Celestial Coordinates. Homework Q & A. Junior Navigation Chapter 5. Objectives: ■ Define declination, hour circle, Greenwich hour angle (GHA), and local hour angle (LHA).
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Q & A
■ Define declination, hour circle, Greenwich hour angle (GHA), and local hour angle (LHA).
■ Understand how the geographical position (GP) of a body is located by celestial coordinates, and the relationships among hour angles.
■ Calculate LHA and declination from Nautical Almanac data
Follow the Student Manual for guidance
a. the arc distance, measured in an easterly direction, from the observer\'s meridian to the hour circle of a GP.
b. the arc distance, measured in a westerly direction, from the Greenwich meridian to the hour circle of a GP.
c. the arc distance, measured in a westerly direction, from the observer\'s meridian to the Greenwich meridian.
d. the arc distance, measured in an easterly direction, from the Greenwich meridian to the observer\'s meridian.
Ref: ¶ 8, 9
3. The declination of the sun is 15°19\'N. Its GHA is 343°41\'. What are the latitude and longitude of the GP of the sun?
L = 15°19\'N
Latitude of the GP of a body is equal to the declination of the body.
Dec = 15°19\'N; therefore L of GP = 15°19\'N.
Lo = 16° 19\'E
If GHA of a body is greater than 180°, longitude of the GP of the body is equal to 360° - GHA of the body, and the GP is in east longitude.
Longitude of GP = 360° - 343°41\' =16°19\'E.
Ref: ¶ 8 - 11
Greenwich meridian hour circle of GP
the GP of the body
c. LHA is measured from ___________________ to
c. LHA is measured from ______________________
the local meridian of
the observer to hour circle of the GP
Ref: ¶ 8, 12 & 13
a. south from the north pole and north from the south pole.
b. from the pole of the hemisphere in which the observer is located.
c. north or south from the equator.
d. north or south from the observer\'s parallel of latitude.
Ref: ¶ 12
Ref: ¶ 13
a. GHA sun + _____ = LHA sun.b. GHA sun - _____ = LHA sun
Ref: ¶ 19
8. Given: Lo 150°E; ZT 0600; GHA sun is 120°. Find: UT and LHA of the sun.
UT = 2000 previous day
Solution:Compute ZD = Lo divided by 15 and rounded.ZD = Lo 150° / 15° = 10. Since Lo is E, the ZD is negative, ZD = -10Compute UT: ZT 0600
ZD -10 UT -0400 +2400 - 1 day
UT 2000 previous day
LHA sun = 270°
Solution:GHA computation: Since the mean sun will be over Greenwich at 1200 UT, GHA at 1200 will be 000°. At UT 2000, it\'s 8 hours later and the sun will have moved 8 hr x 15°/hr making the GHA 120°.GHA Sun 120°LoE +150°
LHA Sun 270°
Ref: ¶ 19 - 21
9. Given: UT 2300, 29 Mar; Lo 115°30\'E; GHA sun 163°50.8\'.a. Find: ZT and date.
0700 30 Mar
b. Find the LHA of the sun.
Solution:Compute ZD:ZD = Lo 115°30.0\' / 15 = 7.70 which rounds to 8. Lo is east so ZD is negative. ZD = -8.Compute ZT:UT 2300
ZD -8 (rev)
ZT 3100 29 Mar
Corr -2400 + 1 day
ZT 0700 30 Mar
GHA Sun 163°50.8\'
Lo E + 115°30.0\'LHA Sun 279°20.8\'
Ref: ¶ 19 - 21
10. The sun is directly overhead at your position. You note the time, and look up in your Nautical Almanac for this date and time that the GHA of the sun is 75° and its declination is 21° N.
What do you know about your position?
Ans: Since the sun is directly overhead, you are at the GP of the sun; your position is L 21° N, Lo 75° W
Ref: ¶ 23 - 25
End Of Homework Q & A