Lee Spring

Stock Compression and Extension Tolerances
and Engineering Notes


Compression and Extension
.057” to .094” O.D. ± .003” .102” to .250” O.D. + .003” -- .005” .300” to .500” O.D. ± .008” .540” to .850” O.D. ± .015” .875” to 1.125” O.D. ± .020” 1.150” to 1.218” O.D. ± .025” 1.250” to 1.460” O.D. ± .030” 1.480” to 1.687” O.D. ± .040” 1.937” to 2.000” O.D. ± .055”		1.45 to 2.36mm ± .08mm 2.59 to 6.10mm + .08mm -- .13mm 7.62 to 12.70mm ± .20mm 13.97 to 21.59mm ± .38mm 22.23 to 28.58mm ± .51mm 29.21 to 30.94mm ± .64mm 31.75 to 37.08mm ± .76mm 37.59 to 42.85mm ± 1.02mm 49.20 to 50.80mm ± 1.40mm
Spring rate: ± 10% Solid height: + 5%, no lower limit

Metric Conversion Table

FORCE

  To convert newtons to ......Kilograms.....Pounds
  multiply by ................0.102.........0.22487
  To convert kilograms to ....Newtons.......Pounds
  multiply by ................9.807.........2.2046
  To convert pounds to........Newtons.......Kilograms
  multiply by ................4.448.........0.4536

RATE

  To convert kg./mm. to.......lb./in. ......N./mm.
  multiply by ................55.998........9.807
  To convert lb./in. to ......Kg./mm. ......N./mm.
  multiply by ................0.017858......0.175133
  To convert N./mm. to .......Kg./mm. ......lb./in.
  multiply by ................0.101968......5.7099

LENGTH

  To convert inches to .......Meters ...Feet ......Millimeters
  multiply by ................0.0254 ...0.08333 ...25.4
  To convert millimeters to ..Meters....Feet ......Inches
  multiply by ................0.001 ....0.003281 ..0.0393701


COMPRESSION SPRINGS
To find the load at any working length, when free length and rate are given, use the formula: P = R x F where P is the load in lbs.; R is the rate in lbs. per inch; F is the deflection from free length. Example: (Lee Stock Spring Catalog #LC-032C-8) — Given a free length of .750” and a rate of 22 pounds per inch, find the load at 1/2” working length. P = 22 x .250 = 5.5lbs.
EXTENSION SPRINGS
To find the load at any working length, when the free length, rate and initial tension are given, use the formula: P= (R x F) + I.T. where P is the load in lbs.; R is the rate in lbs. per inch; F is the deflection from free length; I.T. is the initial tension. Example: (Lee Stock Spring Catalog #LE-031C-1) — Given a free length of 1”, a rate of 6.9 pounds per inch, and .7 pounds initial tension, find the load at 1.500”. P = (6.9 x .500) + .7 = 3.45 + .7 = 4.15 lbs.