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  • Originally posted by Ferric Toes View Post
    Hi Skippy, The can I have measures 0.07mm in the middle section, including paint and any inner coating. It contained Southern Comfort, Lemonade and Lime. The can says it was produced in Belgium. I'll look out for a discarded Coke can; don't drink it myself.
    I can detect a 10mm square of the 0.07mm stuff at 5in with the Vallon and slightly discernable at 6in. I cut a piece 9mm square and couldn't detect it.

    The Hocking Sigma 2000 Conductivity Meter has an inductive probe running at 250kHz so paint, corrosion, etc do not have any effect. One of the calibration pieces has a plastic film over it. You have to get the same reading as another piece without the film. Both read 100 as they are annealed copper which everything else is referenced to as a %.

    Eric.
    I tested a 9x9mm can side awhile back and got 2usec for the TC.

    Comment


    • A quick bit of Googling reveals that the wall thickness is commonly 0.097mm to 0.12mm, though clearly it's going to vary a bit around the world. It's not unreasonable to assume that the development of better alloys will result in thinner walls.
      I understand the problem with higher-strength alloys is that they don't extrude well, they crack, etc. So the search is on for a soft hard malleable strong alloy......

      Comment


      • Thickness of the sample makes a big difference.

        I found that a square of 1in wide copper tape used in EMC applications gave the same conductivity reading as a 1in square aluminium off a drink can. The copper tape is a bit thinner at 0.05mm and I have a whole roll of the stuff which is easy to cut up. I cut five identical 1in squares and first measuring one, I then stuck the next one on top, and so on, until five were stuck together. The tape is adhesive backed. Results with the Hocking probe are as follows.
        1. 7.67% IACS... 0.05mm thickness
        2. 27.00........... 0.10
        3. 45.50........... 0.15
        4. 53.40........... 0.20
        5. 53.70........... 0.25

        Adding more caused little change. If I get time over the next few days, I will plot the log/lin decay of similar samples. Also interesting would be to do the same exercise with 10mm squares of copper tape.

        Eric.

        Comment


        • Originally posted by Ferric Toes View Post
          I found that a square of 1in wide copper tape used in EMC applications gave the same conductivity reading as a 1in square aluminium off a drink can. The copper tape is a bit thinner at 0.05mm and I have a whole roll of the stuff which is easy to cut up. I cut five identical 1in squares and first measuring one, I then stuck the next one on top, and so on, until five were stuck together. The tape is adhesive backed. Results with the Hocking probe are as follows.
          1. 7.67% IACS... 0.05mm thickness
          2. 27.00........... 0.10
          3. 45.50........... 0.15
          4. 53.40........... 0.20
          5. 53.70........... 0.25

          Adding more caused little change. If I get time over the next few days, I will plot the log/lin decay of similar samples. Also interesting would be to do the same exercise with 10mm squares of copper tape.

          Eric.
          Hi Eric


          Does the adhesive between layers affect the conductivity and eddy currents similar to laminations in a transformer?


          Thank you and have a good day,
          Chet
          Last edited by Chet; 05-27-2016, 05:49 PM. Reason: spelling

          Comment


          • Originally posted by Chet View Post
            Hi Eric
            Does the adhesive between layers affect the conductivity and eddy currents similar to laminations in a transformer?

            Chet
            Interesting point, but it appears not. The reason is that the magnetic flux is perpendicular to the surface, while in a transformer it is along the laminations. Each layer of the copper foil acts as a parallel circuit damping the one above and slowing down the decay in the case of pulse excitation. I'm not sure how the conductivity meter works though as that is sine wave. Probably measuring phase shift.

            Comment


            • Copper foil tests

              Following the previous tests on copper foil and the fact that 5 layers brought the IACS readings to a respectable level, I was interested to see now how small a piece the Vallon would detect. I cut a five layer sandwich until is was down to 4mm x 4mm. This I could still detect at 3 - 4in. I then plotted the decay curve which came out as follows. The delay readings are from TXoff to start of first sample and were done on a viscosity meter usually to measure soil samples. The delays are micro controlled and accurate to a few nS. I got a max reading of 850 arbitrary units at 10uS and which falls to 245 at 15uS and so on.

