Recalculation of the columns

Hello all, I wanted to know if c'? a way to recalculate in an exact way,
wanting to introduce a new point B. a new column on the basis of the increments of the original map.
Let me explain:

See Tab. 1 lambda

having logged the car after the closing of the egr, and some other retouching I find a greater deal? of air to the MAF.
The BP. lambda ends at 1200 mg/str. on the sw. ori - I should insert 1300 mg/str. in the last column.
How do I ricalcolarmi all the values in the new column with B. P. 1300?

Might go if the percentage recalculation of the values of b.p. and the respective column? More I would like to know which column should I remove saw that the last 3 are all the same.
Thanks

Hello...I try to help you....given that I climbed all the edc15...then including the lambda map....
So, I guess that your problem is scaling the bp of the intake air...which stops precisely at 1200...well...the map in question we know that ? the map AFR, the cio? returns the relationship between the air fuel...so..***** and you will comprehend the value on the Z-axis not ? so that the division between the intake air (measured in mg) and the diesel (always in mg)...the number that jumps out ? your afr...now, if you set the mg of the air intake to 1,300, knowing the afr (which revenues from the other columns...please refer to one of the 1,200 mg), you know that the values of diesel to enter in the column...
example
1,200 mg AFR16--> avrai75mg of diesel...
Now, by inserting in the bp air 1.300 econservando the afr of 16---->81.25 mg of diesel
I hope I was clear

Response to novices, THANKS!!!

however, I doubt is as ever I have 4 columns the same, the same values? 900 - 1000 - 1100 - 1200

* 2,000 rpm if I have 900:15,71= 57,288

* 2,000 rpm if I have 1.000:15,71= 63,653

* 2,000 rpm if I 1.100:15,71= 70,019

* 2,000 rpm if I 1.200:15,71= 76,384


Which I agree to remove to the new 1300 of the 4? I think that by 1,200, and how values could also leave the same in the table of 1,200.

* 2,000 rpm if I 1.300:15,71= 82,74

After that, redo log, and I see the changes that you say?

On the curve you want to remove I is always that of the 1200;)
ah, I would do also also the curve of the 1,300 equal to the other of the 900...

Well thank you thank you :rolleyes:,
let's say that this is the case ? pretty simple, but if I were to recalculate a column for the entire
how do I do that?
I explain I have the B. P. of the times that make a jump (see attached table) 60 mm3/str to 100 mm3/str have values in the column are different.
If I wanted to insert a B. P. 80 mm3 and a second B. P. 110 removing and then 2 BP with values lower e.g. 31,50 and 41,50 how do I do that?
Here's the thing, it becomes difficult, the tables from what I know' using the linear interpolation, but I don't really know how to do it since I don't have
a calculation of departure on which to elaborate.
Help please

Not c'? a formula or some useful reference to be able to create the new columns?

hello megamind...I don't really understand...xk? I can't see any attachments...however, if you always reported to the lim fumes...? easy..you can remove the bp that you want the formula ? always that! set the afr and knowing the air intake....you calculations for the diesel...

No I am referring to the duration of injection
TABLE: Injector Duration Table 1

bar mm3/str
0.00 0.10 0.80 1.20 2.00 2.50 3.20 5.00 8.00 15.00 23.00 28.00 31.50 35.00 38.00 41.50 46.00 52.00 60.00 100.00
1800.0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1750.0 0 175 175 175 175 175 222 272 307 369 437 494 538 577 610 650 749 905 1094 1899
1600.0 0 175 175 175 175 175 233 280 316 384 457 518 565 608 648 698 811 975 1161 1985
1500.0 0 175 175 175 175 178 241 286 322 395 478 538 583 630 671 734 859 1022 1206 2070
1400.0 0 175 175 175 175 182 251 293 332 403 498 560 604 655 700 770 902 1073 1261 2155
1300.0 0 175 175 175 175 194 262 302 341 418 521 586 634 683 735 825 956 1127 1314 2250
1200.0 0 175 175 175 175 228 274 310 351 436 548 613 662 717 772 880 1021 1184 1391 2334
1000.0 0 175 175 175 196 267 293 329 381 480 604 676 732 800 894 1019 1164 1331 1551 2565
800.0 0 178 178 178 232 290 316 354 423 548 686 773 848 964 1080 1207 1350 1550 1796 3070
600.0 0 197 197 197 296 321 345 412 493 646 810 920 1050 1207 1324 1455 1619 1846 2137 3631
400.0 0 224 224 304 364 397 434 528 636 833 1043 1236 1421 1588 1728 1894 2120 2442 2805 4717
300.0 0 310 310 368 456 502 551 651 781 1017 1300 1545 1730 1970 2154 2355 2645 3000 3460 5855
250.0 0 372 372 428 526 572 624 745 892 1188 1626 1895 2160 2405 2605 2855 3180 3610 4165 6000
150.0 0 465 465 532 646 705 770 916 1114 1530 2121 2534 2867 3207 3451 3768 4167 4714 5357 6000
125.0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0.0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

The table here on the forum to me is the whole offset :mad:
it keeps the rows and columns:confused:

TABLE 16x20 grommet:
Injector Duration Table
x=mm3/str
y=bar
z=µsec.

Quote:

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Originally Written by megamind

The table here on the forum to me is the whole offset :mad:
it keeps the rows and columns:confused:

TABLE 16x20 grommet:
Injector Duration Table
x=mm3/str
y=bar
z=µsec.

