6lgede do\u011fal sapma 3W oldu\u011funa g\u00f6re o andaki pruva derecesi i\u00e7in yapay sapma miktar\u0131 nedir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: Soruda G\u00fcne\u015fin meridyenden ge\u00e7i\u015fi s\u0131ras\u0131nda dedi\u011fi i\u00e7in True ( T ) 180 olur.<\/span><\/h4>\n

      C\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 M\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 V\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 T<\/span><\/h4>\n

      190\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ?\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ?\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3W\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 180<\/span><\/h4>\n

      C\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 M\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 V\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 T<\/span><\/h4>\n

      190\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 7W+<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 183\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3W+\u00a0\u00a0\u00a0\u00a0\u00a0 180\u00df Cevap: yapay sapma 7W<\/strong>\u00a0<\/span><\/h4>\n

      <\/h4>\n

      <\/h4>\n

      <\/h4>\n

      14 Ocak 2003 seyir yapt\u0131\u011f\u0131m\u0131z b\u00f6lgeye ait haritan\u0131n pusula g\u00fcl\u00fc \u00fczerinde 0.35\u2019 ( 1981 ) 5\u2019W ibaresi olan bulunmaktad\u0131r. Manyetik pusula ile 225 rotas\u0131na seyrederken bu rotadaki yapay sapma de\u011feri 4 derece W ise gemimizin hakiki rotas\u0131 nedir?<\/span><\/h4>\n

      Variation hesab\u0131: 2003 \u2013 1981=22 y\u0131l\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 22 x 5= 110\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 110= 1 derece 50 dk W 1.50 \u2013 0.35= 1.15 W bulunur.<\/span><\/h4>\n

      C\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 M\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 V\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 T<\/span><\/h4>\n

      225\u00e04 W\u00e0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 221\u00e0 1.15W\u00e0219.45 bulunur.<\/strong><\/span><\/h4>\n

      20 Ocak 2003 tarihinde seyir yap\u0131lan b\u00f6lgede bulunan iki burnun transiti harita \u00fczerinde 055 dir. Ayn\u0131 transit geminin manyetik pusulas\u0131 ile 065 olarak \u00f6l\u00e7\u00fcld\u00fc. Haritan\u0131z\u0131n pusula g\u00fcl\u00fc \u00fczerinde 3.20 W\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (1985) 5\u2019 W ibaresi var. Manyetik pusulan\u0131n yapay sapmas\u0131 ne kadard\u0131r?<\/span><\/h4>\n

      Variation hesab\u0131: 2003 \u2013 1985=18 y\u0131l\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 18 x 5 =90\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 90= 1.30 W eder. 3.20 W + 1.30 W= 4.50<\/span><\/h4>\n

      Variation= 4.50 W<\/span><\/h4>\n

      C\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 D\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 M\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 V\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 T<\/span><\/h4>\n

      065\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5.10W<\/strong>\u00a0\u00a0 59.50\u00a0\u00a0\u00a0\u00a0 4.50W\u00a0\u00a0 055\u00df sa\u011fdan sola gidilince W i\u015faretliler toplanacak (+) Cevap:5.10W<\/strong><\/span><\/h4>\n

      Kerteriz problemleri:<\/strong><\/span><\/h4>\n

      \u0130skele taraf\u0131m\u0131zdaki bir cisimden Nispi kerteriz al\u0131rsak; Form\u00fcl; Hakiki kerteriz \u2013 rota<\/span><\/h4>\n

      Sancak taraf\u0131m\u0131zdaki bir cisimden Nispi kerteriz al\u0131rsak; Form\u00fcl; Nispi kerteriz + rota.<\/span><\/h4>\n

      Hakiki kerteriz kuzeyden itibaren saat y\u00f6n\u00fc istikametinde 000 \u2013 360 derece aras\u0131nda \u00f6l\u00e7\u00fcl\u00fcr. Nispi kerteriz geminin pruvas\u0131ndan itibaren Sancak \/ \u0130skele ye do\u011fru 1 \u2013 180 derece aras\u0131nda \u00f6l\u00e7\u00fcl\u00fcr.<\/span><\/h4>\n

