Case II 



TO PROCEED FROM ONE RELATIVE POSITION TO ANOTHER ON A SPECIFIED COURSE 



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

 COURSE TO BE USED BY THE MANEUVERING UNIT. 



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



Example; — Guide on course 230°, speed 10.0 knots. Ship "M", now 4.0 miles on the starboard beam of the Guide 

 is ordered to take station 10.0 miles broad on the port quarter of the Guide, steering course 180° en route. 



Required. — (a) Speed of "M". (b) Time for "M" to reach new position. (See fig. 4.) 



Procedure. — Plot Guide at "Gr" and locate initial and final positions of Maneuvering Unit at Mi and M 2 , respectively. 

 Join Mi ... . M 2 . 



Draw vector "e . . . . g", representing course and speed of "G". From "e" lay out course line specified for "M". 



Transfer slope "Mi .... M 2 " to "g", intersecting the course line of "M" at "m". The vector "e . . . . zn" repre- 

 sents the course and speed of the Maneuvering Unit. 



Divide the Relative Distance. "Mi .... M 2 ", by the Relative Speed, "g . . . . zn", to obtain the time required 

 to reach the new station on the specified course. This is easily done on the Logarithmic Scale. 



Answer. — (a) 8.82 knots, (b) 98 minutes or 1 hour 38 minutes. 



NOTE. — Should the Maneuvering Unit be a plane, the wind vector, "e . . . . w", would appear in the Vector Diagram. The specified 

 ground course is laid off from " e", as above. By reference to " w" the Air Speed and Air Course corresponding could be found. If the Air Course 

 were specified, this direction would be laid off from " w" instead of from " e", and the proper Air Speed determined by the intersection of this course 

 line with the transferred slope " Mi .... M" laid off from "g". Time in either event would be obtained in the manner illustrated. 



