Announcement

Collapse
No announcement yet.

Ground Balance Theory

Collapse
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • Ferric Toes
    replied
    Here are two superimposed plots of Oz Rock (green) and Tiva Tuff (red). Both plots commenced at 1000 units and placed on top of each other so only seen as one at the start point. The curve fit for the Oz rock was too far out in relation to the rest of the points, so ceases just below the 1000 dot. This indicates a faster initial decay for the Oz rock. The linear decay plot shows the true difference between both curves which gradually come together at late times. I get a slope calculation of -1.094 for Oz compared to -1.016 for Tiva for the portion between 10 and 20uS. You can see how the red dots drift below the line between 20 and 50uS and how some red dots drift above in the Tiva plot. This is to do with the opposing changes of slope that have been noted with these two different materials.

    Does it matter? Not for most ground, but Oz ground is still a challenge that needs to be addressed.

    Eric.
    Attached Files

    Leave a comment:


  • Ferric Toes
    replied
    Hi Aziz,

    Thanks for doing all this work. You had better summarise in words though, so we can all understand what you have found. I will re-measure some Oz rock samples as accurately as I can and post them shortly. I don't think a detailed magnetic study has been done before on Oz goldfield ironstone, so its properties will not be in any paper or textbook. Pity the sample I gave to Yoga Das as few years ago was not investigated. I will get in touch with someone else I know who might be able to get a mineralogical analysis done.

    Eric.

    Leave a comment:


  • Aziz
    replied
    Now lets see the best fitting results:

    Red Hill Soil:
    Click image for larger version

Name:	Decay_Eric_RedHill02.gif
Views:	1
Size:	22.4 KB
ID:	336214

    The freakin Oz Rock:
    Click image for larger version

Name:	Decay_Eric_OzRock02.gif
Views:	1
Size:	23.2 KB
ID:	336215

    Tiva Tuff:
    Click image for larger version

Name:	Decay_Eric_TivaTuff02.gif
Views:	1
Size:	25.3 KB
ID:	336216

    And the freakin Excel tables (zipped, three tables):
    ExcelTables01.zip

    Bugger me!, keep digging every signal!!!
    (Could be a hotrock! *LOL*)

    Cheers,
    Aziz

    Leave a comment:


  • Aziz
    replied
    Aziz's Magic Formula G(t) = B*(t+p)^b


    My magic formula is the best fitting and approximating formula so far!
    G(t) = B*(t+p)^b


    I'll show you all the three formulas in comparison soon.

    Aziz

    Leave a comment:


  • Aziz
    replied
    The Freakin Oz Rock

    Now look at the freakin Oz Rock this time. And you will understand, that the Oz Rock is difficult to tame and it will screw your GB. All the nice presented formulas in the scientific papers can be thrown into the garbage bin now.

    Click image for larger version

Name:	Decay_Eric_OzRock01.gif
Views:	1
Size:	17.5 KB
ID:	336212

    Excel table (zipped):
    Decay_Eric_OzRock01.zip

    I can't get better approximations with less errors. Even my own solver software does not deliver them.
    Maybe Eric makes a more accurate measurement once again? (like the Tiva Tuff)

    Cheers,
    Aziz

    Leave a comment:


  • Aziz
    replied
    Look at this.
    G2(t) approximates the decay behaviour of Tiva Tuff much better (less error).

    Click image for larger version

Name:	Decay_Eric_TivaTuff01.gif
Views:	1
Size:	20.3 KB
ID:	336209

    Excel table (zipped):
    Decay_Eric_TivaTuff01.zip

    Aziz

    Leave a comment:


  • Aziz
    replied
    Thanks for the clarification guys.
    I'll update the data for the Tiva Tuff.

    BTW, the Oz Rock can't be modelled (fitted) with less error.
    Both approximation functions
    G1(t) = B*t^b and
    G2(t) = -B * (Exp(-t/T1) - Exp(-t/T2)) / (t*Log(T2/T1))
    doesn't fitt well in the measured data. The error at the beginning and at the end of the samples (early and late samples) is quite high.

    Cheers,
    Aziz

    Leave a comment:


  • Ferric Toes
    replied
    Just out of interest I tried log log graphing in sections of the Tiva plot. i.e 1,2,3, - 2,3,4, - 3,4,5 etc and getting the graph calculation for each. I got 1.014, 1.023, 1.037, 1.039, 1.053 ..... which are not far removed from my original results. I still think the effect is genuine, particularly as the RH Soil and Oz rock go the other way.