      \u00d6rnek: A gemisinin rotas\u0131 135 derece olup B gemisi sancak 35 derece de kerteriz edilmi\u015ftir. B gemisinin hakiki kerterizi nedir?<\/span><\/h4>\n

      Cevap: 35 + 135=170 derece<\/strong><\/span><\/h4>\n

      Rotas\u0131 070 olan A gemisinin vardiya zabiti B gemisini \u0130skele 40 derecede kerteriz etmektedir. B gemisi vardiya zabiti ise A gemisini iskele 110 derecede kerteriz etmektedir. Bu durumda B gemisinin rotas\u0131 nedir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: A gemisi B gemisini Hakiki = 070 + 40 = 110 da kerteriz ediyor.<\/span><\/h4>\n

      B Gemisi A gemisini Hakiki = 030 + 180= 210 da kerteriz ediyor.<\/span><\/h4>\n

      B gemisinin rotas\u0131: 210 + 110=320 bulunur.<\/strong><\/span><\/h4>\n

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      262 rotas\u0131na seyreden bir A gemisinin hakiki olarak 082 de g\u00f6rd\u00fc\u011f\u00fc B gemisinin nispi kerterizi nedir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm. Hakiki kerteriz= Nispi kerteriz ( – \/ + ) rota. Bu soruda A gemisi 262 ye giderken 082 hakiki kerterizde B gemisini g\u00f6rd\u00fc\u011f\u00fcne g\u00f6re; B gemisi A gemisinin k\u0131\u00e7 taraf\u0131ndad\u0131r.<\/span><\/h4>\n

      082= Nispi Kerteriz + 262\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Nispi kerteriz= 262 \u2013 82\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 180\u00a0\u00a0\u00a0\u00a0 Nispi kerteriz 180 derece<\/strong> bulunur.<\/span><\/h4>\n

      172 rotas\u0131na ilerleyen bir gemiden 150 derece hakiki kerterizinde bulunan bir fenerin nispi kerterizi ka\u00e7 derecedir?<\/span><\/h4>\n

      172 rotas\u0131na giderken fener\u00a0 150 derece hakiki kerteriz edilmi\u015f. Buna g\u00f6re fener iskele taraftad\u0131r.<\/span><\/h4>\n

      150= Nispi kerteriz \u2013 172\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Nispi kerteriz= 150 \u2013 172=22 derece iskelede bulunur.<\/span><\/h4>\n

      Veya Nispi kerteriz = 360 \u2013 022= 338 derece<\/strong> bulunur. Bu soru cevab\u0131 338<\/strong> dir.<\/span><\/h4>\n

      Rotas\u0131 340 derece olan bir A gemisi B gemisini Sancak 85 de g\u00f6rmektedir. B gemisi A gemisini hakiki ka\u00e7 olarak g\u00f6rmektedir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: A gemisi Hakiki kerterizi= Nispi kerteriz + rota<\/span><\/h4>\n

      A gemisi Hakiki kerterizi= 85 + 340\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 425 -360= 65<\/span><\/h4>\n

      B gemisi Hakiki kerterizi= 180 + 65= 245\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 B gemisi hakiki kerterizi 245<\/strong> bulunur.<\/span><\/h4>\n

      Rotas\u0131 040 olan bir A gemisi B gemisini \u0130skele 70 de g\u00f6rmektedir. B gemisi A gemisini hakiki ka\u00e7 derece g\u00f6r\u00fcr?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: A gemisi hakiki kerterizi= 040 \u2013 70=30\u00a0\u00a0 360 \u2013 30= 330<\/span><\/h4>\n

      B gemisi hakiki kerterizi: 330 \u2013 180= 150 bulunur.<\/span><\/h4>\n

      Rotas\u0131 340 olan bir A gemisi B gemisini sancak 60 derece g\u00f6rmektedir. B gemisi A gemisini hakiki ka\u00e7 derecede g\u00f6r\u00fcr?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm: A gemisi hakiki kerterizi : 60 + 340= 400\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 400 \u2013 360= 040<\/span><\/h4>\n

