Case XXVH 



TO SCOUT OUT AND IN ALONG GIVEN RELATIVE LINE AT MINIMUM SPEED IN GIVEN TIME, RETURNING 



TO ORIGINAL RELATIVE POSITION 



GIVEN: COURSE AND SPEED OF GUIDE, INITIAL RELATIVE POSITION (WHICH IS ALSO FINAL 

 RELATIVE POSITION), DIRECTION AND LENGTH OF RELATIVE LINE, AND TOTAL TIME INTERVAL. 



TO DETERMINE: COURSES AND SPEED OF SCOUT AND TIME TO TURN. 



Example — Guide on course 320°, speed 12.0 knots, has a scout stationed 4.0 miles broad on her starboard bow. 

 Scout receives orders to scout, at minimum speed, to a distance of 25.0 miles from her present station, maintaining constant 

 bearing on the Guide, leaving formation at 1300 and returning to station at 1600. 



Required.— (a) Speed used by Scout, (b) Courses for Scout, (c) Time to turn to second course. (See fig. 38.) 



Procedure. — Plot the Guide at point G, and locate the initial position and the intermediate point of the scout at S and P 

 respectively. Join S and P. S .... P is the Relative Movement Line on the first course and P .... S is the Relative 

 Movement Line for the second course. 



Lay out the Guide's vector, e . . . . g, and transfer the slope S . . . . P to g, extending in both directions. The use of 

 the regular Fictitious Ship in this case would involve a vector g .... f of length equal to zero, since the initial and the 

 final positions coincide, and yielding an indeterminate result. This is overcome by introducing a second Fictitious Ship whose 

 vector is equal to the length of the perpendicular erected from e to the transferred slope S .... P. Relative to this extra 

 Fictitious Ship the Fictitious Scout travels at the same rate both -going out and returning. Mathematically this constant 

 rate is equal to the Relative Distance divided by the total time available plus the square root of the sum of the squares of the 

 Relative Distance divided by the total time available and of the Relative Speed of the Guide to the Second Fictitious Ship. 

 This process is laborious, so the procedure outlined below will be used as the graphic solution. 



Erect a perpendicular at e to the transferred slope, cutting it at O. Extend this perpendicular beyond the slope line. 

 Along this same perpendicular lay out O . . . . Q equal to the Relative Distance divided by the total time available and to 

 the speed scale in use. Connect Q . . . . g and lay out length Q . . . . g from Q along the perpendicular, locating point R. 

 With O as center and radius equal to O .... R, swing an arc cutting the transferred slope at s x and s 2 . The first course is 

 e .... Si and the second course is e . . . . s 2 . The speed is shown by the length of either e .... sj or e .... s 2 



Time on first leg is found by dividing Relative Distance S .... P by Relative Speed g . . . . Sj. 



Answer. — (a) 22.0 knots, (b) First course 342°; second course 207V2 o . (c) 1508. 



NOTE. — The solution of this case using aircraft is shown in example XXVII-B. 



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