SHIP'S MAGNETISM. 



305 



which is produced at the compass box by the magnetization of 

 the A-bar, is proportional * to H' or equal to k^H' . When the 

 ship is headed east, as shown in Fig. 15, the full value of H' acts 

 to magnetize the B-bar, and the magnetic field T 2 , which is pro- 

 duced at the compass box by the magnetization of the B-bar, is 

 proportional to ff or equal to k^H 1 . Therefore, if T t = T v 

 then k = k v The letter k will be used in what follows for k 

 and k z . 



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

 as shown in Fig. 16. The component of H' which magnetizes 

 the A-bar is H' cos a, and the 

 magnetic field T a which is pro- 

 duced at C by the magnetiza- 

 tion 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_ and 7! is 



Fig. 16. 



+ r 2 = kH' V cos 2 a + sin 2 a = kH' 



(0 



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

 since T a = kH' x cos a and T b = kH' x sin a, it is evident 

 that the resultant of T a and T b is always opposite to H' in 

 direction, so that the actual field at the compass box is constant 

 in value and always parallel to H f , or, in other words, the 

 compass error due to the temporary magnetism of the ship is 

 zero on all headings of the ship when T v in Fig. 14 is equal to 

 T 2 in Fig. 1 5. 



The quadrantal correctors. The quadrantal error of the ship's 

 compass is eliminated (that is to say, compensated) by means of 



* Because the magnetization of the A-bar is proportional to If, and the field T 

 is proportional to the magnetization of the A-bar. 

 21 



