Q0 POWER DISTRIBUTION FOR ELECTRIC RAILROADS. 
the feeder copper could obviously be reckoned as if we 
were dealing with a 1000 volt transmission through a com- 
plete metallic circuit. For the same percentage loss of 
energy the copper required will be rather less than half 
that needed on the 500 volt system. The caseis slightly 
  
  
  
  
  
  
-+ C=1000 
§< 0 et . L=500 K. W. 
F1c. 52. 
different from that of a lighting circuit since in the latter 
we are comparing two complete metallic circuits, one of 
double the voltage of the other, whilein the former we are 
comparing a very good ‘‘grounded’’ circuit with a return cir- 
cuit of double the voltage. In other words the track, which 
as a return conductor serves a very important purpose, as a 
““neutral ’’ is in use only so far as the system is unbalanced 
and to serve the purpose of a local conductor between 
cars. ‘Toillustrate by a concrete case, suppose a load of 
-t C =500 
L=250 K.W. 
=+ C=0 Track 
9 L=250 K. W. 
  
  
FI16G. 53. 
500 k. w isto be operated at a distance of 5000 ft. from the 
station. For simplicity we will suppose it to consist of a 
mass of cars bunched on a double track. With the ordi- 
nary system we have the state of things shown in Fig. 52. 
Using the constant 13 in our stock formula, the 
total area of copper comes out 1,300,000 c. m. As a three- 
wire system in complete balance we have the conditions set 
forth in Fig. 53. 
Here we employ the ordinary formula for metallic 
circuits and, of course, the constant 11. ‘The copper con- 
sequently amounts to 550,000 ¢. m. in area and since both