Message:

<p>[QUOTE=&quot;calcol, post: 2789999, member: 77639&quot;]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.&nbsp; It has to be an area and can&#039;t be just a point or a line.&nbsp; A point or line, which has no area, would have infinite resistance.</p><p><br /></p><p>Contact area often involves some assumption and approximation.&nbsp; 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).&nbsp; This is assumed even though in practice, the probe tips (or clips) usually make contact on the sides of the wire near the ends.</p><p><br /></p><p>I wasn&#039;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.&nbsp; So, I started searching the net and found the problem isn&#039;t trivial.&nbsp; It was solved in 2000 by Kirk McDonald, a Princeton physicist.&nbsp; 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.&nbsp; The other dimension of the contact area is the thickness of the disk.</p><p><br /></p><p>Prof. McDonald&#039;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.&nbsp; Expressed using resistivity, ρ, rather than conductivity, the result is:</p><p>R = ( 2ρ/(πt) ) ln(2d/σ).&nbsp; 2/π is about 0.64, which is smaller than the 0.88 factor in the [USER=74849]@NSP[/USER] equation.&nbsp; However, the 0.64 would be multiplied by ln(2d/σ), which will be greater than one.&nbsp; So in some cases, the [USER=74849]@NSP[/USER] equation would be pretty close.</p><p><br /></p><p>Like the [USER=74849]@NSP[/USER] equation, Prof. McDonald&#039;s equation is &quot;independent of the radius &#039;a&#039; of the disk&quot;.&nbsp; 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&#039;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.&nbsp; A pair of rectangular blunt tips with dimensions t by δ would do.</p><p><br /></p><p>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.&nbsp; I&#039;ll use r to represent the radius of the disk, so the diameter is 2r.&nbsp; If the probes are 180 degrees apart, the resistance, Rmax = ( 2ρ/(πt) ) ln(2*2r/σ) = ( 2ρ/(πt) ) ln(4r/σ).&nbsp; Suppose, the probes are placed 90 degrees apart.&nbsp; The distance between them will be the hypotenuse of a right triangle with sides of r, which is sqrt(2r^2).&nbsp; So, R90 = ( 2ρ/(πt) ) ln(2*sqrt(2r^2)/σ) =</p><p>( 2ρ/(πt) ) ln(2*sqrt(2)*r/σ).&nbsp; 2*sqrt(2) is 2.83, which is less than 4, so R90 is less than Rmax as expected.</p><p><br /></p><p>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.</p><p><br /></p><p>Cal</p><p><br /></p><p>Link to Prof. McDonald&#039;s derivation:</p><p><a href="http://www.physics.princeton.edu/~mcdonald/examples/resistivedisk.pdf" target="_blank" class="externalLink ProxyLink" data-proxy-href="http://www.physics.princeton.edu/~mcdonald/examples/resistivedisk.pdf" rel="nofollow">www.physics.princeton.edu/~mcdonald/examples/resistivedisk.pdf</a>[/QUOTE]</p><p><br /></p>

Your name or email address:

Do you already have an account?

• No, create an account now.
• Yes, my password is:
• Forgot your password?

Stay logged in