Chart work basics and Concepts for plotting

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A chart is a diagramatic and systematic representation of a particular description or subject,so that the chartered information can be easily understood and can ve viewed at a glace.Charts are the sea maps

Navigational charts: It represents features of earth on a flat paper and includes details of sea (underwater and surface),coast and land, which may be useful for navigational purpose.Charts maybe of different types ie Mercator charts, Gnomonic charts, Plan charts etc.

Mercator Projection: 

• It is said to be a cylindircal orthomorphic projection

• Derived mathematically

• Meridians are represented as parallel staright lines at right angles to the equator.

• Meridians divide the equator into 360 parts.

• Equator and parallel of latitudes appear as horizontal parallel lines at selected distances.

• Distances between successive latitudes increases towards the poles,this is to overcome the east west distortion( as on the earths surface the meridians converge towards the poles,but on the mercator projection they are represented as parallel staright lines, therefore a delibrate distortion is carried on the north south direction.)

  Gnomonic Projection: 

• Constructed on gnomonic or tangential projections.

• Considering the earth as sphere, all points on the surface of a sphere is projected from center of shpere to a plane that is tangential to the sphere.

• The tangent point is usually chosen around the center of the area to be represented.

• If tangent point isone of the poles it is called polar gnomonic chart.

• Great circls appear as straight lines on gnomonic charts,therefore meridians appear to converge towards the poles.

• Compass roses are not shown on gnomonic charts as they are valid for a particular location(because the meridians are converging).

• Rhumb line courses and bearings cannot be laid off as they appear as curved lines

• Covers all areas of the world and great circle courses can be easily laid down as staright lines.

Gnomonic or tangential projection

TYPES OF CHARTS

1.DECCA CHARTS 

Normal basic navigational charts with the appropriate Decca lattice superimposed on them, and can be used in the place a corresponding basic navigational chart. 

(The Decca system work with minimum three shore based transmitter stations called ‘Chains’ operating within the frequency range of 70-130 kHz .)

Prefixed with L(D) and suffixed with Decca chain number.

2.CONSOL CHARTS

Mainly used in air navigation, but maybe used an aid to ocean navigation also, they show great circle bearings of console stations (as per ALRS).

3. LORAN CHARTS

Used for ocean navigation and cover most of northern hemisphere and parts of central Pacific Ocean. British admiralty Loran charts cover only the North Atlantic.US oceanographic publishes charts for the Pacific Ocean

4.ROUETING CHARTS

These charts give information for planning of passages across the oceans. Separate charts recommending routes for different months of the year are published for each ocean and give recommended tracks and distances between ports, average meteorological and ice conditions and ocean currents.

5.VARIATION CHARTS

These charts show isogonic lines. (isogonic lines are those lines joining points having the same variation.)

LETS SEE THE LAYOUT OF CHART AND ITS SAILIENT FEATURES.

Details of the chart including Datum and corrections to be applied for GPS positions, unit of depth

Heights units and reference for heights given, IALA system A or B, type of projection and sources 

of the chart.

Location and authority of publishing the chart

Chart corrections

THINGS TO REMEMBER WHILE USING CHARTS

• Always use the largest scale charts available for the area.

• Note carefully the units in which soundings are given.

• Familiarise yourself thoroughly with graduations on the chart before reading latitude and longitude.

• When measuring distance, along the latitude scale, the dividers should be used along the latitude scale.

• When using compass rose, the ruler must pass through the centre of compass rose and 180 deg on the opposite direction.

• If in doubt about a Cocked hat always assume ship is closest to danger.

(COCKED HAT-if you plot three position lines, when there is an error it will show up, because the lines will cross in a triangle. This is known as Cocked hat; the size of the cocked hat gives an indication of the level of accuracy of that fix.)

• Always keep the chart dry. Keep bottles and pens away from the charts. Use soft black pencils and soft erasers. Never use copying pencils.

POSITION FIXING METHODS

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1.BEARING OR POSITION LINE

2 RANGE OR POSITION CIRCLE

3 POSITION LINE AND POSITION CIRCLE

4 CELESTIAL OBSERVATION (GIVES POSITION LINES)

5 SOUNDING LINES OR DEPTH CONTOURS.

6 HORIZONTAL SEXTANT ANGLE(HSA) (IT GIVES A POSITION LINE OR A POSITION CIRCLE)

7 VERTICAL SEXTANT ANGLE (VSA) GIVES A RANGE IE. POSITION CIRCLE.

8 RANGE OF LIGHTS OR OBJECTS (GEOGRAPHICAL RANGE, NOMINAL RANGE, PRESENT LUMINOUS RANGE)

9 SECTOR OF LIGHTS.

10 TRANSIT BEARINGS (CALLED AS RANGE LIGHTS IN U.S.A) (GIVES A POSITION LINE)

RANGE OF LIGHTS

In your plotting exercise you will be required to use a range, the examiner would give a choice just to confuse the candidate, let’s go through some basics first.

Geographical range: The geographical range of a light is the range at which an observer sees it considering only height of object and height of observer, provided the light is sufficiently powerful and the weather is clear. This depends on the curvature of the earth. 

You can calculate the geographical range of a light using the formula above.

Light is said to be raised when the light is first sighted on the bridge of a ship, it is said to be dipped when it is seen for the last time before it dips below the horizon. Hence raising and dipping distances will be at maximum range for the particular light. The distance of the visible horizon will depend upon the height of observer’s eye and also height of the light.

LUMINOUS RANGE: Luminous range takes into account the weather conditions and the intensity of the light.

It is also called present luminous range as it changes with time and weather conditions.

