Recent Posts

Pages: « 1 2 3 4 5 6 7 8 9 10 »
81
General Discussion / Re: S1R9A9M9 18HP Briggs riding mower engine ran on water( 2008)
« Last post by Login to see usernames on June 09, 2024, 05:36:39 am »
Currently hooking up the capacitor70 toroid pulse transformer circuit in conjunction with isolated DC to AC to DC power converter, so as can discharge 6 amps through spark plug using the stepped up higher, high voltage . There are 3 spark gaps total. The high volts is to fire through the outer section of N.P. filter capacitor+ protective Varistor , and the 2 diodes in the bridge rectifier. The Cap70 circuit is to allow NO series diodes protective bank that loses 21 volts at 30kv, and would greatly increase the power input from battery. Statements from the Internet said others had used this Cap70 circuit successfully. The (2) toroid cores FT240-31 are isolated transformers at each end of circuit. Now Nathren, S1R9A9M9, had earlier used thin copper washer under spark plugs with wire attached, so as can have a direct return line back to power source, without going through the engine block. I have a spring steel squeeze 3/4" hose clamp opened with vice grips and #12 scraped magnet wire underneath, so as there is a direct line bottom of plug going to power supply ground. It's like the high volts pulse goes through certain section of P.S. secondary side and when at plug top firing ignition, the amps also follows. The P.S. first toroid has high volts 30kv cable on opposite sides of toroid core. There are several types of isolation here. Way back in time, experimenters were shorting out their Inverter boxes. They need the old style ones that have no circuit board on the secondary side. Mine is one better, and is the cheaper,  China model open board 500Watt PLAIN INVERTER. The existing SMALL transformer is removed. The original inductance of 1/2 primary was 31uh on inductance test meter. At the 20khz frequency preset, you just add about 4 turns+4 turns centertap windings for primary of Toroid transformer. The end filter capacitor is the low ESR type as non polarized, polypropylene 20uf  At using high frequency, you get to use very low count of hand windings.The lower the primary windings, the better is the ratio, for less turns on secondary side needed.  Final output should be about 12-18 volts DC, (instead of 35vDC to overcome voltage loss). Current through the spark plug should be 6-7 amps., at about 24kv. The adjustments of final high volts come from the section of PF series capacitors + the gap distance of 1st adjustable spark gap., and gap of engine plug. The L + C together is also the frequency of primary firing. Use computer calculator for series capacitors formula, as about 600pf total with L inductance of primary.  When testing spark plug under compression, use Snap-on MT-2700 neon bulbs high volts voltmeter. The proper higher volts will suffice for PLASMA PATH to allow the amps to pass through electrolyte at plug gap. During the 4 cycle waste spark, volts  may be higher as way less moisture at that time.  With spark in air, plug is 3kv per mm. Protection of bridge diodes is using Varistor 800 volts at spark discharge across PS, and also spark gap across PS set about .01" for 750 volts , as less than the diodes 1kv. The Briggs engine under compression would be 2-5. times higher volts applied. I personally prefer the  isolated power supply put together, rather than the pre-made  Boost DC converter as non isolated, easy to mount, that requires a 2nd battery, because the grounds are the same at both ends. I hope you can use some of this data that relates to the duplication of S1R9A9M9 lawn mower18HP engine that ran on water to Hydrogen during summer of 2008 YouTube videos.
83
Sebosfato / Re: My new approach
« Last post by Login to see usernames on June 08, 2024, 23:50:59 pm »
So the missing reading is All six in parallel so I can find the capacitor value to be used split between them

84
Sebosfato / Re: My new approach
« Last post by Login to see usernames on June 08, 2024, 23:36:47 pm »

If I have a set of coils with 39,5mh each and when I connect them in series their inductance is 148,6mh what is the mutual inductance and what would be the inductance when I connect them in series with fields opposing ? Their resistance is 18 ohms each and it’s air core

To find the mutual inductance (\( M \)) of the coils and the inductance when connected in series with fields opposing, we can use the following equations.

### Inductance with Fields in Series (Aiding):

When two inductors are connected in series with their fields aiding, the total inductance (\( L_{aiding} \)) is given by:
\[ L_{aiding} = L_1 + L_2 + 2M \]
where:
- \( L_1 \) and \( L_2 \) are the inductances of the individual coils.
- \( M \) is the mutual inductance.

