Ultra Rare $20 Gold Piece disguised as Zinclon

[david680]

Found this on the desk today. I believe this is a ultra rare experimental $20 Gold piece disguised as a zinc Lincoln cent. It weighs exactly the same as a zinclon, but I think its actually a ounce of gold compressed into a cent sized piece. Not sure why they did this, but you can look at it and just tell. Any ideas?

 

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[ARC]

Looks like copper to me...

Did you test this ?

 

[Loco-Digger]

I've got of few of hose I'll sell ya.

 

[DirtStalker]

Amazing I found one of those this weekend too. If you look close beside Abe you'll see his metal detecto or sand scoop

 

[relic nut]

I sold mine. Used the money for some ocean front property in Kentucky

 

[diggummup]

I'll take some of whatever your smoking... I didn't know it was legal in Indiana yet!

 

[david680]

Looks like copper to me...

Did you test this ?

Oh no way. 'Cause if I did and the tests didn't come back positive, you all might tell me I'm wrong, and I don't want you all posting any negativity in my thread.

 

[kingskid1611]

I have a few still in sets. They look good

 

[mojjax]

Weight ?

 

[david680]

2.5 grams. Amazing how they got a ounce of gold down to that.

 

[huntsman53]

2.5 grams. Amazing how they got a ounce of gold down to that.

Just think, if you took the coin to Jupiter, you would be really rich as Jupiter's gravity is 2.5 times that of Earth's gravity.


Frank

 

[villagenut]

I think you should throw it into a beach campfire and form a gold fire nugget from it....

 

[G.I.B.]

I thought they did that to the 2008 series only?

So I saved all this '08 gold for nothing?

 

[Crispin]

Have you dropped it on the floor to run the sound test?

 

[david680]

Have you dropped it on the floor to run the sound test?

It gave a dull thunk with a ding. I think the ding was the copper coating.

 

[Crispin]

It gave a dull thunk with a ding. I think the ding was the copper coating.

Hmmm, very interesting. I think what you need to do is have a friend drop the coin from a height such that when it hits the ground it will have achieved terminal velocity. That will give off the true sound. Make sure you calculate wind resistance, humidity, and barometric pressure as variables. In case you forgot gravity accelerates at 9.8 meters per second squared.

Mathematically, terminal velocity—without considering buoyancy effects—is given by

V_t= \sqrt{\frac{2mg}{\rho A C_d }}

where

V_t is terminal velocity,
m is the mass of the falling object,
g is the acceleration due to gravity,
C_d is the drag coefficient,
\rho is the density of the fluid through which the object is falling, and
A is the projected area of the object.

In reality, an object approaches its terminal velocity asymptotically.

Buoyancy effects, due to the upward force on the object by the surrounding fluid, can be taken into account using Archimedes' principle: the mass m has to be reduced by the displaced fluid mass \rho\mathcal{V}, with \mathcal{V} the volume of the object. So instead of m use the reduced mass m_r=m-\rho\mathcal{V} in this and subsequent formulas.

On Earth, the terminal velocity of an object changes due to the properties of the fluid, the mass of the object and its projected cross-sectional surface area.

Air density increases with decreasing altitude, ca. 1% per 80 metres (260 ft) (see barometric formula). For objects falling through the atmosphere, for every 160 metres (520 ft) of falling, the terminal velocity decreases 1%. After reaching the local terminal velocity, while continuing the fall, speed decreases to change with the local terminal velocity.
Derivation for terminal velocity

Mathematically, defining down to be positive, the net force acting on an object falling near the surface of Earth is (according to the drag equation):

F_{net} = m a = m g - {1 \over 2} \rho v^2 A C_\mathrm{d}

At equilibrium, the net force is zero (F = 0);

m g - {1 \over 2} \rho v^2 A C_\mathrm{d} = 0

Solving for v yields

v = \sqrt\frac{2mg}{\rho A C_\mathrm{d}}

I am positive this will yield the true nature of the gold and not harm the coin in any way.

 

[Scrappy]

They're worth about $500 each . Interested in a future investment ?

Sheez

 

[david680]

Hmmm, very interesting. I think what you need to do is have a friend drop the coin from a height such that when it hits the ground it will have achieved terminal velocity. That will give off the true sound. Make sure you calculate wind resistance, humidity, and barometric pressure as variables. In case you forgot gravity accelerates at 9.8 meters per second squared.

