Upload
others
View
44
Download
8
Embed Size (px)
Citation preview
Version:11.25.19 ©2019PurdueUniversityAllRightsReserved Page|1
Go-KartActivityForcesinGoKartRacing
StudentName:
DeterminingtheKartCenterofGravity/Mass(CG/CM)GuidingQuestion:Howcanthecenterofgravity(CG)ofanevGrandPrixkartbesetsothatthereisa47%/53%weightdistributionfromfronttobackanda50%/50%fromsidetoside?
Whatismeantbythecenterofgravity?Centerofgravity,alsoknownascenterofmass,isthatpointatwhichasystemorbodybehavesasifallitsmasswerecenteredatthatpoint.Wheretheweight,andalsoallaccelerativeforcesofacceleration,brakingandcorneringactthroughit.
Centerofgravitylocationcanbedefinedas:-Thebalancepointofanobject-Thepointthroughwhichaforcewillcausepuretranslation-Thepointaboutwhichgravitymomentsarebalanced(seeFig.1)-Thepointwhichifthebodyishangedfromitwillstaybalanced(leveledasitisontheground).
Fig.1Summationofmomentsofpartsweightsaroundanypointisequaltothemomentofthesummationofweightsaroundthispoint.W.XCG=(W1.X1+W2.X2+W3.X3+W4.X4+W5.X5+……..)XCG=Σ(W.X)/Σ(W),whereXiisthedistanceinxdirectionbetweenthepointiandthatpoint.WhatistheimportanceoftheCG?Whenmakingananalysisoftheforcesappliedonthekart,theCGisthepointtoplacethekartweight,andthecentrifugalforceswhenthekartisturningorwhenacceleratingordecelerating.AnyforcethatactsthroughtheCGhasnotendencytomakethekartrotate.
Thecenterofmassheight,relativetothetrack,determinesloadtransfer(relatedtoweighttransfer)fromsidetosideandcausesbodylean.Whenthetiresofakartprovideacentripetalforcetopullitaroundaturn,themomentumofthekartactuatesloadtransferinadirectiongoingfromthekart's
Version:11.25.19 ©2019PurdueUniversityAllRightsReserved Page|2
currentpositiontoapointonapathtangenttothekart'spath.Thisloadtransferpresentsitselfintheformofbodylean.Bodyleancanbecontrolledbyloweringthecenterofweightorbywideningthekarttrack,itcanalsobecontrolledbysprings,anti-rollbarsortherollcenterheights.
Fig.2IncreasingtheheightoftheCGordecreasingthewidthofthekarttrackwillcausethekarttotopple.
Thecenterofmassheight,relativetothewheelbase,determinesloadtransferbetweenfrontandrear.Thekart'smomentumactsatitscenterofmasstotiltthekartforwardorbackward,respectivelyduringbrakingandacceleration.Sinceitisonlythedownwardforcethatchangesandnotthelocationofthecenterofmass,theeffectonover/understeerisoppositetothatofanactualchangeinthecenterofmass.
Alowercenterofmassisaprincipalperformanceadvantageofsportscars,comparedtosedansand(especially)SUVs.Somecarshavebodypanelsmadeoflightweightmaterialspartlyforthisreason.
ObtainingthepositionofthekartCG:Akartisnotsymmetricalinshapeormassfromfronttorear.Manykartsaresymmetricallefttorightinshapebutnotinmass.
Fig.2Weight(W),Reaction(R) Fig.3TheCGlocationinthesideandfrontview
ThedifferencebetweenweightWandReactionR:FromFig.2,Wistheweightofthetire(alwaysverticalanddownward),whileRisthereactionfromtheground(alwaysperpendiculartothegroundsurfaceandawayfromit).Thetireisbalancedintheverticaldirectionunderitsweight(W)andthegroundreaction(R),thatmeansthesummationofforcesinthatdirectioniszero(W–R=0),whichgivesW=R.
Inouranalysiswemeasuretheweights,butwhenstudyingthekartbalanceweusethereactionssupportingthekartfromtheground(where:Wf=Rf,Wr=Rr,Ww=RwandWL=RL).
