3.4 Vector Loops of a Mechanism

The main difference between freely moving bodies and the moving links in a mechanism is that they have a constrained motion due to the joints in between the links. The links connected by joints form closed polygons that we shall call a loop. The motion analysis of mechanisms is based on expressing these loops mathematically.

In kinematic analysis we shall assume that all the necessary dimensions of each link is given and link length dimensions (i.e. the distance between the joints or the angles) can be determined from the given dimensions using the geometry of the link.

In Section 2.1.2, we have seen that it is sufficient to represent the position of each link (rigid body) by describing the position of any two points on that link. One way of selecting these two points on a link is to use the permanently coincident points. It is obvious that in such a procedure, the origin of a vector will be defined by the previous vector and thus the number of parameters to define the link positions will be decreased.

Let us consider a four-bar mechanism as shown above as a simple example. In this mechanism A0 ,is a permanently coincident point between links 1 and 2, A is peranently coincident point between links 2 and 3, B between 3 and 4 and B0 between 1 and 4. Let us disconnect joint B. In such a case we will obtain two open kinematic chains A0AB (links 2, 3) with two degrees of freedom and A0B0B (links 1, 4) with one degree of freedom (see figure). To determine the positions of the links we must have a reference frame. One obvious choice is to select the fixed pivots A0, B0 as one of the co-ordinate axes and select A0 or B0 as the origin. Next,in order to define the position of link 2, we must define angle θ12, which is related with the degree of freedom of the joint between links 1 and 2. To determine the position of link 3, since the location of the permanently coincident point A between 2 and 3 can be determined when θ12 defined, we must now define θ13 , which is related to the freedom of the joint between links 2 and 3. Similarly θ14 must be defined to determine the position of link 4. Hence we need 3 parameters (θ12, θ13 and θ14 ) which are all related to the joint freedoms for the open kinematic chains obtained when we disconnect a joint to eliminate a loop.

Each link can be defined by a vector fixed on that link, let us select the permanently coincident points between the links as the tips of these vectors and define vector A0A (for link 2), AB (for link 3), B0(for link 4) and A0B0 (for link 1). Except A0B0 , the other three vectors will be a function of time (since the distances between the two points on the same link are fixed, the magnitudes will remain constant but the directions of these vectors will change in time). Since the mechanism contains revolute joints only, The magnitude of the vectors are constant link lengths (a2 = |A0A|, a3 = |AB|, a1 = |A0B0| and a4 = |B0B|). The angular orientation of these vectors will be rotation variables (θ12, θ13 and θ14 ). When the joint at B is disconnected, B3 and B4 may not be coincident. For the open kinematic chain, the position of point B may be defined in two different forms as:

A0A + AB = A0B3 (1, 2, 3 open loop)

A0B0 + B0B = A0B4 (1, 4 open loop)

However, at every instant the revolute joint between links 3 and 4 must exist and point B must remain a permanently coincident point for different values of the position variables if the system we are considering is a mechanism. Therefore the vector A0B3 and A0B4 obtained from the two equations using the two open kinematic chains must be equal and this results with the vector equation:

A0A + AB = A0B0B0B

This vector equation must be valid for all positions due to the permanetly coincident points. If this vector equation can nott be satisfied for a given input angle, then that position cannot exist (mechanism cannot be assembled at that position).

In a four-bar mechanism there is a single loop formed and the vector equation describes the closure of this loop mathematically.The equation(s) that describes the closure of the loop(s) formed in the mechanism are known as loop closure equation(s). The variables in the loop closure equations are always related by the joint freedoms and we can solve for two position variables from any loop equation. In plane the vector equation will correspond to two scalar equations. In the four-bar example there are three variables (θ12, θ13 and θ14) which we shall call “position variables“. If one of the position variable (say θ12), the other position variables (θ13 and θ14) can be solved from this vector loop equation. The number of independent parameters that are required will always be equal to the degree-of-freedom of the mechanism. The relation between the position variables is a nonlinear, trigonometric relation.

