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Re: E-Tesla = Medhurst
Hi Bob,
At 01:40 PM 6/11/00 -0400, you wrote:
>Hi Terry.
>
>After a little more thought or was it a lot more thought.
>
>For a large parallel C the voltage a long the coil approaches a linear
>function regardless of what the distribution is with out parallel C. ie the
>current flowing in the distibuted intrinsic C hardly effects the voltage
>across the coil because the coil current is much higher than the intrinsic C
>current.
>
>Hence the additional current flowing in to the distributed intrinsic C is
>only a function of the voltage and C. This is even true for low H/D (<0.25)
>with almost unity coupling (auto transformer action).
>
>Under such condition the distibuted intrinsic C can be refereed to one end
>by simple summing the currents. Because the currents are proportional to
>the voltage a voltage weighted intrinsic C is equal to a single C you must
>put in parallel with the coil to have the same total current flow that the
>distibuted intrinsic C had.
>
>Hence if you run E-Tesla with a linear voltage and on an isolated coil it
>will produce med C. This is a theoretically valid comparison and should
>correlate to approximately. 1% in C not F. A good test of the theory and
>model.
Ok. I stripped out E-Tesla5.50 to put a linear voltage on the secondary
and stick it in free space and all that. The program is at:
http://users.better-dot-org/tfritz/site/programs/ETMED.ZIP
I then ran the program with a 200 grid on 10 inch diameter coil of various
lengths. I then compared the results to Medhurst.
H/D Cmed C ET5 calc Error %
10.0 1.32 1.2056 -8.66
9.0 1.22 1.1238 -7.89
8.0 1.12 1.0482 -6.41
7.0 1.01 0.96745 -4.21
6.0 0.92 0.88716 -3.57
5.0 0.81 0.80313 -0.85
4.5 0.77 0.75848 -1.50
4.0 0.72 0.71428 -0.79
3.5 0.67 0.66897 -0.15
3.0 0.61 0.62244 2.04
2.5 0.56 0.57424 2.54
2.0 0.5 0.52573 5.15
1.5 0.47 0.4758 1.23
1.0 0.46 0.42783 -6.99
The graph is at:
http://users.better-dot-org/tfritz/MedComp.gif
Note that both graphs have a bend at 5.0!?! Must be something magical
happening at that point. Below 5.0 Medhurst tracks very well until the
coils get really small. I wonder if the error is him or us? :-) ET5 looks
like two straight lines while the Medhurst graph is showing obvious signs
of a little experimental error. The program did about 300 billion more
calculations than Medhurst did so it should be straight! :-)) I wonder if
I dare say that the computed numbers "may" be more accurate than Medhurst's
in the 2.5 to 5.0 range?
Looks like Bob is again right!!
>
>If you run it with a ground plain it will produce med C with a ground plain.
>But as Medhurst did not produce such a table you can not make any valid
>comparison. The true C / TM equation is again theoretically valid for an
>isolated coil.
>
>When I say valid for..... I dont mean its not valid for other conditions it
>means only that I can only prove its valid for that condition/s.
>
>So E-Tesla can be said to be theoretically accurate for coils with
>approximately 1000 turns (low turn to turn C effects), for large top loads
>and with a ground plain sufficiently below the coil (at least several
>diameters) that the inductance is valid.
>
>Note E-Tesla must include some internal C (turn to turn C) effects because
>the C is determined with a voltage profile as opposed to a unity profile
>(the correct way to calculate intrinsic C) then weighting the result with
>the voltage profile. However as the internal C has mostly internal
>current I don't know if it can be refereed to the one end and I don't if how
>you treat the internal coupled flux is valid or correctly produces a
>reffered C to one end. That will require more thought but as I believe its
>a small effect I am not motivated to try. If you try it both ways and the
>answer (refereed C or med C) is significantly different then I will be
>motivated to try.
>
>Now what about the self resonant case. When the only current flowing in the
>coil is due to the distributed intrinsic C both the voltage and current are
>not linear so the above method of referring the intrinsic C to one end is
>not valid. A 5% to 10% error in F is 10% to 20% error in C or L.
I am working on such things to further increase accuracy...
Cheers,
Terry
>
>Regards Bob
>
>You can put this on the list if you think anybody else is interested
>