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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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Finding Ship's Distance Off The Second Bearing And When Abeam

A ship on course of 253° T at 14 knots. At 2329 a lighthouse was observed bearing 282° T. At 2345 the same lighthouse bears 300° T. Find the ship’s distance off the second bearing and when abeam?


Given:

              Course = 253°
               Speed = 14 knots
     First Bearing = 282°
 Second Bearing = 300°

Find:
             A.) Ship’s distance off the second bearing
             B.) Ship’s distance when abeam

Solution:

1. Find first Angles  A, B, C:

       a.)Co = 253° T
        Brg1 = 282° T 
    Angle A = 29° R

      b.) Co = 253° T
         Brg2 = 300° T
     Angle B = 47° R

     c.) Angle B = 47° R
          Angle A = 29° R
          Angle C = 18° R 

2. Then solve for AB (Distance Run Between 1st & 2nd Observation)

               AB = ( Time 2 – Time 1 ) x Speed
                     = ( 2345 – 2329 ) x 14 knots
                     = 16 minutes x 14 knots
                AB = 3.73 nautical miles

 3. Solve for Distance off at second bearing (BC): (By SINE Law)

                BC = Sin A x AB 
                              Sin C
                 BC = Sin 29° x 3.73
                              Sin 18°
                 BC = 5.9 nautical miles (Distance off at 2nd bearing)

4. Solve for Distance at Abeam (CD) : (By SOH – CAH – TOA)

                CD = BC x Sin B
                CD = 5.9 x Sin 47°
                CD = 4.3 nautical miles ( Distance at Abeam)

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Question From Readers #20 Calculate The New KG


A vessel's displacement is 2,400 tons and her KG is 10.8 meters. What is the new KG if a weight of 50 tons already on board is raised 12 meters vertically?

Given:

Displacement = 2, 400 tons
       Distance = 12 meters
         Weight = 50 tons
               KG = 10.8 meters

Find:
          New KG

Solution:

1. Shifting of Weights

                              GG1 = Weight x Distance
                                            Displacement
                                      = 50 tons x 12 m
                                            2, 400 tons
                                      = 600 tons meter
                                            2, 400 tons
                               GG1 = 0.25 m

2. The problem states that a weight is shifted vertically to 12 meters, so we can see that the weight is shifted upwards because our KG is only at 10.8 m. From this given situation we can conclude that KG will increase. So, what we are going to do this is to add our answer in step 1 to the old KG.

                           Old KG = 10.8 m
                               GG1 =   0.25 m
                          New KG = 11.05 m will be the height of the center of gravity

Thank you for submitting this question. If you have more or you want clarification, feel  free to raise it. Once again, Thanks.

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Solving For ETA And LZT Of Arrival

You are on a voyage from New York, USA to San Francisco, USA. The distance from pilot to pilot is 5,132 miles. The speed of advance is 13.5 knots. You estimate 32 hours of bunkering at Colon and 14 hours for the Panama Canal transit. If you take the departure of 0600 hours (ZD +4) on May 16, what is your ETA (ZD +7) at San Francisco.

Given:

           Distance = 5, 132 miles
               Speed = 13.5 knots
         Bunkering = 32 hours
Panama Transit = 14 hours
  LZT Departure = 0600 16th May

Solution:

              LZT departure = 0600 16th May
                ZD departure = (+)4 (W) +            (E – / W +)
            GMT departure = 1000 16th May
          + Steaming Time = 2009 15days      (Steaming Time: always + )
                GMT arrival = 3009 31st May
                   ZD arrival = (+)7 (W) –         (E + / W –)
                 LZT arrival = 2309 31st May
                   Bunkering = 32 +                 (Any Delays: always +)
 Panama Canal Transit = 14 +                 (Any Delays: always +)
                 LZT arrival = 2109 02nd June is the final answer

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Question From Readers #19: Metacentric Height

Metacentric Height

A vessel with a beam of 72 feet, has an average rolling period of 18 seconds. What would be her metacentric height?


Given:

                    Beam = 72 feet
        Rolling period = 18 seconds

Solution:

Metacentric Height = ((Beam x 0.44) ÷ Rolling Period)²

I used 0.44 to multiply the ship's beam because the unit is feet.

                       GM = ((B x 0.44) ÷ RP)²
                             = ((72 ft x 0.44) ÷ 18)²
                             = (31.68 ft ÷ 18)²
                             = (1.76)²
                             = 3.09 ft will be the metacentric height(GM)

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