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Author Topic: Stanley Meyer Explained  (Read 450968 times)

ramset

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Re: Stanley Meyer Explained
« Reply #45 on: March 20, 2009, 02:22:17 PM »
Loner
Injectors can handle this[diesel] direct combustion chamber injection
I don't know that we should get the splitting hairs efficiency that user spewing claims, but some where in between just dumping HHO In the intake and direct injection we could probably run a generator on less HH0 with this method [allowing ambient air to purge the exhaust refill the cylinder ,,just inject what we need]

 I know this is not the finale intent here [brute force][ But it could yield quick results ,as the other tech becomes better understood

Chet
PS and yes Loner I know you ran a genset on HH0 years ago
God speed  Loner
PPS something on brute force and HHo from user Ironheads research group, running a generator
http://www.youtube.com/watch?v=VteoVsK93Mg
« Last Edit: March 20, 2009, 03:43:18 PM by ramset »

Digits

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Re: Stanley Meyer Explained
« Reply #46 on: March 20, 2009, 02:44:09 PM »
Hi you all

I have read this and to me it makes sense What H2O power says.
I also was trying to move my efforts to the gas gun instead of the water fuel cell and it was great coming on this morning and finding this plethora of info thanks H2O

You have answered my question on the higher energy fields cause that was a stumbling block but now it does make sense.
I was going to build a voltage pump my one friend build allot of these,it's a piece of mesh wire that stripes electrons off nice big arches comes of these,so I reckoned the water must be in a gaseous form before I can utilize this like the gas gun, so when I read the comments this morning it all made sense thanks

Does some one have plans for a setup so that we can start work?
In the following week I will draft my experiments and start working on them I will happily share when finished

H2O you said Stan made 7l/min to run his buggy, I achieved 12l/min with the Boyce setup but as I can get from what you have posted is that is is not the quantity so much but the how much you excite the gas (an higher energy state) so this is the new goal right?
And can some one help me with the LED or Lazar thing, is this a normal red led or a real Lazar we are talking about.

Thanks guys enjoyed this,I'll wait for the replies

Digits

Mark69

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Re: Stanley Meyer Explained
« Reply #47 on: March 20, 2009, 03:20:31 PM »
Also, I m looking for the patent for the Gas Processor; reading H2o's post on the other site state that the gas processor is the key.  Can someone tell me, post a link, or pdf of the patent for the gas processor?  Oh and the VIC?  (still reading the posts on the other site)

Thanks!
Mark
« Last Edit: March 20, 2009, 03:48:51 PM by Mark69 »

ramset

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Re: Stanley Meyer Explained
« Reply #48 on: March 20, 2009, 05:37:09 PM »

                  HHO as A CAR FUEL COMPARED TO GASOLINE
         
                        From user  Creativity    Very good reading

                ,and reason for a gas processor or as Stan Meyer said an

                                            """"" ACCELERATOR ''''''''''

Let us calculate how much an engine needs to run.Goal is to calculate an amount of air and fuel used.I will calculate the air needs in liters as we are interested here in volumes to compare it later to the volumes of H2 or H2,O2 gases needed to run an engine.
First of all who still has no knowledge of AFR (air fuel ratio) please read:

http://en.wikipedia.org/wiki/Air-fuel_ratio

1)Now our good running engine uses around 14,7 g of air per 1g of gasoline (heptan,octane mixture),that comes out of AFR calculation of perfect burning.I will make an example how to calculate the AFR for a given fuel:

C7H16 + 11O2 ->7C02 + 8H20

heptane has a moll mass = 7*12+16*1=100 [g/mol]
we are using air as an oxidant so we have also nitrogen present and it is 79 % of volume of air.It will give us around 41 molecules of N2 in this situation.So our air mole mass=11*16*2 + 41*14*2=1.500 [g/mol]
 now 1.500/100 = 15 this is our AFR for heptane.

2)just how much is this 14,7 g of air in volume?

Air has density of 1,2 g/liter.Volume is then =12,25 liter

3)let assume that our engine uses 10l of gasoline to drive our car 100km/h in 1hour.So an engine of around 2-2,5 liters of displacement,rated with around 150HP( i assume ,but i feel it can be also quite good calculated out of engine efficiency and gasoline caloric value).
How much air volume will it use?

