Numerical Integrator. Graphics

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1 1 Introuction CS229 Dynamics Hanout The question of the week is how owe write a ynamic simulator for particles, rigi boies, or an articulate character such as a human figure?" In their SIGGRPH course notes, Davi araff an nrew Witkin o an excellent job of aressing particle an rigi boy simulations. This hanout will supplement those notes by escribing how to set up a simulation of an articulate character. 2 System for Dynamic Simulation state, time Numerical Integrator Compute Derivatives state erivative new state Graphics Figure 1: lock iagram for a passive ynamic simulation. Figure 1 shows a block iagram for a ynamic simulation of a passive system. The user provies an initial state of the system as input, an numerical integrator churns away to fin the state of the system at future points in time. s the integrator works on the problem, it will perioically ask for erivatives of the state vector, given a state an a time. Occasionally, the state information is passe to a graphics subsystem so that the results can be isplaye. Let's look at each of these pieces in a little more etail. The numerical integrator, although it seems to take central place in the figure, will probably be the simplest element. For example, assume the state of our system is a one-imensional position an velocity [x; v]. We coul write a very basic numerical integrator as follows: ffl sk for erivatives given the start state [x 0 ;v 0 ] an time t 0. ffl Derivatives are returne. For example, if we have a particle in a gravitational fiel with acceleration g, the results will be [v 0 ;g]. Velocity v 0 is the erivative of position x 0 an acceleration g is the erivative ofvelocity v 0.

2 ffl ssume the erivatives are constant for the entire timestep an return [x 1 ;v 1 ] = [x 0 + v 0 t; v 0 + g t] as the next state. This is calle Euler integration, an it oes not work especially well. Runga-Kutta is not much more complex an prouces better results by requesting erivatives as several ifferent points within an integration timestep. Implicit integration is somewhat more computationally expensive, but can ramatically improve stability if you have a stiff system. Stiff systems can result from spring-mass simulations, from penalty metho computations use for collision response, or from characters with active control systems. (We'll iscuss control systems later on in class.) For more information, practically presente, see Numerical Recipes, Chapter 16 on Integration of Orinary Differential Equations. The graphics subsystem must take a state vector an isplay the correct character or object configuration on the screen. To make this work, the user must efine the geometry an renering style of the object an specify camera placement. The initial inputs to the system are an initial state vector an an integration timestep. character may begin an action from a neutral pose, with zero velocities, for example. The integration timestep is supplie to control the accuracy, stability, an spee of the simulation. That leaves the erivative computation. In most cases this is largely an exercise in computing forces an torques exerte on the system ue to interaction between boies or between boies an the environment. Forces an torques may come, for example, from: ffl Gravity ffl Collisions ffl rtificial force fiels esigne to steer objects in esire irections (e.g. away from collisions) ffl Drag (e.g. ue to velocity in flui) ffl Spring forces in a spring mass system The actual esign of forces an torques in your system is up to you as a esigner. Note that we are assuming that any of these forces can be compute base only on the current state of the system. To compute spring forces, for example, we only nee to know the current istances between particles. To compute rag forces, we only nee to know velocity information. 1 ssuming that an appropriate mechanism for computing forces an torques is in place, we are face with the problem of computing state erivatives base on these forces an torques.

3 θ x r x r y θ x Figure 2: Two rigi boies, connecte by a single rotational joint. 3 rticulate Mechanisms - n Example We will look at the very simple articulate mechanism shown in Figure 2. The mechanism is a pair of rectangular rigi boies that are free to translate an rotate in the plane. The boies are connecte by a single joint, an they can also rotate about that joint. Gravity acts in the ^y irection. In this section, we will go through computation of the state vector an erivatives for this system. 3.1 State Vector What state vector shoul we use for this example? If we follow araff's lea, we will have the following information for each of the two boies an. There are six total parameters for each boy, giving us a state vector of size 12: ffl Position (^x an ^y) ffl Linear momentum (in both ^x an ^y irections) ffl Orientation (about ^z) ffl ngular momentum (in the ^z irection) 1 It is not always the case that forces an torques can be compute base on state alone. One example is computation of actively controlle joint torques (think muscle forces) for an animate character. If a character is to walk across a room in a coorinate way, itmay nee to carry aroun more information than just the current state of the physical system at each point in time! ut, we'll get to that in a couple of weeks.