              Click image for larger version

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              Personally, I am quite happy with this level of small object sensitivity (4 x 4mm x 0.25mm). Even for gold nugget hunting, as chasing smaller pieces is not of great interest. I believe the Vallon would really shine on a Civil War battle site as bullets and miniballs are readily detectable at good ranges. I will post some figures in due course.

              Comment


              • Originally posted by Ferric Toes View Post
                Following the previous tests on copper foil and the fact that 5 layers brought the IACS readings to a respectable level, I was interested to see now how small a piece the Vallon would detect. I cut a five layer sandwich until is was down to 4mm x 4mm. This I could still detect at 3 - 4in. I then plotted the decay curve which came out as follows. The delay readings are from TXoff to start of first sample and were done on a viscosity meter usually to measure soil samples. The delays are micro controlled and accurate to a few nS. I got a max reading of 850 arbitrary units at 10uS and which falls to 245 at 15uS and so on.

                [ATTACH]36321[/ATTACH]

                Personally, I am quite happy with this level of small object sensitivity (4 x 4mm x 0.25mm). Even for gold nugget hunting, as chasing smaller pieces is not of great interest. I believe the Vallon would really shine on a Civil War battle site as bullets and miniballs are readily detectable at good ranges. I will post some figures in due course.
                [Yes, I have started to do similar tests myself and it will be interesting to compare results. I am using my Magnetic Viscosity Meter to plot decay from 10uS - 50uS in 5uS steps. I will do a log signal vs time as you do. I have some reservation about aluminium from cans. It is very springy and obviously an alloy. Aluminium varies tremendously in conductivity depending on what the other alloy components are, but maybe all Coke cans are the same. I cut up a can from another drink assuming it would be the same and I can detect a 10mm square with the Vallon. I will report back in more detail later.] reply#174

                Wondering why you charted linear scales. Do you see something on the linear chart that makes it better?

                Comment


                • Originally posted by green View Post

                  Wondering why you charted linear scales. Do you see something on the linear chart that makes it better?
                  I was intending on doing a log amplitude scale but my computer kept coming up with an error signal. For magnetic soil samples I always use a log/log scale which then gives a straight line with a slope of -1.
                  Having said that, a linear scale gives a good visual idea of the scale of the difference in signal amplitude that the preamp sees at later delays. I will try and fix the computer error so that we are comparing like for like.

                  I did advocate the use of wide band log preamps some years back but other than a few experiments, didn't do anything further about it. By differentiating the linear slope you get a constant amplitude that gets noisier with time for non-ferrous. Iron still looks like an exponential.

                  Comment


                  • Sorted the problem. Here is the log/linear plot. Click image for larger version

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                    Comment


                    • Originally posted by Ferric Toes View Post
                      Sorted the problem. Here is the log/linear plot. [ATTACH]36329[/ATTACH]
                      To get the target time constant off a chart I've been using the decay time between 8 and 3 or time between (2 and .1)/3 for 1 TC. Wasn't getting the same TC as I got when I charted your data when I noticed the fine lines on your chart aren't even increments. My math is really rusty, what would be the correct way using your chart?

                      Comment


                      • The lin. vs log. scale is tricky to use. You can't interpolate accurately on the log scale, so don't try. The linear scale can easily be interpolated, if need be. The best way is to choose two points on the straight line that cross through convenient points on the chart. Choose them a decent distance apart for accuracy.
                        Example, from Erics chart: V=90 at t=19.5 us, and V=8 at t=30.5 us.
                        Your basic decay equation is:

                        V/Vo = e^(-t / RC) for the more familiar R & C charge/discharge circuit. RC = time-constant, (which also = L/R for inductive decay).
                        So in this case: V/Vo = 8/90, and t = (30.5 - 19.5) = 11 us.
                        Take natural ( Naperian) logs of each side:
                        ln(8/90) = -11 us / (time-constant)

                        or, re-arranging:

                        time-constant = (-11 x 10^-6) / ln (8/90)

                        which I calculate as 4.54 microsecs for this test object.

