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Author Topic: HHO - Joe Cell - Implosion Engine - New Ideas  (Read 248 times)

Offline Martin Levac

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HHO - Joe Cell - Implosion Engine - New Ideas
« on: October 13, 2016, 06:51:43 AM »
Hi, first post, just registered to write it. I'm not an expert, I'm just some guy. I have no hands-on experience, I just read and watch stuff and put things together in my mind to explain things to myself. So, don't take any of this too seriously.

It's about HHO, Joe Cell, implosion engine, you guys know all about those things, so I don't need to re-iterate anything. I'll just jump straight to my new ideas so maybe you guys can get some progress going or something.

1. Premise - We're dealing with a type of water called "EZ water" with molecular structure H3O2, check Gerald Pollack and his work for more details
2. Premise - The gas is not HHO, it's HHHO, or more appropriately H3 + O, or some other arrangement of these same 4 atoms
3. Premise - The inter-conversion from liquid-to-gas and vice versa is probably not equal, we start with a different molecule than we end up with

OK, so the overall sequence is like this:

We start with a total of:
3x H2O (liquid)

Then, from first stage electrolysis (actually, self-induction or self-ionization, however you want to call it), transduce these into:
H3O2 + H3O (both liquid)

Then, from second stage electrolysis:
H3O (liquid) -> (H3 + O) (gas)

The H3O2 is not electrolyzed, it remains in the liquid water and forms a lattice matrix which holds a charge, which can be measured directly. This lattice matrix grows and shrinks according to input energy, and/or discharge. Bubbles of (H3 + O) gas form, surrounded by a lattice matrix of H3O2, which forms the bubble surface. H3O2 is not exactly liquid, it's more like a gel. It's negatively charged, so it's more appropriate to call it H3O2-. H3O is positively charged, so it's more appropriate to call it H3O+, therefore it's more appropriate to call the gas of these same atoms (H3 + O +) or (HHHO+).

The reverse conversion from "combustion", or "implosion", is like this:

We start with a total of:
6x H + 2x O, or 2x (H3 + O), or (HHHHHHOO) (gas)

Then combine from -> into:
(H3 + O) -> (H2O + H) (some liquid, some gas)

2x (H3 + O) -> (2x H2O) + (2x H)

There's 2x free H, so we draw 1x O from somewhere, maybe from atmosphere, or from H3O2 on the bubble surface, collapsing the bubble as we do so. If from atmosphere, we end up with more water than we started with, and maintain existing H3O2, which just falls back into the water as the bubble collapses. If from the bubble, bubble collapses, but charge is correspondingly reduced because transferred to 1x H2O. Anyways, we end up with:

3x H2O + 1x H3O (because -1x O from 1x H3O2)

3x H2O (because -1x O from atmosphere)

I don't think it really matters for now if it's either-or above, it's just to show they're both possible. Now on to power generation in a standard internal combustion engine converted to run on HHO (or more appropriately HHHO gas).

1. Premise - It's a function of pressure differential between top of piston and underneath piston, to produce a net pressure to push piston, regardless of whether it uses conventional combustion or implosion
2. Premise - Implosion from conversion from gas-to-liquid produces work by creating vacuum top of piston, then atmospheric pressure underneath piston pushes piston upward (where normally combustion pushes piston downward)
3. Premise - Implosion is many times more efficient than combustion because of the pressure differential between atmospheric and vacuum (which is effectively zero), for example:

1,000 psi vs 15 psi = 67 power coefficient
15 psi vs 0 psi = infinite power coefficient

Yes, I know it's a bit absurd to describe it this way but I don't know the math and I looked but couldn't find the equation that would illustrate this pressure differential even for a standard internal combustion engine. It's like they don't even know about it, yet it's obvious it's acutely important to determine actual engine efficiency. Do a search for "ping pong ball vacuum launcher" to get an idea of what we're dealing with here. Experimenters get insane muzzle speeds (upwards of about 1,800 mph) with just ~80 psi behind the ball, and a vacuum in front of the ball, launched from a ridiculously short tube. The problem of piston/crank/connecting-rod stress with implosion compared to combustion is a null argument, these things are already stressed tremendously in a combustion engine, an implosion engine would actually generate less stress on these components. Paradoxically, it appears that an implosion engine produces much more output power than a combustion engine at the same RPM, go figure, I don't get it either, but I'm betting it's all about this pressure differential.  So, to you math geeks, start calculating to get that all-important efficiency/power coefficient between the two engine designs.

