Ok, so I would like to see if there are any math geniuses out there. I'm a bit stumped on this one. Say you have a piece of gold plated copper/ nickel that's 1/8" x 1/8" . About 1/32" thick. There is a 50 micron gold plating all around, ( says so on the label). How many of these pieces once refined would it take to accumulate 1 cubic cm of pure gold?? :roll: I think I may have done the math wrong, but I came up with 1600 pieces. Can anyone help please? :lol: thanks in advance.
Gold calculation by micron thickness
- Thread starter jonn
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NobleMetalWorks
Well-known member
Here is an online calculator for figuring total Au weight of sheet or wire, it works for plating so long as you know how many micro inches the plating is.
http://www.saltlakemetals.com/SWCalcAu.htm
There are also calculators for Ag, Pd, Pt and Rh
Scott
http://www.saltlakemetals.com/SWCalcAu.htm
There are also calculators for Ag, Pd, Pt and Rh
Scott
John,
Goldsilverpro has posted formulas for calculating the gold on plated items if you know the dimensions and plating thickness. Search his posts with the keywords 'plating calculation' for the information.
Welcome to the forum.
Steve
Goldsilverpro has posted formulas for calculating the gold on plated items if you know the dimensions and plating thickness. Search his posts with the keywords 'plating calculation' for the information.
Welcome to the forum.
Steve
Thank you brown and Steve. I went to salt lake website that you sent a link to Brown and I came up with this:
Length .125
Width .125
Thickness .001968
Gold weight .0097 grams. There are 2 sides so that should double to be .0194 grams each piece.
Since there are 16 pieces on each unit, that would be .3104 grams per unit.
Could this be right??
The items I refer to are Cat 5 cable ends labeled as 50 micron gold plate thickness.
Each end has 8 tabs on it and 2 ends per cable for a total of 16 tabs about 1/8" x 1/8" that's what I refer to as a unit above.
So each Cat 5 cable has about .3104 grams? That's about $15.83 a piece?? Somehow I must have miscalculated again.
I did place 8 pieces in a nitric bath and have nice gold foils left in the bottom. I'm just not sure if I did the math right again. If I did, I wonder how much I should be able to sell these for?? : :roll:
Length .125
Width .125
Thickness .001968
Gold weight .0097 grams. There are 2 sides so that should double to be .0194 grams each piece.
Since there are 16 pieces on each unit, that would be .3104 grams per unit.
Could this be right??
The items I refer to are Cat 5 cable ends labeled as 50 micron gold plate thickness.
Each end has 8 tabs on it and 2 ends per cable for a total of 16 tabs about 1/8" x 1/8" that's what I refer to as a unit above.
So each Cat 5 cable has about .3104 grams? That's about $15.83 a piece?? Somehow I must have miscalculated again.
I did place 8 pieces in a nitric bath and have nice gold foils left in the bottom. I'm just not sure if I did the math right again. If I did, I wonder how much I should be able to sell these for?? : :roll:
A
Anonymous
Guest
Maybe it's me,but that sounds like a huge number for cat5 connectors.Can you post a picture of the label that states "50 micron gold plate thickness"?Can you also give us the manufacturers name?jonn said:
Geo
Well-known member
i seriously doubt it would be 50 microns. millspec is measured in micro-inches, there is a big difference in thickness between micron and micro-inches.
Belkin, Apc and any rj45 . Do a google search under Cat 5 50 micron gold. I know it sounds like a lot but that's how it's labeled. I have some and the micron symbol is listed with 50 in front of it. Google searches say minimum requirements of 50 micron gold thickness. Did I stumble onto a gold mine?
Geo
Well-known member
theres a lot of info about these on the internet , and all say 50 microns. but i cant find a spec sheet on them. 50 microns = 1950 micro inches or 1950 millionths of an inch thick.
if it is 50 microns thick, it would indeed be a good find.
if it is 50 microns thick, it would indeed be a good find.
