Gold Content of Scrap Items Series: 3COM ISA Network Cards

[catfish]

Hey Aflac:

Us old farts get confused some times and our fingers and “other extremities” don’t always follow our brain. I think I lost him when I made a typo on .00975square inches instead of .00975 cubic inches.

Just always remember “old age and treachery will always prevail over youth and enthusiasm”. :x

Keep the posts coming; I look forward each day to see what you are going to think up next. :lol: :lol:

Catfish

 

[mike.fortin]

Catfish--GSP---thanks for the replys. Ill just keep trying. Those URLs were shure helpful. Mike.

 

[aflacglobal]

Auhhhhh !!! The duck. What’s he up to now. I have spent the last two days learning about RF waves. Or more deeply, magnetic flux. Basically that's all that energy is. From the atom, to the volt, to the magnet, radio waves, microwaves, light, matter, energy, you get the idea. Their is more to cover that one would think. People don't realize the role it plays in every day life. If you have something electronic, it is based on some magnetic Principal . Ever wonder how them chips get that coating of gold on the wafer. Plasma fields.

One particular point of interest is crystal growing. :idea:
Water can be acted upon by magnetic waves to create forces on the atomic level that are just crushing in effect and pressure. You think ultra Sonics is something. Send a magnetic wave ( not sound, magnet flux density ) thru a solution at a few MHZ or GHZ and the whole show changes. Physics is not just about controlling chemical processes with chemicals. As variables and parameters change, suddenly the impossible become the possible. At least more likely. lol

Just a report.

 

[lmills148]

Now for the yield information:

The 214 cards yielded 600 grams of fingers and 125 grams of gold plated pin tips from the 16 pin AUI connectors. The fingers are in the large bag in the photo above and the AUI gold tips are in the quart sized bag. The small bag contains the unplated portions of the AUI pins and weighs 64 grams.

Every 5 cards weighed 15.8 ounces (~1 LB) and yielded 14 grams of clean cut fingers.

28 grams (1 ounce) of fingers (10 cards worth) produced the foils seen in the jar in the upper right corner of the photo above. Here's a closeup of the foils:

Steve

this is great information, thanx a few questions tho:

1) the 15.8 lbs per 5 boards is that with foreign attachments, specifically the steel "things" at the end of the board that attach them to the housing?

2) have you processed the pins yet? what kind of yeild do you expect?

3) do you bother with the rj45 jacks?

4) was there anything else that you harvested such as flat packs or mono.cap. that could contribute to the overall value of scrap.

from what I see so far it seems a good "scrap value" would be around 40cents per lb (allowing for cost of recovery refining and @50% margin).


.
Lloyd

 

[lmills148]

There is no answer to "why" this works.

GSP
I applied this to a memory stick I had sitting around and it looks like this ...
1.05sq in.x.066x 20u = $1.38 does this sound right?
I hope so I have about a thousand of these.
I didn't think the yeild was that hi.

.
Lloyd

 

[goldsilverpro]

NOTE: Totally ignore this post. I used the wrong assumptions and the wrong math. I should have used 100,000, instead of 10,000. Everything about this post is wrong.

What is the 20u? Is this the number of individual fingers or the estimated thickness? In the following, I assume it is the thickness - 20 microinches?

You forgot to multiply times the gold spot and divide by 10000.

Dollar value of gold plating = Spot price divided by 10000 X thickness in micro" X area in square inches.

1.05 X .066 X 20 = 1.386. Now, multiply this times 659 (gold spot) and divide by 10000.

1.386 X 659, divided by 10000 = $.09

 

[lmills148]

What is the 20u? Is this the number of individual fingers or the estimated thickness? In the following, I assume it is the thickness - 20 microinches?

yes thickness 20 microns

You forgot to multiply times the gold spot and divide by 10000.

that is the .066 ($660 / 10000=.066)

Dollar value of gold plating = Spot price divided by 10000 X thickness in micro" X area in square inches.

1.05 X .066 X 20 = 1.386. Now, multiply this times 659 (gold spot) and divide by 10000.

1.386 X 659, divided by 10000 = $.09

sorry I wasn't clear this is what I did

1.05sq in.(area)x $.066(spot price/10000) x 20 microns(estimated thickness)

Lloyd

 

[lazersteve]

1) the 15.8 lbs per 5 boards is that with foreign attachments, specifically the steel "things" at the end of the board that attach them to the housing?

2) have you processed the pins yet? what kind of yeild do you expect?

3) do you bother with the rj45 jacks?

4) was there anything else that you harvested such as flat packs or mono.cap. that could contribute to the overall value of scrap.

• The 15.8 Ounces per 5 boards is for complete boards, just as they come out of the pc.
  
• No. I expect between 0.5 - 1.0 grams per pound.
  
• When I run out of high and mid grade scrap I'll look into the low grade stuff. These particular RJ45 jacks are the type that are only partially plated. If they were fully plated I would process them as mid grade scrap.
  
• See part one of answer #3.

Steve

 

[goldsilverpro]

I really screwed up, mathwise. If you have used my NEW method, you have been evaluating the gold plating at 10 times too high. I made a decimal point error. I hope I didn't cost anyone any money. What can I say?

