Running rich?
Running rich? I am just getting familiar with scanning and tuning, so bare with me. I have been trying to get the Wideband in HP reading the same as the gauge, and it APPEARS to be pretty close. So here's the scenario;
My wife's C5 has all the bolt on's, headers, vararam, rocker arms etc. and we had gotten a mail order tune last year. The PE values at 4000 rpm and above are set at 1.188, but when I logged a couple of WOT runs, it logged from a low of 11.1 to 11.7 AFR. Does this sound right and I can just make some small adjustments to the PE values for now, or should I look for something else.
I did a few mins. of LTFT logging, and had mostly 1's and 2's, but didn't get any long term values.
NOTE; I am going to try to tune (as I learn) every aspect of the car, but I was just trying to address a "miss" or "popping" that she had at WOT along with some TC issues.
Thanks
My wife's C5 has all the bolt on's, headers, vararam, rocker arms etc. and we had gotten a mail order tune last year. The PE values at 4000 rpm and above are set at 1.188, but when I logged a couple of WOT runs, it logged from a low of 11.1 to 11.7 AFR. Does this sound right and I can just make some small adjustments to the PE values for now, or should I look for something else.
I did a few mins. of LTFT logging, and had mostly 1's and 2's, but didn't get any long term values.
NOTE; I am going to try to tune (as I learn) every aspect of the car, but I was just trying to address a "miss" or "popping" that she had at WOT along with some TC issues.
Thanks
Since your PE value is 1.188 then your AFR should be around 11.9 so 11.7 isn't too far off. To me, around 12.5-13.0 is a good range for AFR under PE and WOT. However, thats were I like it and other people differ. If you want to run a little leaner set the PE value to about 1.130 that should give you an AFR of about 12.76. If you want to know about to get the AFR this is the formula that I use:
(2 - PE Value) * 14.67 = AFR
EX:
(2 - 1.188) * 14.67 = 11.91204
(2 - 1.130) * 14.67 = 12.7629
I got this formula from reading Greg Banish's book. Seems to work pretty well for me but maybe someone else can chime in on this.
(2 - PE Value) * 14.67 = AFR
EX:
(2 - 1.188) * 14.67 = 11.91204
(2 - 1.130) * 14.67 = 12.7629
I got this formula from reading Greg Banish's book. Seems to work pretty well for me but maybe someone else can chime in on this.
Great; I'll try that and check it again. She rarely goes WOT for more than a few seconds, and she doesn't spray, so I think around 12.76 should work well for her car.
I was simply dividing the command AFR by the PE value like FROST is showing in his formula, and getting 12.3, so it seemed to be off quite a bit.
I was a little curious of whether or not the vararam was having any affect on the AFR at higher speeds.
I was simply dividing the command AFR by the PE value like FROST is showing in his formula, and getting 12.3, so it seemed to be off quite a bit.
I was a little curious of whether or not the vararam was having any affect on the AFR at higher speeds.
Launching!
Joined: Jan 2002
Posts: 299
Likes: 41
From: Detroit, Murder City
Frost has it right. Lambda*Stoich=AFR and PE (in units of phi, which is 1/lambda) is enrichment ratio. Therefore:(Stoichiometric AFR) / (PE, in phi) = Target AFR ...or...
14.67 / 1.188 = 12.3484848...
14.67 / 1.130 = 12.98230
There is no [(2-x)*anything] function in there.
I came up with that from the book because the PE value is actually Phi, which is the inverse of Lambda. So therefor, if PE is 1.25 then lambda is .75, (says that in the book under GM Tuning) you get the same thing if you take two (2) subtract it from Phi(1.25) and you get .75, take that number and multiply it by Stoich (~14.67) and you get your AFR:
So,
2-1.25(phi) = .75 (lambda)
.75(lambda) * 14.67 (Stoich) = 11.0025 (AFR)
I could be wrong, but it corresponds with the AFR calcs in the book.
So,
2-1.25(phi) = .75 (lambda)
.75(lambda) * 14.67 (Stoich) = 11.0025 (AFR)
I could be wrong, but it corresponds with the AFR calcs in the book.
I came up with that from the book because the PE value is actually Phi, which is the inverse of Lambda. So therefor, if PE is 1.25 then lambda is .75, (says that in the book under GM Tuning) you get the same thing if you take two (2) subtract it from Phi(1.25) and you get .75, take that number and multiply it by Stoich (~14.67) and you get your AFR:
So,
2-1.25(phi) = .75 (lambda)
.75(lambda) * 14.67 (Stoich) = 11.0025 (AFR)
I could be wrong, but it corresponds with the AFR calcs in the book.
So,
2-1.25(phi) = .75 (lambda)
.75(lambda) * 14.67 (Stoich) = 11.0025 (AFR)
I could be wrong, but it corresponds with the AFR calcs in the book.
1.25 in EQ is 11.736 in AFR ...... 14.67 / 1.25 = 11.736
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Joined: Jan 2002
Posts: 299
Likes: 41
From: Detroit, Murder City
Yes, phi is the inverse of lambda. However, that doesn't mean that it is always 2-x. The further you get away from 1.0 the more apparent this becomes. For an extreme (but very possible during cat protection) example of lambda = 0.65:
lam = 0.65
phi = 1 / (0.65) = 1.53846
If stoichiometry is 14.64, that's (0.65)*(14.64) = (14.64) / (1.53846) = 9.516:1 AFR
Using your formula:
2 - (0.65) = 1.35 ...an error of over 18% enrichment!
(14.64) / (1.35) = 10.8444:1 AFR This is a difference of over 1.3 air-fuel ratios!
The 1/x function does not equal 2-x. You must do the math the right way if you want repeatable and correct results. I can assure you that the AFR calcs in my book are all done using the first (correct) method shown here in this post. Hopefully this will help you in your future tuning efforts.
lam = 0.65
phi = 1 / (0.65) = 1.53846
If stoichiometry is 14.64, that's (0.65)*(14.64) = (14.64) / (1.53846) = 9.516:1 AFR
Using your formula:
2 - (0.65) = 1.35 ...an error of over 18% enrichment!
(14.64) / (1.35) = 10.8444:1 AFR This is a difference of over 1.3 air-fuel ratios!
The 1/x function does not equal 2-x. You must do the math the right way if you want repeatable and correct results. I can assure you that the AFR calcs in my book are all done using the first (correct) method shown here in this post. Hopefully this will help you in your future tuning efforts.
I dont mind being wrong, just trying to be innovative and understand a little more. Thanks for the correction, much appreciated.
To Bottle Rocket, sorry for the confusion man, just trying to help.
To Bottle Rocket, sorry for the confusion man, just trying to help.
Launching!
Joined: Jan 2002
Posts: 299
Likes: 41
From: Detroit, Murder City
Don't feel bad. I don't mean to single you out by any means. There's no shame in discovery and learning. We (as a tuning community) used to do a lot of sketchy/scary stuff back in the day just to get some of the "more interesting" combos to run. That was back before we really understood the system like we do today.
Taking the time to learn the right math behind the systems' functions has gone a long way toward making everything easier and more accurate. It also makes it a LOT quicker to tune these things when it counts.
Taking the time to learn the right math behind the systems' functions has gone a long way toward making everything easier and more accurate. It also makes it a LOT quicker to tune these things when it counts.
