Find The Initial Metacentric Height

Vessel of 11,000 tons displacement has a moment of statical stability of 500 tons/meters when heeled 5°. Find the initial metacentric height.

Given:

Displacement = 11,000 tons
 Angle of heel = 5°
             MSS = 500 tons/meters

What is asked:

Initial metacentric height

Solution:

Metacentric height is also known as GM.

GM =  Moment of Statical Stability 
              Displacement x Sin Ɵ
   
       =     500 tons/meters      
           11,000 tons x Sin 5°
       = 500 tons/meters
           958.71 tons 
       =  0.522 meters is the initial metacentric height

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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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Calculating Area Of A Circle With A Sector Removed

What is the area of a circle with a diameter of 2 ft after a sector of 60° has been removed?

Given:

           Diameter = 2 ft

Sector removed = 60°

What is asked:

Area of a circle after 60° has been removed

Solution:

1. Solving for radius.

              Radius = Diameter ÷ 2
                          = 2 ft ÷ 2
                          = 1 ft

2. Now this is the formula in getting the area of a circle.

Area of a circle = π r² × (360° - 60°)

                                            360°

                        = 3.1416 × (1 ft)²  × (360° - 60°)
                                                              360°
                        = 3.1416 × 1 ft² × (300°)
                                                      360°
                        = 3.1416 ft² × 0.83333
                        = 2.62 ft² is the area of a circle with 60 sector removed
                     

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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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Find The Weight Of Sea Water In A Drum

A cylindrical drum, 1.6 meters high has a diameter of 90 cm. Find the weight of sea water it contains if filled to capacity.

Given:

    Height = 1.6 m
Diameter = 90 cm

What is asked:

Weight of sea water it would contain

Solution:

1. Convert first the 90 cm diameter into meter, and then get the radius.

          90 cm ×  1 m      = 0.90 m
                         100 cm

         Radius = Diameter ÷ 2
                     = 0.90 m ÷ 2
                     = 0.45 m

2. We have to solve for the volume of cylindrical drum.

            Volume of cylinder = (πr²) × height
                                             = ( 3.1416 × (0.45 m)² × 1.6 m
                                             = 3.1416 × 0.2025 m² × 1.6 m
                                             = 1.01787 m³

3. We can now proceed to solve for the weight of sea water it would contain.

                                  Weight = Density × Volume
                                              = 1.025 tons/m³ × 1.01787 m^{3
                                              = 1.04 tons}

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Calculating Rope's Diameter

If the circumference of a mooring rope is 250 mm, what is the diameter?

Given:

Rope circumference = 250 mm

What is asked:

Diameter of a rope

Solution:

Circumference = π × Diameter

Circumference = π  ×Diameter 
          π                        π 
Circumference = Diameter
          π
          250 mm  = Diameter
          3.1416
             80 mm = Diameter 

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What Other Bearing That Gives Best Fix

If you take a bearing of 086° to a lighthouse, what other bearing of another prominent objects would give the best fix?

Given:

Bearing of a Lighthouse = 086°

What is asked:

Other bearing of prominent object that gives the best fix.

Solution:

The BEST fix is obtained when the angle between the lines of position is 90°. If the circumstance does not provide, you take a bearing of an object that is near of 90°. In the case of our problem, the answer would be 000°.

Other bearing of prominent object that gives the best fix.

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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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Converting Time To Arc

What is the equivalent of 8 min 56 sec in arc units?

Given:

Time = 8 min 56 sec (8.93 min)

What is asked:

Arc unit equivalent

Solution:

15 degress = 1 hour

8.93 min × 15°        = 2.2325°
                 60 min    - 2          
                                0.2325° ×  60'  = 14'
                                                  1°

Therefore, 8 min 56 sec is equivalent to 2° 14' of an arc.

Time to Arc conversion

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Converting Arc to Time

What is the equivalent of 1° 53' in time units?

Given:

Arc = 1° 53'  (Also equivalent to 1.88333 degrees)

What is asked:

Time units

Solution:

The earth rotates 15 degrees per hour.

1.88333° ×  1 hr  = 0.12555 hr ×  60 min  = 7.533 min
                  15°                              1 hr       -7              
                                                                  0.533 min × 60 sec  = 32 sec
                                                                                       1 min 
So 1° 53' is equal to 7 min 32 sec.

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Calculate How Many Board Feet

You are to load a consignment of lumber. Each piece measure 3-inches thick, 12-inches wide and 16-feet long. There are 30,000 pieces in the shipment. How many board feet would be listed on the Bill of Lading?

Given:

Measurement of each stick: 3 in 12 in 16 ft
Total number of sticks = 30,000 pieces

What is asked:

Number of board feet

Solution:

One board feet is equal to 1 ft lenght × 1 ft width × 1 in height.

