CS 283Advanced Computer GraphicsSimulation BasicsJames F. O’BrienAssociate ProfessorU.C. Berkeley2•A solid object that does not deform•Consists of infinite number of infinitesimal mass points... •...that share a single RB transformation•Rotation + Translation (no shear or scale)•Rotation and translation vary over time•Limit of deformable object as A Rigid BodyxW= R · xL+ tks→∞Tuesday, November 24, 20093A Rigid BodyIn 2D: In 3D: Translation 2 “directions”Rotation 1 “direction”Translation 3 “directions”Rotation 3 “direction”3 DOF Total6 DOF Total2D is boring... we’ll stick to 3D from now on...Translation and rotation are decoupledCenter of mass4Translational MotionvJust like a point mass:˙p = v˙v = a = f/mNote: Recall discussion on integration...Tuesday, November 24, 20095Rotational MotionvωRotation gets a bit odd, as well see...Rotational “position” Rotation matrix Exponential map QuaternionsRotational velocity Stored as a vector (Also called angular velocity...) Measured in radians / second Rω6Rotational MotionvωKinetic energy due to rotation:“Sum energy (from rotation) over all points in the object”E =�Ω12ρ˙x ·˙x duE =�Ω12ρ([ω×]x) · ([ω×]x) duTuesday, November 24, 20097Rotational MotionvωAngular momentum Similar to linear momentum Can be derived from rotational energyHFigure is a lie if this really is a sphere...H =�Ωρ x ×˙x duH =�Ωρ x × (ω × x) duH =��Ω··· du�ωH = Iω“Inertia Tensor” not identity matrix... H = ∂E/∂ω8Inertia TensorI =ZΩρy2+ z2−xy −xz−xy z2+ x2−yz−xz −yz x2+ y2duSee example for simple shapes at http://scienceworld.wolfram.com/physics/MomentofInertia.htmlCan also be computed from polygon models by transforming volume integral to a surface one. See paper/code by Brian Mirtich. Tuesday, November 24, 20099˙HW=˙RILRTωW+ RIL˙RTωW+ RILRTαWRotational MotionvωHFigure is a lie if this really is a sphere...HW= IWωWConservation or momentum:˙HW=0˙R = ω × RHW= RILRTωWαW=(RILRT)−1(−ωW× HW)In other words, things wobble when they rotate.10Rotational MotionvωHFigure is a lie if this really is a sphere...Take care when integrating rotations, they need to stay rotations.αW=(RILRT)−1�(−ωW× HW)+τ�τ = f × x˙R =[ω×]R˙ω = αTuesday, November 24, 200911Couples•A force / torque pair is a couple•Also a wrench (I think)•Many couples are equivalentτfτf12Constraints•Simples method is to use spring attachments•Basically a penalty method•Spring strength required to get good results may be unreasonably high•There are ways to cheat in some contexts...Tuesday, November 24, 200913Constraints•Articulation constraints•Spring trick is an example of a full coordinate method•Better constraint methods exist•Reduced coordinate methods use DOFs in kinematic skeleton for simulation•Much more complex to explain•Collisions•Penalty methods can also be used for collisions•Again, better constraint methods exist14A Simple Spring•Ideal zero-length spring•Force pulls points together•Strength proportional to distance fa→b= ks(b − a)fb→a= −fa→bTuesday, November 24, 200915A Simple Spring•Energy potentialfa→b= ks(b − a)fb→a= −fa→b-2-1012-2-101202468-012fa= −∇aE = −�∂E∂ax,∂E∂ay,∂E∂az�E =1/2 ks(b − a) · (b − a)16A Simple Spring•Energy potential: kinetic vs elastic-6 -4 -2 2 4 6-1-0.50.51E =1/2 ks(b − a) · (b − a)E =1/2 m(˙b −˙a) · (˙b −˙a)Tuesday, November 24, 200917Non-Zero Length Springsfa→b= ksb − a||b − a||(||b − a|| − l)Rest lengthE = ks(||b − a|| − l)2-2-1012-2-10120123-01218Comments on Springs•Springs with zero rest length are linear•Springs with non-zero rest length are nonliner•Force magnitude linear w/ discplacement (from rest length)•Force direction is non-linear•Singularity at ||b − a|| =0Tuesday, November 24, 200919Damping•“Mass proportional” damping•Behaves like viscous drag on all motion•Consider a pair of masses connected by a spring•How to model rusty vs oiled spring•Should internal damping slow group motion of the pair?•Can help stability... up to a pointf = −kd˙af˙a20Damping•“Stiffness proportional” damping•Behaves viscous drag on change in spring length•Consider a pair of masses connected by a spring•How to model rusty vs oiled spring•Should internal damping slow group motion of the pair?fa= −kdb − a||b − a||2(b − a) · (˙b −˙a)Tuesday, November 24, 200921Spring Constants•Two ways to model a single springl∆l∆l/2l/2l/2∆l/222Spring Constants•Constant gives inconsistent results with different discretizations•Change in length is not what we want to measure•Strain: change in length as fraction of original lengthks� =∆ll0Nice and simple for 1D...Tuesday, November 24, 200923Structures from Springs•Sheets•Blocks•Others24Structures from Springs•They behave like what they are (obviously!)This structure will not resist shearingThis structure will not resist out-of-plane bending either...Tuesday, November 24, 200925Structures from Springs•They behave like what they are (obviously!)This structure will resist shearingbut has anisotopic biasThis structure still will not resist out-of-plane bending26•They behave like what they are (obviously!)Structures from SpringsThis structure will resist shearingLess biasInterference between spring setsThis structure still will not resist out-of-plane bendingTuesday, November 24, 200927•They behave like what they are (obviously!)Structures from SpringsThis structure will resist shearingLess biasInterference between spring setsThis structure will resist out-of-plane bendingInterference between spring setsOdd behaviorHow do we set spring constants?28Edge SpringsBridson et al. / Simulation of Clothing4. An Accurate Model for BendingThe physics of cloth bending are poorly understood. The dy-namics of anisotropic fibers twined together and woven intoa sheet of fabric constantly interacting with massive defor-mations and friction is certainly more difficult to model witha two-dimensional continuum than for example steel. How-ever, several basic qualitative properties of such a model canbe identified that are essential for a plausible simulation, andwithout these a model is incorrect.In order to handle unstructured triangle meshes and getfiner, more robust control over bending than in
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