Lesson 3 Definitions • – – – – – – Your textbook provides you with a short section on some of the fundamental definitions in Kinematics. Review this section and familiarize yourself with the following definitions: Rigid bodies Mechanisms Kinematic analysis Kinematic synthesis Kinematic chains Kinematic joints A summary of some of these topics is provided in the following sections. Reference Frames • – – In a multibody system, we need to define: One non-moving (also called global, absolute, or inertial) frame: x-y (in 2D); x-y-z (in 3D) One body-fixed frame (also called local) per body: (in 2D); (in 3D); A body-fixed frame carries the corresponding body index (subscript). z Array (Vector) of Coordinates • Array of coordinates for a particle i (point) is defined as: – 2D: 3D: z x i r = y i z i ri i x ri = i y i • • ri x i , yi and z i are the Cartesian coordinates of the particle For n particles: i = 1, ..., n r1 r= r n • Array of coordinates for a rigid body can be defined as: – 2D: 3D: xi qi = yi i • • • xi y i zi qi = i i i i i z Oi i Oi i i i r ri i xi , yi and zi are the Cartesian coordinates of the origin of the body-fixed frame Rotational coordinates in 3D, such as i , i and i will be discussed later in this course Array of coordinates for a multibody system can be defined in different ways. For example for this planar four-bar we can define B 1 A q = 2 3 3 y 2 Or, we may define 1 q1 q = q 2 x q 3 where qi ; i = 1, 2, 3 are either 3-vectors (2D) or 6-vectors (3D) Degrees of Freedom • • An important characteristics of any mechanical system is its number of degrees-of-freedom (DoF) In most planar systems we can determine the number of DoF intuitively, however, in spatial systems we may have a hard time to do the same. In most cases, we could apply an analytical formula to find the number of DoF! A triple pendulum has 3 DoF A four-bar mechanism has 1 DoF Constraints • The dependency between the coordinates of a mechanism is called a constraint. For the 4-bar mechanism if the array of coordinates is defined as 3 1 2 q = 2 2 3 3 Then we can write the following position constraints 1 1 cos 1 + 2 cos 2 + 3 cos 3 a = 0 1 b 1 sin 1 + 2 sin 2 + 3 sin 3 b = 0 a Note: No. of DoF = No. of coordinates – No. of independent constraints Kinematic Constraints • Assume a multibody system having k degrees-of-freedom: – – An array of n coordinates is defined as q – The Jacobian matrix is defined as an m x n matrix D q – Velocity constraints are expressed as m equations D q = 0 – Acceleration constraints are expressed as m equations q = 0 D q + D Position constraints are expressed as m equations, where k = n - m (q) = 0 Driver Constraint • – For kinematic analysis of a multibody system with k degrees-of-freedom we must define k driver constraints. Most mechanisms have only one DoF! Driver constraints at position, velocity, and acceleration levels are of the form (d ) ( q) f (t) = 0 (d ) (d ) D q f (t) = 0 (d ) • • (d ) q + (d ) D Dq f (t) = 0 Driver constraints normally have very simple forms A driver constrain is normally a function of a single coordinate Example: Slider-Crank Mechanism • A 2D example; Driver constraint and the derivatives – Case 1: crank (i) is the driver (d,1) i 0i 0i t 12 t 2 = 0 (d,1) i 0i t = 0 i = 0 (d,1) – Case 2: slider (j) is the driver (d,1) x j a cos t = 0 x j + a sin t = 0 (d,1) x j + a 2 cos t = 0 (d,1) Kinematic Analysis • At t = i t solve the following equations: – Position analysis: Solve the following m + k nonlinear algebraic equations for q (q) = 0 (q) f (t) = 0 (d ) – Velocity analysis: Solve the following m + k linear algebraic equations for q D q = 0 (d ) – D q f (t) = 0 Acceleration analysis: Solve the following m + k linear algebraic equations for q q = 0 D q + D (d ) • (d ) (d ) q + (d ) D Dq f (t) = 0 The time variable t is incremented from an initial time to a final time at reasonable increments (the size of the increments t may vary from one problem to another). The position, velocity, and acceleration constraints are solved at every time-step t.