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How To Solve Gyro Error

While your vessel is proceeding down the channel you notice a range of lights in line with your vessel's masts. If your vessel is on course 001° per gyro compass and the charted value of the range of lights is 359° T, what is the gyro compass error?

Given:
          Gyro course = 001°
           True course = 359°

What is asked?
     
           Gyro compass error

Solution:

To solve gyro error we have to subtract the gyro and true course.

PGC =   001°    (add 360° to the pgc)
T/Co = - 359°      
 G/E =      2° W

                     * The gyro error is named west because PGC is greater than the true course. In case you forget how, always remember this that "Compass best, error west. Compass least, error east."

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Calculating Distance Off The Light When Abeam

While underway you sight a light 11° on your port bow at a distance of 12 miles. Assuming you make good course, what will be your distance off the light when abeam?

Given:
           Sin θ = 11°
      Distance = 12 nm

What is asked?

               Distance off abeam          

Solution:

In this problem you can figure out a plane triangle. 11° as your angle, 12 miles as your hypotenuse, and the distant off abeam as the opposite side.

So we can use the sine trigonometric function in solving this problem.


              Sin θ = opp
                          hyp

               Opp = hyp × Sin θ

Distant abeam = 12 × Sin 11°
                      = 12 × 0.190808
                      = 2.3 miles


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Finding Arrival Longitude


You depart LAT 50° 06’ N, LONG 153° 06’ E and steam 879 miles on course 090°. What is the longitude of arrival?

Given:
           LAT 50° 06’ N
           LONG 153° 06’ E
           Distance = 879 miles
           Course = 090°

What is asked?
                        Longitude of arrival

Solution:

Step 1. Solving for the difference of longitude (Dlo). We can use the formula of parallel sailing in getting Dlo.

             Distance = Dlo × Cos Lat
                    
                     Dlo = Dist        
                               Cos Lat
                     Dlo = 879             
                               Cos 50° 06’
                     Dlo = 1370.33 miles (divide this by 60 to convert it into degrees)

                     Dlo = 1370.33 ÷ 60             
                     Dlo = 22° 50’ 20” E  (the sign is east because our ship is moving easterly, course  090°)

Step 2. Add Dlo to the longitude of departure (long1) to get longitude of arrival (long2).
                  
                  Long1 =   153° 06’ E
                      Dlo = + 22° 50’ 20” E  (add because Long1 and Dlo have the same name)
                  Long2 =  175° 56’ 20” E  is the longitude of arrival
                                 
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How Much Compass Variation Should You Apply

The chart indicates the variation was 3° 45' W in 1988, and the annual change is decreasing 6'. If you use the chart in 1991, how much variation should you apply?

Given:

           Compass variation = 3° 45' W in 1988
                Annual change = 6'

What is asked?
     
            Compass variation

Solution:

              Step 1. Get the year difference. And then multiply the difference by the annual change.                
                              
                          1991
                        - 1988
                               3 
                             ×6'
                              18'


              Step 2. Subtract 18' to 3° 45' because the annual change is decreasing.   

                         3° 45’ W
                      -      18'      
                         3° 27' W is the variation to apply.


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Solving Circumference At The Equator

Assuming the equatorial radius of the earth is 3444 nm, find the circumference of the equator.

Given:

           Equatorial radius = 3444 nm

What is asked?

           Circumference of the earth at the equator.     

 Solution:

           This is the formula for the circumference;

           Circumference = 2π r
                                  = 2 × 3.1416 × 3444 nm
                                  = 21,639.3 nm is the circumference of the earth at the equator.
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Finding Ship's Arrival Latitude

A ship sails from Lat. 34° 54' N and steams going south (180) for a distance of 372 miles. What is her arrival latitude?

Given:
            Latitude = 34° 54' N
           Distance = 372 nm

What is asked?

