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# Inductance (fwd) [correction]

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From:  Malcolm Watts [SMTP:MALCOLM-at-directorate.wnp.ac.nz]
Sent:  Sunday, April 05, 1998 4:57 PM
To:  Tesla List
Subject:  Re: Inductance (fwd) [correction]

Hi Alan,

> From:  Alan Sharp [SMTP:AlanSharp-at-compuserve-dot-com]
> Sent:  Thursday, April 02, 1998 3:50 AM
> To:  INTERNET:tesla-at-pupman-dot-com
> Subject:  Inductance (fwd) [correction]
>
> Malcolm wrote:
> >Hi Alan,
> >         Sorry to say this but there is a formulaic error in your snip<
>
> Its more than possible - the alagbra circuits could have got
> fried along with the calculus neurons.
>
> but I just did it again with pencil and paper - rather than
> directly on the screen and it checks out, and I looked again
> at Bylund he also comes to h=0.9r.
>
> There is a typo in the original script N = b / d is N = b * d
> but Erik got to the right result. Must have been a copy
> error. But however we have got there, we have now two
> results:
> Maximum L when h = 0.9 * r
> or h/d = 0.45
> And you supply:
> Maximum Q when h= 2 * r
> or h/d = 1
> Is this result from theory or observation?

Both. As it turned out, the typo was inconsequential. h/d = 0.45 is
the same as h/r = 0.9

Perhaps I might elaborate a little on H/D ratios since I've just dug
out the results I measured on lots of different coils for a different
purpose.

SPacewound Coils:
Highest Q:  obtained in a space wound coil with *no* topload of h/d =
1.  If a topload is added, h/d around 0.5 gives the higher Q of the
two but can't match the unloaded h/d = 1 coil. This is a result
measured with a high degree of coil isolation. The coil was 6.6" in
diameter. Allowing for signal generator internal impedance, the Q of
this coil nearly hit 500.

Closewound Coils:
Highest Q without a topload is obtained with h/d around 4. With a
topload, h/d = 3 gives the highest Q in general. Also measured but
was theoretically derived by me several years ago. Thickness of the
wire cf skin depth had a lot to do with this result. The tradeoff was
L vs Rac.

> Is this because while h = 0.9 * r   minimises the
length of > wire and therefore the resistance but it is going to have
> a higher voltage rise on each turn - increasing the effects
> of inter turn capacitance?
>
> Presumably then h = 2 * r gives the best comprimise between
> wire resistance and the effects of inter turn capacitance.

Not so much interturn capacitance but sheet capacitance to ground.
For a given diameter, the h/d factor in Medhurst's Cself formula
bottoms out when h/d = 1.

Cheers,
Malcolm

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