Chapter 6.2-Fuoyancy Force Generators

Following code snippet resides in the buoyancy force generator.

 1 void ParticleBuoyancy::updateForce(Particle* particle, real duration)
 2 {
 3    // Calculate the submersion depth.
 4    real depth = particle->getPosition().y;
 5    // Check if we’re out of the water.
 6    if (depth >= waterHeight + maxDepth) return;
 7    Vector3 force(0,0,0);
 8    // Check if we’re at maximum depth.
 9    if (depth <= waterHeight - maxDepth)
10    {
11    force.y = liquidDensity * volume;
12    particle->addForce(force);
13    return;
14    }
15    // Otherwise we are partly submerged.
16    force.y = liquidDensity * volume *
17    (depth - maxDepth - waterHeight) / 2 * maxDepth;
18    particle->addForce(force);
19 } 

      The formula of ‘d‘ is computed in the below manner:

       where y_o is the y coordinate of the object and y_w is the y coordinate of the liquid plane (assuming it is parallel to the XZ plane).

       Why we compute ‘d‘ in such a manner?

       The author only leaves the formula to the reader without any further explanation, which once frustrates me for quite a long time.

       if you have some experience of 3D programming, especially doing some so-called post-effect or mapping texture coordinates from normal device coordinate to screen space. Then this will be familar to you because it also works this way.

       Now let‘s take a look into deeper mechanism of this formula. Trust me, it quite simple:P

       We see that:

       when y₀ = y_w + s,       we have d = 0;     (1)
       when y₀ = y_w ,            we have d=-1/2; (2)
       and when y₀ = y_w - s, we have d=-1;     (3)
       in fact ‘d‘ is a linear function of y₀. Then d = f(y₀) = ky₀+b.

       So all we need to do is to find k and b. We choose (1) and (3) to perform the processing.

       The point (y_w + s, 0) and point(y_w - s,-1) are located on the line. So k = (-1 - 0)/[(y_w - s) - (y_w + s)] = 1/2s;

       Plug k into (1), 0 = 1/2s*(y_w + s) + b, sovel this equation we have b = -(y_w + s)/2s;

       Now all the puzzle are fixed.

       d = 1/2s*y0-(y_w + s)/2s

       d = (y₀ - y_w - s)/2s