Difference between revisions of "Darwinbots3/Physics"

Revision as of 16:28, 28 April 2009

Basic concepts

Forces' linear affects

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:

__________
|        |
|        |  
|   X    |  <--- 
|        |
|________|  <--- 

 Diagram 1 

Where X is the center of mass for the body. is exactly centered, so it produces no torque. The change in acceleration of the body's center of mass is given by .

Let have the same magnitude and direction as . 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 .

Forces' angular affects

Consider Diagram 1 again. will not produce any change in angular acceleration for the body, because it is centered. will produce change in angular acceleration, because it is off center. In general, the torque () produced by a force is given by:

And the change in angular acceleration is given by:

Where:

• is the scalar torque term.
• is the vector Force term.
• is the vector perpendicular to the vector from the body's origin to the place is acting on the body.
• is the scalar angular acceleration
• is the body's scalar moment of inertertia.

Simple collision

__________      __
|        |     /  |
|        | ___/   |
|   X    |P___  Y |
|        |    \   |
|________|     \__|
 
 Diagram 2 

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.

The change in angular and linear velocity for body X is given by:

where:

• is the change in linear velocity.
• is the change in angular velocity.
• is the scalar impulse term applied to the body at point P to correct its velocity from the collision.
• is the scalar mass for the body.
• is the scalar moment of inertia for the body.
• 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
• is the vector perpendicular to the vector from the center of mass of body X to the collision point P.

Body Y likewise, but the changes are opposite in sign (equal and opposite reaction).

We can define a relationship between the velocities before and after the collision using the coefficient of restitution. Which is basically a fractional scalar value between 0 (for inelastic collisions) and 1 (for perfectly elastic collisions).

where:

• are the final velocities of bodies X and Y at point P.
• is the coefficient of restitution for the equation
• are the initial velocities of bodies X and Y at point P.

To find the velocity of a body at a given point, use the formula:

where:

• is the velocity at a certain point on the body.
• is the body's angular velocity.
• is the vector perpindicular to the vector from the body's center of mass to point P.

Using all of the equations above, we can find by the following algorithm:

suppose we are supplied with:
  a contact point P
  a collision normal n
  two bodies in collision, bodyX and bodyY
  a coefficient of restitution e

Vector VelocityAtPoint(Body body, Vector point)
{
  return body.Velocity + body.AngularVelocity * (point - body.Position);
}

Scalar ResistanceFromBody(Body body, Vector point, Vector n)
{
  Vector rPerp = (point - body.Position).Perp();
  Scalar a = n.LengthSquared() * body.InverseMass; // The resistance to linear acceleration
  Scalar q = Squared(rPerp.DotProduct(n)) * body.InverseMomentOfInertia; // The resistance to angular acceleration
  
  return a + q;
}

Scalar vXP = VelocityAtPoint(bodyX, P);
Scalar vYP = VelocityAtPoint(bodyY, P);

Scalar b = -(1 + e) * (vXP - vYP).DotProduct(n);

Vector rXPNorm = (point - bodyX.Position).Perp();
Vector rYPNorm = (point - bodyY.Position).Perp();

Scalar resistanceX = ResistanceFromBody(bodyX, P, n);
Scalar resistanceY = -ResistanceFromBody(bodyY, P, n); //equal but opposite

Scalar j0 = b / (resistanceX - resistanceY);

return j0;

Multiple collisions

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:

Failed to parse (unknown function "\Large"): \Delta\vec{v} = \Large{\frac{\sum_{i=0}^{N-1} j_i \cdot \vec{n}}{m}}

Failed to parse (unknown function "\Large"): \Delta\omega = \Large{\frac{\sum_{i=0}^{N-1} j_i \cdot \vec{r_{\perp}^{Pi}} \cdot \vec{n}}{I}}

where:

• is 0 if that contact point isn't part of that body's collision.

We can now form N equations (one for each collision), and solve for N unknowns (each collision impulse ).

The equation for collision is given by:

Failed to parse (unknown function "\large"): \large{V_{BPi} - V_{APi} = b_i}

Failed to parse (unknown function "\large"): \large{V_{XPi} = \sum_{k=0}^{k=N-1} (aX_k + qX_k)} \cdot j_k

Failed to parse (unknown function "\large"): \large{aX_k = \frac{\vec{n_i} \cdot \vec{n_k}}{m_X}}

Failed to parse (unknown function "\large"): \large{qX_k = \frac{ (\vec{r^{XPi}_\perp} \cdot \vec{n_i}) \cdot (\vec{r^{XPk}_\perp} \cdot \vec{n_k})}{I}}

Failed to parse (unknown function "\large"): \large{b_i = -(1 + \epsilon) ((V_{BP} - V_{AP}) \cdot n_i)}

Failed to parse (unknown function "\large"): \large{V_{XP} = V_X + \omega_X \cdot r^{XP}_\perp}

All terms in the above quantities are known at the beginning of the collision except for . So we can represent the system of equation as:

Failed to parse (unknown function "\large"): \large{\mathbf{A} \mathbf{j} = \mathbf{b} }

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). If we take the equation for above, are replace with commas, we can form the system of equations into a matrix:

Failed to parse (unknown function "\large"): \large{ \begin{bmatrix} V_{BP0} - V_{AP0} \\ V_{CP1} - V_{BP1} \\ V_{CP2} - V_{BP2} \\ V_{DP3} - V_{CP3} \\ \end{bmatrix} \begin{bmatrix} j_0 \\ j_1 \\ j_2 \\ j_3 \\ \end{bmatrix} = \begin{bmatrix} b_0 \\ b_1 \\ b_2 \\ b_3 \\ \end{bmatrix} }

Which can then be solved in .