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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.
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| − | * <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 | |
| − | * <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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| − | | | ___/ | | |
| − | | 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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| − | 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}{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:
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| − | 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.DotSquared(n) * body.InverseMass; // The resistance to linear acceleration
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| − | Scalar q = rPerp.DotProduct(n) * body.InverseMomentOfInertia; // The resistance to angular acceleration
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| − | q *= q;
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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).DotProduct(n);
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| − | Scalar vYP = VelocityAtPoint(bodyY, P).DotProduct(n);
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| − |
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| − | Scalar b = -(1 + e) * (vXP - vYP);
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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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