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Re: An Interesting Problem


From: 	John H. Couture[SMTP:couturejh-at-worldnet.att-dot-net]
Sent: 	Saturday, October 25, 1997 10:39 PM
To: 	Tesla List
Subject: 	Re: An Interesting Problem   

At 09:24 PM 10/16/97 +0000, you wrote:
>
>
>---------- Forwarded message ----------
>Date: Fri, 17 Oct 1997 08:47:50 +1200
>From: Malcolm Watts <MALCOLM-at-directorate.wnp.ac.nz>
>To: Tesla List <tesla-at-pupman-dot-com>
>Subject: Re: An Interesting Problem  
>
----------------------------------------------------------------

  Malcolm -

  I have delayed an answer to this post until I received the waveform photos
that you sent to Richard Hull. Thank you for making these photos available
to coilers in the USA. I am referring below to photo #3 which Richard said
was the waveform you used to calculate the transfer efficiency.

  The transfer efficiency you refer to cannot be measured as you suggest.
This transfer of energy from the TC primary coil to the secondary coil is by
induction and is always a complete transfer (100%). That is the amount of
electrical energy leaving the primary coil is always completely transferred
to the secondary coil. None of this energy is lost in the transfer. The only
factor affecting this TC transfer is the coupling. Reducing the coupling
will increase the time to transfer the energy from the TC primary coil to
the secondary coil.

  However, do not confuse this with the losses that occur independently in
the primary and secondary coils. These losses cause the reduction of the
amplitude in the waveform to which you refer. These losses have to do with
the log decrement, Q factor, and effective resistance (R) in the following
equations.

    Q = Xl/R    Log Dec = ln(a1/a3)     Q = 3.1459 /Log Dec

  If the reduction of the waveform amplitude is about 80% as you suggest the
log dec is .223 and the Q factor is 14. The resistances (R) would depend
upon the Xl of the primary and secondary coils. A similar example is shown
in my Tesla Coil Notebook.

  Also, do not confuse the above with the amount of energy that leaves the
primary coil. This can be anything from zero to 100%. Regardless of the
amount of energy in the primary this amount is what is transferred to the
secondary coil with no losses. In this way the law of conservation of energy
is satisfied. Many coilers and people working in electronics have trouble
understanding this interesting application of the energy law. Why is it that
this cannot be called perpetual motion ?

  There is much more to Tesla coil operation.

  Comments welcomed.

  John Couture

---------------------------------------------------------------

>Hello John,
>
>> From: "John H. Couture" <couturejh-at-worldnet.att-dot-net>
>> To: Tesla List <tesla-at-pupman-dot-com>
>> Subject: Re: An Interesting Problem  
>> 
>> At 03:31 AM 10/15/97 +0000, you wrote:
>> <SNIP>
>
>> >Around 300 - 350kV based on an estimated transfer efficiency of 70 - 
>> >80%. It won't be far off and I now have a means of measuring this 
>> >efficiency pretty accurately.
>> ------------------------------------------------------
>>   How do you measure the efficiency?
>
>Answer: by comparing secondary energy at the end of second ringup to 
>energy at the end of the first ringup. This is not completely accurate
>and should be done on the primary waveform as the two ringups 
>mentioned don't measure the first ringup loss. In observing the 
>secondary waveform with a scope aerial you can't measure absolutes 
>but you can measure relatives. Since Energy is proportional to V^2 
>you can see how the method works. For example, if V second ringup = 
>1/2V first ringup you can safely say that between the two peaks, 3/4
>of the energy has gone. Since you are interested in the loss over _1_
>transfer, you halve that loss figure to get an approximation 
>(probably more like 2/3). BTW, a coil that loses to this degree is 
>scrapheap material.
>
>Malcolm
>
>