THE RETURN CIRCUIT. 29 
may properly take up the return circuit piecemeal and see 
what the actual state of things may be. 
First as to the rails. Mild rail steel is a very fair 
conductor. Weight for weight it is, comparing the com- 
mercial metals, just about one-seventh as good a conductor 
as copper. Now a copper wire weighing one pound per 
yard has an area of about 110,000 C. m.; hence an iron 
bar weighing one pound per yard is equivalent to about 
16,000 c. m. of copper, very nearly equal to No. 8 B 
& S gauge. 'This enables us at once to get the equivalent 
conductivity of any rail neglecting the joints. 'Theresult is 
somewhat startling, for since ordinary rail runs sixty 
pounds per yard or more, the conductivity ofa pair of such 
rails is equivalent to about 1,900,000 c. m. of copper, in 
some cases nearly ten times the cross section of the out- 
going circuit. 
The resistance of a copper wire of 16,000 c. m. is 
roughly six-tenths of an ohm per thousand feet. Hence the 
resistance of any single rail in ohms is, per thousand feet 
R = < where W is the weight 
W 
per yard. Or since two rails form the track 
3 
W 
That is, if the rail used weighs sixty pounds per yard the 
track resistance is approximately {5 ohm per thousand feet. 
For convenience the relation between weight of rail and 
equivalent copper is plotted in Fig. 17, 
These relations enable one to figure the drop in the 
track, neglecting joints, by the formule already given. 
For this purpose the distance in the formula should be, of 
course, the actual length of track, not the double length as 
when a return circuit of copper is figured. Thus one 
would separate the outgoing and return circuits and com- 
pute the drop in them separately. For simplicity it is 
however desirable to make allowance if possible for the 
return circuit, incorporating it in the constant of the 
original formula so as to make but a single calculation. 
R—="