Re: Helical cap
> Also, I believe your result adds the plate lengths, for 16 feet total.
> This would be correct if both plates were conducting the full current load
> from one end to the other. However, when you think about how a plate
> capacitor charges, with charge being distributed evenly over the surface of
> the plate, the last inch of the plate on the end away from the tap will pass
> nearly zero ccurrent as it charges, because no charge needs to pass
> through that point to get to points on down the line. The first inch
> after the tap, however, would pass nearly all of the current it would
> take to charge the cap. If this reasoning is correct, the current along
> the length of the plate would decrease in linear fashion from capacitor
> input current at the tap end to zero at the far end. Since the plates
> are, as you surmised, arranged so that one tap is at the center and the
> other is on the outside, the sum of the two plates' current-paths should
> add to resemble the length of the plates when stacked, for inductance
> calculations, and be uniform over the length of the plates, if their
> widths are uniform. Also, since I was rolling foil and PE, my coil width
> very low, about 1/4 inch total when rolled tightly.
> Dia=7.75 in.
> Circumference=24.3 In.
> Turns= 3.95 calculated, 3.75 actual.
Yes, I get it. I made a foolish error in the calculation, I dropped a
value of pi (. . . duh).
One way to look at the current distribution is as individual vectors at
every point on the plate.
The direction of the vectors is always oriented to the discharge tab
from every point. And the
magnitude of the vectors is inversely proportional to the distance from
the discharge tab.
This is much easier to see in an unrolled or plate type cap. The net
current vector is just
the average of all the vectors and is therefore equal to the vector at
the geometric center of the
plate. And the average current path is from this point to the discharge
But the plates are discharging in opposite directions, with opposite
therefore their vectors ADD. This results in a net (equivalent) current
path equal to the
distance between the two tabs; i.e. a flat spiral in your case, and a
helical cone if the tabs
are on top-outside, bottom-inside.
Of course, this only matters if the cap thickness is non-negligible;
otherwise it behaves as
a cylindrical coil essentially.
Nice work Reinhard, when I get a chance, I'm going to try to reduce
those sines and
cosines to x/r and y/r values and see if the equations collapse a bit
Wells' cap/coil is tapped on opposite ends of the plates, but at the
same level; making
h = 0.
> Original Poster: "Reinhard Walter Buchner"
> Hello Wells, Bryan, Malcom, all
> Now, I AM thoroughly confused. I thought Wells was looking
> for the CONICAL inductance equation, which would be:
> (please use courier font to view picture)
> -- o / o
> | o / o
> | o N turns / o
> o Z / o
> h o / o /
> o / o /
> | o / o / Angle = X
> | o \ o /
> -- o o ------------
> | w | R |
> |<-- W -->|
> Center | Line
> Z = Coil Width (hypotenuse length)
> X = Angle of Cone
> h = Z*sin(X) Effective vertical Height
> w = Z*cos(X) Effective horizontal Width
> W = R + w/2 Average horizontal Radius
> L1 = W^2*N^2/(9*W+10*h) (Vertical Inductance Component)
> L2 = W^2*N^2/(8*W+11*w) (Horizontal Inductance Component)
> L = SQRT[(L1*Sin(X))^2 + (L2*cos(X))^2]
> And for the guys, who work it backwards (like me), here is
> the equation that gives the number of turns for a specific
> inductance value. Very special thanks go to JIM MONTE, who
> re-wired the equation for me (I had tried to re-wire it, but it only
> resulted in "garbage in, garbage out") (:o))
> N = sqrt(L) /(W * ((sin(X)/(9*W+10*h))^2 + (cos(X)/(8*W+11*w))^2 )^0.25)
> >> still trying to find the actual equation. It seems to me
> >> that somebody has it posted on thier web site.
> >> Bryan Kaufman
> >I think it should be (r^2 * n^2)/(8a + 10h + 11r) where r = mean
> Hope this helped and was what your were looking for.
> Coiler greets from germany,