#12 solid wire

Discussion in 'General Technical Questions and Answers' started by WB7DMX, Feb 1, 2006.

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  1. WB7DMX

    WB7DMX Guest

    think nothing about it, you got to admit that this is a very interesting subject, and I have had fun with it for many years, you can bet I am going to print up this thread, so in the future I will have something to back up the fact that it is there, sort of, you guys almost had me convinced that I was nuts about this.
    yes it is inconceivable from a normal everyday point of view, that is why it made such a impression on me when I first had to work it out, I thought they were nuts.

    and I want to thank everyone for all the input,

    now shall we add some rf to the thought ?

    NOT
     
  2. KE5FRF

    KE5FRF Ham Member QRZ Page

    Aha!!

    I'm Back from the Jackson, MS. Hamfest, and DMX got his answer!

    Good job, OM, I knew you would find somebody to support it with a formula.
     
  3. K7JEM

    K7JEM Ham Member QRZ Page

    Pretty good formula, but again I don't buy it.

    Here's why.

    The formula is linear. The only variable is resistance, everything else stays the same.

    A #6 guage wire 4 inches long will have the same resistance as a one inch piece of 12 guage wire, a #0 wire will be 16 inches long for the same resistance.

    All of these wires will exhibit the same capacitance if you use this formula, since the formula depends on resistance only.

    You could get to a point where the wire or rod is several inches in diameter, and several feet long, and you would come up with the same answer. This just not seem correct when dealing with a homogeneous element.

    Joe
     
  4. KE5FRF

    KE5FRF Ham Member QRZ Page

    Man, you are a hard guy to please!!

    [​IMG]

    But I admire your attention to detail.
     
  5. WB7DMX

    WB7DMX Guest

    there are two variables in the problem, the circular-mills and the resistance will vary for differant sizes of wire. but I don't think it would be enough to make much differance to the results.
     
  6. SM0AOM

    SM0AOM Ham Member QRZ Page


    I'm afraid that you have to buy it, as dimensional analysis of the equation shows that the result has the dimension of capacitance.

    The resistance of a wire is usually expressed in terms of conductivity, length and cross-section, and if you put these into the formula and perform a new dimensional analysis, the result still will be expressed in capacitance with the unit F.

    Changed lengths or cross-sections only result in scale factors that are cancelled out.

    However, we are splitting hairs here, as we long ago crossed the borders for any physical significance of the "parasitic capacitor" associated with this piece of wire.

    This certainly has been a refreshing discussion about the fundamentals of electrical networks.

    73/

    Karl-Arne
    SM0AOM
     
  7. WB7DMX

    WB7DMX Guest

    to say the least, I am very impressed, you sound like some one I really would like to talk to on the air.

    also the numbers are really climbing on this thread for one simple question.
     
  8. K8ERV

    K8ERV Ham Member QRZ Page

    Youse guys are making a mountain out of a ----- what is that other lump called?

    TOM K8ERV Montrose Colo
     
  9. WB7DMX

    WB7DMX Guest

    its called a piece of #12 wire that is 1 inch long.
     
  10. K7JEM

    K7JEM Ham Member QRZ Page

    In the equation you are using the "free space" value of permittivity. Shouldn't this be the permittivity of copper? After all, the capacitance we're talking about here is "through the copper" not through "free space".

    Back to the drawing board.

    Joe
     
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