4 We can then write the state vector as: Y (t) =[x ; P ; ; L ; x ; P ; ( + ); L ] T (1) Let's look at each of the components of Y : ffl x is the position of the center of mass of boy in the ^x an ^y irections. ffl P is the linear momentum of boy. This is simply m _x, where m is the mass of boy an _x is the velocity ofboy. ffl is the orientation of boy. ffl L is the angular momentum of boy. This is just I _, where I is the inertia of boy in the ^z irection about its center of mass an _ is the angular velocity ofboy. I is a scalar. 2 ffl x is the position of the center of mass of boy in the ^x an ^y irections. ffl P is the linear momentum of boy. This is simply m _x, where m is the mass of boy an _x is the velocity ofboy. ffl ( + ) is the orientation of boy. ffl L is the angular momentum of boy. This is just I ( _ + _ ), where I is the inertia of boy in the ^z irection about its center of mass an ( _ + _ ) is the angular velocity of boy. Note that 's angular velocity epens on the sum of changes in angles an over time. This is a little confusing, but oes follow the kinematics iscussion in Watt an Watt. 3.2 State Vector Derivatives To compute the erivatives of the state vector, we nee to know all the forces an torques acting on each of the boies an. If the boies are passively falling, each is subject to gravitational forces, which we can easily calculate. There is also an unknown force, however, that acts to hol the boies together at the joint. We represent this force explicitly for both boies in the Free oy Diagram of Figure 3. We make use of Newton's thir law here: for every force there is an equal an opposite reaction force. Now we can represent erivatives using the two basic equations relating force to linear momentum an torque to angular momentum: f(t) = t P (2) fi(t) = t L (3) 2 fter all the iscussion in class about inertia tensors, why shoul I be a scalar? You can convince yourself that it is true by working out the inertia tensor for a rectangular boy an examining how it changes with orientation of that boy.

5 f j y r r x m g -f j m g Figure 3: Free boy iagram of the two rigi boies. Term by term: t x =_x = 1 P m (4) t P = m ẍ = f j m g^y (5) t = _ = 1 L I (6) t L = I = r f j () t x =_x = 1 P m (8) t P = m ẍ = f j m g^y (9) t ( + ) =( _ + _ ) = 1 L I (10) t L = I ( + ) = r ( f j ) (11) This is all wonerful, except for the fact that we have no iea about the value of f j on the various right han sies of the equations above Solving for ccelerations To compute the erivative of the state vector, we nee to either solve for constraint force f j or somehow eliminate it from the equations above. Our unknowns (expresse now as scalars) are: ẍ, ÿ,,ẍ,ÿ,, f j;x, f j;y.

6 From the above expressions of erivatives, we have six equations in these eight unknowns: m ẍ = f j;x (12) m ÿ = f j;y m g (13) I = (r f j ) ^z (14) m ẍ = f j;x (15) m ÿ = f j;y m g (16) I ( + ) = (r f j ) ^z (1) The value of r an r in these equation is (from Figure 3): r = [l cos( );l sin( )] T (18) r = [ l cos( + ); l sin( + )] T (19) n the final piece of information that we nee to solve the system is the connection between boies an. This equation is the only place where we are explicitly escribing the type of constraint between boies an : (20) x = x + r r (21) Substituting in for r an r in the constraint equation (Equation 21) an ifferentiating twice give us the following (try it!): ẍ =ẍ l cos( ) _ 2 l sin( ) l cos( + )( _ + _ ) 2 l sin( + )( + )(22) ÿ =ÿ l sin( ) _ 2 + l cos( ) l sin( + )( _ + _ ) 2 + l cos( + )( + )(23) Ok. Now we combine all of this information to expan equations 12 through 1 an eliminate the unknown f j;x an f j;y.we eliminate ẍ an ÿ while we're at it, since they can be calculate from Equations 22 an 23. This gives us a mess of equations that is best represente as a matrix equation: M(q) q + V (q; _q)+g(q) =0 (24) Here q is our vector of remaining, unknown accelerations: 2 3 ẍ q = 6 ÿ 4 5 (25)

