Analytical Modeling of Kinematic Linkages, Part 2 ME 3230 R. Lindeke, Ph.D.

Topics: Slider Crank Modeling Inverse Slider Crank System Modeling RPRP – 4 vectors RRPP – 3 vectors Compound Mechanisms ME 3230 9/22/2018

General Approach to Slider Crank Linkages: rP = r1 + r4 = r2+r3 Just like 4-Bar mechanisms Must be satisfied throughout the entire motion of the linkage r1 has fixed angle but variable length 4 = 1 + /2 rad (both constants!) ME 3230 9/22/2018

Looking at Solutions: Constants in Slider Crank: Thus, Depending on which of the 3 remaining terms is the Input to the system we will solve the component equations for 2 and/or 3 and/or r1 in terms of the one which is the input ME 3230 9/22/2018

Velocity and Acceleration Again, we need to take the derivatives (1st and 2nd w.r.t time) of the solutions from our position equations (for 2 of the 3 values in terms of the 3rd value) When we work the solutions, we must specify an assembly mode value () to match the solution we desire! Finally, Since we will encounter ‘Sqrt’ functions too, we must recognize that negative values indicate geometries that can’t be assembled (remembering the A-B-C rules we encountered earlier) Tables 5.4 and 5.5 include summaries of equations for Slider-Crank with angular or Linear inputs ME 3230 9/22/2018

Try One? The venerable 350CI V8 Solve with 2 as input Currently 2 is at 55 Crank is turning at 8300 RPM (869.17 rad/s) Determine the velocity and acceleration of X1 ME 3230 9/22/2018

Starting Point – Finding 1 and Magnitude of X2 ME 3230 9/22/2018

Solutions: Positions Drawing equations from Table 5.4: r1 = 7.1319” 3 = .7340 rad (42.06) ME 3230 9/22/2018

Solution: Velocity The velocity of the piston is 347.35”/s downward ME 3230 9/22/2018

Solution: Acceleration Crank Acceleration is constant! Piston is accelerating downward at 1.44 million”/s2! ME 3230 9/22/2018

Looking at the Inverse Slider-Crank These are Mechanisms (seen in Backhoes, Boom Trucks, etc) where linear actuators are used to move the linkage and this “Slider” is not direct connected to an end of the mechanism as in a Slider-Crank Analytically they consist of revolute to slider joint connected by a link Again we have 4 links driving Pt. P Solution is found by the link closure eqns: rP = r2 = r1+r3+r4 ME 3230 9/22/2018

Positional Models: r1 (base vector) constant in direction & magnitude r2 and r4 constant in magnitude (directions change) r3 variable in both magnitude and direction Therefore: r1, r2, r4, 1 4 constants ME 3230 9/22/2018

Solutions: Again, we need to solve models for position, velocity and acceleration depending on the input “variable”: one of 2, 3, or r3 These models are summarized in Table 5.6 (2 as input) Table 5.7 when 3 is the input value Table 5.8 when r3 is the input value ME 3230 9/22/2018

RPRP Mechanisms A common mechanism using it is the Rapson slide used in steering gear for ships Closure Equations are again “4-bar” The constants are: r1, r2, 1 and 4 ME 3230 9/22/2018

Models: ME 3230 9/22/2018

Solution follows as before: Select which of r3, r4, 2 or 3 should be chosen as input – and with 3 equations we can solve the system Practically since 2 and 3 are directly related only one makes sense as a choice. This last fact typically means that only one angle or one length is chosen as the input When Angular Input is used the trajectory equations are summarized in Table 5.9 Linear Input r4 are summarized in Table 5.10 Linear Input r3 are summarized in Table 5.11 Example: Problem 5.7 ME 3230 9/22/2018

The RRPP Linkage A common application is the Scotch Yoke mechanism One finds them in places where we index or move items using a rotary input drive. These mechanisms are typically 3-vector sets ME 3230 9/22/2018

Solution In this mechanism: r2, 1 and 3 are constants Additionally: where  is also constant Typically 2 is chosen for input but r1 or r3 can also be chosen Solution for Trajectory is found in Tables: 5.12 if 2 is input 5.13 if either r1 or r3 is chosen See Problem 5.18 as an example ME 3230 9/22/2018

Elliptic Trammel (RRPP) and Oldham Mechanisms (RPPR) Both are 3 link mechanisms as seen above as mechanisms and models ME 3230 9/22/2018

Elliptic Trammel (RRPP) and Oldham Mechanisms (RPPR) These mechanisms are used to generate elliptical coupler curves for function generation Analytically: Constants are r3, 1, 2 (and: 1 = 2 + ) Variables are: 3, r1, and r2 (one must be known) Trajectory Models are seen in Tables 5.14 (angular input) or 5.15 (r1 input) See Problem 5.5 as an example ME 3230 9/22/2018

Elliptic Trammel (RRPP) and Oldham Mechanisms (RPPR) An Oldham coupler is a method to transfer torque between two parallel but not collinear shafts. It has three discs, one coupled to the input, one coupled to the output, and a middle disc that is joined to the first two by tongue and groove. The tongue and groove on one side is perpendicular to the tongue and groove on the other. Often springs are used to reduce backlash of the mechanism. The coupler is much more compact than, for example, two universal joints. – Wikipedia Definition Originally developed for paddlewheel steamers but now being used in Compressors, Disc Brake Schemes, etc. ME 3230 9/22/2018

Elliptic Trammel (RRPP) and Oldham Mechanisms (RPPR) R1 and 1 are constants Additional model (beyond vector state components): 4 = 2 +  ( is constant) Variable (one must be input): 2, 4, r2 & r4 Using the 3 equations we can solve the mechanism! Table 5.16 includes trajectory eqns. For ’s as input Table 5.17 includes trajectory eqns. When r’s as inputs ME 3230 9/22/2018

Addressing Multiple loops Using Systematic Loop equations with universal reference frame and vectors drawn from joint to joint Treat sliders as two vectors: one in slider direction, one normal to the slider path After sketching the mechanism to ‘near-scale’ determine known angles and lengths and variable angles and lengths ME 3230 9/22/2018

Following up on Compound Mechanisms: Write all separate “loop” paths as state equations (x-components and Y-components) Write any auxiliary equations (typically angles or perhaps lengths) Call n the total number of equations for the mechanism = 2*#loops + Aux. Eqns. Determine the DOF of the mechanism (f) The mechanism can be uniquely solved if: n + f = # of Unknown lengths or angles ME 3230 9/22/2018

Following up on Compound Mechanisms: These position solution involve non-linearity's (transcendental – trigonometric – equations in the unknowns) Solution requires guess and iterate techniques (Newton-Rapson methods) – or Separation of mechanism into simpler loops – or Separation into a series of 2-Equation/2-Unknown systems For Velocities, take derivatives of the position equations (these are linear in the unknowns!) So they can be directly solved using linear solvers (matrix methods based on Gaussian elimination) For Acceleration take derivatives of the velocity equations – while involved and extended, they are also linear in the unknowns ME 3230 9/22/2018