A
Anonymous
Guest
http://www.amazon.com/100-pack-Connector-Solid-Micron-nickle/dp/B000067ONL
100 pack for $19? Something is not right.They would not be selling several grams of gold for $19.
But then there is this website http://www.cat-5-cable-company.com/faq-goldplating-RJ45.htm
100 pack for $19? Something is not right.They would not be selling several grams of gold for $19.
But then there is this website http://www.cat-5-cable-company.com/faq-goldplating-RJ45.htm
50 microns = 0.005 centimeters
Specific Gravity of Gold is 19.3 grams per cubic centimeter (cc)
1 inch = 2.54 centimeters
1 cm x 1 cm x 50 microns = 0.005 cc
19.3g/cc x 0.005 cc = 0.0965 grams of gold per square centimeter @ 50 microns thickness.
a) Measure the plated surface area in centimeters (1 mm = 0.10 cm) and multiply the length by the width. For cylinders and irregular objects sum the plated surface areas of the entire item.
b) Divide the plating thickness in microns by 10,000 (ie: move the decimal four places to the left; eg: 10 microns = 0.001 cm, 30 microns = 0.003 cm, etc.).
c) Multiply the result of a. times b. above.
d) Multiply the result of c. above by 19.3 g/cc to estimate the weight of the gold plating in grams.
Steve
Specific Gravity of Gold is 19.3 grams per cubic centimeter (cc)
1 inch = 2.54 centimeters
1 cm x 1 cm x 50 microns = 0.005 cc
19.3g/cc x 0.005 cc = 0.0965 grams of gold per square centimeter @ 50 microns thickness.
a) Measure the plated surface area in centimeters (1 mm = 0.10 cm) and multiply the length by the width. For cylinders and irregular objects sum the plated surface areas of the entire item.
b) Divide the plating thickness in microns by 10,000 (ie: move the decimal four places to the left; eg: 10 microns = 0.001 cm, 30 microns = 0.003 cm, etc.).
c) Multiply the result of a. times b. above.
d) Multiply the result of c. above by 19.3 g/cc to estimate the weight of the gold plating in grams.
Steve
For your item:
0.3175 cm x 0.3175 cm x 0.005cm = 0.00050403125 cc
19.3g/cc x 0.00050403125 cc = 0.00972780313 g =~ 0.0097 grams each
100 x 0.0097 g = 0.97 grams per 50 double sided items.
16/50 = 0.32
0.32 x 0.97 = 0.3104 g per cable
At todays spot each gram is :
$1593.60 / 31.1 = $51.24
$51.24 per gram x 0.3104 grams = Estimated value is $15.90 worth of Au in the cable ends at today spot.
Perhaps all sides of the item are not plated the same thickness? Perhaps only the contacting edges are plated at 50 microns?
Steve
0.3175 cm x 0.3175 cm x 0.005cm = 0.00050403125 cc
19.3g/cc x 0.00050403125 cc = 0.00972780313 g =~ 0.0097 grams each
100 x 0.0097 g = 0.97 grams per 50 double sided items.
16/50 = 0.32
0.32 x 0.97 = 0.3104 g per cable
At todays spot each gram is :
$1593.60 / 31.1 = $51.24
$51.24 per gram x 0.3104 grams = Estimated value is $15.90 worth of Au in the cable ends at today spot.
Perhaps all sides of the item are not plated the same thickness? Perhaps only the contacting edges are plated at 50 microns?
Steve
A
Anonymous
Guest
Thank you Steve.I thought there was a mistake somewhere.
Thank you Steve. That's .97 g of gold for 50 items. There are 8 items per end and 2 ends per cable. I think our math is the same or very close. 50/16= 3.125 cords. $51.00 per gram x .97 = $49.47 divided by 3.125 = $15.83 each cable. I suspect you may correct in stating that only the contact end is 50 microns plated. If that is the case then the equation would differ greatly. I will recalculate based upon that assumption without giving any credit to the sides.
Check this page out:
RJ45 Plating Problems
Note the list as plating thickness in microinches not microns in section 5.1.1 Gold Surface Layer.
