Case XXIV 



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



ANOTHER RELATIVE POINT IN GIVEN TIME AT GIVEN SPEED 



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

 CHART LINE TO BE SCOUTED, SPEED OR SPEEDS OF SCOUT, AND TOTAL TIME OF OPERATION. 



TO DETERMINE: COURSES OF SCOUT, LENGTH OF CHART LINE SCOUTED, AND TIME TO TURN. 



Example. — Carrier C, now on course 500° at a speed of 20.0 knots, is 60.0 miles, 110° from an air station. A plane 

 takes off from the air station at 0600 to scout a chart line in direction 140° as far as possible, landing on the carrier at 1000, 

 using an air speed of 85.0 knots for the entire period. The true wind is from 340°, velocity 25.0 knots. 



Required. — (a) Air courses for the plane, (b) Length of chart line scouted, (c) Time plane turns to head for carrier, 

 (d) Bearing and distance of carrier from plane at (c). (See fig. 34.) 



Procedure. — Lay out e .... cande .... w, vectors of carrier and wind, respectively. With w as center, draw plane's 

 air speed circle, radius 85.0 knots. 



Plot the air station at any convenient point, AS, and locate the present relative position of the carrier at C. 



Lay off first chart course of plane from e in the specified direction, 140°, intersecting the plane's speed circle at pi. Join 

 w . . . . pi and c . . . . p\. The first air course for the plane is w .... p!. The vector c . . . . pi indicates the slope 

 of the Relative Movement Line of the plane on its first leg. 



Join AS .... C and transfer slope to c. Along this slope lay off rate AS .... C divided by total time 4.0 hours, 

 locating point /. The Relative Speed and course of the Fictitious Ship is indicated by this vector c . . . . /. Draw Time 

 Line pi .... f .... p 2 , intersecting the plane's speed circle at p 2 . w . . . . p 2 is the second air course of plane. 



Transfer the slope c . . . . pi to AS and the slope c . . . . p 2 to C. These slopes intersect at P, the turning point for 

 the plane. The time on the first leg is found by dividing the Relative Distance AS by the Relative Speed c . . . . pi or else 

 by using the Time Line and the formula. Time on first leg equals f . . . . p 2 divided by pi ..../... . p 2 times Total 

 Time Allotted. The length of chart line covered is found by multiplying the Ground Speed, e . . . . p u by time on first leg. 



Answer.— (a) First course 134l/ 2 °; second course 358°. (b) 162 miles, (c) 0729. (d) 349° distant 123 miles. 



NOTE. — To illustrate the scouting of a line in a given chart direction by a surface ship, example XXIV-B is appended. 



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