DIRECT-CURRENT DYNAMOS AND MOTORS. 
quantities known are the ampere-turns required and the 
voltage by which the shunt winding is supplied—that is, 
the voltage of the machine. One of the simplest methods 
is based upon the fact that the resistance of one mil-foot 
of commercial copper—that is, the resistance of one foot 
of a copper wire, one mil, or one-thousandth of inch, in 
diameter—is almost exactly 1/2 ohms at 60° C., which is 
about 35° C. above the average machine-room tempera- 
ture. Hence, the resistance of any conductor of com- 
mercial copper at 60° C, is 
Rp- 2? 7 ah 
in which /’ is the length of the wire, in feet, and d’ its 
sectional area in circular mils. Or, since 12 xX l/l’ is the 
length of the wire in inches, we can write 
in which 7’ is the length of the wire expressed in inches. 
The current flowing in the above wire is, by Ohm’s Law, 
ti Seer ge oe 
OF gon aaa 
It is also evident that the ampere-turns in any shunt 
winding are numerically equal to the amperes that 
would result if a single turn of the magnet wire were 
supposed to be subjected to the given voltage, because 
two turns would have twice the resistance and take one- 
half the current; three turns would have three times the 
resistance and therefore take one-third the current; and 
so on for any number of turns. Since thecurrent flowing 
in ashunt circuit is the terminal EK. M. F. divided by the 
shunt resistance, we have therefore: | 
Voltage 
Resistance of One Turn’ 
Expressing the resistance according to formula (4'7), 
this becomes: 
Armpere-turns = 
Voltage x Circular Mils 
Length of One Turn in inches’ 
Ampere-turns = 