NOMINAL RANGE: The nominal range of a light is considered at a range at which the light will be seen when visibility is 10 nm.

MET VISIBILTY: Depends on current meteorological conditions.

SOME IMPORTANT THINGS TO REMEMBER.

During day time we will not use the power of light in any question. During daytime object will only be seen at the meteorological visibility.

To asses if its daytime or night.

0600-1800 hrs –Daytime

1800-0600 hrs-  Night

There are 2 choices in every question:

GR (GEOGRAPHICAL RANGE)

Or

PLR (Present luminous range)

Always compare both the values, and use the least value as the range in your question.

When Geographical range is used terms like – RAISING AND DIPPING WILL BE USED.

When Present Luminous range is used terms like- FIRST SEEN/LAST SEEN or (FIRST SIGHTED/LAST SIGHTED) WILL BE USED.

During night time(1800-0600hrs)

USE EITHER GR OR PLR (whichever is least)

During day time (0600-1800hrs)

USE EITHER GR OR MET VISIBILTY

(we don’t use PLR during the day as the light emitted can’t be seen)

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IMPORTANT PRINCIPLES AND CONCEPTS FOR CHART WORK

1. Lowest astronomical tide (LAT) is used as a zero reference for 3 things:

• Chartered depth

• Drying height

• Height of tide

2. All heights of objects are measured from Mean high water spring (MHWS) or Highest astronomical tide(HAT), this is will be given in description of the chart.

3. Use only the latitude scale on a Mercator chart for distance measurement ie 1 minute= 1 nm.

4. Simultaneous fix: means 2 or more position fixing tools used at the same time.

Running fix: means two or more position fixing tools at different times.

5.Chartwork triangle:

Whenever you proceed to solve a chart work question note down these 6 components, this enables you to determine the type of question, whether it is an allowing or counteracting method etc.

The 6 components are:

1. Course steered, course to steer.

2. Engine speed.

3. Set

4. Rate

5. Speed made good (i.e. for 1 hour)

6. Course made good.

To complete a chart work triangle, you need a minimum of 4 components out of these 6 to proceed.

Symbols used:

• The Course to steer is represented by one arrow

• The course made good is represented by 2 arrows

• The set and drift is represented by 3 arrows.

Make sure you show these arrows in a triangle, you may lose marks if you don’t do so.

Case 1:  A ship at 0800 hours following a course of 040T at a speed of 15 kts. Experiencing set 130T at rate 2 kts. Also we are given a radar fix at 0900 hours. Let’s see each step in constructing a chart work triangle. 

Note down the given data:

1.CTS: 040T

2.Engine Speed: 15kts

3.Set: 130T

4.Rate:2kts

5.SMG:?

6.CMG:?

Step1: From the fix at 0800 hrs draw the course for 1 hour (i.e. 040T X 15 miles). The position so obtained is the DR position at 09000hrs (i.e. without allowing for set and rate)

Step 2: From the DR position apply the set and rate for one hour i.e. 130T and 2nm. The position arrived is the estimated position at 0900hrs.

Step 3: Now join the initial fix at 0800 hrs to the EP 0900 hrs, this line represents the CMG to reach the EP 0900hrs.

Step 4: As it is given in the scenario we have a fix at 0900hrs, this is the actual position of the vessel, join a line from 0800hrs to 0900 hrs fix, this line represent the actual course made good by the vessel.

Step 5: Join a line from DR 0900hrs to fix at 0900hrs. Measuring this line would give us the actual set and drift experienced by the ship.

This method is an allowing method.

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Case 2:  Effect of wind: A ship at 0800 hours following a course of 040T at a speed of 15 kts. Experiencing set 130T at rate 2 kts. Wind coming from North at 3 degrees. Also we are given a radar fix at 0900 hours. Let’s see each step in constructing a chart work triangle

1.CTS: 040T

2.Engine Speed: 15kts

3.Set: 130T

4.Rate:2kts

5.SMG:?

6.CMG:?

Note: Wind from N ,3 degrees.

Step1: Since the wind is coming from north and ship is on a course of 040T the wind will push the vessel more to starboard therefore the effective course due to wind is 043T.From the fix at 0800 hrs draw the course after applying wind i.e. 043T X 15nm.The position we reach is an EP (known as EP wind) 0900hrs

Step 2: From the DR position apply the set and rate for one hour i.e. 130T and 2nm. The position arrived is the estimated position at 0900hrs.

Step 3: Now join the initial fix at 0800 hrs to the EP(wind+current) 0900 hrs, this line represents the CMG to reach the EP 0900hrs.Also since fix is given at 0900hrs by radar join initial fix to final fix this would give you the actual CMG and SMG.

Step 4: Join a line from DR 0900hrs to fix at 0900hrs. Measuring this line would give us the actual set and drift experienced by the ship.

This is an allowing method.

Note: If you are asked to find the vessels position in between 0800hrs and 0900 hrs , the position would always be on the CMG.

The vessels heading would be same as the CTS.

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Case 3: Counteracting the current A ship with engine speed 15 kts, given set 130T at 2 kts. The ship is to reach a desired fix; we need to find the course to steer (CTS).

1.CTS: ?

2.Engine Speed: 15kts

3.Set: 130T

4.Rate:2kts

5.SMG:?

6.CMG: Can be obtained by joining initial and final fix.

Step1: Join a line from the initial fix to the desired fix.

Step 2: From the initial fix apply the current in the direction it occurs for one hour (since the triangle is for one hour)

Step 3: From point B measure the ships speed on the compass and cut an arc on the CMG track(i.e. line joining initial and final fix).Join B to the point where it cuts the CMG track.

Therefore, we can get the CTS to reach the desired fix in the presence of current.