    Eric.

    Leave a comment:


  • Ferric Toes
    replied
    Originally posted by Midas View Post
    Hi Eric,

    That method for determining exponents actually only holds true for an exponent of 1. Anything else and it delivers incorrect results, try it on a generated curve and see.
    Aziz I'm not sure what your method is but I'm pretty sure its also not correct. Nice work on the least error fit though, better than mine.

    My method pretty simple exploiting the fact that the slope of the log\log plot is equal to the exponent. So, using the first set of Redhill points as an example: a=(log(204)-log(134))/(log(30)-log(20))

    Applying it to the Tuff there's not much of a trend. It does go up broadly to start with but very erratically and the final exponent is actually less than the first. I'm starting to lean towards measurement error.
    Well it's all good clean fun, but all I need is the slope to 2 decimal places, so I might as well stick to my graphing software exponent calculation. 2DP is an order greater than has been done before, and I have eliminated many of the inaccuracies in previous instrumental results. Anything further may be of interest to environmetal magnetism scientists, but the primary interest here is knowledge for effective ground balancing in a PI metal detector.

    -1.05 it is.

    Eric.
    Attached Files

    Leave a comment:


  • Midas
    replied
    Originally posted by Ferric Toes View Post
    Brilliant! That is just what I was hoping someone would do. My graphing does a best fit, so the exponent is averaged over the whole curve, but the fit is pretty good. I always noticed that the 10uS sample was a bit higher in amplitude than it should be and wondered if it was an instrumental problem, but I could not see what was going on further down i.e. change of slope. Now with your results and what I have measured over the past two days, I see that it is not an instrument problem, and I will explain why.

    Firstly there are no timing errors throughout the range of delays. At the shortest delay (10uS) the clock is running at 100kHz and measured today on a precision frequency meter with an oven controlled reference, is accurate to 1Hz. Likewise, the TX and sample widths are all as accurate. This being so if measurements are taken at 10uS, 20uS, 40uS and 80uS there should be a factor of 2 between each succeeding pair for a slope of -1. e.g. say 500/10uS, 250/20uS, 125/40uS, and 62.5 at 80uS. I also did the intermediate values 15 - 30, 25 - 50uS etc. Having done this I also plotted a downward trend in exponent as you have for the Oz Rock and Red Hill soil. Basically, I took the shorter delay reading, halved it, and divided it by the longer delay reading. e.g. (1000/2)/494 = 1.0121 for the first reading of the Tuff discussed below.

    Now, it could be argued that there is some, as yet, undiscovered error in the instrument - but for the following. I plotted the results for Tiva Canyon Tuff, which although derived from volcanic ash has some interesting properties. The main one being a very high frequency dependence of susceptibility, >20% resulting in a high viscosity reading relative to its dc susceptibility. With this material the trend of the exponent goes the other way i.e. it increases at late times. To my thinking, this rules out an instrumental error.

    To improve the S/N figure for the Red Hill and Tiva Canyon material I used 20gm samples instead of 10gm. This almost doubles the signal throughout the range. Theoretically it should double exactly but the bigger pot is poking slightly out of the sensor coil where the flux density will be a bit less. I repeated the measurements a few times and it is consistent.

    As you say, we are dealing with very small changes here. The Tuff goes from a slope of -1.0121 between 10&20uS to -1.055 between 50&100uS. Plotting out the full results for the Tuff and doing a best fit curve, it almost looks as though the exponent has its own power law of +0.026.

    Eric.
    Hi Eric,

    That method for determining exponents actually only holds true for an exponent of 1. Anything else and it delivers incorrect results, try it on a generated curve and see.
    Aziz I'm not sure what your method is but I'm pretty sure its also not correct. Nice work on the least error fit though, better than mine.

    My method pretty simple exploiting the fact that the slope of the log\log plot is equal to the exponent. So, using the first set of Redhill points as an example: a=(log(204)-log(134))/(log(30)-log(20))

    Applying it to the Tuff there's not much of a trend. It does go up broadly to start with but very erratically and the final exponent is actually less than the first. I'm starting to lean towards measurement error.

    Click image for larger version

Name:	Tuff.jpg
Views:	1
Size:	39.5 KB
ID:	336199

    Leave a comment:


  • Ferric Toes
    replied
    Originally posted by Dave J. View Post
    Electrically conductive ferrite material.

