Solder Wire and calculating it's length

AaronRollens · April 7, 2022, 10:01pm

Solder Wire and calculating it’s length.

If you have been searching for a spool of solder and have gotten frustrated because they do not tell you the length of the product on the spool, there is a reason for that. The Industry does not normally sell solder by the length, they sell it by the weight instead.

If you have trouble visualizing the amount of product you are going to receive, this value can still be calculated. There are three things we will need to know.

The weight of the product (not including packaging)
The diameter of the wire.
The composition of the wire (so we can find its Density)

Var	Description	Units
W	Weight	g
D	Density	g/cm³ or Kg/m³
d	Diameter	mm
L	Length	cm

Equation for finding L

L=\frac{W/D}{π \Bigl(\frac{d/2}{10} \Bigr)^{2}}

WAIT! Don’t go just yet. While it’s admittedly true that the equation does look a little scary, I promise it looks worse than it is. You can do this.

Below is a real world example that we can run through from beginning to end.

Example One:

MFR PART #	4900-35G
DIGI-KEY PART #	473-1116-ND
MANUFACTURER	MG Chemicals
DESCRIPTION	Lead Free No-Clean Wire Solder Sn96.5Ag3Cu0.5 (96.5/3/0.5) 20 AWG, 21 SWG Tube, 0.60 oz (17g)
DATASHEET	Click Here

First Lets get these variables out of the way.


W	(Weight)	17g

This was in the description of the part above. Because solder is usually sold by weight it can often be found there.


d	(diameter)	0.8mm

While the description calls out a 20AWG wire, we need this to be in the mm diameter format for the calculation. I took a look at the data sheet and the value was listed there.


D	(Density)	7.38 g/cm³

All elements have a consistent Density. This means that as long as we know how much of something is in something, we could calculate the density. We won’t have to that because in the case of these Alloys, that has already been done for us. Before we find it, lets understand what we are looking for.

This alloy is listed as Sn96.5/Ag3.0/Cu0.5. It is made from 96.5% Tin, 3.0% Silver, and 0.5% Copper. Anytime we have an object with this exact combination of these elements, the density will be the same. It’s physics.

So where do we find the density you ask. Well it is usually in the data sheet but if it is not, a quick google search of the alloy with the word density usually helps us find a resource that will give us that number. With that said, if you keep scrolling down I have a list of our most common solders and their densities below.

Now that we have the values we need lets plug them into our equation and calculate the length of this product.

L=\frac{17/ 7.38}{π \Bigl(\frac{0.8/2}{10} \Bigr)^{2}}

Lets start by simplifying this equation.

\frac{17}{7.38} = 2.3

\frac{0.8}{2}=0.4

L=\frac{2.3}{π \Bigl(\frac{0.4}{10} \Bigr)^{2}}

and now just a little bit more.

\frac{0.4}{10} = 0.04

0.04^2 = 0.0016

L=\frac{2.3}{π *0.0016}

π *0.0016 = 0.00502654824

\frac{2.3}{0.00502654824}

L = 457.57 cm (4.57 m)

If you would like to convert to feet all you need to do is multiply that answer by 0.033

L = 15 ft

Example Two:

MFR PART #	4865-454G
DIGI-KEY PART #	473-1108-ND
MANUFACTURER	MG Chemicals
DESCRIPTION	Leaded No-Clean Wire Solder Sn63Pb37 (63/37) 20 AWG, 21 SWG Spool, 1 lb (454 g)
DATASHEET	Click Here


W	(Weight)	454g
d	(diameter)	0.81mm
D	(Density)	8.40 g/cm³

L=\frac{454/8.4}{π \Bigl(\frac{0.81/2}{10} \Bigr)^{2}}

L = 10484 cm =104 m = 346 ft

Common Values

I have compiled a list of known Alloy compositions and their densities below that you can use in your calculations.

Composition		Density
Alloy Composition		g/cm³
Bi57Sn42Ag1		8.57
Bi58Sn42		8.57
In100		7.31
In52Sn48		7.3
In97Ag3		7.38
Pb60Sn40		9.28
Pb70Sn30		9.72
Pb85Sn15		10.47
Pb88Sn10Ag2		10.73
Pb93.5Sn5Ag1.5		11.02
Pb95Sn5		11.03
Pb97.5Ag1.5Sn1		11.26
Sn50Pb50		8.87
Sn60Pb40		8.51
Sn62Pb36Ag2		8.42
Sn62Pb38		8.37
Sn63Pb37		8.4
Sn95Ag5		7.4
Sn95Sb5		7.25
Sn96.5Ag3.0Cu0.5		7.38
Sn96.5Ag3.5		7.37
Sn96.5Ag3Cu0.5		7.37
Sn96Ag4		7.38
Sn97Ag3		7.36
Sn97Cu3		7.33
Sn99		7.28
Sn99.3Cu0.7		7.26

I would like to take a moment to point out that this does not only work for solder. This equation can work for many different scenarios. If we manipulate the equation a little bit we can change the value we are searching for.

Equation for finding L (Length)

L=\frac{W/D}{π \Bigl(\frac{d/2}{10} \Bigr)^{2}}

Equation for finding D (Density)

D=\frac{W}{Lπ \Bigl(\frac{d/2}{10} \Bigr)^{2}}

Equation for finding W (Weight)

W= DLπ \Bigl(\frac{d/2}{10} \Bigr)^{2}

Click here to see formulas for use in database applications like Microsoft Excel

To find Weight

=Density*Length*PI()*POWER(((Diameter/2)/10),2)

To find Density

=Weight/(Length*PI()*POWER(((Diameter/2)/10),2))

To find Length

=(Weight/Density)/(PI()*POWER(((Diameter/2)/10),2))

Example 3

Here is a different type of example. I am going to run some PVC pipe on a boat I am retrofitting. I do not want to exceed the weight limitations so I am trying to keep a running tally of anything I put it it. If I had a piece in front of me I could weigh a length of it and then do the math to figure out how much it would weight for the appropriate length but I don’t have any right now. I also don’t have a great scale to use for this. What can I do? Obviously, I could go buy some and do it OR…

I could do google search for the density of PVC and find that it has a density of 1.38/cm3. I have calculated that I am going to need to run approximately 100ft of 2" pipe.

I found a chart that gave me the information here:

Source: www.pvcfittingsonline.com

which gave me these dimensions:
OD = 2.375 in (60.325 mm)
ID = 2.047 in (51.9938 mm)

Var	Description	Value	Units
W	Weight	W	g
D	Density	1.38	g/cm³ or Kg/m³
d	Diameter	60.325	mm
L	Length	3048	cm

W= DLπ \Bigl(\frac{d/2}{10} \Bigr)^{2}

W= (1.38)(3048)(π \Bigl(\frac{60.325 /2}{10} \Bigr)^{2})

The calculated weight of the pipe so far is:

W = 120220 g
W = 120.22 Kg
W = 265.04 Lbs

If that seems a little heavy that is because I cannot not to forget that the pipe is empty on the inside. I have to do the calculation one more time using the inside diameter value and subtract it from the value above.

W = 89307g
W = 89.30Kg
W = 196.88 Lbs

The calculated weight of the pipe is:

W=30913 g
W=30.91 Kg
W=68.15 Lbs

100 ft of 2" PVC would weigh approximately 68.15 lbs. Granted, the chart above tells us that the weight per/ft is 0.720 (lbs?) and that would give us 72 Lbs at 100 ft. It would be interesting to know which value is closer to the real weight.