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Modeling a magnifier
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From: John H. Couture [SMTP:couturejh-at-worldnet.att-dot-net]
Sent: Tuesday, March 10, 1998 1:25 PM
To: Tesla List
Subject: Re: Modeling a magnifier
At 03:28 PM 3/9/98 -0600, you wrote:
>
>----------
>From: Malcolm Watts [SMTP:MALCOLM-at-directorate.wnp.ac.nz]
>Sent: Sunday, March 08, 1998 7:25 PM
>To: Tesla List
>Subject: Re: Modeling a magnifier
>
>Hi John,
>
>> From: John H. Couture [SMTP:couturejh-at-worldnet.att-dot-net]
>> Sent: Sunday, March 08, 1998 3:16 PM
>> To: Tesla List
>> Subject: Re: Modeling a magnifier
>>
><snip>
>> ------------------------------------------------------------------
>>
>> Note that the true TC uses dampened waves instead of CW and the peak
>> voltage is higher. The peak voltage of a dampened wave can be much greater
>> than a CW wave with the same energy. This could mean a longer spark. Also
>> note the peak voltage does not occur at the same time as the sine wave. JC
>>
>> -------------------------------------------------------------------
>
> Tesla might have had some difficulty agreeing with that first
>sentence. Peak output is basically limited to what the ROC of the
>system will allow no matter how it is driven.
> I think the second sentence is invalid. If a CW oscillator could
>put as much energy into the resonator in as many cycles as set k
>allows a particular Ec to get here, output voltage should be
>identical. Output voltage for a particular resonator is dependent on
>how much electrostatic (vs electromagnetic) energy resides in it at
>any moment in time
> Would you please explain exactly what the third sentence means?
>
>Malcolm
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Malcolm, All -
Why would Tesla have difficulty agreeing with the first sentence? He never
used CW for a TC. What does ROC mean? The peak voltage in the pri/sec coils
is definitely higher using dampened waves compared to a continuous sine wave
of the same frequency and energy.
However, you would be correct as far as charging the sec term is
concerned. The end voltage using either the dampened or sine wave is
according to the equation
J = .5 Cs x Vs^2 or Vs = sqrt(2 x J/Cs) . With J and C unchanged
the V is unchanged.
The second sentence is not invalid. It is obvious that some of the cycles
of the dampened wave are greater in amplitude than a sine wave with the same
frequency and energy.
The third sentence refers to a little known fact of dampened waves. The
peak amplitude of a dampened wave always lags the peak amplitude of a sine
wave of the same frequency. The amount of lag depends on the amount of
dampening. Only at the zero axis are the two waveforms in synchronism. Does
this mean that the rotary gaps should be in synchronism with the Utility
sine wave or the TC dampened wave?
You can learn a lot about dampened waves using a spreadsheet with graphing
capabilities such as Quattro Pro, etc. Just superimpose the two waveforms at
the same frequency on the graph and you will see many differences between
the two waveforms.
John Couture