Phrase of the week

May 26, 2013 at 10:06 AMMichele Mottini

A committee is an animal with four back legs.

John le Carré in ‘Tinker, Tailor, Soldier, Spy'.

Posted in: Opinion


Billiard simulation–part 5: friction

May 26, 2013 at 9:03 AMMichele Mottini

The velocity of the balls progressively decreases due to three different effects: loss of energy during collisions, friction with the table (rolling friction or rolling resistance) and friction with the air (drag).

The loss of energy during collisions is already accounted for by the coefficient of restitution used in the collisions formulas.

The general formula for the rolling friction is:

$$ F_{rr} = c_{rr}N $$

where \(F_{rr}\) is the resulting rolling friction force, \(c_{rr}\) is the coefficient of rolling friction – that depends on the materials of the ball and the table – and \(N\) is the downward force exerted by the ball on the table. \(N\) is equal to the mass \(m\) of the ball times the gravitational acceleration \(g\): \(N = mg\). The deceleration \(a_{rr}\) of the ball due to the rolling friction is the friction force divided by the mass of the ball: \(a_{rr}=F_{rr}/m\). Combining these two equation the rolling friction deceleration is:

$$ a_{rr} = c_{rr}g $$

The formula for the air drag is:

$$ F_d = \frac{1}{2}\rho_a v^2 c_d A $$

where \(F_d\) is the resulting drag force, \(\rho_a\) is the air density, \(v\) is the ball speed, \(c_d\) is the drag coefficient and \(A\) is the cross-sectional area of the ball. The deceleration \(a_{d}\) of the ball due to the air drag is the air drag force divided by the mass of the ball:

$$ a_d = \frac{F_d}{m} = \frac{F_d}{\frac{4}{3}\pi\rho_b r^3} $$

where \(\rho_b\) is the density of the ball and \(r\) is its radius. The cross-sectional area of the ball is \(A=\pi r^2\) – putting all this together the air drag deceleration is:

$$ a_d = f_d \frac{v^2}{r}$$

where the air drag factor \(f_d\) is defined as

$$ f_d = \frac{3\rho_a c_d}{8\rho_b} $$

The drag coefficient \(c_d\) for a sphere varies quite a bit with the velocity of the object – see here for example, but overall the air drag is quite small, so even using a constant would not change the result much; hence given the air density and the ball density \(f_d\) can be considered a constant as well.

Assuming that the time interval used to update the simulation is small compared with the rate of change of the ball velocities, the speed after the interval is simply the original speed decremented by the deceleration times the interval:

$$ v_t = v_0 – (a_{rr}+a_d)t $$

The resulting code is:

   * Updates the ball velocity, applying rolling resistance and air drag for a specified time interval
   * @param t the time interval to use. Only first-order effects are considered, so the time interval 
   * must be small compared to the rate of change of the velocity
   * @param airDragFactor total air drag factor computed from the ball and air densities and the 
   * air drag coefficient (assumed to be constant)
   * @param rollingResistanceDeceleration total rolling resistance deceleration computed from the 
   * rolling resistance coefficient and the gravitational acceleration 
  updateVelocity(t: number, airDragFactor: number, rollingResistanceDeceleration: number) {
    var speed2 = this.v * this.v + this.w * this.w;
    if (speed2 > 0) {
      var airResistanceDeceleration = airDragFactor * speed2 / this.radius;
      var totalDeceleration = airResistanceDeceleration + rollingResistanceDeceleration;
      var speed = Math.sqrt(speed2);
      var newSpeed = speed - totalDeceleration * t;
      if (newSpeed <= 0) {
        // The ball stopped
        this.v = 0;
        this.w = 0;
      } else {
        // Update the speed, keeping the velocity direction the same (the air drag and rolling resistance 
        // forces are in the exact opposite direction of the velocity)
        this.v = this.v * newSpeed / speed;
        this.w = this.w * newSpeed / speed;
  } // updateVelocity

Posted in: Physics | Programming

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Phrase of the week

May 19, 2013 at 7:24 AMMichele Mottini

 Apple Computer is now unlikely to survive its current crisis

Brad de Long in The Corporation as a Command Economy (originally written in 1997)