    --Dave J.
    Hi Dave,

    Nice try, but the rods that exhibit viscosity are >20M end to end, while the ones that I use as PI probe cores measure about 3k.

    Tiva canyon Tuff is interesting and there is quite a bit of research being done by various organisations. The Institute for Rock Magnetism has a couple of good articles for download starting vol.14 No.3, 2004. I was lucky enough to get some a few years ago when starting some magnetic soil measurements.

    Eric.

    Leave a comment:


  • Ferric Toes
    replied
    Originally posted by Aziz View Post
    Hi Eric,

    ok, I have checked the point, that you fit the exponent between two readings (Amplitude(t) = A1, Amplidute(2*t) = A2). Is this correct? Or how do you calculate the exponents in the last picture?

    If I have understood it correct,
    (1) B*t^b = A1
    (2) B*(2*t)^b = A2

    solved 1+2 for b:
    b = log2(A2/A1) = log(A2/A1) / log(2)

    But if I put the parameters into the formula however, I get different values.
    Aziz
    The exponents in my data are arrived at by taking the reading at 10uS delay and for convenience, setting it at a round figure with the sensitivity control. Usually 1000, 500, or 200 for weaker specimens. There is a both a rotary switch for delay and a toggle switch that exactly doubles whatever delay is set on the rotary switch. The sample is removed, which causes the unit to auto-zero. The toggle switch is then set to x2 (20uS) and the new reading taken. If the slope is exactly -1 then the second reading should be exactly half the first i.e. 500, 250, or 100, in the case of the above initial settings. In the case of the Tuff with a first reading of 1000 at 10uS, then a second reading of 495 at 20uS, we get 1000/495, then divided by 2 again gives 1.01 which is the slope. Repeat at 30uS delay and first reading is 325 and second (60uS) 155. 325/155/2 = 1.048, so we can see that the exponent has increased. This process is repeated for all delays on the rotary switch and then doubled on the toggle switch. Dead easy really with no complex maths. Only realised a couple of days ago that I could do this easily for a quick test; A 10 - 20uS calculation, 30 - 60, and a 50 - 100. This will give 3 data points to show if the exponent reduces, increases, or stays flat across the measurement range.

    One point to note is that the sample widths are always the same as the delay. i.e. 10uS delay, 10uS sample, 15uS delay, 15uS sample and so on such that the 100uS delay has a 100uS sample. Pulse frequency has also dropped proportionally so that integration time remains the same. The delay figure should strictly be the mid point of the sample pulse rather than the time to the start edge. 10uS delay to the start edge is really 15uS to the mid point a.s.o.

    Eric.

    Leave a comment:


  • Dave J.
    replied
    Originally posted by Ferric Toes View Post
    I have just done some viscosity measurements on a piece of ferrite rod of a grade that is used for radio antennas. This has been suggested in the past as a possible substitute for magnetically viscous ground. Compared to the Red Hill soil, which I tend to use as a medium viscosity standard, the ferrite gives only 27% signal. i.e. a spot reading of 40 instead of 140. The problem though is that the slope is different from any natural soil or rock that I have measured; -1.27 compared to a range of -0.99 to -1.05 for the latter. Red Hill soil is -1.03.

    Eric.
    Electrically conductive ferrite material.

    --Dave J.

    Leave a comment:


  • Dave J.
    replied
    Originally posted by Ferric Toes View Post
    I have always run into problems when joining ferrites. Had one today where I was building a coil whose sole purpose is to detect nothing. In other words I wound a coil on a full pot core to stop any external field. By the way, this is a grade that only has a small viscosity. Can't see it on the VM but just see it on scope when the coil is in the core. All is OK until I bolt the two halves together, when I get a sustained ringing. If I put a layer of double sided Sellotape between the coil halves then bolt up, the ringing is gone. I have seen this before when joining rods or rings end to end - always need a very thin mechanical damper between sections.

    Eric.
    Magnetostriction.

    --Dave J.

    Leave a comment:


  • Dave J.
    replied
    Eric, for reasons I am not at liberty to disclose, I regard your interpretation of the Tiva Tuff data as correct. (However, this is a poorly understood area of magnetic physics and I may be wrong.)

    --Dave J.

    Leave a comment:

Working...
X