The map informed delta 1.6 mjt

[tranky]

can you please post the one that I used.. so much for the info..

[giuseppe89]

thanks tranky me you already sent your excel sheet of calculation and it's exactly what I'm "building" my to get a clearer view of the advances I'm first calculating the duration for the map ori from 1500 rpm on, and the iq from 5mm3 to rise so as to know in which window do I need to stay for the iq increase, and do so nice work, clean...for the little time I have and for the many calculations to be performed are still in the middle of work but I'll make it!!

[tranky]

The formula to remember here so it is easier to make calculations of comparison,premising that I write literally "square root"instead of srqt in order not to confuse any person,as well as the symbol " x "to indicate a multiplication instead of the symbol "*".

k=(the square root of the prex rail) x tinj/Qi
Oi=(square root of the prex rail) x tinj/k
Tinj=(Qi x k) : square root of the prex rail.

Taking the example of a edc15c6 tells me that the curve ori corresponding to 1350bar that the duration is Us1230 to 70mm3,obviously the injectors are original.

then,
k=(square root of 1350) x 1230/70=645,61551..
Qi=(square root of 1350) x 1230/645,61551..
Tinj=(70 x 645,61551...) : (the square root of 1350bar)=1230

Now,to equal the k-factor, and rail pressure,is that I use the formula,whether I face proportion, the result does not change.I have always a linear increase.

Formula :
T inj= (85 x 645,61551..) : square root of 1350=1493,..(In the formula, we see that the only information that varies is the amount of the mm3 of fuel,then mathematically will be a proportion as a (85 :70) x 1230=1493,..as it is.

Munro I tried the formulas as you defined, but does not return anything...
then.. I try to calculate K (I assume you meant advances with a K) then I:
QI static = 100mm3
so here the calculation only in the column of the new IQ of the bp changed without the other iq at the moment and I calculate the time and the rail pressure
Tinj on prx.rail set in the table:
1000bar = 1394.8
1150bar = 1311.2
1300bar = 1220.4
1450bar = 1126
1500bar = 1062.53
1600bar = 1030.8
1700bar = 995
1800bar = 959.2
then having this can I:
prx.rail on rpm
1500rpm = 1114
1750rpm = 1229
2000rpm = 1343
2250rpm = 1457
2500rpm = 1571
3000rpm = 1686
3500rpm->5000rpm = 1800 (here the value is static to 1800bar by 3500rpm at the end of the scale)

to 1500bar, k = (rq 1500) x (1062.53/100)
38.72 x 10.62 = K = 411.41 <-- seems to count as a value, no?
at this point xo if I divide instead of multiply, I get:
38.72 : 10.62 = K = 3.64 <-- 1500rpm ori 1.8 on 70mm3 I would say that seems correct..
It is the wrong formula or something is missing to calculate the advance payments ?

[giuseppe89]

Tranky the formula posted by munro used to calculate the new Tiny is if you want to increase it,and if you must shorten it to a raising of the rail. Here's an example :if we want to calculate the K-factor curve of 1600 bar take from the map time tiny for 70 mm3,in my map corresponds to 923 us if I'm not mistaken. From here we take the rd of 1600 x 923 /70 and we find the k that corresponds to 527,4285714286 . Now, if we want to raise the iq to 100 mm3, and let the prex rail 1600 bar we multiply k x the new iq 100 and divide it by the square root of 1600 bar... so we 1.318 us for the 100 mm3. Then, if we want to bring the prex rail 1800 bar in the last part of the calculation we replace 1800 to 1600 and get 1.243 us... with this calculation we get the tiny which we then have to convert to angles of rotation of the motor shaft and have the duration in had regard to whom then subtract the degrees of advance to know where it ends l injection

[tranky]

Tranky the formula posted by munro used to calculate the new Tiny is if you want to increase it,and if you must shorten it to a raising of the rail. Here's an example :if we want to calculate the K-factor curve of 1600 bar take from the map time tiny for 70 mm3,in my map corresponds to 923 us if I'm not mistaken. From here we take the rd of 1600 x 923 /70 and we find the k that corresponds to 527,4285714286 . Now, if we want to raise the iq to 100 mm3, and let the prex rail 1600 bar we multiply k x the new iq 100 and divide it by the square root of 1600 bar... so we 1.318 us for the 100 mm3. Then, if we want to bring the prex rail 1800 bar in the last part of the calculation we replace 1800 to 1600 and get 1.243 us... with this calculation we get the tiny which we then have to convert to angles of rotation of the motor shaft and have the duration in had regard to whom then subtract the degrees of advance to know where it ends l injection

and luckily you were behind me.. I'm the caper of Joseph!!! you that you're someone!!!
Thanks for the clarification, missed the part of converting from degrees in fact.. I thought it was included in the formula that, in fact, was not returned..
thanks friend

[giuseppe89]

I'm glad to helped you be that way!and anyway, up until now, I've always taught you many times and then you're still in advantage hahhaha it is always a pleasure to help!

[munro]

Bravo Joseph...

[giuseppe89]

Thanks munro!