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Topic: Bitcoin cannot be filled with Tungsten - page 3. (Read 8631 times)

legendary
Activity: 3318
Merit: 4606
diamond-handed zealot
September 19, 2012, 08:14:53 PM
#60
Here is a way to test without ultrasound equipment.

First dip gold in water to find the amount of displacement.  The displacement will give you a very accurate volume of the bar.  Now using a scale measure the weight of the bar.  Using the volume and the atomic weight of gold you can also calculate weight.  If the calculated weight and measured weight are the same it's a real bar.  If it is lighter then it might have tungsten in it.  Something like that.

Incorrect, the specific gravity test used to be very safe, modern fakes using a mixture of tungsten and osmium precisely match gold's specific gravity.  Sad but true
legendary
Activity: 1400
Merit: 1013
September 19, 2012, 08:01:43 PM
#59
measuring the volume is trivial with something as small as a 1 oz bar. Just drop  it into a graduated cylinder of water and note how much the level changes.

It's theoritically trivial but in practice finding a graduated cylinder of sufficient accuracy is not.
member
Activity: 295
Merit: 98
September 19, 2012, 07:48:51 PM
#58
measuring the volume is trivial with something as small as a 1 oz bar. Just drop  it into a graduated cylinder of water and note how much the level changes.
hero member
Activity: 784
Merit: 500
September 19, 2012, 05:09:48 PM
#57
there are https://www.casascius.com these coins Tongue
hero member
Activity: 763
Merit: 500
September 19, 2012, 05:06:18 PM
#56
Heh.  No.
ah, well, in my head, i mixed their order, ... but basically i'm still right.
also, instead of platinum, you can use http://en.wikipedia.org/wiki/Osmium#Price (or iridium, maybe)
legendary
Activity: 2352
Merit: 1064
Bitcoin is antisemitic
September 19, 2012, 03:31:48 PM
#55
Jim Sinclair’s Commentary

My opinion is this is disinformation. The diamond district in NYC is Hassidic Jewish.

Nobody cons those guys.

http://www.jsmineset.com/2012/09/19/in-the-news-today-1314/
sr. member
Activity: 250
Merit: 250
September 19, 2012, 03:12:28 PM
#54
Bad news: I just checked one of my Bitcoins. I opened it up, and inside I actually found a LiteCoin!!!  This is a serious problem. I'm worried if I check my other Bitcoins the same thing will happen :/  Should I tell Gavin?
loool! That was a good one Smiley
sr. member
Activity: 451
Merit: 250
September 19, 2012, 03:11:18 PM
#53
First dip gold in water to find the amount of displacement.  
as written above, this is pretty hard.

second, and as an additional point why this doesn't work: you can mix the tungsten with small amounts of some other very cheap element to get the exact density of gold. then, both parts (pure gold, and the tungsten+something mix) have exactly the same density.

Heh.  No.

no he's right. you could toss a little platinum in there to offset the weight to have a net equivalent density of gold in the same volume.

in the end you'd really have to go to more complex testing eg ultrasound or conductivity
Platinum? Let's keep things cheap while we're ripping people off. A proper ratio mixture of lead and tungsten could have the exact density of gold.

Platinum is cheaper than gold (Crazy, right?) and lead is 1/2 the density. Pt is the cheapest element that has a higher density than gold.

The equation that would need to be fulfilled is:

p_Au*f_Au + p_Pt*f_Pt + p_W*f_W = p_Au, for the bar to pass the archimedes (effective density) test.

where p_x is the density of element x and f_x is the volume fraction of element x.

p_Au = 19.282, p_Pt = 21.46 and p_W = 19.25. This equation is subject to the constraint f_Au+f_Pt+f_W = 1.

The price of W is $45/kg, which is $1.28/oz. Gold price = $1771/oz., Platinum price = $1638/oz.

Now we can establish the price of the bar as a function of the volume fractions.

The price per oz. of the forged bars is given by P_bar(f_Au, f_W, f_Pt) = P_Au*f_Au+P_Pt*f_Pt+P_W*f_W, which we would like to minimize given the above density and volume constraints. This arrives at an edge solution (max. amount of W, no Au in bar), at f_W = (p_Au-p_Pt)/(p_W - p_Pt) = 98.4vol.% W, 1.6vol% Pt.

This bar would cost $27.40/oz. and have exactly the same apparent density as a gold bar.

Of course, it would not look like gold, but basically you could add any amount of Pt/W in this proportion (98.4% W, 1.6% Pt) to the gold bar, and you would not change the effective density.
sr. member
Activity: 504
Merit: 250
September 19, 2012, 02:51:08 PM
#52
It's worse: http://www.exohuman.com/wordpress/2011/02/fake-gold-bars-sold-to-china/

possible 1.3 million Tungsten bars (for 400 ounce gold bars) ordered to fake Fort Knox bars and later sold of to China. The FED apparently knowing about this!

Is this really true? Are such stories spread for gold market manipulation?
legendary
Activity: 1274
Merit: 1004
September 19, 2012, 02:44:15 PM
#51
First dip gold in water to find the amount of displacement.  
as written above, this is pretty hard.

second, and as an additional point why this doesn't work: you can mix the tungsten with small amounts of some other very cheap element to get the exact density of gold. then, both parts (pure gold, and the tungsten+something mix) have exactly the same density.

Heh.  No.

no he's right. you could toss a little platinum in there to offset the weight to have a net equivalent density of gold in the same volume.

in the end you'd really have to go to more complex testing eg ultrasound or conductivity
Platinum? Let's keep things cheap while we're ripping people off. A proper ratio mixture of lead and tungsten could have the exact density of gold.

Show work please, then apologize.

