[Lowfer] UWL on

[Zack Widup]

I am most used to seeing that formula used with C in pF, f in MHz and
L in uH. In this case, I get the same answer as Garry, with

C = 25330/[(0.185302^2)(6900)] = 106.9 pF

I just find the decimal points easier to manage with these units!

73, Zack W9SZ


On Wed, Mar 17, 2010 at 2:34 PM, Eric Smith <esmithmail at gmail.com> wrote:
> Thanks, Gary -- I used C=25330.3 / f^2 L  (which I found on the internet, so
> it's gotta be right :)
>
> See -- I did this (WRONG):
> ---------
> [eric at gum ~]$ bc -l
> bc 1.06
>
> 25330.3/(185302^2)*.0069
> .00000000509013394280
> ---------
>
> So I screwed up on order of operations.  When I do it like the below, I get
> what everyone else did:
>
> ---------
> [eric at gum ~]$ bc -l
> bc 1.06
>
> 185302^2
> 34336831204
> .*.0069
> 236924135.3076
> 25330.3/.
> .00010691312629299468
> ---------
>
> (The dot above means "the last answer")
>
> Thanks to you and all for setting me straight!
>
> And to Bill --
>
>> That's not the only error in your ways, Eric :)
>
> I know what you are thinking, Bill, but I just can't get excited about
> trying a loop, OR windows (has been since 1993 that I kicked _that_ like a
> bad habit :).
>
> But regarding the SVM -- yes, you are right, and I know I can speak for Mike
> too when I say that we've been checking the mailbox from time to time for
> the last -- what has it been -- 8, 10 years now?  And no sign of one yet
> ....
>
> :D
>
> Eric KD5UWL / WD2XFX
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