[rrauscher at nni.com]

>
>It's the total event time that is in question here. Hence
>all that math. . . Typically your injectors are sync'd
>to ignition events, we need to know how many cylinders
>and how often the injector in question fires relative to
>the ignition pulses.
>
>Bruce, I think you a playing with a '148? For the 3.8l
>Boosted-Buick? What is the injector firing relative
>to the ignition events? IE: batch fired? every other
>ignition event?
>
>>
>>Duty cycle, comparison of On to Total event time,  On/Total.   Can't be over

>
>>100%.
>>
>>So if the injectors fire once per revolution, at 3.450, with an on injector

>
>>time of 11.6  would be 11.6/17.2   (About 68%).

I gotta' learn how to read, yep, you got it Bruce (me too). If
the injector fires once per rev, it's the 68% you have.

BobR.

>
>Well, the 17.2ms is the time for one rev, but how many
>times does a ignition event occur during that time? And
>how many injector firings happen?
>
>>
>>Bob has a slight error as I see it of adding the two injector times, ie must

>
>>have been thinking dual throttle body with 2 injectors, in which case the

>>total time that both injectors could be on would be 2x as long as compared

>
>>to a single injector (or batch strategy),
>>Grumpy
>
>This is true, the 'event window' is twice as large if
>the injectors alternate on ignition events.
>
>>
>>
>>> I think you're trying to hard, Bob.  The way I understand it, Duty cycle

>
>>is
>>> simply the injector "on" time (11.6 ms) divided by the time it takes for

>
>>the
>>> engine to complete one rev -er, ahem, _cycle_ (34.5 ms)- irregardless of

>
>>> injection type, synch or asynch mode, or number of cylinders, yada, yada,

>
>>> yada...
>>>
>>> That said, I'd like to bashfully change my previous answer from 67% to
>>33.6%
>>> Duty Cycle, remembering the old rule that you have 20 ms to play with at

>
>>> 6000 RPM...
>
>That is correct if you are talking about a single cylinder.
>On a v8, you have eight ignition/injectors events within
>that 20ms. This is where it becomes confusing.
>
>20 ms / 8 = 2.5msec between each ignition event. On a
>system that fires the 'other' injector on each ignition
>event, you have a total of 5ms before that same injector
>will be fired again.
>
>Or a system that fires all injectors every second ignition
>event.
>
>BobR.
>
>>>
>>> Oh, alright, who's got the real answer?
>>>
>>> Jeremy
>>>
>>> >This is what I came up with on a v8 tbi setup
>>> >(such as '747):
>>> >
>>> >3475 rpm @ 11.6 msec inj pulse.
>>> >
>>> >-----------
>>> >
>>> >3475rpm / 60 = 57.91 rev's per sec
>>> >
>>> >1 / 57.91 = 0.01726 sec's per rev, or 17.2 milli-sec per rev
>>> >
>>> >v8 = 4 spk's per rev, tbi alternates, so 2 inj cycles
>>> >for each injector per rev.
>>> >
>>> >17.26 / 2 = 8.62 milli-secs between inj cycles.
>>> >
>>> >11.6 / 8.62 = 134% duty cycle.
>>> >
>>> >For a v6, it's 3 sprk's per cycle. For alternate inj firings
>>> >per sprk (or bank fired on every other sprk):
>>> >
>>> >3 / 2 = 1.5 inject cycles per rev.
>>> >
>>> >17.26 / 1.5 = 11.5 milli-secs between inj cycles.
>>> >
>>> >11.6 / 11.5 = 100% duty cycle.
>>> >
>>> >
>>> >Corrections welcome. . .
>>> >
>>> > BobR.
>>> >
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