Distorted Android

// Let (v.x,v.y) be point P (the current vertex).

// Let vPoint[effect].xy be point S (the center of effect)

// Let vPoint[effect].xy + vRegion[effect].xy be point O (the center of the Region circle)

// Let X be the point where the halfline SP meets a) if region is non-null, the region circle b) otherwise, the edge of the bitmap.

//

// If P is inside the Region, this function returns |PX|/||SX|, aka the 'degree' of point P. Otherwise, it returns 0.

//

// We compute the point where half-line from S to P intersects the edge of the bitmap. If that's inside the circle, end. If not, we solve the

// the triangle with vertices at O, P and the point of intersection with the circle we are looking for X.

// We know the lengths |PO|, |OX| and the angle OPX , because cos(OPX) = cos(180-OPS) = -cos(OPS) = -PS*PO/(|PS|*|PO|)

// then from the law of cosines PX^2 + PO^2 - 2*PX*PO*cos(OPX) = OX^2 so

// PX = -a + sqrt(a^2 + OX^2 - PO^2) where a = PS*PO/|PS| but we are really looking for d = |PX|/(|PX|+|PS|) = 1/(1+ (|PS|/|PX|) ) and

// |PX|/|PS| = -b + sqrt(b^2 + (OX^2-PO^2)/PS^2) where b=PS*PO/|PS|^2 which can be computed with only one sqrt.

//

// the trick below is the if-less version of the

// return degree of the point as defined by the Region

// Currently only supports circles; .xy = vector from center of effect to the center of the circle, .z = radius

  return max(sign(D),0.0) / (1.0 + 1.0/(sqrt(DOT*DOT+D*one_over_ps_sq)-DOT));  // if D<=0 (i.e p is outside the Region) return 0.

  return max(sign(D),0.0) * min(1.0/(1.0 + 1.0/(sqrt(DOT*DOT+D*one_over_ps_sq)-DOT)),E);  // if D<=0 (i.e p is outside the Region) return 0.