Sometimes a question can arise where the navigator not only has to find the course to steer to make a particular waypoint but is also told when he must be at the waypoint.
In cases such as these he has not only to determine the course to steer but also has to calculate what speed the vessel has to move through the water to get there in time.
These type of questions are very similar to those discussed in lesson 13, however the main difference is now we know what time we have to be at the destination.
Again the navigator will have to contend with the effects of tide and wind, but this time we can work out exactly how much tide we will experience in the time allowed to get to the destination.
Consider the following
A vessel at position A in the diagram below has to reach position B.
A tidal stream is setting due south at two knots and a southerly wind is expected to cause 5 of leeway.
Find the course to steer and the speed required to reach position B in two and a half hours.
To solve this type of problem we need to adopt a similar approach to that in the previous lesson.
We have 2.5 hours to cover the distance from A to B.
Lets assume that the vessel remains stopped at position A for the whole 2.5 hours. The tide would set the vessel 5 nm to the south of the original position. (i.e. at position C)
However at the end of the time period we need to be at position B. Previously we drew a circle, with a radius equal to the ships speed, centred on the end of the tide vector and marked an arc on the ground track. This told us where the vessel would end up on the ground track after one hour, if the water track was taken as the line joining the end of the tide vector and the scribed position on the ground track.
The difference in this new case is that we know exactly where we have to be after 2.5 hrs.
To find the water track this time we simply join the end of the tide vector to the destination B and read the direction off the compass rose.
The length of the water track CB gives the total distance the vessel has to move through the water, in the 2.5 hrs allowed to reach B.
Therefore if we divide the length of the water track by the time allowed we can calculate the speed required through the water.
From the chart we can determine the direction of the water track (CB) from the compass rose.
In this particular case the water track required is 265 (T).
The length of the water track gives the total distance to be steamed through the water, which is 9.8 nmls.
Therefore the true course to steer to reach B in 2.5 hours is
Water Track = 265 (T)
Leeway = - 5 (applied against the wind direction)
True Ships Head = 260 (T)
Speed required to reach B in 2.5 hrs is
Water Track Distance/ Time allowed = 9.8 / 2.5 knots
= 3.92 knots.
A vessel is in the North East bound lane of the Dover Straits Traffic Separation Scheme and obtains a fix which puts her in position A at 1600 hrs.
The vessel wishes to reach position B by 1730 hrs.
A tidal stream is estimated to be setting 109 (T) at 2 knots and a North Westerly wind is estimated to cause 4 of leeway.
Find :
a) the true course to steer to reach position B at the required time
b) the speed the vessel must make through the water, to make the ETA.
On the chartlet below positions A and B are marked.
1) Join A and B with a straight line to obtain the required ground track
2) From A lay off 1.5 hrs worth of tidal stream i.e. 3 nmls to obtain position C
3) From the end of the tide vector (C) draw a straight line to B and obtain the required water track.
4) Determine the true direction of the water track
Water Track = 053.5 (T)
5) Apply the leeway (into the wind) to obtain the true ships head
Leeway = - 4
True Ships Head = 049.5 (T)
6) Measure the length of the water track CB using the latitude scale to get the distance steamed through the water in the allowed time
Distance CB = 12.9 nmls
Time = 1.5 hrs
Required Speed = 12.9 / 1.5 = 8.6 knots
Occasionally there is an additional question that may be asked in a course to steer type problem.
You may be asked to find the time and position when a charted object will be abeam.
In this case you MUST remember that beam bearings must be referred to the true ships head, not the water or ground track. Failure to do this will result in an error of principle in an examination.
Consider the problem given above.
As an additional question you could be asked to find the position and time the vessel will be abeam of Dunkerque light buoy.
In this case we have already seen that the true ships head has been determined as 049.5 (T).
If the vessel is on this heading then bearing of objects abeam will be either
049.5 (T) +90 = 139.5 (T) ( if they are to the south of the ground track)
049.5 (T) – 90 = 319.5 (T) (if they are to the north of the ground track)
In this case Dunkerque light buoy is to the south of the ground track and will be abeam when it is bearing 139.5 (T).
On the chart lay off a bearing of 139.5 (T) through the Dunkerque light buoy and extend it until it cuts the ground track at position D.
This is where the vessel will be when the light buoy is abeam (to Starboard in this case) The position can be read directly off the latitude and longitude scales on the chart.
To find the time when the vessel will be abeam, measure the distance AD on the chart. This is how far the vessel moves up the ground track before the buoy is abeam.
Now measure the distance AB which is how far the vessel must move in 1.5 hrs and calculate the ground speed.
Divide AD by the ground speed to get the time elapsed since the vessel was at A.
From the chart AB = 14.8 nmls
Therefore ground speed = 14.8 / 1.5 = 9.9 knots
From the chart AD = 6.9 mls
Therefore time interval = AD / Ground Speed = 6.9 / 9.9
= 0.697 hrs = 42 minutes
Hence time abeam = 1600 hrs
+ 42
= 1642 hrs
(See Chartlet below)
NOTE:
The same problem could also be asked in a straight forward course to steer type question. However the approach to the solution would be exactly the same.
Lesson 10.doc ETA Problems DGR1999
6
dariusz.lipinski