ohm meter to test for real from fake siler

That's why I used the thickness of the coin, not the diameter. The calculation across the diameter is more complex and a bit "calculusy."

 

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Out of boredom I decided to calculate a formula for resistance if you put the leads directly across from each other on the edge of the coin (e.g.- 12 o'clock and 6 o'clock positions). It certainly did get very calculusy and there was even an inverse hyperbolic sine that appeared at the end, but the result was surprisingly simple. I ended up with R = 0.88137ρ/t, where ρ = resistivity and t = the thickness of the coin.

The units cancel properly and I double checked my math, so I think I did it correctly. I was surprised that the resistance was completely independent of the coin's diameter, but I suppose since the length (L) and cross sectional area (A) in the R=ρL/A formula are both dependent on diameter that the diameter terms canceled out.

Math is weird, especially when you use it to unnecessarily complicate coin collecting.

 

According to my college transcripts, there was a time when I could do that kind of derivation. It clearly didn't stick.

 

Very nice.

And if the probes are positioned on the edge, but aren't positioned directly across from each other? In other words, how will measured resistance change as the degrees of arc between the probes change? We know two values at present, the one you calculated for 180 degrees, and we know it's zero for zero degrees (i.e. probes touching).

Cal

 

Actually, doing it obverse to reverse ain't that simple to model. The probe tip contact area would be very small compared to the face area of the coin. There would curved lines (or shells) of current from probe tip to probe tip passing through the metal cylinder. Resistance is not measured directly but by application of a fixed voltage and measurement of current flow in the instrument (or in some cases, a fixed current is applied and voltage drop is measured). Try checking a 10 mA fuse with a handheld multimeter; even if it was good, it won't be after the probes are applied.

Misalignment of the tips would cause variation in the measured resistance. Even if the tip apposition is perfect, variation in positioning of the tips relative to the axis of the cylinder would probably affect the measurement. I say "probably" based on intuition, but a mathematical proof might show otherwise. Then there is the fact that coins aren't perfect cylinders. Modern coins have excellent circularity (except for some error coins), but the thickness obverse to reverse can vary quite a bit at various points on the face (i.e. cylinder ends), depending on design elements, strike and wear. So where you place the probe tips on obverse and reverse would have a significant effect on micro-ohm (or nano-ohm or pico-ohm) measurements.

Cal

 

Or, you could have someone XRF them. It's becoming common enough that better coin dealers employ the technology now.

I will never again believe myself capable of overcomplicating things. Not after that.

 

Haha, glad to see that everyone enjoyed my mathematical mess! I'm not sure about how to calculate it if the leads aren't directly across from each other, but I'll think about it. Maybe the Math Gods will speak to me....

 

Love it!

 

OK I withdraw my comments. I misread the resistivity table as being ohms/meter. I was wrong, sorry.

 

I'm not going to say it can't be done, but I will say it can't be done with the type of ohm meter most electricians & electronics technicians use. The meter has 2 settings 2000 ohms and 200k ohms. As you can see in the photo at 200k ohms the meter can not distinguish copper from nickel from clad from silver from fake gold.

If anyone has a Simpson Ohm meter collecting dust, break it out and see if there is a difference in the Meg Ohm range...

 

Only if you're waving the probes at the coin from a distance.

The Keysight U1252B I just picked up measures down to 0.01 ohm on the 500-ohm range -- but it's +/- 10 counts in that range, meaning it's only accurate to 1/10 ohm. Doesn't matter; the resistance across an ASE would be in the microohm range.

On the other hand, it does measure microvolts. So if I hook the ASE up to my car battery with jumper cables, and run a few hundred amps through it, I could probably get a pretty robust voltage reading before the cables melt. (I expect the cables would melt before the ASE would, because they'd be dissipating a lot more heat. They'd almost certainly spot-weld themselves to the coin's surface, though.)

 

You'd have to measure the conductivity in Seimens.

This would take a pair of jaws sorta like these, along these lines, but, fully insulated from each other, the fixture of phenolic or bakelite, and the dead soft fully annealed copper jaws formed to fit the reeding well, and have means to connect adequately sized leads.



Then hook each jaw up to the leads of a DC tig welder while measuring the voltage drop across the coin while your tig unit displayed B+ voltage and current flow, which I believe would require 100's of amps, enough to DC weld aluminum (yes you can with stick and not AC high freq) and color the coin.

