CHARTWORK - CONCEPTS OF PLOTTING
CHARTWORK
Chartworknumericals
are a combination of one or more concepts.So it is necessary to understand
these concepts.
Position
Line (PL) means ship is anywhere on this line.
Position
circle (PC) means ship is anywhere on this circle.
Position
fix can be obtained by
-
Intersection
of 2 or more such PL or PC at the same time.
-
Lat/long
-
Position
can be given by chart datum depth on chart, e.g. 9m.
I)
Position
Lines(PLs) or Position circles (PCs) can be found by one of the below concepts.
The
following concepts give a PC
1.
HSA(Ѳ)= 90º or HSA(Ѳ)= <90º or
HSA(Ѳ)= >90º
2.
VSA
4.
Doubling
the angle on the bow –
i)PC at time of 1st
angle ii)PC at time of double
the angle
5.
Special
angles
6.
Raising
or Dipping
7.
First
sighted or Last sighted
The
following concepts give a PL
1.
Bearing
of a Lt Ho. or Object.
2.
PL
of a celestial body (Lat by mer.Alt., Long by chron, Intercept, Ex-meridian,
Polaris).
3.
Transit
bearing of Lights.
4.
HSA
= 0° (Same as transit bearing).
5.
HSA
= 180° (PL is in between boththe Light Houses or Objects).
6.
When
a light has different sectors of lights e.g. WRG
7.
When
a light has an obscured sector.
8.
When
2 lights are equidistant.
9.
Depth
contour on chart
a) When a light has different sectors of
lights e.g. WRG
i)
PL is when Lt changes
from W to R or R to W orW to G or G
to W,
ii)
PL is when R Lt
becomes visible or when G Lt becomes visible.
b) When Light has an obscured sector.
i)
PL is when light was
obscured & now becomes visible.
ii)
PL is when light was
first visible & now becomes obscured
c) When 2 lights are equidistant.
-
The perpendicular
bisector is the PL.
(Take
a distance of more than ½ between the 2 lights & cut on each side. The line
joining the 2 intersections is the perpendicular bisector i.e. the PL)
d) Horizontal Sextant Angle (HSA)between
2 Lt Ho or objects gives us a PC or PL
1. When Compass bearings of Lt. Ho. are
given
a)
If
the Ship’s Compass Heading/course
is NOT given and the Compass bearings of Lt. Ho. is given,
it is a HSA method.
-
The
side of the Ship’s position can be
found by roughly plotting the compass bearings or is opposite to land.
-
When
3 compass brgs are given. First arrange the Lt Hos. compass bearings from port
to Stbd (clockwise) with respect to position.
-
Join
the 2 lights.
-
Do
the same for the 2nd& 3rdCompass bearings
i)
If HSA(Ѳ) <
90°,
drawAngle = (90° – Ѳ)
towards
ship’s positionand
centre
of PC
will be the intersection of the 2 angle
lines at C,
ii)
If HSA(Ѳ) >
90°,
drawAngle = (Ѳ
- 90°)on the opposite side
to ship’s position&
centre
of PC
will be the intersection of the 2 angle
lines at C.
iii)
If HSA(Ѳ) =
90°,
centre
of the PCis
the mid-point(C) between the 2
lightsfound by using perpendicular
bisector method.
- Take a distance of
more than ½ between the 2 lights & cut on each side.
iv)
If HSA(Ѳ) =
0°, We get a PL.
e.g. brg of A is
300º(C) & of B is also 300º(C)
The compass error can be
obtained by comparing the True brg from the chart and compass brg.
e.g. brg of A is
270º(C) & of B is also 090º(C)
b)
If the Ship’s compass heading is given, and the compass bearings of
the 2 Lt.Ho.is given, then it is NOT HSA
method.
-
First
find the compass Error using the ship’s Compass heading, then convert compass
brgs to true brgs and plot directly on the chart. The intersection of the PLs
will give the ship’s position.