1. We have to calculate how many board feet in one stick.

      Board feet = L × W × H  
                                12
                       = 3 × 12 × 16 
                                 12
                       = 48 board feet in one stick

2. Since there are a total of 30,000 pieces of the same kind of stick, so

       30,000 sticks  ×  48 board feet  = 1,440,000 total board feet
                                  1 stick 

                 

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Formula For Sheave Diameter

What is the sheave diameter to be used with a 3 inches manila rope?

Given:

Rope circumference = 3 inches

What is asked:

Sheave diameter

Solution:

This is the formula for solving sheave diameter.

Sheave diameter = 2 × Circumference
                            = 2 × 3 in
                            = 6 inches is the sheave diameter

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Determine Speed Of The Ship If It Has 60,000 Astern Power

If a ship of 15,000 tons deadweight fitted with steam turbine has an astern power of 60,000 knot-tons, what is the speed of the ship?

Given:

   Deadweight = 15,000 tons
Astern Power = 60,000 knot-tons

What is asked:

Speed of the ship

Solution:

            Speed = Astern Power  ÷ (¹/₃)
                               Speed
                       = 60,000 knot-tons  ÷ (¹/₃)
                           15,000 tons 
                       = 4 knots ÷  1 
                                          3
            Speed = 12 knots

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Calculating Ship's Astern Power

Normally a ship fitted with steam turbines has an astern power, one third of ahead power. If a ship is 7,000-ton deadweight with speed of 15 knots, what is the astern power?

Given:

Deadweight = 7,000 tons
        Speed = 15 knots

What is asked:

Astern power

Solution:

Step 1. Solving for ahead power

            Ahead Power = Weight × Speed
                                  = 7,000 tons × 15 knots
                                  = 105,000 knot-tons is the ahead power

• 
  

Step 2. Solving for astern power. We are going to get one third of 105,000.

            Astern power = 105,000 knot-tons ×  1  = 35,000 knot-tons
                                                                      3

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Finding Radius Of The Parallel Of latitude

The mean radius of the earth is 3440 nautical miles, find the radius of the parallel of latitude of Manila (Lat. 14° 30' N approximately).

Given:

Latitude 14° 30' N

Earth's radius = 3440 nautical miles

What is asked:

Radius of the parallel at latitude 14° 30' N

Solution:

1. We are to solve first for the circumference of the parallel at latitude 14° 30' N before we can solve its radius.

              Circumference = 2πr × Cos Lat
                                        = 2 × 3.1416 × 3440 × Cos 14° 30'
                                        = 20925.74 nm

2. Since we have now the value of the circumference of the earth at Latitude 14° 30' N, we can now 
solve the radius of this parallel of latitude. Cross multiplying the formula of getting the 
circumference we can solve for the radius.
            
            Circumference = 2πr 
                            Radius = Circumference 
                                                     2π         
                            Radius = 20925.74 nm   
                                               2 × 3.1416
                                        =  20925.74 nm   
                                                 6.2832
                                        = 3330.42 nm  

Finding Circumference Of A Parallel Of Latitude

With an equatorial radius of the earth of 3444 nm find the circumference of a parallel of latitude 30° N.

Given:

Latitude 30° N
Earth's radius = 3444 nautical miles

What is asked:

Circumference of a parallel at latitude 30° N.

Solution:

Circumference = 2πr × Cos Lat 
                           = 2 × 3.1416 × 3444 nm × Cos 30°
                           = 18740.2 nm

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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Finding Gyro Error

While your vessel is proceeding down the channel you notice a range of lights in line with your vessel's mast. If your vessel is on course 001° per gyro compass and the charted value of the range lights is 359° True, find your gyro compass error.

Solving for Gyro Error:

Normally gyro compass points to the true north. However, there are times when it does not coincide with true if you compare gyro to the chart. So there must be a gyro error, and that's what we are going to solve in here.

Gyro error = 001° - 359°
                 = 2° West         (west because the gyro reading is greater)

"Compass best, error west"
"Compass least, error east"

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Calculating True Course Made Good

Your vessel is steering course of 197° psc, variation of the area is 7° East and the deviation is 4° West. The wind is from the west producing a two-degree leeway. What true course are you making good.

Given:

Compass Course = 197°
             Variation = 7° East
            Deviation = 4° West
               Leeway = 2°

What is asked?

Course made good

Solution:

In correcting a compass, that is compass to true:

+  if east
-  if west

Compass Course = 197°
            Deviation =  - 4° W
Magnetic Course = 193°
             Variation =  + 7° E 
       True course = 200°
              Leeway =   - 2°     (wind is from starboard, so subtract)
                              198° T is the course made good

Course Made Good

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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"

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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