Case XXV 



TO SCOUT A GIVEN RELATIVE LINE AS FAR AS POSSIBLE SO AS TO REACH A GIVEN CHART POINT IN GIVEN 



TIME AT SPECIFIED SPEED 



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

 TION OF RELATIVE LINE TO BE SCOUTED, SPEED OF SCOUT, AND TOTAL TIME INTERVAL. 



TO DETERMINE: COURSES OF SCOUT, LENGTH OF RELATIVE LINE COVERED, LOCATION OF TURNING 

 POINT, AND TIME OF TURNING TO SECOND COURSE. 



Example. — Lexington, on course 315° at speed 25.0 knots, at 0615 bears 210°, distant 70.0 miles from a Naval Air Station. 

 At 0615 a plane is launched to scout a relative line in direction 340° from the carrier, as far as possible, returning to the Naval 

 Air Station in 4.0 hours. Plane has air speed 70.0 knots; and the wind is from 100° with a velocity of 22.0 knots. 



Required. — (a) Air courses for the plane, (b) Relative length of line covered, (c) Bearing and distance of Lexington 

 when plane starts second leg. (d) Time plane starts second leg. (See fig. 35.) 



Procedure. — Lay out vectors of Lexington and wind, e .... I and e ..... -w, respectively, and draw 70.0 knot speed 

 circle with w as center. Plot position of Lexington at any point, L, and locate Naval Air Station at N, 70.0 miles bearing 

 030° from Lexington's 0615 position. 



From /, lay out the slope of the required Relative Movement in direction 340°, intersecting the plane's air speed circle at 

 p 1# The first air course for the plane is w . . . . p t . 



DrawL .... N, and transfer slope to e, since L .... N represents an instantaneous chart position at 0615. Layout 

 along this slope the rate obtained by dividing distance L . . . . N by total time, 4.0 hours, locating /. e .... /is the vector 

 of the Fictitious Ship. 



Fromp! draw a Time Line through/, intersecting the plane's speed circle at p 2 . w . . . . p 2 is the course for the second leg. 



Using the 0615 position of Lexington, the charted position N, and the courses made good over the ground by the plane, 

 e .... pi and e . . . . p 2 , the chart track of the plane is from L to P and back to N. 



Maintaining L as the fixed position of the Guide, the position of N must be offset an amount equal to the Relative Move- 

 ment of N during the 4.0 hours allowed. This is in direction 1 . . . . e and locates N', the position of the Naval Air Station 

 relative to Lexington at 1015. Transfer the slope of / . . . . pj to L and the slope of 1 . . . . p 3 to N'. These two slopes intersect 

 at T, the turning point for the plane. L .... T is the relative length of line covered before changing course. T . . . . L 

 is the bearing and distance of Lexington from plane at turning point. 



Time on first leg is found by dividing L .... T by rate / .... pi, L .... P by ground speed e . . . . Pi, or by use of 

 the proportions of the Time Line pj .... f .... p 2 . This time added to 0615 gives the time to turn. 



Answer. — (a) First course 347°; second course 114^°. (b) 95.5 miles, (c) Bearing 160°, distant 95.5 miles, (d) 0755. 



NOTE. — The solution of this example would be the same for a surface scout except that the courses would be referred to e instead of w. 

 Also speeds need not be the same on each leg. 



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