2. When Gyro bearings of Lt. Ho. are
given
If the Gyro brgsof 2 Lt Hos.are
given and Gyro error is NOT given, then
we have to use the above HSA concept.
e.g. Gyro brg of A is 160º(G) &of B is 230º(G), so HSA(Ѳ) = 70º
3. When Horizontal Sextant Angle (HSA)
between Lt. Ho. are directly given
e.g. HSA between Lt A & Lt
B is 60º, so HSA(Ѳ)=60º
e.g.Vessel at anchor, 3 Compass
brgs of Lt Hos. are given:
P – 280º(C), Q - 355º(C), R - 095º(C). Find Compass Error.
Solution: P – 280º(C)
Q - 355º(C),
R - 095º(C).
P –
280º(C)
Q -
355º(C),
HSA(Ѳ) = 75º
As HSA(Ѳ) < 90º, Angle = 90º- Ѳ = 90º-75º = 15º
Q -
355º(C),
R - 095º(C).
HSA(Ѳ) = 100º
As HSA(Ѳ) > 90º, Angle = Ѳ - 90º = 100º - 90º =
10º
Compass brgs P = 280º(C), Q
= 355º(C), R = 095º(C).
If from plot T. brgs of P = 282º(T), Q = 358º(T) R
= 096º(T)
Compass Error of P =
2ºE Q = 3ºE R
= 1ºE
Average Compass Error = (2º + 3º + 1º) =
2ºE
3
e) Vertical Sextant Angle (VSA), gives PC
Distance off
in n.miles = Height of Lt House
x 1.854
VSA in minutes
Ht
of Lt Ho=
the ht of Lt Ho given on the chart +/- the diff betwn MHWS & Present water
level
(+ve if present level is below MHWS
&-ve if present water level is
above MHWS)
VSA
=
Sextant angle +/- I.E
f) Raising or Dipping (Geographical Range
– G.R), gives PC
As the earth is a sphere, the
line of sight depends upon the Ht of Eye (H) &ht of Lt Ho (h). When the light is raised, the light
will be just in the line of sight (vessel goes towards the light).
When
the light is dipped,the
light just about goes out of line of sight(vessel goes away from light).
G.R(distance from Lt Ho)=
2.095√H+2.095√h)
WhereH - Ht of Eye, h – ht of Lt Ho.
g) First sighted or Last sighted, gives
PC
-
Find Geographical
Range (G.R), if
H = 12m & h = 25m
G.R = 2.095√H + 2.095√h
= 2.095√12 + 2.095√25= 7.3 + 17.7 = 25nm
-
Find Luminous Range
(L.R) from
the Luminous Diagram using Nominal Range(given for 10nm met.Visibility)from
chartandpresent meteorological visibilitycurve (e.g.5nm met. Visibility& NR
of Lt Ho. is 20nm)
From
LR diagram, LR = 11.8nm
-
Compare
the G.R & L.R and use whichever is less to draw the PC
-
LR
< GR, so draw PC of LR = 11.8nm
h) Special angles(Ѳ&)
AC -
Distance off at the time of 1st angle (Ѳ),
gives a PC
BD–Beam distance
AB – distance run between A
& B
By
Trigonometry,
When CotѲ
- Cot = 1, where
Ѳ =1st
angle on the bow &=2nd
angle on the bow
Then Run = Beam distance, i.e. AB = CD
AC =
Run / Sin
Ѳ
i) Doubling the angle on the bow,gives PC
i)
PC at the time of double the angle
BC
-
Distance off at the time of double the
angle (2Ѳ), gives a PC
AB – distance
run between A & B
AB
= BC (doubling the
angle on bow method)
ii)PC
at the time of 1st angle
AC
–
Distance off at the time of 1st
angle (Ѳ), gives a PC
CD - Beam
distance
AB – distance
run between A & B
AB = BC (doubling the angle on bow method)
CD = BC x Sin
2Ѳ
AC = CD / Sin Ѳ= BC
x Sin 2Ѳ / Sin Ѳ
AC
= (Run x Sin 2Ѳ) / Sin
Ѳ
j) Depth contour on chart, gives a PL
e.g. When a ship crosses a10m
contour
II)
Allowing
for Current and/or Leeway
The size of
the triangle depends on the time in the question
a)
Allowing
for Current (when CTS, EngSpd, Set & Rate given)
E.g. a 1 hour
triangle
- Plot
the CTS from the posn&Cut the Engspd
- From
the DR plot the Set & Rate gives EP
- Join
the Posn& EP gives us CMG & SMG
- Eng
dist = Engspd x (Time interval in mins)
60
- Drift
= Rate x (Time interval in mins)
6
b) Allowing for Current & Leeway (CTS, EngSpd, leeway, Set & Rate
given)
E.g. a 1 hour
triangle
-
First
plot the CTS
-
Then
apply leeway track opposite to wind direction
-
Cut
Engspd on the Leeway track
-
From
DR apply Set & Rate, gives EP
- Join
the Posn& EP gives us CMG & SMG
- If
triangle is for > or < 1 hour than we use Engdist& drift.
