ANALYTICAL LINKAGE SYNTHESIS
Imagination is more important than knowledge ALBERT EINSTEIN 5.0 INTRODUCTION
With the fundamentals of position analysis established. We can now use these techniques to synthesize linkages for specified output positions analytically .The synthesis tech-niques presented in Chapter 3 were strictly graphical and somewhat intuitive .The analytical synthesis procedure is algebraic rather than graphical and is less intuitive .However, its algebraic nature makes it quite suitable for computerization.These analytical synthesis methods were origiated by Sandor and further developed by his students.Erdman,Kaufman,and Loerch et al.
5.1TYPE OF KINEMATIC SYNTHESIS
Erdman and Sandor define three types of kinematic synthesis ,function ,path,and motion generation,which were discussed in Section 3.2.Brief definitions are repeated here for your convenience,
FUNCTION GENERATION is defined as the correlation of an input function with an output function in a mechanism.Typically. a double –rocker or crank-rocker is the result ,with pure rotation input and pure rotation output. A slider-crank linkage can
be a function generator as well ,driven form either end ,i.e,rotation in and translation out or vice versa.
PATH GENERATION is defined as the control of a point in the plane such that it fllows some prescribed path .This is typically accomplished with a fourbar crank-rocker or double-rocker,wherein a point on the coupler traces the desired output path .No attempt is made in path generation to control thelink which contains the point of interest.The coupler curve is made to pass through a set of desired output point. However,it is common for the timing of the arrival of the coupler point at particular locations along the parth to be defined. This case is called path generation with pre-scribed timing and is analogous to function generation in that a particular output func-tion is specified.
MOTION GENERATION is defined as the control of a line in the plane such that it assumes some sequential set of prescribed positions.Here orientation of the line con-taining the line is important.This is typically accomplished with a fourbar crank-rocker or double-rocker,wherein a point on the coupler traces the desired output path and the linkage also controls the angular orientation of the computer link containing the output line of interest.
5.2 PRECISION POINTS
The points ,or positions ,prescribed for successive locations of the output link in the plane are generally referred to as precision points or precision positions.The number of precision points which can be synthesized is limited by the number of equations available for solution.The fourbar linkage can be synthesized by closed-form methods for up to five precision points for motion or path generation with prescribed timming and up to seven points for function generation .Synthesis for two or three precision points is relatively straightforward,and each of these cases can be reduced to a system of linear simultaneous equations easily solved on a calculator.The four or more posion synthesis problems involve the solution of nonlinear, simultaneous equation systems ,and so are more complicated to solve,requiring a computer.
Note that these analytical synthesis procedures provide a solution which will be able to ‘’be at’’the specified precision points, but no guarantee is provided regarding the linkage’s behavior between those precision points,It is possible that the resulting linkage will be incapable of moving form one precision point to another due to the presence of a toggle position or other constraint.This situation is actually no different than that of the graphical synthesis cases in Chapter 3,wherein there was also the possibility of a toggle position between design points.In fact,these analytical synthesis methods are just an alternate way to solve the same multiposition synthesis problems.One should still build a simple cardboard model of the synthesized linkage to observe its behavior and check for the presence of problems,even if the synthesis was performed by an esoteric ananlytical method. 5.3 TWO-POSITION MOTION GENERATION BY ANALYTICAL SYNTHESIS
Figure 5-1 shows a fourbar linkage in one position with a coupler point located at a first precision position P1,It also indicates a second precision position to be achieved by the rotation of the input rocker,link2,through an as yet unspecified angleP2.Note also that the angle of the coupler link 3 at each of the precision positions is defined by the angles of the position vectors Z1 and Z2.The angle φ corresponds to the angle θ3 of link3 in it first position.This angle unknown at the start of the synthesis and will be found.The angle α2 represents the angular change of link3 form position one to position two.This angle is defined in the problem statement .
It is important to realize that the linkage as shown in the figure is schematic.Its dimensions are unknown at the outset and are to be found by this synthesis techniques.Thus,for example,the length of the position vector Z1 as shown is not indicative of the final length of that edge of link3,nor are the length or angles of any of the links as shown predictive of the final result. The problem statement is:
Design a fourbar linkage which will move a line on its coupler link such that a point p on that line will be first at P1 and later at P2 and will also rotate the line through an angleα2 between those two precision positions.Find the lengths and angles of the four links and the coupler link dimensions A1P1 and B1P1 as shown in Figure 5-1.
