IC.1 basics

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Instant Centers of VelocitiesA simple method to perform velocity analysis in planar systems is the instant center (IC) method. This is a graphical method that requires drawing lines, finding intersections, finding ratios, etc.In this presentation we will review the followings:1.Definition of instant centers2.Number of IC’s for a given mechanism3.Kennedy’s rule4.Finding IC’s5.Finding velocitiesDefinition of an instant centerBetween every two bodies in planemotion there is a common point that has the same instantaneous velocity in each body. This common point is called the instant center (IC) of velocity.An instant center may be an actual common point between two bodies,such as the pin joint shown in (a), or it may be a virtual common point as shown in (b).An instant center between links (bodies) i and j is labeled as I i,j or I j,i .Note:For convenience we agree to reserve “1” as a label for the non -moving body; i.e., the ground.I i,j(a)(i)(j)I i,j(b)(j)(i)Obvious instant centers Some instant centers are very easy to identify:1.Two bodies connected by a pinjoint : The center of the pin joint isthe IC between the two bodies asshown in (a). Since the pin jointremains between the bodies asthey move, this IC is a permanentcenter as well.2.Two bodies connected by a slidingjoint: Since the two bodies rotatetogether; i.e., there is no relativerotation between them, the instantcenter is in infinity on an axisperpendicular to the axis of slidingas shown in (b). Note: ωi= ωjI i,j(a)(i)(j) (i)(j)I i,j(b)Number of instant centersFor a mechanism containing n links, there aren (n −1) ∕2instant centers. For example for a fourbar mechanism, there are4 (4 − 1) ∕2 = 6instant centers. The same is true for or a slider-crank mechanism.Book keeping:A small circle will help us keep track oflocating the instant centers. On the circumference of the circle we put as many marks as the number of links. For example if a system contains 4 links and they are numbered 1, 2, 3 and 4, on the circumference we put four marks. Each time we find a center between two links, we draw a line between the corresponding marks on the circle. 12 43►Instant Centers of VelocitiesKennedy’s rule Kennedy’s ruleBetween every three links (bodies), there are 3 (3 − 1) ∕2 = 3 instant centers. These three IC’s lie on the same straight line.For the three links that are labeled in the figure as i, j and k, the three centers are shown schematically to be on a straight line. The three centers may appear in different orders when the orientation of mechanism changes.We will use this rule extensively to find non-obvious IC’s in mechanisms as will be seen in other presentations.I i,j(i)(j)I k,i(k)I j,kO 2 = I 1,2=I 4,1A = I 2,3B = I 3,4Finding instant centersIn other presentations on instantcenters, we will see how to find all the IC’s for a given mechanism. We first locate all the obvious IC’s. We then apply Kennedy’s rule to find the non -obvious IC’s.As an example, for the slider-crank mechanism shown, the obvious IC’s are the three pin joints and the sliding joint.The non-obvious IC’s are found by applying Kennedy’s rule.1243I 1,3I 2,4►►Finding velocitiesIn the following slides we review several cases for determining velocities. These cases are presented in a general form. In all these cases always two moving bodies are involved.Known:It is assumed that for one body (body i) we know either its angular velocity or the velocity (absolute) of a point on that body.Unknown: We want to find either the angular velocity of the second body (body j) or the velocity (absolute) of a point on that body.Third body:In any analysis we need to get a third body involved. We use the ground (body 1) as the third body.Instant centers:Between bodies, 1, i and j, we have three IC’s: I 1,i , I 1,j and I i,j . These are the three centers to use!I i,j(i)(j)I 1,iI 1,jProblem 1Assume ωi is known and we need to find ωj.Solution: We solve this problem in two steps.ωi(i)(j)I1,jI i,j I1,iStep 1:The instant center I i,j is a point on link i. Link i is pinned to the ground at I 1,i (real or imaginary). Since we know ωi we can determine V I i,j (magnitude and direction):V I i,j = ωi ∙ R I i,j I 1,iIts direction is obtained by rotating R I i,j I 1,i 90°in the direction of ωi .For the next step we no longer need vector R I i,j I 1,i and the IC I 1,i .►(i)(j)ωiV I i,jR I i,j I 1,iI i,jI 1,iStep 2:The instant center I i,j is also a point on link j. Link j is pinned to the ground at I 1,j (real or imaginary). Since we know the velocity of I i,j , we can determine ωj :ωj = V I i,j ∕R I i,j I 1,jThe direction of the angular velocity is found by rotating R I i,j I 1,j 90°to match the direction of V I i,j .►(i)(j)R I i,j I 1,jωjV I i,jI i,jI 1,jProblem 2Assume that instead of ωi , we have been given the velocity of point P on link i and we need to find ωj .(i)(j)I i,j I 1,iI 1,jP V PSolution: We first note that the axis of V P is not arbitrary. This axis must be perpendicular to R P I 1,i . We measure the length R P I 1,i then we compute ωi as:ωi = V P ∕R P I 1,iNow that we have determined ωi , we can continue as in Problem 1.(i)(j)ωiI i,j I 1,iI 1,jP V PR P I 1,i►►Problem 3In this problem we add one more unknown to Problem 1 (or Problem 2). Determine the velocity of point B on link j.(i)(j)ωiI i,jI 1,iI 1,jBSolution: We start solving this problem as in Problem 1 (or Problem 2) until we have ωj . After that we can easily find the velocity of point B:V B = ωj ∙ R BI 1,j(i)(j)ωiI i,jI 1,iI 1,jB►ωj►R BI 1,j V BInstant Centers of Velocities General commentsGeneral commentsThe procedures we have seenin this presentation for findinginstant centers and then usingthe centers to find velocitiescan be applied to any planarmechanism.We will see the process forfour-bar and slider-crankmechanisms in detail in thecoming presentations.。