Rigid Body Simulation

0 前言

声明:此篇博客仅用于个人学习记录之用,并非是分享代码。Hornor the Code

刚体动力学仿真的实现,方法来自Games103。

同时因为之前学习Jonh Crag的《机器人学》,感觉自己的动力学部分差的很多,所以也有很多动力学的知识来自于Cambridge David Tong: Lectures on Dynamics and Relativity, 这个讲义的相对论那边没怎么看。

参考了《Physically Based Modeling by David Baraff Pixar Animation Studios》 只用到了前面一章的内容 Unconstrained Rigid Body Dynamics。这也是给出的课后阅读材料,大致70页左右。

现在David Baraff 是斯坦福 cs448b 的 Course Speakers https://graphics.stanford.edu/courses/cs448b-00-winter/papers/phys_model.pdf 又有一本全面的大概300页的物理仿真的Lecture。

有关姿态更新的地方,王华民老师的课件里给出的是不包含torque-free-motion的角速度更新方程。torque-free-motion的核心就是,因为角动量守恒,在不受力矩的情况下,具有初始的角速度,因此刚体的姿态会发生变化,导致角速度也得变化。

碰撞响应的部分,并没有同PBM给出的回溯时间的方法,当有顶点侵入到物体内部,时间回溯到碰撞之时,重启ode-solver。因为刚体碰撞,也就是说,两个物体之间是有相对速度的,碰撞的时候,速度会发生较大的改变,可以说在那瞬间,速度并不是连续的。

本课程给出的碰撞响应,是通过计算冲量的方式,用一个简化方式。\(J = F \Delta{t} = Ma \Delta{t} = M \Delta{v}\)

当让能够称之为等号的是 \(dv = adt\), 上式其实只能算近似 \(\Delta{v} \approx a \Delta{t}\)

这就导致了位置一直是连续的,所以哪怕是侵入了墙体之后,兔子也不可能瞬间出来,反而会在里面待一会儿。

image

1 核心技术

1.1 Semi-implicit Euler

\[\begin{cases}\mathbf{v}^{[1]}=\mathbf{v}^{[0]}+\Delta tM^{-1}\mathbf{f}^{[0]}&\longleftarrow\text{Explicit}\\[2ex]\mathbf{x}^{[1]}=\mathbf{x}^{[0]}+\Delta t\mathbf{v}^{[1]}&\longleftarrow\text{Implicit}\\[2ex]\end{cases} \]

这个如何理解呢,关键就是对于每一帧的渲染。我们只有三个量

  • \(\mathbf{v}^{[0]}\)
  • \(\mathbf{x}^{[0]}\)
  • \(\Delta{t}\)

也就是说,前一帧的位置和速度信息,以及一个时间差。

需要我们计算当前的位置与速度,并且渲染出来。

半隐式欧拉这玩意有1阶精度。但更稳定一点。

怎么说呢,就以课程代码给出的时间间隔t = 0.0015f, 那么就算是2阶精度 t^2 = 0.00000225。就以纯累加算,1秒60帧,一分钟3600帧,已经导致将近 0.0081的误差(似乎也不能么算)。

事实上,我试过只有torque-free-motion的仿真,5分钟直接爆炸,numerical instability

1.2 刚体模拟

image
图片来源: Games103 lecture3 Rigid Body Dynamics

Inertia_tensor 的从 body 到 world 的变换可以看 PBM。这个tensor会随着姿态的变化而变化, body frame 的坐标系原点是质心。

  • 2.10 The Inertia Tensor Page.13

角速度的更新不考虑torque-free-motion,这个可以看 PBM。

  • C.2 Angular Accelerion Page.60 具体推导。

四元数的更新可以看 PBM。PBM中给出的四元数的乘法,说是为了与矩阵乘法类似,是左乘的过程。而Unity给出的是右乘。这是需要注意的点。

  • 4 Quaternions vs. Rotation Matrices Page.20
  • Appendix B Quaternion Derivations Page.58

ODE相关的知识可以去看 MIT的 Matrix Calculus 2023年10月23日 鸡汤来喽。里面关于导数的一些认识,特别有启发。

1.3 Collision

image
图片来源: Games103 lecture4 Rigid Body Contacts

𝑎 should be minimized while meeting Coulomb’s law, 𝑎是决定切向速度的改变量。最小是0,也就是说,一碰,切向的速度就成了0,因为比较大的摩擦力的缘故。

