Has anyone ever scrapped these got AU?
- Thread starter cyberdan
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Anonymous
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The pins on the outside yield but I guess you know that already.
Inside is a chip that may also contain gold- if you break one open you'll see the chip. It can be similar to the type of chip you find on a RAM board. They don't yield very much though from my previous experience.
One thing in your favour is that the older ones often contain more gold than the newer ones and 32Mb/128Mb/256Mb is older in the scheme of these cards.
Inside is a chip that may also contain gold- if you break one open you'll see the chip. It can be similar to the type of chip you find on a RAM board. They don't yield very much though from my previous experience.
One thing in your favour is that the older ones often contain more gold than the newer ones and 32Mb/128Mb/256Mb is older in the scheme of these cards.
rickbb
Well-known member
Most of the gold is what you see, not much more inside. I've scrapped a few, just mixed them in with trimmed fingers from RAM and PCI cards. Yield is pretty much the same as the trimmed fingers.
I measured and weighed an SD card in my camera. I assume they're all the same. Figuring only the fingers and assuming 30 micro" gold thickness, I estimate about $10.90/pound at a $1211 spot. They weigh about 2g each (227 parts/pound) and the plated area is about 0.13 in2.
(.000030 in3 Au/in2)(10.17 tr.oz.Au/in3)($1211/tr.oz.Au)(.13 in2 Au/part)(227 parts/pound) = $10.90/pound:
The units are (in3)(tr.oz.)($)(in2)(parts) divided by (in2)(in3)(tr.oz.)(parts)(pound). Everything cancels out but $/pound.
When using U.S. weights and measures, 10.17 is always the constant for gold in this formula (for Ag, the constant is always 5.5 - Cu is 4.7). It should be noted that the gold thickness number is always the same as the in3 of gold at that thickness - 0.000030" x 1" x 1" = 0.000030 in3.
Below in bold type is the simplified formula for gold in $, inches, and pounds. For different units, such as microns and/or kg, the 10.17 factor would be different, but it is easily calculated. For example, if you were still using inches and wanted the answer in $ per kilogram, the factor of 10.17 would be multiplied by 2.2, the number of pounds per kilogram. A change in currency would not alter the factor but the results would be in the new currency per unit weight. A change from in, in2, and in3 to cm, cm2, and cm3 would require a new factor.
$ value of gold per pound of parts = (thickness of gold expressed in inches) X (10.17) X (gold spot in $/tr.oz.) X (gold surface area per part in square inches) X (number of parts per pound)
I've gone over this a jillion times but I feel that everyone should know this formula. If too confusing, ask specific questions.
(.000030 in3 Au/in2)(10.17 tr.oz.Au/in3)($1211/tr.oz.Au)(.13 in2 Au/part)(227 parts/pound) = $10.90/pound:
The units are (in3)(tr.oz.)($)(in2)(parts) divided by (in2)(in3)(tr.oz.)(parts)(pound). Everything cancels out but $/pound.
When using U.S. weights and measures, 10.17 is always the constant for gold in this formula (for Ag, the constant is always 5.5 - Cu is 4.7). It should be noted that the gold thickness number is always the same as the in3 of gold at that thickness - 0.000030" x 1" x 1" = 0.000030 in3.
Below in bold type is the simplified formula for gold in $, inches, and pounds. For different units, such as microns and/or kg, the 10.17 factor would be different, but it is easily calculated. For example, if you were still using inches and wanted the answer in $ per kilogram, the factor of 10.17 would be multiplied by 2.2, the number of pounds per kilogram. A change in currency would not alter the factor but the results would be in the new currency per unit weight. A change from in, in2, and in3 to cm, cm2, and cm3 would require a new factor.
$ value of gold per pound of parts = (thickness of gold expressed in inches) X (10.17) X (gold spot in $/tr.oz.) X (gold surface area per part in square inches) X (number of parts per pound)
I've gone over this a jillion times but I feel that everyone should know this formula. If too confusing, ask specific questions.
cyberdan
Well-known member
goldsilverpro said:
Thanks GSP, I came up with 232 parts/pound so we are talking the same thing. The guy i bought them from knows his business. We settled on $10lb.
Best to sell on feeBay. Price is all over the place $2.00 - $4.00 & even $10.00 each (sold prices)
I have enough to flood the market, so will have to list them intellegently.
Also mixed in were over 300 smart media cards. One went for $265 and I have dozens of these 128's and this guy sold 7 already. I have already ordered a card reader so I can test & clear. No telling what is on those cards, could be millions of personal photos. :shock:
http://www.ebay.com/itm/128MB-SMARTMEDIA-CARD-FOR-OLYMPUS-C-4000-ZOOM-C-3100-C-3040-/140635193884?pt=LH_DefaultDomain_0&hash=item20be82c21c&nma=true&si=0RXwJhb6E1Ndhq1nCtvFyjU1pdA%253D&orig_cvip=true&rt=nc&_trksid=p2047675.l2557
Those SD cards are handled by the users often and they surely will have to withstand a lot of insertions. Also, some abuse is likely. Therefore, the gold thickness might be thicker than 30 micro", but I wouldn't bank on it. I would guess that 30 micro" would be the minimum.
I've never opened one myself, but there is a high probability of bond wires inside too. Probably not more than 10 as there are only 9 pads.
Göran
Göran
An inch of gold bonding wire is worth from about $.005 to $.01 in gold, depending if its diameter is .0007" or .001". So, in an SD card with 10 bonding wires, there might be about 1.5 cents to 3 cents worth of gold. That could raise the total to about $14-$17/pound.
