Techniques Used in Case IX 



TO DRAW A LINE ACROSS THREE BEARING LINES SO INCLINED THAT INTERCEPTS WILL BE PROPORTIONAL 



TO TIME INTERVALS BETWEEN BEARINGS 



Although a trial and error solution of this problem is possible by the use of parallel rulers and dividers, the three 

 methods shown below have been found to be more expeditious. In each method the same data is used. Three bearings, 

 O .... a, O .... b, and O . . . . c, are taken by an observing vessel Oof a second vessel, with an interval of time ti 

 between the first and second bearings and an interval of time r 2 between the second and third bearings. The angular difference 

 between bearing O .... a and O .... b is given by angle X, while angle Y indicates the difference between bearings 

 O . . . . b and O . . . . c. These methods are shown sketched on a Maneuvering Board because of the convenience of the 

 attached scales, but this is not a requirement. 

 First Method (Sketch a). (See fig. 13.) 



At any point P on bearing line O .... a erect a perpendicular cutting bearing line O . . . . b at B. Along this perpen- 



P B t 



dicular lay off B .... Q so that **'*.- = — ■ Erect a perpendicular at Q, cutting O . . . . c at C. 



B .... y r 2 



Connect B and C and extend this line to O .... a, intersecting at A. The required slope is A .... B .... C. 



Second Method (Sketch b). 



At any point P on bearing line O . . . . b erect a perpendicular and lay out P .... Q and P .... R so 



P O t 



that n * ' * * p = — • Erect perpendiculars at Q and R, cutting O .... a and O .... c at A and C respectively. 

 P . . . . R £2 



Connect A and C, cutting the bearing line O .... b at B. The required slope is A .... B .... C. 

 Third Method (Sketch c). 



Along O .... a lay out any convenient distance O .... A. Along O . . . . c lay out length O . . . . C obtained 

 by the formula: 



o....c=o....AX^x sinang i e * 



t x sin angle Y 

 The required slope is A .... B .... C, obtained by connecting A and C. In case both angles X and Y are less than 

 16°, the numerical values may be used instead of the sines. 



TO DRAW A LINE OF GIVEN LENGTH PARALLEL TO ANOTHER LINE, BETWEEN TWO DIVERGENT LINES 



(Sketch d) 



Let O .... a t and O . . . . a 2 be two divergent lines and P . . . . Q a given slope. It is required to draw a line of 

 given length, parallel to P .... 0, between O .... a! and O . . . . a 2 . 



Along P .... Q, extended if necessary, lay out the given length P . . . . R. From R draw a line parallel to O . . . . a u 

 intersecting O a 2 at C. Through C draw A .... C parallel to P .... Q. 



A .... C is the required line, the proof of which is apparent, from the parallelogram law. 



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