8/6/2019 Injector Pulse Width

1/3

sinceijustwentthoughthis,ithoughtishouldwritedownwhatilearnedbeforeiforgotit.hopefullyitwillhelpsomepeople...

pleaseacceptmyapologiesbeforehandforhowlongandramblingitis...soheregoes...

theinjectortimingtableisbasedoncrankdegrees,andspecifiesthedegreeat

whichtheinjectorpulseshouldend.soatwhatweisawaytofigureoutaprocessthatgivesus,foragivenrpmandload,whatcamdegreetheinjectorpulseshouldend.notethatcrankdegreesgofrom0-720insteadoffrom0-360becausethecrankrotates2timesforeachrevolutionofthecam.

whenthereislittleornooverlapbetweentheclosingoftheexhaustvalveandtheopeningoftheintakevalve,itmaybedesirabletoactuallystartinjectingthefuelbeforetheintakevalveopens.inthiscase,thefuelissquirtedontothebackoftheclosedintakevalve,becausethatmayhelpatomizethefuel.

butifthereisalotofvalveoverlap,thenthatapproachisnotasdesirable,becausewhentheintakevalveopens,someofthefuelthatwaswaitingtogointo

thecylinderwillgetmixedinwiththeexhaust(andejectedfromthecylinder)andthereforenotbeburned,potentiallyresultinginaleancondition.

soafactorindecidinghowearlytostarttheinjectorsmaybehowmuchoverlapthereisinyourcam.

herearemythoughtsonaprocesstodeterminewhatvaluestoputintheinjectortimingtable...

beforewecanpickanendingdegree,weneedawaytodeterminehowmanydegreestheinjectorneedstoprovidetherightamountoffuel.inordertodothat,weneedawaytofigureouthowmuchfuelisneeded.onceweknowthat,wecandeterminehowmanymillisecondstheinjectorneedstobeopentoprovidethefuel.

thenneedtobeabletofigureout,foragivenrpm,howmanydegreesthecamshaftwillturnduringthetimetheinjectorisopen.then,knowingthedegreesatwhichthevalvesopenandclose,wecanfinallypickreasonableendingdegreesfortheinjectortiming.

foracertainrpmandload,wecanestimatewhattheinjectorpulsewidthwouldbe.forexample,supposewehada410with42#injectorsspinningat3000rpmunder50%load.whatwouldtheestimatedpulsewidthbe?

onecylinderhasamaximumvolumeof410/8,or51.25ci.sincethevolumetricefficiencyoftheengineisnot100%,forourexample,let'suse.85astheVEfactor,whichmeansthatagivencylinderwillendupwithamaximumof51.25*.85=

43.5625ciofair.sinceourexampleisat50%load,thatwouldbe43.5625*.50=21.78125ci.thisistheestimatedamountofair(incubicinches)eachcylindershouldreceivepercycle.

thereisaconversionfactorthatsaysonecubicfootofaircontains.075lbsofair.soinourexample,wehave21.78125/(12*12*12)=0.01260489cubicfeetofofair.sowehave0.01260489*.075=0.00094536675lbsofair

nowthatweknowwehave0.00094536675lbsofairinthecylinder,wecandeterminehowmuchfuelweneed...usingthestandardair/fuelratioof14.67,wewouldneed0.00094536675/14.64=.0000645742316lbsoffuel.

sinceinjectorsaregiveninlbsperhour,andpulsewidthsareinmilliseconds,

weneedtoconverttolbsperms.sofor42#injectors,wehave42/(60minutes*60seconds*1000milliseconds)=0.0000116666667lbsoffuelperms.

