Re: 1/4 wave theory/cite the variance?

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

Original poster: "Terry Fritz" <twftesla-at-qwest-dot-net>

Hi Mike,

So we have 882 turns, 8.125 inches diameter, 30 inches long.

We need to know the inductance.  Using:

L=N^2 x R^2 / (9 x R + 10 x H)

L = inductance of coil in microhenrys (µH)
N = number of turns
R = radius of coil in inches (Measure from the center of the coil to the middle
of the wire.)
H = height of coil in inches

We get:

882^2 x 4.0625^2 / (9 x 4.0625 + 10 x 30) = 38.147mH

Now to E-Tesla6.  I had to guess at some of the numbers but it will be pretty
close.

Thinking....

 E-TESLA  V6.11C  November 21, 2000 
  
 Grid Size = 300.000000
 Scale Multiplier = 1.000000
 Terminated Coil
 Ceiling Distance = 100.000000
 Wall Distance = 50.000000
 Primary Height (Inner Turn) = 10.000000
 Primary Inner Diameter = 12.000000
 Primary Outer Diameter = 24.000000
 Primary Height (Outer Turn) = 10.000000
 Strike Rail Height = 0.000000
 Strike Rail Diameter = 0.000000
 Secondary Base Height = 10.000000
 Secondary Diameter = 8.125000
 Secondary Length = 30.000000
 Secondary Inductance (mH) = 38.146999
 Terminal-1 Height (center) = 50.000000
 Terminal-1 Cord Diameter = 8.000000
 Terminal-1 C-C Diameter (toroid only)= 14.400000
 Terminal-2 Height (center) = 50.000000
 Terminal-2 Cord Diameter = 0.000000
 Terminal-2 C-C Diameter (toroid only)= 0.000000
 Top Voltage = 1.000000
 Pass     Ccalc (pF)          Change            Fcalc (Hz)
  1000   C=2.67114       Err%=267014.312500       F=498587.47
  2000   C=8.13491       Err%=204.547852       F=285702.22
  3000   C=13.84733       Err%=70.221054       F=218981.27
  4000   C=18.26176       Err%=31.879305       F=190685.92
  5000   C=21.49665       Err%=17.714010       F=175753.73
...
 56000   C=34.16609       Err%=0.001828       F=139409.48
 57000   C=34.16650       Err%=0.001219       F=139408.63
 58000   C=34.16683       Err%=0.000965       F=139407.95
 59000   C=34.16703       Err%=0.000569       F=139407.56

Fo will be 139.4kHz

Cheers,

        Terry





At 07:16 PM 2/16/2002 -0500, you wrote: 
>
> Ok, quick fun for all you math whizzes: 
>    
>   I am going to wind a coil with the following specs: 
>
> Coil Diam.     8.125" 
> winding length     30.0" 
> wire gauge    20 
> assumed turns per inch   29 
> Toroid   8.0" x 22.44" 
>
> what is the  "LC"  resonant freq?   what is the 1/4 lambda freq? 
> (if you wanna get picky,  a precise conversion factor for that is  F =
> 245892.764/wire length = F in kHz) 
> Mike 
>
>>
>> Paul, Terry, et al, 
>>
>> Perhaps your modern view and Tesla's view are both right?  The arguments 
>> being brought up in favor of an adjustment to quarter wave are based on 
>> tightly wound coils with self-capacitance and tightly coupled inductance 
>> between the wires. 
>>
>> If you'll look carefully at all the pictures in CSN, you will see that all 
>> of Tesla's coils are space wound, thus eliminating the effects you're 
>> mentioning.  Also, Tesla utilizes space winding on his large flat spiral 
>> secondary.  This should be a significant consideration when discussing 
>> quarter wave in relation to Tesla's work. 
>>
>> Dave 
>