- Engdist
= Engspd x (Time interval in mins)
60
- Drift
= Rate x (Time interval in mins)
60
III) Counteracting Current and/or leeway
- This concept
is used to find the CTS when CMG is
given.
CMG
may be given by one of the methods.
i.Vessel
needs to pass ‘x’ miles off a Lt Ho. or object from a posn.
- Draw an arc
x’ miles from Lt Ho.
- Draw a tangent to this arc from
posn. This is the CMG
ii.Frompresentposnvessel
needs to go to another posne.g. A pilot station.
iii.
Three point bearings.
a)
Couteracting
for Current(Engspd,
set & Rate given)
- Find
CMG
- First
from present posn ‘A’ apply Set &cut Rate at ‘B’
- From
‘B’ cut Engspd on the CMG at ‘C’
- Join
‘B’ & ‘C’, this is the CTS.
- ‘AC’
is the SMG
- ABC
is a 1 hour triangle
b)Counteracting
for Current & leeway
- Find
CMG
- First
from present posn ‘A’ apply Set & cut Rate at ‘B’
- From
‘B’ cut Engspd on the CMG at ‘C’
- Join
‘B’ & ‘C’, this is the LEEWAY TRACK.
- ‘AC’
is the SMG
- From
‘B draw apply leeway & CTS towards the wind
- ABC is a 1 hour triangle
IV) Transfer of PL or PC
-
A PL can be
transferred from any point along the line.
-
A PC can be
transferred only by transferring the centre of the circle
a)
Transfer
of PL without current, when 2 brgsgivenat different times from the same Lt Ho.
-
Draw
both PLs from the Lt Ho
-
Anywhere
along the 1st PL, from ’A’ draw the CTS & cut Engspd at ’B’.
-
Transfer
1st PL to ‘B’
-
The
PLs will intersect at ‘C’, this is the 2ndposn
-
Reverse
plot the CTS from ‘C’
-
CTS
intersects at ‘D’, this is the 1stposn
b)
Transfer
of PL with current, when 2 brgsgivenat different times from the same Lt Ho.
-
Draw
both PLs from the Lt Ho
-
Anywhere
along the 1st PL, from ’A’ draw the CTS & cut Engspd at ’B’.
-
From
‘B’ draw the Set & Rate to ‘C’
-
Transfer
1st PL to ‘C’
-
The
PLs will intersect at ‘D’, this is the 2ndposn
-
Reverse
plot the CMG from ‘D’
-
CMG
intersects at ‘E’, this is the 1stposn
c)
Transfer
of PL with current, when 2 brgsgivenat different times from the different Lt Hos.
-
Draw
the PLs from Lt Ho.1 & Lt Ho.2
-
Anywhere
along the 1st PL, from ’A’ draw the CTS & cut Engspd at ’B’.
