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if I were increased 5 points in fuel over a certain rom range what would the a/f be on a dyno??

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if I were increased 5 points in fuel over a certain rom range what would the a/f be on a dyno??

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If you were at 13.0 : 1 and added 5% on a 4 injector bike (no upper injectors) it should bring you to about 12.2 : 1

if I were increased 5 points in fuel over a certain rom range what would the a/f be on a dyno??

How did I get this you may ask?

A good buddy of mine tought me something actually recently that when tuning something using Lambda numbers instead of air fuel seems to be WAYYYY easier to tune bikes.

if you dyno a bike and say its making your peak power at 13.0:1 which is .88 Lambda and your rich at say 12.6:1 = (.86 lambda) you subtract 2% which is taking .86 and subtracting .02 (2%) you should be at .88 lambda (13.0:1)

Make sense?

For expression as an air/fuel ratio you have to flip this upside down

New air/fuel ratio = Old air/fuel ratio x (Old PC correction factor / New PC correction factor)

The correction factor is 1 + whatever the number in the cell is divided by 100

e.g. if you have "5" in a particular cell then the correction factor is 1.05, and if you have "-10" in a particular factor then the correction factor is 0.9

If you had air fuel ratio 12.0 originally with no power commander, and you apply "-10" in that cell, then effectively the old correction factor was 1, so the new air/fuel ratio will be 12.0 x (1.0 / 0.9) = 13.3...

All of this is making an assumption that the fuel injectors have a delivery rate that is linear with injector duration (it should be very close unless you have a restriction in the system somewhere) and that your air/fuel ratio meter is displaying accurately and you are not going outside of the range in which it displays correctly (which is not guaranteed ...)

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DIY A/F meter is that possible, bcause those thing are quite expensive.....

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For expression as an air/fuel ratio you have to flip this upside down

New air/fuel ratio = Old air/fuel ratio x (Old PC correction factor / New PC correction factor)

The correction factor is 1 + whatever the number in the cell is divided by 100

e.g. if you have "5" in a particular cell then the correction factor is 1.05, and if you have "-10" in a particular factor then the correction factor is 0.9

If you had air fuel ratio 12.0 originally with no power commander, and you apply "-10" in that cell, then effectively the old correction factor was 1, so the new air/fuel ratio will be 12.0 x (1.0 / 0.9) = 13.3...

All of this is making an assumption that the fuel injectors have a delivery rate that is linear with injector duration (it should be very close unless you have a restriction in the system somewhere) and that your air/fuel ratio meter is displaying accurately and you are not going outside of the range in which it displays correctly (which is not guaranteed ...)

if you go by the lambda method which what Garth said it should be 13.5 and not 13.3

Btw what is the difference between 4 injector and 8 injector set ups would be? Guessing with 8 injectors -10% would act like -20% on 4 injector set up?

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