另外冲量的计算是遍历所有的顶点,找出其中处于碰撞的点,然后取平均值,基于平均的速度和位置,计算冲量。

2 实现

using UnityEngine;
using System.Collections;

public class Rigid_Bunny : MonoBehaviour 
{
	bool launched 		= false;
	float dt 			= 0.015f;
	Vector3 v 			= new Vector3(0, 0, 0);	// velocity
	Vector3 w 			= new Vector3(0, 0, 0);	// angular velocity
	
	float mass;									// mass
	Matrix4x4 I_ref;							// reference inertia

	float linear_decay	= 0.999f;				// for velocity decay
	float angular_decay	= 0.98f;				
	float restitution 	= 0.5f;                 // for collision Mu_N


	// Mine Var
	Vector3 g = new Vector3(0, -9.8f, 0);
	float Mu_T = 0.8f;


	// Use this for initialization
	void Start () 
	{		
		Mesh mesh = GetComponent<MeshFilter>().mesh;
		Vector3[] vertices = mesh.vertices;

		float m=1;
		mass=0;
		for (int i=0; i<vertices.Length; i++) 
		{
			mass += m;
			float diag=m*vertices[i].sqrMagnitude;
			I_ref[0, 0]+=diag;
			I_ref[1, 1]+=diag;
			I_ref[2, 2]+=diag;
			I_ref[0, 0]-=m*vertices[i][0]*vertices[i][0];
			I_ref[0, 1]-=m*vertices[i][0]*vertices[i][1];
			I_ref[0, 2]-=m*vertices[i][0]*vertices[i][2];
			I_ref[1, 0]-=m*vertices[i][1]*vertices[i][0];
			I_ref[1, 1]-=m*vertices[i][1]*vertices[i][1];
			I_ref[1, 2]-=m*vertices[i][1]*vertices[i][2];
			I_ref[2, 0]-=m*vertices[i][2]*vertices[i][0];
			I_ref[2, 1]-=m*vertices[i][2]*vertices[i][1];
			I_ref[2, 2]-=m*vertices[i][2]*vertices[i][2];
		}
		I_ref [3, 3] = 1;
	}
	
	Matrix4x4 Get_Cross_Matrix(Vector3 a)
	{
		//Get the cross product matrix of vector a
		Matrix4x4 A = Matrix4x4.zero;
		A [0, 0] = 0; 
		A [0, 1] = -a [2]; 
		A [0, 2] = a [1]; 
		A [1, 0] = a [2]; 
		A [1, 1] = 0; 
		A [1, 2] = -a [0]; 
		A [2, 0] = -a [1]; 
		A [2, 1] = a [0]; 
		A [2, 2] = 0; 
		A [3, 3] = 1;
		return A;
	}

	// In this function, update v and w by the impulse due to the collision with
	//a plane <P, N>
	void Collision_Impulse(Vector3 P, Vector3 N)
	{
		Mesh mesh = GetComponent<MeshFilter>().mesh;
		Vector3[] vertices = mesh.vertices;
		Vector3 x = transform.position;
		Matrix4x4 R = Matrix4x4.Rotate(transform.rotation);
		Matrix4x4 Inertia_tensor = R * I_ref * R.transpose;

		Vector3 avg_pos = new Vector3(0, 0, 0);
		Vector3 avg_vol = new Vector3(0, 0, 0);
		bool isCllison = false;
		int count = 0;

		for (int i = 0; i < vertices.Length; i++)
		{
			Vector3 Rv_i = R.MultiplyVector(vertices[i]);
			Vector3 world_vertice = x + Rv_i;
			float SDF = Vector3.Dot(world_vertice - P, N);

			if (SDF < 0.0f)
			{
				Vector3 world_volocity = v + Vector3.Cross(w, Rv_i);
				float vSDF = Vector3.Dot(world_volocity, N);

				if (vSDF < 0.0f) 
				{
					count++;
					avg_pos += vertices[i];
					avg_vol += world_volocity;
					isCllison = true;
				}

			}	
		}


		if (isCllison) 
		{
			avg_pos /= count;
			avg_vol /= count;

			Vector3 Rv_i = R.MultiplyVector(avg_pos);
			// Compute the wanted v_new^i
			Vector3 v_N = Vector3.Dot(avg_vol, N) * N;
			Vector3 v_T = avg_vol - v_N;
			float a = Mathf.Max(1 - (Mu_T * (1 + restitution) * v_N.magnitude / v_T.magnitude), 0.0f);
			v_N = -restitution * v_N;
			v_T = a * v_T;
			Vector3 v_new = v_N + v_T;
			float massinv = (1.0f / mass);
			Matrix4x4 massinv_Indentity = new Matrix4x4(massinv * Matrix4x4.identity.GetColumn(0),
				massinv * Matrix4x4.identity.GetColumn(1),
				massinv * Matrix4x4.identity.GetColumn(2),
				Matrix4x4.identity.GetColumn(3));
			Matrix4x4 K = M_minus_M(massinv_Indentity,
				Get_Cross_Matrix(Rv_i) * Inertia_tensor.inverse * Get_Cross_Matrix(Rv_i));
			Vector3 j = K.inverse.MultiplyVector(v_new - avg_vol);
			Debug.Log(K);
			Debug.Log(j);
			// update volocity
			v = v + massinv * j;
			w = w + Inertia_tensor.inverse.MultiplyVector(Vector3.Cross(Rv_i, j));
		}
	}

