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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}/ft The 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
COATING The 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 
Reduction in Area: Step reduction, drawing wire from d_{i1}to d_{i} (8) Examples :drawing from step 3 to step 4: r_{4}[1(0.090/0.102)^{2}] X 100 drawing 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 100 drawing 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 DoubleDraw SingleBlock Frame: Total required reduction at the second hole (13) = 100%  

Graph for Formula Four INCLUDING DIE ANGLE, degrees Work Sheet for designing tooling geometry to prevent central bursting. 
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