Figure 2-2: The output of a sine representation in .NET

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Figure 2-2: The output of a sine representation
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So, the result is a set of dots ( . ) starting at (0,100) and then as x goes from 0,1,2,3,4, . . . 500, y produces cosine values. Since we know from trigonometry that a cosine will always be between 1 and 1, we multiply by 50 to make y a value between 50 and 50. Obviously, the resulting numbers place the points along the path of a curve. These numbers are sample values that result from the x and y parameters. The more we decrease the distance between consecutive i values, the more precise the path is. If we want to shorten the curve in the x direction, then we need to do the following adjustment to the x coordinate:
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1 2 3 4 5 for(int i=0; i<5000; i++){ int x = i/10; int y = (int)(50. * cos(PI/180.* i) point(x , y+50); }
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The resulting points will be placed along a full circle when i goes from 0 to 360, so, in our case, 5000 will result in 5000 / 360 = 13.8 full circles. We don t want x to go to 5,000 because the screen is only 400 pixels long and it will be drawn outside of the visible screen. So we divide i by 10, and therefore x will go to only 500, resulting in an image like that shown in Figure 2-3.
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Figure 2-3: The output of a cosine representation Web Forms qr generatoron .net
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If we reverse the values of x and y, as is done in the following code, then we can obtain a rotated curve, as shown in Figure 2-4.
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1 2 3 4 5 for(int i=0; i<5000; i++){ int x = (int)(50. * cos(PI/180.* i) int y = i/10; point(x+50 , y); } );
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Figure 2-4: Reversing the direction of a cosine representation
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Now, if we combine the two, alternating sine and cosine, as in the following code:
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1 2 3 4 5 for(int i=0; i<5000; i++){ int x = (int)(50. * cos(PI/180.* i) int y = (int)(50. * sin(PI/180.* i) point(x+50 , y+50); } ); );
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this will result in the unexpected (perhaps) output shown in Figure 2-5.
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Figure 2-5: The output of the combination of a sine and cosine representation
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A circle! We will use this technique later on to rotate objects in the screen, because basically what we are doing here is forcing x and y to move on the perimeter of a circle. Or, to be precise, we force x and y to rotate around a center 13.8 times, since the counter goes from 0 to 5,000. If the counter is going from 0 to 180, as shown here: Web Service Crystal barcode 3 of 9 integratingfor visual
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1 2 3 4 5 for(int i=0; i<180; i++){ int x = (int)(50. * cos(PI/180.* i) int y = (int)(50. * sin(PI/180.* i) point(x+100 , y+100); } ); );
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we would have created a half-circle (see Figure 2-6), because i is the number of degrees of the rotation angle.
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Figure 2-6: A half-circle
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The equations used to construct a circle or portions of it are defined through the following formulas:
x=r*cos(i), y=r*sin(i)
where i is the counter or any parameter that changes in an orderly fashion. Thus, such a set of equations are referred to as parametric, where the parameter
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here is i and, consequently, the resulting circle is a parametric one. However, in analytic geometry, the equation for generating a circle is different: x2 + y2 = r2 Such an equation denotes that for every point on a plane, only the points that satisfy the above equation are part of a circle of radius r and center (0,0). The following code generates such a circle (shown on the left in Figure 2-7):
1 2 3 4 for(int x = -50; x<50; x++) for(int y=-50; y<50; y++) if(x*x + y*y == 25*25) point(x+50,y+50);
Line 3 in the above code can be replaced with the following statement:
3 if(x*x + y*y > 25*25 && x*x + y*y < 26*26)
In this case, a circle is produced by selecting a series of points that fit the preceding two inequalities. This circle has a radius that ranges between 25 and 26 (shown on the right in Figure 2-7).