# Pre-Calculating Wire-Feed Speed, Travel Speed and Voltage

It is possible for manufacturers that are welding with solid wire or flux-cored wire to calculate the starting wire-feed speed and travel speed? These calculations will get you close, but some tweaking may have to be done to achieve your desired weld.

A shop would like to decrease the development time for new MIG welds. Is there a way to pre-calculate wire-feed speed, travel speed and voltage, to get them close before they strike the first weld?

Yes, it is possible to calculate starting wire-feed speed and travel speed. This is a very common question from manufacturers welding with solid wire or flux-cored wire. Most welding professionals know the wire-feed speed (WFS), where a process runs well based on their experience, or can quickly get WFS from the manufacturer’s recommended procedures.

However, determining how fast to travel for a particular size weld bead ends up being an iterative, time-consuming process. By understanding a few concepts and doing some math with a few simple formulas, we can determine at least a good starting point for a welding procedure that produces the desired weld.

**DEPOSITION RATE**

It is essential to recall that Deposition Rate is directly proportional to the speed at which a particular wire diameter emerges from a welding gun during welding. Deposition rate has nothing to do with how fast the gun is traveling, nor the voltage setting on the machine. Deposition rate is simply a measure of how many pounds of wire come from the welding gun in a certain amount of time, typically measured in lb/hr.

If wire-feed speed increases, deposition rate increases. We also understand that if we maintain the wire-feed speed and change to a larger diameter wire, deposition rate will increase as well. Armed with this understanding, calculating deposition rate ends up being a very powerful exercise that gives you a number that can be used to calculate key welding parameters. Let’s look at the formula and an example:

**Deposition Rate Calculation**

Deposition rate (lb/hr) = 13.1 × (Wire diameter)^{2} × (Wire-feed speed) × (Efficiency)

**—** Wire diameter in inches (in)

**—** Wire-feed speed in inches per minute (ipm)

**—** Efficiency (1.0 for solid wire, 0.85 for cored wire)

**—** This calculation is for steel only

**—** *e.g.*: Wire diameter = 0.045 in (1.2 mm) solid wire, WFS = 300 ipm

Deposition rate = 13.1 × (0.045)^{2} × (300) × (1.0) = 7.96 lb/hr

**Calculating Travel Speed With Deposition Rate**

Knowing the deposition rate, we can calculate the travel speed in inches per minute (ipm) for a particular weld. Let’s say we want to make a 3/8 in steel fillet weld (assume 10 percent reinforcement or 0.4125 in leg) using 0.045 in solid wire at 300 ipm, the weight of weld metal per foot can be calculated by multiplying the density of steel (0.283 lb/in^{3}) by the volume of weld metal per foot as follows:

**Weld Weight Per Foot Calculation**

Volume of weld metal/ft = 1/2 × b × h × 12 in = 1/2 × 0.4125 in × 0.4125 in × 12 in = 1.02 in^{3}

Weight of weld metal/ft of 3/8 in fillet weld = (0.283 lb/in^{3}) × (1.02 in^{3}) = 0.2887 lb/ft

From the calculation below, we see that the travel speed for a one-pass, 3/8 in fillet weld would be 5.52 ipm, 11.03 ipm for a two-pass fillet weld, or 16.55 ipm for a three-pass weld.

**Travel Speed Calculation**

Travel speed = (deposition rate) × (# of passes)/5 × (weight of weld metal per foot {lb/ft}) =

Travel speed = {7.96 x 1} / {5 x 0.2887} = 5.52 ipm

**CALCULATING WIRE-FEED SPEED WITH DEPOSITION RATE**

Let’s assume a requirement to make fillet welds at a rate of 12 lb/hr using 0.045 in welding wire. We can calculate the WFS using the formulas below and Weight of Weld Wire per foot in **Table 1**.

