ii6 ELEMENTARY LESSONS ON [CHAP. n. 



calculated if the dip is known. 1 Or if the horizontal 

 and vertical components of the force are known, the 

 total force and the angle of the dip can both be cal- 

 culated. 2 The horizontal component of the force, or 

 " horizontal intensity," can be ascertained either by the 

 method of Vibrations or by the method of Deflexions. 

 The mean horizontal force of the earth's magnetism at 

 London in the year 1880 is -18 dynes, the total force (in 

 the line of the dip) is '47 dynes. The distribution of 

 the magnetic force at different points of the earth's 

 surface is irregular, and varies in different latitudes 

 according to an approximate law, which, as given by 

 Biot, is that the force is proportional to Vi + 3 sin 2 /, 

 where / is the latitude. 



139. Magnetic Maps. For purposes of conveni- 

 ence it is usual to construct magnetic maps, on which 

 such data as these given in the Table on p. 115 can be 

 marked down. Such maps may be constructed in 

 several ways. Thus, it would be possible to take a map 

 of England, or of the world, and mark it over with lines 

 such as to represent by their direction the actual 

 direction in which the compass points ; in fact to draw 

 the lines of force. A more useful way of marking the 

 map is to find out those places at which the declination 

 is the same, and to join these places by a line. The 

 Magnetic Map of England which forms the Frontis- 

 piece to these Lessons is constructed on this plan. At 

 Bristol the compass-needle in 1888 will point 19 to 

 the west of the geographical north. The declination at 

 Torquay, at Stafford, at Leeds, and at Hartlepool, will in 

 that year be the same as at Bristol. Hence a line joining 

 these towns may be called a line of equal declination^ or 

 an Isogonic line. It will be seen from this map that the 

 declination is greater in the north-west of England than 



1 For if H = Horizontal Component of Force, and I = Total Force, and 6 

 = angle of dip, I = H -r- cos 0. 



2 For H3 + V2 = 12, where V = Vertical Component of Force. 



