IF
THE SUN’S DECLINATION
IS 15º 30’N AND INCREASING, CALCULATE
THE SUN’S SHA
ASSUMING OBLIQUITY OF
THE ECLIPTIC TO
BE 23º 26.7’.
GIVEN:
SUN’S DECLINATION
15º 30’N AND INCREASING
OBLIQUITY OF
THE ECLIPTIC 23º 26.7’.
SINE
AB = TAN BX TAN
[90-A]
=
TAN 15º
30’ TAN 66º
33.3’
AB = 39º
45.2’
R. A. OF SUN= 39º 45.2’
SHA OF SUN = 320º
14.8’ ANS
PROBLEM
No 21
IF
THE EQUATION OF
TIME WAS + 04M 06S,
WHEN RAMS WAS
14H 32M 15S, CALCULATE THE
SUN’S DECL.
GIVEN:
EQ OF
TIME = 04M 06S
R.A.M.S. = 14H 32M 15S
EQ
OF TIME = MEAN
SUN - TRUE SUN [
WESTERLY]
= TRUE
SUN - MEAN SUN [
EASTERLY]
= R.A.T.S. - R.A.M.S.
04M 06S = R.A.T.S. - 14H 32M 15S
R.A.T.S. = 14H 36M 21S
= 219º
05’ 15’’
MC = 219º
05’ 15’’ - 180º
= 39º
05’ 15’’
OBLIQUITY = 23º
26.7’
SINE
MC = TAN MT TAN
[90-C]
SINE 39º
05’ 15’’ = TAN MT TAN
[90- 23º 26.7’]
MT = 15º
17.2’
SUN’S
DECL = 15º
17.5’ S ANS
PROBLEM
No 22 H.W.
SHAMS
16H 06M 10S , EQUATION
OF TIME [-] 02M 48S,
OBLIQUITY OF THE
ECLIPTIC TO BE 23º
26.7’. CALCULATE THE
SUN’S DECLINATION
EQ
OF TIME = MEAN
SUN - TRUE SUN [ WESTERLY]
= SHAMS - SHATS
[-] 02M 48S = 16H 06M 10S - SHATS
SHATS = 16H 08M 58S
= 242º
14.5’ - 180º
CE = 62º
14.5’
OBLIQUITY = 23º
26.7’
SINE
CE = TAN ET TAN
[90-C]
SINE 62º 14.5’ = TAN ET TAN
[90- 23º 26.7’]
= TAN ET
TAN 66º 33.3’
ET = 20º
59.7’
SUN’S
DECL = 20º
59.7’ N ANS
PROBLEM
No 23
GHA
ARIES γ 06º 13’, GHA
SUN 270º 43’,
SUN’S DECLINATION 23º
20.9’N. CALCULATE OBLIQUITY
OF ECLIPTIC.
FROM
γ TO TRUE
SUN
RATS = 360º - 270º
43’ + 06º
13’
γTS = 95º
30’
MT = DECL 23º 20.9’
SINE
MC = TAN MT TAN
[90-C]
SINE 84º
30’ = TAN
23º 20.9’ TAN [90- C]
90-C = 66º
33.3’
C = 23º
26.7’
OBLIQUITY = 23º
26.7’ ANS
PROBLEM No 24
GIVEN
DECLINATION OF THE SUN 10º 15’N
AND DECREASING, SHAMS
206º 36.8’. CALCULATE
THE VALUE OF
EQUATION OF TIME.
SINE
AC = TAN AB TAN
[90-C]
= TAN
10º 15’ TAN [90- 23º 26.7’]
AC = 24º
38.6’
SHATS = 180º + 24º
38.6’
= 204º
38.6’
GIVEN
SHAMS = 206º
36.8’
EQ
OF TIME = MEAN
SUN - TRUE SUN [
WESTERLY]
= SHAMS - SHATS
= 206º 36.8’ - 204º
38.6’
= 1º
58.2’
= [+]
07M 52.8S ANS
PROBLEM No 31
AT A CERTAIN TIME, THE
SUN’S RA WAS
03H 56M 40S
AND GHA γ WAS
312º 48.4’. ASSUMING
OBLIQUITY OF THE ECLIPTIC
23º 27’. CALCULATE
THE GP OF THE SUN.
