Re: Inductance of a conical coil

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Barton B. Anderson by way of Terry Fritz <teslalist-at-qwest-dot-net>" <classictesla-at-netzero-dot-com>

Hi Matt,

Yes, sorry, the x 2 on the end should not have been included. I have a 
spreadsheet that I quickly did a copy and paste from. One cell calcs the 
formula and another cell shows the formula in text. It's the text cell that 
I copied which mistakenly has the x 2 multiplication on the end. I 
apparently didn't update the text cell way back when.

You are also correct on reducing the formula. At the time, I simply popped 
the idea into the spreadsheet to see it perform and rarely use that 
particular spreadsheet anymore. I always figured someone would find a 
solution other than applying this correction factor (I'm sure someone will).

Anyhow, we have:
factor = 1/sqrt(sina+cosa)

Take care,
Bart


Tesla list wrote:

>Original poster: "by way of Terry Fritz <teslalist-at-qwest-dot-net>" 
><Mddeming-at-aol-dot-com>
>
>In a message dated 2/23/03 10:39:43 AM Eastern Standard Time, 
>tesla-at-pupman-dot-com writes:
>
>
>
>>This formula weights the horz and vert Wheeler equations, but is poor in
>>accuracy. I've added my own "weighting" to this of which the final L above
>>is multiplied by a factor based on the angle.
>>
>>I then have:
>>L = [SQRT[(L1*Sin(X))2 + (L2*cos(X))2] ] * my factor
>>
>>First I find the cosine and sine of the angle in radians and denote them as
>>"sina" and "cosa". My factor is then: ((cosa^2+sina^2)/SQRT(cosa+sina))*2
>
>
>Hi Bart,
>        Using the elementary trigonometric identity: S^2(a) + C^2(a) = 1, 
> the formula immediately reduces to :
>1/(sqrt(sina +cosa)*2 = 2/sqrt(sina+cosa)
>However, for the values you gave:sina = 0.939693, cosa = 0.342020
>2*(!/sqrt(0.939693 +  0.342020)) = 1.766585254 exactly double what you stated:
>
>My factor then = 0.883292708
>
>Think you need a slight revision
>
>Matt D.
>
>
>
>