Table of MIG Wire Size - Feet on a Spool

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How Many Feet of MIG Wire on a Spool?Once in a while I need the answer, so here it is.I provide one sample calculation (that you can ignore) followed by a Table for the common MIG wire sizes in steel, stainless steel (SS), aluminum and for reference - copper.Material Density------gm/cm^3----lbs/in^3Iron/Steel------------7.7--------0.27812Stainless Steel ------8.0--------0.28896Aluminum--------------2.7--------0.09752Copper----------------8.9 -------0.32147Sample CalculationHow many feet of 0.035 inch diameter mild steel wire in a 2 lb spool of MIG wire?Density conversion factor: 1 gm/cubic cm = 0.03612 lbs/cubic inch Density of steel = 7.7 gm/cubic cm or 7.7 x 0.03612 = 0.27812 lbs/cubic inch2 lbs of steel wire has a volume of 2 lbs / 0.27812 lbs/cubic inch = 7.191 cubic inches.Wire of 0.035 inch diameter (D) has a cross-sectional area of = pi/4 x DxD = 3.14156/4 x 0.035 x 0.035 = 0.000962113 square inches.Wire Volume = Wire Length x Wire Area.Wire Length = Volume/Area = 7.191 / 0.0009621 = 7,474.3 inches or 622.8 feet.Formula: Wire Length = [weight of wire (lbs) / density (lbs/in^3)] / [3.14159/4 x Diam^2 (inches^2)]TABLE: FEET OF MIG WIRE ON A SPOOL (Solid Wire, not Flux Core)Wire Diam---Spool--Steel----SS------Aluminum---Copper(inches)---(lbs)---(feet)--(feet)---(feet)-----(feet)0.023--------1-------721------694-----2056-------6240.023--------2------1442-----1388-----4112------12480.023--------10-----7210-----6940----20560------62400.023--------44----31724----30536----90464-----274560.030--------1-------424------408-----1209-------3670.030--------2-------848------816-----2418-------7340.030--------10-----4240-----4080----12090------36700.030--------44----18656----17952----53196-----161480.035--------1-------311.4----300------888-------269.40.035--------2-------623------600-----1776-------5390.035--------10-----3114-----3000-----8881------26940.035--------44----13703----13189----39080-----118550.045--------1-------188.4----181------537-------1630.045--------2-------377------362-----1074-------3260.045--------10-----1884-----1810-----5373------16300.045--------44-----8289-----7978----23640------71720.060--------1-------106------102------302--------91.70.060--------2-------212------204------604-------1830.060--------10-----1060-----1020-----3022-------9170.060--------44-----4662-----4488----13298------4034Rick V
Reply:Good info Rick.Based on wire feed rate and arc time, or on weld joint volume, you can figure how much wire you're using in lbs or feet.
Reply:Hey Pulser, your are right... and once in a while folks tell us tales of how many lbs per hour they were laying down.Not only that, the Table is interesting/useful if a fellow takes the time to measure (calibrate) the wire-speed at various settings of his welding machine.e.g.Here is the Wire Feed Measurements from my Lincoln Mig PAK 15 (SP-175T)Setting       inches/minMin…………….  481……………….  512……………….  903……………….1404……………….1905……………….2656……………….3207……………….3808……………….4409……………….50010……………..560Max…………...563Question: If my welder was loaded with a new 2 lb spool of 0.023 inch diameter steel MIG wire, I could weld continously at setting #4 for how long?According to the Table, 2 lbs of 0.023 = 1442 feet or 12 x 1442 = 17304 inches.Setting #4 = 190 inches/minuteTherefore, number of minutes = 17304 / 190 = 91 minutes or about 1.5 hours.How much gas would I use?  If I were running a typical gas flow of 15 cubic feet per hour, I would use/need 1.5 hours x 15 ft^3/hour = 22.5 cubic feet.The range of my welder settings, maximum speed to minimum gives me from 1/2 hour to 6 hours continuous welding from a 2 lb spool of 0.023 solid wire.Rick V
Reply:

Originally Posted by Rick V

How Many Feet of MIG Wire on a Spool?Sample CalculationHow many feet of 0.035 inch diameter mild steel wire in a 2 lb spool of MIG wire?Density conversion factor: 1 gm/cubic cm = 0.03612 lbs/cubic inch Density of steel = 7.7 gm/cubic cm or 7.7 x 0.03612 = 0.27812 lbs/cubic inch2 lbs of steel wire has a volume of 2 lbs / 0.27812 lbs/cubic inch = 7.191 cubic inches.Wire of 0.035 inch diameter (D) has a cross-sectional area of = pi/4 x DxD = 3.14156/4 x 0.035 x 0.035 = 0.000962113 square inches.Wire Volume = Wire Length x Wire Area.Wire Length = Volume/Area = 7.191 / 0.0009621 = 7,474.3 inches or 622.8 feet.Formula: Wire Length = [weight of wire (lbs) / density (lbs/in^3)] / [3.14159/4 x Diam^2 (inches^2)]
Reply:Always wonderd how the estimators estimate the cost of wire and rods, now I got a pretty good idea
Reply:Would it be easier to just cut 10 ft off. wad it up and weigh it on electronic scales or like my beam style powder scale for reloading. With that you could calculate how many ft is on a full, unused spool. Or if you know the tare weight of an empty spool you could calculate the number of feet on a used spool.Ol' Stonebreaker  "Experience is the name everyone gives to their mistakes"Hobart G-213 portableMiller 175 migMiller thunderbolt ac/dc stick Victor O/A setupMakita chop saw
Reply:You would be assuming that the tare weight of one spool is the same as another spool... That might be a bit optimistic... Using his values, 1 ft of wire would weight 1/721 lbs, which means 9.70873786408 grains... 1 lb = 721 ft=> 1/721 lb = 1 ft=> 1/721 lb * 7000 gr/lb = 1 ft=> 9.70873786408 gr = 1 ftMost electronic powder scales will give you at least 1/10th of a grain precision...The OP was last on here back in 2018, so he might not be actively following the site anymore...Last edited by NavyVet1959; 5 Hours Ago at 10:37 PM.
Reply:Getting a little extreme with digits to the right of the decimal point!!Ol' Stonebreaker  "Experience is the name everyone gives to their mistakes"Hobart G-213 portableMiller 175 migMiller thunderbolt ac/dc stick Victor O/A setupMakita chop saw