CaseXXIH 



TO SCOUT A GIVEN RELATIVE LINE AS FAR AS POSSIBLE FROM A GIVEN INITIAL RELATIVE POSITION, 

 RETURNING TO ANOTHER RELATIVE POSITION IN SPECIFIED TIME AT SPECIFIED SPEED 



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

 RELATIVE LINE, SPEED OF SCOUT, AND TOTAL TIME INTERVAL. 



TO DETERMINE: COURSES OF SCOUT, LENGTH OF RELATIVE LINE COVERED, AND TIME TO TURN 

 TO FINAL COURSE. 



Example. — Fleet is on course 228°, speed 15.0 knots. Lexington, now located 85.0 miles due west from the Guide, re- 

 ceives orders to launch a plane at 1200 for purposes of scouting a relative line 180° from the Lexington as far as possible, using 

 air speeds of 90.0 knots and landing on the Saratoga at 1700. Saratoga is stationed 4.0 miles 42° abaft the starboard beam 

 of the Guide. Wind is from 120°, force 20.0 knots. 



Required. — (a) Air courses for the plane, (b) Time plane changes course to head for Sara toga, (c) Length of relative 

 line covered by plane, (d) Length of chart line covered on first leg. (See fig. 33.) 



Procedure. — Lay out e . . . . g, the course and speed of the Guide, which is also the course and speed of Saratoga. 

 Lay out wind's vector e . . . . w. 



Plot Fleet Guide at any convenient point, G, and locate relative positions of Lexington and Saratoga at L and S respec- 

 tively. Lay out first Relative Line to be run by plane in direction 180° from L of indefinite length and transfer this slope to g. 



Join L and S, the distance run by the Fictitious Ship. Transfer this slope to g and lay out g . . . . / equal to the rate 

 found by dividing L .... S by the 5.0 hours available. 



With w as center and radius equal to 90.0 knots, draw a circle of the plane's air speed, intersecting the first slope from g 

 at pi. w . . . . pi is the first air course of the plane, which makes good a course indicated on the chart by e . . . . pi. 



Draw a time line from pi through / to the plane's speed circle, intersecting at p 2 . w . . . . p 2 is the second air course for 

 the plane. 



Join p 2 . . . . g and transfer slope to S, intersecting outgoing relative line from L at P, which is the Relative Position 

 reached by the plane when course is changed to head for Saratoga. 



Time plane is on first leg is found by dividing Relative Distance L . . . P by Relative Speed s . . . pi, which time may 

 also be found from the Time Line pi ..../.,.. p 2 as previously explained. 



Length of relative line covered is L .... P. Length of chart line covered is Ground Speed, e . . . . p u multiplied by 

 the time on first leg. 



Answer .—(a) First course 176'/ 2 ; second course 0311/ 2 °. (b) 1451. (c) 198 miles, (d) 230 miles. 



NOTE. — Solution for surface ships is the same except that wind vector and air speed circle are not used. The required speed circles are drawn 

 about e, which is the origin of the courses to be steered. The speed on both legs need not be the same, but the Time Line must pass through /. 



The action of the wind results in the necessity of steering a course to the left of the ground course while on the first leg and the steering of a 

 course to the right of the ground course on the second leg. 



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