Module 10 - Chartwork
Worked EP and Course-to-Steer Example
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.
- EP predicts the result of a chosen course through the water
- CTS solves for the water track needed to make good a chosen ground track
- The EP is about 5.10 M on 101°T from the start
- The CTS water track is about 078°T with SOG about 4.9 kn
- Apply leeway and compass corrections only after the true water-track solution
| 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 |
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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