            Arrival latitude

Analysis:

We are in latitude 34° 54' N and is going south for a distance of 372 nm. It’s important to remember that equator is equal to zero latitude. So since our ship is in the northern hemisphere and going south, we can expect that our arrival latitude would be lesser.

Solution:

Step 1. Convert the distance into degrees. We can do it by dividing it by 60. Perhaps you asked why I divide it by 60. I divide it by 60 because one degree is equal to 60 minutes. And one minute of an arc is equal to one nautical mile.
                    1° = 60’
                    1’ = 1 nm
  
     372 nm ÷ 60 = 6° 12’ S - This could serve now as our Dlat (difference of latitude).

Step 2. Solving for arrival latitude.
             
              Rule in finding arrival latitude (L2):
                 + If signs are the same
                  - If signs are opposite
                 And then affix the sign of the greater number.
                  
                     L1 = 34° 54' N
                 -
                   Dlat = 6° 12’ S
                      L2 =  26° 42’ N is the ship’s arrival latitude.
    
Arrival Latitude

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Solving Distance By Parallel Sailing



Your vessel receives a distress call from a vessel reporting her position as LAT 35 01’ S; LONG 018 51’ W. Your position is LAT 35 01’ S; LONG 021; 42’ W. Determine what will be your True Course and distance to the vessel in distress by parallel sailing method.

Given:
             λ1 = 021° 42’ W
 λ2 = 018° 51’ W
Lat = 35° 01’ S

What is asked?
           
            True course and distance by parallel sailing

Analysis:

            It is vital to read carefully when answering a question especially when taking a board exam. As of this particular problem we are given a hint of what kind this problem is because it tells us that by parallel sailing method we are to answer it. But even if, supposing, the problem does not mention parallel sailing, we can know it is a parallel sailing by looking the given latitude. From our ship’s latitude of 35 01’ S we go to the vessel in distress’ latitude of 35 01’ S. Our ship’s and the vessel in distress latitude is the same. Therefore, what we have here is a parallel sailing.

Solution:

  •  Solving for the distance.

          Step 1. We have to solve first for the difference of longitude (dlo). This is the rule  in getting Dlo:

                   +  If the longitude are opposite
                   -   If the longitude are the same

             So,

                         λ1 = 021° 42’ W
                      + λ2 = 018° 51’ W
                        Dlo = 002° 51’ E   (The sign of Dlo is east because from
                                                      λ1 to λ2, the direction is eastward)
                                  ×    60           to convert it into miles
                        Dlo = 171 miles

          Step 2. We can now proceed to this formula in getting distance by parallel sailing.

                        Distance = Dlo × Cos Lat
                                      = 171 × Cos 35° 01’
                                = 140 miles
  •   To determine the true course:
            Speaking of a parallel sailing, it is important to note that the true course is either 090° or 270°.  To  determine it is by looking the sign of Dlo. Our dlo here is east, therefore, our true course is 090°.

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Finding Compass Error



You have just taken a satellite fix on your ship and from there the Master traced a new course line on the chart which is 130° True. The variation for the locality is 10° West the deviation is 6° East. Find the compass error and then solve for the compass course to steer.

Given:

            130°T = Course
            10° W = Variation
              6° W = Deviation

What is asked?

            Compass error and compass course to steer

Solution:
           
            In problem like this I’m reminded of the mnemonic that says, “True virgin makes dull companion”. It would be easy if we do the tabulation.

True                          True course = 130° 
Virgin                             Variation = 10° W
Makes                           Magnetic =
Dull                               Deviation = 6° E
Companion         Compass course =                

When converting true course to compass course, west is to be added and east will be minus. So since our variation is west, we have to add it to True course. And we have to subtract the magnetic course by the deviation because it is east.


True                          True course =    130°
Virgin                             Variation = +  10° W
Makes                           Magnetic =   140°
Dull                               Deviation = -     6° E 
Companion         Compass course =   134° per standard compass (psc)

Now let us solve for the compass error.

