FourInversionsofatypicalFourBarLinkage
Note:
Iftheaboveconditionwasnotmetthenonlyrockingmotionwouldbepossibleforanylink..
MechanicalAdvantageof4barlinkage
Themechanicaladvantageofalinkageistheratiooftheoutputtorqueexertedbythedrivenlinktotherequiredinputtorqueatthedriverlink. ItcanbeprovedthatthemechanicaladvantageisdirectlyproportionaltoSin(β)theanglebetweenthecouplerlink(c)andthedrivenlink(d),andisinverselyproportionaltosin(α)theanglebetweenthedriverlink(b)andthecoupler(c). Theseanglesarenotconstantsoitisclearthatthemechanicaladvantageisconstantlychanging.
Thelinkagepositionsshownbelowwithanangleα=0oand180ohasanearinfinitemechanicaladvantage. Thesepositionsarereferredtoastogglepositions. Thesepositionsallowthe4barlinkagetobeusedaclampingtools.
Theangleβiscalledthe"transmissionangle". Asthevaluesin(transmissionangle)becomessmallthemechanicaladvantageofthelinkageapproacheszero. Intheseregionthelinkageisveryliabletolockupwithverysmallamountsoffriction. Whenusingfourbarlinkagestotransfertorqueitisgenerallyconsideredprudenttoavoidtransmissionanglesbelow450and500.
Inthefigureaboveiflink(d)ismadethedriverthesystemshownisinalockedposition. Thesystemhasnotogglepositionsandthelinkageisapoordesign
Freudenstein'sEquation
Thisequationprovidesasimplealgebraicmethodofdeterminingthepositionofanoutputleverknowingthefourlinklengthsandthepositionoftheinputlever.
Considerthe4-barlinkagechainasshownbelow..
Thepositionvectorofthelinksarerelatedasfollows
l1+l2+l3+l4=0
Equatinghorizontaldistances
l1cosθ1+l2cosθ2+l3cosθ3+l4cosθ4=0
EquatingVerticaldistances
l1sinθ1+l2sinθ2+l3sinθ3+l4sinθ4=0
Assumingθ1=1800thensinθ1=0andcosθ1=-1Therefore
-l1+l2cosθ2+l3cosθ3+l4cosθ4=0
and..l2sinθ2+l3sinθ3+l4sinθ4=0
Movingalltermsexceptthosecontainingl3totheRHSandSquaringbothsides
l32cos2θ3=(l1-l2cosθ2-l4cosθ4)2
l32sin2θ3=(-l2sinθ2-l4sinθ4)2
Addingtheabove2equationsandusingtherelationships
cos(θ2-θ4)=cosθ2cosθ4+sinθ2sinθ4) and sin2θ+cos2θ=1
thefollowingrelationshipresults..
Freudenstein'sEquationresultsfromthisrelationshipas
K1cosθ2+K2cosθ4+K3=cos(θ2-θ4)
K1=l1/l4 K2=l1/l2 K3=(l32-l12-l22-l24)/2l2l4
Thisequationenablestheanalyticsynthesisofa4barlinkage. Ifthreepositionoftheoutputleverarerequiredcorrespondingtotheangularpositionoftheinputleveratthreepositionsthenthisequationcanbeusedtodeterminetheappropriateleverlengthsusingthreesimultaneousequations...
VelocityVectorsforLinks
Thevelocityofonepointonalinkmustbeperpendiculartotheaxisofthelink,otherwisetherewouldbeachangeinlengthofthelink.
OnthelinkshownbelowBhasavelocityofvAB=ω.ABperpendiculartoA-B." Thevelocityvectorisshown...
Consideringthefourbararrangementshownbelow.Thevelocityvectordiagramisbuiltupasfollows:
∙AsAandDarefixedthenthevelocityofDrelativetoA=0aanddarelocatedatthesamepoint
∙ThevelocityofBrelativetoaisvAB=ω.ABperpendiculartoA-B.Thisisdrawntoscaleasshown
∙ThevelocityofCrelativetoBisperpediculartoCBandpassesthroughb
∙ThevelocityofCrelativetoDisperpediculartoCDandpassesthroughd
∙ThevelocityofPisobtainedfromthevectordiagrambyusingtherelationshipbp/bc=BP/BC
Thevelocityvectordiagramiseasilydrawnasshown...
VelocityofslidingBlockonRotatingLink
ConsiderablockBslidingonalinkrotatingaboutA.TheblockisinstantaneouslylocatedatB'onthelink..
ThevelocityofB'relativetoA=ω.ABperpendiculartotheline.ThevelocityofBrelativetoB'=v.Thelinkblockandtheassociatedvectordiagramisshownbelow..
AccelerationVectorsforLinks
Theaccelerationofapointonalinkrelativetoanotherhastwocomponents:
∙1)thecentripetalcomponentduetotheangularvelocityofthelink.ω2.Length
∙2)thetangentialcomponentduetotheangularaccelerationofthelink....
∙Thediagrambelowshowshowtotoconstructavectordiagramfortheaccelerationcomponentsonasinglelink.
Thecentripetalaccelerationab'=ω2.ABtowardsthecentreofrotation. Thetangentialcomponentb'b=α.ABinadirectionperpendiculartothelink..
Thediagrambelowshowshowtoconstructanaccelerationvectordrawingforafourbarlinkage.
∙ForAandDarefixedrelativetoeachotherandtherelativeacceleration=0(a,daretogether)
∙TheaccelerationofBrelativetoAaredrawnasfortheabovelink
∙ThecentripetalaccelerationofCrelativetoB=v2CBandisdirectedtowardsB(bc1)
∙Thetangent