When we apply the CTS from initial fix the vessel is slowly pushed by the current to reach to the desired fix, this can be seen in the diagram below.

Case 4: Counteracting the current in presence of wind SE X 3 degrees ,A ship with engine speed 15 kts, given set 130T at 2 kts.The ship is to reach a desired fix; we need to find the course to steer (CTS).

Step1: Join a line from the initial fix to the desired fix.

Step 2: From the initial fix apply the current in the direction it occurs for one hour (since the triangle is for one hour)

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Step 3: From point B measure the ships speed on the compass and cut an arc on the CMG track(i.e. line joining initial and final fix).Join B to the point where it cuts the CMG track.

The course so obtained is the leeway course and not the course steered or course to steer. Since wind is acting from SE at 3 degrees, the ship has to head into the direction from which the wind is coming to finally move on the leeway track so as to reach the desired fix.

Therefore the CTS will be leeway track + 3 degrees in this case.

Now the ship starts from the initial fix using the CTS and due to the effect of wind it is pushed more to port and follows the leeway track and finally due to current it reaches the desired fix.

NOTE: If in question they are asking the CTS, you should write down the leeway course applied for wind and not the leeway course.

6. 

Variation changes with place and with time.

Deviation changes with ships heading.

It is for compass heading that the deviation cards are made.

Compass error = Variation + Deviation 

7.RUNNING FIX

Case 1: CTS 040T and Engine speed 15kts.

Remember running fix occurs at 2 different times.

Bearing 1= 010 T at 0800hrs

Bearing 2= 320T at 0900 hrs

Find the ships position at 0800hrs an 0900 hrs?

Step 1: Draw the two bearings 

Step 2: Take an arbitrary point on the first bearing, let this be point A.

Step 3: Draw the course to steer and cut the distance covered in one hour. Let this be point B (since the interval between the observations is 1 hour, apply the same distance using ship speed)

Step 4: Transfer the first bearing (Position line) to the point B. This transferred bearing cuts the second bearing at a point, this point is the fix at 0900 hours.

Step 5: Now transfer the Course back on to the fix obtained, this cuts the initial bearing, this point is the fix at 0800hrs.

Case 2: CTS 040T and engine speed 15 kts

Vessel experiencing a current of 130T at 2 kts rate. 

Find the ships position at 0800hrs an 0900 hrs?

Bearing 1= 010T at 0800hrs

Bearing 2= 320T at 0900 hrs

Step 1: Draw the two bearings 

Step 2: Take an arbitrary point on the first bearing, let this be point A.

Step 3: Draw the course to steer and cut the distance covered in one hour.

Step 4: From B draw the current i.e. 130T X 2 nm (for 1 hour)

Step 5: From join point A to point C, this is the CMG.

Step 6: Now transfer the initial bearing to point C, this transferred position line cuts the second bearing at a point, which is the fix at 0900 hrs.

Step 7: Transfer the CMG line to the fix at 0900 hrs , this line cuts the initial bearing at a point , this is the fix at 0800 hrs.

Therefore, you can find the position at 0800hrs and 0900 hrs.

Case 3: CTS 040T and engine speed 15 kts

Vessel experiencing a wind from north at 3 degrees. 

Find the ships position at 0800hrs an 0900 hrs?

Bearing 1= 010T at 0800hrs

Bearing 2= 320T at 0900 hrs

Step 1: Draw the two bearings 

Step 2: Take an arbitrary point on the first bearing, let this be point A.

Step 3: Since the wind is from north It will push vessel more to starboard. Therefore, apply wind to CTS. Thus we get leeway course as 043T. From point A draw this leeway course and cut out the engine speed for 1 hour. Let this be point B.

Case 4: Transfer the first bearing to the point B, this cuts the second bearing at a point, this is the fix at 0900 hrs.

Case 5: Transfer the leeway course to the fix at 0900hrs, this meets the initial bearing at a point, this point is the fix at 0800 hrs.

Thus we obtain position at 0800 hrs and 0900 hrs.

Case 4: CTS 040T and engine speed 15 kts

Vessel experiencing a current 130 T at 2 kts rate and wind from north at 3 degrees. 

Find the ships position at 0800hrs an 0900 hrs?

Bearing 1= 010T at 0800hrs

Bearing 2= 320T at 0900 hrs

Step 1: Draw the two bearings 

Step 2: Take an arbitrary point on the first bearing, let this be point A.

Step 3: Since the wind is from north It will push vessel more to starboard. Therefore, apply wind to CTS. Thus we get leeway course as 043T. From point A draw this leeway course and cut out the engine speed for 1 hour. Let this be point B.

Step 4: From B draw the current 130 T X 2nm (for one hour). 

Step 5: Join A to C, this is the CMG.

Step 6: Transfer the first bearing to the point C, this position line meets the second bearing at a point, this point is the fix at 0900hrs.

Step7: Now transfer the CMG line to the fix obtained at 0900 hrs, this will meet the first bearing at a point, this point is the fix at 0800 hrs.

Case 5: CTS 040T and engine speed 15 kts

Vessel experiencing a current 130 T at 2 kts rate and wind from north at 3 degrees. 

Range 1= 5 miles at 0800hrs

Range 2=  8 miles at 0900 hrs

When we are given with the ranges so we have position circles at two different times. The intersection of two position circles gives a fix , but since the same light is observed here the position circles do not intersect.

Step 1: Draw the range circles.

Step 2 : In these type of questions you have to apply the triangle on the lighthouse which we are observing. Since there is wind and since it’s a allowing method the leeway course would be 043T. Draw the leeway course and cut of the distance steamed for 1 hour

Step 3: Now draw the current 130T for 1 hour i.e. 2nm

Step 4: Now join a line from lighthouse to the point B, this is the CMG.