8/6/2019 Injector Pulse Width

2/3

sotoget.0000644421779lbsoffuel,wewouldopentheinjectorsforapulswidthof.0000645742316/0.0000116666667=5.53493412ms.butsincethereisasmalldelay(about.05ms)beforetheinjectorsactuallystartsquirtingthefuel,letsadd.05tothepw,gettingabout5.58forthedesiredinjectorpulsewidth.

noticethatifwefigureoutthepwforthecasewhereloadis100%(WOT),thenwecanjustmultiplythatbytheloadlatertosimplifythings.let'sdothathe

re:

maxairinlbs:(51.25*.85*.075)/(12*12*12)=0.00189073351lbsairfuelneededformaxairabove:0.00189073351/14.67=0.000128884357lbsfuelfuelinlbsperms:(42/360000)=0.0000116666667msneededtodelivertheabovefuel:(0.000128884357/0.0000116666667)=11.05msnotethatwewillneedtoaddthe.05msdelaytothepulsewidthafterfactoringintheload...

so11.09msisneededtodeliverthemostfueltheinjectershouldhavetodeliver.ofcourse,thiswillbeslightlyhigherifwewanttheafrtobelower.like

whenatWOTforexample.

wewillmultiplythistotheloadlater...

ok,wearehalfwaythere...nowwegettofigureouthowmanydegreesthecrankwillturnduringthat5.58ms.thatdependsonhowfasttheengineisturning.

inourexample,theengineisturning3000rpm.sincethereare360degreesinarevolution,60secondsperminuteand1000mspersecond,thenwehavethecrankturningat(360*3000)/(60*1000)=18degreesperms.soin5.58ms,thecrankturns18*5.58=100.44degrees.

noticethattheonlyvariableaboveistherpm,soforagivenrpmwecandothi

stogetthedegreesperms:

360*(rpm)/(60*1000)=360*rpm/60000=rpm*(360/60000)=rpm*0.006

we'llusethislatertoo...

solet'ssaythatourcamhasanintakelobecenterof110,alsaof112,adurationof230at.050"and290totalduration.sohowdoesthattranslateintocrankdegrees?here'show:

bydefinition,crankdegrees0-360refertothefirstrevolutionofthecrank,anddegrees360-720refertothesecondrevolutionofthecrank.

intakeeventsareoffsetsfromtopdeadcenter(TDC)orbottomdeadcenter(BDC)ofthecylinderduringthephaseinquestion.fortheintakevalveevents,TDCisatdegree360.

sincetheintakelobehasacenterof110,thatmeansthevalveisopenitshighestat360+110=470degrees.adurationof290meansthatthevalveopensat470-(290/2)=325andclosesat470+(290/2)=615degrees.similarly,thevalveisat.050"orhigherfromdegree(470)+/-(230/2),or355and585degrees.

alsaof112meansthattheexhaustlobecenteris112*2degreesbeforetheintakelobecenter.

8/6/2019 Injector Pulse Width

3/3

sowithalsaof112,ourexhaustvalveisfullyopen(theexhaustlobecenter)at470-(112*2)=246degrees.doingthesamecalcsasabove,wehaveopen/closeat101and391,andat.050"theopen/closeeventsare131and361degrees.

nownoticethattheexhaustclosesat391,andtheintakeopensat325.sobothvalvesareopenfor391-325=66degrees.thisisthetotaloverlap.similarly,at.050"thereisanoverlapof6degrees.inthiscase,iwouldprobablywantt

hefueltostartinjectingwhentheexhaustvalvecloses,orataboutdegree390.

nowwearealmostthere...foragivenloadandrpm,wecanfigureouthowmuchmsofinjectorweneed,andhowmanydegreesthatmsneeds.nowallwehavetodoistiethattothecameventsforourcamandwecanfillinthetablewithvaluesthatshouldbefairlyreasonable.

tofinishoffourexample,withthe410ciengineand42#injectors,50%loadat3000rpmwithourcamasabove,thecalculationforthatcellwouldbe...

theneededinjectorpwwouldbe:11.05*.50+.05=5.595ms

thecamrotationfor5.595msat3000rpmwouldbe:3000*.006*5.595=101degrees(rounded)ifthepwstartswhentheexhaustvalvecloses...theinjectorfiresfromdegree390thru491(390+101)

so491shouldbeafairlyreasonablevalueforthatcell.