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Old 06-10-2008, 04:14 PM   #13 (permalink)
UNISA JECS
 

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How did you get 12800???

"[ 6969 x 288 / (2500rpms x 50 / 256) / 6 cylinders = TP ]


So 2007072 / 12800 = 156.8 - and since this is the total amount of fuel needed for the whole engine, we divide it by the number of cylinders (156.8 / 6cyl). That gives us our TP Value of 26dec (or 1Ahex) . How significant is this??? If we know the MAF voltage and the rpm we can calculate the TP and find the exact points being accessed on our Timing and Fuel Maps! So lining up the TP value of 26dec (1Ahex) and the rpm of 2500 we can find the corresponding block (or block of 4 points if it does not fall exactly on 1 point) on the maps under those engine conditions. With a data logger like a Conzult we can quickly find and adjust or fix points on the maps. Even without data logging we can still find points using a good voltage tester and remembering rpms."
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