                        Comment


                        • Originally posted by Skippy View Post
                          The lin. vs log. scale is tricky to use. You can't interpolate accurately on the log scale, so don't try. The linear scale can easily be interpolated, if need be. The best way is to choose two points on the straight line that cross through convenient points on the chart. Choose them a decent distance apart for accuracy.
                          Example, from Erics chart: V=90 at t=19.5 us, and V=8 at t=30.5 us.
                          Your basic decay equation is:

                          V/Vo = e^(-t / RC) for the more familiar R & C charge/discharge circuit. RC = time-constant, (which also = L/R for inductive decay).
                          So in this case: V/Vo = 8/90, and t = (30.5 - 19.5) = 11 us.
                          Take logs of each side:
                          ln(8/90) = -11 us / (time-constant)

                          or
                          time-constant = (-11 x 10^-6) / ln (8/90)

                          which I calculate as 4.54 microsecs for this test object.
                          The fine lines aren't even increments. Try counting 10, 20, 30, 40, 50, 60, 70, 80, 90, 100. Thanks for the formula.

                          Comment


                          • I don't really understand your statement...
                            But there is something peculiar with the scaling, it looks like there's 12 steps per decade, but there's not... is that what you mean.?
                            Maybe Eric had better chime in here, in case there's something he's done wrong.
                            The pale grey lines for 10, 100, 1000 are there to indicate where those voltages are on the vertical axis, but the actual scaling of the true decade graduations are unfathomable, it looks like the decade starts at about 9, then there's 10 steps to get to 90, etc.
                            That's what happens when you let a computer do the work. If I'd plotted this, I wouldv'e done it on paper, with a ruler.

                            So...ignore my earlier readings from the chart, they are incorrect.
                            Based on the 'computer best fit' data that says "b = -0.22627", we can calculate -1/b as 4.42 , this is the time-constant in microsecs, as Eric has obviously entered the time figures as 10, 20 etc rather than 10 x 10^-6, 20 x 10^-6, etc.
                            So the targets time-constant is 4.42 microsecs.

                            Comment


                            • Dang that's bizarre. It looks like the 'decade' is in 11 steps of about '9', starting at 93.

                              Change of topic: I've got hold of a number of new ally cans, standard diameter (65mm), energy-drink sized (55mm) and standard and tall height. They were all from 0.095mm to 0.105mm thickness. This makes Eric's thin-wall beverage can look unusual.
                              I'll try and get an IACS resistivity reading for one of them, just to see if there's any real difference to my previously posted test result, and the 'datasheet ' value.

                              Comment


                              • Originally posted by Skippy View Post
                                I don't really understand your statement...
                                But there is something peculiar with the scaling, it looks like there's 12 steps per decade, but there's not... is that what you mean.?
                                Maybe Eric had better chime in here, in case there's something he's done wrong.
                                The pale grey lines for 10, 100, 1000 are there to indicate where those voltages are on the vertical axis, but the actual scaling of the true decade graduations are unfathomable, it looks like the decade starts at about 9, then there's 10 steps to get to 90, etc.
                                That's what happens when you let a computer do the work. If I'd plotted this, I wouldv'e done it on paper, with a ruler.

                                So...ignore my earlier readings from the chart, they are incorrect.
                                Based on the 'computer best fit' data that says "b = -0.22627", we can calculate -1/b as 4.42 , this is the time-constant in microsecs, as Eric has obviously entered the time figures as 10, 20 etc rather than 10 x 10^-6, 20 x 10^-6, etc.
                                So the targets time-constant is 4.42 microsecs.
                                I can't understand the scale. Using your formula, looks like a good way would be divide the time change for 1 decade decrease in amplitude by 2.3 or 2 decades by 4.6. Using the same line in each decade it wouldn't matter how many lines there were.

                                Comment

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