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Showing posts with label Sailing. Show all posts
Showing posts with label Sailing. Show all posts

Question From Readers #12: Dead Reckoning


Dead Reckoning

gud eve... i need you help with this. SHIP AT 0800 OCT 10 2011 LAT 9 2 N LONG 116 49.5E TRUE COURSE 38 SPEED 16 KTS FIND DEADRECKONING POS. AT 1600 OCT 10 2011


Solution:

Step 1. Get the steaming time.

           Time of arrival = 1600 10 Oct
     Time of departure = 0800 10 Oct 
                                   8 hrs is the steaming time

Step 2. Solve for the distance traveled for 8 hours.

          Distance = Speed x time
                       =  16 kts x 8 hrs
                       = 128 nautical miles

Step 3. Solving for Dlat.

                Dlat = Cos 38° x  128 nm    
                       = 100.86 nm
                           ÷   60          to convert it to degrees and minutes
                Dlat = 1° 40.86' N  North because the ship is northward

Step 4. Add Dlat and your latitude at position 1 to get the latitude of your new position.

                Dlat = 1° 40.86' N
       Latitude 1 = 9°  2' N         (Add because same name)

       Latitude 2 = 10° 42.86' N

Step 5. Solving for the departure and Dlo.

       Departure = Sin 38° x Distance
                       = Sin 38° x 128 nm
                       = 78.80 nautical miles
                          ÷ Cos 10° 42.86'    to get Dlo. But if in equator, Dep = Dlo
                           80.2 nm 
                        ÷ 60                  to convert it to degrees and minutes
                           1° 20.2' E is the Dlo

Step 6. Solving for the new longitude.

          Long 1 = 116° 49.5' E
               Dlo =     1° 20.2' E     same sign; add
          Long 2 = 118° 9.7' E

    So the new position of the ship is 10° 42.86' N, 118° 9.7' E.                                            

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Question From Readers #10: ETA Costa Rica to Los Angeles

 ETA: Costa Rica to Los Angeles

 Here's a question being asked to me. No matter how I tried my answer is different from his  review material. My answer is 0136H 08 FEB. I have to deferred my solution for a time being for I want to know your answer to this question. If you may, why not show your solution? Hope you do.

You are on a voyage from Limoy, Costa Rica , to Los Angeles USA. The distance pilot to pilo is 3150 miles. The speed of advance is 14 knots. YOu estimate 24 hours for the Panama CANAl Transit. If you take departure at 1836 hours (ZD+6) on 28 January, find your ETA (ZD +8) at LOs Angeles?

ANS: 1336, 08 Febuary
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Question From Readers # 2: Parallel Sailing: Calculate Latitude

Parallel Sailing Formula

  A ship "A" is on the equator steering 090°T at 16 knots ; while a ship"B" is on the parallel of north latitude, steering 270°T at 12 knots. When "A" makes Dlo of 1', "B" makes a Dlo of 48'. Calculate the latitude of "B".

Given:

Ship A
               Course = 090°
                Speed = 16 knots

Ship B

               Course = 270°
                Speed = 12 knots

What is asked?

Latitude of ship B

Analysis:

The relationship between ships is that when "A" makes Dlo of 1', "B" makes a Dlo of 48'.

This is not quite an easy problem compared to other parallel sailing questions. However, if we could solve for a distance traveled by ship A when she makes a Dlo of 1 nautical mile at the equator, then that will be our key to solving for the latitude of ship B.

This is how...

Solution:

1. Solving for a distance traveled by ship A at the equator when she makes a dlo of 1'.

            Distance = Dlo x Cosine Latitude
                          = 1 nm x Cos 0°               (equator is zero latitude)
                          = 1 nm x 1
            Distance = 1 nm

2. Second, we have to get how much time when ship A travel for 1 nm with the speed of 16 kts.

   Steaming Time = 1 nm               (1 knot is equal to 1 nm/hr)
                             16 nm/hr
                         = 1 nm        
                            16 nm/hr   
                         = 0.625 hr       (convert it to minutes)
                         = 0.625 hr x 60 min = 3.75 mins = 3.75 mins
                                              1 hr       1
                         = 3.75 mins

   So, for ship A with the speed of 16 knots, she could travel 1 nautical mile at the equator for 3.75 mins and makes a  Dlo of 1'. How about ship B?

3. Now for the distance traveled for ship B, we have to use the same steaming time 3.75 mins because of this relationship that when "A" makes Dlo of 1', "B" makes a Dlo of 48'. In parallel sailing, distance is equal to the departure.

           Distance = Speed x Time
                         = 12 x 3.75
                         = 45            is the departure of ship B. Dep = Dist
     
                                               Distance = Dlo x Cos Lat
                                               Depature = Dlo x Cos Lat

4. And now for solving for the latitude of ship B, we have to divide the departure by Dlo, and then inverse cosine the answer.

