Electrical igniter for gas engines A keystone to understanding by Magluvin

[stprue]

@delboy and definetly Tito

I think it's time to release the ou info you claim to have or at the very least show us some of your ou setups/rigs!!!

[Goat]

@ all

Assuming the interrupt switch is of negligible draw.

How do we use HV #1 and HV #2 ? 

Is it one wire to the receiver or is it 2 wire to the next stage of the circuit?

How do we NOT kill the dipole on this part of the circuit?

[ramset]

Sooo.........

The "class" awaits master Teecho -> <- [aka Teets]

A handsome fellow ,  very white teeth         ^      [I think he uses "whitener"]

Hopefully the "class" will behave while Master Teecho gets his lesson together.
I hear he can be "lazy"[he said so],I personally won't care if he Snores during class
as long as we get something to "study".   
Chet

[forest]

Summary as I see this topic :
Everybody think that Tesla igniter is for creating HV without consuming much current from primary.
Am I correct ? Yep it's small step forward.
Any other concepts ?
I think it's amplification of power we should see here.Hmm...
We should see something like : put small 1.5 battery as power source and after a while get kW from the output.

[delboy]

@goat and @forest

This is Tesla design. Look at picture. SW1 and SW2 are switches for operating receiver. SW1 can be any Fast recovary diode fo example BY-329 and SW2 can be Thyristor or Triac which will periodicaly discharge condenser to receiver. Capacity on receiver should be bigger, for example 100uF/400V.
Two spheres are secondary capacity neccesary for resonant effect in secondary! But design of secondary should be like Tesla said, length of secondary quater of wave, spiral form to secure minimal parasitic capacity and resistance low as posible for securing high Q factor.
For example , if you want 300 V/m on 50kHz then length of secundary should be 1500m and voltage on secondary should be 450kV. If you take that Q factor of secondary is Q=500 then primary have to give to secondary only 900V. If you are using 12V baterry then you must have 900/12=75 Q factor of input inductivity Taking losses in count you will take Q=100
Problem here is to hit the resonance in secondary because of very small bandwidth = 50kHz/500 =100Hz (for example). I must be inside this bandwidth 50000+/-100 Hz if I want at least half of available power!
On one receiver you would pick up 300V on 1 meter length and charge condenser, then discharge it on receiver. If frequency of discharge is 100Hz then power delivered to receiver will be P=900W There can be more than 2 receivers You calculate is it OU if I input only 1kW for operating circuit controller!