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Analytical Modeling of Kinematic Linkages

Analytical Modeling of Kinematic Linkages. Vector Chain Analysis ME 3230 R. R. Lindeke, Ph.D. Topics of Discussion. The fundamental Models of closure Position Velocity Acceleration Applications to Various “Simple Systems” 4-Bar Linkages Slider–Cranks and their Inversion RPRP systems

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Analytical Modeling of Kinematic Linkages

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  1. Analytical Modeling of Kinematic Linkages Vector Chain Analysis ME 3230 R. R. Lindeke, Ph.D.

  2. Topics of Discussion • The fundamental Models of closure • Position • Velocity • Acceleration • Applications to Various “Simple Systems” • 4-Bar Linkages • Slider–Cranks and their Inversion • RPRP systems • RRPP systems • Application to “Complex Systems” • Multi-loop mechanisms • Higher Order Joint Types ME 3230

  3. Representing a “Chain of Points” as a “Set of Vectors” For a Mechanism • We define all angle using RH Rules from +X direction (CCW is positive) • We will be interested in resolving vectors into X (i related terms) and Y (j related terms) components which can then be solved for defining position and motion of target geometries from fundamental locations • While it is often possible (and always desirable), the vectors of “Resolution” need not correspond to the links of the mechanism at hand! ME 3230

  4. Referring to the Image of Vector Sets: • We desire a positional (and trajectory) model of Point Q (or S & P & Q) • Defining: RQ = R1 + R2 + R3 • In general we say: RQ = Rn (all n-vectors in set) • These can be resolved into Directional Components (Xi & Yj) ME 3230

  5. In Mathematical Terms: ME 3230

  6. Considering Velocity • We build a velocity equation be differentiating our positional vector equation • Thus: ME 3230

  7. Closely examining these ideas: • When we take a derivative of a vector, the magnitude and direction can both change • We must use the chain rule to differentiate the vectors • Thus for each vector: ME 3230

  8. Putting These into X and Y Component Models and generalizing: This is a general velocity model – it would be solved over the closure chain for a mechanism ME 3230

  9. Stepping over to Acceleration: • We can differentiate our Velocity model • Again, as with velocity models we apply the chain rule since both direction of acceleration and magnitude of acceleration can change: ME 3230

  10. Solving the Model for a given vector: ME 3230

  11. Leads to this general Acceleration Model: ME 3230

  12. Using these 3 general models, we can analysis any linkage • When one studies Chapter 5 we see a series of solutions for specific mechanisms (5 are single-loop) • 4-bar • Slider-Crank and Its inversions • RRPP • RPRP • Each uses the general models we have derived here and completes closures of simple or complex structural loops ME 3230

  13. One Special Case (both points defining a vector on the same link) ME 3230

  14. 4-Bar Linkage Done Analytically! • Note the definition of each vector angle: R.H. Rule (CCW is positive) • The Closure equation is:rP = r2 + r3 = r4 + r1 • Mathematically: ME 3230

  15. By Components: • These closure Equations must be satisfied throughout the motion of the linkage • The Base Vector is constant (r1 & 1 are constant) • Depending on which link is the input, we solve the above “component equations” for the other two angles (their position, Vel. and Accel.) recognizing that the vectors lengths are not changing so the ‘ri-dot’ terms in the general equations will be zero ME 3230

  16. Focusing on the Case when Link 2 is Input: 2, and its derivatives, are known throughout motion • Follow along in the text … we see that we isolate 3 by movingthe 2 terms to the RHS of our component equations (2 and 1 known's) • Square the two component equations, add them together and eliminate 3 by using Sin2X + Cos2X = 1 ME 3230

  17. Leads to: ME 3230

  18. A couple more ideas: • Because of the “Square root” in 4 solution we may find a complex solution – this is a configuration that can’t be assembled (Type 2 Grashof linkages at certain geometries) • Dealing with the Inverse tangent: • Over time a ‘special function’ has been developed to give us absolute angle using quadrant identity • This is the ATAN2(Numerator, Denominator) model *but be careful when entering the numbers* • Using this function, the SIGN of the Numerator and Denominator functions are maintained and we solve the angle as a specific quadrant value • If we find ATAN2(+,+) it is in the 1st quadrant • If ATAN2(+,-) it is 2nd quadrant – actual angle is 180 - tan-1(ratio) • If ATAN2(-,-) 3rd quadrant –actual angle is 180 + tan-1(ratio) • If ATAN2(-,+) it is 4th Quadrant – angle is 360 - tan-1(ratio) • Note (ratio) considered without signs! ME 3230

  19. Solving for 3 – Back Substitute 4 in Component Equation • Then divide Y-component by X-component to get: Tan(3)={Sin3/Cos3} *** Be Very Careful of the software you are using – the original ATAN2 function used AcrTan[Denominator, Numerator] rather than (Numerator, Denominator) to define the “ratio” *** ME 3230

  20. Finding Velocity Equations: ME 3230

  21. Working on Acceleration: ME 3230

  22. Since these equation are only in terms of two Unknowns ( and ) we can solve them: All trajectory eqns. are summarized in Table 5.1 for 4-bar Linkage ME 3230

  23. Application: Try Problem 5.23 • A person doing a pushup is a “quasi-4Bar” Linkage • Link 1 is the floor • Link 2 is the Forearm • Link 3 is upper arm – the coupler as a driver – driven from Link 2 • Link 4 is the back/legs • Find 4 and 4 ME 3230

  24. We Can solve the Table 5.1 Equations: • I used ‘Mathematica’ for Position: ME 3230

  25. And Continuing with Values: Note:  is set to -1 for the correct assembly ME 3230

  26. This is Good – ‘cause it agrees with my scale model! ME 3230

  27. Attacking Velocity: • Here: ME 3230

  28. Solving in Mathematica: 1st term is 4 ‘dot’ in rad/sec ME 3230

  29. Attacking Acceleration • Here: ME 3230

  30. Simplifying and Into Matrix Form: We will solve this matrix in Mathematica as well! ME 3230

  31. Solution from Mathematica ME 3230

  32. Getting Information about “Other Points” on a rigid body • This means points not directly on the vector loops • like coupler points on plate members (or P on the Luffing Crane) • Considering a Pt C – that is related to 2 other points (A & B) on a Link as a vector at an angle  from the line AB – we’ll call AC link 6 • We can find L. position, velocity and acceleration of Pt. C if: ME 3230

  33. As Seen Here: ME 3230

  34. Mathematically: • We can also proceed if we know pos., vel, and acc. of Pts A and B but not the angle 5 • Uses ArcTan for finding angle 5; then angle velocity by solving vel. Eqn. of rB; finally angle accel. by solving accel. Eqn. for rB • See table 5.3 ME 3230

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