Posted in: Opinion


Billiard simulation–part 4: more collisions, and a problem

May 19, 2013 at 6:07 AMMichele Mottini

The last part covered the collisions between balls, what about collisions with the sides of the table? If the balls are not spinning and there is no ball-table friction the component of the ball velocity that is parallel to the table side remains the same, and the component perpendicular to the side changes direction and intensity based on the coefficient of restitution. Using primed letters for the velocity after the collisions, the formula for a collision with the left or right side is:

$$ v’ = –cv $$

$$ w’ = w $$

where \(c\) is the ball-table coefficient of restitution (that in general is different from the ball-ball one). The formula for a collision with the ‘top’ (as seen on the screen) or ‘bottom’ side is:

$$ v’ = v $$

$$ w’ = -cw $$

With this last piece, the complete update code is:

  update(dt: number) {
    while (dt > 0) {
      var firstCollisions = this.detectCollisions(dt, 0, 400, 0, 300);
      if (firstCollisions.t > 0) {
        // The balls move freely up to the time of the first collision: update their positions
        for (var i = 0; i < this.balls.length; i++) {
          var ball = this.balls[i];
      // Compute the new velocities after the collisions
      for (var i = 0; i < firstCollisions.collisions.length; i++) {
        var collision = firstCollisions.collisions[i];
        switch (collision.type) {
          case "x":
            collision.b1.v = -collision.b1.v * this.parameters.sideRestitution;
          case "y":
            collision.b1.w = -collision.b1.w * this.parameters.sideRestitution;
          case "b":
            collision.b1.collide(collision.b2, this.parameters.ballRestitution);
      // Continue with the remaining time
      dt -= firstCollisions.t;
  } // update

where ‘detectCollisions’ is a function that uses the code described previously to detect the first collision – or collisions – that occur within the specified interval. It returns an object containing the time of the collision and an array of the balls that collide. Each element of this array has two members ‘b1’ and ‘b2’ containing the colliding balls (‘b2’ is null for collisions with the sides) plus a ‘type’ member that indicates if the collision is with the left or right sides (“x”), with the top or bottom sides (“y”) or between two balls (“b”).

Here is the ‘detectCollisions’ code:

  private detectCollisions(dt: number, minx: number, maxx: number, miny: number, maxy: number) {
    var result = {
      t: dt,
      collisions: <{ type: string; b1: Ball; b2: Ball; }[]>[]
    var addCollision = (t, collision) => {
      if (t === result.t) {
        // The new collision happens at the exact same time of the current one, it has to be added to the list
      } else {
        // The new collision happens before the current one, so it replaces the entire list
        result.collisions = [collision];
        result.t = t;
    for (var i = 0; i < this.balls.length; i++) {
      // Collisions with the sides
      var ball = this.balls[i];
      var t = ball.sideXCollisionTime(minx, maxx);
      if (t && t <= result.t) {
        addCollision(t, { type: "x", b1: ball, b2: <Ball>null });
      t = ball.sideYCollisionTime(miny, maxy);
      if (t && t <= result.t) {
        addCollision(t, { type: "y", b1: ball, b2: <Ball>null });
      // Ball-ball collisions
      for (var j = i + 1; j < this.balls.length; j++) {
        var otherBall = this.balls[j];
        t = ball.collisionTime(otherBall);
        if (t && t <= result.t) {
          addCollision(t, { type: "b", b1: ball, b2: otherBall });
    return result;
  } // detectCollisions

As promised in the title this code has a problem though: it works fine in most cases but fails when there are multiple ball-ball collisions at the same time.

Have a look at this: the ball coming from the left hits the two balls at rest at the same time, but the code evaluates the collisions one at a time: it first consider the collisions with the ball at the top, updating the velocity of the moving ball to move down, and then uses this velocity to compute the second collision. The result is clearly wrong: the moving ball should bounce straight back, and the result should not depend on the order of the balls.

I have no solution for this at the moment being.

Posted in: Physics | Programming

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Billiard simulation–part 3: collision between two balls

May 13, 2013 at 1:27 PMMichele Mottini

..or more formally: elastic collision of spheres in two dimensions.

Let’s start with a simple case: ball 1 moving with velocity \((\bar v_1, \bar w_2)\) hitting ball 2 at rest. To make things even easier ball 1 and ball 2 are aligned horizontally at the moment of impact – i.e.  they have the same \(y\) coordinate:


