SHIP'S COMPASS. 



Ill 



north, east, south or west. // the value of T is the same in Figs. 

 73 and 74, it can be shown that the temporary magnetism of the 

 ship does not tend to deflect the compass, whatever the direction of 

 the bow of the vessel. 



To prove this proposition, we will assume that the iron of the 

 ship is equivalent to two long slim horizontal bars of iron, one 

 parallel to the ship's keel (the A -bar) and the other at right angles 

 thereto (the .B-bar). When the ship is headed north, as shown 

 in Fig. 75, the full value of H f acts to magnetize the A -bar, and 

 the magnetic field 7\, which is produced at the compass box 

 by the magnetization of the A -bar, is proportional* to H' or 

 equal to k\H f . When the ship is headed east, as shown in Fig. 

 76, the full value of H f acts 

 to magnetize the .B-bar, and 

 the magnetic field T 2 , which 

 is produced at the compass 

 box by the magnetization of 

 the .B-bar, is proportional to 

 H' or equal to k^H'. There- 

 fore, if r!=r 2 , then fe = fe. 

 The letter k will be used in 

 what follows for ki and & 2 . 



Consider the ship when it is headed a degrees east of north 

 as shown in Fig. 77. The component of H' which magnetizes 

 the A -bar is H'cosa, and the magnetic field T a which is pro- 

 duced at C by the magnetization of the A -bar is k X H' cos a. 

 The component of H' which magnetizes the .B-bar is H' sin a, 

 and the magnetic field T b which is produced at C by the 

 magnetization of the .B-bar is k X H' sin a. The resultant of 

 T a and T b is 



Fig. 77. 



VX 2 + TV = kH' l/cos 2 a + sin 2 a = kH' (i) 



Therefore the resultant of T a and T b is constant in value, and, 



* Because the magnetization of the A -bar is proportional to H f , and the field 

 Ti is proportional to the magnetization of the A -bar. 



