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INJECTION TIMING 4 cylinder

Diesel Tuning Fundamentals

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Discussion and questions related to the course Diesel Tuning Fundamentals

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Hello friends ,

I have a 4-cylinder diesel engine, i would like to know if the formula described in the courses applies to the 4-cylinder engine?

For example :

1.7ms duration time injector

120 / 1600= 75ms 720 /75 = 9.6. 9.6 /1.7 = 16.3

50% before 50% after = 16.3x0.5=8.5

Many thanks

Yes, you're correct. the formula works for any 4 stroke diesel engine irrespective of number of cylinders.

Big thanks André !!! 😁💪

On the mathematics you did above, I have a confusion.

you have 9.6 degrees and 1.7ms

Here you divide 9.6 /1.7 = 16.3, but the result is multiplication, I mean 9.6/1.7 = 5.647 (division) / multiplication is 9.6*1.7 = 16.3

The result is ms or degrees before and after TDC??

For example :

1.7ms duration time injector

120 / 1600= 75ms 720 /75 = 9.6. 9.6 /1.7 = 16.3

50% before 50% after = 16.3x0.5=8.5

Many thanks

The 9.6 is degrees per MS.

His example is at 1600rpm. So to change that it rev's a second would be /60, but to make that cycle time (two revolutions on a 4 stroke) you can divide by 120 which equals 75

as there are 720 degrees in a cycle, we then divide this by the time, so 720/75 is 9.6degrees per ms.

9.6 times 1.7ms means the injection period is 16.3 degrees (yes, it should have been multiplication instead of division)

He has then gone on a 50% burn after TDC it has been split, for an injection timing of 8.5 degrees BTDC