Density of gasoline 737,22 [g/liter] shows us that we used 7.372g of it for our trip.
So air used=14,7*7.372=108.368 g.
It is then =108.368/1,2= 90.307 liters of air.

4)How much of air was used per minute?

90.307/60=1505 liters

5)how much fuel per minute was used?

7.372/60=123 g

6) and when the fuel was evaporated how much space it occupied?

Gasoline expands around 220 times in volume when evaporated(comes from comparison of density ratio of liquid gasoline with the gaseous form) , so we used 2200liters of evaporated gasoline.
It is around 2200/60=37 [liters/minute].

OK let as got some attention to our big question of using H2,O2.From a SAE technical paper i have a graphic describing the effects of adding H2 into the internal combustion engine runned on CNG.Graphic speaks about the possibility of further shift of lean mixture AFR limit when H2 was feed to the fuel.Unfortunately i can`t put this graphic here because it is copy righted.But i think copyright allows me to say what is the effect observed Smiley Lean limit of pure gasoline is around AFR =25:1 according to this source.Adding of 8% of H2 into the combustion shifts AFR to lean limit of 26 :1.So it is a shift of 4%.Further addition of around 20% of H2 shifts AFR to 27,5:1 resulting in relative 9% shift.So it makes some sence to add H2 to the engine.

This effect was described for addition of pure H2,however i believe if H2,O2 was used effect could be slightly different.Result of electrolysis in form of a H2,O2 mix has 33% of volume made of O2.I could think of two situations:
a)extra oxygen would lean the charge even more and counteract the H2 effect
b)oxygen would assist the burning of the fuel by adding extra heat to burn even leaner charge.

Now let me try to calculate the H2 burning in air.I will try to make analogical calculations as for heptane,just to see how much of the H2 would we need to run our can in a situation as above.

2H2 + O2 ->2H2O

AFR (2*16+3,76*2*14)/(2*2)=34,25 :1

it is 34,25/1,2=28,5 liters of air per 1g of H2.

This time we need to know how much energy was released in gasoline engine,because we want to make our hydrogen car to run at least as well as our gasoline car.In the first example we used 7.372g of gasoline for our trip,this would translate to 327MJ produced.I won`t take into consideration the efficiency of the engine,i just assume it won`t change substantially under pure H2.I make this assumption to make first approximation of amount of H2 required to run an engine.Later we can add efficiency to the calculation.
327MJ translates to 327[MJ]/130[MJ/Kg]= 2.517 g of hydrogen used to make the same work.
Now volumes:

0,09 g/liter is the density of H2 gas.

0,09*2.517=27.967 liters of H2 gas

As in our previous example all of it was used in 1h time for our trip.It is 27.967/60 = 466 liters of gaseous H2 per minute!.
Norm amount to produce from on-board electrolysis alone,no?But that is not the end now.Just compare the numbers:

Our car on gasoline used 90.307 liters of air + 10 liters of not yet evaporated fuel.This all went for the trip.I use 10l of gasoline because when it was feed to the engine most of its evaporation process took place inside of the cylinder.
Our car would need to use 27.967 liters of H2 + 2.517*28,5 liters of air in total to burn it for the trip.IT is 99.701/90.317= 1,1 times more volume of gases would have to go through the engine(so volumetric efficiency would have to improve).We simply suck not enough oxygen from the air to burn our H2.Our engine would have to be supercharged or rpm would have to go higher for the same power demand.

Situation would be different if we used 2H2,O2 mixture(as from electrolyser).In that case AFR would be a bit more tricky to calculate.We have here an extra oxygen feed and exactly as much as we need,In this case oxygen from the air is not needed at all and a result will be that we run lean mixture.Problem gets not easier when we feed more H2,O2.In that case we get more power and hot running engine,feeding even more will result in big explosions and only way i can see would be inject water to cool down the engine and slow down the burning.Maybe dumping of some oxygen from the electrolyser outside of the engine could help,just to achieve a stoichiometric mixture for H2 without adding more of H2 and ruining the engine.