7 M is a generalize mass matrix, which epens on the current state. Using S( ) for sin( ) an C( ) for cos( ) wehave the following: m + m 0 m l S( ) m l S( + ) m l S( + ) 0 m + m m l C( )+m l C( + ) m l C( + ) m l S( ) m l C( ) I 0 m l S( + ) m l C( + ) I I (26) Look at the iagonal terms. These shoul make sense. For example, (m + m )ẍ is total mass times acceleration in the ^x irection. In the absence of constraint forces, this woul be zero for a rigi boy in free-fall. ll of the aitional terms arise from the constraint at the connecting joint. lso notice that every non-zero term in the matrix M either represents a mass or an inertia. V (q; _q) isavelocity term reflecting centrifugal an coriolis forces. These seem very mysterious, but they pop up any time you have circular motion. They reflect the iea that a mass rotating at constant angular velocity about an axis actually has a linear acceleration, an the irection of that acceleration is along the vector from the mass to the axis of rotation. 2 3 V (q; _q) = 6 4 m l C( ) _ 2 m l C( + )( _ + _ ) 2 m l S( ) _ 2 m l S( + )( _ + _ ) n, finally, G(q) is a term reflecting forces ue to gravity: G(q) = (m + m )g m l C( )g m l C( + )g 3 5 (2) 5 (28) Now we can solve the system of equations above for the unknown accelerations, plug them into the expresstion for _ Y (t) in Equations 4 through 11, an return _ Y (t) to the numerical integrator to continue the simulation! ing pplie Forces The above erivation only allows us to calculate erivatives for a jointe boy in free fall. We also woul like tohave the ability to apply torque at the joint between boies (e.g. if there is a motor at that joint) an to apply external forces to the boies (e.g. collision forces). How can we a these extra forces an torques? The secret can be foun in the free boy iagrams. Simply a in the new forces, noting that torque applie to one boy at the joint will result in the opposite reaction torque on the secon boy (Figure 4). 3 5

8 f,e τ f j f,e r r,e y x r,e m g -f j r τ m g Figure 4: Free boy iagram with ae forces. Now the original Equations 12 through 1 can be written as: m ẍ = f j;x + f ;e;x (29) m ÿ = f j;y m g + f ;e;y (30) I = (r f j ) ^z + r ;e f ;e fi (31) m ẍ = f j;x + f ;e;x (32) m ÿ = f j;y m g + f ;e;y (33) I ( + ) = (r f j ) ^z + r ;e f ;e + fi (34) Propagating these aitional forces an torques through the erivation of the equations of motion gives us a new version of Equation 24: M(q) q + V (q; _q)+g(q) =F (q) (35) where M(q), V (q; _q), an G(q) are all as before! The new force term F (q) is: 2 f ;e;x + f ;e;x f ;e;y + f ;e;y F (q) = 6 4 r f ;e ^z + r ;e f ;e ^z fi r f ;e ^z + r ;e f ;e ^z + fi 3 5 (36) The first two terms in this matrix shoul seem intuitive if you think about how the applie forces an torques will affect linear acceleration. The secon two terms will probably not seem intuitive, because the last two equations in this system mix linear an angular accelerations.

9 4 Implementation Notes There are a couple of things that shoul be consiere when actually implementing a simulation such as that escribe in this hanout. The first is that the state vector Y (t) that we are using is reunant. We coul, for example, eliminate x an P, because these terms can be compute from other state parameters (e.g. see Equation 21). In fact it is probably an excellent iea to remove x an P from the state vector, because numerical errors coul otherwise give us inconsistent values for x. In other wors, the two boies coul appear to slowly rift apart. Secon, inverting a matrix every time we nee to compute erivatives is expensive, an these erivatives will have to be compute many many times uring the course of a particular simulation. On the other han, the equations of motion for a boy o not change uring the simulation. This means it is well worth the time to symbolically solve the system in Equation 24 to avoi oing matrix inversion. Thir, you might ask how o we a aitional egrees of freeom to the system?" n actual character has many more joints than the one iscusse here. Well, the basic principle remains the same. For every new joint ae to a tree, we will have new state parameters for the egrees of freeom at that joint. Specifically, there will be two new parameters: the joint angle for the new joint n, an the angular momentum of the new boy about the axis of rotation of that joint L n. These parameters are ae to state variable Y (t). free boy iagram rawn for the new boy can help you figure out how to compute the erivatives. cceleration of the new joint angle must be ae to the matrix equation 24 to solve for motion at the new joint. This technique oes get cumbersome very quickly, however, an it may be helpful to refer to some of the classic references on the topic Kane, T. R. an Levinson, D.., Dynamics: Theory an pplication, McGraw-Hill, New York, Featherstone, R., Robot Dynamics lgorithms, Kluwer caemic Publishing, 198.

Vectors in two dimensions

Vectors in two dimensions Vectors in two imensions Until now, we have been working in one imension only The main reason for this is to become familiar with the main physical ieas like Newton s secon law, without the aitional complication