162 POWER DISTRIBUTION FOR ELECTRIC RAILROADS. 
per cent drop, the equivalent current is fifty amperes and 
we may proceed as follows: 
From formula (1) Chap. I 
G == 22 90 200,000 20 209,000 = 110,000. 
1000 
This gives the size of wire necessary for a single phase cir- 
cuit. If the circuit is two phase, half of the energy is sent 
over each circuit, which must be then of 55,000 ¢. m. wire. 
From formula (5) 
W —33X50X40,000 _ 66,000 1bs. 
1000 : 
This amount is the same for both the two phase and single 
phase circuits, in the way usually employed for operating 
two phase circuits, i. e., a complete and independent circuit 
for each phase. Sometimes the two phases have a common 
wire which modifies the amount of copper required, but 
this method of interconnection is seldom used on a large 
scale, since on long lines and at high voltages it involves 
serious practical difficulties. 
The three phase system requires a special, though very 
simple, calculation for the line. As ordinarily installed 
the three phases are mutually interconnected, so that the 
line consists of only three wires. ‘This combination of cir- 
cuits so utilizes the wire that for a given amount of energy 
delivered with a given maximum voltage between lines 
and at a given loss, the copper required is just seventy-five 
per cent of that necessary for an eqnivalent single phase 
line. 
This means that since the three phase line consists of 
three equal wires stretching from station to station, each of 
these wires must be of half the cross section needed for a 
single phase line wire under similar circumstances. If the 
single phase line consists of two wires each weighing 1000 
1bs. per mile, the three phase line will consist, for the same 
loss, of three wires each weighing 500 1bs. per mile. 
There are, of course, divers ways of taking ac- 
count of this saving in the formulae, but the author has