# Need help from a mathamagician!



## glondor (May 6, 2014)

Hey all, I am looking for a tutorial and assistance on calculating the gold yield on a certain type of pin. I have a spec sheet with the pertinent numbers and a bit more info but need some help getting a grip on the math. Any help appreciated. Here is the part. Each piece weighs 2.35g. 193 pcs per pound. The question is, via math, what would the expected yield be per pound of pins? Assume 50 micro-inches gold thickness. Thanks!


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## TomVader (May 7, 2014)

I would calculate like this: Measure the length and diameter of one pin, I would use metric. Determine the surface area of one pin: Diameter x 3.14 x length. Convert micro-inches to millimeters. 50 micro-inches looks like .00127 mm. Multiply surface area(in mm) by .00127. You'll get the amount of gold in cubic mm. There are 1000 cubic mm in one cubic centimeter. One cc of gold weighs 19.3 grams.
I can't make any sense of the drawing you posted so I'll make up values to illustrate. Pin is 20mm long, 3mm thick. 3 x 3.14 = 9.42 x 20 =188.4 x .00127 =.239268 cubic mm Au/pin.
19.3 divided by 1000 = .0193g per cubic mm. .0193 x .239268 = .0046178 grams of gold per pin. I hope this helps. PLEASE if I made a mistake, someone point it out!


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## TomVader (May 7, 2014)

I realize that the pins are not a uniform cylinder, I would use the smaller diameter for my estimate.


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## mls26cwru (May 7, 2014)

figuring out the exact surface area would entail some pretty hefty calculus... I would say just estimate the area by approximating the pin as a cylinder of what ever diameter you are comfortable with using. also, it looks like the drawing is calling for an area that is 50 micro and 30 micro... since I cannot see the schematic very clearly, you may want to really examine that point to determine what you are thickness you are using.


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## etack (May 7, 2014)

I think I ran the female style to those pins. They ran about 2.1g#

Also do you have a link to the PDF or a part number I cant see the pic well.

AMP may have a CAD file for this pin if so it would be real easy to find the surface area.

Eric


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## joubjonn (May 7, 2014)

If I still had solidworks I could draw that in 5min and tell you. My job changed and I no longer have a seat. That would be the easiest way to do it.


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## glondor (May 8, 2014)

Hey guys, Thanks for the pointers. I did manage to calculate via Tom Vaders method @ 424.735 square mm. I did have to fudge a bit for the round corners and a missing dimension, but I think it is close enough to make a reasonably accurate yield figure. .53 cubic mm gold per piece. Here is the PDF and TIFF files... http://www.te.com/commerce/DocumentDelivery/DDEController?Action=selcritrslt&searchby=part&searchfor=449379-2&doctypes=All&docformats=All&doclangs=All&TCPN=449379-2 Cheers and Thanks Mike.


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## Geo (May 8, 2014)

Did you factor in the gold on the inside of the barrel?


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## goldsilverpro (May 8, 2014)

The dimensions are given in both mm and inches. I kept everything in inches, including the gold thickness, since it's easier for me to think in those terms. There are 2 different gold thickness possibilities given, .000030" and .000050". Also, there are parts with 2 different lengths, .845" and .931".

To do this accurately, I count 14 different calculations that would have to be made. If I were buying, that's the way I would do it. Instead, I made a guess as to the average diameter, taking into consideration those flange areas on the sides on the larger diameter sections. I also cut the lengths down a small amount to take into consideration the rounded ends.

I used .185" as the average diameter, which may be high. That is the approximate diameter of that center section with a width of .090". I cut the lengths down to .830" and .915".

