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Showing posts with label Celestial Navigation. Show all posts
Showing posts with label Celestial Navigation. Show all posts

Finding Latitude Of The Observer At Meridian Transit

What is the latitude of a place where the sun is at the zenith of observer at local apparent noon of June 21 or 23?

Given:
            Date: June 21 & 23

Find:
            Latitude of the observer

Solution:

              June 21 or 23 is summer solstice, and the declination of the sun on this date is 23° 27' N.

        If the Sun is at the Zenith it means that Altitude( Ho) = 90°
                                                                                       -  90° 
                                                                                ZD  =  0°
                           Declination of  the Sun on June 21/23    = 23° 27’ N 
                                                                                Lat  = 23° 27’ N

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Calculate Time For A Star To Rise

Your vessel is at the equator at midnight on 1 January, and a star is observed rising. At what time will this same star rise on 1 February, assuming your vessel's location is still at the equator?


Solution:

One should know how many hours is one sidereal day, and of course a solar day.

             1 solar day = 24hr
         1 sidereal day = 23h 56h 4s   
                                    0h  3m 56s        is the time difference
                                  x            31        since there a 31 days from Jan. 1 to Feb. 1
                                  2h 01m 56s        total difference
       Midnight 1 Jan =  - 24h 00m 00s
                                 21h 58m 04s on 1 February that the star will rise

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Finding Observed Altitude

At meridian passage, upper passage, the observer's latitude was found to be 43° 34.7' N, Dec. is 1° 46.3' N. Find observed altitude.

 Given:
            Lat = 43° 34.7' N
            Dec = 1° 46.3' N

What is asked?

           Observed altitude (Ho)

Solution:

           Lat = 43° 34.7' N
           Dec = 1° 46.3' N         (Always reverse sign)

            Lat = 43° 34.7' N
           Dec = 1° 46.3'  S      (Subtract because it has different name)
             ZD = 41° 48.4' N
                  - 90°                 (Always minus 90° to get observed altitude)
           Ho = 48° 11.6' N

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Find The Declination Of The Sun

Find the declination of the sun as it rises and bears 115° 11' at Lat 42° 27' N.

Given:

Azimuth = 115° 11'
Latitude = 42° 27' N

What is asked?

Declination of the sun

Solution:

1. Solve for the amplitude.

                  Azimuth = 115° 11'
                                 -  90°          
               Amplitude = E 25° 11' S

2. With the answer in step 1, we can now solve for the declination.

         Sin Amplitude = Sin Declination
                                   Cos Latitude   (cross-multiply to derive formula for declination)   

        Sin Declination = Sin Amplitude x Cos Latitude
                               = Sin 25° 11' x Cos 42° 27'
                               = 0.42552 x 0.73787
        Sin Declination = 0.31398
              Declination = 0.31398 inv Sin
                               = 18.3 South is the declination

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Find The True Azimuth At Sunrise

At Lat 55° 53' N, Long 130° 18 W, sun's declination is 12° 40' S. Find the true azimuth (Zn) at sunset.

Given:

     Latitude = 55° 53'
Declination = 12° 40' S

What is asked?

True azimuth at sunset

Solution:

 1. We have to get first the amplitude. This is the formula for amplitude:

                Sin amplitude = Sin Declination
                                         Cos Latitude
                                     = Sin 12° 40' 
                                        Cos 55° 53'
                                     = 0.21928
                                        0.56088
               Sin Amplitude = 0.39096
                                     = 0.39396 inv Sin 
                     Amplitude = W 23 S             (West because the sun is setting)
                                                                   (3rd quadrant)
2. Solving for true azimuth. Since the sun is setting, we have to subtract 270° by our amplitude.

                            270°
                          -   23°       (minus because it is in the 3rd quadrant)
                            247° T is the true azimuth of the sun at sunset

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Finding Sun's Amplitude

On Nov. 21, the sun bore 101° psc to an observer in DR Position Lat 41° 29' N, Long 120° 30' E, declination is 14° 32.2' S. Find the amplitude.

Given:

    Latitude = 41° 29' N
declination = 14° 32.2' S

What is asked?