                Dep = Dlo x Cos Lat

 Dlo x Cos Lat = Dep

 Dlo x Cos Lat = Dep 
        Dlo             Dlo

 Dlo x Cos Lat = Dep 
        Dlo             Dlo

         Cos Lat = Dep    
                        Dlo

         Cos Lat = 45
                        48
                     = 0.9375
                Lat = 0.9375 inv Cos
                Lat = 20° 22' N   is the latitude of ship B

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At What Speed Must You Proceed To Reach Your Next Port

Your vessel arrive in port with sufficient fuel to steam 1175 miles at 19 knots. If you are unable to take on bunkers, at what speed must you proceed to reach your next port, 1341 miles distance?

 Given:

   Old distance = 1175 miles
New distance = 1341 miles
     Old speed = 19 knots

  What is asked?

 New speed

Solution:
 
  New Speed = √ Speed2 x Old Distant
                              New Distance 

                          = √192 kts x 1175 mi 
                               1341 mi 

                    = √361 kts x 1175 mi
                             1341 mi

                    = √424175
                          1341

                    = √316.31

 New Speed = 17.8 knots
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Middle Latitude Sailing: Solving For True Course And Distance

A vessel at Lat. 28° 00' N, Long. 116° 00'W heads for a destination at Lat. 34° 00' N, Long. 123° 40' W. Determine the true course and distance by Middle Latitude Sailing.

Given

L1 = 28° 00' N
L2 = 34° 00' N
λ1 = 116° 00' W
λ2 = 123° 40' W

What is asked?

True course and distance by Middle Latitude Sailing.

Solution

1.  Find for DLAT:

            L2 = 34° 00' N
            L1 = 28° 00' N  
       DLAT = 06° 00' N   (Sign of DLAT is north because ship's direction is going north)
                       x  60        (to convert DLAT into miles)
                     360 miles

2. Find for DLO:

            λ2 = 123° 40' W
            λ1 = 116° 00' W   
        DLO =     7° 40' W  (Sign of DLO is west because ship's direction is going west)
                    x 60              (multiply 7° by 60 to convert into miles)
                     420
                    + 40  
        DLO = 460 miles

3. Solving for MLAT. We are going to add the latitudes and then divide it by 2.

            L2 = 34° 00' N
            L1 = 28° 00' N 
                     62° 00'
                   ÷  2            
      MLAT = 31°

4. Solving for true course by Middle Latitude Sailing.

Tan Course = Cos MLAT x DLO      
                             DLAT   
                   = Cos 31° x 460 miles  
                             360 miles
                   = 1.09526
       Course = Inv Tan 1.09526
       Course = N 47.6° W       (N-W because: DLAT is North; DLO is West)
                                         To covert this course angle into True Course, we have to subtract it to 360°.
                       360°
                     -  47.6° 
True Course = 312.4°

5. Solving now for the distance:

     Distance = DLAT    
                      Cos Co
                   = 360 miles  
                      Cos 47.6°
    Distance = 533.88 miles

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Mercator Sailing Formula

A vessel at LAT 18° 54' N, LONG 073° E heads for a destinations at  LAT 13° 12', LONG 054° E. Determine the true course and distance by Mercator sailing.

 Given:

Postion 1:   LAT 18° 54' N, LONG 073° E
Position 2:  LAT 13° 12' N, LONG 054° E

What is asked:

Course and distance by Mercator sailing

Solution:

1. Solving for difference in latitude (DLAT)

                Lat1 = 18° 54' N
                Lat2 = 13° 12' N  (same sign, so subtract)
              DLAT =  5° 42' S (Sign of DLAT is south because the direction of the ship is going south)
                             x  60     to convert it into miles
                              342 miles

2. Solving for the difference in longitude (Dlo)

             Long1 = 073° 00' E
             Long2 = 054° 00' E  (same name: subtract)
                  Dlo =  19° W       (Sign is west because the direction of the ship is westward)
                          x  60       to convert it into miles
                             1140 miles

3. Solving for Meridional Parts in each latitude and the difference in meridional parts (DMP). This is the formula in getting meridional parts.

             Meridional Parts = (Latitude ÷ 2) + 45 = Tan Log Ans × 7915.7 - Sin latitude × 23.2 

             So, Lat 18° 54' N = 1147.6 MP
                  Lat 13° 12' N =  793.8 MP

And to get DMP, we have to subtract the meridional parts because our latitudes are of the same sign.

                            MP1 = 1147.6
                            MP2 =   793.8 
                            DMP =   353.8

4. We can now solve for the course. This is the formula in solving course by Mercator sailing.

                  Tan Course = Dlo    
                                        DMP
                                     = 1140 
                                        353.8
                         Course = 72.76°
                                     = S 72.76° W  (southwest, that is in the third quadrant. You might ask, why SW? What is the rule in naming? South because that's the sign of your DLAT; and West because that's the sign of your Dlo).

