28 
DIRECT-CURRENT DYNAMOS AND MOTORS. 
23. 
24. 
The latter result can also be obtained approximately from 
column 8, thus: 56.55 X .92 X .6 X 25.8 = 805 wires, 
which is a few per cent. in excess of the actual value, 
owing to the use of .6 instead of 3 x .197 = .591 as the 
height of the winding space. 
The length of one complete turn on this armature is about 
9 feet, and includes 2 wires; consequently the weight 
. 
of wire, by column 4, is = xX 9 X .1002 = 358 lbs., and 
its total resistance, all in series, at 20° C., by column 4, is 
m xX 9 X .000313 = 1.12 ohm. 
The resistance can also be obtained in the following man- 
ner: The cubical contents of the winding space is 50.55 
x .591 X 25 = 835 cubic inches. Doubling this figure, 
to allow for the heads and commutator Scnecctans. we 
have, from column 9: 835 X 2 X .000673 = 1.124 ohm 
as the approximate total resistance R, of the wire on oe 
armature (see Par. 39). 
Approximate Size of Armature Conductors.— 
It follows from Par. 20 that the approximate cross3-sec- 
tion of the armature conductor for a given case is found 
by multip! ying the current flowing through it by the 
specific cross-section suited for that case. Having in this 
manner found the number of circular mils of area required, 
the size of the wire is taken from column 3, Table (3. 
Example of Bipolar Armature.— What size of con- 
ductor will be required on the armature of a5 Hi. P. 
bipolar motor for 110-volt circuit ? 
Solution.—5 H. P. is equivalent to 5 x 746 = 3,730 
watts output. Assuming 85% efficiency as a fair value 
for this size, the watts input are 3,730 +.85 = 4,388 watts. 
Thecurrentis 4,388 + 110 = 40 amperes, hence each con- 
ductor must carry one-half of 40, or 20 amperes, which 
at 600 circular mils per ampere requires a wire of 12,000 
cir. mils. Referring to ‘Table 13, the nearest size ig 