ThesymbolsinFig.3aredonatedto:W-isthekartweight,
KartdoesnottoppleoverKarttopples
Version:11.25.19 ©2019PurdueUniversityAllRightsReserved Page|3
Rf-isthegroundreactionofthefrontalaxleweightRr-isgroundreactionoftherearaxleweightRR-isgroundreactionofthekartrightwheelsweightRL-isgroundreactionofthekartleftwheelsweightL-isthekartwheelbase(distancebetweenthefrontandrearkartwheels/axles)T-isthekarttrack(distancebetweenthecenterofthewheelsonthesameaxle)a-isthelocationoftheCGbehindthefrontaxleb-isthelocationoftheCGinfrontoftherearaxlex-isthelocationoftheCGawayfromtherightwheelsy-isthelocationoftheCGawayfromtheleftwheels
MaterialsRequired:• 4DigitalBathroomScales• Assembledgokart• Z-axisBlocks(InstructionsIncluded)
SetupandProcedures:Step1–Putscalesunderneatheachofthekarttiresandmakesurethatallthetirescreateaflatplanethatbalancestheweightoftheunloadedshelfevenly.Dothisbyensuringthatthesurfaceonwhichthescalesaresettingislevel.Havethedriversitintheseatwithhandsonthesteeringwheelandfeetonthepedals.
Fig.4Measurethefrontandrearaxleweights(Wf,Wr),andthewheelbaselength(L)
Fig.5Thecaronalevelsurface,Listhewheelbase,thefrontandrearaxlesweights(Wf,Wr)and(aandbareunknownlongitudinaldistancesoftheCG)
Step2–Measurethewheelbase,L(thedistancebetweenthecenterofthefrontandrearwheels),Fig.3.RecordthisvalueontheKartCenterofGravityCalculationsForm.
L a b= + (1)
a L b= − (2)
Version:11.25.19 ©2019PurdueUniversityAllRightsReserved Page|4
( ) 0f rW R R− + = (3)
Then ( )f rW R R= + (4)
Step3–MeasurethekartfrontaxleweightWf=Rf(Fig.3)byaddingthereadingsofthescalesunderthefrontwheelsofthekartandrecordthesevaluesintheKartCenterofGravityCalculationsForm.
Step4–MeasurethekartrearaxleweightWr=Rr(Fig.3)byaddingthereadingsofthescalesundertherearwheelsofthekartandrecordthesevaluesintheKartCenterofGravityCalculationsForm.
Step5–DeterminethetotalkartweightW(Equation4)byaddingthefrontaxleweightandtherearaxleweighttogetherandrecordthisvalueintheKartCenterofGravityCalculationsForm.
Inordertolocatethecenterofgravityinrelationtotherearaxleyouwillneedtodeterminethe“moment”aroundtherearaxle(E)usingthefollowingformula
0fR L Wb− =
fR L Wb= ,then
( )fR
b LW
= (5)
Step6–Useeq.(5)tofindthedistanceb.RecordthisvalueatthebottomoftheKartCenterofGravityCalculationsForm.
Step7–DeterminethepercentageofweightonthefrontaxleofthekartandtherearaxleofthekartandrecordthesevaluesintheKartCenterofGravityCalculationsForm.Calculatethepercentageusingthefollowingequation.
Axle Weight 100 %Total Kart Weight⎛ ⎞
× =⎜ ⎟⎝ ⎠
Step8–Substitutethevalueofbineq.(2)togetthedistancea.RecordthisvalueatthebottomoftheKartCenterofGravityCalculationsForm.NOTE:InEq.5,theunitsofRfandWcanbeexpressedinunitforce(N,lb)orunitmass(kg),theunitsofLandbcanbeexpressedin(m,cm,mm,orft,in).
Fig.8FindingtheCGpositioninthefrontalview,distancex,y.
Version:11.25.19 ©2019PurdueUniversityAllRightsReserved Page|5
Step9–Measurethewheeltrack,T(thedistancebetweenthecenterofthefrontwheels),Fig.8.RecordthisvalueontheKartCenterofGravityCalculationsForm.
