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Chapter Three Part I

Chapter Three Part I. Output Primitives. Outline. In this chapter we will cover the following topics: Part I: Line drawing algorithms. Circle generating algorithms Part II: Pixel addressing and object geometry. Filled-Area primitives. Setting frame-buffer values

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Chapter Three Part I

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  1. Chapter ThreePart I Output Primitives

  2. Outline • In this chapter we will cover the following topics: Part I: • Line drawing algorithms. • Circle generating algorithms Part II: • Pixel addressing and object geometry. • Filled-Area primitives. • Setting frame-buffer values • This chapter will show the discrete nature for computer graphics.

  3. Line drawing • What is a line? • Is the path between two end points. • Any point (x, y) on the line must follow the line equation: y = m * x + b, where • m is the slope of the line. • b is a constant that represent the intercept on y-axis.

  4. Line drawing (cont.) • If we have two end points, (x0, y0) and (x1, y1), then the slope of the line between those points can be calculated as: m = y / x = (y1 – y0) / (x1 – x0) • To draw a line, we can use two algorithms: • DDA (Digital Differential Analyzer). • Bresenham’s Line Algorithm.

  5. Line drawing (cont.)

  6. Line drawing – DDA algorithm • (DDA) is a scan-conversion line algorithm based on calculating either y or x. • y = m * x • x = y / m • we have many cases based on sign of the slope, value of the slope, and the direction of drawing. • Slope sign: positive or negative. • Slope value: <= 1 or >1. • Direction: (left – right) or (right – left)

  7. DDA – case 1 • Positive slope and left to right: • If slope <= 1 then: xk+1 = xk + 1 yk+1 = yk + m • If slope > 1 then: xk+1 = xk + 1/m yk+1 = yk + 1

  8. DDA – case 2 • Positive slope and right to left: • If slope <= 1 then: xk+1 = xk – 1 yk+1 = yk – m • If slope > 1 then: xk+1 = xk – 1/m yk+1 = yk – 1

  9. DDA – case 3 • Negative slope and left to right: • If |m |<= 1 then: xk+1 = xk + 1 yk+1 = yk – m • If |m | > 1 then: xk+1 = xk + 1/m yk+1 = yk – 1

  10. DDA – case 4 • Negative slope and right to left: • If |m | <= 1 then: xk+1 = xk – 1 yk+1 = yk + m • If |m | > 1 then: xk+1 = xk – 1/m yk+1 = yk + 1

  11. Bresenham’s Line Algorithm • Is an efficient algorithm for line drawing. • When we have a point (xk, yk), then we must decide whether to draw the point (xk+1, yk) or (xk+1, yk+1). • Note that, at all cases, we move to xk+1, still we must decide to move to yk or yk+1. d2 y d1 yk+1 yk Xk+1

  12. Bresenham’s Line Algorithm (cont.) • After calculating d1 and d2 we will choose yk or yk+1. y = m (xk + 1) + b d1 = y – yk = m (xk + 1) + b - yk d2 = (yk+1) – y = yk+1 - m (xk + 1) - b d1 – d2 = 2m*xk + 2m – 2yk + 2b -1 d1 – d2 = 2m (xk + 1) – 2yk + 2b -1

  13. Bresenham’s Line Algorithm (cont.) • Now we will define pk (decision parameter) as: pk = ∆x (d1 – d2) = 2 ∆ y * X k- 2 ∆ x * yk+ c, where c = 2 ∆ y + ∆ x(2b – 1) , k represents the kth step • If pk < 0 (i.e. d1 < d2)  we plot the pixel (xk+1, yk) Otherwise we plot the pixel (xk+1, yk+1)

  14. Bresenham’s Line Algorithm (cont.) • to get the next pk pk+1 • pk+1 = pk+2 ∆y - 2 ∆x (yk+1 – yk) • p0 = 2 ∆ y - ∆ x

  15. Bresenham’s Line Algorithm (cont.) • Input two end points and store (x0, y0) in the frame buffer. • plot (x0, y0) to be the first point. • Calculate the constants ∆ x, ∆ y, 2 ∆ y, and 2 ∆ y – 2 ∆ x, and obtain the starting value for the decision parameter as p0 = 2 ∆ y - ∆ x. • At each xk along the line, starting at k = 0, perform the following test. If pk < 0, plot (xk+1, yk) and pk+1 = pk+2∆y Otherwise, plot (xk+1, yk+1) and pk+1 = pk+2∆y - 2∆x. • Perform step 4 ∆x – 1 times.

  16. Bresenham’s Line Algorithm ( Example) • Note: Bresenham’s algorithm is used when slope is <= 1. • using Bresenham’s Line-Drawing Algorithm, Digitize the line with endpoints (20,10) and (30,18). • y = 18 – 10 = 8 • x = 30 – 20 = 10 • m = y / x = 0.8 • plot the first point (x0, y0) = (20, 10) • p0 = 2 * y – x = 2 * 8 – 10 = 6 , so the next point is (21, 11)

  17. Example (cont.)

  18. Example (cont.)

  19. Bresenham’s Line Algorithm (cont.) • Notice that bresenham’s algorithm works on lines with slope in range 0 < m < 1. • We draw from left to right. • To draw lines with slope > 1, interchange the roles of x and y directions.

