math help

[Anonymous]

Calculating gold content of plated area. I know that I have seen topics on this but I am just getting some crazy numbers when I figure the gold content of plated materials.

Example 12 micron plating .75 inches long by .375 inches wide.
I calculate the thickness as .00047 * .75 * .375 =.0001329 cubic inches.
10.17 troy ounce to the cubic inch, yields .000135 tr oz, yields $1.56 for this tiny piece of gold, I just can not see what I am getting wrong here.

thanks
jim

 

[goldsilverpro]

James,

Your math technique is right. The problem is that gold plating is never 12 microns thick, at least on any electronic parts that I've seen made in the last 60-70 years. About the thickest I've seen recently was on some military pins that I ran an assay on. These ran about .0001" or 2.5 microns. That's nothing to sneeze at since their gold value was about $280/#. Fingers typically run about .00003" or .75 microns, more or less. The plating on many CPU packages runs .00004" - .00006" = 1 to 1.5 microns. Twelve microns is getting into the realm of electroforming. With most gold plating baths, 12 microns is beyond their capabilities. Although you'll very rarely see it plated that thick, the hard gold plating baths used for fingers and pins is usually limited, technically, to about 3 to 5 microns, due to the internal stress of the deposit.

Maybe the gold is 12 microinches thick rather than 12 microns. The difference is 1/40. This would make the gold worth about $.04, for that 3/8" X 3/4 piece, assuming it really is 12 microinches thick.

An easy way to determine the dollar value of 1 sq.in. of 1 microinch thick (1/40 micron) gold is to simply divide the spot gold price by 100,000 (to blow my own horn, as far as I can determine, I was the first to notice that relationship, many years ago). This is an estimate but, if I remember right, it is only mathematically about 1.7% in error, assuming the density of plated gold is 19.3 g/cc (actually, though, the density is usually lower that this, except for special types of plating baths). Once you do that calculation, you can estimate the value (in $/#) of the parts in question by knowing the sq.in. of plated area, by estimating the gold thickness typically used for that particular type of part, and by knowing the weight of the part. If you're buying, be conservative with your estimate. If you're selling, be liberal.

Chris

 

[Anonymous]

Thanks,
evidentally, the micron and micro inch are the source of my problem.

I am still surprized as even at 1.5 microns I get nearly $.20.

any idea how a guy could figure out the thickness? my scale is in now way going to measure the gold weight if I scrape it off.

when I look at this under a 30x hand held microscope I can clearly see the thickness of the plated layer, if it was 12 micro inches, 12 millionths of an inch right? do you think I could see the layer as a clear and seperate layer?

jim

 

[goldsilverpro]

PLATING THICKNESS MEASUREMENT

In the plating industry, there are several different types of equipment used to measure thickness. In each, accurately plated standards are needed for calibration. Different standards are needed for different thickness ranges and different substrates (the metal or metals underneath the gold).I haven't kept up with this, but I think that x-ray equipment is most commonly used today.

In the dark ages, when I was in plating, we mainly used beta backscattering equipment, such as the Betascope or the Microderm. They were quite accurate but they required the use of several radioactive isotopes. I'm sure they wouldn't pass OSHA's muster today.

We also had another type, which is still used, made by Kocour. It uses a specific solution and current to actually dissolve about an 1/8" dia. spot of gold plating (or, whatever type plating you're measuring). The thicker the gold, the longer it takes to dissolve through it. When the solution penetrates through the gold and into the metal layer below, the current changes abruptly, the machine automatically shuts off, and the time is registered. With the right standards, it can be quite accurate.

You might get lucky and pick up one of these old machines for a song. Here again, though, without proper standards, they are worthless.

Sectioning and Mounting. Another common way is to plate a thick backup layer of something hard, like nickel, onto the gold, cut the part into sections, mount and polish a section, and measure the thickness with an optical comparator scale built into the lens of a high power metallurgical microscope. I've done a lot of that.

For determining gold values, the problem with all the above methods, is that you're only measuring the gold in one area. Plating is never 100% uniform. It is always thicker on the edges and thinner in the recesses, although, with some types of baths, this effect is minimized. To determine the value of a part, you need the average thickness. All the above methods suffer from that problem, although each can be used to test at multiple places on the part and come up with some sort of a half-a**ed average thickness.