T x y= + (6)
x T y= − (7)
Then ( )R LW R R= + (8)
Step10–MeasurethekartrightsideweightWR=RR(Fig.8)byaddingthereadingsofthescalesunderthewheelsontherightsideofthekartandrecordthisvalueintheKartCenterofGravityCalculationsForm.
Step11–MeasurethekartleftsideweightWL=RL(Fig.8)byaddingthereadingsofthescalesunderthewheelsontheleftsideofthekartandrecordthesevaluesintheKartCenterofGravityCalculationsForm.
Inordertolocatethecenterofgravityinrelationtotherearaxleyouwillneedtodeterminethe“moment”aroundtherightsidewheelsusingthefollowing
0RR T Wy− =
RR T Wy= ,then
( )RRy TW
= (9)
Step12–DeterminethepercentageofweightontherightsideofthekartandtheleftsideofthekartandrecordthesevaluesintheKartCenterofGravityCalculationsForm.Calculatethepercentageusingthefollowingequation.
Side Weight 100 %Total Kart Weight⎛ ⎞
× =⎜ ⎟⎝ ⎠
Step13–Useeq.(9)tofindthedistancey.RecordthisvalueatthebottomoftheKartCenterofGravityCalculationsForm
Step14–Substitutethevalueofyineq.(7)togetthedistancex.RecordthisvalueatthebottomoftheKartCenterofGravityCalculationsForm.NOTE:InEq.9,theunitsofRRandWcanbeexpressedinunitforce(N,lb)orunitmass(kg),theunitsofTandycanbeexpressedin(m,cm,mm,orft,in).
Mostkartsareasymmetricinweightdistributioninthefrontview(RRisnotequaltoRL).SothattheCGpositionwillnotbeinthemiddle.
DeterminetheheightabovegroundofCG:Theweightoftherearaxle(Wr1)willbeweighedwhilethefrontpairofwheelsareraisedupquiteasmalldistanceH(orh1)(asshowninfigure6).
Step15–Constructtwo9inchby7inchZ-axiswheelblocksbycuttingfour7inchpiecesof2x6,two6inchpiecesof2x6,andtwo7inchpiecesof2x2lumber.
Version:11.25.19 ©2019PurdueUniversityAllRightsReserved Page|6
Step16–Makearectangularboxbyplacingtwoofthe7inchpieces,paralleltooneanotherand6inchesapart.Placetwo6inchpiecesbetweenthem.Screwtogetheratcorners.Predrillingtheseholeswillmakeiteasiertoattachthesetogether.
Step17–Placetheremainingtwo7inchpiecesof2x6materialonthetopofthenewlyconstructedboxandscrewintoplace.
Step18–FinishoftheZ-axiswheelblocksbyattachingthetwo7inch2x2piecesateachendoftheboxtop.NOTE:yourZ-axiswheelblockshouldlooksomethingliketheimagetotheright.
Fig.6Measuretherearaxleweight(Wr1)andthedistanceoffrontraise(H)
Fig.7ThefrontwheelsareraisedupasmalldistanceH(h1),risthewheelradius,(histheunknowndistanceoftheCGheight)
Step19–PlaceZ-axiswheelblocksonfronttirescales,thenplacefronttiresofkartonZ-axisblocks.Havethedriversitintheseatwithhandsonthesteeringwheelandfeetonthepedals.
Step20–MeasuretheheightoftheZ-axiswheelblocksandrecordthisvalueinthebottomoftheKartCenterofGravityCalculationsForm.
Step21–MeasuretheelevatedfrontaxleweightWf1=Rf1(Fig.7)bysubtractingtheweightoftheZ-axiswheelblocksfromthereadingsofthescalesundertheZ-axiswheelblocksandrecordthisvalueintheKartCenterofGravityCalculationsForm.
Inordertolocatethecenterofgravityinrelationtothechassisofthekartyouwillneedtodeterminethesummationofverticalforcesintheydirectionusingthefollowing.