One simple and concise form of writing the vector loop equations is to use complex numbers. for example, If the length of the vector A0is a2 and if the vector makes an angle θ12:

A0A = a2cosθ12 + ia2sinθ12

or, using Euler’s equation:

A0A = a2e12

In a similar fashion if the link lengths are denoted as ai olarak (a1= |A0B0|, a2=|A0A|, etc.) the vector loop equation in complex numbers can be written as:

a2e12 + a3e13 = a1 + a4e14

If required, the equation can be written in cartesian coordinates as:

a2cosθ12i + a2sinθ12j + a3cosθ13i + a3sinθ13j = a1i + a4cosθ14i + a4sinθ14j

x and y components can be equated separetely two yield two scalar equations in the form:

a2cosθ12 + a3cosθ13 = a1 + a4cosθ14
a2sinθ12 + a3sinθ13 = a4sinθ14

In case of a four-bar, the vectors in the loop closure equation have fixed magnitudes. However, the angular inclinations of the three vectors representing the moving links will change. Hence, there are three position variables (θ12, θ13 and θ14). If one of these variables is defined, the remaining two variables can be solved from the vector equation. If we refer to the definition of the degree-of-freedom of a mechanism, the variable that must be defined is the input variable and for a constrained motion the number of input variables must be equal to the degree-of-freedom of the joints involved. In case of a four-bar, since all the connections are revolute joints, the variables are all rotation variables. In case of a prismatic joint, the variable will be the magnitude of a vector or a vector component. Consider a slider-crank mechanism as shown in Fig. A. Let us disconnect the revolute joint at B (Fig. B). In order to determine the positions of links 2 and 3 we must define θ12 and θ13. To locate the position of link 4 its displacement along the slider axis must be known and the position variable s14 must be defined. The resulting loop closure equation is:

AoA + AB = AoB

Again there are 3 variables (θ12, θ13 and θ14) one of which must be specified as the input. In this case the vectors AoA and AB have fixed magnitudes and varying directions. The vector AoB has a fixed y component (length c) and a changing x component (s14). Depending on the applications either θ12 (i.e. in pumps) or s14 (i.e. internal combustion engines) is the input. In complex numbers the vector loop equation will be:

a2e12 + a3e13 = s14 + ic

The vectors defined and the variables used in the loop closure equations are not unique. For example, for the slider crank mechanism, rather than disconnecting the revolute joint at B, one can as well disconnect the revolute joint at A between links 2 and 3 (Fig C). We must now define the angle θ13′ = ∠xBA instead of the angle θ13 o determine the position of link 3. Note that the angles. θ13 and θ13 differ by a constant angle (In this case by 180o). The resulting loop equation is:

AoA = AoBBA

or in complex numbers:

a2e12 = s14 + ic + a3e13

Example: The loop closure equations of a six link mechanism

Referring to the four-bar mechanism above, one can write a vector equation in the form:

AoAAB AoB

Considering the vector AoB the magnitude and direction of this vector are variables (or its x and y components) and these variables are not related with the joint freedoms.we can solve for the vector AoB provided that the magnitudes and the directions of the other two vectors are known. Such an equation will not help us for the solution of position variables. Although it is a valid vector equation, it is not a loop closure equation. One can identify such loops by noting the variables involved are not related with the joint freedoms and these equations are not obtained by disconnecting the joints.

A similar argument can also be made for the vector equations:

AoA + AC = AoC         (i)

AB + BC = AC         (ii)

In the later case, all three vectors are on the same link and they have a fixed orientation with respect to each other. If the origin and the angular orientation of one of the vectors is known, due to rigidity the orientation of the other vectors will be known.

In certain other cases we may have to use instantaneously coincident points or other points on the links as the tips of the vectors, as shown in the figure. The variable involved in such a case is the relative displacement of one link with respect to another link. The vector BoA can be separated into two components: BoC and CA so that one component is of constant magnitude and the magnitude of the other vector is related to the displacement of the prismatic joint between links 3 and 4. The loop closure equation can than be written in the form:

AoA = AoBo + BoC + CA

or in complex numbers:

a2e12 = a1 + a4e14 + s43ei(θ14 + α4)

Points A2 ve A3 are permanently coincident points. Point A4 is instantaneously coincident with point A3 when the mechanism is moved from this position the two points will be displaced by a distance Δs along the slider axis relative to each other. Points A2 and A3 will be two other different points that will be coincident with (A′) (Fig. C).

