URGENT! WATER AS FUEL DISCOVERY FOR EVERYONE TO SHARE

[djzissis]

Hi everybody..!!

A couple of hours I came back from my home-village.
I have good and semi-good news.!!
The good news is that the plasma circuit is now fully installed on "Unity"
It was encouraging that the exhaust fumes, running on gasoline, had no smell at all.

Check at the video I captured. (really sorry for the shaking camera)

http://www.youtube.com/watch?v=piR4fqOVDDg


Now the other news.... ( more important )
After this video,
I found a plastic bottle, and filled with tap water. Took out the carburetor, and put some water into the intake.
THE ENGINE STARTED WITH THE FIRST PULL
and after 25-30 revs it died. (Just like Capacitor's 70 video.) The combustion chamber was full of water and the engine stuck.... 
Immediately i took out the spark-plug, and emptied the combustion chamber and ran it again with gasoline until any water boiled and the engine was clean.
Everything OK now.
The bad thing is that I didn't have enough time to do it again and record it to video.
I promise next time to make it run, record it, and maybe I will use the carburetor to merge the water intake and not stuck it again. 

That's all for now.!!
More progress to come....
I want to thank you all from my entire heart, for all the work we are doing together here at overunity.
85% of the job is already done from all of us.!!


Thanks again,
All the best to you all,
Zissis

[demios]

@djzissis

Excellent results buddy! Maybe you overflowed the engine with water, maybe you should have used the I.V. dripping system, like gmeast did. That would have significantly reduced the amount of water in the engine. Either way, experimenting is more worth than any theory. That's why your work is precious to all of us. Thank you again!

@Alpeko

This is how I see things. If I'm wrong, please correct me. Thank you again

FORMULA (taken from http://www.electronics2000.co.uk/calc/calcchrg.php):

Charge: Q = CV where C is the capacitance in Farads, V is the voltage across the capacitor in Volts and Q is the charge measured in coulombs (C)

I calculated that my dump cap has 880uC (micro Coulombs) of charge 400V x 2.2uF = 880uC
And we know that 1 Ampere is 1 Coulomb per second (1A=1Cx1s), so I calculated that at 5000 rpm the spark fires 10.000 times, which means it fires 10.000/60=167 times every second. The total charge released in one second at 5000 rpm is 167x880uC=0.147A. So this is a sort of peak power for most engines. Barely anyone uses higher rpm. Anyhow, we have Amperage and Voltage (remember, my dump cap is 400V), so by multiplying the numbers we get 400V x 0.147A = 58.8W.

So, we need one resistor (commonly known as R1, here on forum) that has a power rating of approx 50-60W.
500 Ohms is chosen, I believe to limit the current to less than 1A >>> 450V (capacitor bank) / 500 Ohm = 0.9A (max current going to dump cap - C1).

Again, this is how I interpret the diagram gmeast shared with us. The final verdict will come from those who know more than I do about electronics.

[gotoluc]

Great job Zissis

Looking forward to your next video

Thanks for sharing.

Luc

[djzissis]

Hello demios and Luc..!!

Until May I am full of homework for school because it's the last year at school and I must study as much as I can.
This is the reason I don't have enough time for tests.

Thank you very much for your encouragement.
The best are to come....

Next progress videos and photos are going to be in better quality, and posted as files. (not youtube)
Anyway, thanks again,
Keep up the good work..!!

Regards,
Zissis!!

[gmeast]

@djzissis

Excellent results buddy! Maybe you overflowed the engine with water, maybe you should have used the I.V. dripping system, like gmeast did. That would have significantly reduced the amount of water in the engine. Either way, experimenting is more worth than any theory. That's why your work is precious to all of us. Thank you again!

@Alpeko

This is how I see things. If I'm wrong, please correct me. Thank you again

FORMULA (taken from http://www.electronics2000.co.uk/calc/calcchrg.php):

Charge: Q = CV where C is the capacitance in Farads, V is the voltage across the capacitor in Volts and Q is the charge measured in coulombs (C)

I calculated that my dump cap has 880uC (micro Coulombs) of charge 400V x 2.2uF = 880uC
And we know that 1 Ampere is 1 Coulomb per second (1A=1Cx1s), so I calculated that at 5000 rpm the spark fires 10.000 times, which means it fires 10.000/60=167 times every second. The total charge released in one second at 5000 rpm is 167x880uC=0.147A. So this is a sort of peak power for most engines. Barely anyone uses higher rpm. Anyhow, we have Amperage and Voltage (remember, my dump cap is 400V), so by multiplying the numbers we get 400V x 0.147A = 58.8W.

So, we need one resistor (commonly known as R1, here on forum) that has a power rating of approx 50-60W.
500 Ohms is chosen, I believe to limit the current to less than 1A >>> 450V (capacitor bank) / 500 Ohm = 0.9A (max current going to dump cap - C1).

Again, this is how I interpret the diagram gmeast shared with us. The final verdict will come from those who know more than I do about electronics.

Hi demios,

That is precisely correct at every level.  For a 4-cylinder engine, a 2uF cap at 5,000 RPM and 4 time constants for the RC calculation you only need 750 Ohms and about 27 watts (400 VDC)... but why skimp ... if I want to put a 3 uF cap in it then R1 = 500 ohm won't starve C1 and charge it short.

Nice job,

Greg