In our model (at the moment being at least) balls are not spinning, and there is no ball-ball friction, so the force between balls in such a collision is only along the horizontal line connecting their centers. This means that the vertical components of the velocities before and after the collision are the same:

$$ \bar{w}’_1 = \bar{w} $$

$$ \bar{w}’_2 = 0 $$

- where the primed values are the ones after the collision. The horizontal components of the velocities on the other hand are related by the ball-ball coefficient of restitution \(c\):

$$ \bar{v}’_2 - \bar{v}’_1 = c(\bar{v}_1 - \bar{v}_2) = c\bar{v}_1  $$

(\(\bar{v}_2\) is zero). The horizontal momentum must be the same before and after the collision:

$$ m_1\bar{v}’_1 + m_2\bar{v}’_2 = m_1\bar{v}_1 $$

- where \(m_1\) and \(m_2\) are the masses of the two balls. Combining these two equation gives:

$$ \bar{v}’_1 = \frac{m_1 – cm_2}{m_1+m_2}\bar{v}_1 $$

$$ \bar{v}’_2 = \frac{(1+c)m_1}{m_1+m_2}\bar{v}_1 $$

that is the result for this simple case: the velocities after the collision in terms of the (known) velocities before the collision.

What abut the general case in which both balls move and are not aligned?


This second case can be converted in the simple one with a coordinate transformation: subtract the velocity of the second ball from all velocities – i.e. use the frame of reference of the second ball - and rotate the axis by the angle \(\alpha\):

$$ \bar{v} = (v-v_2)\cos\alpha + (w-w_2)\sin\alpha $$

$$ \bar{w} = -(v-v_2)\sin\alpha + (w-w_2)\cos\alpha $$

where \((v, w)\) is an arbitrary velocity and \((\bar{v}, \bar{w})\) is the corresponding transformed velocity. The angle \(\alpha\) is*:

$$ \alpha = \arctan \frac{y_1– y_2}{x_1 – x_2} $$

where \((x_1, y_1)\) and \((x_2, y_2)\) are the positions of the two balls.

The inverse transformation is:

$$ v = \bar{v}\cos\alpha - \bar{w}\sin\alpha + v_2 $$

$$ w = \bar{v}\sin\alpha + \bar{w}\cos\alpha + w_2 $$

Now the procedure to solve the general case is: transform the original velocities in their ‘bar’ equivalent; apply the formula of the simple case; apply the inverse transformation to go back to the velocities in the original coordinate system (without ‘bar’). These steps produce the following results:

$$ v’_1 = \frac{1}{m_1+m_2}\Big[\big(m_1-m_2(c\cos^2\alpha-\sin^2\alpha)\big)(v_1-v_2)-(1+c)m_2\sin\alpha\cos\alpha(w_1-w_2)\Big]+v_2$$

$$ w’_1 = \frac{1}{m_1+m_2}\Big[-(1+c)m_2\sin\alpha\cos\alpha(v_1-v_2) + \big(m_1-m_2(c\sin^2\alpha-\cos^2\alpha)\big)(w_1-w_2)\Big]+w_2$$

$$ v’_2 = \frac{(1+c)m_1}{m_1+m_2}\cos\alpha\Big[(v_1-v_2)\cos\alpha + (w_1-w_2)\sin\alpha\Big] + v_2 $$

$$ w’_2 = \frac{(1+c)m_1}{m_1+m_2}\sin\alpha\Big[(v_1-v_2)\cos\alpha + (w_1-w_2)\sin\alpha\Big] + w_2 $$

The first two equations are quite unwieldy, but adding and subtracting \(v_1\) from the first one and \(w_1\) from the second one, and then doing some algebra produces:

$$ v’_1 = –m_2a\cos\alpha+ v_1 $$

$$ w’_1 = –m_2a\sin\alpha+ w_1 $$

$$ v’_2 = m_1a\cos\alpha+ v_2 $$

$$ w’_2 = m_1a\sin\alpha+ w_2 $$


$$ a = \frac{1+c}{m_1+m_2}\Big[(v_1-v_2)\cos\alpha + (w_1-w_2)\sin\alpha\Big] $$

A nice symmetric solution – and correct to boot.

Assuming that all the balls have the same density, their mass is a fixed constant times their radius cubed. In the formulas above the masses appear always as a quotient, so they can be replaced with the balls’ cubed radiuses.

