Example XXVH-B 



Carrier C, on course 140°, speed 20.0 knots, launches a plane at 1115 to scout a relative point bearing 210°, distant 100.0 

 miles, returning in 2.5 hours. Wind is from 170°, velocity 25.0 knots. The maneuvers to the Carrier to launch the plane may- 

 be disregarded. 



Required. — (a) Air speed of plane, (b) Air courses for plane, (c) Time to turn to homing course. (See fig. 39.) 



Procedure. — Plot position of Carrier at any point C, and the relative point to be reached by the plane at P. 



Lay out the Carrier's vector at e .... c and the wind's vector at e .... w. Transfer the slope C . . . . P to c, 

 extending in both directions. 



Since the wind is the controlling factor with all aircraft problems, erect a perpendicular to the transferred slope of 

 C . . . . P from w, intersecting the transferred slope at O. Extend w . . . . O indefinitely. 



From O, lay out O .... equal to the Relative Distance, 100.0 miles, divided by the 2.5 hours available, and to the 

 speed scale in use. Join Q and C and lay out Q .... R equal to Q .... C in length. With O as center and radius equal 

 to O .... R, swing an arc cutting the transferred slope at p x and p 2 , respectively. 



The air speed of the plane is given by the length of w .... pi or w ... . p 2 . First air course is indicated by the 

 direction of w . . . . pi and the second air course by the direction as w . . . . p 2 . 



Time on first leg is found by dividing the Relative Distance C .... P by the Relative Speed c . . . . pi. This time 

 added to 1115 gives the time to turn. 



Answer. — (a) 95.0 knots, (b) First air course I88J2 ; second air course 051°. (c) 1252. 



NOTE. — It is immaterial whether the perpendicular extends toward e or away from e as the same points p\ and p^ will result in either case. 



The ground courses and speeds of the plane are indicated by vectors e . . . . pi and e . . . . p 2 . If it were required that a plane guard trail 

 the plane, it would therefore take the courses indicated by e . . . . pi and e . . . . p%. 



In case the transferred slope passes through e for surface vessels or w for aircraft, the perpendicular from e or w respectively would have zero 

 length, but it would still have direction. The procedure outlined above is still carried out, considering that the point O would coincide with e or w. 



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