FROM lmills148: ....sorry I wasn't clear this is what I did

1.05sq in.(area)x $.066(spot price/10000) x 20 microns(estimated thickness)

You calculated this to be worth $1.38, using my formula, exactly.

I looked at the gold area, of about 1 sq.in., and a thickness of 20 micro", and I knew it couldn't be worth $1.38. Here's the right formula:

The Correct Formula
Dollar value of gold plating = Spot price divided by 100,000 X thickness in micro" X area in square inches.

lmills148, back to your material. The calculation would be:

$660/100000 X 1.05 sq.in. X 20 micro" = $.138.

I have only given my NEW, FANCY method in 1 other place. I will edit all of my posts that contain this error and will add to each post a note, in blue, explaining what I did. Give me a couple of days days. Sorry, again.

 

[aflacglobal]

1.05sq in.(area of my home)x $.066(spot price/10000)( My yearly salary) x 20 microns(estimated thickness of the IRS agents head doing my Audit)

Hey, think i will try that formula on my taxes this year.

 

[lmills148]

The Correct Formula
Dollar value of gold plating = Spot price divided by 100,000 X thickness in micro" X area in square inches.

lmills148, back to your material. The calculation would be:

$660/100000 X 1.05 sq.in. X 20 micro" = $.138.

I have only given my NEW, FANCY method in 1 other place. I will edit all of my posts that contain this error and will add to each post a note, in blue, explaining what I did. Give me a couple of days days. Sorry, again.

I made a mistake as well I was using microns instead of micro inches.

 

[Anonymous]

Hi everybody ,

Hey Mike:

Let me see if I can weigh in on this.

One pad is 2mm x 5mm+10 sq mm

10sqmm = .0155 sq in. per pad.

Each board has a total of 49 pads on each side for a total of 98 pads per bd.

.0155sq in x 98= 1.519 sq in per bd.

Today’s price of gold is 672.70 per oz.

.672 x 1.519 x.30=.306 or roughly 31 cents per bd.

214 bds x .31= 66.34/21.63 per oz of gold = 3.06 grams gold per Silverpros method.

My method is as follows:

10mm=.0155sqin X 98=1.519sq in per bd

214 bds x 1.519=325.066sq in at 30 u in thick would be .00975 cu in of gold

.00975 sq in = .1598 cc
19.3 grams per 1 cc= 19.3x.1596=3.08 grams of gold if the pads was 30 micro inches thick

If they were 20 micro inches then the expected gold yield would be 2.05 grams

It appears the gold thickness is about 25 micro inches

Catfish

19.3 grams per 1 cc
0.00193 grams per 1 micro metar on sq centimetar
http://en.wikipedia.org/wiki/Micrometre

If they were 20 micro metar


a. 20 (micro metar) x 0.00193 (grams per 1 micro metar / sq centimetar) x 9.8 (sq centimetar per bd) = 0.3782 (grams per bd) =
0.3782 (grams per bd) x 214 ( bd) = 80.95 grams (grams per 214 bd) NICE but WRONG why :?:

b. 2.5 (grams per 214 bd) / 214 = 0.01168 (grams per 1 bd) / 9.8 ( sq centimetar on 1 bd) =
0.001192 (grams per 1sq centimetar on 1 bd) / 0.00193 ( grams of gold per 1 micro metar on sq centimetar) = 0.6176 microns of foil on the bd WRONG or... :?:
http://en.wikipedia.org/wiki/Gold_plating#Electronics

 

[catfish]

Hi Bonzo:

I am not sure I understand just what your point or question is!

You may want to refer back to the site you posted as reference;

http://en.wikipedia.org/wiki/Gold_plating#Electronics

Please note that the reference states that: Quote;

(The resulting layer is typically only 97% pure and thin, typically (0.5-0.75 µm)).

Then if you will refer to;
http://www.sciencemadesimple.net/length.php

You will note that .5um or microns equal .000020 inches or 20 micro inches.
.75um or microns equals .000030 inches or 30 micro inches.

You will find that most consumer electronics equipment manufactured in the past 10 years, the gold plating will be within that range. Note there are still some exceptions though, some more and some less.

I think where you are getting confused is the difference between microns (micro meters) and micro inches. There is a big difference. All Federal Trade Commissions (FTC) standards for gold plate thickness are in micro inches or either microns.

I believe if you will go back and refigure the problem by using either .000020 to .000030 micro inches or either using .5 micro meters (Microns) to .75 micro meters (microns) you will come up the same answer I did. Please remember that this is only one part of the equation. Thickness times the squared area should equal the cubic volume. Then apply the rule of 19.3 grams of gold equals 1 cubic centimeter or 1cc

By the way, GSP’s method is accurate enough for an estimate on the gold quantity in a circuit board or etc. with a flat surface and even if the surface is round, you can still figure the square surface area and come up with the same numbers. His way is much simpler and easier to use than my method.

If you will go back on the forum and read some of the earlier posts last year, you will find that several of the members made several test cases on how to determine the quantity of gold versus the actual amount of gold recover from the items. They came out very close.

Hope this helps,

Catfish


PS If you would like to discuss this futher, we need to move this discussion to a different thread on another subject, other that assaying.

 

[Anonymous]

Thanks catfish,

yes, yes I was confused with micro inches and microns. :shock:

This help very much to me.