Solder Wire and calculating its 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
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 Let’s 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 can calculate the density. We won’t have to do that because, in the case of these Alloys, that has already been done for us. Before we find it, let’s 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 datasheet 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 let’s plug them into our equation and calculate the length of this product.
Lets start by simplifying this equation.
and now just a little bit more.
If you would like to convert to feet all you need to do is multiply that answer by 0.033
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 |
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)
Equation for finding D (Density)
Equation for finding W (Weight)
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 in. 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 weigh 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? I could go buy some and do it OR…
I could do a 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:
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 |
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. Knowing which value is closer to the real weight would be interesting.
If for some reason you cannot find a density for a particular alloy or find that you do not trust what is presented, there is a good chance it can still be calculated.
Here is a link to a fantastic post from Senior Technologist Ron Lasky with the Indium Corporation. Please click the link below to view the original post in its entirety
Solder Alloy Densities: Comparing Calculated to Actual
Here is a derivation of the correct density formula:
Many people incorrectly assume that if you have an alloy of x % tin and y % silver, that the density of this alloy would be 0.xDensity tin +0.yDensity silver. This intuitive linear formula is incorrect however, as density has two units (mass and volume). An easy way to understand the derivation of the correct formula (proposed by Indium Corporation engineer Bob Jarrett) is to consider a 96% tin, 4 % silver example.
Lets assume I have 1 g of this alloy, 0.96 g is tin and 0.04 g is silver.
The volume of the tin is 0.96 g/7.31g/cc = 0.131327cc
The volume of the silver is 0.04g/10.5g/cc = 0.00381cc
So 1 g of the alloy has a volume of 0.131327 + 0.00381 cc = 0.135137 cc
Hence it’s density is 1g/0.135137cc = 7.39989g/cc
Hence, the general formula is:
1/Da = x/D1 + y/D2 + z/D3
Da = density of final alloy
D1 = density of metal 1, x = mass fraction of metal 1
same for metals 2, 3
The formula continues for more than 3 metals.
He has developed an Excel spreadsheet that calculates density automatically. If anyone wants a copy, send him an email at rlasky@indium.com