Case XX 



TO PROCEED FROM ONE RELATIVE POSITION TO ANOTHER AT KNOWN SPEED, ARRIVING AT GIVEN TIME, 



GUIDE CHANGING COURSE DURING OPERATION 



3 .: " ■ 



GIVEN: COURSES AND SPEEDS OF GUIDE AND TIME AT WHICH GUIDE WILL CHANGE COURSE, 

 INITIAL AND FINAL RELATIVE POSITIONS, SPEED OF MANEUVERING UNIT, AND TIME OF ARRIVAL AT 

 FINAL POSITION. 



TO DETERMINE: COURSE OF MANEUVERING UNIT AND TIME OF LEAVING INITIAL RELATIVE 

 POSITION. 



Example. — At 0800 Guide is on course 140°, speed 14.0 knots, and will change to course 070°, speed 16.0 knots, at 0930. 

 Ship M, now stationed 8.0 miles on the port beam of the Guide, is ordered to take station 12.0 miles on the starboard beam 

 of the Guide, arriving at 1200, time of departure optional. M decides to maintain present station as long as possible so as to 

 carry out his orders at 14.0 knots. 



Required. — (a) Course of M. (b) Time M leaves present station, (c) Minimum speed and corresponding course avail- 

 able to M. (See fig. 29.) 



Procedure. — Lay out e . . . . gi and e . . . . g 2 , the first and second vectors of the Guide. Join gt and g 2 . 



Plot the position of the Guide at any point G. Locate M lf the initial position of M, which is 8.0 miles bearing 050° from 

 G. In a similar manner locate M 2 bearing 160° and distant 12.0 miles from G. 



From the rules listed on the preceding page, the vector of the Fictitious Guide coincides with the vector of G's first leg, 

 and the final position of M must be offset for the time that the real Guide and the Fictitious Guide are not together. 



Offset M 2 in the direction parallel to gf . . . . g 2 , a distance equal to the Relative Run for 150 minutes at Relative Speed 

 gi . . . . g 2 , as found from the Logarithmic Scale, locating M 2 . Join M : and M 2 , and transfer the slope of Mi ... . M 2 

 to gi cutting the 14.0 knot circle at m. The course for M at 14.0 knots is indicated by e .... m. 



Using the Relative Speed gf . . . . m and the Relative Distance M x . . . . M 2 , the time required for M to complete 

 the maneuver at 14.0 knots is found. This time, subtracted from 1200, gives the time at which M should first head for his 

 final position. 



If the minimum speed is to be used, M must start at once. His course and speed would either be found by a simple 

 Navigational Plot, or, using the offset position previously plotted, the Relative Speed would be the distance M x .... M 2 ' 

 divided by the 4.0 hours available. This Relative Speed, gf . . . . m' , indicates that the minimum speed and the corre- 

 sponding course is given by the vector e . . . . m' . 



Answer— (s.) 098°. (b) 163 minutes before 1200 or at 0917. (c) 13.2 knots on course 111°. 



NOTE. — The procedure would be the same were the course and not the speed specified, the determining intersection being that of the slope 

 from gf parallel to M\ .... M 2 with the specified course line. 



In case e . . . . m' is not normal to gf . . . . zn, it is possible to use a lower speed where these conditions obtain. However, the problem 

 will have to start earlier than 0800 in order to utilize this lower speed. 





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