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—="