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
9/17/2018

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
9/17/2018

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
9/17/2018

5. In Mathematical Terms:
ME 3230
9/17/2018

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

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
9/17/2018

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
9/17/2018

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
9/17/2018

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

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

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
9/17/2018

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

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
9/17/2018

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
9/17/2018

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 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

17. Leads to:
ME 3230
9/17/2018

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 tan-1(ratio)
If ATAN2(-,-) 3rd quadrant –actual angle is tan-1(ratio)
If ATAN2(-,+) it is 4th Quadrant – angle is tan-1(ratio)
Note (ratio) considered without signs!
ME 3230
9/17/2018

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
9/17/2018

20. Finding Velocity Equations:
ME 3230
9/17/2018

21. Working on Acceleration:
ME 3230
9/17/2018

22. 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

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
9/17/2018

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

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

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

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

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

29. Attacking Acceleration
Here:
ME 3230
9/17/2018

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

31. Solution from Mathematica
ME 3230
9/17/2018

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
9/17/2018

33. As Seen Here:
ME 3230
9/17/2018

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
9/17/2018