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| − | == Basic concepts ==
| + | * [[Darwinbots3/Physics/Response | Collision Response]] - Article covers math behind responding to collisions |
| − | | + | * [[Darwinbots3/Physics/Detection | Collision Detection]] - Article covers methods of broad and narrow phase collision detection for simple 2D shapes |
| − | === Forces' linear affects ===
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| − | Any force acting on a body, at any point on a body, applies the same change in acceleration to the body's center of mass. Consider the diagram below:
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| − | __________
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| − | | X | <--- <math>\vec{Force_1}</math>
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| − | | |
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| − | |________| <--- <math>\vec{Force_2}</math>
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| − |
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| − | ''' Diagram 1 '''
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| − | | |
| − | Where X is the center of mass for the body. <math>\vec{Force_1}</math> is exactly centered, so it produces no torque. The change in acceleration of the body's center of mass is given by <math>\Delta \vec{a} = \frac{\vec{Force_1}}{Mass}</math>.
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| − | Let <math>\vec{Force_2}</math> have the same magnitude and direction as <math>\vec{Force_1}</math>. However it's applying its force at a different point on the body, and will produce torque. Even though it's off center, the change in acceleration for the body's center of mass is still <math>\Delta \vec{a} = \frac{\vec{Force_1}}{Mass}</math>.
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| − | | |
| − | === Forces' angular affects ===
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| − | Consider Diagram 1 again. <math>\vec{Force_1}</math> will not produce any change in angular acceleration for the body, because it is centered. <math>\vec{Force_2}</math> ''will'' produce change in angular acceleration, because it is off center. In general, the '''torque''' (<math>\tau</math>) produced by a force is given by:
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| − | | |
| − | :<math>\tau = \vec{F} \cdot \vec{r_{\perp}^{P}}</math>
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| − | | |
| − | And the change in angular acceleration is given by:
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| − | | |
| − | :<math>\Delta \alpha = \frac{\tau}{I}</math>
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| − | | |
| − | Where:
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| − | * <math>\tau</math> is the scalar torque term. | |
| − | * <math>\vec{F}</math> is the vector Force term.
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| − | * <math>\vec{r_{\perp}^{P}}</math> is the vector perpendicular to the vector from the body's origin to the place <math>\vec{F}</math> is acting on the body.
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| − | * <math>\alpha</math> is the scalar angular acceleration
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| − | * <math>I</math> is the body's scalar moment of inertertia.
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| − | | |
| − | == Simple collision ==
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| − | __________ __
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| − | | | / |
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| − | | | ___/ |
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| − | | X |P___ Y |
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| − | | | \ |
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| − | |________| \__|
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| − |
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| − | ''' Diagram 2 '''
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| − | | |
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| − | Consider a collision between two bodies: body X and body Y. They collide at point P. We assume that the collision takes 0 time. That is, the bodies "instantly" resolve their collision.
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| − | The change in angular and linear velocity for body X is given by:
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| − | : <math>\Delta\vec{v} = \frac{j_0 \cdot \vec{n}}{m_X}</math>
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| − | : <math>\Delta\omega = j_0 * \frac{\vec{r_{\perp}^{XP}} \cdot \vec{n}}{I_X}</math>
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| − | | |
| − | where:
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| − | * <math>\Delta\vec{v}</math> is the change in linear velocity.
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| − | * <math>\Delta\omega</math> is the change in angular velocity.
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| − | * <math>j_0</math> is the scalar impulse term applied to the body at point P to correct its velocity from the collision.
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| − | * <math>m_X</math> is the scalar mass for the body.
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| − | * <math>I_X</math> is the scalar moment of inertia for the body.
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| − | * <math>n</math> is a vector representing the "normal" to the colision. In the case of the vertex-on-edge collision in Diagram 2, n would probably be <math><1, 0></math>
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| − | * <math>\vec{r_{\perp}^{XP}}</math> is the vector perpendicular to the vector from the center of mass of body X to the collision point P.
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| − | Body Y likewise, but the changes are opposite in sign (equal and opposite reaction).
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| − | We can define a relationship between the velocities before and after the collision using the [http://en.wikipedia.org/wiki/Coefficient_of_restitution coefficient of restitution]. Which is basically a fractional scalar value between 0 (for inelastic collisions) and 1 (for perfectly elastic collisions).
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| − | : <math>v_{XPf} - v_{YPf} = -(1 + \epsilon) * (v_{XPi} - v_{YPi})</math>
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| − | where:
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| − | * <math>v_{XPf}, v_{YPf}</math> are the final velocities of bodies X and Y at point P.
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| − | * <math>\epsilon</math> is the coefficient of restitution for the equation
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| − | * <math>v_{XPi}, v_{YPi}</math> are the initial velocities of bodies X and Y at point P.
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| − | To find the velocity of a body at a given point, use the formula:
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| − | : <math>v_P = v + \omega * r^P_\perp</math>
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| − | | |
| − | where:
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| − | * <math>v_P</math> is the velocity at a certain point on the body.