( )1 1 0f rR R W+ − =
Version:11.25.19 ©2019PurdueUniversityAllRightsReserved Page|7
Then: 1 1f rR W R= −
Becausethekarthasbeenelevated,thehorizontaldistancebetweentheCGandtherearaxlehasbecomeslightlysmaller(SeeFig.7).ThecalculateddistanceofbcannowberepresentedbyalinethatconnectspointsAandC.Theelevatingofthefrontofthekartrotatesthecenterofgravityaroundtherearaxle,causingtheCGtomoveupandtowardstherearaxle.IfwedrawaverticallinefromtheCGtothehorizontalplane(ArrowW)thenewdistancebetweentherearaxleandtheCGwouldberepresentedbyalineconnectingpointsAandB.
Knowingthatthesummationofmomentsaboutanypointmustbeequal0,then:
1( cos ) ( ) 0fR L W ABθ − =
1( cos ) ( )fR L W ABθ = (11)
Where:
AB AC BC= − ,
cosAC b θ= ,and
( )sinBC ED h r θ= = − ,
Then:
cos ( )sinAB AC BC b h rθ θ= − = − −
SubstitutingthevalueofABformtheaboveequationinEq.(11)andrearrangingtoisolateheight,then
1( cos ) ( cos ( )sin )fR L W b h rθ θ θ= − −
1( cos ) cos ( )sinfR L Wb W h rθ θ θ= − −
1( )sin cos ( cos )fW h r Wb R Lθ θ θ− = −
1[ ( )]cotfRh r b LW
θ− = − (12)
1[ ( )]cotfRh b L rW
θ= − + (13)
where:
1sin ( )HL
θ −= (14)
Version:11.25.19 ©2019PurdueUniversityAllRightsReserved Page|8
Step22–Determinetheanglethekartisraisedusingequation(14)whereHistheheightoftheZ-axiswheelblocksandListhewheelbaseofthekart.
Step23–DeterminetheheightofCenterofGravityfromthegroundusingequation(13).RecordthisvalueonthebottomoftheKartCenterofGravityCalculationsForm.(h-r)isthedistanceofCGabovetheaxleplane,Eq.(12)histhedistanceofCGabovetheground,Eq.(13)NOTE:InEq.13,theunitsofRf1andWbothcanbeexpressedinunitforce(N,lb)orunitmass(kg),theunitsofb,L,randhcanbeexpressedin(m,cm,mm,orft,in).
Step24–VerifyallcalculationsusingtheExcelprogramfiletitled“KartCenterofGravityCalculator”toobtainthecenterofgravitylocation(a,b,x,yandh).ThisfileisincludedinExperimentalResources
Step25–RepeatprocessofdeterminingCGofyourkartaminimumof3times.
DataAnalysis1. Whereisthecenterofgravityforyourkart?2. Whywasthekartelevatedduringtheactivity?HowdidthishelpinidentifyingtheCG?3. Didtheweightregisteredbythescaleschangefromonetrialtoanother?Why?4. Didyouobtaina47%-53%ratioofweightdistributionfromfronttobackonthekart?5. Whatneedstobedonetothekarttoimprovethefronttobackandsidetosideweight
distribution?
Resources
http://www.thecartech.com/subjects/auto_eng/Center_of_Gravity.htm
Version:11.25.19 ©2019PurdueUniversityAllRightsReserved Page|9
Go-KartActivityForcesinGoKartRacingStudentName:
WheelTrack: Wheelbase:
Left Right Total Percentage
Back
Front
Total 100%
Percentage 100%
TrialTWL
Totalweightleft
TWRTotalweight
right
TWFTotalweight
front
TWBTotalweight
back
EHElevatedheight
ϴTW
Totalkartweight
ETWFElevatedtotalweightfront
1
2
3
4
5
TWFb wheelbaseTW
a wheelbase b
⎛ ⎞= ⎜ ⎟⎝ ⎠
= −
1sin EHlength
θ − ⎛ ⎞= ⎜ ⎟
⎝ ⎠
TWRy wheeltrackTW
x wheeltrack y
⎛ ⎞= ⎜ ⎟⎝ ⎠
= −
( )( )cos ( )( )cos _( )sin
TW b ETWF wheelbaseheight wheel diameterTW
θ θθ
⎛ ⎞−= +⎜ ⎟⎝ ⎠
a= b= x= y= height=