Given:
\[ L_{aiding} = 148.6 \text{ mH} \]
\[ L_1 = L_2 = 39.5 \text{ mH} \]

Plugging these into the formula, we get:
\[ 148.6 \text{ mH} = 39.5 \text{ mH} + 39.5 \text{ mH} + 2M \]
\[ 148.6 \text{ mH} = 79 \text{ mH} + 2M \]
\[ 2M = 148.6 \text{ mH} - 79 \text{ mH} \]
\[ 2M = 69.6 \text{ mH} \]
\[ M = \frac{69.6 \text{ mH}}{2} \]
\[ M = 34.8 \text{ mH} \]

So, the mutual inductance (\( M \)) is 34.8 mH.

### Inductance with Fields Opposing:

When the fields oppose, the total inductance (\( L_{opposing} \)) is given by:
\[ L_{opposing} = L_1 + L_2 - 2M \]

Using the values we have:
\[ L_{opposing} = 39.5 \text{ mH} + 39.5 \text{ mH} - 2 \times 34.8 \text{ mH} \]
\[ L_{opposing} = 79 \text{ mH} - 69.6 \text{ mH} \]
\[ L_{opposing} = 9.4 \text{ mH} \]

Therefore, the inductance when the coils are connected in series with their fields opposing is 9.4 mH.

Each inductance has a value of 86mh and in parallel the 3 have 39,5 what is the mutual inductance please?

Given the following data:

1. Each coil has an inductance \( L = 86 \text{ mH} \).
2. When three identical coils are connected in parallel, the total inductance is \( L_{parallel3} = 39.5 \text{ mH} \).

We need to determine the mutual inductance \( M \) between the coils.

### Inductors in Parallel:

For three identical inductors in parallel, considering mutual inductance \( M \), the formula for the equivalent inductance \( L_{parallel3} \) is:
\[ \frac{1}{L_{parallel3}} = \frac{1}{L + M} + \frac{1}{L + M} + \frac{1}{L + M} \]

Given:
\[ L = 86 \text{ mH} \]
\[ L_{parallel3} = 39.5 \text{ mH} \]

### Mutual Inductance with Parallel Configuration:

\[ \frac{1}{L_{parallel3}} = \frac{3}{L + M} \]

So:
\[ L_{parallel3} = \frac{L + M}{3} \]

Given \( L_{parallel3} = 39.5 \text{ mH} \):

\[ 39.5 = \frac{86 + 2M}{3} \]

Solving for \( M \):

\[ 3 \times 39.5 = 86 + 2M \]

\[ 118.5 = 86 + 2M \]

\[ 118.5 - 86 = 2M \]

\[ 32.5 = 2M \]

\[ M = \frac{32.5}{2} \]

\[ M = 16.25 \text{ mH} \]

Therefore, the mutual inductance \( M \) is 16.25 mH.


The coupling coefficient (\( k \)) is a measure of how effectively two inductors are magnetically coupled. It is defined as:

\[ k = \frac{M}{\sqrt{L_1 \cdot L_2}} \]

Given that each coil has an inductance \( L_1 = L_2 = 86 \text{ mH} \), and the mutual inductance \( M \) has been calculated as \( 16.25 \text{ mH} \):

\[ k = \frac{16.25}{\sqrt{86 \cdot 86}} \]

First, calculate the product of \( L_1 \) and \( L_2 \):

\[ L_1 \cdot L_2 = 86 \cdot 86 = 7396 \text{ mH}^2 \]

Next, take the square root of the product:

\[ \sqrt{7396} = 86 \text{ mH} \]

Now, calculate the coupling coefficient \( k \):

\[ k = \frac{16.25}{86} \]

\[ k \approx 0.189 \]

Therefore, the coupling coefficient \( k \) in this case is approximately 0.189.


And in the case of having them in series 148mh and in parallel 9,4mh what is the coupling being the mutual inductance 34,8

To find the coupling coefficient (\( k \)) given the series and parallel inductance values, we will use the provided mutual inductance (\( M \)).

### Given:
- Series inductance: \( L_{series} = 148 \text{ mH} \)
- Parallel inductance: \( L_{parallel} = 9.4 \text{ mH} \)
- Mutual inductance: \( M = 34.8 \text{ mH} \)

### Inductance of Each Coil:
From the series inductance formula, we know:
\[ L_{series} = L_1 + L_2 + 2M \]

Given \( L_1 = L_2 = L \) (assuming identical coils):
\[ 148 = L + L + 2 \times 34.8 \]
\[ 148 = 2L + 69.6 \]
\[ 2L = 148 - 69.6 \]
\[ 2L = 78.4 \]
\[ L = 39.2 \text{ mH} \]

So, the inductance of each coil is \( L = 39.2 \text{ mH} \).