-
From
‘B’ draw the Set & Rate to ‘C’
-
Transfer
1st PL to ‘C’
-
The
PLs will intersect at ‘D’, this is the 2ndposn
-
Reverse
plot the CMG from ‘D’
-
CMG
intersects at ‘E’, this is the 1stposn
d) Transfer of PL with current, when 1 PL
& 1 PCgiven at different times from the different Lt Hos.
-
Draw
the PL from Lt Ho.1 & PC from Lt Ho.2
-
Anywhere
along the 1st PL, from ’A’ draw the CTS & cut Engspd at ’B’.
-
From
‘B’ draw the Set & Rate to ‘C’
-
Transfer
1st PL to ‘C’
-
The
transferred 1stPL will intersect the PC at ‘D’, this is the 2ndposn
-
Reverse
plot the CMG from ‘D’
-
CMG
intersects the 1st PL at ‘E’, this is the 1stposn
e)
Transfer
of PC with current, when 2 PCsare given at different times from the different
Lt Hos.
-
Draw
the PC from Lt Ho.1 & PC from Lt Ho.2
-
From
the centre of the PC ’A’, draw the CTS & cut Engspd at ’B’.
-
From
‘B’ draw the Set & Rate to ‘C’
-
Take
the radius of PC1 anddraw from ‘C’ the transferred PC1
-
The
transferred PC1 will intersect the PC2at ‘D’, this is the 2ndposn
-
Reverse
plot the CMG from ‘D’
-
CMG
intersects the PC1at ‘E’, this is the 1stposn
f)
Transfer
of PC with current, when1 PC & 1 PL is given at different times of
different Lt Ho.
-
Draw
the PC from Lt Ho.1 & PL from Lt Ho.2
-
From
the centre of the PC ’A’, draw the CTS & cut Engspd at ’B’.
-
From
‘B’ draw the Set & Rate to ‘C’
-
Take
the radius of PC anddraw from ‘C’ the transferred PC1
-
The
transferred PC will intersect the PL at ‘D’, this is the 2ndposn
-
Reverse
plot the CMG from ‘D’
-
CMG
intersects the PC at ‘E’, this is the 1stposn
g) Transfer
of PL by allowing for Current
h) Transfer of PC with current, when PC1
(first sighted)& PC2 (Lt Dipped) is given at different times of same Lt Ho.
-
Draw
the PC1& PC2 from Lt Ho.1
-
From
the centre of the PC ’A’, draw the CTS & cut Engspd at ’B’.
-
From
‘B’ draw the Set & Rate to ‘C’
-
Take
the radius of PC1 anddraw from ‘C’ the transferred PC1
-
The
transferred PC1 will intersect the PC2at ‘D’, this is the 2ndposn
-
Reverse
plot the CMG from ‘D’
-
CMG
intersects the PC1at ‘E’, this is the 1stposn
V)
Find
the Actual Set & Drift
when
actual posn (D) after sometime is not on the CMG.
- First
by Counteracting method find the CTS
- Than
draw the planned leeway track from A & cut the Eng dist for the required
time interval.
- Join
the DR & the Posn2 ‘D’. This is the Actual Set & Drift.
DR
VI)
Find
the posn& time when the light will be abeam.
- First
by Counteracting method find the CTS
- Then
from the Lt Ho draw the beam bearing (=CTS +/- 90º).
- The
beam Brg will intersect the CMG at ‘D’.
- Measure
AD = DTMG
- TTG
= DTMG/SMG x 60’
VII) RAISING or FIRST SIGHTING or DISTANCE
OFF – THE LIGHT θº ON THE PORT or STBD BOW
(whenEngine Speed is given&without
current)
- Find:AC = (GR-
Raising)or (GR or LR whichever less - First Sighting) or (Distance off)
-
Draw
the PosnCircle from the Light with AC as radius.
-
Find:Beam distance(BC)= AC x SinѲ
-
Draw
a tangent to the beam distance arc, from present position. This is the CTS
-
At
posn‘A’, the light will be Ѳº on the port bow.