Converting quadrantal form to the new form:

                          Course = S 72.76° W
                                           +  180°         (because that's in the 3rd quadrant)
                  True Course = 252.76°

5. Solving now for the distance.

                        Distance =    DLAT         
                                        Cosine Course
                                      =    342        
                                         Cos 72.76°    (not 252.76 to avoid negative answer)
                                      = 1153.9 nautical miles is the distance
           
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Calculate Radar Contact's Speed

You are underway at 10 knots. At 1800 you note a radar contact dead ahead at a range of 10 miles. At 0812 the contact is dead ahead at a range of 8 miles. The estimated speed of the contact is ____ .

 Given:

Own ship speed = 10 knots
           Distance = 10 miles at 1800H
                          = 8 miles at 0812H

What is asked:

Speed of the contact

Solution:

1. We have to get first the distance traveled and the time interval.
             
                                10 miles at 0800H
                               -  8 miles at 0812H 
                  Distance= 2                 12 min = Time interval

2. We can now proceed to the formula in getting contact's speed.

     Contact's Speed = (Distance ÷ Time) × 60) +/- Own Speed
                                = (2 miles ÷ 12 min) × 60) - 10 knots  (minus because head-on & reciprocal course)
                                = 10 knots - 10 knots
     Contact's Speed = 0 knot. That means the contact is dead in the water.

                      
Radar contact speed

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Calculate The Speed Of The Vessel

A vessel in latitude 55° 12' N sails on course 270° T and make a Dlo of 21° 36.6'. If the time taken was 3 days  2 hours, find the vessel's speed.

Given:

                              Time = 3 days 2 hrs or 74 hrs
                         Latitude = 55° 12' N
Difference Of Longitude = 21° 36.6'

What is asked?

Vessel's Speed

Solution:

1. We have to solve first for the distance traveled.
                       Distance = Dlo × Cos Latitude
                                      = 21° 36.6' × Cos 55° 12'
                                      = 12.33 miles

2. Now solving for the vessel's speed.
                          Speed = (Distance ÷ Time) × 60
                                     = (12.33 ÷ 74) × 60
                                     = 0.16662 × 60
                                     = 10 knots

             

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When Ship Is "On Range"?

I have a classmate during my review that during his board examination for OIC-NW licensure he encountered the "on range" question. It's almost a bonus question because it's so easy to answer.The following is the example problem.

A vessel is following a range to keep herself in a narrow channel. The chart indicates the range to be to be 041°. The vessel is "on range" when she is steering a course of ________. 

 Data on the chart and the figures are made in observance of the true north. That is why if the chart indicates the range to be 041°, the ship must be steered to 041° True, not 041° psc.

If ever you have any corrections to my solutions or you wanted to comment on something, I would love to welcome that. Thanks for visiting this blog of mine.

The vessel is "on range"


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Formula For Solving Meridional Parts

It is important to know how to get meriodnal parts in solving a problem on Mercator Sailing. So since I will be featuring mercator problem next, I think it would be necessary for me to write the formula for getting the meridional parts.

So this is the formula for meridional parts:

Meridional Parts = (Latitude ÷ 2) + 45 = Tan Log Ans × 7915.7 - Sin latitude × 23.2

The application for this formula will be dealt with in my next post on mercator sailing.

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Solving Upper Vertex Latitude

A ship following a great circle track crosses the equator at long. 135° 00' E on course 051° T. If she continues along this great circle track, find the latitude and longitude of the upper vertex.

Given:

    Course = 051° T
Longitude = 135° 00'

What is asked?

Latitude and longitude of the upper vertex

Solution:

In solving latitude of the upper vertex, we have to use the course to subtract to 90° or 270° depending to the direction where the ship is going.

Upper vertex latitude = 90° - 051°
                                 = 38° 12'

For the longitude, just copy the figure and change the name of the longitude.



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Determining Latitude Of Vertex

The great circle distance from Lat. 35° 08' S, Long. 019° 26' E to Lat. 33° 16' S; Long. 115° 36' E is 4559 nautical miles and the initial course is 121° T. Determine the latitude of the vertex.

Given:

             Lat1 = 35° 08' S
             Lat2 = 33° 16' S
          Long1 = 019° 26' E
          Long2 = 115° 36' E
Initial Course = 121° T

What is asked?

Latitude of vertex

Solution:

This is the formula to getting the latitude of vertex...

Cos Lat Vertex = Cos Lat1 × Sin I.C.

Cos Lat Vertex = Cos 35° 08' × Sin 121°
Cos Lat Vertex = 0.70100
       Lat Vertex = Inv Cos 0.70100
       Lat Vertex = 45° 30' S

Note:
         Name the Latitude vertex according to the name of Lat1.                        

Latitude Of Vertex


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