SINE AB = TAN BC TAN [90-A]
SINE 59º
10’ = TAN
BC TAN [90- 23º 27’]
= TAN
BC TAN [66º 33’]
BC = 20º
25.7’ [LAT OF BODY] ANS
GHA γ = 312º
48.4’
Gγ [Easterly] = 47º 11.6’
RA☼ = 59º
10.0’
LONG = 106º
21.6’ E [LONG
OF BODY] ANS
WHEN
GHA ARIES γ WAS 212º
14’, THE EASTERLY
HOUR ANGLE OF THE
TRUE SUN
WAS 35º TO
AN OBSERVER IN
LONG 35º WEST.
FIND THE R.A.
OF THE TRUE
SUN
EHA
IS MEASURED EASTWARD
FROM OBSERVER TO TS
R.A. IS MEASURED EASTWARD
FROM γ TO TS
GHA
γ = RATS
= 212º
14’
= 14H 08M 56S ANS
FIND
THE TRUE SUN’S
SHA AT THE
INSTANT WHEN THE
FIRST POINT OF
ARIES CROSSED THE
MERIDIAN OF 85º E,
IF ON THAT
DAY THE SUN’S
GHA WAS 60º
12’ WHEN GHA γ WAS
255º
GIVEN:
GHA
TS 60º 12’
GHA γ 255º
LONG 85º E
FROM γ
TO OBSERVER = 360º - 255º - 85º
= 20º
FOR γ TO
COME TO OBSERVER’S
MERIDIAN IT REQUIRES TO CROSS
20º
360º--------------- 23H 56M 04S
[SIDEREAL TIME]
20º
--------------- 01H 19M
47S
WHEN
ARIES COMES ON
OBSERVERS MERIDIAN ,
SUN MOVES WESTWARD
BY AN AMOUNT
01H --------------- 15º [
SOLAR TIME ]
01H 19M 47S --------------- 19º 56.7’
SHA TS =GHA
TS +LONG+ADDITIONAL AMOUNT TO
MERIDIAN
= 60º
12’ +85º +
19º 56.7’
= 165º
08.7’ ANS
PROBLEM
No 6
TO
AN OBSERVER ON
A SHIP AT
ANCHOR IN A
NORTHERN LATITUDE, THE
SUN ROSE AT
0559 SMT AND
SET AT 1806
SMT. CALCULATE THE EQUATION OF
TIME
SUNRISE 05H 59M
SUNSET 18H 06M
= 12H
02M 30S
LAT
OF MERPASS = 12H 00M
00S
EQ
OF TIME = 02M
30S
= MEAN
TIME - APPARENT TIME
= 12H
02M 30S- 12H 00M 00S
= [+] 02M 30S ANS
PROBLEM
No 29
FOR A VESSEL ON A CONSTANT HEADING AT THE
SAME POSITION, THE SUN ROSE
BEARING 086˚[C] AND SET
BEARING 292˚[C]. IF THE DEVIATION
OF THE COMPASS WAS 2˚E, FIND THE VARIATION
GIVEN;
SUNRISE 086˚
SUNSET 292˚
VARIATION 2˚E
SUNRISE 086˚
SUNSET 292˚
= 189˚
LAT
AT MERPASS = 189˚
COMPASS
ERROR = 189˚ [C] - 180˚[T]
= 9˚ W
DEVIATION = 2˚ E
VARIATION = 11˚ W ANS
PROBLEM
No 9
IN
LATITUDE 50˚N, A STAR WITH
AN SHA OF
146˚ 10’ HAD A
TRUE ALTITUDE OF
51˚ 36;, WHEN
BEARING TRUE EAST.
FIND THE LOCAL
SIDEREAL TIME
ZX 38˚ 24’
PZ 40˚ 00’
SIN
PZ = TAN [90-P] TAN ZX
TAN [90-P] = SIN 40˚
= 10H
49M 43S ANS
