Author Topic: Stanley A Meyer Gas Production Problem  (Read 1229 times)

0 Members and 1 Guest are viewing this topic.

Offline Login to see usernames

  • Moderator
  • Hero member
  • ****
  • Posts: 789
Stanley A Meyer Gas Production Problem
« on: June 28, 2017, 17:50:24 pm »
Stan Meyer says you have to ask the right question. So let's determine how much gas is produced by the vertical cluster array tube in
the various demonstrations.
The approach is to determine the amount of hho gas needed to cause a specific change in pressure. The moving pressure gauge is seen in
the videos, so possibly by seeing a change in pressure in a closed vessel, the amount of hho gas needed to cause such a changem ight be able
to be calculated.

Step 1

What is the volume of air in the vertical cluster array tube above the water level.?

By looking at the available images, there's a couple parts to the calculation:

1.  The volume of air in a cylinder portion of the volume
 2. The volume of air in the cone shaped portion of the volume
3. The volume of the air in the pipes to the pressure gauge

The sum of the above volumes should represent the volume of gas above the water level

Formula for volume of cylinder   Volume (v) = area base(b) times
Height (h)

So now to get some measurements

1.What is the inside diameter of  vertical cluster array  demonstration tube?

 Dynodon had some information on this  (See attachment)

Inside diameter of demo cell = 4.75 inches

The area of the base is 1/2 D  squared. Times pi
 therefore:
The base of the cylinder of gas is  2.75 squared times pi = 23.76 square inches.
This is the of the base of the cylinder that will be used in the calculations.

------------
It struck me that the height of the cylinder that we are calculating is very small
perhaps only a few inches. SO A SMALL AMOUNT OF GAS WOULD CAUSE A
LARGE INCREASE IN PRESSURE, If the volume of the air was lets say a gallon
the same amount of HHO  introduced would cause a much smaller increase in
pressure.  So if one wanted to show a high pressure of HHO the amount of air above
the column of water can be made small

It also occurred to me that if the pressure increased to 15 psi  which is about  1 atm 14.7 lbs per square inch
a double amount of air would double the pressure   

If air was being introduced and the pressure doubled in one minute then the volume of air
being introduced/ minute would equal the volume of air that was in the vessel to begin with

so lets say for example it taes 15 seconds to go from 0 to 15 psi  , then4 times the volume of
air in the cell initially if being produced   so if we can determine the time to reach 14,7 pounds
then that is the amount of air produced

If Stan flushed the cell of air by leaving the gas valve open at the start of HHO gas production and then subsequently shut it, then the math is
considerably simplified.  If the volume of space above the water level is determined and if we assume standard 14,7 pounds per square inch
then how long does it take to go from 0 psi to 14.7, then  the  volume of HHO that is produced in that amount of time is equal to the
volume that is equal to the space above the water

so no lets go back to the video in the environmental tape and dealership tapes that have views of the gauge see how long it takes to produce  a given increase in pressure
pounds/square inch  The application of Boyle's Law would indicate that a doubling of the pressure with a constant volume means that twice the mass
of gas in the same volume will double the pressure. So if we go from 1 ATM to 2 ATM the amount of HHO  produced  means that a vooumme of hho equal to the void was produced
 
1 Dealership Tape

Pressure in psi 
11     to 14.75 
  time elapsed in sec
35 sec

3.75 change in psi / sec  = .1071

therefore  14.7 / .1071 =  137 sec/??
to generate  gas to fill  void ?


2 Environmental  Tape full

ok so let compare that to the environmental tape gasproduction rate  rate

Pressure psi              time in sec


For an estimate of the height a comparison between
Diameter and the distance from the water to the
Bottom of the conical  cavity should suffice
Several frames of the video were selected to compare the height of the air column
from  top surface of the water to the bottom of the conical top cap

Then volume of air in the cyclinder part of the cavity
Is calculable

The estimate the volume of the air in the conical
Portion of the cavity requires another formula

Formula for.volune of cone

V=. Pi times r squared times(. H divided by 3)

Now in this case the base of the cone is a little smaller
And a different height needs to.  Be determined

Lastly the volume of gas up until the shutoff valve needs to determined

Once the three major  volumes are calculated they are totaled and converted to liters
This represents the volume of the air cavity above the water

The temperature of the air is assumed to be 70 degrees   Fahrenheit
Air pressure is.   Assumed to be. 1 ATM Standard.
« Last Edit: February 09, 2021, 23:13:15 pm by jim miller »

Online Login to see usernames

  • Administrator
  • Hero member
  • ****
  • Posts: 4529
    • water structure and science
Re: Stan Meyer Gas Production Problem
« Reply #1 on: June 28, 2017, 19:45:07 pm »
My goodness.
Thats really back in time for 10 years, Jim.
Ill guess you and Max are not progressing.
You still think electrolysis will do the job?

I ran two engines on HHO and i learned that the type of isotopes in the gas are the reall topic. Not the amount of it.

Why do you think Herman Anderson and Horvath are now hot topic?

Good luck

Online Login to see usernames

  • Member
  • **
  • Posts: 280
Re: Stan Meyer Gas Production Problem
« Reply #2 on: June 28, 2017, 20:21:55 pm »
In my opinion, the tubular cell with alternator is only the polarization process, amp restricted.
But by the way is always good to start from the earlier Meyer devices.
If we start from the resonant cavity without knowing how it REALLY works, better to give up.
I think that is time to put the cards on the table.
People who have unreleased info or access to them, like Jim and Max Miller, Tony Woodside and the other guy who visited the Meyer state at same time that Dynodon did, better to release that, instead of throwing "crumbs" and act like they are helping.
If we want to do something, we need to get together and work like a team, respecting each other and not hidding anything from the others.

Offline Login to see usernames

  • Jr. member
  • *
  • Posts: 39
Re: Stan Meyer Gas Production Problem
« Reply #3 on: June 29, 2017, 08:01:38 am »
Quote
People who have unreleased info or access to them, like Jim and Max Miller, Tony Woodside and the other guy who visited the Meyer state at same time that Dynodon did, better to release that, instead of throwing "crumbs" and act like they are helping.

They also troll around under different usernames.  They have tried everything they can to bait those of us who know stuff. That square VIC thing making all kinds of hho just cracks me up, they don't even know what it is!.. Hey guys lets run 220v to the square vic and tell everyone we figured it out. hahahaha.  Then watching them all in fight with each other over nothing.  Oh and lets not forget that drunken Max and Ed video proclaiming they figured it out. Then we have the radar guy, he is full of it also, coming up with his con to keep that site going.  I wish I had investors stocking my shop with equipment and gadgets.

Anything left of the Meyer estate or any video's do not reveal what Stan was doing. This guy had national security stuff on him over the true technology. You can dig one of them up in the estate photo's.

If your trying to calculate gas and pressure on an insignificant demo cell, your grasping at straws. You can not comprehend what was being done or how it was being done.

Quote
I ran two engines on HHO and i learned that the type of isotopes in the gas are the real topic. Not the amount of it.

Very True.

As for the patents they are "good for electrolysis" in Stan's own words, as given to me by someone who was with him for 4 yrs.