INJ ms calculating correctly?

[Gernby]

I have an '08 S2000, and it looks like my INJ values are incorrect in the datalogs. The values range from about 9 to 12 ms, which seems too high. Unless I'm calculating something wrong, a full rotation of the crank only takes about 7 ms at 8500 RPMs, so I would expect the max injector pulse duration to be about 3.5 ms.

BTW, I'm still using 1.1.3.4.

[Gernby]

Forgot to include the datalog ...

[Hondata]

Since the engine is 4 stroke, each cylinder fires every second revolution. In any case I've checked it with a 500MHz scope.

From your datalog: 8500 rpm = 141.7 rev/second = 70.8 cycles/second = 14.12 ms / cycle. An injector duration of 12.5 ms would give a duty cycle of 88.5%.

[Gernby]

I believe you are dividing by 2 when you should be multiplying by 2.

At 8500 RPMs, the crank / flywheel is rotating 8500 times per second, right? If so, then cylinder 1 is firing 4250 times per second, right?

4250 combustions per second = 14 ms per combustion cycle (4 full strokes)

For a 4 stroke engine, the intake stroke would be 1/4 of the 14 ms, which would be 3.5 ms.

[Hondata]

RPM = revolutions per minute.

8500 RPM = 141.7 rev/second

Divide by 2 to get how often each cylinder fires = 70.8 Hz

Convert to ms 1/70.8 x 1000 = 14.1 ms

[Gernby]

RPM = revolutions per minute.

8500 RPM = 141.7 rev/second

Divide by 2 to get how often each cylinder fires = 70.8 Hz (this is for all 4 strokes)

Convert to ms 1/70.8 x 1000 = 14.1 ms (also for all 4 strokes)

Now that you've calculated that all 4 strokes occur in 14.1 ms, you now need to divide again by 4 to get just the intake stroke (3.5 ms).

[Gernby]

It might be easier to see if you work it backwards.

3.5 ms for 1 stroke = 14 ms for all 4 strokes
1/14 ms = 71 combustions per second
71 combustions per second = 4285 combustions per minute
4285 combustions per minute = 8571 revolutions per minute

[Hondata]

Now that you've calculated that all 4 strokes occur in 14.1 ms, you now need to divide again by 4 to get just the intake stroke (3.5 ms).

Why are you dividing by 4 here? The injector event does not occur just in the intake stroke - the full cycle of the engine is used for injection. Also see the definition of 'duty cycle'.

[Gernby]

Now that you've calculated that all 4 strokes occur in 14.1 ms, you now need to divide again by 4 to get just the intake stroke (3.5 ms).

Why are you dividing by 4 here? The injector event does not occur just in the intake stroke - the full cycle of the engine is used for injection. Also see the definition of 'duty cycle'.

Are you saying that the injectors do most of their spraying while the intake valves are closed? That seems like it would just create a puddle of fuel on top of the closed valves. If that's the case, then I really appreciate you straightening me out on this.

[Hondata]

Yes, the injector can fire when the intake valve is closed. Every fuel injected engine does this - once the engine is warm, the fuel stays as vapour in the intake manifold. Going to a bigger injector and reducing the injector duration doesn't really make any difference in power, so it is not a bad thing. The ECU does try to fire the injector so that the intake valve is open by the time the fuel reaches the value (depending on the calculated intake velocity), but it also tries to avoid the fuel arriving in the cylinder during valve overlap, which is bad for emissions and economy.

[Gernby]

Wow. Thanks! I thought the whole benefit of port injection over the old throttle body injection was so that the fuel would be sprayed straight into an open intake valve.

[Hondata]

It does 99% of the time. At full load the injector event exceeds the valve opening time.