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1 Homework Q & A Junior Navigation Chapter 12 Wind & Current.

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1 1 Homework Q & A Junior Navigation Chapter 12 Wind & Current

2 2 Objectives: ■ Distinguish between leeway and current. ■ Know the meaning of the terms: DR track, intended track, speed of advance, course and speed over ground, course and speed made good, drift angle. ■ Determine course to steer, allowing for current/leeway. ■ Determine set and drift of current by observation. ■ Determine leeway by observation. ■ Understand elements of estimated reckoning.

3 3 1. The actual path of a boat over the bottom is: a. course made good. b. course over ground. c. DR track corrected for drift angle. d. course over ground corrected for drift angle. Ref.: Table 12 - 1

4 4 2. Current is: a. the horizontal motion of water. b. the horizontal motion of a vessel through the water. c. the net difference between COG and CMG. d. the net difference between COG and DR track. Ref.: ¶ 18

5 5 3. Given the same wind, leeway will be greatest for: a. a deep-keeled boat sailing downwind. b. a deep-keeled boat sailing with the wind abeam. c. a shallow-draft boat sailing downwind. d. a shallow-draft boat sailing with the wind abeam. Ref.: ¶ 19

6 6 4. Speed over ground is best measured by: a. knot meter b. GPS c. measuring the distance run between fixes an hour apart. d. measuring the distance run between fixes several hours apart. Ref.: ¶ 25

7 7 5. The navigator of a boat crossing the Gulf Stream plots a DR position and a fix at 1130. The distance measured between the two positions represents: a. the drift angle. b. the drift of the Gulf Stream current over the last hour. c. the average drift of the Gulf Stream current since the last fix. d. the net distance that the boat has been set off the DR track by current, wind, and/or other factors. Ref.: ¶ 44

8 8 6. You wish to maintain a rhumb line course of 010°T, and anticipate a current setting 040° at 1.5 kn. Your optimum cruising speed is 9 kn. a. What is the true course to steer? b. What would be the course to steer allowing for 4° of leeway in a strong northwest wind? Drawing a current diagram and measuring with a protractor shows a drift angle of 5° and CMG of 010°T. To maintain a course of 010°T, steer a course of 010° - 5° = 005°T. Wind will set the vessel to the east of the desired course. To offset leeway, the steering course is 005° - 4° = 001°T. Ref.: ¶ 46, 47

9 9 7. Consult the pilot chart from your student material kit. Departing from the entrance to the Straits of Gibraltar (L 35° 56'N, Lo 6° 15'W) enroute to Anegada Pass. a.What is the initial true course? Measuring a rhumb line on the pilot chart from Gibraltar on the route to Anegada Pass, the initial steering course is 265°T.

10 10 Problem 7. Continued from previous slide b. Allowing for current (set 180°, drift 0.5 knots), what would be the true course to steer at a boat speed of 5.5 knots? Drawing a current diagram indicates a drift angle of 5° to the south. To offset the current, the course to steer is 265° + 5° = 270°T.

11 11 Problem 7. Continued from previous slide c. Allowing for 3° of leeway, what would be the magnetic course to steer assuming no deviation? The pilot chart suggests the leeway would be southerly. Adding leeway, the magnetic course to steer would be 270° + 3° + 6° = 279° M (Variation is 6°W, from the pilot chart). Ref.: ¶ 34 – 39 & the pilot chart

12 12 8. The course steered from 0800 to 1200 has been 095°M; the course made good from an 0800 fix to a 1200 fix has been 103°M. If you want to maintain a rhumb-line course of 090°T, what magnetic course do you steer now? Variation is 5°W. _____________ Solution: Steering CourseCourse Made Good T090°T098° V 5°WV 5°W M095°M103° Working with true courses, drift angle is 098° – 090° = 8° to the south. To offset this drift angle, steering course should be 090° – 8° = 082°T. Convert this to magnetic course. T082° V 5°W M087° Ref.: ¶ 27 – 29 & Advanced Piloting 087° M

13 13 9. Your steering course is 075°. From the pilot chart you see that the average current in this area is setting to the northeast at 0.3 knot. The wind is steady from the southwest at Force 4. Your speed by knot meter has been 6.0 knots but regular plots of the DR track and fixes indicate SMG is 5.3 to 5.5 knots. What is the most probable explanation? a. The current is stronger than expected. b. The wind is setting up a surface current that offsets the normal average current. c. The knot meter is inaccurate. d. There is an unanticipated current setting to the southeast. Ref.: ¶ 34, 35

14 14 10. Make a similar analysis for the same passage as Example 3 in reverse, from Cristobal to Puerto Rico. Assume you have a motor vessel with optimum cruising speed in calm water of 9 knots, making negligible leeway at this speed. a. What current would you probably encounter first upon departure from Cristobal? The first major current you would encounter after departure from Cristobal would be setting to the east at about 0.8 knot.

15 15 Problem 10. Continued from previous slide b. From a point on the course at Lo 75°W, what compass course would you have to steer to stay on the course? For this part of the problem, use the average current determined in Example 3, and consider only this current and average boat speed. At about 75°W, you would begin to encounter the strong west-going current flowing between the Caribbean islands and the north coast of South America. The average current would have the same set and drift as determined in Example 3: Set 270°, Drift 0.9 knot (See Figure 12-5 with Example 3, paragraph 32). Graphically, this results in a current offset of 3° and a course to steer of 054° + 3° = 057°T

16 16 10. Continued from previous slide c. Given the resulting SOA, how long will it take to cover the remaining distance from this point to Puerto Rico? Answer: 62.65 hr or 2 days, 15 hr Solution:Click to view Fig. 12 – 8 Draw a current vector diagram for this set and drift, with boat speed of 9 knots. (See Figure 12–8) Speed of Advance (SOA) is 8.3 knots. From the point at which the course crosses the 75ºW meridian, the distance to Puerto Rico is approximately 520 nautical miles by measurement on the pilot chart. Therefore, the time required to reach Puerto Rico from this point is: T = D ÷ S T = 520 nm ÷ 8.3 kn T = 62.65 hr or 2 days 15 hr

17 17 Fig. 12-8

18 18 Problem 10. Continued from previous slide d. Discuss the probable effects of wind on the boat. How will leeway and waves affect the course and speed you could make over the ground? The wind roses at L 12º30´N, Lo 72º30´W, and L 17º30´N, Lo 67º30´W on the pilot chart show that over 90% of the time winds can be expected to be from the east or northeast at Force 4 to 5 (11 to 21 knots). You could expect to be heading into moderately high waves much of the time. Indeed wave heights ranging up to 12 feet can be expected occasionally in part of this area. These waves may require a reduction of speed or a course change to achieve an easier ride. Expect to use more fuel than usual and have an uncomfortable passage. When the wind is from the east, the boat would probably make some leeway, particularly at slower speeds. Ref.: ¶ 35 - 39

19 19 End Of Homework Q & A Junior Navigation Chapter 12 Wind & Current


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