Strip and Weigh. With a scale that will measure down to about .00001 grams, you might be able to get an half-a**ed estimate of the average thickness by measuring and calculating the plated surface area, completely dissolving everything away from the gold, and then collecting the gold and weighing it. It's easy to see, though, how this method is potentially fraught with very severe difficulties. I don't think I need to list them. Due to the small quantities of gold involved when only doing one part, this is best done on a large sample. With a large sample, you could also get away with a lesser scale.

Gold Recovery. With a large sample and, if the parts are all the same, you could calculate the surface area and then recover, refine, and weigh the gold. The average thickness could then be quite accurately calculated. This might be the best way of all.

when I look at this under a 30x hand held microscope I can clearly see the thickness of the plated layer, if it was 12 micro inches, 12 millionths of an inch right? do you think I could see the layer as a clear and seperate layer?

Are you talking about first peeling the gold off and then looking at the edge or, are you dissolving everything but the gold foil and then looking at the edge? Whether you can see it or not, this would be a very poor way of judging the thickness. Very, very subjective. I could easily see a 100-200% error in doing this. Sectioning and mounting, as discussed above, would be the only way to do this with any accuracy at all.

 

[g_axelsson]

I went through some current data sheets for modern connectors and found that most platings were either 15 or 30 micro inches ( 0.36 um or 0.76 um).
http://goldrefiningforum.com/phpBB3/viewtopic.php?f=45&t=6924

Example 12 micron plating .75 inches long by .375 inches wide.
I calculate the thickness as .00047 * .75 * .375 =.0001329 cubic inches.
10.17 troy ounce to the cubic inch, yields .000135 tr oz, yields $1.56 for this tiny piece of gold, I just can not see what I am getting wrong here.

I'm going to turn this into metric and use the thickness for plating I found....
Example 15 micro inches plating .75 inches long by .375 inches wide. (0.36 um, 1.9 cm by 0.95 cm)
I calculate the thickness as 0.000036 * 1.9 * 0.95 = 0.000065 cubic cm.
Density of gold is 19.30 g/cubic cm which gives 19.30 * 0.000065 = 0.00126 g or 1.26 mg.
Todays spot is 1179/ tr oz. = 34.6 dollar/g which gives 0.00126*34.6 = $0.043 for your gold.

If you are looking at gold on pcb:s then it is most probably covered by enig=Electroless Nickel Immersion Gold and that have a thickness of 5-12 microinches of gold. For gold fingers the thickness could be more. (30 u inch from one manufacturer)

/Göran

 

[MMFJ]

Taking key bits though staying in context...

Fingers typically run about .00003" or .75 microns, more or less. The plating on many CPU packages runs .00004" - .00006" = 1 to 1.5 microns.

This would make the gold worth about $.04, for that 3/8" X 3/4 piece, assuming it really is 12 microinches thick.

An easy way to determine the dollar value of 1 sq.in. of 1 microinch thick (1/40 micron) gold is to simply divide the spot gold price by 100,000 This is an estimate but, it is only mathematically about 1.7% in error [acceptable], assuming the density of plated gold is 19.3 g/cc.

Once you do that calculation, you can estimate the value (in $/#) of the parts in question by knowing the sq.in. of plated area, by estimating the gold thickness typically used for that particular type of part, and by knowing the weight of the part. If you're buying, be conservative with your estimate. If you're selling, be liberal.

With all this in mind, I've been trying to do some calculations on some telecom equipment I purchased in a lot (mostly intended to sell, but some will be scrapped). However, in doing the 'lazy' calculation v. the 'extended version', I get totally different numbers so I think I'm missing something in my process. Any help in identifying my error(s) is appreciated!

First, is this telecom backplane (from a Cisco router - without power supplies, these aren't worth much... - I have 4 of them) - plastic removed around the pins to see all the 'pretty stuff'...

in counting the pins (yes, I did...), I got up to 7490 and lost count :x with several more on the top and all the back (not too many there) to go, I just called it 8000 (I love 'lazy' math...

Trying to keep the calculations as close as possible, withing a reasonable margin of error (note, I kept the full digits in the calculator, but only rounded to save space in the post, so there is perhaps a couple small rounding differences below - I am more interested in the PROCESS of the math than the exact numbers at this point);

These 8000 pins are 3/4" long and 1/48" wide (estimating that one - just how much is the width of one line drawn on a tape measure???)

