ohm meter to test for real from fake siler

I'm not going to give you a pass on the above. Considering which 'actual electrical testers' you may be implying along with the rest of the sentence had my eyebrows doing funny things.

Specificity in aisle 20 please!

 

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Fine, I'll go dig around for the thread as I should've done in the first place.

 

"actual electical testers" and "those testers", do they have an actual name and what actual type of tester/meter they are?

Several types of electrical and electronic test equipment, not coin specific, work "sorta" like your description.....

 

Okay, here's one thread mentioning Sigma Metalytics:

https://www.cointalk.com/threads/sigma-metalytics-modern-5-half-eagles.296929/

I thought there was another, but the one I was thinking about mentioned this contact tester instead, which (a) seems to be limited to gold and platinum and (b) seems to require polishing or scratching the test piece.

 

It's funny, a new member just started a thread about a quarter weighing 5.7g that a Sigma machine "verified" as 90% silver. I'm ready to argue that there's no way a Sigma will be able to distinguish between silver-plated clad and 90% silver, but since I don't have either a silver-plated quarter or a Sigma in hand, take it for what it's worth.

 

As Doug mentioned, checking physical characteristics of the metal won't distinguish genuine from fake coins cast or struck from the "real stuff". Even sophisticated metal analysis, like XRF which gives a trace metal readout, won't work if original metal is used, e.g. striking fake 1889CC Morgans from metal obtained from cheaper, worn CC Morgans. So, physical characteristics are useful for detecting wrong-metal fakes only.

Electrical characteristics, like resistance, capacitance, inductance, are among the worst physical measurements to use to evaluate the metals in coins for several reasons. As has been mentioned, these characteristics don't differ a lot among metals used in coins (copper, silver, nickel, gold, etc.), so very expensive test gear would be needed. It was also mentioned that the test leads have to have a very good attachment to the DUT (device under test aka coin), which means the oxide (or sulfide or carbonate or ...) layer has to be pierced. Tough to do that without damaging the coin. Finally, electrical characteristics are dependent on geometry. The resistance of a coin in the 6-12 digit accuracy needed to distinguish metallic composition would be influenced by seemingly minor variations in surface contamination, wear, strike, diameter and thickness. Positioning of the test leads would be critical too.

Variation in distribution of metals within a coin would affect resistance even if the overall composition was identical. As an extreme example, imagine two coins that are 90% copper and 10% silver. In one coin, the silver is all in a disc in the middle. In the other coin, the silver is all in a ring that runs around the periphery of the coin. If you attach the test leads to the edges of the coins, the resistance of the latter coin would measure less than the former.

Better physical methods for metal analysis range from the simple like specific gravity and magnetic characteristics to sophisticated like XRF, ultrasound transmission, x-ray / gamma-ray absorption, etc.

Cal

 

This meter will do.
Whites 6000DI Pro detector.

 

Please post it here, I would love to read it.

 

I hope he didn't get irradi-tated

 

At the ohm levels you are trying to measure the SLIGHTEST surface oxidation is going to throw off your results so much as to be worthless.

I believe you are off by a factor of 1,000,000. 1.59e-8 ohms/meter times 2.74 meters is .000000044 ohms. That is 44 picoohms. 44 milliohms is .044 ohms.

The same length in copper would be 47 picoohms. So you are talking about 3 BILLIONTHS of an ohm over a distance of 2.74 meters. Since the thickness of a coin is in the neighborhood of 2 mm You are talking about trying to measure a difference of .006 billionths of an ohm.

 

I don't think so. The 1.59e-8 ohms/meter is for something with a cross-section of 1 square meter. The cross section of the wire is 1 millionth of that, which goes in the denominator of the equation for the resistance based on resistivity, length, and cross-sectional area.