The math for the .830" length. The 10.17 factor is the number of troy oz of gold per in3. The 1289 is the gold spot.
Area = .185 X pi X length = .185 X 3.14 X .830 = .48 in2
$ value @ 50 micro" = .000050 X 10.17 X 1289 X .48 = $.31/pin; $/pound = .31 X 193 = $59.50/pound
$ value @ 30 micro" = .000030 X 10.17 X 1289 X .48 = $.186/pin; $/pound = .186 X 193 =$36/pound

For the .915" length, these values are .915/.830 = 1.10 times greater than the shorter length.
Therefore, for 50 micro", it would be 59.50 X 1.1 = $65.45/pound
For 30 micro", it would be 36 X 1.1 = $39.60/pound

Looking at those medium-heavy weight pins, I don't think these numbers are far off. If anything, I would cut the values down at least 10-20% in my mind for buying negotiation purposes. I would also assume the 30 micro" thickness, unless you know the exact part number and know that all the pins are the same. I always try to be on the low side when refining or buying. I don't like surprises. Selling is another story.

Etack's female version runs about $87/pound, but they are lighter and have more surface area than the males. Also, etacks are probably the 50 micro" variety.

______________________________________________


The area you gave was 425 mm2/pin. Doing some math, this would give an average diameter of about .250". I don't think that's possible since the largest diameter in the drawing is .214". If you had a fire assay setup or an AA, you would know exactly what they're worth within about 3 hours. Without either of these, you could take about 10 pieces, dissolve the base metal with hot nitric (assuming they are copper base), and weigh the rinsed, dried foils with at least a 3 place scale, although a 4 place would be better. The retention clip should dissolve also. When only the foils are left, I would pour the nitric off carefully, put on fresh nitric, and heat near boiling for at least 20 minutes. This would only put you in the ballpark but it would be a tighter ballpark than the math method above.


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## goldsilverpro (May 8, 2014)

Geo said:


> Did you factor in the gold on the inside of the barrel?


What barrel? Glondor's pins seem to be male, whereas etack's are female.


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## Geo (May 8, 2014)

Silly me, I refreshed the page and was seeing Eric's pictures. Please pay no attention to my absent mindedness.


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## glondor (May 8, 2014)

Thanks a million GSP, Math of this type has always been a challenge for me, but I do try.
Your numbers have put the material right smack into the range for the actual yield, which is a great relief for me. With this information I may prevent a good customer from making a bad deal. Hat's off to all of you and a big thanks. Mike


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## Anonymous (May 8, 2014)

Mike I wouldn't bother trying to calculate it mate. 

I'd process a sample batch and see the real figures.


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## glondor (May 8, 2014)

Hi Spaceships, a sample was done and did not correlate to the claim of the seller, so I looked to other methods to confirm or deny the validity of the sample process. The yield and math are far closer to reality than the seller's claim of expected yield. Just trying to verify the recovery process via math and other comparison methods to justify a proper buy price. 

This was a great exercise in validation methods for me, and I thank all of those who participated ! It also validated I need to sharpen my math skills  Cheers Mike


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## Anonymous (May 8, 2014)

No worries Mike, and thanks for explaining.

Incidentally don't you think that the seller should be the one to back up his expectation? 

Jon


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## goldsilverpro (May 8, 2014)

Mike, what was the actual yield and/or the sample results?


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## glondor (May 8, 2014)

Hey GSP, my sample was relatively small, 234 g The yield was .34 g which puts my thinking towards 65-70 grams per 100 pounds. The seller told the potential buyer 160-180 g per hundred pounds.

My eyes told me right off .7 g per pound based on much similar material and the huge amount of base metal involved. To me, anything that chunky is going to be < 1 g per pound for average non specialty pins. 50 micro-inches IS a healthy coat of gold, but it can not over come the large mass of copper carrying it. 

I tried to account for any small lot losses, as even a stray prill in a melting dish or a slight dusting of settlement in a solution a week later can add up to significant value when multiplied by 100 or 1000 times, but the solution was barren and clean after a week of settling, and if there was a "flyer" in the melt, it could not come close to the almost 3 x difference shown by the actual yield. 

That being said, I do think my yield is low, but i always feel that way. I always think I could have done it better somehow. But we get what we get. 