Area of a cylinder (close on this, though actually not plated on the bottom, but margin of error is small);
2(pi)r² + 2(pi)rh
(2*3.1415926 * 0.0104167²) + (2 * 3.1415926 * 0.0104167 * .75)
(2*3.1415926 * 1.0850694e-4) + (2 * 3.1415926 * 0.0104167 * .75)
6.81769e-4 + 0.0490875
= 0.0497693 <== surface area in² of each pin
* 8000 pins
=============
398.1545 in² of gold (outside) on board

First (SIMPLE) calculation from above (SPOT / 100,000) would yield (@ $1624.80 spot)
398.1545 * 1624.80 /100000 = $6.47 worth of gold on this board (presuming 1 microinch thick) - seems like a low number, but if true, it is sure easy to work with.

However, in calculating the cubic inch total, we got a wildly different number
398.1545 * .00004 = 0.01592618 in³ of gold total
0.01592618 * 10.17 (troy oz to in³) = 0.1619692506 troy oz * $1624.80 spot = $263.51 - now, that's a number I LIKE for these boards!

I thought I just got a bit punch drunk on the numbers and should divide the in³ by the troy oz, but that is even worse!
0.01592618 / 10.17 (troy oz to in³) = 0.001566 troy oz * $1624.80 spot = $2.54

I've got another board I want to run this math process on, but I'm getting lost in the numbers a bit. I know I'm missing something simple, just not seeing it.

Thanks for your input.

 

[samuel-a]

MMFJ

My website calculator returns a figure of: 6.57$ worth of gold with the data you gave.
This figure is slightly higher then GSP's calculation and is to be expectet.
You can say that my calculator is based on perfect condition, where GSP's calculation is a more 'realistic' one. the difference will always be around 1.5% between the two.

In your numbers:
1 microinch is expressed in inches like so: 1 microinch = 0.000001 inch

 

[MMFJ]

MMFJ

My website calculator returns a figure of: 6.57$ worth of gold with the data you gave.
This figure is slightly higher then GSP's calculation and is to be expectet.
You can say that my calculator is based on perfect condition, where GSP's calculation is a more 'realistic' one. the difference will always be around 1.5% between the two.

In your numbers:
1 microinch is expressed in inches like so: 1 microinch = 0.000001 inch

THANKS! Boy, was that a simple thing (and big oversight on my part!)

I was taking the 'easy' out and just copying from GSP's answer of ".00004" - .00006" = 1 to 1.5 microns" and took .00004 = 1 - didn't check the 'micron' v 'microinch' - big difference!

Now, though I see the correct equation and, again, proven that GSP "knows all"... 8)

So, for anyone running across this post (and wanting to know the right way to calculate the amount of gold on pins), the correction is only on the next-to-last line

398.1545 * .00004 = 0.01592618 in³ of gold total
should be
398.1545 * .000001 = 0.0003981545 in³ of gold total

then, we get
0.0003981545 * 10.17 (troy oz to in³ ) =
0.004049231265 troy oz * $1624.80 spot = $6.579190959372 <== same as samuel-a, depending on how you round all the numbers.

Even though I know know the 'exact' method, I think I'll stick with GSP's "lazy" way! Now, if we could only come up with a lazy way to get the in³ number, it would be a snap! Guess that'll be something to work toward.

 

[goldsilverpro]

This may be a simpler way to look at it.

12 microinches expressed in inches is .000012". The volume of 1 square inch of .000012" thickness is: 1" x 1" x .000012" = .000012 cubic inches. Note that, since you are multiplying the thickness x 1 x 1, the volume of 1 square inch always equals the thickness.

There are about 10.17 tr.oz. of gold per cubic inch. Therefore, the thickness x 10.17 = the number of tr.oz. of gold in 1 sq.in. of that thickness.

Therefore, the dollar value of 1 sq.in. = thickness expressed in inches x 10.17 x gold spot

Therefore, at a $1658 market, the dollar value of 1 sq.in of 12 microinch gold = .000012 x 10.17 x 1658

I don't have my calculator handy but you can take it from there.

BTW, for silver, the factor is 5.5 instead of 10.17, since 1 cu.in. of silver weighs 5.5 tr.oz.

To summarize:

$ value of 1 in² of gold = thickness in inches x 10.17 x gold spot

$ value of 1 in² of silver = thickness in inches x 5.5 x silver spot