 

*Chiming in* I appreciate your discussion/debate. About a year ago, I crunched the numbers on the theoretical side and then did the best with what resources I had to apply this idea. After the data was analyzed that I had gleaned from the most precise voltmeter I could find, at the smallest amount of ohm (the SI I decided to use - you know, kept it numbers and orders of magnitude as much as possible: 10^-9, 10^-12, etc.), it became apparent that access to a high-energy physics lab would be the most efficient, cost-effective route to the tool needed. Or, that is where I assumed I would most likely find the right tool for the nob without spending MY money. The other option being to ping the bar or coin with EM waves and record the wavelengths/frequencies that are reflected back (like an IR photospectrometer). Around my neck of the woods, the cheapest tool for personal or commercial use is $20K, give or take a few. I would rather spend that on coins or the bullion instead. Maybe there is a large bullion purchasing concern near you that already has one of these non-destructive and "non-invasive" machines? Until you make friends with the owner, just drop by with an ounce and ask for a qoute because you're shopping around to sell. Your cost should just be gas and time (time being every mortals' most valuable resource, though), and you will have your answer(s) as to content. Downside: yeah, I know, i get it...it's not as fun. Helluva lot cheper. Make friends and speak the language, you might get to use it yourself after you become part of the "club", so to speak. And there aren't any dues and costs were just previously mentioned.

Sent from my SM-G930P using Tapatalk

 

*job

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I apologize for misspellings, etc. Just awoke in a pile of Jeffersons. CRH, some habits die hard, but I scored a pretty nice find - 1954-S/D in at least gem quality with a nice, subtle, natural Cu/Ni toning. Here lately it seems as if some heir(ess) decided to spend at least some of their inheritance.

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I think I agree with you. Given length L=2.736 m, cross sectional area A=1x10^-6 m^2, and resistivity ρ=1.59x10^-8 Ω•m:
R=ρL/A
R=(1.59x10^-8 Ω•m)(2.736 m)/(1x10^-6 m^2)
R=(1.59x10^-8 Ω)(2.736)/(1x10^-6) (canceling out meters)
R= 0.0435 Ω= 43.5 mΩ

 

1.59e-8 ohms/m is ohms per meter (a length function) not ohms per square meter (an area function)

In NSP's calculation the second equation R=(1.59X10^-8 Ω/m)(2.736 m)/(1x10^-6 m^2) Carry out the multiplication in the numerator and you get .0000000435 Ω (The meters cancel here) Do the division and you get .0435 Ω/m^2

You can't cancel the meters out as he did in the third equation

 

The units for resistivity are Ω*m, not Ω/m, and the units for the resulting resistance should be in Ω, not Ω/m^2.

 

No, he did it exactly right. There are two meter units in the numerator, and meter-squared in the denominator; m times m over m squared is 1.

Resistivity is given in ohm-meters, not ohms per meter. Think of it this way: resistance goes up as distance (length) increases, but goes down as cross-sectional area increases. A fatter conductor has less resistance for a given length; a thinner one has more.

So, imagine you've got a 1mm square cross-section wire with a resistance of one ohm per millimeter. A one-meter length of that wire would have a resistance of one thousand ohms.

Stack ten one-meter lengths side-by-side, and the resistance of the bundle would be one hundred ohms -- ten 1Kohm resistances in parallel.

Stack one hundred one-meter wires in a ten-by-ten bundle, and the resistance of the bundle would be ten ohms.

The resistance of the whole thing goes up with length (meters), but down with cross-section (meters squared). So, to state the resistivity of the material, you measure its resistance (in ohms), divide by its length (in meters), and multiply by its cross-section (in square meters). The resulting unit is ohm-meters. (Not to be confused with the measuring device...)

 

So, what happens if your probe placement on the edges is off a couple of degrees from being on a diameter chord? Measured resistance should be less than for a diameter chord. Right? I'm assuming resistance would be measured edge to edge as opposed to obverse to reverse. Geometrical misalignment would be more likely in the latter case, plus resistance would be much lower than edge to edge through a diameter chord. And what do you do if the coin isn't a perfect cylinder, like many ancient and error coins?

Cal