

- 

 Case XII 



TO OPEN TO GIVEN RANGE IN MINIMUM TIME 



GIVEN: COURSE AND SPEED OF GUIDE, INITIAL RELATIVE POSITION, OWN SPEED AVAILABLE, 

 AND RANGE TO BE ATTAINED IN MINIMUM TIME. 



TO DETERMINE: COURSE OF MANEUVERING UNIT AND TIME TO REACH SPECIFIED RANGE. 



Example. — Guide on course 300°, speed 14.0 knots, has vessel Mnow bearing 000° and distant 15,000 yards. M, which 

 has power available for 21.0 knots, receives orders to open out to 30,000 yards from the Guide as quickly as possible. 



Required. — (a) Course for M. (b) Time required for M to reach the given range, (c) Relative bearing of M from G 

 when given range is reached. (See fig. 16.) 



Procedure. — Plot Guide at any point G, and locate initial position of Maneuvering Unit at M. 



About G draw a circle with a radius of 30,000 yards. 



From G, in the direction of the Guide's course, lay out G .... X equal to ^ — 5 — jr-r>X30,000 yards=20,000 yards. 



apeed ot M 



From X, draw a line through M, intersecting the 30,000-yard circle at Y. M .... Y is the Relative Movement Line of M. 



The explanation for the formula is appended below. 



Transfer the slope X .... M ... . Y to & in the Vector Diagram, intersecting the 21.0 knot speed circle at m. e . . . . m 

 represents the course of M at 21.0 knots to open to the required range in minimum time. 



The time required is found from the Logarithmic Scale, using Relative Distance M . . . . Y and Relative Speed g . . . . m. 



Bearing of M from G at 30,000 yards is G .... Y. 



Answer.— (a) 046°. (o) 23.7 minutes, (c) 106° relative. 



NOTE. — Should the speed of G be greater than the speed available to M, the point X will Ue outside the specified range circle and X . . . . M, 

 as extended, will cut the range circle twice. The point Y should be so chosen, in this case, that M lies between points X and Y. 



Explanation.' — The geometric construction used in the solution of this problem is based on the fact that the shortest 

 distance from a point within a circle to its circumference is along a radius passing through that point. Therefore, if M is to 

 reach the 30,000-yard circle from G in the minimum time, M must run along the radius of a circle whose center is the navi- 

 gational position of G at the instant the circumference is reached by M. 



Consider that the time required for this evolution is r hours. In that length of time the Guide will have traveled 14 r 

 miles, plotted as the distance G . . . . G', along course 300°. From G' draw a line through M equal in length to the specified 

 range. This line is G' .... M .... A, and M . . . . Ais equal to the distance traveled by M in r hours or 21 r miles. Com- 

 plete the parallelogram whose adjacent sides are G . . . . G' and G' .... M .... A. It will readily be seen that the triangles 

 XGY, XG'M, and YAM are similar since their sides are parallel. The proportionality follows: 



(G X):(A Y) = (G Y):(M A) 



or(G....X) = (^....y)X§^^==(G....y)X^-^- 



Now G . . . . Fis equal to the limiting distance and G . . . . G' and M .... A are drawn equal respectively to 14 r and 



21r. 



14r 

 Therefore G . . . . X= limiting distance X^r- 



= limiting distance X 



21r 



speed of Guide 



" speed of Maneuvering Unit' 



This problem is possible of solution by the trial and error method but much laborious work may be avoided by the geo- 

 metric method described above. 



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