  20. Circle Generating Algorithms • What is a circle? • It is a set of points that are all at a given distance r from center position (xc, yc). • The distance relationship equation of a circle is expressed by the Pythagorean theorem in Cartesian coordinates as: ( x – xc)2 + ( y – yc)2 = r2

  21. Circle Generating Algorithms (cont.) • We can re-write the circle equation as: y = yc ± ( r2 – ( x – xc)2 )0.5 • By substitution with x , xc and yc we can get y. • Two problems with this approach: • it involves considerable computation at each step. • The spacing between plotted pixel positions is not uniform, as demonstrated below

  22. Circle Generating Algorithms (cont.) • Polar coordinates (r and ) are used to eliminate the unequal spacing shown above. • Expressing the circle equation in parametric polar form yields the pair of equations • x = xc + r cos • y = yc + r sin

  23. Circle Generating Algorithms (cont.) • When a circle is generated with these equations using a fixed angular step size, a circle is plotted with equally spaced points along the circumference. • The step size chosen  depends on the application and the display device. • Computation can be reduced by considering the symmetry of circles. The shape of the circle is similar in each quadrant. • We can take this one step further and note that there is also symmetry between octants.

  24. Circle Generating Algorithms (cont.)

  25. Mid-point circle algorithm • A method for direct distance comparison is to test the halfway position between two pixels to determine if this midpoint is inside or outside the circle boundary. • This method is more easily applied to other conics, and for an integer circle radius. • we sample at unit intervals and determine the closest pixel position to the specified circle path at each step.

  26. Mid-point circle algorithm (cont.) • For a given radius r and screen center position (xc, yc), we can first set up our algorithm to calculate pixel positions around a circle path centered at the coordinate origin ( 0, 0 ). • Then each calculated position (x, y) is moved to its proper screen position by adding xc to x and yc to y. • Along the circle section from x = 0 to x = y in the first quadrant, the slope of the curve varies from 0 to - 1. • Therefore, we can take unit steps in the positive x direction over this octant and use a decision parameter to determine which of the two possible y positions is closer to the circle path at each step.

  27. Mid-point circle algorithm (cont.) • Positions in the other seven octants are then obtained by symmetry. • To apply the midpoint method. we define a circle function: fcircle(x, y) = x² + y² – r² • Any point (x, y) on the boundary of the circle with radius r satisfies the equation fcircle( x, y ) = 0. • If fcircle( x, y ) < 0, thepoint is inside the circle boundary , If fcircle( x, y ) > 0, the point is outside the circle boundary, If fcircle( x, y ) = 0, the point is on thecircle boundary.

  28. Mid-point circle algorithm (cont.)

  29. Mid-point Circle Algorithm Calculating pk First, set the pixel at (xk , yk ), next determine whether the pixel (xk + 1, yk ) or the pixel (xk + 1, yk – 1) is closer to the circle using: pk= fcircle (xk + 1,yk – ½) = (xk + 1)²+(yk – ½)² – r²

  30. Calculating Pk +1 • Pk +1 = fcircle (xk + 1 + 1,yk + 1 – ½) • =[(xk + 1) + 1 ]² + (yk + 1 – ½)² – r² Or • Pk + 1 = pk + 2(xk + 1) + (y²k+ 1 – y²k ) – (yk+ 1 – yk ) + 1

  31. Calculating p0 • p0 = fcircle (1, r – ½) = 1 + (r – ½)² – r² or • p0 = 5 /4 – r ≅ 1 – r

  32. Circle-Drawing Algorithms Mid-point Circle Algorithm - Steps • Input radius rand circle center (xc, yc ). set the first point (x0, y0) = (0, r ). • Calculate the initial value of the decision parameter as p0 = 1 – r. • At each xk position, starting at k = 0, perform the following test: If pk < 0, plot (xk + 1, yk ) and pk+1 = pk + 2xk + 1 + 1, Otherwise, plot (xk+ 1, yk– 1 ) and pk+1 = pk + 2xk+1 + 1 – 2yk+1, where 2xk + 1 = 2xk + 2 and 2yk + 1 = 2yk– 2.

  33. Circle-Drawing Algorithms Mid-point Circle Algorithm - Steps • Determine symmetry points on the other seven octants. • Move each calculated pixel position (x, y) onto the circular path centered on (xc, yc) and plot the coordinate values: x = x + xc, y = y + yc • Repeat steps 3though 5until x  y. • For all points, add the center point (xc, yc )

  34. Mid-point Circle Algorithm - Steps Now we drew a part from circle, to draw a complete circle, we must plot the other points. We have (xc + x , yc + y), the other points are: (xc - x , yc + y) (xc + x , yc - y) (xc - x , yc - y) (xc + y , yc + x) (xc - y , yc + x) (xc + y , yc - x) (xc - y , yc - x) Circle-Drawing Algorithms 34

  35. Mid-point circle algorithm (Example) • Given a circle radius r = 10, demonstrate the midpoint circle algorithm by determining positions along the circle octant in the first quadrant from x = 0 to x = y. Solution: • p0 =1 – r = – 9 • Plot the initial point (x0, y0 ) = (0, 10), • 2x0 = 0 and 2y0 =20.  • Successive decision parameter values and positions along the circle path are calculated using the midpoint method as appear in the next table:

  36. Mid-point circle algorithm (Example)

  37. Mid-point circle algorithm (Example)

  38. Mid-point Circle Algorithm – Example (2) • Given a circle radius r = 15, demonstrate the midpoint circle algorithm by determining positions along the circle octant in the first quadrant from x = 0to x = y. Solution: • p0 = 1 – r = – 14 • plot the initial point (x0 , y0) = (0, 15), • 2x0 = 0 and 2y0 = 30. • Successive decision parameter values and positions along the circle path are calculated using the midpoint method as:

  39. Mid-point Circle Algorithm – Example (2)

  40. Mid-point Circle Algorithm – Example (2)

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