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Ah here...want to know how to calculate the times for a IQ different from the one that you find in the bp...well, look at it ? a speech a p? complicated...why? you must take into account the pressure of the rail...I advise you to keep as much as possible and the bp of the original, where you need to go up with the amount? IQ see some of the other ecu's similar to yours, and copy all of the time...but if you have to calculate them....I can go to the formula that I use...explained in another forum (I can do it without breaking the rules?)

But the formula, I think it is covered by a copyright...if you are afraid there are always the pm ;)

hello...here is the formula that I use...Tinj=(IQ*(K)/√Prail

as you can see, the times are related both to the rail pressure that the K-factor....of course the K te the revenues with inverse formula...

Quote:

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Originally Written by mariodarkblue

hello...here is the formula that I use...Tinj=(IQ*(K)/√Prail

as you can see, the times are related both to the rail pressure that the K-factor....of course the K te the revenues with inverse formula...

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Good' but I don't know' what' the factor k, where I go to extrapolate?

the K-factor ? a constant that makes revenue through inverse formula...for example...you're on the curve of the 1,400 bar...you know that for 80mm3 you need 1200us...you calculate the K of the curve...and then you will be? easy to calculate the new tinj given k, the pressure and the iq

I need an example prarico and real, I come back a little complex the explanation..:confused:

Thanks for the tips

Then..? simple and I don't need to see the table...
For example, I have that:
PRAIL 1400bar
IQ 70mm3
Tinj 1240us
K= (√Prail * Tinj)/ IQ
K=662.8
I've calculated the K-factor for the curve ? 662.8...
Now, to calculate the new tinj, user? that K...
Suppose that you enter 80mm3 in bp
Tinj=(IQ*(K)/√Prail = (80*662.8)/sqrt1400=53024/37.41=1417

here you go;)

Quote:

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Originally Written by mariodarkblue

Then..? simple and I don't need to see the table...
For example, I have that:
PRAIL 1400bar
IQ 70mm3
Tinj 1240us
K= (√Prail * Tinj)/ IQ
K=662.8
I've calculated the K-factor for the curve ? 662.8...
Now, to calculate the new tinj, user? that K...
Suppose that you enter 80mm3 in bp
Tinj=(IQ*(K)/√Prail = (80*662.8)/sqrt1400=53024/37.41=1417

here you go;)

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Sorry mariodarkblue, are ignorant of rimappature but I know I can do a bit of accounts and I have made many first to tell you that with your formula if you increase the 10% of the IQ, the increase of the time will be? always 10 %

Quote:

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Originally Written by CHIP TUNNING

Sorry mariodarkblue, are ignorant of rimappature but I know I can do a bit of accounts and I have made many first to tell you that with your formula if you increase the 10% of the IQ, the increase of the time will be? always 10 %

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To me that doesn't add up. Practically you say that if you increase to a 10% IQ ricalcoleresti the whole column in the times by adding a 10%?
But the column if you see the increases in data from the house does not correspond to a fixed percentage, but variable.

sorry eh...but if the formula ? the one posted by mario I will go to impelagare with the percentage??

I think that if we talk about IQ, we increase in percentage, the formula I think is valid for the rail pressure

Quote:

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Originally Written by munro

sorry eh...but if the formula ? the one posted by mario I will go to impelagare with the percentage??

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Hey Burt I I exercise :)

Quote:

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Originally Written by megamind

To me that doesn't add up. Practically you say that if you increase to a 10% IQ ricalcoleresti the whole column in the times by adding a 10%?
But the column if you see the increases in data from the house does not correspond to a fixed percentage, but variable.

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I know that the increments the data from the home are not a fixed percentage, but the formula above gives us always the same percentage

why? prediamo of the case values just to make an example

For example, I have that:
PRAIL 1400bar
IQ 70mm3
Tinj 1000us
K= (√Prail * Tinj)/ IQ
K=534,42
the calculated K-factor for that curve ? 534,42
Now, to calculate the new tinj, we use the formula of mario
Suppose that you insert 70 + 10%= 77mm3 in bp
Tinj=(IQ*(K)/√Prail = (77*534,42)/sqrt1000=37352/37,41= 1100 cio? 1000+ 10%= 1100

ANOTHER EXAMPLE example I:
PRAIL 2000bar
IQ 90mm3
Tinj 1000us
K= (√Prail * Tinj)/ IQ
K=496,90
the calculated K-factor for that curve ? 496,90
Now, to calculate the new tinj, we use the formula of mario
Suppose that you insert 90 + 20%= 108mm3 in bp
Tinj=(IQ*(K)/√Prail = (108*496,90)/sqrt2000=53665,62/44,72= 1200 to cio? 1000+ 20%= 1200



If you then changed the rail pressure then the thing remains the same, but with rail unchanged, the formula gives you a result that is always the same percentage that increased Your IQ

Check a little bit with different values? But I think I have reason.
Hello to all

Quote:

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Originally Written by franco75

I think that if we talk about IQ, we increase in percentage, the formula I think is valid for the rail pressure

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If you increase the rail pressure then, but apressione rail invarita not need

If you can be useful in annex I place a tiny little program to calculate the time of injection, I don't know if when andrette to ziparlo there will still be the formulas. Let me know?
Hello to all

Give us a look and let me know, why? if all goes well as soon as I have time I improved it better.
Hello