Step 5: Taking B as centre cut out an arc measuring the same as the first range i.e. 5 miles. This cuts the second range circle at a point this is the fix at second observation i.e. 0900hrs.

Step 6: Transfer the CMG line to the fix at 0900 hrs, this meets the first range circle at a point, this point is the fix at 0800hrs.

8.DOUBLING THE ANGLE AT BOW.

This is also a type of running fix problem. In these questions you would be given the ships course(CTS) and ships speed. Also you would be given 2 bearings in such a way that if you see the angle at the bow at one time would be double the angle at another time.

We use the property of an isosceles triangle in solving such problems.

NOTE: Current and leeway won’t be given in such problems.

Let’s see how to go ahead with such problems with an example.

A ship sailing on a course of 090T at a speed of 10 kts observes a light at a bearing of 060T at 0800 hrs, after sailing for one hour the ship then observes the bearing of the lighthouse to be 030T.

Step 1: Draw the two bearings.

Step 2: Take an arbitrary point A on the first bearing, From A draw the ships course 090T for one hour mark this point as B.

Step 3: Transfer the first bearing onto B, this meets the second bearing at a point, this point is the fix at 0900hrs.

Step 4: Transfer the CTS back to the fix at 0900 hrs, this meets the initial bearing at a point, this point is the fix at 0800hrs.

Therefore you got the fixes at 0800hrs and 0900hrs, as you can see this approach was similar to the running fix method.

As you can see here the angle at bow at first observation is 30 degrees and after one hour at second observation the angle at bow is 60 degrees, thus the angle has doubled.

The triangle so formed i.e. triangle OMC is an isosceles triangle where sides OM and MC are equal.

This concept is useful in day’s work problem also.

9.HORIZONTAL SEXTANT ANGLE

How to identify HSA type questions: Headings will not be given in these types of questions. Questions would consist of 2 or more compass bearings and no heading given. 

Let’s say we have the following question.

The following bearings was observed.

Christianso Lt                            - 057 C

Svaneke Lt                                 - 167C

Hammerodde LT                       - 284C

Find the ships position and deviation for ships head if variation was 6 W.

As you can see the ships heading isn’t given anywhere and you are asked to find the deviation. Remember deviation is for a ship heading at the time of observation. All we have is 3 compass bearings and using this we have to find the deviation.

This can be done by determining the ships position, from which we can find the true bearings to each of the lights, this would enable us be determine the compass error, and since variation is given you can find the deviation.

Case 1: Compass bearing to light house A = 330 C

              Compass bearing to light house B= 030 C

Looking at the bearing we can conclude that the observer is South of both the lights.

Step 1: Find the difference between the two bearings.

Now 330-30 =300, but this cannot be used, as HSA cannot be greater than 180 degrees.

So the difference would be (360-330) +30 = 60 degrees

Let this angle be θ

We need to find α, where

α=90- θ

α= 90- 60

α=30 degrees.

Step 2: Draw a straight line joining both the lighthouses. (i.e. from one lighthouse to another) 

Step 3: From each of the lighthouse measure an angle of 30 degrees (i.e. α) keeping the line joining as the two lighthouses as the reference while using protractor.

Step 4: Now taking C as centre and the radius as line joining C to any one of the lighthouse make a circle. This circle passes through both the lights.

Therefore, we get a single position circle, the ship could be anywhere on this position circle, to get a fix we need one more position circle or a position line observed at same time.

Case 2: Compass bearing to light house A = 330 C

              Compass bearing to light house B= 030 C

              Compass bearing to light house C= 150 C

Step 1: Find the difference between the two bearings.

Now 330-30 =300, but this cannot be used, as HSA cannot be greater than 180 degrees.

So the difference would be (360-330) +30 = 60 degrees

Let this angle be θ1

We need to find α, where

α1=90- θ1

α1= 90- 60

α1=30 degrees.

For the second set i.e. θ2, 150-30= 120 degrees.

Then, α2=90-120 degrees.

α2= -30 degrees.

The negative sign implies that the angle should be drawn away from the vessel.

Step 2: Draw a straight line joining the first and second lighthouses and another line joining the second lighthouse and light vessel.

Step 3: Measure out the angles using the line joining the lighthouse and light vessel as reference using a protractor.

In the second case since α2= -30 degrees the angles are measured away from the ship, as you can see in the above diagram that the angles are measured along towards the land.

Step 4: Keeping C1 and C2 as centres and measuring the radius as R1 and R2 draw two circles, these circles pass though the lighthouses .

As you can see above that both the position circles meet at a point, this is the fix 

Step 5: From the fix draw lines to each of the lighthouse and the light vessel  and measure these on the compass rose, the values you obtain are the true bearings of the lighthouses, and since you are given in question the compass bearings you can find out the compass error by comparing the true and compass bearings.

Note: Ideally you have to get the same compass error in each case

For e.g.:

If the true bearings on this case were 335T,025T and 155T, then the compass error would be 5 degrees E in every case.

But if the bearings were to be 335T,024T,156T, then when you see the error individually you get 5 E,4E,6E; in such cases it is advisable to take the average of the three, i.e. in this case 5.3 E, this is an accuracy error and arises due to some inaccuracy.

Case 3: Compass bearing to light house A = 330 C

              Compass bearing to light house B= 060 C

This is a case where θ= 90 degrees ((360-330) +60)

When θ=90, then α=0

When α=0, that means the center of the position circle would be the exact center of the lines joining the lighthouse and light vessel.

Step 1: Draw a line joining the lighthouse and light vessel.

