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, 不过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》 不过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。

不过现在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，兔子页不可能瞬间出来，反而会在里面待一会儿。

1 核心技术

1.1 Semi-implicit Euler

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 似乎有个更有趣的名字 leapfrog method，这玩意有二阶精度。但实际上，二阶精度也是不够看。

怎么说呢，就以课程代码给出的时间间隔t = 0.0015f, 那么 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

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。

• MIT 18.S096 Matrix Calculus

1.3 Collision

图片来源: Games103 lecture4 Rigid Body Contacts

? should be minimized while meeting Coulomb’s law, ?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)

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