Coming back to our car trip  Grin.We still want to produce the same energy as in pure gasoline or H2 case.This time we have extra oxygen available so volumetric efficiency won`t suffer,no supercharging or higher revving will be needed.We can then stick to gasoline base case  90.317 liters of charge introduced to the engine during the trip.The reason is that we can manipulate the amount of oxygen supplied to burn H2.We have an excess of oxygen so we will dump all the oxygen that could cause lean burn.We use as in a pure H2 case  27.967 liters of H2.The  90.317 - 27.967=62.350 liters will have to contain all the oxygen we need.Only air won`t support enough oxygen (as shown in pure H2 usage case),supplying of all the O2 from electrolyse,next to ordinary air will give us too much oxygen.Wear have to find how much of the O2 we need to burn stoichiometric(the best).
 coming back to :

2H2 + O2 ->2H2O

What we see here is we use 1 volume of O2 for every 2 volumes of H2 to have a nice burn.We have 27.967 liters of H2 so we need the half of it in volume of O2.

1-{[62.350 -27.967/2])/62.350}= 0,224

With above formule i calculated what part of the total air and O2 supplied has to be the oxygen.In easy words we need upgrade air to have 22,4% of oxygen.The rest of the oxygen we don`t need anymore.With this i offer you ,the one who had a long road to read through all this calculations  Grin  Cool my respect Smiley.As a reward i can bring u step closer to the solution of an on-board hydrogen production.

                            BELOW IS MR PASZKOWKI"S  THEORY[on going]


I am an author of an idea as follows:
As u see oxygen release from the electrolysis is not what we want.We can make a small amount of O2 (2,5% of air volume sucked to the engine) but it costs us a big penalty of energy used to release this oxygen from OH bond.Sure we can Strip it and get this one extra hydrogen,that is what you all do in electrolysis.I say it is not the way.Use this energy to strip another water molecule of the only one Ht.As well as combustion engine,fuel cell can also use oxygen from the air.What i see, is usage of the low energy electrolysis with minimal OH bond braking rate.This cell will produce almost only hydrogen that can be feed to the fuelcell to produce electricity with air oxygen.This system has a chance to become overunity.The secret of the not stripping of the OH bond will stay for a moment here.I have to finalise my long research on it first.

all the best,
Bartosz Paszkowski


ramset

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Re: Stanley Meyer Explained
« Reply #49 on: March 20, 2009, 05:59:10 PM »

Mark69

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Re: Stanley Meyer Explained
« Reply #50 on: March 20, 2009, 06:01:17 PM »
Thanks Chet

h20power

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Re: Stanley Meyer Explained
« Reply #51 on: March 20, 2009, 06:44:52 PM »
What Stanley Meyer says makes a lot of sense for he says 4 or more electrons have to be stripped off of the oxygen atoms. When I made my energy content comparisons at the 4th energy level is when the reaction over takes the reaction of gasoline. And if you keep on going to oxygen's 8th level it is 44k-108k barrels of oil, depending on the grade of the gasolin. So, Stanley Meyer had to have made these same calculations I have, he just never put any of this in his patents. So, 2 H-H bonds formed 872 kJ/mol + 84078.0 kJ/mol = 84,950 kJ/mol(1BTU/1.055 kJ)(1mol/.018L)(3.78541L/gal)= 16,933,679.8BTUs/gal. 1US gallon gasoline = 115,000 Btu's, so it takes 147.249 gallons of gasoline(19.2gal of gasoline/1 barrel of oil) so 7.67 barrels of oil needed for one gallon of water's energy content without recycling any of the water. Now Meyer must have added the fact that you can put the water back into the tank, so my guess is water is being consumed in some small amount with this reaction.

As for the VIC transformer all of the coils must hit resonance at the same time, that is to say each bobbin cavity must hit resonance with the primary coil at the same time. There is more than one way to do this, what I noticed Stanley Meyer doing was to alter the wire sizes too match inductances of every bobbin coil to that of the primary coil. Thus when the primary coil hits resonance so does every bobbin cavity of both the secondary, and dual chokes.  This way is most effeicnt use of the VIC transformer, noteing that there are two types of transformers and the use of each type is another story.

The energy content calculations are answering what the scientist have been asking about Stanley Meyer's technology for all of these years. It is a ionic reaction that is short lived, that is saying we only have a short period of time to conduct the react with the unstable oxygen atoms, and that time, as far as I can tell, is 0.74 seconds. If you calculate the gas speeds inside of the intake system you will find that that is plenty of time to get the unstable oxygen atoms into the combustion chamber and run the ionic reaction which Stanley Meyer calls Thermo Explosive Energy or The Hydrogen Fracturing Process. One thing that is truly needed in working with Stanley Meyer patents is a definition of terms of Stanley Meyers' words. For example it took me well over a year to figure out what the "bidirectional wrap" ment in the VIC injectors primary coil. Now I ask everyone reading this to tell me what a Bidirectional wrap is, after you all give it a shot I will tell you what it means. Stanley Meyer's patents are full of words like this, and it took me a long time to learn to talk as he did.