Step 2: As said earlier the center of position circle is the center of the line joining both the lighthouse and the light vessel. To determine the center of a line, just take a measurement of little more than the half from the lighthouse and light vessel using your compass, then cut arcs both above and below the line, you can see 2 cross marks, join the center of these crosses, this passes through the center of the line.

Step 3: Taking C as center and r1 or r2 as radius (as both are equal) draw a circle, this circle passes through the lighthouse and light vessel.

This is the position circle; the vessel could be on any point on this position circle.

Case 4: Compass bearing to light house A = 350 C

              Compass bearing to light house B= 170 C

This is a case where θ= 180 degrees ((360-350) +170)

When θ=180 means that the vessel is observing 2 opposite lighthouses, so the line joining the two lighthouses would give us a position line, the vessel could be anywhere on the position line, the exact fix can be found if we have another position line or position circle.

The bearing to light house C at same time is a true bearing and shown here for understanding purpose only, so as to understand that the fix can be obtained if we have 2 P/L’s.

Case 5: Compass bearing to light house A = 030 C

              Compass bearing to light house B= 030 C

This is a case where θ= 0 degrees 

This means that the observation is a transit bearing, transit bearings can be obtained by joining the two lighthouses in a line, and on finding the true bearing the error can be deduced.

9.FINDING A COURSE TO STEER TO GET A LIGHTHOUSE AT A CERTAIN ANGLE AT THE BOW.

In these questions current and leeway would not be given. A range would be given which could be a radar range, rising, first sighted etc.

Let’s see an example.

Find the CTS (course to steer) to sight a lighthouse 30 degrees on the starboard side at a distance off 12 miles, also find the position when the lighthouse s 30 degrees on port side. We are given a fix at 0800 hrs.

Step 1: The below calculations should be done in your rough book, before getting it onto the chart.

Taking the lighthouse as center draw and arc of 12 miles, the radius of this arc is 12 miles.

Step 2: From the fix at 0800 hrs draw a line to the arc at a point where the radius touches the arc, extend this line further.

Step 3: This forms a triangle , where we need to find the value of x.

Step 4: Considering triangle ABC we see that it’s a right angled triangle and the hypotenuse is 12’ and the angle ABC is 30 degrees.

So sin 30= opp/hyp

             ½=opp/hyp

            Opposite side(AC)= ½*12’= 6 miles

Step 5: Now on the chart keeping 6 miles as radius draw a circle.

Step 6: From the fix at 0800 hrs draw a tangent to the circle. This is the CTS.

Step 7: Keeping the lighthouse as the center cut an arc measuring 12 ‘on the course to steer.

The point where the arc touches the course would be the fix at which the ship would sight the lighthouse at 30 degrees on the port side.

10.FINDING COURSE TO STEER(CTS) TO GET A LIGHTHOUSE/ OBJECT RIGHT AHEAD WITH CURRENT AT A PARTICULAR DISTANCE. ( ALSO KNOWN AS CROSSING THE BOW)

In these questions the engine speed, set and rate and range of the light (radar range, rising, first seen etc.) should be given.

Given data:

CTS:?

SPEED:12kts

SET: 130T

RATE:2kts

CMG:?
SMG:?

It is said that the range of the light via radar is 18 miles. Also a fix given at 0800hrs.

Step 1: Draw a circle of 18 miles keeping the lighthouse as the centre.

Step 2: Now we need to determine the interval/duration of the chart triangle. Since the range is 18’ an ships speed is 12’, to find the time the ship covers 18’, we divide 18’/12=1.5hrs. So our entire chart triangle should be for a duration of 1.5 hours

This is because we apply the triangle from the lighthouse itself and within its range i.e. 18’.

Step 3: From the light house draw the current 130T X 3   miles (because the triangle would be for 1.5 hours)

Step 4: Now join a line from the fix at 0800hrs to the point on the current vector where we have marked off 3 miles, this is the CMG by the vessel.

Step 5: You can see that the CMG cuts the range circle at a point. Now to determine the CTS to sight the light right ahead simply draw a line from the point where CMG cuts the range circle to the light house.,This is the CTS.

So the vessel would move in this CTS so as to sight the vessel right ahead

Note: If you are asked to find the CPA and TCPA it should be measured along the CMG and not the CTS as CMG is actual track of the vessel, for CMG just draw a perpendicular from CMG to the object and measure to TCPA using the SMG and not ships speed.

2. If you are asked to find the beam bearing it should be measured using the CTS and not the CMG.

Beam bearing=CTS +/- 90 degrees.

11: 3 POINT BEARING/ 3 BEARING

This is also a type of running fix problem.

• This method gives us only the CMG.

• 3 bearings can be from a single lighthouse or from 2 lighthouses.

• With the given data choose between allowing or counteracting method.

• Instead of one bearing, a fix maybe given in the question.

• Usually only 3 data will be given out of the 6 data required for a chart work triangle, in case 2 data is only given you will be given with a fix.

• All three bearings should intersect at one point.

 CASE 1: 

Lets take the following example:

Hook head light bore 020T at 0800hrs , and at 0820 hrs it bore 350T and at 0900 hrs it bore 300 T, fins the vessels position at 0800hrs,0820hrs and 0900hrs.

Given data:

CTS: 040T

SPEED :12kts

SET: 130T

RATE: ?
CMG: ?
SMG: ?

Step 1: Draw the three bearings.

Step 2: Draw a perpendicular to the middle bearing passing through the light as shown below.

