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 RRPP systems Application to “Complex Systems” Multi-loop mechanisms Higher Order Joint Types ME 3230 9/17/2018

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 9/17/2018

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 9/17/2018

In Mathematical Terms: ME 3230 9/17/2018

Considering Velocity We build a velocity equation be differentiating our positional vector equation Thus: ME 3230 9/17/2018

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 9/17/2018

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 9/17/2018

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 9/17/2018

Solving the Model for a given vector: ME 3230 9/17/2018

Leads to this general Acceleration Model: ME 3230 9/17/2018

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 9/17/2018

One Special Case (both points defining a vector on the same link) 9/17/2018

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 9/17/2018

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 9/17/2018

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 moving the 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 9/17/2018

Leads to: ME 3230 9/17/2018

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 9/17/2018

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 9/17/2018

Finding Velocity Equations: ME 3230 9/17/2018

Working on Acceleration: ME 3230 9/17/2018

All trajectory eqns. are summarized in Table 5.1 for 4-bar Linkage 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 9/17/2018

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 9/17/2018

We Can solve the Table 5.1 Equations: I used ‘Mathematica’ for Position: ME 3230 9/17/2018

And Continuing with Values: Note:  is set to -1 for the correct assembly And Continuing with Values: ME 3230 9/17/2018

This is Good – ‘cause it agrees with my scale model! ME 3230 9/17/2018

Attacking Velocity: Here: ME 3230 9/17/2018

Solving in Mathematica: 1st term is 4 ‘dot’ in rad/sec ME 3230 9/17/2018

Attacking Acceleration Here: ME 3230 9/17/2018

Simplifying and Into Matrix Form: We will solve this matrix in Mathematica as well! ME 3230 9/17/2018

Solution from Mathematica ME 3230 9/17/2018

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 9/17/2018

As Seen Here: ME 3230 9/17/2018

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 9/17/2018