Stanley Meyer > Stan Meyers system 3

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Dynodon:
to calculate maximum time for injection,devide 120,000 by the rpm,and you get milli seconds.
eg. 120,000/6000 rpm = 20 ms.
I use to do these numbers all the time when I was tuning pulse width for programable computers in cars.
Don

sebosfato:
Nice dynodon

But what is this number? And where it comes from ? Can you explain to us ? thank you

I have an explanation for the difference between your number and mine

I considered the four cycles (1°down intake)  (2°up compression) (3°down explosion) and up for exhaust and theorized you should only inject the gas during the intake

at 3000 rpm you have 1500 complete cycles in a minute, witch = 25 complete cycles per second, if you consider a complete cycle 4 semi-cycles you might find why i found this number

25 complete cycles take 1 sec to occur right so 1sec / (100 quarters of cycle) or (25 * 4 semi cycles)
thats how i estimated 10ms absolute maximum injection time for 3000rpm

Actually I think the timing for the injection should be fix at about 4 ms and that the amount of hydrogen per injection should remain the same, the car should accelerate and de-accelerate by only controlling the amount of air that comes in and or also the exhaust gas re-entering the combustion chamber... the hydrogen will be generated on demand and maintained under min working pressure. If you think about when you mix more exhaust the motor will lower its rpm automatically lowering the consumption of hydrogen as its consume is dictated by number of  injections . Probably meyer regulated the max rpm mixing with the air and the minimum adding also exhaust gases. What do you think??????? HAPPY NEW YEAR!!!

outlawstc:
hmmmm i always figured injection time changes when revolution time changes (rpm) and the changes are made proportionately  with rpm.. having a predetermined variable being horsepower.

seb check my math above once more please.. pretty sure i have it right...   so it is 7.4 ml per cylinder for 50 hp motor amounting to 2.6 liters per hour at 65mph and 2000rpm? and a average hydrocarbon fuel gets around 6.5 liters a hour at 65mph and 2000rpm due to the fact that there is 2.5 more power in water from having a higher quantity of hydrogen per 1:1 by volume.

Dynodon:
sebosfato,that number is 120k ms. (milliseconds)
1 second = 1000 ms
1 minute = 60 seconds
60 seconds x 1000 = 60,000 ms
It takes two revolutions to complete one cycle,so that means two 2 prm = 120 seconds
120 seconds x 1000 ms = 120,000 ms
Don

sebosfato:

--- Quote from: outlawstc on January 03, 2010, 16:18:21 pm ---hmmmm i always figured injection time changes when revolution time changes (rpm) and the changes are made proportionately  with rpm.. having a predetermined variable being horsepower.

seb check my math above once more please.. pretty sure i have it right...   so it is 7.4 ml per cylinder for 50 hp motor amounting to 2.6 liters per hour at 65mph and 2000rpm? and a average hydrocarbon fuel gets around 6.5 liters a hour at 65mph and 2000rpm due to the fact that there is 2.5 more power in water from having a higher quantity of hydrogen per 1:1 by volume.

--- End quote ---

Now you corrected it right? Attention because you still confusing micro liters with milliliters. But now the results seems right! I don't understand why you talk about 2000 rpm? Do you mean for 6 cylinders? still not clear...
however
stan said 7.4 micro liters per injection cycle for a 50hp 4 cylinders running at 3000rpm and 65mph
I believe he used the mixing of the gas to control the motor speed i mean he used more or less exhaust recirculation accordingly with the rpm desired. I say this because he talked about the speed of the explosion of the hydrogen witch would be to great alone. So to reduce the speed or the rpm (MEYER TRANSLATION) he substituted the air in for exhaust gas. why? Because is much easier (KISS) to make a pulse of fix width that changes only in frequency (rpm) than is to create a pulse that changes frequency and also width right ?

If the pressure of the cell is fix and the quenching circuit works the way i proposed it becomes obvious that he stated 7.4ul per injection cycle because for every injection cycle you will inject the same amount of hydrogen. The amount of gas injected will be controlled by the rpm since as you add exhaust gas the RPM will change REDUCE automatically.

He probably had a sensor on the valves that every time the intake valve open the solenoid opens for 4ms injecting the gas  as i said 13,3ml of hydrogen oxygen gas per injection...
than on the intake he probably used a kind of three way switch that has the air input and also the exhaust return... he than just regulated the speed rotating the switch... Isn't it very simple?

All is needed here now is to calculate the pressure needed to be maintained on the cell for having this amount of gas passing inside the quenching circuit in this predetermined time at a velocity greater than as he stated 350cm/second.
here is a link with a calculator i'm studying :
http://www.pipeflowcalculations.com/airflow/index.htm

pressure in the cell is free so you can use it so no need for gas pump.

Right you also understood well the 2,5 times more powerful comparison is (by) volume but or also more probably or precisely (BY) weight...