**Wire-Feed Speed Calculation**

Wire-feed Speed = (deposition rate)/5 × (weight of wire per foot {lb/ft}) = (12)/5 × (0.0054) = 444.4 ipm

Of course, the travel speed for a one-pass 3/8 in fillet weld at a 12 lb/hr deposition rate would be 8.31 ipm as calculated below:

Travel Speed = (deposition rate) × (# of passes)/5 × (weight of weld metal {lb./ft}) = (12) × (1)/5 × (0.2887) = 8.31 ipm

**MAKING IT EASIER**

The Bartonian Conversion Factor (see **Table 2**) makes things a bit easier for fillet welds. The example below uses the conversion factor to calculate travel speed at 5.57 ipm for that same 3/8 in fillet weld using a 0.045 in solid wire.

Travel Speed = 7.96 × 0.7 = 5.57 ipm (.2887)

Weight of weld metal per foot can be calculated for any joint type by calculating the volume and multiplying by the density of the weld metal (eg. 0.283 lb/in^{3} for steel). However, the values shown in **Table 3 through Table 6** eliminate the need to do the calculation. These values are taken from Table 12-1 in *The Procedure Handbook of Arc Welding* by the Lincoln Electric Co., and show the Weight of Weld Metal per Foot for several common joint types welded with steel. The following sample calculations use **Tables 3 through 6**.

**SAMPLE CALCULATION NO.1**

1/2 in plate Vee groove weld with 90 deg included angle and 1/8 in reinforcement using 0.052 in solid wire, 90%Ar/10%CO_{2} shielding gas. Manufacturer recommends WFS at 325 ipm and 30 volts.

Deposition Rate (lb/hr) = 13.1 × (0.052)^{2} × (325) × (1.0) = 11.51 lb/hr

Travel Speed for fill and cap passes = (11.51) × (6)/5 × (0.849 + 0.199) = 13.18 ipm

**SAMPLE CALCULATION NO.2**

3/8 in plate square butt weld into a backing, with a 3/16 in gap and 1/8 in reinforcement using 1/16 in cored wire, 75%Ar/25%CO_{2} shielding gas. Our process runs great with WFS at 285 ipm and 26 volts.

Deposition rate (lb./hr) = 13.1 × (1/16 in)^{2} × (285) × (.85) = 12.39 lb/hr

Travel speed for fill and cap passes = (12.39) × (1)/5 × (0.239 + 0.053) = 8.49 ipm

**SAMPLE CALCULATION NO.3**

What should my wire-feed speed be if I want to make a fat, 1/4 in fillet weld at 20 ipm travel speed using 0.045 in solid wire and 90%Ar/10%CO_{2} shielding gas?

Re-arranging the Travel Speed calculation above to solve for Deposition Rate we get:

Deposition rate = 5 × Travel speed × (Weight of weld metal {lb/ft})/(# of passes) = 5 × 20 ipm × (0.165)/1 = 16.5 lb/hr

Re-arranging the Deposition Rate calculation above to solve for Wire-feed Speed we get:

Wire-feed speed = Deposition rate/13.1 × (Wire diameter)^{2} × (Efficiency) = 16.5/13.1 × (.045 in.)^{2} × (1) = 622 ipm

**SUMMARY**

Using math and these simple formulas can reduce your weld-development time, and improve your pre-WPS documentation. These calculations will get you close, but some tweaking may have to be done to achieve your desired weld. Having a good starting point for welding procedure development will decrease time and guesswork on the shop floor.

## 15 Comments

I want to find out how to work back to the weld deposition rate with relation to wire speed, amps, volts . . . any calculation there is to trace out how much deposition is done in one job. For example, we are generating data like 27.5 volts, 310 amps, and 336 mm/min wire speed. Can you help me?

Prakash Karkera

Welding Engineer

Model Infra Corporation pvt ltd

Marmagao, India

I would like to know if travel speed is referring to an automated process or is it required when welding manually/freehand.