Compass error = Compass course - True course
                       = 134° - 130°
                       = 4° west

  • The error is “west” because the compass course is greater than the true course.

Compass best, error west.
Compass lest, error east.


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How To Calculate Time When Abeam Of The Light


You are steering 163°T and a light was picked up dead ahead at a distance of 11 miles at 0142. You change course to pass the light 2 miles off abeam to starboard. If you are making 13 knots, what will be the time when abeam of the light?



Given:

                   163°T = Course

                 0142H = Time during observation

      11 nm = Distance of the ligh at 0142H

        2 nm = Distance off abeam of the light



What is asked?



            Time when abeam of the light.



Solution:



            Step 1. We have to solve for the distance run to abeam by using Pythagorean theory.



                         Distance = √112 - 22

                                      = √121 – 4

                                      = √117

                                      = 10.8 nm





Step 2. Divide the 10.8 nm by the speed to get the time interval.



                       TI = (10.8 nm ÷ 13 knots) x 60

                           = 49.84 or 50 mins



           

Step 3. Solving for time to abeam of the lighthouse. Add Time Interval to the time during observation.



            0142H

        +     50 min

               92 min

        +1hr-60 min

           0232H is the time when abeam of the light
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Height Of Center Of Gravity (KG)



A vessel displacing 29,500 long tons has a KG of 60 feet. A weight of 500 long tons is added at a VCG of 150 feet. What is the change in KG?

Given:

            Vessel’s displacement = 29,500 tons
                       Weight of load = 500 tons                                      
                                     VCG = 150 ft
                                       KG = 60 ft

What is asked?

           The change in KG

Solution:

Step 1. We have to solve for the moment. It is done by multiplying weight and distance.
            
             Weight   x   Distance   =   Moment
            
                  29,500 tons x 60 ft = 1,770,000 ft/tons  
                     500 tons x 150 ft = 75,000 ft/tons

Step 2. We have to add all weights and all moments because the problem tells that we are loading. But if unloading of cargoes, we have to subtract weights and also subtract the moments.

                Weight   x   Distance   =   Moment
            
                     29,500 tons x 60 ft = 1,770,000 ft/tons  
                   +   500 tons x 150 ft = +  75,000 ft/tons
Total Weight = 30,000 tons           1,845,000 ft/tons = Total Moment

Step 3. Solving for the new KG. We are going to divide total moment by total weight.

              New KG = TM ÷ TW
                           = 1,845,000 ft/tons ÷ 30,000 tons  
                           = 61.5 ft is the new KG

Step 4. Solving for the change in KG:

              Change in KG = New KG – Old KG
                                   = 61.5 ft – 60 ft
                                   = 1.5 ft upward - is the final answer

*** The change in KG is upward because it is going higher from 60 ft to 61.5 ft.

Height of Center of Gravity


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Rolling Period Formula



If your vessel has a GM of one foot and a breadth of 50 feet, what will be your full rolling period?

Given:
            GM = 1 ft
      Breadth = 50 ft

What is asked:
           
Rolling period

Solution:

            The necessary thing to remember in solving for the rolling period is to note the unit of the ship’s breadth. This is so because our formula in getting the rolling period depends on it. Supposing, if the unit of breadth is feet, we have to multiply the breadth by 0.44. On the other hand, if the unit of breadth is meter, we have to multiply the breadth by 0.797.

            Now since the unit of breadth in the problem is feet, the rolling period formula therefore would be 0.44 times breadth over the square root of GM.

Rolling Period = (0.44 x Breadth) ÷ √GM
                     = (0.44 x 50 ft) ÷ √1
                     = 22 ÷ 1
                     = 22 seconds

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How Many Tons Of Cotton in Bales Having a Stowage Factor of 85

The lower hold of your vessel has a bale capacity of 60,000 cubic feet. How many tons of cotton in bales having a stowage factor of 85 can be stowed in the lower hold assuming a broken stowage of 20%?