### Coupling Coefficient:
The coupling coefficient (\( k \)) is given by:
\[ k = \frac{M}{\sqrt{L_1 \cdot L_2}} \]

Since \( L_1 = L_2 = L \):
\[ k = \frac{34.8}{\sqrt{39.2 \cdot 39.2}} \]
\[ k = \frac{34.8}{39.2} \]
\[ k = 0.887 \]

Therefore, the coupling coefficient (\( k \)) is 0.887.

So the result was exact as I designed to be and expected

The coils have a low coupling between them allowing to lower the inductance when connect them in parallel

The coupling between them side by side is high as I expected 88% this couples the resonance together

Is possible that the screw inside holding it may affect some the readings but is just to get the idea..







85
Sebosfato / Re: My new approach
« Last post by Login to see usernames on June 08, 2024, 22:18:06 pm »
I forgot to think first how to connect the cell so now I’m wondering how to make it…
The inner electrode is easy but the outer I will need to make some design

I should have made it before all and leave it connected to the copper foil and maybe even apply resin to avoid leak

The problem of the electrodes is that they need to be only where the water generates voltage otherwise they act as a load to the voltage generated

Of course we still get amp restriction but much lower than would be possible

So is important that they are only within the high field

This is absolutely important otherwise you may see very little effect on voltage and amps


86
Sebosfato / Re: My new approach
« Last post by Login to see usernames on June 08, 2024, 16:03:05 pm »
All this info came from carefully analyzing the circuit and what should happen!

What intrigues me is that no one before did such simple basic electric calculations

Meyer intention to mislead people into thinking the Vic was the only thing really worked

87
Sebosfato / Re: My new approach
« Last post by Login to see usernames on June 08, 2024, 15:39:48 pm »
Water resistance as seeing from the circuit will will rise as much as 20 or more times

So if we start with 2kohm is possible that it will increase up to 400kohm that is an open circuit for the input transformer basically

In a manner making the secondaries in series as I described cuts the possibility of tuning to the resonant peak

This along with reducing the q by loading the coils and adjusting the phase and current to water makes easier to tune only to the cavity specific frequency and not to the resonance itself for ex


88
Sebosfato / Re: My new approach
« Last post by Login to see usernames on June 08, 2024, 08:44:06 am »
I decided to make the secondaries divided in two sections

One well coupled one isolated by the gap of the core this make its leakage great

This makes this secondary behave as a very light load for the core and actually its voltage drops a lot when any current flow and it get even reversed when the other secondary is in series with it

In the middle of it goes the feedback

So now it’s clear how the phase will zero when current drops to zero while will be showing almost opposite phase when out of tune

It looks almost as if the dot is in the wrong size of the coil if read on oscilloscope

So this secondary may have another 435 v or even higher but will become zero volts or even negative because of phase inversion I just mention

The resonant charging coils are on the Vic

The resonant chokes are another assembly

So there is a total of 8 coils and the gap between the secondaries control the frequency to an extent and also the imaginary inductance in series with the secondaries

The secondary I mention over the resonant coils will give a voltage that is 90 degrees off from its current so it not really a good way of driving the cell actually but it helps reduce its Q out of tune and that will help finding the tune

to make this voltage in phase a capacitor is needed and maybe a inductor to fine tune it before rectification

Adding a series capacitor put the current in phase an inductor would sum another 90 degrees making it 180 that is fine

In that way you make the phase have a fixed relation at that frequency range

So this is the tune to the dielectric properties of water because this capacitor is actually water and it’s where we want the voltage to get in phase with the current of the coils

So resuming the tune to the dielectric water is done by making the phase match of voltage reaching water vs the current on resonance 

Of course all that being driven at the cavity f







89
Sebosfato / Re: My new approach
« Last post by Login to see usernames on June 08, 2024, 06:13:10 am »
Building capacitors is a lot fun

I think I’m going to create a capacitor and coil business of it since I already build the machine and know how to do it

Is incredible out a piece of plástic to get amps to flow thru it

Another way to make it even more non inductive is to connect the leads in the middle of the foil

Or almost the middle if rolled

90
Sebosfato / Re: My new approach
« Last post by Login to see usernames on June 08, 2024, 00:09:14 am »
Now I ask to you what happens if you make the capacitor coiled on a transformer?

It will have a voltage induced to it together with the voltage because of the charge in it

Well meyer talk about an electron bounce phenomenon

I could bet has to do with that

A capacitor may be connected  with non inductive orientation

The sinonimia of bounce is jump


?



Pages: « 1 2 3 4 5 6 7 8 9 10 »