      B gemisi hakiki kerterizi= 180 + 40= 220 derece<\/strong> bulunur.<\/span><\/h4>\n

      Sahildeki bir fenerin cayro ripiterinden al\u0131nan kerteriz 310 derecedir. Geminin hakiki mevkiinden bu fenerin haritadan \u00f6l\u00e7\u00fclen kerterizi 308 derecedir. Cayro hatas\u0131 nedir?<\/span><\/h4>\n

      \u00c7\u00f6z\u00fcm. CE= Haritadan \u00f6l\u00e7\u00fclen kerteriz \u2013 ripiterden \u00f6l\u00e7\u00fclen kerteriz. \u0130\u015fareti \u2013 ise W, + ise E olur.<\/span><\/h4>\n

      CE= 308 \u2013 310= -2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -2 bulunur. \u0130\u015fareti ( – ) oldu\u011fu i\u00e7in W olur. Cevap: 2 W<\/strong><\/span><\/h4>\n

      Atlantik okyanusunda 156 rotas\u0131na seyrederken kar\u015f\u0131la\u015ft\u0131\u011f\u0131n\u0131z cayro hatas\u0131 0 olan bir gemi ile yapm\u0131\u015f oldu\u011funuz pusula kontrol\u00fcnde, siz gemiyi 356 da kerteriz ettiniz. Di\u011fer gemi sizi 179 da kerteriz etti\u011fini bildirdi\u011fine g\u00f6re. Cayro pusula hatan\u0131z nedir? \u00d6nce pusula hatas\u0131 0 olan gemiye g\u00f6re rota bulunur. 179 + 180= 359\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 C.E= 359 \u2013 356= 3 derece bulunur. \u0130\u015fareti + oldu\u011fu i\u00e7in E dir. Cevap:<\/strong> 3E<\/strong><\/span><\/h4>\n

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      <\/h4>\n","protected":false},"excerpt":{"rendered":"

      G\u00fcne\u015fin do\u011fu\u015f zaman\u0131n\u0131n hesaplanmas\u0131: 4 May\u0131s 1999 g\u00fcn\u00fc 38 derece N \u2013 026 derece E mevkiinde bulunan bir g\u00f6zlemci i\u00e7in G\u00fcne\u015f\u2019 in do\u011fu\u015f zaman\u0131 UT ( GMT ) nedir? ( Ek cetveller 07 \u2013 08 \u2013 38 \u2013 39 ) \u00c7\u00f6z\u00fcm: ek cetvel 38\u2019 e gir N 35 \u2013 40 derece enlemler aras\u0131ndaki g\u00fcne\u015fin do\u011fu\u015fu […]<\/p>\n","protected":false},"author":1,"featured_media":266,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[16],"tags":[67,93,91,92],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/denizcisin.com\/index.php\/wp-json\/wp\/v2\/posts\/263"}],"collection":[{"href":"https:\/\/denizcisin.com\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/denizcisin.com\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/denizcisin.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/denizcisin.com\/index.php\/wp-json\/wp\/v2\/comments?post=263"}],"version-history":[{"count":3,"href":"https:\/\/denizcisin.com\/index.php\/wp-json\/wp\/v2\/posts\/263\/revisions"}],"predecessor-version":[{"id":269,"href":"https:\/\/denizcisin.com\/index.php\/wp-json\/wp\/v2\/posts\/263\/revisions\/269"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/denizcisin.com\/index.php\/wp-json\/wp\/v2\/media\/266"}],"wp:attachment":[{"href":"https:\/\/denizcisin.com\/index.php\/wp-json\/wp\/v2\/media?parent=263"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/denizcisin.com\/index.php\/wp-json\/wp\/v2\/categories?post=263"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/denizcisin.com\/index.php\/wp-json\/wp\/v2\/tags?post=263"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}