91121543 roti (1)

1 | P a g e W h a t i s R O T I ( R a t e O f T u r n I n d i c a t o r ) ?
For more details please visit: www.captyashpal.blogspot.com
RATE OF TURN INDICATOR
(R.O.T.I.)
Unlike road vehicles a ship does not turn sharply. With lesser friction in water and under the
influence of momentum a vessel continues on her initial course for some time, beginning to turn
slowly and then rapidly later. Vessel thus traverses a curved track which can be treated as an arc of
a circle.
For small ships such arcs are small and can be executed without much problem. But for large sized
merchant vessels these arcs assume greater radii and need to be treated differently. This becomes
all the more important when vessel is altering courses in restricted waters or in close proximity to
navigational hazards.
For same reasons roundabouts are charted in traffic separation schemes.
Under navigation watch keeping principles, vessel is required to do berth to berth passage
planning. It is also required to lay courses in curved segments as well, where required and to mark
the wheel over positions.
In view of the above, vessels having 50,000 GT and above are mandatorily required to be fitted
with ROTI (Rate Of Turn Indicator) as per Chapter V, SOLAS.
ROTI assists OOW in planning, executing and monitoring vessel’s progress along curved segment
of charted course.

The purpose of ROTI is to provide rate of turn to port and to starboard side of ship. The indicator is
usually in the form of a circular dial with zero at top. Port turn is indicated on left of zero and
starboard turn on right of zero. Graduations are provided to indicate ROT up to at least 30
degrees/minute on either side. ROTI can be self-contained, or it may derive information from other
equipment or it may form a part of the other equipment. However, the design is such as to preclude
degradation of other equipment to which ROTI is connected, irrespective of weather ROTI is in
operation or not.
DERIVATION OF FORMULA FOR ROT:
ROT (/t) is expressed in degrees per minute.
Consider following diagram:
PB = Initial course
BN = Final course
AC = Curved segment of charted course;
A is commencement of turn and C is completion of turn.
Length of the curved segment = d
 = Amount of alteration in degrees (Angle MBN or Angle AOC)
C
=  in radians
R = Radius of the circle of which arc forms the part
t = Time

Now, we know that, by definition, radian is the angle subtended at the center of a circle by an arc
equal in length to the radius of the circle. Thus, we have:
C
= d/R = (V x t)/R
Or, /57.3o
= (V x t)/R (Note:  is in degrees)
Or, /t = (V x 57.3)/R in degrees /hour
Or, /t = (V x 57.3)/60R in degrees /minute
Using approximation and cancelling 60 and 57.3 with each other, we have:
/t = V/R in degrees /minute
Thus, we have ROT (/t) given by:
EXAMPLE: Let us assume that a vessel’s initial course is 000 (T) and final course is 060 (T). She
is steaming at 15 knots and intends to negotiate a turn about an islet keeping a distance of 1.5
miles. What will be the ROT and how long will she take to complete the turn?
 = V (degrees per minute)
t R
So, ROT = V/R = 15 / 1.5 = 10 degrees.
Thus vessel will turn at the rate 10 degrees per minute, while maintaining a distance of 1.5 miles
from the islet.
Now amount of alteration = 60 derees
Thus vessel will take 60/10 = 6 minutes to complete the turn
CONSTANT RADIUS TURN:
In this method radius R is kept constant.
We have seen that rate of turn (ROT) is given by (/t) = V/R
Or, R = V/ (/t)
If radius R is to be kept constant, the expression on the RHS will have to be kept constant. But as
vessel turns, velocity V reduces. Thus ROT (/t) will have to be varied proportionately so as keep
R as constant. Thus in constant radius turn ROT does not remain same and changes as vessel
t R

negotiates the turn. Constant radius turn is negotiated in the following two ways, depending on
weather the object ‘O’ is visible or imaginary.
 Object ‘O’ available as visually or radar conspicuous feature:
1. With O as center and R as radius draw an arc. (The value of R will be decided by
OOW/Master and will depend on vessel’s size, draft, weather conditions, proximity to
other dangers etc,).
2. Draw AB and BC as tangents to the arc, representing initial and final courses
 Object ‘O’ not available as visually or radar conspicuous feature:
1. Draw initial and final courses first.
2. Choose suitable value of R depending on size, loading condition of vessel, weather
conditions etc.
3. Calculate AB = BC = R tan /2
4. Draw small arcs, with radius R, from A and C to cut at O.
5. Now O as center and R as radius, draw arc AC. This is the curved segment of the
planned course.

MONITORING THE TRACK ALONG CURVED SEGMENT:
 Object O is visually conspicuous:
We know that any radius to the point of contact of a tangent is perpendicular to the
tangent. This means that for vessel to remain on the curved track, the object O must
remain abeam or very nearly abeam. If the object is falling abaft the beam, it means that
vessel is going outside the arc (or she is turning slowly) and ROT needs to be increased
by giving greater helm.
Conversely, if object is moving ahead of beam, vessel is going inside the arc (or she is
turning too fast) and ROT needs to reduced by easing the helm.
 Object O is radar conspicuous:
In this case VRM along with parallel indexing techniques is utilized for keeping the
vessel on the curved segment of the track. Both RM and TM modes are equally suitable
for the procedure.

