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Case XIX-B 



DETERMINING RATE OF CHANGE OF RANGE ON VARIOUS BEARINGS 





GIVEN: COURSE AND SPEED OF MANEUVERING UNIT AND OF TARGET, AND BEARINGS ON WHICH 

 RCR IS REQUIRED. 



TO DETERMINE: RCR ON GIVEN BEARINGS AND BEARING WHEN RCR IS ZERO. 



Example. — A Maneuvering Unit, M, has the target, T, bearing 090°. M is on course 040°, speed 24.0 knots, and T is 

 on course 200°, speed 18.0 knots. 



Required. — (a) RCR on present bearing, 090°. (b) RCR when M is abeam of T. (c) RCR when T is abeam of M. 

 (d) Bearing of T from M when RCR is zero. (See fig. 26.) 



Procedure. — Lay out e . . . . m and e . . . . t, the vectors of M and T respectively. For convenience, plot M at e. 

 From e, lay out the given bearing 090°, extending as necessary. From m, drop a perpendicular to e . . . . b u intersecting 

 the 090° bearing line at rri\. Similarly, from t, drop a perpendicular to this same line, intersecting at fr. The length of 

 ti .... mi measures the rate of closing on this bearing or the RCR. Since the range is closing, it is marked minus. 



When M is abeam of 3", it v/ill bear 290° from the latter. Lay out the reverse of this bearing as e . . . . b 2 and drop a 

 perpendicular from M to this bearing, intersecting e . . . . b 2 at m 2 . Then e . . . . m 2 measures the RCR when M is 

 abeam of T. 



Similarly, when T is abeam of M, the RCR is found by dropping a perpendicular from t to the beam bearing from M, 

 e . . . . b 3 . Under these conditions the RCR is indicated by e . . . . t 3 . 



For the RCR to be zero, it is necessary that there be no component from either m or r on the bearing line, 

 is found by connecting m and r, and drawing the bearing from M normal to m . . . . t. 



Answer. — (a) Minus 720 yards per minute, range decreasing, (b) Minus 274 yards per minute, range decreasing, 

 (c) Plus 205 yards per minute, range increasing, (d) 12 1%°. 



NOTE. — The perpendiculars dropped from m and t to the bearing lines make intercepts which are added together to find the RCR if they 

 are on opposite sides of e. In case these intercepts are on the same side of e, the shorter is subtracted from the longer to find the RCR. Thus, 

 when the RCR is zero, the perpendiculars dropped from both m and r intersect the bearing line at the same point and the difference between the 

 length of the intercepts becomes zero. 



As long as the target bearing is on the approach side of the normal to the relative line, nil .... t, (when the bearing line is moving towards 

 the normal from M to the Relative Movement Line), the RCR will be minus or the range will be closing. When the target bearing has passed this 

 normal from M to the Relative Movement Line, the RCR is plus and the range is opening. 



This bearing 



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