Collision Resolution -Game Physics Engine Development总结

The velocity of a point

The velocity of a point on an object depends on both its linear and angular velocity:

$$\dot{q} = \dot{\theta} \times (q - p) + \dot{p} \qquad \qquad [1.0]$$

where $\dot{q}$ is the velocity of the point, $p$ is the position of the point in world coordinates,$p$ is the position of the origin of the object, and $\theta$ is the angular velocity of the object.

Impulse

An instantaneous change in velocity.In the same way that we have

$$f = m\ddot{p} \qquad \qquad [1.1]$$

for forces,we have

$$g = m\dot{p} \qquad \qquad [1.2]$$

一般时使用p而不是使用g,但是为了防止和位置(p)冲突所以使用了g。这个impulse和force是等价的。一些作用力不只是通过加速度来体现出来，有些是通过速度而不是加速度体现的。其实 $\dot{p}应该是速度的改变吧，此处有疑问$

Impulsive Torque

.

The linear component is given by

$$\ddot{p} = \frac{1}{m}f \qquad \qquad [1.3]$$

and the angular component by the torque

$$\tau = P_{f} \times f \qquad \qquad [1.4]$$

where the torque generates angular acceleration by

$$\ddot{\theta} = I^{-1}\tau \qquad \qquad [1.5]$$

In the case of the collision it stands to reason that the collision will generate a
linear change in velocity (the impulse) and an angular change in velocity. An instantaneous
angular change in velocity is called an “impulsive torque”.

$$u = I\dot{\theta} \qquad \qquad [1.6]$$

where the $u$ is the impulsive torque. $I$ is the inertia tensor, and $\theta$ is the angular velocity.该公式对应了上面的[1.2] (个人感觉此处应该是angular velocity的改变值)

Impulses behave just like forces. In particular for a given impulse there will be both a linear component and an angular component.The impulsive torque generated by an impulse (impluse也会产生impulsive torque) is given by

$$u = P_{f}\times g \qquad \qquad [1.7]$$

for collisions the point of application $P_{f}$is given by the contact point and the origin of the object:

$$P_{f} = q - p \qquad \qquad [1.8]$$

where $q$is the position of the contact in world coordinates and $p$ is the position of
the origin of the object in world coordinates.

VELOCITY CHANGE BY IMPULSE

Impulses cause a change in velocity both angular and linear.

The Linear Component

The linear change in velocity for a unit impulse will be in the direction of the impulse, with a magnitude given by the inverse mass:

$$\Delta\dot{P}_{d} = m^{-1} \qquad \qquad [1.9]$$

For collisions involving two objects, the linear component is simply the sum of the two inverse masses:

$$\Delta\dot{P}_{d} = m_{a}^{-1} + m_{b}^{-1} \qquad \qquad [1.10]$$

The Angular Component

First, equation 1.7 tells us the amount of impulsive torque generated from a unit
of impulse:

$$u = q_{rel} \times \hat{d} \qquad \qquad [1.11]$$

where d is the direction of the impulse (in our case the contact normal).
Second, equation 1.6 tells us the change in angular velocity for a unit of impulsive
torque:

$$\Delta \dot{\theta} = I^{-1}u \qquad \qquad [1.12]$$

And finally, equation 1.0 tells us the total velocity of a point. If we remove the
linear component, we get the equation for the linear velocity of a point due only to
its rotation:

$$\dot{q} = \dot{\theta}\times q_{rel} \qquad \qquad [1.13]$$

So we now have a set of equations that can get us from a unit of impulse, via the
impulsive torque it generates and the angular velocity that the torque causes, through
to the linear velocity that results.

Vector3 torquePerUnitImpulse = relativeContactPosition % contactNormal; (参考:1.11)
Vector3 rotationPerUnitImpulse = inverseInertiaTensor.transform(torquePerUnitImpulse); (参考:1.12)
Vector3 velocityPerUnitImpulse = rotationPerUnitImpulse % relativeContactPosition; (参考:1.13)