Re: New Coil: From Scratch

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

Original poster: "by way of Terry Fritz <twftesla-at-uswest-dot-net>" <Mddeming-at-aol-dot-com>

In a message dated 4/22/01 10:47:32 AM Eastern Daylight Time, 
tesla-at-pupman-dot-com writes: 



>
> Original poster: "Matt Skidmore by way of Terry Fritz <twftesla-at-uswest-dot-net> 
> " <fox-at-woozle-dot-org> 
>
> Hi all, 
> I have been very impressed with the continuing success of my first mini 
> coil and then my 4.5 inch coil that I have created (specs on foxxz-dot-net). 
>
> Mistakes have been made and much has been learned since then and I've 
> decided to again start on another coil, putting my best effort forward to 
> make it look nice and getting more output. Thinking about what I wanted to 
> do last night caused me to get very little sleep. so I've come here for 
> some suggestions as to how I might design this one. 
>
> I know some people design their coils around the toppled and some have 
> other ways, but I usually like to start with the secondary. I'm thinking of 
> something like 8 to 12 inches in diameter. I've seen coils such as terry's 
> 10 inch diameter that are wound 10 inches, and some longer. I had thought 
> that the rule of diameter x 4 was pretty good but once again there are so 
> many variances its hard to decide which kinds might be the most 
> successful. 



Hi Matt S. 

      With respect to coil length vs diameter, an optimal value depends on 
just what you decide to optimize. For example, if you want maximum 
inductance, L for a given length of wire (x) in a close-wound coil: 
using L=(nr)^2/(9r+10h) 
doing these substitutions: 
N=x/(2pi*r) and Area A=2pi*rh = x* diam(wire)=constant ---> h=A/(2pi*r) 
therefore L=r^2*[x/(2pi*r)]^2 / [9r+10A/(2pi*r)] 
with a little algebra, L=x^2/{36pi^2[r+10A/(18pi*r)]} 
Now with Z=constant, A=constant: 

L=Const. * 1/[r+(10A/18pi*r)] and L is a maximum when the denominator is a 
minimum. 
Let denominator=Y=r+10A/18pi*r then  dY/dr= 1-10A/(18pi*r^2) 
setting the derivative dY/dr= 1-10A/2pir^2 = 0 
            (10*2pi*rh)/(18pi*r^2)=1 
            20pi*rh=18*pi*r 
            10h=9r or h=0.9r 

For a given length of primary wire (tubing) in a pancake coil, where 
R=(rmax+rmin)/2 and W= rmax-rmin, 
L=(Rn)^2/(8R+11W) 
and by similar manipulation,  the maximum inductance is achieved at W=8R/11. 

Whether or not these maximum L values are in some way "Optimal" depends on 
your  goals and definitions, but it is a way of defining and then maximizing 
some value. The voltage difference per turn of wire (or inch of height) may 
make this solution impractical. The validity of the formula may even be 
suspect at short coil heights (lengths), but it is a way of proceeding. 

Matt D. 

"If you don't know where you're going how do you know you haven't already 
gotten there?" 
                                                                              
-attributed to Bob Dylan