Areas for Copper and Iron Lightning-Rods. 337 



of copper is about six times as great as that of iron; but the 

 melting-point of iron is about 50 per cent, higher than that of 



copper; therefore y^ =4 is the ratio for the sectional area of 



iron to copper. 



This simple treatment of the problem, however, is incom- 

 plete, because it neglects to take three most important factors 

 into consideration — namely (1) the influence of the rise of 

 temperature in increasing the electrical resistance of the metal, 

 (2) the difference between the "thermal capacity" or " specific 

 heat" of copper and iron, and (3) the fact that, the iron rod 

 being made several times more massive than the copper rod, 

 it will require a proportionately greater quantity of heat to 

 increase its temperature. These omissions introduce an enor- 

 mous error in the result. 



The effect of the passage of a discharge of lightning through 

 the rod will be to raise its temperature. 



The temperature (T) to which a given length of the rod 

 will be raised will depend on : — 



(1) The quantity of heat developed by the discharge. 



(2) The mass of the rod. 



(3) The specific heat a of the metal composing the rod. 

 This may be expressed mathematically as follows: 



TT 



T= const — ; 

 Gin 



where m is the mass of a unit length of the rod, which we 

 shall assume to be uniform in sectional area throughout its' 

 length, and H is the quantity of heat developed by the dis- 

 charge. 



We may take <r = 0*1013 for copper, and a = 04218 for 

 iron. These figures were only verified, by Dulong and Petit, 

 up to 300° C. It is probable, however, that their ratio, with 

 which we are only here concerned, would not greatly alter at 

 higher temperatures. At any rate, comparing the specific 

 heats between 0° and 100° C. with those between 0° and 300° 

 C, we infer that any alteration would be in favour of iron, i. e. 

 that the specific heat of iron would increase in a quicker ratio 

 than that of copper. 



Adopting the centimetre as the unit of length, the mass of 

 one centimetre of the rod = pa, where a is the sectional area 

 of the rod in square centimetres, and p = 8'9 for copper, and 

 p = 7'S for iron. 



Further, assuming the quantity and duration of the dis- 

 charge to be constants, H = const x R, where R is the resist- 

 ance of the unit length of the conductor. 



