Question 66 FCP01 - First Class Pilot

The Correct Answer is C --- ### 2. Explanation for Option C (222°T) The problem requires calculating the Course Made Good (CMG) by vectorially adding the vessel's intended course (Course Steered, CS) and speed (Speed Through Water, STW) with the environmental influence (Current Set and Drift). **Vector Analysis:** 1. **Vessel Vector (CS/STW):** 215°T @ 12 knots. 2. **Current Vector (Set/Drift):** 000°T @ 2.3 knots. The vessel is heading Southwest (215°T). The current is coming directly from the North (000°T). This means the current is hitting the starboard quarter of the vessel, effectively pushing the vessel's stern southward and its bow to the **West**. Because the vessel is being pushed to the West, the Course Made Good must be a higher bearing than the Course Steered (i.e., $> 215^\circ$T). Using vector addition (solving the triangle of velocities): * **Resultant Latitude Component (North/South):** $Y = (12 \cdot \cos 215^\circ) + (2.3 \cdot \cos 0^\circ)$ $Y = (12 \cdot -0.8192) + (2.3 \cdot 1) \approx -9.83 + 2.3 = -7.53$ (South) * **Resultant Departure Component (East/West):** $X = (12 \cdot \sin 215^\circ) + (2.3 \cdot \sin 0^\circ)$ $X = (12 \cdot -0.5736) + 0 \approx -6.88$ (West) **Calculate the Angle of the Resultant Vector:** The vessel is moving South and West, placing it in the third quadrant. $$\text{Ref Angle} = \arctan \left( \frac{|X|}{|Y|} \right) = \arctan \left( \frac{6.88}{7.53} \right) \approx 42.4^\circ$$ **Calculate the Course Made Good (CMG):** $$\text{CMG} = 180^\circ + \text{Ref Angle} = 180^\circ + 42.4^\circ = 222.4^\circ\text{T}$$ Rounding to the nearest whole degree gives **222°T**. --- ### 3. Explanation of Incorrect Options **A) 209°T** This option implies that the current pushed the vessel to the East (a smaller bearing than 215°T). Since the current is coming from 000°T and the ship is heading 215°T, the current hits the starboard side and pushes the ship to the West. Therefore, 209°T is geometrically impossible. **B) 217°T** This result suggests a deviation of only 2 degrees. While the course must be greater than 215°T, a 2.3-knot current acting nearly perpendicular to the vessel's track at that specific angle would cause a greater change in bearing, especially relative to a 12-knot speed. This deviation is too small. **D) 232°T** This result suggests a deviation of 17 degrees, which is too large. Such a significant change in bearing would occur only if the current were much stronger (e.g., closer to 5 or 6 knots) or if the vessel's speed were much slower (e.g., 5 or 6 knots).