Step 3: Now we need to select a ratio to cut on the perpendicular drawn in the last step, this ratio could be of time or the distance, since distance is unknown here we will do it with time

In terms of time 

For first stretch i.e. 0800 hrs to 0820 hrs the time interval is 20’

For the second stretch i.e. from 0820 hrs to 0900 hrs the interval is 40’

So the ratio is 20:40 or 1:2,

To get a clearer and less clustered working take ratio which is large, here we took 3:6.

The first value (3’) should be cut on the perpendicular at that time interval i.e. between 0800hrs and 0820 hrs, and second value (6’) should be cut on the perpendicular between 0820 hrs and 0900 hrs.

Step 4: Transfer the centre bearing onto these cut arcs on the perpendicular.

These meet the first and last bearing at a point, draw a line from the point on the first bearing to the last bearing, this is the CMG.

So we can conclude that we always get the CMG using 3 point bearing method.

Step 5: Now draw the CTS from the point on the first bearing, and cut out the engine speed for the total duration of the observation, in this case it is 1 hr.

Step 6: From the point where you have cut off the engine speed draw the current i.e. 130T , this current vector meets the CMG track at a point as shown above, if you measure the current vector now you get the rate of current.

Step 7: Now transfer the first bearing to the point C. This position line cuts the last bearing at a point, this is the fix at 0900hrs.

Step 8: Now transfer the CMG track to the fix at 0900hrs, this line meets the 2nd bearing at a point and 1st bearing at a point, this is the fix at 0820 hrs and 0800 hrs respectively.

Therefore we get the fix at 0800hrs, 0820hrs and 0900 hrs.

CASE 2: When CTS is not given.

Lets take the following example:

Hook head light bore 020T at 0800hrs , and at 0820 hrs it bore 350T and at 0900 hrs it bore 300 T, fins the vessels position at 0800hrs,0820hrs and 0900hrs.

Given data:

CTS: ?

SPEED :12kts

SET: 130T

RATE: 2kts
CMG: ?
SMG: ?

Step 1: Draw the three bearings.

Step 2: Draw a perpendicular to the middle bearing passing through the light as shown below.

Step 3: Now we need to select a ratio to cut on the perpendicular drawn in the last step, this ratio could be of time or the distance, since distance is unknown here we will do it with time

In terms of time 

For first stretch i.e. 0800 hrs to 0820 hrs the time interval is 20’

For the second stretch i.e. from 0820 hrs to 0900 hrs the interval is 40’

So the ratio is 20:40 or 1:2,

To get a clearer and less clustered working take ratio which is large, here we took 3:6.

The first value (3’) should be cut on the perpendicular at that time interval i.e. between 0800hrs and 0820 hrs, and second value (6’) should be cut on the perpendicular between 0820 hrs and 0900 hrs.

Step 4: Transfer the centre bearing onto these cut arcs on the perpendicular.

These meet the first and last bearing at a point, draw a line from the point on the first bearing to the last bearing, this is the CMG.

So we can conclude that we always get the CMG using 3 point bearing method.

Step 5: Since the CTS is not given this is a counteracting method, so we need to apply the current at the point on the first bearing. So draw the current vector 130T X 2 nm.

Step 6: With C as centre measure 12 miles i.e. ships speed on the compass and cut and arc on the CMG line (extend the CMG line if required). 

Step 8: Transfer the first bearing on to point D , this meets the last bearing at a point, this is the fix at 0900 hrs.

Step 9: Transfer the CMG line onto the fix at 0900 hrs, this cut the first bearing at a point, that’s the fix at 0800hrs and the second bearing at a point, that’s the fix at 0900hrs.

CASE 3: When SET is not given, but rate given.

Lets take the following example:

Hook head light bore 020T at 0800hrs , and at 0820 hrs it bore 350T and at 0900 hrs it bore 300 T, fins the vessels position at 0800hrs,0820hrs and 0900hrs.

Given data:

CTS: 040T

SPEED :12kts

SET: ?

RATE: 2kts
CMG: ?
SMG: ?

A hint would be given in these type of question whether the current is a supporting one or an opposing current, we will see how to go about then.
Step 1: Draw the three bearings.

Step 2: Draw a perpendicular to the middle bearing passing through the light as shown below.

Step 3: Now we need to select a ratio to cut on the perpendicular drawn in the last step, this ratio could be of time or the distance, since distance is unknown here we will do it with time

In terms of time 

For first stretch i.e. 0800 hrs to 0820 hrs the time interval is 20’

For the second stretch i.e. from 0820 hrs to 0900 hrs the interval is 40’

So the ratio is 20:40 or 1:2,

To get a clearer and less clustered working take ratio which is large, here we took 3:6.

The first value (3’) should be cut on the perpendicular at that time interval i.e. between 0800hrs and 0820 hrs, and second value (6’) should be cut on the perpendicular between 0820 hrs and 0900 hrs.

Step 4: Transfer the centre bearing onto these cut arcs on the perpendicular.

These meet the first and last bearing at a point, draw a line from the point on the first bearing to the last bearing, this is the CMG.

So we can conclude that we always get the CMG using 3 point bearing method.

Step 5: Now draw the CTS from the point on the first bearing, and cut out the engine speed for the total duration of the observation, in this case it is 1 hr,let this point be A.

Step 6: From point A as centre and measuring the rate for 1 hour i.e 2 nm cut an arc, this arc cuts the CMG track at two points, this is where you have to decide if the question says opposing or supporting current, if the current is opposing the vessel would steam less distance and if supporting more distance. This is a simple way to determine it. In this case we have considered an opposing case.

Step 7: From point A draw a line to the point marked as opposing on the CMG track, if we measure out this line on compass rose we get the SET.

Step 8:Transfer the first bearing onto the point marked as opposing, this position line will meet the last bearing at a point, this is the fix at 0900hrs.