The final code is this:

   * Updates the velocities of this ball and another one after a collision
   * The coordinate of the balls must be at the collision point.
   * @param otherBall second colliding ball
   * @param restitution coefficient of restitution for a ball-ball collision
  collide(otherBall: Ball, restitution: number) {
    var dx = this.x - otherBall.x;
    var dy = this.y - otherBall.y;
    var dv = this.v - otherBall.v;
    var dw = this.w - otherBall.w;
    var alpha = Math.atan2(dy, dx);
    var sinAlpha = Math.sin(alpha);
    var cosAlpha = Math.cos(alpha);
    var m1 = Math.pow(this.radius, 3);
    var m2 = Math.pow(otherBall.radius, 3);
    var a = (1 + restitution) / (m1 + m2) * (cosAlpha * dv + sinAlpha * dw);
    this.v = -m2 * a * cosAlpha + this.v;
    this.w = -m2 * a * sinAlpha + this.w;
    otherBall.v = m1 * a * cosAlpha + otherBall.v;
    otherBall.w = m1 * a * sinAlpha + otherBall.w;
  } // collide



(*) That ‘naïve’ formula for \(\alpha\) does not work when \(x_1-x_2\) is zero, nor it handles the various possible combination of signs of the numerator and denominator, but in the code all that is handled by Math.atan2().

Posted in: Physics | Programming

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Phrase of the week

May 12, 2013 at 12:47 PMMichele Mottini

It is not a correct deduction from the principles of economics that enlightened self-interest always operates in the public interest. Nor is it true that self-interest generally is enlightened

John Maynard Keynes in 'The end of laissez-faire'

Posted in: Opinion


Phrase of the week

May 5, 2013 at 2:00 PMMichele Mottini

It's usually difficult to make the complicated case easy; that's why it's called the complicated case.

Raymond Chen, in 'Another way to create a process with attributes, maybe worse maybe better'

Posted in: Opinion


Billiard simulation – part 2: detect collisions

May 4, 2013 at 11:07 AMMichele Mottini

First of all, what exactly ‘detect collisions’ means? As explained in the previous post the update algorithm needs the time of the next collisions between two balls and between a ball and the sides of the table – so ‘detecting collisions’ means computing these times for all the balls.

The initial position of the center of the \(k\)-th ball is \((x_k, y_k)\) and its velocity is \((v_k, w_k)\). The position of the center of each ball varies with time, at time \(t\) it is \((x_k+v_kt, y_k+w_kt)\).

A ball collides with one side of the table when the distance between its center and the side equals the ball radius. A ball moves toward the left or right sides with its horizontal speed \(v_k\) – if this speed is \(0\) the ball won’t collide, otherwise the collision times are \((x_{min}+r_k-x_k)/v_k\) and \((x_{max}-r_k-x_k)/v_k\), where \(x_{min}\) and \(x_{max}\) are the horizontal coordinate of the left and right sides (and hence the minimum and maximum values for the \(x\) coordinates). Only one of these values will be positive, and that is the result (if the ball moves towards the right \(v_k\) is positive and it will collide with the right side in the future – collision time positive, and had collided with the left side in the past – collision time negative; vice-versa for a ball moving to the left).

The collisions time with the top and bottom sides are computed in the same way – replacing \(x_k\) with \(y_k\), \(v_k\) with \(w_k\), \(x_{min}\) with \(y_{min}\) and \(x_{max}\) with \(y_{max}\). 

The complete code for collisions between a ball and the table sides is:

   * Returns the time of the first future collision with a side of the table. 
   * Returns null if there is no collision.
   * @param x ball horizontal or vertical position
   * @param v ball horizontal or vertical speed
   * @param min minumum horizontal or vertical position (i.e. co-ordinate of one side of the table)
   * @param max maximum horizontal or vertical position (i.e. co-ordinate of the other side of the table)
  private sideCollisionTime(x: number, v: number, min: number, max: number) {
    if (v === 0) {
      // Not moving towards the sides: no collision
      return null;
    var result;
    if (v < 0) {
      // Moving up/left - check agains the minumum 
      result = (min - x + this.radius) / v;
    } else {
      // Moving down/right - check agains the maximum
      result = (max - x - this.radius) / v;
    if (result >= 0) {
      return result;
    return null;
  } // sideCollisionTime

   * Returns the time of the first future collision with the left or right sides of the table. 
   * Returns null if there is no collision.
   * @param min minumum horizontal position (i.e. x co-ordinate of the left side of the table)
   * @param max maximum horizontal position (i.e. x co-ordinate of the right side of the table)
  sideXCollisionTime(min: number, max: number) {
    return this.sideCollisionTime(this.x, this.v, min, max);
  } // sideXCollisionTime

   * Returns the time of the first future collision with the top or bottom sides of the table. 
   * Returns null if there is no collision.
   * @param min minumum vertical position (i.e. y co-ordinate of the top side of the table)
   * @param max maximum vertical position (i.e. x co-ordinate of the bottom side of the table)
  sideYCollisionTime(min: number, max: number) {
    return this.sideCollisionTime(this.y, this.w, min, max);
  } // sideYCollisionTime

