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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) Formula

Example : The Weight per foot os .090" wire is :
Formula

The length per lb of wire is:
(3)
Formula

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:
Formula

  * Surface Area:
The surface area of one foot of wire is:
(4) a = X 0.090 / 12 = 0.0236 ft2/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 (ft2/ft)
a/w = surface area of one lb of wire (ft2/lb)
d = wire diameter (inch)
w = weight of one foot of wire (lb/ft)
y = specific weight (for steel, 0.283 lbs/in3)

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 ft2lb

* 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/ft2zinc is:
(6) Formula

Example : for 0.090" diameter wire coated 0.3 oz/ft2 :

(6) Formula
The weight of zinc on d" diameter wire coated with p% zinc is :

(7) Formula

Where :
p% = weight percent of zinc(%)
c = zinc coating weight (oz/ft2)
d = wire diameter(inch)

Example: for 0.090" diameter wire, 2.00% zinc coating weights :

Formula

* Refer to Page 301, Volume 1, Steel Wire Handbook
  Reduction in Area:
Step reduction, drawing wire from di-1to di
(8) Formula
Examples :drawing from step 3 to step 4:
r4[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 d0 to di
(9) Formula
Examples :at step 4: r = [1-(d4/d0)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 d0 to df
(10) Formula

Examples :drawing 7/32" to 0.090" :
rf= [1-(0.090/0.219)2] X 100 = 83.11%

NOTES :
1. Reduction in area is not additive, rf is not f1 + r2 + ... + rn
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) Formula

Example:Drawing 7/32" rod 25% reduction:
Formula
Given the small diameter, the large diameter is:
(12) Formula

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:
do =

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% -
Arrow
Graph for Formula Four
Gharph
INCLUDING DIE ANGLE, degrees

Work Sheet for designing tooling geometry to prevent central bursting.


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