I am working this calculation but I come up with a different answer. I come up with .459. Can you explain why I am coming up with a different answer?

(7.96) × (1)/5 × (0.2887) = 5.52 ipm

Owen, did you get a reply? to me there is either something wrong with the math, or something wrong with the formula. This formula is used elsewhere on the internet with the same results.

There is a typo in the formula regarding parenthesis…….. formula should be as follows:

Travel speed = {7.96 x 1} / {5 x 0.2887} = 5.52

The revised calculation has been inserted. Thank you.

Is there such a tool as a “Wire Speed Calibrator (for travel)?

What does 13.1 reference to in the deposition rate formula? Is it the voltage?

13.1 is a derived constant to convert input units to the correct output units. Just a constant!

Mr. Beardsley, what does 5 reference to in the Travel Speed Calculation formula? Thank you very much.

Dear Mr.Kevin,

I am confused with my number I calculated for FCAW welding with wire 1.6 mm diameter for fillet weld 3 mm 90 deg. What shall I consider for welder productivity as travel length per day?

Actually I am making the calculation for new steel structure factory and I am doing the estimation right now where I have a 1034 meter of welding per day. Shall I take the number of travel as it is or I shall consider some waste of time during operation?

Hi Kevin.

Thanks a lot for this useful information.

Can I know why are we using value 5 in

Travel speed = (deposition rate) × (# of passes)/5 × (weight of weld metal per foot {lb/ft})

Can you please help me on that?

Regards,

Gopalakrishna. V

United Motors and Heavy Eqp Pvt

I believe the 5 is a consolidated value derived from 60 / 12. 60 being min. per hour and 12 being inches per foot.

Very cool article. This is exactly what I was looking for.

I tried your calculations versus some published numbers I found and the travel speed matched up very nicely for all but one….0625″ plate, .035 wire, wfs = 185 ipm. I got a travel speed of 74 ipm versus the suggested speed of 25. I used the same formulas in a spread sheet and all of the other sets (thicker sheet sizes) matched up. Is it possible that the approximations don’t scale to such a thin sheet or more likely that the published suggestions I’m looking at have an error?

I’d love to see these calculations done for a single v bevel at 32.5* on a 4″ sch 40 pipe to be welded in a positioner with .045″ diameter wire using a TIG and cold wire feeder (max feed rate 237 ipm).

.045″ wire @ 237 ipm = ____ lbs/hr Deposition Rate

((237″/12)x .0054 {weight per ft of .045″ wire from table above}) x 60 min = 6.399 lbs/hr

Travel Speed = ___ (seems there would be a window based on the rpms of the positioner and the Deposition Rate limited by the Cold Wire Feeders max feed rate)…

Travel Speed = (6.399 DR x 1 {# of passes})/(5 x .345286 {weight of weld}) = 3.70649 ipm

Volume of the weld metal to fill the Single V Bevel at 32.5* with a 1/8″ gap = ___ cu in

My approach was to find the area of the gap (.237″ wall thickness x .125″ gap = .029625 sq in)

Using a right triangle calculator to get the height (.237″) and base width (.15099″) of the two triangles formed by the beveled areas on each side then multiplying them I get an area of .03578463 sq in.

In my mind unrolling the pipe circumstance as if it were plate… 4.5″ x Pi = 14.137 linear inches.

Find the area of the weld reinforcment or cap pass ((.10599×2 base lengths)+.125 gap)=.33698 wide. LxW (.33698x.125)/2 gets us close… .2106125 sq in for the cap.

My mind says find an equivalent square for the areas added together… Sqrt ( weld cap + gap area + triangle areas) = sqrt (.08647088 sq in) = .294059 side lengths.

Now multiplying the two equivalent sides by the circumference yields a volume of 1.222 cu in.

Weight of weld metal (useful for how much filler to purchase)

Steel density (.283 lb per cu in) x the weld metal volume 1.222 cu in = .345286 lbs