IW brass

I have spools of IW brass I bought years ago. Can anyone tell me how to
calculate the length I have on the spools by weighing them? I used to
have the specifications, but the website seems long gone.

Many thanks.

You might find the information here.

David

This could be difficult if you have no idea what an empty spool weighs!

You need to measure the diameter also, in addition to the weight to estimate the length.

| Dennis

I have spools of IW brass I bought years ago. Can anyone tell me how to
calculate the length I have on the spools by weighing them? I used to
have the specifications, but the website seems long gone.

Many thanks.

Dear Dennis

There are a few calculations required here, and yes, you will need to know the weight of an empty spool!

Basically,

length = mass / (cross section * density)

To determine the cross section,

cross section = (d/2)^2 x π

For density, you can use the figures from Rose & Law: US Yellow Brass is 8404kg/m³, Red Brass slightly more at 8564kg/m³.

wire mass = (density x volume) - spool mass

Also,

volume = length x (d/2)^2 x π

Plumb all that into the first formula, and convert to the Imperial system that Lutz sold his wire in, and you will see that an unused 4oz spool of .014˝ Yellow Brass would contain 135.9m of wire.

(Credit to Weicong Li who provided these formulae on my request a few months ago.)

Regards

Carey

Dear Dennis:
Please note that the densities Carey kindly posted from Rose & Law are a bit different from the ones I had years ago from Lutz:
Yellow Brass 8670, Red Brass 8760.
This suggests that without knowing the exact density, your calculations can easily be off the mark by about 2%: not a big deal perhaps.

I have now derived a full simple formula for your needs. It is now past 1.20 am. Will test it tomorrow with a numerical example before posting it here.

CALCULATING SPOOL LENGTH FROM ITS NET WEIGHT, STRING DIAMETER AND ALLOY DENSITY.
Aw well, let me post my formulae now.
You need to know the string diameter, density and net spool weight (not including any plastic holder or envelope).

  • measure the string diameter (D) in millimetres.
  • measure the net weight (W) the spool in grams.
  • find out the string density (Y): I suggest average values between Rose&Law and Lutz: you will be good with 8540 for Yellow Brass and 8660 for Red Brass.
  • after some algebra I got the formula for the total (T) length as follows.
    T = 4000 x W / (D^2 x π x Y)

CALCULATING SPOOL LENGTH FROM SPOOL AVE. DIAMETER AND NUMBER OF CIRCLES

Perhaps less accurate (error up to 5%) but much, much easier measurements and simple formula: this is the method I have always used, and it fits the purpose, at least for me.

  • count the number (N) of “circles” in the spool.
  • measure the average spool (D) diameter in millimetres.
  • compute the total (T) length in millimetres:
    T = Pi x D x N

Considering that the Weight method has easily an error of up to 3%, for an accurate result with an error below 2%, you may use both methods and average the results.

Needless to say, my “CALCULATING SPOOL LENGTH FROM SPOOL AVE. DIAMETER AND NUMBER OF CIRCLES” is good only for relatively small coils, where you can count the number of circles. It is pretty useless for large spools.

Except the archived website gives the length of spools of brass, and 0.014 is 100 feet (~30m) for large spools. Some wrong conversion factor?

Dear All

Oh, I do hope my posted formulae are correct! The archived Instrument Workshop page you link to is for the small and large coils of brass which were sold by specific length, rather than the spools of wire which were sold by weight.

As @CDV mentions, his simple formula gives approximate results for small coils where it is possible to count the number of revolutions, not appropriate for spools.

Regards

Carey

Claudi, Carey, David & all: many thanks, I’m all set. In the meantime, I
remembered I’d figured out years ago an Excel formula that worked for my
spools of P-wire, but when I did manage to find it on my computer, I
didn’t have the density data for IW’s brass that would allow me to use
it (besides which I had a hard time making sense out what I’d done at
the time). So, thank you.

.