The two-position analytical motion synthesis procedure is as follows:
Define the two desired precision positions in the plane with respect to an
arbitrarily chosen global coordinate system XY using position vectors R1 and R2 as shown in Figure 5-1a.The change in angleα2 of vector Z is the rotation required of the coupler link.Note that the position difference vector P21 defines the displacement of the output motion of point P and is defined as: P21=R2-R1
The dyad W1Z1 defines the left half of the linkage.The dyad U1S1 defines the right half of the linkage.Note that Z1 and S1 are both embedded in the rigid coupler and both of these vectors will undergo the same rotation through angleα2 form position 1 to position 2 .The pin-to-pin length and angle of link 3 is defined in terms of vectors z1 and s1.
V1=Z1-S1
The ground link 1 is also definable in terms of the two dyads. G1=W1+V1-U1
Thus if we can define the two dyads W1,Z1,andU1,S1,we will have defined a linkage that meets the problem specifications.
We will first solve for the left side of the linkage and later use the same procedure to solve for the right side.To solve for W1 and z1 we need only write a vector loop equation around the loop,starting with W2.
W2+Z2-P21-Z1-W1=0
Now substitute the complex number equivalent for the vectors . wej(θ+β2)+zej(φ+α2)-p21ejδ2-zejφ-wejθ=0
The sums of angles in the exponents can be rewritten as products of terms. wejθejβ2+zejφejα2-p21ejδ2-zejφ-wejθ=0 Simplifying and rearranging: wejθ(ejβ2-1)+zejφ(ejα2-1)=p21ejδ2
Note that the lengths of vectors W1 and W2 are the same magnitude w because they represent the same rigid link in two different positions.The same can be said about vectors Z1 and Z2 whose commen magnitude is z.
Equations 5.5 are vector equations,each of which contains two scalar equations and so can be solved for two unknowns.The two scalar equations can be revealed by substituting Euler’s identity and separating the real and imaginary terms as was done in Section 4.5. Real part:
[wcosθ](cosβ2-1)-[wsinθ]sinβ2+[zcosφ](cosα2-1)-[zsinφ]sinα2=p21cosδ2 Imaginary part:
[wsinθ](cosβ2-1)+[wcosθ]sinβ2+[zsinφ](cosα2-1)-[zcosφ]sinα2=p21sinδ2
There are eight variables in these two equations:w, θ, β2,z, φ, α2,p21,andδ2.We can only solve for two.Three of the eight are defined in the problem statement,namely α2,p21,andδ2.Of the remainning five,w, θ, β2,z, φ, we are forced to choose three as “free choices” in order to solve for the other two.
One strategy is to assume values for the three angles,θ, β2,z, φ,on the premise that we may want to specify the orientationθ, φof the two link vectors W1 and Z1 to suit packaging constraints,and also specify the angular excursionβ2of
the link 2 to suit some driving constraint.This choice also has the advantage of leading to a set of equations which are linear in the unknowns and are thus easy to solve.For this solution,the equations can be simplified by setting the assumed and specified terms to be equal to some constants. In equations 5.6a, let
A=cosθ(cosβ2-1)-sinθsinβ2 B= cosφ(cosα2-1)-sinφsinα2
C= p21cosδ2
And in equationa 5.6b let:
D= sinθ(cosβ2-1)-cosθsinβ2 E= sinφ(cosα2-1)-cosφsinα2 F= p21sinδ2 Then: Aw+Bz=C Dw+Ez=F
And solving simultaneously,
W=(CE-BF)/(AE-BD); Z=(AF-CD)/(AE-BD)
A second strategy is to assume a length z and angleφ for vector Z1 and angular excursionβ2 of link 2 and then solve W1. This is a commonly used approach. Note that the terms in square bractets in each of equations 5.6 are respectively the x and y compoents of the vectors W1and Z1. W1x=wcosθ; z1x=zcosφ W1y=wsinθ; z1y=zsinφ Substituting in equation 5.6,
W1x(cosβ2-1)-w1ysinβ2+z1x(cosα2-1)-z1ysinα2=p21cosδ2 W1y(cosβ2-1)-w1xsinβ2+z1y(cosα2-1)+z1xsinα2=p21sinδ2
Z1xand Z1y are k own from equation 5.8a with z andφ assumed are free choiceS.To futher simplify the expression,combine other known terms as: A=cosβ2-1; B=sinβ2; C=cosα2-1 D=sinα2; E=p21cosδ2; F=p21sinδ2 Substituting:
AW1x-BW1y+CZ1x-dz1y=E AW1y+BW1x+CZ1y+dz1x=F And the solution is:
W1x=[A(-CZ1x+DZ1y+E)+B(-CZ1y-DZ1x+F)]/-2A W1y=[A(-CZ1y+DZ1x+F)+B(-CZ1x-DZ1y-E)]/-2A
Either of these strategies results in the definition of a left dyad W1Z1 and its pivot locations which will provide the motion generation specified.
