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Case XIX-D 



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TO BRING TARGET ON A GIVEN BEARING WITH A SPECIFIED RCR ON THAT BEARING 



GIVEN: COURSE AND SPEED OF TARGET, INITIAL RANGE AND BEARING, AND BEARING ON WHICH 

 SPECIFIED RCR IS TO OBTAIN. 



TO DETERMINE: COURSE AND SPEED OF MANEUVERING UNIT AND TIME INTERVAL UNTIL RCR 

 CONDITIONS ARE FULFILLED. 



Example.- — Maneuvering Unit, M, with 24.0 knots speed available, has Target Vessel, T, now bearing 320°, distant 10.0 

 miles. T is known to be on course 340° at a speed of 15.0 knots. M wishes to bring T bearing 000° as soon as possible so 

 that the RCR on this bearing is ( — ) 200 yards per minute, and from this initial point to close T, maintaining this same RCR. 



Required. — (a) Course and speed for M to reach this initial point and RCR. (b) Course and speed for M to close T, 

 maintaining this RCR. (c) Length of time required to reach initial point, (d) Distance from T when required bearing is 

 reached. (See fig. 28.) 



Procedure. — Lay out e .... t, the vector of T, and from e draw the 000° bearing line e . . . . b. From t, drop a 

 perpendicular r .... t' to e .... b. From t' , lay out along the bearing line t ' .... m' equal to 6.0 knots or a 

 RCR of 200 yards per minute, and draw m' . . . . r perpendicular to e .... b. The line m' .... r is the locus of 

 all courses for M which will yield the required RCR. 



With e as center and 24.0 knots as radius, swing a circle, cutting m' .... r at m^ M's maximum speed is utilized since 

 it is desired to reach the bearing and RCR as soon as possible and e . . . . m^ is the vector for M to reach the initial point. 



From t, draw a line parallel to e .... b, intersecting m' .... r at m 2 . Vector e . . . . m 2 represents the course 

 and speed for M to close T while maintaining the required RCR. 



From T, draw T .... X in a direction the reverse of the required bearing. Transfer the slope r . . . . mi to M, 

 intersecting the line T .... X at M' '. M' is the initial point to be reached by M when the required bearing and RCR obtain. 

 By means of the Logarithmic Scale, the time required for M to reach M' at Relative Speed t . . . . m, is found. 



Answer.— (a) Course 327°, speed 24.0 knots, (b) Course 346°, speed 20.7 knots, (c) 48 minutes, (d) 2.85 miles. 



NOTE. — In problems involving RCR only, the range to the Target is immaterial, since RCR is purely a velocity function found from the Vector 

 Diagram. In the above problem, time to reach a desired bearing was added; and, therefore, the range of the Target was necessary. 



Although the position of M, in the examples shown, has been assumed at e, the Relative Plot does not have to coincide with the Vector Diagram. 













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