What about collisions between balls? This is a bit more complex – two balls collide when their distance is equal to the sum of their radiuses:


The square of the distance between balls \(k\) and \(l\) at time \(t\) is:

$$s^2(t)=(x_k + v_kt - x_l - v_lt)^2 + (y_k + w_kt - y_l - w_lt)^2 = (\Delta x + \Delta vt)^2 + (\Delta y + \Delta wt)^2$$

where \(\Delta v = v_k – v_l\), \(\Delta w = w_k – w_l\), \(\Delta x = x_k – x_l\) and \(\Delta y = y_k – y_l\). Equating this to the squared sum of their radiuses produces:

$$(r_k + r_l)^2 = (\Delta x + \Delta vt)^2 + (\Delta y + \Delta wt)^2$$

that is a quadratic equation in \(t\). Solving it gives the collision time:

$$t_{kl} = \frac{-p \pm \sqrt{q}}{\Delta v^2 + \Delta w^2}$$

where \(p=\Delta v\Delta x + \Delta w\Delta y\) and  \(q = (\Delta v^2+\Delta w^2)(r_k+r_l)^2-(\Delta v\Delta y - \Delta w\Delta x)^2\).

If \(q\) is negative the equation has no (real) solutions, that means that the balls do not collide. If \(q\) is zero the balls have only one collision: they ‘glance’ each other. If \(q\) is greater than zero there are two ‘collisions’: the balls touch, then go through each other and touch again on the other side. In the complete simulation after the first collision the balls change course, so we are interested only in the first – i.e. minimum – value. Finally, one or both solutions can be negative – corresponding to collisions that happened in the past; we want only the collisions in the future, so the result is the minimum non-negative solution.

With this the algorithm is complete, but there is still a problem: it all works fine if the balls never overlap, but if they do we will detect spurious collisions.

If the balls do not overlap to begin with, and the simulation detect the collisions and make the balls bounce away when they touch, how is it possible to even have overlaps? The answer is floating-point arithmetic: the actual code uses floating-point values, that introduces rounding errors in the computation.

What can happen (and actually DO happen) is:

  1. two balls move toward each other,
  2. they reach the collision point,
  3. the simulation updates their velocity so that they ‘bounce’,
  4. due to rounding errors there is a slight overlap in their position,
  5. this cause a new immediate (spurious) collision to be detected, with a new update in the balls velocities

…and so on. The final effect is that sometimes balls touch and then stick together, or start to bounce inside each other – sort of fun, but definitely not billiard.

The solution for this is to check the value of \(p\): only if it is negative the balls are moving toward each other, otherwise are moving apart (or parallel); so when \(p\) is not negative we always return ‘no collision’, this excludes the spurious collisions at step (5) above and fixes the simulation.

The complete code for collisions between two balls is:

   * Returns the time of the first future collision between this ball and another ball. 
  *  Returns null if the balls do not collide
   * @param other the other ball
  collisionTime(other: Ball) {
    var dv = this.v - other.v;
    var dw = this.w - other.w;
    var dx = this.x - other.x;
    var dy = this.y - other.y;
    var p = (dv * dx + dw * dy);
    if (p >= 0) {
      // The balls are moving apart or parallel
      return null;
    var dv2 = dv * dv + dw * dw;
    var r = this.radius + other.radius;
    var q = dv2 * r * r - Math.pow(dv * dy - dw * dx, 2);
    if (q < 0) {
      // No real solution: the balls do not collide
      return null;
    q = Math.sqrt(q);
    if (p <= q) {
      return (-p - q) / dv2;
    return (-p + q) / dv2;
  } // collisionTime

How comes that the sign of \(p\) indicates if the balls are moving apart or not? The time derivative of the distance between the balls is

$$\frac{ds}{dt}\bigg|_{t=0} = \frac{\Delta x\Delta v + \Delta y\Delta w}{s} = \frac{p}{s}$$

\(s\) is positive, so if \(p\) is negative the derivative is negative and the distance is decreasing – i.e. the balls are moving towards each other. 

Another way to prove it is noticing that \(s\) is the scalar product of the relative position and speed of the balls, hence the sign of \(p\) is the sign of the cosine of the angle \(\varphi\) between those two vectors. If \(p\) is negative the angle is greater than \(90^\circ\), that means that the direction of the relative velocity is opposite of the direction of the relative position – i.e. the balls are moving towards each other.


[Calculus agrees with geometry – that is good]

Posted in: Programming

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