1878.] Areas for Copper and Iron Lightning lS,ods. 193 



Therefore T = Const. , , ,^ for copper 



and T = Const. ., .-^^ — r r-. for Iron 



0-1218 X 7-8 X A^ 



Thus T = Const. — ^— for copper 



, _ „ ^ 112-63 „ . 

 and T = Const. — -r- tor iron 

 A^ 



Now putting T = the temperature of fusion in each case 



1400 = Const. — -^— for copper 



«^^^ r. . 112-63 ^ . 

 2000 = Const. — — - for iron 

 A^ 



^, . ,^^. 1400 112-63 



Therefore 



(^)'= 



2000 * 11 09 

 = 0-7 X 10-16 



= 7-112 



Whence A = 2*7a about 



Q 



= — a about 

 o 



or the sectional area of an iron rod should be to the sectional area of a 

 copper rod in the ratio of 8 to 3. 



