ME451 Kinematics and Dynamics of Machine Systems Driving Constraints 3.5 September 30, 2013 Radu Serban University of Wisconsin-Madison

2 Before we get started…

Driving Constraints 3.5

4 Driving Constraints The context Up until now, we only discussed time invariant kinematic constraints Normally the mechanism has a certain number of DOFs Some additional time dependent constraints (“drivers”) are added to control these “unoccupied” DOFs Physically, these represent actuators whose inputs describe time history of some coordinates or relative positions You thus control the motion of the mechanism For Kinematics Analysis, you need NDOF=0

5 Driving Constraints: Types Absolute Drivers Absolute x-coordinate driver Absolute y-coordinate driver Absolute angle driver Absolute distance driver Relative Drivers Relative x-coordinate driver Relative y-coordinate driver Relative angle driver Relative distance driver Revolute-rotational driver Translational-distance driver

Absolute Drivers 3.5.1

7 Absolute Coordinate Drivers (1) Indicate that the coordinate of a point expressed in the global reference frame assumes a certain value that changes with time

8 Absolute Coordinate Drivers (2)

9 Absolute Distance Driver

Relative Drivers 3.5.2

11 Relative Coordinate Drivers Indicate that the difference in a certain coordinate of points on the two bodies, expressed in the global reference frame, has a specified time evolution.

12 Relative Distance Driver (1)

13 Relative Distance Driver (2)

14 Revolute-Rotational Driver

15 Translational-Distance Driver (1)

16 Translational Distance Driver (2)

17 Driver Constraints: Conclusions (1) What is after all a driving constraint? We start with a kinematic constraint, which indicates that a certain kinematic quantity should stay equal to zero Rather than equating this kinematic quantity to zero, we allow it to change with time: or equivalently:

18 Driver Constraints: Conclusions (2)

19 [handout] Example: Relative Distance Drivers Generalized coordinates: Prescribed motions: Derive the constraints acting on the system Derive the linear system whose solution provides the generalized velocities

20 MATLAB: How to Handle Arbitrary Motions Problem: Given a string that represents the expression of some function of time t, how do you evaluate the function, as well as its first two derivatives? % Assume the string ‘str’ contains the expression of the function % (e.g. as read from the ‘fun’ property of a driver constraint in the ADM file) fun_str = ‘(1.5 * sin(t) + 3 * t^2)^2’; % Declare a symbolic variable for time syms t; % Evaluate the given string as a symbolic expression. % NOTE: we must play a trick here to deal with the case where ‘fun_str’ % represents a constant function; i.e. there is no explicit dependency % on t. In this case, ‘eval’ by itself would return a double, not a sym! fun_sym = sym(eval(fun_str)); % Symbolically differentiate the function. funD_sym = diff(fun_sym); funDD_sym = diff(fnuD_sym); % Create Matlab function handles from the three symbolic functions above fun_handle = matlabFunction(fun_sym, ‘vars’, t) funD_handle = matlabFunction(funD_sym, ‘vars’, t); fundDD_handle = matlabFunction(funDD_sym, ‘vars’, t);

Position, Velocity, and Acceleration Analysis 3.6 Carl Gustav Jacob Jacobi (1804 – 1851)

22 Kinematic Analysis: Steps

23 Kinematic Analysis

24 Position Analysis (1) Framework: Somebody presents you with a mechanism and you select the set of nc generalized coordinates to position and orient each body of the mechanism: You inspect the mechanism and identify a set of nk kinematic constraints that must be satisfied by your coordinates: Next, you identify the set of nd driver constraints that move the mechanism: NOTE: You must end up with nc = nk + nd

25 We end up with this problem: given a time value t, find that set of generalized coordinates q that satisfy the equations: What’s the idea here? Set time t=0, and find a solution q by solving above equations Then advance the time to t=0.001 and find a solution q by solving above equations Then advance the time to t=0.002 and find a solution q by solving above equations Then advance the time to t=0.003 and find a solution q by solving above equations … Stop when you reach the end of the interval in which you are interested in the position What you do is find the time evolution on a time grid with step size  t=0.001 (in this example) You can then plot the solution as a function of time and get the time evolution of your mechanism Position Analysis (2)

26 Two issues associated with the methodology described on previous slide: The first issue: related to the fact that you are solving nonlinear equations. Does a solution exist? Example: x 2 +4=0 (no real number x will do here) Is the solution unique? Example: x 2 -4=0x (both 2 and -2 are solutions) The second issue: The equations that we have to solve at each time t are nonlinear. How do you actually solve them? For instance, how do you find the solution (x=-1.2) of this nonlinear equation: Deal with this issue next week (discuss Newton-Raphson method) Position Analysis (3)

27 Implicit Function Theorem (IFT)

28 IFT: Implications for Position Analysis

29 Velocity Analysis (1)

30 Observations: Note that as long as the constraint Jacobian is nonsingular, you can solve this linear system and recover the velocity The reason velocity analysis is easy is that, unlike for position analysis where you have to solve a nonlinear system, now you are dealing with a linear system, which is easy to solve Note that the velocity analysis comes after the position analysis is completed, and you are in a new configuration of the mechanism in which you are about to find out its velocity Velocity Analysis (2)

31 Acceleration Analysis (1)

32 Acceleration Analysis (2)