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Case XIX-A 



BASIC CONSIDERATION OF RATE OF CHANGE OF RANGE 



GIVEN: COURSE AND SPEED OF MANEUVERING UNIT, COURSE AND SPEED OF TARGET, AND 

 INSTANTANEOUS RELATIVE POSITIONS. 



TO DETERMINE: RATE OF CHANGE OF RANGE BETWEEN UNITS. 



Example. — Maneuvering Unit on course 020°, speed 11.5 knots, has the Target Vessel bearing 080° and distant 11,500 

 yards. The Target Vessel is known to be on course 340°, at speed 8.0 knots. 



Required.- — (a) Instantaneous rate of change of range between units. (See fig. 25.) 



Procedure. — Locate the Maneuvering Unit at any point M, and from this position plot in the position of the Target 

 Vessel at T. . I 



Draw e . . . . m, the vector of M, and e . . . . t, the vector of T. From the principles explained earlier in this volume, 

 t . . . . m represents the travel of M relative to T and m . . . . t represents the travel of T relative to M. This Relative 

 Speed may be resolved into two components, one along the present line of Relative Bearing and one normal thereto. The 

 former, t' . . . . m', represents the speed at which the two units are approaching each other on the present bearing or the 

 speed by which the range is affected. This speed, converted into yards per minute, is known as the rate of change of range 

 (RCR). When the RCR tends to decrease the range, it is marked with a minus sign; when the RCR tends to increase the 

 range between the two units, it is marked with a plus sign. 



For a different picture of the same problem, draw M . . . . M' representing the vector of M and T . . . . T" representing 

 the vector of T, from their respective initial positions. Resolving M . . . . M' into its two components along and at right 

 angles to the line of bearing, it is seen that M is approaching T at a rate equal to M . . . . M" and tending to pull ahead of 

 T at a rate equal to M" .... M'. Resolving T . . . . 7" into its two similar components, it is seen that T is approaching 

 M at a rate equal to T .... T" and tending to pull ahead of M at a rate equal to T" . . . . T'. The two vessels are there- 

 fore approaching each other at a rate equal to the sum of M . . . . M" and T . . . . T" , which is the same as the rate found 

 in the Vector Diagram, f . . . . m'. The rate of change of bearing, not required in this set-up, is equal to the difference 

 between M" . . . . W and T" . . . . T'. 



Answer.- — (a) RCR is minus 238 yards per minute. 



NOTE. — When two vessels are on converging courses, the RCR is minus and a maximum when these vessels are on collision courses. When 

 not on collision courses, but on converging courses, the value of the RCR decreases until it becomes zero, after which point the two vessels tend 

 to diverge and the RCR is plus and increasing in value. 



A convenient scale to use for all RCR problems is to have one division equal to 3.0 knots, which is 100 yards per minute. 





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