50 µ in = 1.27 µm = 0.000127 cm
So our gold drops to :
0.3175 cm x 0.3175 cm x 0.000127 cm = 0.0000128023937 cc
0.0000128023937 cc x 19.3 g/cc = 0.000247086198 g per item side
32 sides x 0.000247086198 = 0.00790675834 grams per cable
Mystery solved. The previously quoted website confused microinches with microns (micrometers). The site linked above also states gold density of 17 g/cc so you are likely dealing with plating that is an alloy. But items number 2 and 3 on the plating section seem to be contradictory?
Steve
RJ45 Plating Problems
Note the list as plating thickness in microinches not microns in section 5.1.1 Gold Surface Layer.
50 µ in = 1.27 µm = 0.000127 cm
So our gold drops to :
0.3175 cm x 0.3175 cm x 0.000127 cm = 0.0000128023937 cc
0.0000128023937 cc x 19.3 g/cc = 0.000247086198 g per item side
32 sides x 0.000247086198 = 0.00790675834 grams per cable
Mystery solved. The previously quoted website confused microinches with microns (micrometers). The site linked above also states gold density of 17 g/cc so you are likely dealing with plating that is an alloy. But items number 2 and 3 on the plating section seem to be contradictory?
Steve
Just the contact top would equal .0023 per item. X 16 items per cable = .0368g per cable. That works out to be about $1.88 per cable. So a box with 500 cables in it would have $940.00 worth of gold in it. This does not calculate any wall thickness plating at all. Does this look correct to you Steve? That's $188.00 per 100 cables or 200 ends.
Check out my post above your last one.
So, it is micro inches, not microns. Ok so 1 cable would have .00790675 grams of gold. Got it. So 500 cables would have 3.953375 grams of gold. That's $201.62 per box of 500 cables or 1000 ends. Or about .201 cents each. Or $20.16 worth of gold per 100 ends. The reason I ask is that I've come across 74 boxes. Let's do the math together on this one. 74 boxes x 500 cables = 37000 cables. That's 74000 ends at .2016 cents each = $14918.40 or just shy of 10 oz. Troy ? What do you think Steve? Is it correct ?
A
Anonymous
Guest
Jonn,
the phrase "If it's too good to be true,it probably is" ,applies to recovering gold,as much as it does to everything else.That's not to say that you won't be surprised(in a good way),at the yield of some items,but I highly doubt those items will include Cat5 cable ends.When I said it sounded like a huge number,I didn't say it because I was trying to be mean.I said it because I have been doing this a long time,and I guarantee that no person,or company,ever made it by losing money.You are not going to find $1000 worth of gold,in $50 worth of product off the shelf.There is one guaranteed method to figure out how much gold is there,and that is to process a set amount of the material.
the phrase "If it's too good to be true,it probably is" ,applies to recovering gold,as much as it does to everything else.That's not to say that you won't be surprised(in a good way),at the yield of some items,but I highly doubt those items will include Cat5 cable ends.When I said it sounded like a huge number,I didn't say it because I was trying to be mean.I said it because I have been doing this a long time,and I guarantee that no person,or company,ever made it by losing money.You are not going to find $1000 worth of gold,in $50 worth of product off the shelf.There is one guaranteed method to figure out how much gold is there,and that is to process a set amount of the material.
Thank you Mic. These items were not purchased off the shelf, and thank you for the apology, but there was no offense taken. I am simply trying to calculate the value of the material that I have come across. I have itemized my math calculations above according to lazersteves figures. I hope they are correct and if so, I may be listing some for sale to anyone who would be interested in refining them. The cables I purchased are attached with a 1' cable that I will be clipping and sending to the scrap yard. The ends have some value and I would simply like to know what to expect from a sale or refining. I suppose I could send them all off to a refiner and pay all their fees and hope that they are honest. Either way, it's always good to do research and I greatly value and respect all the fine folks here at the forum. Hats off to you gentlemen and thank you for all your input.
