Case XXI 



TO PROCEED FROM ONE RELATIVE POSITION TO ANOTHER AT KNOWN SPEED, STARTING AT KNOWN 



TIME, GUIDE CHANGING COURSE AND SPEED DURING OPERATION 



GIVEN: COURSES AND SPEEDS OF GUIDE AND TIME ON FIRST LEG, INITIAL AND FINAL RELATIVE 

 POSITIONS OF MANEUVERING UNIT, SPEED OF MANUEVERING UNIT, AND TIME MANEUVER STARTS. 



TO DETERMINE: COURSE OF MANEUVERING UNIT AND TIME TO ARRIVE IN FINAL POSITION. 



Example. — Guide is on course 320°, speed 15.0 knots, and has signaled that at 1400 her course and speed will be changed 

 to 280° and 12.0 knots. Ship M, now located 18.0 miles broad on the port beam of the Guide, receives orders at 1130 to take 

 position 10.0 miles bearing 055° from the Guide as soon as possible using 15.0 knots. 



Required. — (a) Course for M. (b) Time required to reach final position. (See fig. 30.) 



Procedure. — Plot the Guide at any point G, and locate the initial and final positions of M at Mi and M 2 , respectively. 

 Measure Mi ... . M 2 . 



Draw e .... ^i and e .... ^2, the vectors of G's first and second legs. 



A rapid inspection of the plotted positions will indicate that it will be impossible for M to reach its new station in the 

 2.5 hours that G is on the first leg, since M is held to the same speed as G and must shift from 18.0 miles on one beam to 

 10.0 miles slightly abaft the other beam. This may be checked by transferring the slope M x .... M 2 to g u cutting the 

 15.0 knot circle at g\ and m'. Relative Speed gi . . . . m' is too low to permit the change of station while G is on her first 

 leg, so a Fictitious Guide must be used. 



Following the rules formulated in the introductory pages of this section, the Fictitious Guide takes the second vector 

 of G, so that e . . . . gf coincides with e . . . . g 2 , and the initial position of M is offset for the time that the real Guide 

 and the Fictitious Guide are not together. 



By means of the Logarithmic Scale, using the time G is on the first leg, and the Relative Speed gi . . . . gf, the distance 

 Mx is to be offset is obtained, thus locating M/. Connect M/ .... M 2 and transfer this slope to g 2 , cutting the 15.0 knot 

 speed circle at m. e .... m is the course for M. 



By means of the Logarithmic Scale, the time required for the maneuver is found. 



Answer. — (a) 347°. (b) 195 minutes or 3.25 hours. 



NOTE. — Had minimum speed for M been a requirement instead of minimum time at 15.0 knots, the vector for M would have been found 

 by dropping a perpendicular from e to the slope g 2 . . . . m. 



Had the speed for M been between e .... £2 and the minimum speed as explained above, and M was not required to be in position as soon 

 as possible, two solutions would result through the transferred slope Mi .... M 2 crossing the specified speed circle twice. 



A Navigational Plot, to reduced scale, is inserted on the diagram to give a graphic picture of the tracks of the real Guide, the Fictitious Guide, 

 .and the Maneuvering Unit. 



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