for the Distribution of Electric Energy. 189 
Then we have 
A=fU /l700xl200xT 
V 7-5xl0 7 xT 
._C 
~6' 
or, when C is reckoned in amperes, 
A =ro < 2 > 
This gives 60 amperes per square centimetre of section as 
the proper relation between the current and the section of the 
conductor. It must be remembered, however, that this calcu- 
lation is based on the assumption of bare copper conductors 
and on a particular estimate of the cost of energy, both of 
which niay have to be greatly modified in particular cases. 
If, for example, covered cable costing, say, five times as 
much as bare copper be used, the proper ratio of the section 
of the conductor to the current would be 
A_ 1 _ 1 . m 
C~6(\/5~134' W 
or for every 134 amperes of current there should be one 
square centimetre of section in the copper conductor. 
Safety. 
We have next to consider whether the size of the conduc- 
tors given by the above considerations will be sufficient for 
safety. The answer to this question depends on the rise of 
temperature which will be produced by the greatest current 
which is to be sent through the conductor. (This greatest 
current must not be confounded with C in equation (1), as the 
strength of the current may occasionally greatly exceed that 
which it is proper to use according to that equation.) It 
is easy to calculate to a considerable degree of approximation 
the rise of temperature which will be produced in a con- 
ductor by an electric current when we know the specific 
resistance of the material, the emissivity of the surface, and, 
in the case of covered wire, the thermal conductivity of the 
covering. 
If we assume the rate of cooling for any conductor per unit 
length per unit difference of temperature to be r, we have for 
the total rise of temperature t the equation 
2 S + a£ ± . c 2 S ,., 
cr — i — =tr, or t= — t ~k~- , . . (4) 