 Object O is imaginary:
In this case vessel’s position has to be closely monitored at more frequent intervals to
ensure that the vessel stays at the curved segment of the track.
 With ECDIS onboard:
ECDIS has brought a revolutionary change in the way we navigate ships. The way
computers have made many old human skills irrelevant; ECDIS is also likely to play the
same role. As per IMO performance standards for “route planning and monitoring” on
ECDIS, it should be possible to draw both straight and curved segments of planned courses.
Thus, executing and monitoring ship’s progress along any curved segment is very easy. The
fact that the courses drawn can be seen against the background of chart on the screen itself
makes the procedure a very simple task. ECDIS provides real time fixing. Hence, OOW
knows where the vessel is at any moment rather than where she was few moments ago.
CONSTANT RATE TURN:
In this method ROT (/t) is kept constant.
We know that rate of turn (ROT) is given by (/t) = V/R
If ROT is required to be kept constant, the expression on the RHS will have to be kept constant.
But as vessel turns, velocity V reduces. Thus radius R will have to be varied proportionately so as
to keep ROT as constant. Thus in constant rate turn radius R does not remain same and changes as
vessel negotiates the turn.
Practically ROT is calculated for a mean value of vessel’s speed (mean of speed at the start of
curved segment and speed at the end of curved segment). This value corresponds nearly to vessel’s
position midway on the arc. At other locations on arc value of radius will differ from R. But these
variations are small and within practical and tolerable limits.
WHEEL OVER POINT (WOP):
It is the point on initial course at which wheel is put over to initiate the turning of the vessel. It is
obtained by intersection of initial course by wheel over line.
The distance between the WOP and the ship commencing its turn is denoted by F and depends on:
 Size of vessel
 Loaded/ballast condition
 Trim
 Type of vessel etc.
WHEEL OVER LINE:
It is a line drawn parallel to the final course. The point at which it cuts the initial course line is the
wheel over point. The distance at which wheel over line is to be drawn parallel to the final course
is given by following formula:
F Sin + R (1 - Cos)

F= distance between WOP and the point when vessel begins to turn.
R= radius of turn
 = difference between initial and final course
PROCEDURE:
1. Take F as 0.1, 0.15 or 0.2 miles depending on weather vessel is small, medium or large in
size.
2. Take R as radius of turn, depending on your vessel’s size, draft and external factors.
3. Calculate the distance between WOL and final course by the formula given above.
4. Take any arbitrary point on final course and draw on arc, towards initial course, at the
distance calculated.
5. Draw a tangent to the arc parallel to the final course, cutting the initial course.
6. The point of intersection is the wheel over point.
For illustration purpose only

EXAMPLE 1:
A small vessel is coming out of Aberdeen harbour at a course of 056 (T). Her next course is 120
(T). With the Racon Girdle Ness as centre and 0.52 miles as radius, find the WOP.
If the vessel is steaming at 10.4 knots, find the rate of turn and time taken to complete the turn.
SOLUTION:
!Distance between final course and WOL is given by:
F Sin + R (1 - Cos)
Substituting the values, we have:
0.1Sin 64 + 0.52 (1- Cos 64)
= 0.38 miles.
Now, take any arbitrary point on final course and draw an arc at a distance of 0.38 miles. (Refer to
the photo above).
Draw a tangent to the arc parallel to the final course to cut the initial course at T. Now, the parallel
line drawn is the wheel over line and T is the wheel over point.
Now,
Thus, rate of turn = 10.4/0.52 = 20 degrees/minute
Time taken to complete the turn = 64/20 = 3.2 minutes.
EXAMPLE 2:
A small vessel is leaving Aberdeen harbour and is steaming a course of 056 (T) at a speed of 13
knots. Her next course is going to be 190 (T). Draw the courses with curved segment using the
Racon Girdle Ness as centre and 0.52 miles as radius. Also, show the WOP.
How long will she take to arrive at the point of final course?
SOLUTION:
Distance between final course and WOL is given by:
F Sin + R (1 - Cos)
Substituting the values, we have:
0.1Sin 134 + 0.52 (1- Cos 134)
= 0.95 miles.
Now, take any arbitrary point on final course and draw an arc at a distance of 0.95 miles. (Refer to
the photo below).
Draw a tangent to the arc parallel to the final course to cut the initial course at T. Now, the parallel
line drawn is the wheel over line and T is the wheel over point.
Now,
t R

Thus, rate of turn = 13.0/0.52 = 25 degrees/minute
Time taken to complete the turn = 134/25 = 5.36 minutes.
For illustration purpose only
t R

91121543 roti (1)

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