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//xSide is 1 when _c1 is to the right of _c2, //and -1 when it's to the left of _c2. //ySide is 1 when _c1 is below _c2, //and -1 when it's above _c2 _c1.xPos > _c2.xPos xSide = 1 : xSide = -1; _c1.yPos > _c2.yPos ySide = 1 : ySide = -1; //Reposition _c1 out of the collision. //Use xSide and ySide to make the overlap vector either //negative or positive, depending on the side of the collision _c1.setX = _c1.xPos + (collision_Vx * xSide); _c1.setY = _c1.yPos + (collision_Vy * ySide); //Reposition _c2 _c2.setX = _c2.xPos + (collision_Vx * -xSide); _c2.setY = _c2.yPos + (collision_Vy * -ySide); //We ve moved the circles out of the collision. Now we //need to bounce them away at the correct angle //_c1's motion vector var v1:VectorModel = new VectorModel ( _c1.xPos, _c1.yPos, _c1.xPos + _c1.vx, _c1.yPos + _c1.vy ); //_c2's motion vector var v2:VectorModel = new VectorModel ( _c2.xPos, _c2.yPos, _c2.xPos + _c2.vx, _c2.yPos + _c2.vy ); //Project v1 onto v0 and v0.ln var p1a:VectorModel = VectorMath.project(v1, v0); var p1b:VectorModel = VectorMath.project(v1, v0.ln); //Project v2 onto v0 and v0.ln var p2a:VectorModel = VectorMath.project(v2, v0); var p2b:VectorModel = VectorMath.project(v2, v0.ln);
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//Mix and match the results of p1 and p2 to //find the correct bounce vectors //Bounce c1 //using p1b and p2a _c1.vx = p1b.vx + p2a.vx; _c1.vy = p1b.vy + p2a.vy; //Bounce c2 //using p1a and p2b _c2.vx = p1a.vx + p2b.vx; _c2.vy = p1a.vy + p2b.vy; } A collision is detected when the magnitude of the vector between the circles is less than their combined radii. This is exactly the same way a collision was detected in our first example. But because both circles are moving, we have a new problem. When only one circle was moving, it was easy to know which circle to reposition after the collision: the moving one. Now that both circles could be moving, which circle do we reposition, and by how much To figure that out, we need to do the following:
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1. Find out the amount by which the circles overlap when they collide.
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var overlap:Number = totalRadii - v0.m; This is the same code that we used to find the overlap in the example with the stationary circle.
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2. Create a collision vector. This will give us the vx and vy values we need to separate
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the circles. This vector must be divided in half so that we can share it between the two circles. One half will go to circle 1, and the other half will go to circle 2. var collision_Vx:Number = Math.abs(v0.dx * overlap * 0.5); var collision_Vy:Number = Math.abs(v0.dy * overlap * 0.5); The vector is found by scaling v0 s magnitude to the amount of overlap. By multiplying the overlap by v0 s dx and dy, we have a new vector inclined at the same angle as v0 but with the same magnitude as the overlap. Figure 3-22 shows how this vector is found. The vx and vy values need to be absolute (without a plus or minus sign). We need these values to be neutral because we want the flexibility to decide whether to add or subtract this vector to each circle s position (you ll see how in the next step).
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COLLISIONS BETWEEN CIRCLES
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Figure 3-22. Find the collision vector, divide it by half, and make it absolute.
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3. Is the first circle above, below, to the right of, or to the left of the second circle We need
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to know this information so that we can correctly add or subtract the overlap vector to each circle s position. The easiest way to track this is by creating variables that are assigned 1 or -1, depending on where the circles are in relation to each other. Why we need to do this will become clear in the next step. var xSide:int; var ySide:int; _c1.xPos > _c2.xPos xSide = 1 : xSide = -1; _c1.yPos > _c2.yPos ySide = 1 : ySide = -1;
This bit of code illustrates the use of the ternary operation. It s a shorthand style of writing if / else statements. This line of code
_c1.xPos > _c2.xPos xSide = 1 : xSide = -1;
is the same as writing this:
if(_c1.xPos > _c2.xPos) { xSide = 1; } else { xSide = -1; }
For short, simple conditional tests, the ternary operation is very helpful. It takes up much less space and will make your code more readable once your eyes get used to the syntax. Try it!
4. We need to push the circles apart to resolve the collision. For example, on the x axis,
we ll need to push one circle to the left and the other to the right. The collision vector that we calculated in step 2 will be the correct direction for one of the circles, but not the other. However, we know that the directions the circles need to move in will be the polar opposites of each other. That means we can use the xSide and ySide variables (which will be 1 or -1) to correctly invert one of the vectors. //Move _c1 out of the collision _c1.setX = _c1.xPos + (collision_Vx * xSide); _c1.setY = _c1.yPos + (collision_Vy * ySide); //Move _c2 out of the collision _c2.setX = _c2.xPos + (collision_Vx * -xSide); _c2.setY = _c2.yPos + (collision_Vy * -ySide); The orientation of the circles will always be changing, so we can never know which circle s overlap vector to invert. Luckily, the xSide and ySide variables take care of keeping track of that for us automatically. Figure 3-23 shows how the circles new positions are found.
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