One important aspect is the apparent need for intake and exhaust, or lack thereof, in an implosion engine. If the conversion from gas-to-liquid draws that extra O from the atmosphere, then we need an intake. But, if it needs only a fraction of it compared to a combustion engine, then we should consider a much smaller intake manifold for better throttle control. Otherwise, it's like full-on or full-off, just with a very slight twist of the throttle with a standard relatively large volume/flow intake manifold, and basically that's what's been reported by the various experimenters I've read about. Same with exhaust, we need some, but really not that much, cuz we only need a fraction of the exhaust capacity compared to a combustion engine, but then it shouldn't make much of a difference because it's just the exhaust. However, it should be possible to take advantage of exhaust timing to increase implosion efficiency. In fact, exhaust valve should open much earlier than with combustion engine, because there's no actual combustion stroke but the exhaust valve is still closed so piston fights vacuum from implosion stroke as it moves downward during what is normally the combustion stroke. Like this:

Standard valve timings for combustion engine, when used with implosion cycle:

Stroke 1: Intake open, piston moves down, air drawn in
Stroke 2: Both valves closed, piston moves up, implosion, power stroke (instead of compression stroke)
Stroke 3: Both valves closed, piston moves down, vacuum fights piston motion (instead of combustion stroke)
Stroke 4: Exhaust open, piston moves up, exhaust stroke

Modified valve timings for implosion cycle:

Model 1, 4 strokes:

S1: Intake valve open, piston moves down, air drawn in
S2: Intake closed, piston moves up, implosion, power stroke
S3: Either valve remains open, piston moves down, no power loss, but wasted stroke
S4: Either valve remains open, piston moves up, normally exhaust stroke, now extra wasted stroke

Model 2, 2 strokes:

S1: Intake and exhaust valves open near/after 0 TDC, piston moves down, air drawn in (or out, doesn't really matter so long as there's no fighting vacuum, as we'll see)
S1/2: At some point in-between S1 and S2 near/after 0 BDC, all valves closed to start S2
S2: All valves closed near/after 0 BDC, implosion, power stroke

In this Model 2, intake/exhaust timing should be different to maintain throttle control on the intake, otherwise twisting the throttle has little effect because exhaust effectively acts as an intake. Or, add throttle body on exhaust in conjunction with intake throttle body for full throttle control. Note how with both models there's no compression stroke, and with Model 2 there's a power stroke each revolution instead of every 2 revolutions.

Typically, a standard 4 stroke combustion engine works better with more than 1 piston, because of the atmospheric pressure under the piston. If there's at least 2 pistons that work in opposition, one creates a vacuum in the crankcase volume, while the other pushes against this same crankcase volume. If there's just 1 piston, it's always fighting crankcase volume at higher pressure than if there was a second piston to create a vacuum. In fact, that's precisely how a standard 2 stroke engine draws its air through the crankcase and up the combustion chamber, by using the piston/crankcase as an air pump. So, that's something to consider because we're using that crankcase pressure as the source of power in an implosion engine.

I guess it's possible to get a rough Model 2 by using existing cam profiles, but with a 1:1 timing ratio instead of existing 1:2 timing ratio. The idea would be to keep valves closed from around 90 BTDC and 0 TDC, and open at all other points, with a tendency to open something at pretty much exactly 0 TDC.

Anyways, those are my ideas, take it or leave it, up to you.

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