Download pdf - Determining the Kart Center of Gravity/Mass (CG/CM)evgrandprix.org/forms/curriculum/Determing the Go Kart Center of M… · Go-Kart Activity Forces in Go Kart Racing Student Name:

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


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


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)


( ) 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.


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.


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+ − =


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)


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


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=