According to my back of the napkin calculation, you'd only need to add 1.5% platinum to tungsten by volume to get the same density as gold. That would be a pretty minimal expense.
hero member
Activity: 675
Merit: 502
September 19, 2012, 02:34:59 PM
#50
First dip gold in water to find the amount of displacement.  
as written above, this is pretty hard.

second, and as an additional point why this doesn't work: you can mix the tungsten with small amounts of some other very cheap element to get the exact density of gold. then, both parts (pure gold, and the tungsten+something mix) have exactly the same density.

Heh.  No.

no he's right. you could toss a little platinum in there to offset the weight to have a net equivalent density of gold in the same volume.

in the end you'd really have to go to more complex testing eg ultrasound or conductivity
Platinum? Let's keep things cheap while we're ripping people off. A proper ratio mixture of lead and tungsten could have the exact density of gold.
legendary
Activity: 2142
Merit: 1010
Newbie
September 19, 2012, 01:43:34 PM
#49
Don't worry about gold bars. We have bitcoins.
sr. member
Activity: 451
Merit: 250
September 19, 2012, 01:25:58 PM
#48
First dip gold in water to find the amount of displacement.  
as written above, this is pretty hard.

second, and as an additional point why this doesn't work: you can mix the tungsten with small amounts of some other very cheap element to get the exact density of gold. then, both parts (pure gold, and the tungsten+something mix) have exactly the same density.

Heh.  No.

no he's right. you could toss a little platinum in there to offset the weight to have a net equivalent density of gold in the same volume.

in the end you'd really have to go to more complex testing eg ultrasound or conductivity
kjj
legendary
Activity: 1302
Merit: 1026
September 19, 2012, 01:23:27 PM
#47
First dip gold in water to find the amount of displacement.  
as written above, this is pretty hard.

second, and as an additional point why this doesn't work: you can mix the tungsten with small amounts of some other very cheap element to get the exact density of gold. then, both parts (pure gold, and the tungsten+something mix) have exactly the same density.

Heh.  No.
hero member
Activity: 763
Merit: 500
September 19, 2012, 01:13:12 PM
#46
First dip gold in water to find the amount of displacement.  
as written above, this is pretty hard.

second, and as an additional point why this doesn't work: you can mix the tungsten with small amounts of some other very cheap element to get the exact density of gold. then, both parts (pure gold, and the tungsten+something mix) have exactly the same density.
hero member
Activity: 630
Merit: 500
September 19, 2012, 12:20:18 PM
#45
The funny part is that these polished, stamped, numbered bars tend to fetch a premium over the spot price.  This makes them easy targets, since the owners would be less likely to perform any type of destructive testing on them.


We have that same risk with physical bitcoins.    I hated to peel a 25 BTC coin I had purchased as it holds a premium over the 25 BTC it protected by the hologram.  But I did eventually for two reasons, one -- I wanted to be able to confirm first-hand that the code underneath was truly valid -- funded and matched the code on the outside of the coin.  The second reason is because I'ld nearly lost it several times showing it around (as a prop whenever I'm describing Bitcoin to someone in-person.)  At least now, fully spent, I'm only losing a few dollars worth of value if it gets lost.


Here's the tungsten incident from earlier this year:



Bitcoin, 100% Pure w/No Tungsten - Miners Make Sure Of It

  - http://www.bitcoinminer.com/post/19914661084

I often buy bullion but I shy away from really large bars as there is too much opportunity for fraud. (too much temptation with current prices) I feel safer with bullion coins (American eagles, Canadian Maple Leafs Austrian Philharmonics) from a reputable dealer. I evaluate each coins markings and measure each coins weight and size carefully before It goes in my vault
legendary
Activity: 1764
Merit: 1002
September 19, 2012, 12:07:02 PM
#44
The funny part is that these polished, stamped, numbered bars tend to fetch a premium over the spot price.  This makes them easy targets, since the owners would be less likely to perform any type of destructive testing on them.

devious.  sure sign of a top.
hero member
Activity: 675
Merit: 502
September 19, 2012, 11:47:44 AM
#43
Here is a way to test without ultrasound equipment.

First dip gold in water to find the amount of displacement.  The displacement will give you a very accurate volume of the bar.  Now using a scale measure the weight of the bar.  Using the volume and the atomic weight of gold you can also calculate weight.  If the calculated weight and measured weight are the same it's a real bar.  If it is lighter then it might have tungsten in it.  Something like that.

As others have said, it is difficult to calculate a small volume such as a 10 oz gold bar to an accuracy of 0.25% using the "dip in water" method. You would need to be more accurate than 0.25% to detect tungsten in the bar.
sr. member
Activity: 451
Merit: 250
September 19, 2012, 11:46:52 AM
#42
This is actually more wide spread than you think, I know lots of Chinese factories are producing
these tungsten filled "gold bars" that is flooding the market. Most of the time even the merchant would not
know, since it's difficult to know unless you destroy the bar and reveal what's inside. I think at least
20% of gold bars on the market right now, are tungsten filled.

Don't see any problem if these bars r used as legal tender. Noone should care what is inside. Tungsten is even more valuable than paper inside dollar banknotes.

Tungsten is a common element with little inherent value. If people buy gold bars, they expect to be buying a certain quantity of rare, expensive, precious gold, not common tungsten you can find in every lightbulb.
legendary
Activity: 1304
Merit: 1015
September 19, 2012, 11:45:27 AM
#41
Here is a way to test without ultrasound equipment.

First dip gold in water to find the amount of displacement.  The displacement will give you a very accurate volume of the bar.  Now using a scale measure the weight of the bar.  Using the volume and the atomic weight of gold you can also calculate weight.  If the calculated weight and measured weight are the same it's a real bar.  If it is lighter then it might have tungsten in it.  Something like that.
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