Exercise 5 Answers

Q.1.

Long. 35^o,36’.5W. is in zone +2.

GMT   =  3^h 35^m 12^s

Therefore ZT =  3^h 35^m 12^s – 2^h  = 1^h 35^m 12^s.

Q.2.

Date = 5 July.  ZT = 3^h 20^m 30^s

Long. 145^o 15’.33E is in zone -10.

Therefore GMT =   3^h 20^m 30^s – 10^h

Therefore GD = 17^h 20^m 30^s , 4 July.

Q.3.  Long. 158^o, 2’.38E is in zone -11.

If GD = 2330, 3 July, then ZT = 2330 + 11^h

Therefore ZT = 1030 4 July.

Q.4.

Long. 179^o,59’.9E is in zone -12.

Therefore  ZT = GD + 12^h

Therefore  ZT =  30. April 12^h 01^m + 12^h 

Therefore  ZT = 1 May.  00^h 01^m

 Q.5.

Long. 179^o,59’.9W is in zone +12.

Therefore  ZT = GD – 12^h

Therefore ZT =  30 April 11^h 59^m – 12^h 

Therefore ZT = 29th. April 23^h 59^m

Q.6.

Both ships are positioned on the Equator and are 2 minutes of arc apart in terms of longitude.

Therefore the distance between the ships is 2 x 1855.32 m. = 3710.64 m.

Therefore, although the ships are in dates two days apart, they are only  approx. 3.7 geographical miles apart.

Click here to return to exercise 5

 A fuller explanation of this topic is given in the book ‘Astro Navigation Demystified’.

Useful Link:  The Sun as a time keeper.

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