We must repeatthe process for the right hand dyad,U1S1 and U2S2 of the right dyad.Vector U1 is initially at angleσ and moves through anglesγ2 from position 1 to 2.Vector S1 is intially at angleψ.Note that the rotation of vector S from S1 to S2 is through the same angleα2 as vector Z, since they are in the same link.A vector loop equation similar to equation 5.3 can be written for this dyad.
U2+S2-P21-S1-U1=0
Rewrite in complex variable from and collect terms. Uejσ(ejγ2-1)+sejψ(ejα2-1)=p21ejδ2
When this expanded and the proper angles substituted,the x and y component equations become: Real part:
ucosσ(cosγ2-1)-usinσsinγ2+scosψ(cosα2-1)-ssinψsinα2=p21cosδ2 imaginary part(with complex operator j divided out):
usinσ(cosγ2-1)+ucosσsinγ2+ssinψ(cosα2-1)+scosψsinα2=p21sinδ2 Compare equation 5.10 to euations 5.6.
The same first strategy canbe applied to equations 5.10 as was used for equations 5.6 to solve the magnitudes of vectors U and S .assuming values values for anglesσψ,andγ2.The quanatities p21, σ2,andα2 are defined form the problem statement as before,
In equations 5.10b let:
A=cosσ(cosγ2-1)-sinσsinγ2 B=cosψ(cosα2-1)-sinψsinα2 C=p21cosδ2
And in equations 5.10b let:
D= sinσ(cosγ2-1)+cosσsinγ2 E=sinψ(cosα2-1)+cosψsinα2 F= p21sinδ2 Then: Au+Bs=C Du+Es=F
And solving simultaneously,
U=(CE-BF)/(AE-BD) s=(AF-CD)/(AE-BD)
If the second strategy is used,assuming angleγ2 and the magnitude and direction of vector S1(which will define link3),the result will be: u1x=ucosσ s1x=scosψ u1y=usinσ s1y=ssinψ Substitute in equation 5.10: FLGURE 5-2
Right-side dyad shown in two positions
U1x(cosγ2-1)-u1ysinγ2+s1x(cosα2-1)-s1ysinα2=p21cosδ2 U1y(cosγ2-1)+u1xsinγ2+s1y(cosα2-1)-s1xsinα2=p21sinδ2 Let:A=cosγ2-1; B=sinγ2; C=cosα2-1 D=sinα2; E=p21cosδ2; F=p21sinδ2 Substitute in equation 5.12b AU1x-BU1y+CS1x-DS1y=E AU1y+BU1x+CS1y+DS1x=E And the solution is.
U1x=[A(-CS1x+DS1y+E)+B(-CS1y-DS1x+F)]/-2A U1y=[A(-CS1y-DS1x+F)+B(-CS1x-DS1y+E)]/-2A
Note that there are infinities of possible solutions to this problem because we may
choose any set of values for the three choices of variables in the two-position case.Technically there is in finity of solutions for each free choice.Three choice then give infinity cubed solutions .But since infinity cubed is not any more impressively large than just plain infinity.While not strictly correct mathematically ,we will,for simplicity,refer to all of these cases as having “an infinity of solutions”,regardless of the power to which infinity may be raised as a result of the derivation.There are plenty of solutions to pick from,at any rate.Unfortunately,not all will work.Some will have circuit,branch,or order(CBO) defects such as toggle positions between the precision pints.Others will have poor transmission angles or poor pivot locations or overlarge lilnks.Design judgment is still most important in selecting the assumed values for your free choices.Despite their name you must pay for those “free choices”later.Make a model.
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