I researched and retested everything and even brought out you good math magicians to see if I could have had such a huge margin of error. I also compared the 2.1 g yield quoted by Etack (thanks Etack for the info) and tried to math out for the additional surface area on the interior of the female part. It was more of a mental exercise, and I concluded a 40% advantage to the female part for gold yield, but I did miss the fact that they are also lighter, more per pound, which also swings the number more in my direction. 
As far as accuracy goes..... It is out the window, but it is close enough in general terms to tell my client not to buy at their considered price. 

I really do want to thank all of you for your help. The math put it much closer to my ball park than the sellers, and gave me some new ways to think about these things. Cheers Mike.


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## etack (May 8, 2014)

Sorry I should have put that each of my pieces weighed in at 1.34g each. That is quite a bit in gold:weight ratio.

Eric

I ran into this a few weeks ago a seller wanted to sell me 70# of connectors that I new ranged in the 1.5g-1.7g per pound. He said he was getting $80# for them I told him when his buyer runs out of money I would buy them for a profitable price. Sent them out for assay and they came in at 1.7g# thay recovering all the gold not what would happen on a large scale.

The proof of what is in the pudding is on the buyer always.


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## goldsilverpro (May 9, 2014)

Mike,

To satisfy my curiosity, I decided to calculate every segment from the drawing, including the flange areas. I should have done that to start with. For the segments that had no dimensions given, I measured them on the monitor with an accurate ruler that reads in decimals of an inch and then compared them mathematically with those who did have given dimensions. I found that the pin on the drawing is the longer variety. There were a total of 11 separate calculations.

I ended up with an area of .486 in2 on the longer variety pins and .455 in2 on the shorter ones. As I suspected, the .185" average diameter I used before was high and should have been .176". Therefore, all the dollar figures I gave earlier should be multiplied by .95. At the same gold spot I used before, $1289, the results would be:

Short pins:
@ 50 micro" = $56.53/#
@ 30 micro" = $34.20/#

Long pins:
@ 50 micro" = $62.17/#
@ 30 micro" = $37.62/#

Averages assuming a 50/50 mix of the 2 lengths:
$59.35 (50 micro") & $35.91 (30 micro")

I edited the above since I found I had left out 1 calculation. I was closer than I thought to start with on my original post - only 5% off.

I just noticed a small fly in the ointment. You gave the figure of of 193 pins/# but we don't know if those are the shorter pins, the longer pins, or a mix. Of course, the shorter pins would give more pins/# than the longer ones. This could alter the $/# numbers, but probably just a smidgeon.

Another comment. It is the nature of plating that it is never evenly distributed on the part and that the total gold per part usually varies from part to part. If you fire assayed 5 pins (or, whatever), using 1 pin per assay, you would likely get 5 different numbers. I would think that these pins were plated in a plating barrel. If so, the gold distribution would be better than with rack plating, but there would still be variations. Another problem is that each barrel batch can vary in the total amount of gold deposited. These pins were plated with hard acid gold, which usually plates between 30% to 45% efficiency and the efficiency determines the amount of gold deposited per unit time. The efficiency is determined by many different variables (current density, gold concentration in the bath, concentration of other bath constituents, contamination in the bath, pH, etc.) and can change from day to day or even hourly. What I'm saying with all this is that these thickness numbers in the spec, of 30 and 50 micro" can only be used as a ballpark guide. This is one reason I always reduce my estimate before buying.


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## glondor (May 9, 2014)

Thank you ! GSP, that calculation was very helpful. I wish I was closer, I would come round and buy you a beer or few. 

I did not keep a sample, so you are correct, I do not know the proper length of the ones I had. 

Thanks to everyone. I will let you know the outcome of this if I can. Cheers Mike .


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## Anonymous (May 9, 2014)

Guys can you help me out please? when you say $/# what does the # mean?


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## jimdoc (May 9, 2014)

spaceships said:


> Guys can you help me out please? when you say $/# what does the # mean?




.lb or pound.


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## Anonymous (May 9, 2014)

Thanks Jim, appreciate it.

Jon


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