Re: [TCML] charging reactors

[David Dean <deano@xxxxxxxxxxxxxxxx>]

Hi Bart

First, I suspect it is a typo, but you should have:
 LuH = (31.6 x n^2 x ra^2) / (6 x ra + 9 x L + 10 x (ro-ri))
where ra=(ro+ri)/2

This guy is using a formula I can't figure our where he got:
currentInd = ( diamavg / 1000.0 ) * diamavg * turns * turns / (( 18.0 * 
diamavg ) + ( 40.0 * coilLen ));

I think the /1000 is an error, he is doing mH rather than uH, I would have put 
*1000 in the denominator. Or just rearranged the multipliers in the select.

I copied his page and changed just that. Still screwy. So I made some other 
modifications. Try:
http://deanostoybox.com/temp/indcalc2.html
and see if that is not closer to what you are getting. It runs pretty close with 
my spreadsheet which is also in the temp folder. ballast.odf for openoffice, or 
ballast.xls for msoffice. The differences are due to the fact the spreadsheet 
works by layers, the web page by turns. I do not know which is closer to 
reality, but the spreadsheet has been verified experimentally and has proven to 
be accurate within my ability to measure, +/-10% or better.

BTW I use the formula from the AARL handbook:
L (uH) = 0.8 * a^2 * n^2 / (6*a + 9*b + 10*c )				
where				
a = average radius of windings 				
b = length of the coil				
c = difference between the outer and inner radii of the coil.				
all dimensions in inches.  				


On Thursday 05 March 2009 08:59:40 pm bartb wrote:
> Tesla, well, would have been stuck with cotton covered wire. But if he
> had access to todays magnet wire, probably small and compact as possible.
>
> Interesting inductance script. However, either my calc's are wrong or
> the script calc's are wrong for the inductance.
>
> For multilayer, I'm using the following:
> LuH = (31.6 x n^2 x ri^2) / (6 x ri + 9 x L + 10 x (ro-ri))
> n = total turns
> ri = inside coil radius
> ro = outside coil radius
> L = coil length
> (all in meters)
>
> Quite a discrepancy between the script inductance result and the above.
> For the calculators inputs at 5H, he realizes 51.18 layers. If I use 24
> awg (.0213" in his code) and adjust my layers the same, everything works
> out fine (id, od, dcr, turns, wire length, etc..) except I show 1.75H
> where he shows 5H (2"D, 4.75"L, 24G). Somethings odd here... and can't
> put my finger on it.
>
> Anyone else have a known working multilayer calculation for L?
>
> Take care,
> Bart
>
> Jim Harvey (UDN) wrote:
> > For what its worth, I wondered "What would Tesla do?" If he needed an
> > inductor, he didn't worry too much about space in his lab or the cost of
> > wire.
> >
> > My thought was, hey, just make it bigger in diameter since L goes up
> > (approximately, for large diameters) by the square of the diameter.
> >
> > My initial calculations tell me that larger diameter will break the bank
> > in terms of total wire length.
> >
> > In my search, I found this:
> > http://www.pronine.ca/multind.htm
> >
> > Any way, just thoughts for a snowing morning here in Utah.
> >
> > Jim Harvey, W7YV
>
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