	// Update is called once per frame
	void Update () 
	{
		//Game Control
		if(Input.GetKey("r"))
		{
			transform.position = new Vector3 (0, 0.6f, 0);
			restitution = 0.5f;
			launched=false;
		}
		if(Input.GetKey("l"))
		{
			v = new Vector3 (5, 2, 0);
			launched=true;
		}

		// Part I: Update velocities
		float m = 1;
		Vector3 f_i = new Vector3(0, 0, 0);
		Vector3 toruqe_i = new Vector3(0, 0, 0);
		Vector3 F = new Vector3(0, 0, 0);
		Vector3 Torque = new Vector3(0, 0, 0);

		Matrix4x4 R =  Matrix4x4.Rotate(transform.rotation);
		Matrix4x4 Inertia_tensor = R * I_ref * R.transpose;

		Mesh mesh = GetComponent<MeshFilter>().mesh;
		Vector3[] vertices = mesh.vertices;

		for (int i = 0; i < vertices.Length; i++)
		{
			// Linear
			f_i = m * g;
			F += f_i;

			// Rotation
			toruqe_i = Vector3.Cross(R * vertices[i], f_i);
			Torque += toruqe_i;
		}

		// or v = v + F / mass * dt - (1.0f - linear_decay) * v
		v = v + F / mass * dt;
		Vector3 L = Inertia_tensor.MultiplyVector(w);
		//  + Inertia_tensor.inverse.MultiplyVector(Vector3.Cross(L, w))*dt torque free motion
		w = w + Inertia_tensor.inverse.MultiplyVector(Torque) *dt;

		v = linear_decay * v;
		w = angular_decay * w;



		// Part II: Collision Impulse
		// floor
		Collision_Impulse(new Vector3(0, 0.01f, 0), new Vector3(0, 1, 0));

		// wall
		Collision_Impulse(new Vector3(2, 0, 0), new Vector3(-1, 0, 0));

		// Part III: Update position & orientation
		//Update linear status
		Vector3 x    = transform.position;
		x = x + v * dt;
		
		//Update angular status
		Quaternion q = transform.rotation;
		Vector3 w_update = 0.5f * dt * w;
		Quaternion q_dt = new Quaternion(w_update[0], w_update[1], w_update[2], 0);
		// Rotating by the product lhs * rhs is the same as applying the two rotations in sequence: lhs first and then rhs
		q = Q_plus(q, q * q_dt);

		// Part IV: Assign to the object
		transform.position = x;
		transform.rotation = q;

		
	}

	private Quaternion Q_plus(Quaternion lhs, Quaternion rhs) 
	{
		Quaternion q = new Quaternion(lhs.x + rhs.x, lhs.y + rhs.y, lhs.z + rhs.z, lhs.w + rhs.w);
		return q.normalized;
	}

	private Matrix4x4 M_minus_M(Matrix4x4 lhs, Matrix4x4 rhs) 
	{
		Matrix4x4 A = Matrix4x4.zero;
		for (int i = 0; i < 4; i++) 
		{
			for (int j = 0; j < 4; j++)
			{
				A[i, j] = lhs[i, j] - rhs[i, j];
			}
		}
		// Metion here, if A[3,3] == 0 then the matrix is sigular, which doesn't have a inverse, 
		// however, in the singular case, the inverse is Matrix.zero;
		A[3, 3] = 1;
		return A;

	}
}

X Ref

  1. 2013 David Tong: Lectures on Dynamics and Relativity _ Cambridge
  2. 2021 王华民 GAMES103:基于物理的计算机动画入门
  3. 2023 MIT 18.S096 Matrix Calculus
  4. 2019 CSE 291D Physics Simulation Spring 2019
  5. 2019 SCE 291 Rigid Body Simulation
  6. 2014 CIS563, Spring 2014 (Physics-based Animation)
posted @ 2024-01-01 12:15  Dba_sys  阅读(37)  评论(0编辑  收藏  举报