-
Measure DA = DTG
-
Find: (Time to go) TTG = DTG/Engspd x 60’
VIII) Raising or First Sighting or Distance
Off, the Light RIGHT AHEAD with current)
- Find: AC = (GR - Raising)
or (GR or LR whichever less - First Sighting) or (Distance off)
-
Draw the PC from the Light with
AC as radius.
-
Find:Drift (BC) = AC x (Rate / Engspd)
-
Mark the drift on
the Set at B.
-
Draw
a line from posn to ‘B’. This is the CMG
-
Measure AB= DMG
-
Find: SMG = AB x (Rate/drift)
-
MeasureDA = DTMG
-
Find: (Time to go) TTG = DTMG/SMG x 60’
IX) Raising or First Sighting or Distance
Off, the Light RIGHT AHEAD without current)
- Find:BC = (GR - Raising)
or (GR or LR whichever less - First Sighting) or (Distance off)
-
Draw
the PosnCircle from the Light with BC as radius.
-
Draw
a line from ‘A’ to ‘C’. This is theCTS.
-
Measuredistance
AB = DTG
X)
Find
the Unknown, when from a posn various alterations of courses&/or current
and a PL/PC are given (1 unknown)
e.g. various A/Cs & set is
given. Find drift.
- First plot all the KNOWN
Courses & Set from the present posn.
- The UNKNOWN should always be
plotted at the end.
- Then plot the PL/PC from the
Lt ho.
- Where the Set& PL/PC
intersect the UNKNOWN (Drift) can be found.
- The 2ndPosn is the
point of intersection.
XI) Find the Unknown, when from a posn
various alterations of courses&/or current and 2ndposn is given
(2 unknowns)
e.g. various A/Cs & set is
given. Find drift& End Spdbetween 0800-0830.
- First plot all the KNOWN
Courses from the 1stposn.
- Thenplot the 0800-0830 Course
- Then plot the PL&PC from
the Lt ho to obtain the 2ndPosn.
- Plot Set, reverse from 2ndPosn.
- Then measure the
UNKNOWNs(Engdistbetwn 0800-0830 & Drift).
XII) Find the CTS after a interval of time
to have a Light Abeam on the Port or Stbd side, without current
- First draw a circle with
radius of Engdist for the time interval.
- Then draw tangets on either
side of the circle from the Lt Ho.
- Join the Posn& tangent
pts on either side. These are the CTS
to have the Lt abeam
on port or stbd side of the
ship as shown in the figure.
XIII) Three – Point bearings from Lt
Ho./Point/object
-
This
concept gives us the CMG
-
We
can find 1 unkown& the 3 Positions
-
When
a position is given, we can find 2 unknowns & the other 2 positions
-
If
Ship’s Compass heading and 3 Compass brgs of Lt Ho are given at different
times. Find Compass Error & convert
to true brgs and do the question by 3 point bearings method
-
There
are 4 types of 3–point brgs
-
a)
When 3 brgs at
different times are given from a single Lt Ho./Point/object.
b)
When 2 brgs at
different times are given from a single Lt Ho./Point/object & a position is
given.
c)
When brgs are given
at different times from 2 different Lt Hos., intersect at a point & a
position is given.
d)
When brgs from 3
different Lt Hos. are given at different times intersect at a point.
e)
i) 3 point brgs, CTS, EngSpd& Set are
given. Find DRIFT of current & all 3
posns OR
3 point brgs, CTS, EngSpd&
Rate are given. Find SET of current
& all 3 posns
-
Draw
all 3 brgs from the Lt ho
-
Draw
a line perpendicular to 2ndbrg through the Lt Ho.
-
Find
the ratio of time interval between 1st& 2ndbrg and
betwn 2nd& 3rdbrg.
-
Cut
the ratio from the the Lt Ho. - first at ‘B’, then at ‘C’
-
Draw
a line parallel to 2ndbrg, from ‘B’ to intersect at ‘D’
-
Draw
a line parallel to 2ndbrg, from ‘C’ to intersect at ‘E’
-
Join
‘D’ & ‘E’, this is the CMG.
