# Formulas

Our designing team uses various wire drawing formulas for designing a variety of wires with our wire drawing machines such as weight / length relationship formula, surface area formula, zinc coating formula and formula for reduction in area

FORMULA ONE |
FORMULA TWO |

Weight
/ Length Relationship:The weight per foot of wire is : (2) Example : The Weight per foot os .090" wire is :The length per lb of wire is: (3) Where:w = weight per foot (lbs/ft) d = wire diameter(inch) L = length per lb (ft/lb) Example : The length per lb of .090" wire is: |
* Surface Area:The surface area of one foot of wire is: (4) a = X 0.090 / 12 = 0.0236 ft^{2}/ftThe surface area of one lb of wire is : (5) a/w = 1 / (36 X d X y)Where:a = surface area of one foot of wire (ft ^{2}/ft)a/w = surface area of one lb of wire (ft ^{2}/lb)d = wire diameter (inch) w = weight of one foot of wire (lb/ft) y = specific weight (for steel, 0.283 lbs/in ^{3})Example : The surface are of one lb of 0.090" diameter wire
is : a/w = 1 / (36 X 0.090 X 0.2836) = 1.088 ft^{2}lb* Refer to Page 369, Volume 2, Steel Wire Handbook |

FORMULA THREE |
FORMULA FOUR |

ZINC COATINGThe Percent zinc coating on d" diameter wire coated with coz/ft ^{2}zinc
is:(6) Example : for 0.090" diameter wire coated 0.3 oz/ft^{2}
:(6) The weight of zinc on d" diameter wire coated with p% zinc is : (7) Where :p% = weight percent of zinc(%) c = zinc coating weight (oz/ft ^{2})d = wire diameter(inch) Example: for 0.090" diameter wire, 2.00% zinc coating
weights :* Refer to Page 301, Volume 1, Steel Wire Handbook Graph for Formula Four INCLUDING DIE ANGLE, degrees Work Sheet for designing tooling geometry to prevent central bursting. |
Reduction in Area:Step reduction, drawing wire from d _{i-1}to d_{i}(8) Examples :drawing from step 3 to step 4:r_{4}[1-(0.090/0.102)^{2}] X 100drawing 0.102" wire to 090" r = [1-(0.090/0.102) ^{2} X 100 = 22.15%Total Reduction, drawing from d _{0} to d^{i}(9) Examples :at step 4: r = [1-(d_{4}/d_{0})^{2}]
X 100drawing 7/32" rod to 0.102": r = [1 - (0.102/0.219)^{2}] X 100 = 78.31%Overall reduction, drawing from d _{0} to d_{f}(10) Examples :drawing 7/32" to 0.090" :r_{f}= [1-(0.090/0.219)^{2}] X 100 = 83.11%NOTES :1. Reduction in area is not additive, r _{f} is not f_{1}
+ r_{2} + ... + r_{n}2. Reduction is always positive. 3. Reduction is never larger than 100%. 4. For ferrous wire, step reduction is usually between 7% and 38% (limits depend on the product). 5. For overall reduction greater than 85%, the process has to be carefully designed. 6. Clue : for final diameter of one half the original, reduction is 75%. To Compute the diameter from reduction :(11) Example:Drawing 7/32" rod 25% reduction:Given the small diameter, the large diameter is: (12) Where:r = drawing reduction in area(%) d = wire diameter(inch) Example: For an overall reduction of 83.1%, an 0.090" wire
will require a starting diameter of:d _{o} = The Required Percent Reduction Between the Bottom and Top Block of a
Double-Draw Single-Block Frame:Total required reduction at the second hole (13) = 100% - |