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On 2nd Sept 2008, in DR 40˚ 28’N 064˚ 20’E, the rising sun bore 090˚(C). If variation was 5˚ W, find the deviation of the compass.
LMT Sunrise = 2d 05h 28m 00s
LIT (E) = (-) 04h 17m 20s
GMT Sunrise = 2d 01h 10m 40s
Dec (sun ) 2d 01h 10m 40s
= 07˚ 50.3’ N
As we know that:
LMT of Sunrise or Sunset are taken from the nautical almanac and needs only be interpolated and taken in whole minutes of time. If the time ofobservation is given in the problem, then given time should be used for calculation. It is necessary to remove the ambiquity of chron time.
We know that:

NOTE:
NAMING OF AMPLITUDE: The prefix of True Amplitude is E if the body is rising and W if the body is setting. The suffix is same as that of the declination.
T Az = 079.7˚ (T)
C Az = 090.0˚ (C)
Error = 10.3˚ W
Variation = 05˚ W
Deviation = 5.3 W
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On 1st May 2008, in DR 30˚ 06’N 179 ˚ 45’W, the setting sun bore 285˚(C). If variation was 2˚W, find the deviation for the ship’s head.
LMT Sunrise = 1d 18h 38m 00s
LIT (W) = (+) 11h 59m 00s
GMT Sunset = 2d 06h 37m 00s
As we know that:
LMT of Sunrise or Sunset are taken from the nautical almanac and needs only be interpolated and taken in whole minutes of time. If the time ofobservation is given in the problem, then given time should be used for calculation. It is necessary to remove the ambiquity of chron time.
Dec sun 2d 06h = N 15˚ 30.1’
d (+0.7) = + 00.4’
Dec 2d 06h 37m 00s = N 15˚ 30.5’
We know that :

NOTE:
NAMING OF AMPLITUDE: The prefix of True Amplitude is E if the body is rising and W if the body is setting. The suffix is same as that of the declination.
T Az = 288.0˚ (T)
C Az = 285.0˚ (C)
Error = 03.0˚ E
Variation = 02˚ W
Deviation = 5.0˚E
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Jan 20th 2008, in DR 54˚ 20’S 046 ˚ 27’W, the sun set bearing 234˚(C). If variation was 3˚ W, find the deviation of the compass.
LMT Sunrise = 20d 20h 21m 00s
LIT (W) = (+) 03h 05m 48s
GMT Sunset = 20d 23h 26m 48s
Dec sun 20d 06h = S 20˚ 05.7’
d (-0.5) = – 00.2’
Dec 2d 06h 37m 00s = S20˚ 05.5’
We know that :

NOTE:
NAMING OF AMPLITUDE: The prefix of True Amplitude is E if the body is rising and W if the body is setting. The suffix is same as that of the declination.
T Az = 234.0˚ (T)
C Az = 234.0˚ (C)
Error = NIL
Variation = 03˚ W
Deviation = 3.0 W
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Nov 30th 2008, the sun rose bearing 110˚(C) in DR 49˚ 18’N 110˚ 12’E at 00h 10m 00schron time. If chron error was 05m 10s SLOW and variation was 10˚E, find the deviation of the compass.
LMT Sunrise = 30d 07h 35m 00s
LIT (E) = (-) 07h 20m 48s
GMT Sunrise = 30d00h 14m 12s
As we know that:
LMT of Sunrise or Sunset are taken from the nautical almanac and needs only be interpolated and taken in whole minutes of time. If the time ofobservation is given in the problem, then given time should be used for calculation. It is necessary to remove the ambiquity of chron time.
Dec sun 30d 07h = S 21˚40.0’
d(-0.4) = – 00.1’
Dec 2d 06h 37m 00s = S21˚ 39.9’
We know that :

Rising Az = E 34.5˚S
T Az = 124.5˚ (T)
C Az = 110.0˚ (C)
Error = 14.5˚ E
Variation = 10˚ E
Deviation = 4.5˚E
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Sept 23rd 2008, the sun rose bearing 094˚(C) in DR 00˚ 00’ 120˚ 27’W when chron showed 01h 54m. If chron error was 03m 12s FAST and variation was 2.7˚W, find the deviation for the ship’s head.
LMT Sunrise = 23d 05h 49m 00s
LIT (W) = (+) 08h 01m 48s
GMT Sunrise = 23d 13h 50m 48s
As we know that:
LMT of Sunrise or Sunset are taken from the nautical almanac and needs only be interpolated and taken in whole minutes of time. If the time ofobservation is given in the problem, then given time should be used for calculation. It is necessary to remove the ambiquity of chron time.
Dec sun 23d 13h = S00˚20.7’
d(+01.0) = + 00.8’
Dec 2d 06h 37m 00s = S00˚ 21.5’
We know that :

NOTE:
If the latitude is 00 00’, the value of the amplitude is equal to the value of the declination. Naming of the amplitude is done as usual.
Rising Az = E00.4˚S
T Az = 090.4˚ (T)
C Az = 094.0˚ (C)
Error = 3.6˚ W
Variation =2.7˚ W
Deviation = 0.9˚W
In question number 4 of exercise 29 (RSAz Sun) the LMT should be 30d 07h 33m 00s
LMT should be 30D 07h 35m 58s ……….. and Dec is S 21deg 40.1′