# Technical Information

Useful Formulas
The following are standard methods of determining ODs and calculating weights of components that make up a multiconductor cable.

CONDUCTOR
Conductor Weight (lbs./Mft.) Where: = 3.14
D = Diameter
L = Length
P = Pounds of water/gallon (8.34)
G = Density of copper (8.39), aluminum (2.70)
N = Number of strands
C = Cubic inches of water/gallon (231.0)
K = Weight increase due to stranding

 No. of Strands 19 37 49 133 or more K 1.02 1.026 1.03 1.04

INSUALTION
Insulation Weight (lbs./Mft.) Where:
D = Diameter over insulation (in.)
d = Diamteter over conductor (in.)
K = Constant = 680
G = Specific gravity of insulation

TAPE
Tape Weight (lbs./Mft.) Where:
K = 1364
t = tape thickness
f = multiplying factor
d = diameter of cable under tape (inches)
G = specific gravity of tape

 % Lap 17.5 25 33 50 f 1.25 1.28 1.32 2.0

TWISTING LOSS
Use approximately 3% for all cables OD = diameter of a
single conductor in layer, therefore, or, 25 conductors, each 0.075″ OD will fit around a core
diameter of 0.540″.
3. Diameter over the outer layer
= 0.540″ + (0.075″ x 2)
= 0.690″

SHIELD
Shield  OD = Diameter under shield + addition (inches)

 AWG Size (Braid) 40 38 36 34 32 30 28 Addition 0.14 .018 .022 .028 .035 .044 .056

Shield Weight (Lbs./Mft.) Where:
N = Number of ends per carrier
C = Number of carriers
W = Weight of one shielding strand (Lbs./Mft.)
a = Braid angle

Percent Coverage Where:
F = NPd/Sin a
N = Number of ends per carrier
P = Picks per inch
d = Diameter of one shielding strand
a = Braid angle
tan a  = 2 (D + 2d) P/C
D = Diameter of cable under shield in inches
C = Number of carriers

 *AWG Size 40 38 36 34 32 30 d (inches) .0031 .0040 .0050 .0063 .0080 .0100 W (Lbs./Mft.) .0291 .0481 .0757 .120 .194 .303

JACKET
Jacket Weight (Lbs./Mft.) Where:
D = Diameter over jacket (inches)
d = Diameter under jacket (inches)
K = 680
G = Specific gravity of jacket material

TOTAL WEIGHT (Cabled Conductors)
Weight (Lbs./Mft.) = N x L x W
Where:
N = Number of conductors
W = Weight of one insulated conductor (Lbs./Mft.)
L = Twisting loss (1.03)

AMPACITIES

Assigning a current rating to a wire is really a problem of heat transfer. The watts generated at the conductor must be dissipated through the insulation of the ambient without overheating the conductor or the insulation.

While current ratings for power cables have been officially assigned by such bodies as the Insulated Power Cable Engineers Association and the National Electrical Code, no such ratings have been officially developed for appliance wires or apparatus cables.

Inasmuch as appliance and apparatus designs vary widely, any effort to standardize actual current ratings is just about impossible. The best that can be done is to select a cable insulation that has a thermal rating at least equal to that of the machine and use the nearest NEC rating as a guide for the first approximation of conductor size. If the insulation thermal rating is the same as that of the machine, a current density of the cable equal to that of the machine winding will probably suffice. On the other hand it may be economical to use a cable insulation of higher thermal rating, say 125oC cable in a Class A (105oC) machine. In such a case the cable could be operated at a higher current density than the winding at a savings in both conductor cost and space occupied. Sometimes the space savings will allow thermally upgrading the machine without changing the frame size.

For suggested ampacity ratings, see table below.

Current ratings for different conductor materials may be calculated by multiplying the appropriate copper conductor rating by the following factors: