Module 10 - Chartwork
Worked EP and Course-to-Steer Example
What you will learn
Complete and independently check a one-hour EP and course-to-steer construction using the same vessel and tidal inputs.
Using one set of numbers for both constructions exposes the difference between predicting where a chosen course will take the vessel and solving for the course needed to make good a chosen track.
EP problem: from a 1000 fix, the vessel travels on 090°T through the water at 5 kn for one hour. The DR is therefore 5.0 M east of the fix. The stream sets 180°T at 1 kn, so its one-hour drift is 1.0 M south. Plotting that stream from the DR gives the 1100 EP.
The EP is 5.0 M east and 1.0 M south of the start. Its ground displacement is √(5² + 1²) = 5.10 M and its direction is approximately 101°T. Those values are useful checks on the chart construction, not substitutes for carrying uncertainty in the inputs.
CTS problem: the desired one-hour ground track is 090°T, with the same 180°T stream at 1 kn and boat speed of 5 kn. Plot 1.0 M south from the start, then swing a 5.0 M arc from the end of that vector to the eastbound ground-track line. The required water track is approximately 078°T and the expected ground distance is 4.90 M, so predicted SOG is about 4.9 kn.
Worked example
Boat speed is 5 kn for one hour, the stream sets 180°T at 1 kn, and the intended ground track is 090°T.
- 1EP: plot 5.0 M on 090°T, then 1.0 M on 180°T; label the endpoint EP 1100.
- 2EP check: √26 = 5.10 M from the start, on approximately 101°T.
- 3CTS: plot 1.0 M on 180°T from the start, then swing a 5.0 M boat-speed arc to the 090°T ground track.
- 4CTS result: measure approximately 078°T from the tidal-vector end to the intersection.
- 5Ground check: the eastward distance is √(5² - 1²) = 4.90 M, giving about 4.9 kn SOG for one hour.
Sense check: The EP must fall south of the eastbound DR, while the CTS water track must point north of east to oppose the same stream.
| Construction | Plotted movement | Checked result |
|---|---|---|
| DR | 5.0 M on 090°T from the 1000 fix | 5.0 M east at 1100 |
| EP | Add 1.0 M on 180°T from the DR | 5.10 M on about 101°T from start |
| CTS tide | 1.0 M on 180°T from the start | Southward offset to cancel |
| CTS boat | 5.0 M arc to the 090°T ground track | About 078°T; SOG about 4.9 kn |
Use reciprocal problems as a check
Your plotted EP and CTS both lie south of the intended eastbound track even though the tide sets south.
1. Accept the EP direction
A vessel steered east through the water will be carried south, so its EP belongs south of the DR.
2. Reject the CTS direction
The course-to-steer water track must point north of east to cancel that southward movement.
3. Find the construction error
Recheck vector origin, set direction, arc radius and the line from which the final bearing was measured.
Sense check: The same tidal vector pushes a prediction with the tide but requires a steering solution against it.
Common mistake or limitation
- The EP tide vector starts at the DR; the CTS tide vector starts at the original position.
- The 078°T result is a water track before leeway and compass correction, not automatically the compass heading to steer.
Recap
- EP asks where a chosen through-water course will lead.
- CTS asks which through-water course will produce a chosen ground track.
- Vector direction and simple arithmetic provide independent checks on both plots.
Optional quick check
Section 5 of 7
In the worked EP, where does the 1.0 M tidal vector begin?
Continue studying Chartwork
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