Motion in .NET

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The round() command returns an integer number that is closest to the inputted float number. So, by dividing x or y by 10, we obtain a float number that is rounded to its closest integer which, in turn, is multiplied by 10. The result of this process is shown in Figure 6-6 (right).
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6.4 Interactive Transformations
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So far, you have seen how to move geometrical elements or images in order to produce an impression of motion. This is done by placing the element in a new position and then redrawing the screen in a sequential manner. In the following code, we will use the translate() and rotate() commands to rotate two rectangles around points 20,20 and 50,50, respectively. Consider the following code:
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1 2 3 4 5 6 7 8 void draw(){ translate(20,20); rotate(radians(mouseX*3.6)); rect(0,0,20,10); translate(50,50); rotate(radians(mouseX*3.6)); rect(0,0,20,10); }
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In line 2, we use the translate() command, which takes as parameter the coordinates of the location to move at. In line 3, we use the rotate() command, which takes as parameter an angle to be rotated (in radians). So, we use the mouse s position multiplied by 3.6 and then convert it into radians. The reason that we use 3.6 is that the maximum dimension of the screen is 100, so the maximum angle will be 100 3.6 = 360. Then we use the rect() command to draw a rectangle at the new translated and rotated position. The result can be seen in Figure 6.7 (left): the first rectangle rotates around point 20,20 as expected, but the second rectangle rotates also about 20,20 and not point 50,50. The reason is that the whole scene is never reset to the origin. So, the second translation occurs as the addition of the previous two. This problem is explained in more detail in 8 in section 8.5. To solve this problem, we use the popMatrix() and pushMatrix() commands before and after each transformation. So, the preceding code will be as follows:
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1 2 3 4 5 6 7 void draw(){ pushMatrix(); translate(20,20); rotate(radians(mouseX*3.6)); rect(0,0,20,10); popMatrix(); pushMatrix();
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translate(50,50); rotate(radians(mouseX*3.6)); rect(0,0,20,10); popMatrix();
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The result of this code is shown in Figure 6-7 on the right.
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Figure 6-7: Transformation without matrices reset (left) and with (right)
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In the next code sample, we will use multiple instances of transformations in a loop:
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1 void draw(){ 2 background(255); 3 for(float i=0; i<30; i++){ 4 pushMatrix(); 5 rectMode(CENTER); 6 noFill(); 7 translate(50,50); 8 scale(1/(i/mouseX),1/(i/mouseX)); 9 rotate(radians(i*mouseY)); 10 rect(0,0,50,50); 11 popMatrix(); 12 } 13 }
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Within the loop, we use the pushMatrix() and popMatrix() to reset the scene and then draw a square after transforming it in three ways: first, we translate it to the center of the screen (at point 50,50), then we scale it by a fraction of the mouse s position, and finally we rotate it by the mouse s position, treating it as an angle degree. The result of this interactive transformation can be seen in Figure 6-8.
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Figure 6-8: Interactive transformation of a square
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Suppose now that we have a series of rectangles that are translated in random positions, and we want to rotate them around their center just by sliding the mouse right or left. Since each rectangle will be rotated around its center, we need to keep track of each rectangle s center, and its angle of rotation. So, we will use three arrays to hold this information. The code is shown here:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 float px[] = new float[300]; float py[] = new float[300]; float pr[] = new float[360]; void setup(){ size(300,300); } void draw(){ } void mousePressed(){ for(int i=0; i<mouseY; i++){ px[i] = random(width); py[i] = random(height); pr[i] = random(360); } } void mouseDragged(){ background(255); for(int i=0; i<mouseY; i++){ pushMatrix(); rectMode(CENTER); translate(px[i],py[i]); rotate(radians(pr[i] + mouseX)); rect(0,0,5,500); popMatrix(); } }
In the first three lines, we define three arrays that will hold the x and y coordinates and rotation angles for each rectangle. We define the maximum number of rectangles as 300, then in the mousePressed() section of the code we create random numbers that we use to populate the arrays. Please note that we are not using the maximum number of rectangles, that is, 300, but only a number equal to mouseY. This means that the number of rectangles increases or decreases, depending on the vertical motion of the mouse. Next, in the mouseDragged() section of the code, we go through all the rectangles in the scene and perform the transformations: we translate each rectangle to its center, which is stored in the arrays px[] and py[] and then rotate by an angle, which is the addition of the