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| − | * <math>\omega</math> is the body's angular velocity.
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| − | * <math>r^P_\perp</math> is the vector perpindicular to the vector from the body's center of mass to point P.
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| − | Using all of the equations above, we can find <math>j_0</math> by the following algorithm:
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| − | | |
| − | suppose we are supplied with: | |
| − | a contact point '''P'''
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| − | a collision normal '''n'''
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| − | two bodies in collision, '''bodyX''' and '''bodyY'''
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| − | a coefficient of restitution '''e'''
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| − |
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| − | Vector VelocityAtPoint(Body body, Vector point)
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| − | {
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| − | return body.Velocity + body.AngularVelocity * (point - body.Position);
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| − | }
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| − |
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| − | Scalar ResistanceFromBody(Body body, Vector point, Vector n)
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| − | {
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| − | Vector rPerp = (point - body.Position).Perp();
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| − | Scalar a = n.LengthSquared() * body.InverseMass; // The resistance to linear acceleration
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| − | Scalar q = Squared(rPerp.DotProduct(n)) * body.InverseMomentOfInertia; // The resistance to angular acceleration
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| − |
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| − | return a + q;
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| − | }
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| − |
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| − | Scalar vXP = VelocityAtPoint(bodyX, P);
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| − | Scalar vYP = VelocityAtPoint(bodyY, P);
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| − |
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| − | Scalar b = -(1 + e) * (vXP - vYP).DotProduct(n);
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| − |
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| − | Vector rXPNorm = (point - bodyX.Position).Perp();
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| − | Vector rYPNorm = (point - bodyY.Position).Perp();
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| − |
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| − | Scalar resistanceX = ResistanceFromBody(bodyX, P, n);
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| − | Scalar resistanceY = -ResistanceFromBody(bodyY, P, n); //equal but opposite
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| − |
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| − | Scalar j0 = b / (resistanceX - resistanceY);
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| − |
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| − | return j0;
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| − | | |
| − | == Multiple collisions ==
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| − | Consider the case of collisions between '''N''' bodies simultaneously. We can expand from the last section to handle N contact points. The change in linear and angular velocity of a body is described by the equation:
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| − | : <math>\Delta\vec{v} = \Large{\frac{\sum_{i=0}^{N-1} j_i \cdot \vec{n}}{m}}</math>
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| − | : <math>\Delta\omega = \Large{\frac{\sum_{i=0}^{N-1} j_i \cdot \vec{r_{\perp}^{Pi}} \cdot \vec{n}}{I}}</math>
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| − | where:
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| − | * <math>j_i</math> is 0 if that contact point isn't part of that body's collision.
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| − | We can now form N equations (one for each collision), and solve for N unknowns (each collision impulse <math>j_i</math>).
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| − | The equation for collision <math>i</math> is given by:
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| − | : <math>\large{V_{BPi} - V_{APi} = b_i}</math>
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| − | : <math>\large{V_{XPi} = \sum_{k=0}^{k=N-1} (aX_k + qX_k)} \cdot j_k</math>
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| − | : <math>\large{aX_k = \frac{\vec{n_i} \cdot \vec{n_k}}{m_X}}</math>
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| − | : <math>\large{qX_k = \frac{ (\vec{r^{XPi}_\perp} \cdot \vec{n_i}) \cdot (\vec{r^{XPk}_\perp} \cdot \vec{n_k})}{I}}</math>
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| − | : <math>\large{b_i = -(1 + \epsilon) ((V_{BP} - V_{AP}) \cdot n_i)}</math>
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| − | : <math>\large{V_{XP} = V_X + \omega_X \cdot r^{XP}_\perp}</math>
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| − | All terms in the above quantities are known at the beginning of the collision except for <math>j_i</math>. So we can represent the system of equation as:
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| − | :<math>\large{\mathbf{A} \mathbf{j} = \mathbf{b} }</math>
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| − | For an example, suppose that there are 4 collisions. One between body A and body B (collision 0), two between body B and body C (collision 1 and 2), and one between body C and body D (collision 3).
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| − | If we take the equation for <math>V_{XPi}</math> above, are replace <math>j_k</math> with commas, we can form the system of equations into a matrix:
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| − | : <math> \large{
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| − | \begin{bmatrix}
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| − | V_{BP0} - V_{AP0} \\
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| − | V_{CP1} - V_{BP1} \\
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| − | V_{CP2} - V_{BP2} \\
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| − | V_{DP3} - V_{CP3} \\
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| − | \end{bmatrix} \begin{bmatrix}
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| − | j_0 \\
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| − | j_1 \\
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| − | j_2 \\
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| − | j_3 \\
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| − | \end{bmatrix} = \begin{bmatrix}
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| − | b_0 \\
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| − | b_1 \\
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| − | b_2 \\
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| − | b_3 \\
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| − | \end{bmatrix} }</math>
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| − | Which can then be solved in <math>O(N^3)</math>.
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