Mathematically, terminal velocity—without considering buoyancy effects—is given by

V_t= \sqrt{\frac{2mg}{\rho A C_d }}

where

V_t is terminal velocity,
m is the mass of the falling object,
g is the acceleration due to gravity,
C_d is the drag coefficient,
\rho is the density of the fluid through which the object is falling, and
A is the projected area of the object.

In reality, an object approaches its terminal velocity asymptotically.

Buoyancy effects, due to the upward force on the object by the surrounding fluid, can be taken into account using Archimedes' principle: the mass m has to be reduced by the displaced fluid mass \rho\mathcal{V}, with \mathcal{V} the volume of the object. So instead of m use the reduced mass m_r=m-\rho\mathcal{V} in this and subsequent formulas.

On Earth, the terminal velocity of an object changes due to the properties of the fluid, the mass of the object and its projected cross-sectional surface area.

Air density increases with decreasing altitude, ca. 1% per 80 metres (260 ft) (see barometric formula). For objects falling through the atmosphere, for every 160 metres (520 ft) of falling, the terminal velocity decreases 1%. After reaching the local terminal velocity, while continuing the fall, speed decreases to change with the local terminal velocity.
Derivation for terminal velocity

Mathematically, defining down to be positive, the net force acting on an object falling near the surface of Earth is (according to the drag equation):

F_{net} = m a = m g - {1 \over 2} \rho v^2 A C_\mathrm{d}

At equilibrium, the net force is zero (F = 0);

m g - {1 \over 2} \rho v^2 A C_\mathrm{d} = 0

Solving for v yields

v = \sqrt\frac{2mg}{\rho A C_\mathrm{d}}

I am positive this will yield the true nature of the gold and not harm the coin in any way.

Yes, but how will this be effected if this has a rhodium core, rather than solid gold?

 

[Crispin]

Yes, but how will this be effected if this has a rhodium core, rather than solid gold?

This does throw a twist into the sound test. A rhodium core will have no effect on the terminal velocity but it must be calculated into the decibel level of the dropped coin. Tone will also change but the amount of change cannot be determined by the species Sapien. At this point in time it is best to bring along a dog and a cat. The dog will by able to determine any changes in tonotopic frequency that were not present at the arms length drop sound test. The cat will be able to see any ultraviolet light patterns that are given off at impact. When the coin settles, immediateley shoot it with a Geiger counter for a reading and reference the table I have submitted below:

Naturally occurring isotopes
This table shows information about naturally occuring isotopes, their atomic masses, their natural abundances, their nuclear spins, and their magnetic moments. Further data for radioisotopes (radioactive isotopes) of rhodium are listed (including any which occur naturally) below. Isotope Mass / Da Natural abundance (atom %) Nuclear spin (I) Magnetic moment (μ/μN)
103Rh 102.905500 (4) 100 1/2 -0.08840
IsotopeRelative abundance / %Isotope pattern of rhodium (Rh)9798991001011021031041051061071081090255075100www.webelements.com
98Rh: 0.0 %

In the above, the most intense ion is set to 100% since this corresponds best to the output from a mass spectrometer. This is not to be confused with the relative percentage isotope abundances which totals 100% for all the naturally occurring isotopes.
Radiosotope data
Further data for naturally occuring isotopes of rhodium are listed above. This table gives information about some radiosotopes of rhodium, their masses, their half-lives, their modes of decay, their nuclear spins, and their nuclear magnetic moments. Isotope Mass / Da Half-life Mode of decay Nuclear spin Nuclear magnetic moment
99Rh 98.90820 16 d EC to 99Ru 1/2
100Rh 99.90812 20.8 h EC to 100Ru 1
101Rh 100.90616 3.3 y EC to 101Ru; IT 1/2
102Rh 102.906842 2.9 y EC to 102Ru; β- to 102Pd 6 4.04
104Rh 103.906655 42.3 s EC to 104Ru; β- to 104Pd
105Rh 104.905692 35.4 h β- to 105Pd 7/2 4.45

It is also extremely important that the dog chase the cat around in no less then seven circles prior to this test occurring. Failure to do so may lead in catastrophic damage to the coin.

 

[DirtStalker]

I sold mine. Used the money for some ocean front property in Kentucky

RN I hope you at least hunted the beach area. I'm gonna be up there next week 2 condos down from your property looking for my great grandmother's 12 carat ring she lost in the water while riding her jet ski I hope you'll be home.