Step 9: Now transfer the CMG line onto the fix at 0900 hrs, this will cut the first bearing at a point, which is fix at 0800hrs, and the second bearing at a point which is the fix at 0820 hrs.

CASE 4: When only SET and RATE is given.

Lets take the following example:

Hook head light bore 020T at 0800hrs , and at 0820 hrs it bore 350T and at 0900 hrs it bore 300 T, fins the vessels position at 0800hrs,0820hrs and 0900hrs.

Given data:

CTS: 

SPEED :

SET: 130T

RATE: 2kts
CMG: ?
SMG: ?

In these questions you have only 2 data which is insufficient , so one fix would be given.

Step 1: Draw the three bearings.

Step 2: Draw a perpendicular to the middle bearing passing through the light as shown below.

Step 3: Now we need to select a ratio to cut on the perpendicular drawn in the last step, this ratio could be of time or the distance, since distance is unknown here we will do it with time

In terms of time 

For first stretch i.e. 0800 hrs to 0820 hrs the time interval is 20’

For the second stretch i.e. from 0820 hrs to 0900 hrs the interval is 40’

So the ratio is 20:40 or 1:2,

To get a clearer and less clustered working take ratio which is large, here we took 3:6.

The first value (3’) should be cut on the perpendicular at that time interval i.e. between 0800hrs and 0820 hrs, and second value (6’) should be cut on the perpendicular between 0820 hrs and 0900 hrs.

Step 4: Transfer the centre bearing onto these cut arcs on the perpendicular.

These meet the first and last bearing at a point, draw a line from the point on the first bearing to the last bearing, this is the CMG.

So we can conclude that we always get the CMG using 3 point bearing method.

Step 5: Lets say that we are given a fix at 0900hrs, now transfer the CMG line onto the fix at 0900hrs, therefore we get the fix at 0800 hrs and 0820 hrs.

Step 6: Now  from the fix at 0800 hrs draw the Set and rate for the duration of the observation , since this observation is for 1 hr we draw X 2nm.

Step 7: From point C draw a line that meets the fix at 0900 hrs, this is the CTS, if we measure its length we get the ships speed, note that the value of ships speed for one hour is required, if the observation exceeds one hour or is less than one hour, convert the same to 1 hours speed.

12: OBSCURED AND VISIBLE CASES.

Every light has an obscured sector and a visible sector, which can be seen from the image above.

In these questions you would be given a fix at a time when let’s say the south light is visible, and the vessel steams for some time, let’s say in 1 hour, the north light is visible and after steaming for 30 mins, the south light bore an angle.

This question is a type of Three point bearing question and the approach would be the same.

Let’s take an example and see step by step.

We have a fix at 0800hrs the vessel steams on a course and its observed that at 0900hrs the north light is visible (this means vessel has just come out of the obscured sector of the light). Later the vessel observes the south light to bore 030T again, we are asked to find the vessels position at 0800hrs, 0900hrs and 0930 hrs.

Step 1. Plot the fix at 0800 hrs, and at same time draw the bearing at 0930 hrs and draw a line on the obscured sector of the north line as show below.

Step 2: Join a line from the fix at 0800 hrs to the point of intersection of the bearing and the obscured line as shown below.

Step 3: Now as we did before draw a perpendicular through the centre bearing and at the point of intersection.

Step 4: Since the time duration between 1st and 2nd observation is 60’ and between 2nd and 3rd is 30’, the ratio would be 2:1. Cut down the ratio as shown above.

Step 5: Transfer the centre bearing to the arcs cut out on the perpendicular this will meet the first and last bearing.

Step 6: Draw a line connecting the first and last bearing from the points where the perpendiculars cut the first and last bearing. This would give us the CMG

Step 7: Transfer this CMG to the fix at 0800hrs this will meet the 2nd bearing at a point, this is fix at 0900 hrs and 1st bearing at a point, this is fix at 0930 hrs.

13: VERTICAL SEXTANT ANGLE

The angle between a coastal feature’s height (the centre of a light) and the waterline can be measured using a navigational instrument sextant. This vertical sextant angle can then be used to find the distance off by entering the pre-calculated VSA tables or by trigonometry.

This question would consist of a vertical sextant angle given, the index error may or may not be given, Height of eye correction is not being applied. But if there is any tide data given the tide should be considered.

To calculate the distance off we use the following formula= 

Height of object    X   1.854

VSA (in minutes)

This distance off can be used in the question. 

Questions would come with VSA combined with any other of the concepts such as sighting object right ahead etc.

14: SAILING AROUND AN ARC

Lets learn some basics first:

We have a circle of radius r.

Now let’s mark out a sector and say that the circumference of the sector is r.

Therefore in such cases the angle subtended at the centre is 57.3 degrees

We use this relation in the sailing around an arc concept.

So for any angle b subtended in the sector , the circumference of the sector would be 

Radius       x angle b

57.3

Now in real life on board, let’s say we have this scenario

You are asked to plot a course from point A to B. In this leg the master asks to maintain a 5-mile distance at all times while passing a light. The master asks you the total distance between the point A and B

Step 1:Draw the course line

Step 2: As you can see as per master’s instruction the vessel has to pass Foreland point light at a range of 5 miles, this may be due to the presence of some obstruction to navigation like a wreck.

Step 3: Draw a circle of radius 5 miles with the light as the centre.

Step 4: So the total distance covered by the ship would include the distance along the straight line (rhumb line course) and the around the arc of the circle(which is the circumference of a sector)

Step 5: Draw two lines from the centre of the lighthouse to the point where the circle cuts the straight line. This is the radius of the circle.

