Case XXH 



TO PROCEED FROM ONE RELATIVE POSITION TO ANOTHER IN GIVEN TIME AT MINIMUM SPEED, PASSING 



THROUGH AN INTERMEDIATE RELATIVE POINT EN ROUTE 



GIVEN: COURSE AND SPEED OF GUIDE, INITIAL, INTERMEDIATE, AND FINAL RELATIVE POSITIONS 

 OF MANEUVERING UNIT, AND TOTAL TIME INTERVAL. 



TO DETERMINE: COURSES AND SPEEDS OF MANEUVERING UNIT. 



Example. — Guide on course 040°, speed 12.0 knots, has ship M now stationed 8.0 miles broad on her starboard beam. 

 M receives orders to take station 10.0 miles bearing 060° from the Guide, to pass through a relative point 25.0 miles bearing 

 085° from the Guide while shifting stations, and to complete the maneuver in 5.0 hours, using minimum speed. 



Required. — (a) Courses for M. (b) Speed of M. (c) Required time to reach intermediate position. (See fig. 32.) 



Procedure. — Plot the Guide at any convenient point G, and locate the initial, the intermediate, and the final positions 

 of the Maneuvering Unit at M u M 2 , and M 3 . Join Mi and M 2 , M 2 and M 3 , and Mi and M 3 . 



Lay off e . . . . g, the vector of the Guide. Transfer the slope M x .... M 2 to g and mark it slope (1). Transfer the 

 slope M 2 . . . . M 3 to g and mark it slope (2). 



The distance M x . . . . M 3 will be travelled by the Fictitious Ship in the 5.0 hours allotted. Therefore the Relative 

 Speed of F is equal to Mi ... . M 3 divided by 5, and is plotted as g .... f, parallel to the slope of Mi ... . M 3 . The 

 course and speed of F are represented by vector e . . . . /. 



Pivot a straight edge at /, and so orient it that it cuts slope (1) and slope (2) at m, and m 2 respectively, both of these 

 points of intersection being equidistant from e. e . . . . mi is the first course for M and e . . . . m 2 is the course for the 

 second leg. Speed is indicated by either e . . . . mi or e . . . . m 2 . 



To obtain the time for M to reach the intermediate position, the Relative Distance Mi .... M 2 is divided by the Rela- 

 tive Speed g . . . . mi. Another way to obtain this time is by use of the Time Line mi ..../.... m 2 . Time on first 



leg is equal to — "^ times 5.0 hours. 



mi .... m 2 



Answer. — (a) First course 048K°; second course 318°. (b) Speed 16.6 knots, (c) 4.1 hours. 



NOTE. — Since the minimum speed was specified, speed must be the same on both legs. The course of speed for M on either leg could have been 

 specified, in which case the corresponding speed or course would have been found by the intersection of the specified course line or speed circle 

 with the slope for that leg. 



A rapid method for finding the time on the first leg is to draw any line tn 2 .... A from ot 2 equal in length to the total time on both courses, 

 as measured on any convenient scale. A and mi are connected and / .... Bis drawn from /parallel to mj . . . . A. m 2 .... B then measures 

 to the same scale previously chosen the length of time on the first leg. 



62 