Given:
60,000 ft3 - Lower hold bale capacity
 85 ft3/ton- Stowage factor
       20% - Broken stowage

What is asked:
Weight in tons of cotton.

Solution:
This the formula: Weight = (Hold volume x Efficiency)  ÷ Stowage factor

Step 1. We have to solve first for the efficiency.
Efficiency = (100 - Broken Stowage) ÷ 100
               = (100 - 20) ÷ 100
                = 80 ÷ 100
                = 0.80

Step 2. Multiply the efficiency and the hold volume, and then divide it by the stowage factor.

         Weight = (60,000 ft3  x 0.80) ÷ 85 ft3/tons
                    = 48,000 ft3 ÷ 85 ft3/tons
                    = 565 tons is the answer.




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How To Calculate Time To Finish Loading Cargo

You are in the process of loading 465,000 barrels of cargo oil. At 1030, on 5 November, you gauge the vessel and find that you have loaded 203,000 barrels. At 1200, you find that you have loaded 219,000 barrels. If you continue loading at the same rate, you will finish at approximately:

Given:
465,000 bbls- Total cargo to load.
203,000 bbls- Cargo loaded at 1030H 5 November
219,000 bbls- Cargo loaded at 1200H 5 November

What is asked:
Time to finish loading the remaining cargo.

Solution:

Step 1. Get the difference of time and cargo loaded from 1030H to 1200H.
               1200H - 1030H = 1.5 hrs
              219,000 bbls - 203,000 bbls = 16,000 bbls
           
 * We have loaded 16,000 barrels in 1.5 hrs.

Step 2. Solving for the rate of loading.
              16,000 bbls ÷ 1.5 hr = 10,666.67 bbls/hr

Step 3. We have to solve for the remaining cargo to load.
              465,000  bbls - 219,000 bbls = 246,000 bbls

Step 4. Solving for how many hours to load the remaining cargo.
             246,000 barrels ÷ 10,666.67 barrel/hr = 23.06 hrs

Step 5. Solving for the time to finish loading the remaining cargo.
         ***23.06 hrs is equal to 2304H

             1200H   5   November
          + 2304H_____________
            3504H  5  November
          - 24 hr + 1 day________
           1104H  6 November is the final answer.

Time to finish cargo loading
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Converting Relative Bearing To True Bearing

You are on course 027° T and took a relative bearing of a lighthouse and found to be 220°. What is the True bearing of the lighthouse?

Given:
027° = True course
          220° = Relative bearing of the lighthouse

What is asked:
                True Bearing of the lighthouse

Analysis:
                Our course is 027° T. That means our reference is from the true north. And we have a 220° relative bearing of a lighthouse. Being relative, the bearing is measured from the ship's head. So in getting the true bearing, we have to add the relative bearing and the true course.

Converting Relative to True bearing

Solution:
This is the formula in getting true bearing.

True Bearing = True Course + Relative Bearing
                    = 027° + 220°
                    = 247° T is the true bearing of the lighthouse



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How To Convert Local Mean Time to Greenwhich Mean Time (LMT to GMT)

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

Given:
LMT of sunrise = 0552 H
         Longitude = 099° 15' W

What is asked:
                      GMT of sunrise

Analysis of the problem:

Our longitude is west. So the Greenwich Meridian is located east to us. And since the rotation of the earth around its axis is east to west, therefore, the Greenwich Mean Time is ahead to our Local Mean Time.

Solution:

Step 1. We have to divide our longitude by 15° to get how many hours the Greenwich Mean Time (GMT) is ahead to us. Why divide it by 15°? We have to divide the longitude by 15° because the earth rotates 15 degrees of an arc per hour. This is called converting arc to time.

099° ÷ 15° = 0637H

Step 2. We have to add 0637H to our LMT to get GMT because our longitude is west, that is to say, we are behind in time.

0637H + 0552H = 1229H is the GMT of sunrise.

LMT to GMT

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