Step 6: This satisfies the condition as said above and therefore we can use the formula,  Radius       x angle b , to get the circumference of the arc.

                  57.3

Here angles b is the angle of the sector.

Therefore, we can find the circumference and therefore calculate the total distance.

15: CELESTIAL FIXES

It is a method of obtaining a ships position using astronomical position lines.

Astronomical positions lines are a part of a very large radius position circle.

Since drawing a large position circle is not possible we draw a position line, close to our DR position.

Therefore, a small portion of the position circle can be considered as a position line.

We have 2 types of situations here 1. Simultaneous fix

                                                               2. Running fix.

Case 1: Simultaneous fix.

Let’s consider the following example.

At a DR position the vessel observed 2 intercepts

1. AZ 045T intercept 2’ away.

2. AZ 130T intercept 3’ towards.

Step 1: Plot the Dr position and mark out the first reading.

Step 2: Now mark put the second one.

The two position lines meet at a point; this is the fix at that point.

Case 2: Running fix

In these type you will be given with a astronomical reading at one time , after which the ship would sail a distance for some time and reach another place where the  observer would take another astronomical observation.

They would ask the fix at the second observation, initial DR would be given.

Let’s consider the following example.

At 0800hrs AZ 030T, intercept 3’ towards.

At 0900 hrs AZ 330T, intercept 3’ towards.

Ships course 100T, ships speed =10kts.

Step 1: Draw the DR position and the first observation.

Step 2: From the DR position at 0900 hrs the vessel steams 100T for 10 miles. The arrived position is the DR for second observation

Step 3: Now Draw the second observation at the Dr 0900hrs.

Step 4: Transfer the first observation onto the DR at 0900 hrs, the two position lines would meet at a point, this is the fix at 0900hrs.

SOME SPECIAL CASES

1. At 0800 hrs the following compass bearings were observed.

Brownstown head – 350 C

Hook head              -   050 C

The ship then set on a course of 130 T at a speed of 15 kts. Current was setting south at 2kts rate.

At 0900 hrs the vessel observed Conninberg light at a radar range of 10 miles. Find the vessels position at 0800hrs and 0900hrs.

Step 1: First we will solve the HSA part, here

Θ= (360-350)+50= 60 degrees

α=90- θ = 30 degrees.

Step 2: Draw a straight line connecting the two lighthouses.

Step 3: Now measure an angle of 30 degrees from each lighthouse by using   the protractor keeping the line joining the two houses as reference line. These two lines meet at a point, let this be point C.

Step 4: Taking the radius as R (from C to anyone of the light), draw a circle, this circle passes through both the lighthouses. The circle obtained is the position circle of 0800hrs, the vessel could be anywhere on this circle.

Step 5: In these questions we need to apply the chart work triangle on the point C .Draw the chart work triangle with given data.

Step 6: Now its given the Conninberg light at a radar range of 10 miles, keeping light as center draw a circle of 10 miles.

Step 7 :Keeping point A on the chartwork triangle as center and radius equal to R(first position circle), cut an arc on the second position circle.This would cut at a point, this point is the fix at 0900hrs.

Step 7: Now transfer the CMG onto fix at 0900rs fix, this meets the first position circle at a point this is the fix at 0800hrs.

2.Bill of Portland light and S.Catherines lt are at opposite bearings at 0800 hrs.  Also at 0800 hrs it was observed that LtV channel was 30 degrees on stbd bow. From here the vessel steams for 30 mins and observers the same LtV channel at 60 degrees on stbd bow. Engine speed 12 kts, we are asked to find the ships position at 0800hrs.

Step 1: This is a type of doubling of angle at the bow question. Before you head to the chart you need to do some rough work on your answer sheet.

Step 2: The diagram as per the question would be something like the one below.

There are two light houses in opposite bearing, thus we can conclude that the vessel would be on a position line, this position line is simply the line joining both the lighthouses.

Step 3: Take an arbitrary point on the position line, draw a straight lines downwards from this point. Also from the arbitrary point measure 30 on stbd and join a line to the light vessel as shown below.

Step 4: Now it is said that the vessel steams for 30 mins , so it covers 6 miles as engine speed is 12 kts, so cut out 6 miles on the straight line as shown below.

Step 5: Measure 60 degrees on this point and join a line to the light house.

Step 6: Now extend the straight line and draw a perpendicular to the light house. We get 2 triangles CMF(big) and DMF(small)

Step 7:  In the small triangle DMF we know angle M is 30, so applying sin 30 we get the value of DF as 3 miles as shown below.

Step 8: For calculating the range off from the light vessel at initial point we need to find the hypotenuse of the bigger triangle.

In CMF we know angle M is 60 ,so by applying sin 60 we get CM as 18 miles.

All these to be done in rough

In the chart all you have to do is to join both the light houses. from the light vessel take 18 miles and cut an arc on the position line, this given you the fix at 0800hrs.

3.Finding the CTS after an interval of time to have a light abeam on port or stbd side, without current.

Lets say you are given a fix at 0800hrs, ships speed 12kts. We need to find the course to steer to get the lighthouse abeam on both port/stbd side.

No current or leeway given in these questions.

Step 1: Draw a circle keeping the fix at 0800hrs as centre and radius as the engine speed.

Step 2: From the lighthouse draw 2 tangents that meet the circle.

Step 3: From the fix at 0800 hrs draw lines to the point where tangents meets the circle

Therefore, we get the course to steer to get lighthouse abeam on stbd/port side as shown in the image above.

IMP:

• Some lights would have their characteristic light as White Red Green, i.e. WRG, if you are asked to solve using these lights always take the white lights Nominal range.

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