 

As I studied your very nice derivation, one thing that kept nagging me was that it did not consider contact area for the instrument probes. It has to be an area and can't be just a point or a line. A point or line, which has no area, would have infinite resistance.

Contact area often involves some assumption and approximation. For example, when measuring the resistance of a length of wire, which is a cylinder, the contact area is considered to be the ends of the wire (i.e. its cross-section). This is assumed even though in practice, the probe tips (or clips) usually make contact on the sides of the wire near the ends.

I wasn't sure how to approach this mathematically for measurement of resistance across a disk, which is a cylinder where the height (aka thickness) is much less than the diameter. So, I started searching the net and found the problem isn't trivial. It was solved in 2000 by Kirk McDonald, a Princeton physicist. His solution is the link below, is quite elegant, and does consider contact area, albeit expressed as width of the contact area on the edge rather than area per se. The other dimension of the contact area is the thickness of the disk.

Prof. McDonald's result is: R = ( 2/(πσt) ) ln(2d/δ), where σ is conductivity (reciprocal of resistivity), t is thickness of the disk, d is the distance between the probes, and δ is width of the contact area. Expressed using resistivity, ρ, rather than conductivity, the result is:
R = ( 2ρ/(πt) ) ln(2d/σ). 2/π is about 0.64, which is smaller than the 0.88 factor in the @NSP equation. However, the 0.64 would be multiplied by ln(2d/σ), which will be greater than one. So in some cases, the @NSP equation would be pretty close.

Like the @NSP equation, Prof. McDonald's equation is "independent of the radius 'a' of the disk". Surprisingly, in the McDonald equation, the distance, d, between the probes does not have to be the diameter of the disk; it can be less (but greater than δ; i.e. the probe tips can't touch). In practice, the probe tips would need to be jammed into the edges of the coin so their cross-section makes complete contact which would allow their width to be used as δ. A pair of rectangular blunt tips with dimensions t by δ would do.

Because the McDonald equation applies to any distance, d, between the probes, it can be used to calculate the change in resistance if the probes are less than 180 degrees apart. I'll use r to represent the radius of the disk, so the diameter is 2r. If the probes are 180 degrees apart, the resistance, Rmax = ( 2ρ/(πt) ) ln(2*2r/σ) = ( 2ρ/(πt) ) ln(4r/σ). Suppose, the probes are placed 90 degrees apart. The distance between them will be the hypotenuse of a right triangle with sides of r, which is sqrt(2r^2). So, R90 = ( 2ρ/(πt) ) ln(2*sqrt(2r^2)/σ) =
( 2ρ/(πt) ) ln(2*sqrt(2)*r/σ). 2*sqrt(2) is 2.83, which is less than 4, so R90 is less than Rmax as expected.

Although the theory is a fun exercise (for some), ultimately using resistance to evaluate metallic content of a coin is a crummy method as discussed elsewhere.

Cal

Link to Prof. McDonald's derivation:
www.physics.princeton.edu/~mcdonald/examples/resistivedisk.pdf

 

Dropping the coin on the counter and listening to the ring was good enough for my grandfather.

Could you tech this up by measuring the resonant frequency of the coin in some way?

 

Yes

 

I would venture to say this would be a lot more accurate than measuring resistance.

 

I say that 737 angels can dance on the head of a pin...

 

How many archangels?

 

1135, they are smaller but more powerful!

 

For one-centimeter contact widths, given the formula above, I get a resistance of around 7 microohms. Run 100 amps through the coin, and you'd get a voltage drop of 700 microvolts, possible to measure with a meter like mine -- but not very accurately; copper's resistance would be higher by a factor of about 17/16, or about 740 microvolts. I couldn't take a single reading and get a go/no-go answer, but a comparison between a known-good ASE and a copper one would give a fairly reliable difference.

At that current level and voltage drop, you'd be dissipating 74 milliwatts across the coin itself. It would barely get warm to the touch. Your contacts, on the other hand, would heat up pretty quickly.

But if you wiggle the contacts just a little bit, or if there's just a little bit of skin oil or dust on the rim, your contact area will fluctuate significantly. That will also create hot spots, especially if you're using a constant-current source (which maintains a constant current by adjusting the voltage). Sparks are likely; fireworks are possible. Running 100 amps through a metal plasma involves a very great deal of energy.

A crummy method, but one with a lot of entertainment potential.