The amplitude

Solution:

Sin Amplitude = Sin declination
                        Cos latitude

                     = Sin 14° 32.2'
                        Cos 41° 29'
                     = 0.25099
                        0.74914
Sin Amplitude = 0.33504
      Amplitude = 0.33504 inv Sin
                     = 19.57 or simply
     Amplitude  = E 20 S 

* Why E 20 S? EAST because the sun bears 101°, which means it is rising. And SOUTH because that is the sign of the declination. So, just copy the sign of the declination.

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Finding Latitude Of The Observer

On Sept. 12, at GMT 10h 35m 00s LZT in longitude 057° 58' W, the Ho of star Polaris was 35° 50'. The correction values from the Polaris Tables: Ao = 1° 22.1', A1 = 0.4', A2 = 0.9'. Find the latitude of the observer.

Given:
 
  Observed altitude (Ho) Polaris = 35° 50'
                                         Ao = 1° 22.1'
                                         A1 = 0.4'
                                         A2 = 0.9'

What is asked?

The latitude of the observer

Solution:

Add all the values from Polaris Table to the observed altitude of Polaris. And then minus 1 to get the latitude of the observer.
                  
                                 
  Observed altitude (Ho) Polaris = 35° 50'
                                         Ao = 1°   22.1'
                                         A1 =         0.4'
                                         A2 =         0.9' 
                                                  37° 13.4'
                                               -    1°          
                                                  36° 13.4' N is the latitude of the observer      

You might wonder why the sign of the latitude is N. It must be north because you can not observe Polaris in the southern hemisphere. That is, if you can see or observed polaris, know that you are at northern hemisphere.

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How To Solve For Right Ascension

The GHA in the first point of Aries is 315° and the GHA of a planet is 150°. What is the right ascension of the planet?

Given:
  GHA of Aries = 315°
GHA of Planet = 150°

What is asked:

Right ascension of the planet

Solution:

With this problem we can just simply subtract the GHA of Aries by the GHA of the planet. That is if you have memorized the formula. However, it is much better if we are going to draw a figure in order to have a better understanding on how to  get the right ascension.

Step 1. Draw a circle and then write G on the top of it. Use your compass divider starting at G going westward until 315°. That is the GHA of Aries.


   Step 2. For the GHA of the planet, draw westwardly starting at G again until 150°.


Step 3. Right ascension  (RA) can be measured from the first point of Aries to body in eastward direction.


From the figure that we have made above, we can now visualize how to solve for the right ascension of the planet. So this is the formula in getting right ascension;

RA = GHA of Aries - GHA of planet
      = 315° - 150°
RA = 165°

Converting this to time, we have to divide 165° by 15, since the earth rotates according to textbooks15° per hour. I said "according to textbooks" because I do not believe that the earth rotates and revolves around the sun.

165 degrees x  1 hr              = 11 hrs
                       15 degrees


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Solving Geographical Distance From Observer To A Body

At upper transit, if the zenith distance is 34°, the geographical distance from the observer to a body's GP is          .

Given:

Zenith distance = 34°

What is asked:

Geographical distance from the observer to a body

Analysis:

The difference of latitude between geographic position and your position at a time of upper transit is represented by the zenith distance.

So since the problem asks for a geographic distance, we just have to convert 34 degrees into miles by multiplying it by 60.

Solution:

34 degrees x 60 miles    = 2040 miles
                      1 degree


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Find The Zenith Distance

If the sun's observed altitude is 27° 12', the zenith distance is          . 

 Given:

Observed altitude (Ho) = 27° 12'

What is asked:

Zenith distance (ZX)

Solution:

This is the formula in getting zenith distance,

ZX = 90° - Ho
      = 90° - 27 12'        
     
         89° 60'
       - 27° 12' 
ZX = 62° 48'

Find zenith distance

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Finding Geographical Longitude Of A Body

What is the geographical longitude of a body whose GHA is 149° 30'?

Given:

GHA = 149° 30'?

What is asked?

Geographical longitude of a body

Solution:

You can make a diagram to solve this kind of problem. But I think it's much better to just memorize the rule in answering questions like this because it's faster, thus you can save time during your board examination.

So here's the rule:

If GHA is less than 180°  ; the longitude is equal to GHA and is named "west".
            GHA < 180°  ; GHA = Longitude (W)

If GHA is greater than 180° , you have to subtract GHA to 360° , and the result will be the longitude and is named "east".
                      GHA > 180°
                      360°  - GHA = Longitude (E)



Geographical longitude of a body


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