-
From
‘D’ draw the CTS & cut the Engdist at ‘F’
-
From ‘F’ draw the
Set.
‘FG’ is the Drift
-
If Rate is given
find Drift& from ‘F’ cut Drift at ‘G’. Direction ‘FG’ is the SET
-
Through
G transfer the PL1
-
Where
the Transferred PL1 intersects PL3 is
the Last Posn
-
Draw
a reverse CMG from last Posn& find other 2 Posns.
-
e.g.
ii)
3
point brgs, EngSpd, Set Rate are given.
Find CTS& all 3 posns OR
3 point brgs, CTS, Set&
Rate are given. Find EngSpd& all 3
posns
-
Find
the CMG (DE) as explained above.
-
From
‘D’ draw the Set & cut the Drift at ‘F’
-
From ‘F’ cut the
Engdist on the CMG at ‘G’.Direction of ‘FG’ is the CTS
-
If CTS is given from
‘F’ draw the CTS to intersect the CMG at ‘G’. ‘FG’ is the- Engdist
-
Through
G transfer the PL1
-
Where
the Transferred PL1 intersects PL3 is
the Last Posn
-
Draw
a reverse CMG from last Posn& find other 2 Posns.
iii)
2
point brgs, CTS, EngSpdare given. Find
Set &DRIFT of current &other2posns OR
2 point brgs, CTS, Setare
given. Find Engdist& DRIFT of
current &other2posns
-
Draw
the 2brgs from the Lt ho.
-
Join
the posn to the Lt ho& make it 3 point brgs
-
Draw
a line perpendicular to 2ndbrg through the Lt Ho.
-
Find
the ratio of time interval between 1st& 2ndbrg and
betwn 2nd& 3rdbrg.
-
Cut
the ratio from the the Lt Ho. - first at ‘B’, then at ‘C’
-
Draw
a line parallel to 2ndbrg, from ‘B’ to intersect at ‘D’
-
Draw
a line parallel to 2ndbrg, from ‘C’ to intersect at ‘E’
-
Join
‘D’ & ‘E’, this is the CMG.
-
Transfer
the CMD through Posn at ‘F’.
-
The
CMG will intersect PL1 at ‘J’ (Posn at 0830) & PL2 at ‘I’ (Posn at 0900).
-
From ‘F’ draw the
CTS & cut EngSpd at H (DR).
-
Join
H to I. ‘HI’ is the Set & Rate of
current.
OR
-
If CTS & Set are
given, From ‘F’ draw the CTS &from I draw reverse Set to intersect CTS at
‘H’ (DR).
-
Measure
‘HI’ = (Rate) & ‘FH’ =
(EngSpd).
-
XIV)
(Chart
5072, Spd 12 kts, HE 12m, Dev card No.2, Var 1ºW)
Q.1.
A vessel at anchor observes the following compass bearings
a) Christiano (S) Lt. (55º 19’N 015º
11’E) 065º(C)
b) Svanke Lt. (55º 08’N 015º
09’E) 161º(C)
c) Hammerodde Lt. (55º 18’N 014º 47’E) 284º(C)
Find the vessel’s position and
the deviation on the ship’s head.
Solution:
Christiano (S) Lt. 065º(C) Svanke Lt. 161º(C)
Svanke Lt. 161º(C) Hammerodde Lt. 284º(C)
HSA(Ѳ) = 96º HSA(Ѳ) = 123º
As HAS(Ѳ) > 90º, As
HAS(Ѳ)
> 90º,
Angle = Ѳ
- 90º =
96º - 90º = 6º Angle
= Ѳ - 90º
= 123º - 90º = 33º
Position: 55º 16.3’N 015º 02.7’E
Christiano (S) Lt. Svanke Lt. Hammerodde Lt.
Compass brgs 065º(C), 161º(C) 284º(C).
If from plot T. brgs of 82º(T),
358º(T) 096º(T)
Compass Error of 2ºE 3ºE 1ºE
Average Compass Error = (2º
+ 3º + 1º)= 4ºW
3
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