The vector CA changes both its magnitude and direction. However its orientation with respect to the vector B0C will be fixed, and no new variable is needed. If the orientation of the vector B0C is whown by the variable θ14 measured from the positive x-axis of our reference, the orientation of the vector CA with respect to positive x axis is θ14 + α4 and angle α4 is a constant angle measured on link 4 between two lines BoC and CP (point P is any point on the slider axis on link 4). The position variables in the loop equation will be θ12, θ14 and s43.

Note that the same vector loop equation can be derived for the swinging block mechanism shown above. Therefore, the inverted slider crank mechanism is the same as the swinging block mechanism although their construction is different. Although it is a different construction, if the link dimensions (a2, a4, α4) are the same, the motion of the two mechanisms will be the same.

When writing the vector loop equations one must be sure that the equations are valid for every position of the mechanism. The mechanism may be at a special position such that one or more of the links are collinear as shown above (Although links 2 and 1 are collinear, they will have different orientations at some other instant). In such a case you may redraw the mechanism slightly offset from the critical position or show the variable angle θ12, as shown. In some cases if the constant link angles between two vectors are of a certain simple value (such as 90°), then it is advisable to simplifiy the equations accordingly. For example if the angle α4 of the inverted slider crank mechanism is a right angle, The loop closure equation must be written as:

a2e12 = a1 + a4e14 + is43e14

The solution to the loop equations may not exist for every value of the independent parameter. This will mean that for that particular link lengths the mechanism cannot be assembled at the requested position.

How will you determine the number of independent loops from the number of joints and links?

In planar mechanisms we can write vector loop equations for each loop of the mechanism. This corresponds to L (L= number of independent loops) vector equations or 2L scalar equations, if we equate the x and y components of vectors. The number of parameters involved in these equations will be 2L+ F, where F is the degree of freedom of the mechanism. If we now define F number of variables (independent variables or input parameters (variables)), then theoretically, we must be able to solve for the other variables (dependent position parameteres). We can change the input variable within a given range in certain increments and obtain the values for the dependent variables. For example if the input variable corresponds to the angle that defines the angular position of an input crank, we change this angle from 0 to 360o. If the input is the movement of a piston inside a cylinder, then we change this length of the piston starting from the closed position to the most extended position (the difference is the stroke of the piston).

When we determine the values of all the position variables corresponding to a certain input variable, then we can determine the position of any point on any link of the mechanism.

One need not draw the mechanism with dismantled joints to write the loop equations. After some practice one can conceptually disconect the joints, identify the loops and the write the necessary loop equations. Initially of course, as a visual aid, we have shown the joints are shown disconnected.

In the following examples the necessary loop equations for some mechanisms are written. The loop equations and the variables defined are not unique. You are to determine the joint that is disconnected when writing the given loop equation.

In recent years different package programs are available for the analysis of mechanisms. When using these programs, you must input these loops by telling the program which link is connected to which link by what kind of a joint (i.e. what kind of freedom is permitted by that joint). If you are using mathematical packages such as matlab or mathcad, you must type these equations in one form or another. A slight mistake in the loop equations results with erroneous results. Please keep in mind that these equations define the mathematical model of an existing mechanism. This mathematical model can be solved in different ways as we shall see in the coming sections.

Example I:

(a2 = |A0A|, a3 = |AB|, a4 = |BC|)

A0A + ABBC = A0C  ⇒  a2e12 + a3ei(θ13 − α3) + a4e14 = s15 + ia1

A0A + AD = A0D0 + D0D ⇒  a2e12 + s36e13 = c1 + ib1 − ia6e13

Example II:

(c1 = |PB0|, a2 = |A0A|, a4 = |BC|, a5 = |B0B|)

A0A = A0B0 +B0B + BA  ⇒  a2e12 = c1 − ia1 + a5e15 + s43e14

A0C A0B0B0BBC  ⇒  s16 + ib1 = c1 − ia1 + a5e15 + a4e14

Example III:

B0B + BC + CD = B0A0A0A + AD  ⇒  a3e13 + a4e14 + a6e16 = a1 + ib1 + a2e12 + a7e17

B0B + BC = B0C0C0C  ⇒  a3e13 + a4e14 = −d1 + ic1 + a5e15

Due to the gear pair : r3θ13 = −r212 − α2) (θ12 = α2 when θ13 = 0)