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Solving What Course To Steer To Intercept A Vessel

You are enroute to assist vessel A. Vessel A is underway at 5.5 knots on course 033° T and bears 248° T at 64 miles from you. What is the course to steer at 13 knots to intercept vessel A?



Given:

    Target Course = 033° T
     Target Speed = 5.5 knots
   Target Bearing = 248° T
  Target Distance = 64 miles
Own Ship Speed = 13 knots

What is asked?

Course to steer to intercept  vessel A.

Solution:

Step 1. Solving for bearing of own ship from target.

          Bearing of own ship from target = (Target Bearing + 180°) - 360°
                                                            = (248° + 180°) - 360°
                                                            = 428° - 360°
                                                            = 068°

Step 2. Solving ∠A. Angle A is a difference between the target vessel's true course and the true bearing of your vessel from the target vessel.

           ∠A = Bearing of own ship from target - Target Course
                 = 68° - 33°
                 =  35°

Step 3. Solving for Angle C.

     Sin ∠C = Target Speed × Sin ∠A 
                        Own Ship Speed
     Sin ∠C = 5.5 kts × Sin 35°
                           13 kts
     Sin ∠C = 0.24267
           ∠C = inv Sin 0.24267
           ∠C = 14°

Step 4. Solving for True course to steer.

          True course to steer = Target Bearing ± ∠C
                                           = 248° + 14°
                                           = 262°

   
Course to steer to intercept a vessel


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Finding DLAT And Departure

Find the DLAT and departure made good if a vessel steams for 1936 nautical miles on course 248° T.

Given:

  Course = 248° T
Distance = 1936 nm

What is asked?

Dlat and departure

Solution:

This kind of problem is of a Plain Sailing. What one has to bear in mind when speaking of a Plain Sailing is a plane triangle.

Plane Triangle

Solving For Difference in Latitude

   Cos 68° =    Dlat     
                   Distance 
         Dlat = Cos 68° × 1936 nm
         Dlat = 725.2 nm

Solving For Departure

    Sin 68° = Departure 
                     Distance
Departure = Sin 68° × 1936 nm
Departure = 1795 nm

Plain Sailing

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Determining What Current You Are Experiencing?

Your vessel is making way through the water at a speed of 12 knots. Your vessel traveled 30 nautical miles in 2 hours 15 minutes. What current are you experiencing?

Given:

Distance = 30 nautical miles
   Speed = 12 knots
      Time = 2 h 15 min or 2.25 h

What is asked?

Current you are experiencing

Solution:

1. Divide the distance by time to get the actual speed.
              30 nm ÷ 2.25 h = 13.3 knots

2. subtract 13.3 knots by the ship's speed.
              13.3 kts - 12 kts = 1.3 knots following current

It must be a following current because by our ship's speed of 12 kts steaming for 2.25 hours is expected to have traveled 27 miles. But since the problems says the ship has traveled a distance of 30 miles, so we must be experiencing a following current of 1.3 knots.


         
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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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Converting Local Mean Time (LMT) To Greenwich Mean Time (GMT)

The LMT of sunrise as tabulated in the Nautical Almanac indicates 05 h 52 min. You are at longitude 099° 15' E. What will the GMT at sunrise?

Given:

         LMT = 05 h 52 min
Longitude = 099° 15' E

What is asked?

GMT at sunrise

Solution:

Step 1. Converting arc to time. Divide the longitude by 15 since the earth rotates 15 degrees an hour according to textbooks, though I personally do not believe that the earth rotates and revolves around the sun. For I am a geocentrist.

                  099° 15' ÷ 15 = 06 h 37 min

Step 2. Since the longitude is east, the local time is ahead of Greenwich. So therefore, to get the Greenwich Mean Time (GMT), we have to subtract the Local Mean Time (LMT) by 06 h 37 min which is our answer in step 1.
                   LMT = 05 h 52 min    (We have to barrow 24 hours and add it to LMT)
                             - 06 h 37 min 
                   GMT = 2315 H




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Solving Distance By Great Circle Sailing

Your ship departs Yokohama, Japan from position Lat. 35° 27' N; Long. 139° 39' E bound for San Francisco, California, USA. At position Lat. 37° 48.5' N, Long. 122° 24' W. Determine the distance by Great Circle Sailing.   

 Given:
   Lat1 = 35° 27' N
   Lat2 = 37° 48.5' N
Long1 = 139° 39' E
Long2 = 122° 24' W

What is asked?

Distance by Great Circle Sailing

Solution:

1. Solve first for the difference of longitude (Dlo). Here's the formula in getting Dlo:
    If same name, subtract
    If different name, plus
    Note: Affix name of direction. If Dlo is greater than 180, subtract it to 360 and then affix name of  Long1. 
                
