Why Cycle time to calculate pulsewidth??

Hey guys I’m having a really hard time understanding why we use CYCLE time to calculate injector pulsewidth. In an engine running at 6000 rpm we would traditionally divide by two to get cycles per minute. Then we would divide by 60 the get the number of cycles per second. Then we divide the number of seconds (1) by the number of cycles occurring in that window of time (50) to get how quickly each cycle happening (.02secs). This would then tell us the total potential time the injector would have to operate so we could then apply the required duty cycle to get the afr we want. My issue is the injector IS NOT active during at least half of the time it takes for an engine cycle to occur (power and exhaust strokes) so then why do we include the whole engine cycle as time the injector could be active when indeed it’s half that! Someone please help me understand, I’m sure I’m missing something obvious but I just can’t quite seem to figure it out. HUGE THANKS FOR ANY HELP!!

I' m not sure what is the problem, there, but it is certainly possible for the PORT injector(s) to be operating during the power and exhaust parts of the cycle, and with semi-sequential/batch firing this is definitely occuring.

Even with full sequential, it is certainly possible for the injector(s) to be priming the air held in the intake ports before the air-fuel mixture is drawn into the cylinder - there's even an advantage in this because it gives a little more time for the fuel to vapourise and cool the charge.

For those reasons, there may be a small benefit in experimenting with the timing of the injector(s) firing over the rpm and load ranges of the engine's operation.

Noah,

"My issue is the injector IS NOT active during at least half of the time it takes for an engine cycle to occur (power and exhaust strokes)"

This is why you are having problems understanding it. The injectors can still be active during those times. When you are at 100% duty cycle (which you shouldn't because of the non linearity of the injectors at higher duty cycle), the injectors are wide open for the whole cycle (or the 4 strokes), basically they are not cycling on and off anymore.

Yes I went to bed after making this post LATE last night and woke up and understood it 😅 What I was missing is that the injectors can dispense fuel against the back of the valves even during the other cycles in a port injected engine, which means when the valve finally opens it will have received the proper pulsewidth waiting for it in the port. The duty cycle represents the percentage of the total amount of time the injectors could be open that they must be open to create the proper AFR. That’s usually calculated per one minute, so it was throwing me off how that translates into the pulsewidth required per cycle. The answer is even though the individual engine cycles are a much shorter period of time than the measure used to calculate the pulsewidth (one minute), as long as the same ratio of time active to potential time to be active (duty cycle) is used throughout each individual cycle, the result over a one minute period is the same ratio as the original calculation. For instance, 10% of 10 feet is 12 inches. And 10% of 1 foot is 1.2 inches. However, if you multiply the result of 10% of 1 foot ten times the answer is the same as 10% of ten feet. Like I said I knew I was missing something obvious and just needed to step back for a minute and think on it 😅 Thank y’all for your help regardless 🙏 It’s awesome to have an educated community to ask questions like these to, even if sometimes the questions are dumb 😁😂

BTW not all ECUs use duty cycle as fuel injector management system. Some use time in milliseconds directly...