But I will say this all of us must strive for understanding, for it is very important, for if I just dump a set of plans online what happens when something breaks or malfuctions? You have to make the whole darn thing all over again for you don't know anything about the technology enough to fix it, or come running to me and I will drop what I am doing and find you wherever you are on the globe and trouble shoot it for you. You must have the understanding of how the system works so you can fix it if something goes wrong. That is yet another reason why I set it up as an engineering project for it fources you to understand the technology before you build it.

On the LEDs you should try and match oxygens wavelengths here is a site that give at least 73 of them, but note there are more: http://astro.u-strasbg.fr/~koppen/discharge/
Trying to figure out which ones will work best, your on your own for I am right in the game with you on that one.

Hope this helps everyone to see why I did things this way.
h2opower.

h20power

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Re: Stanley Meyer Explained
« Reply #52 on: March 20, 2009, 06:59:55 PM »
Today marks the 11th year since the death of Stanley Meyer. I for one wish to tell his family members his dream of using water as an fuel source is still alive and well, and "We The People" are doing what we can to make his dream a reality for the whole world.

This should be a site holiday at the least, don't you think?

Best Regards,
h2opower.

Outlawstc

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Re: Stanley Meyer Explained
« Reply #53 on: March 20, 2009, 08:33:00 PM »
what do you think of this h2Opower? is this the bibirectional you speak of?

h20power

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Re: Stanley Meyer Explained
« Reply #54 on: March 20, 2009, 10:05:28 PM »
Here is an example of something that is Bi-Directionaly wraped. I was on the upset side when I finnally found out what that word meant to say the least. http://www.energeticforum.com/renewable-energy/3217-stanley-meyer-explained-7.html#post49552

All it is, is a fancy word for something that is Cross Wrapped. After all this time, huh? Just wrap 1st layer 45 degrees one way and 2nd layer 45 degrees the other way.

For a lot of people this will be the first time you have ever understood what that word meant, why couldn't Stanley Meyer just say it was "Cross Wrapped" your guess is good as mine, but at least now you know. Oh, and the wire is parallel bonded bifilar wire that he used so he could make the coils the same as Tesla's in a way.

Just for everyones information,
h2opower.

PS I am really not liking the small upload this site has.
« Last Edit: March 21, 2009, 01:27:49 AM by h20power »

ramset

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Re: Stanley Meyer Explained
« Reply #55 on: March 21, 2009, 12:56:30 AM »
H20 power
The boss knows how to do this

I will try to summon him

Can't Loner get a smiley face?

Chet

PS Hopefully there will be a Stan Meyer Day [soon]

God speed




h20power

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Re: Stanley Meyer Explained
« Reply #56 on: March 21, 2009, 01:39:27 AM »
Yeah I guess since he was close ;D ;D ;D ;D ;) @Loner

But you will find a lot of things like that in the patent, words that Stanley Meyer used that don't have the same meaning as the rest of the world or are words that have long been forgoten to most people.

Not sure if this was for patent protection or if that was really the way he talked.

h2opower.

kinesisfilms

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Re: Stanley Meyer Explained
« Reply #57 on: March 21, 2009, 07:11:06 AM »
yes that is the correct term for bidirectional wrap.....it should be seen as obvious but i can see where people get confused......it's not just single words that confuse people......it's multiple confusing words in one confusing statement.....hopefully our friend john will pull through in his final winds.....hopefully this technology is about to see the light of day.

ramset

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Re: Stanley Meyer Explained
« Reply #58 on: March 21, 2009, 03:20:48 PM »
Loner
same problem here
I pm'd Stephan yesterday about this problem of H20 not being able to post pics and info here

Chet

kinesisfilms

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Re: Stanley Meyer Explained
« Reply #59 on: March 21, 2009, 06:32:09 PM »
a caduceus coil is very unique but does not play a part in stanley meyers work.....scalar waves do not reflecting on metal surfaces (orany surface really) therefore defeating the prupose of the electrodes in a water fuel capcitor.....we need standard normal waves.