LTCFISHER
Jr. Member
Posts: 71
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« Reply #60 on: November 18, 2008, 01:57:24 pm » |
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Ok Bob,
I'm have'n a bad day.
I have looked all over for your info. on the amount of gas required for my engine.
This is what I have so far: Bore x Bore x Stroke X .7854 X (Number of Cylinders) = Total Volume
ACT: 3.78 x 3.78 x 3.13 x .7854 x 4 = 140.5008 X .016387064 = 2.3 Liter Engine
2.3 / 4 = 0.5750. A little over a 1/2 liter per cylinder per stroke.
Compression Ratio = 9.2:1
Ok now what am I missing? , How much Hyro do I need.
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« Last Edit: November 18, 2008, 02:05:17 pm by LTCFISHER »
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Cowboy
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« Reply #61 on: November 18, 2008, 02:12:05 pm » |
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Mas, you took the long way of doing that...
Pie x Bore x Stroke x no. cyl. = displacement
Curious as to how you came up with your figures though. I'm interested in learning new ways of doing this.
But, displacement still isn't how you need to calculate hydroxy needed. I'm not too sure if there's going to be a hard equation. Too many variables. Total volume, stroke compared to bore, no. cyl., differential gearing, transmission gearing, tire size, power-to-weight ratio... The best I can give is to shoot high and adjust as needed. Your your little guy, try 6 lpm for idle. Get 6 lpm going, then pull the injector fuse.
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Manta
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« Reply #62 on: November 18, 2008, 03:06:44 pm » |
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Wow ! That's so much easier using the Metric system.  Manta |
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Good questions have a sting in the tail.
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Bob
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« Reply #63 on: November 18, 2008, 07:19:17 pm » |
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LTCFisher:
THe reason you cannot find "how I figure how much Hydroxy gas I need" is because there isn't one ... Like Cowboy pointed out...too many variables!
... For a sumplemtary system I figure 2LPM per 1Liter of engine displacement...
for going 100% Hydroxy gas its probly at LEAST double that !...perhaps even tripple
...I don't know yet because I have yet to do it !
...sorry !
...
Bob......
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"Mother always told me "Son the Imposable is only a little bit harder"...and You know ... She WAS RIGHT!"
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LTCFISHER
Jr. Member
Posts: 71
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« Reply #64 on: November 19, 2008, 09:55:33 am » |
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Your right Cowboy I did do it the long way, but I didn't know if any one used pi any more. HA HA!!!
So the guesstamation (new word) is 2LPM to start for every Liter of displacement!
I am going to finish a new cell (tube type) which it should produce an easy 15 to 20 lpm at 13.8 v.dc and 32 amps. targit figures.
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Cowboy
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« Reply #65 on: November 19, 2008, 11:28:15 am » |
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That will be nice production Fisher. Keep us up tp date on it. And yeah, it's Pi, not pie, but pie sounded good at the time. I missed lunch.  |
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Manta
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« Reply #66 on: November 19, 2008, 11:42:36 am » |
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LtCFisher, I was wondering why you didn't just divide your engine capacity by four. Looking forward to seeing the new cell. Manta |
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Good questions have a sting in the tail.
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Manta
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« Reply #67 on: December 06, 2008, 11:49:20 am » |
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LtCFisher,
Just been re-reading this thread. I notice that way back, November 10th, you wrote...
....I see that it is necessary to maintain a constant 13.6 Neg., volts to the anode in the cell. ...
( my emphasis)
As the Anode in a circuit is Positive, and the connector to which you connect the positive lead from your battery automaticly becomes the Anode; why would you want to make this negative ?
Just curious.
Manta
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Good questions have a sting in the tail.
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LTCFISHER
Jr. Member
Posts: 71
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« Reply #68 on: December 06, 2008, 07:27:13 pm » |
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My mistake on that one, Thanks for pointing that out. Senior momment. LOL
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