               Long1 = 139° 39' E
               Long2 = 122° 24' W 
                            261° 63'
                           +   1°-60' 
                            262° 03'
                         - 360°        
                  Dlo =   97° 57' E

2. We can now proceed to solve for Great Circle distance. 
     Note: Plus (+) when not crossing the equator; Subtract (-) when crossing the equator.

          Cos GCD = (Cos Lat1 × Cos Lat2 × Cos Dlo) + (Sin Lat1 × Sin Lat2)    
        Cos GCD  = ( Cos 35° 27' × Cos 37° 48.5' × Cos 97° 57' ) + ( Sin 35° 27' × Sin 37° 48.5' ) 
          Cos GCD= -0.08902 + 0.35548
              GCD    = inv Cos 0.26646
                 GCD =  74.55°
                            ×   60        ( to convert to miles)
                 GCD = 4473 miles
           
Great Circle Sailing
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Loading Lead And Cotton Cargo

You are to load lead, SF 18 and cotton, SF 78. The available deadweight capacity is 1,600 tons of cargo and cubic capacity is 58,800 cu. ft. Disregarding broken stowage, how much of each cargo should be loaded to make her full and down?

Given:

                        Weight = 1,600 tons
               Hold volume = 58,800 ft3
   Lead stowage factor = 18
Cotton stowage factor = 78

What is asked?

Weight in tons of lead and cotton

Solution:


Weight of cotton = Hold volume - (weight × SF lead)
                                SF of cotton - SF of lead
                          = 58,800 - (1,600 × 18)
                                       78 - 18
                          = 30,000
                                60
                          = 500 tons


 
   Weight of lead = Hold volume - (weight × SF of cotton)
                                SF of cotton - SF of lead
                          = 58,800 - (1,600 × 78)
                                   78 - 18
                          = 1,100 tons



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Solving For Ship's Moment Of Statical Stability (MSS)

Find the moment of statical stability when a ship of 10,450 tons displacement is heeled 6 degrees if the GM is 0.5 m.

Given:

               GM = 0.5 m
Angle of heel = 6 degrees
Displacement = 10,450 tons

What is asked?

Moment of Statical Stability (MSS)

Solution:

MSS = Weight × GM × Sin Î¸  
MSS = 10,450 tons × 0.5 m × Sin 6°
MSS = 546.2 ton-meters


Moment Of Statical Stability


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How Much Cargo Remains To Be Loaded

The present draft is 8.20 mtrs fwd and 8.60 mtrs aft. You want to load at your maximum load draft of 8.88 mtrs. If the TPC at this draft is 26, how much cargo remains to be loaded?

Given:

                TPC = 23
        Fwd draft = 8.20 m
          Aft draft = 8.60 m
Maximum draft = 8.88 m

What is asked?

Tons of cargo to be loaded

Solution:

1. We have to solve for the present mean draft by adding forward and aft draft, and then divide the sum by two.

                      Mean draft = (fwd draft + aft draft) ÷ 2
                                        = (8.20 m + 8.60 m) ÷ 2
                                        = 16.80 m ÷ 2
                                        = 8.40 m
2. Subtract the maximum load draft by the present mean draft.
                Maximum draft =   8.88 m 
                      Mean draft = - 8.40 m 
                                             0.48 m
3. Convert 0.48 meters to centimeters so that we could apply the TPC easily.
        0.48 m × 100 cm/m  = 48 cm
                                        × 26 tons per centimeter immersion
                                           1248 tons


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TPC Formula

A ship 128 meters long has a maximum beam of 20 meters at the waterline and coefficient of fineness of 0.85. Calculate the TPC at this draft.

Gicen:

Block coefficient = 0.85
     Ship's lenght = 128 m
             Breadth = 20 m

What is asked?

Calculate for the tonnes per centimeter  immersion (TPC).

Solution:

This is the TPC formula:

TPC = Lenght × Breadth × Block coefficient × Seawater density
                                             100
TPC = 128 × 20 × 0.85 × 1.025
                            100
TPC = 22.3 tonnes




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Finding Displacement Of A Vessel

Find the displacement of a vessel if her GM is 3 feet, the weight to be shifted is 30 tons to a distance of 30 feet across her deck, the angle of list is 1° 43'.

Given:

Distance = 30 ft
   Weight = 30 tons
         List = 1° 43'
         GM = 3 ft

What is asked?

Displacement of a vessel

Solution:

Displacement = Weight × Distance  
                         Tan List × GM
Displacement = 30 tons × 30 ft    
                         Tan 1° 43' × 3 ft  
Displacement = 10,009 tons

Displacement of a vessel

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