CS 130. Scan Conversion. Raster Graphics

Transcription

1 CS 130 Scan Conversion Raster Graphics 2 1

2 Image Formation Computer graphics forms images, generally two dimensional, using processes analogous to physical imaging systems like: - Cameras - Human visual system Images are generated and shown with specialized software and hardware 3 Introduction Synthetic images can be created like those taken by a camera For this class, we will draw directly on the image plane 4 2

3 Scan conversion Image Plane is a N x M grid of pixels This grid is similar to a coordinate system only it is discrete. 5 Scan conversion Scan conversion is taking the graphic from the continuous (thin) line to turning on pixels 6 3

4 Scan conversion Scan conversion algorithms compute integer coordinates for pixels near the line or circle Algorithms are invoked many, many times and so must be efficient Result should be visually pleasing, for example, they should have constant density Obviously, they should be able to draw the gamet of possible lines/circles and remain defined 7 2D math for lines Math Review How do we determine the equation of the line? (X2, Y2) (X1, Y1) 8 4

5 Math Review 2D math for lines Slope-Intercept formula for a line (X2, Y2) Slope = (Y2 Y1)/(X2 X1) ( Y - Y1)/(X - X1) Solving For Y (X1, Y1) (X, Y) Y = [(Y2 Y1)/(X2 X1)]X + [-(Y2 Y1)/(X2 - X1)]X1 + Y1 or Y = m X + b 9 Explicit (functional) representation y = f(x) Math Review y is the dependent, x independent variable Find value of y from value of x Example, for a line: for a circle: y = mx + b x 2 + y 2 = r

6 Math Review Implicit representation f(x,y) = 0 Called a membership function, test value of (x,y) to see if it belongs to the curve x,y treated the same, axis invariant Example, for a line: for a circle: ax + by + c = 0 x 2 + y 2 r 2 = 0 ax + by + c < 0? 11 Math Review Parametric Representation x = x(u), y = y(u) where new parameter u (or often t) determines the value of x and y (and possibly z) for each point x,y treated the same, axis invariant 12 6

7 Math Review Parametric formula for a line (X2, Y2) X = X1 + t(x2 X1) Y = Y1 + t(y2 Y1) for parameter t from 0 to 1 (X1, Y1) (X, Y) Therefore, when t = 0 we get (X1,Y1) t = 1 we get (X2,Y2) Varying t gives the points along the line segment 13 Math Review 2D math for circles Circle centered at origin X 2 + Y 2 = R 2 where R is the radius We can plot this as: R Y = +/- R 2 -X

8 Math Review 2D math for circles Y = R 2 -X 2 includes a fair amount of computation May want to parametric form: x = R cos y = R sin R With theta going from 0 to 360, Then use with a look-up sin/cos 15 Scan conversion Scan conversion is taking the graphic from the continuous (thin) line to turning on pixels 16 8

9 Scan conversion for lines 0 < M < 1 M > 1 17 DDA algorithm for lines Parametric Lines: the DDA algorithm for 0 < M < 1 (digital differential analyzer) Y i+1 = m x i+1 + B = m(x i + x) + B x = (x i+1 -x i ) = y i + m( x) <- must round to find int We increment by 1 pixel in X We turn on [xi, Round(yi)] - or same for Y if m >

10 Scan conversion for lines 19 Fast scan conversion for lines DDA includes Round( ); and this is fairly slow For Fast Lines, we want to do only integer math +,- We do this using the Midpoint Algorithm To do this, lets look at lines with y-intercept B and with slope between 0 and 1: y = (dy/dx)x + B ==> f(x,y) = (dy)x - (dx)y + B(dx) = 0 Removes the division => slope treated as 2 integers 20 10

11 Midpoint Algorithm For lines in the first octant, the next pixel is to the right or to the right and up one By looking at the difference between the line and the midpoint, e, we only need to look at the sign i.e.) no Round( ) needed midpoint e 21 Midpoint: Decision Variable Rewrite such that f(x,y) = ax + by + c (from f(x,y) = (dy)x - (dx)y + B(dx) =0) Then, f(x + 1, y + 1/2) is the next column's midpoint = a(x + 1) + b(y + 1/2) + c = d (this d is called the decision variable) if (d = 0) then the line passes on the midpoint if (d > 0) then the line passes above if (d < 0) then the line passes below 22 11

12 The question is how do we determine the value of d for a given line? For example to start, f(xo + 1, yo + 1/2) = d d = a(xo + 1) + b(yo + 1/2) + c = a(xo) + b(yo) + c + a + b/2 = f(xo,yo) + a + b/2 but f(xo,yo) = 0 = a + b/2 Midpoint: Decision Variable What about d after this first case? 23 Scan Converting Circles What problems arise when we think about scan conversion for circles (ie. when we increment one pixel in x, say) R 24 12

13 Scan Converting Circles What problems arise when we think about scan conversion for circles (ie. when we increment one pixel in x, say) 25 Scan Converting Circles Better approach takes into account the 8 way symmetry of each octant in a circle (Lets call this function circle_points ) 26 13

14 Fast Circles Only consider the first octant of the circle of radius r centered at origin Start from (r,0) and go until x < y (at 45 degrees) The decision is at each step to go up, or go over to the left and up 27 Fast Circles - via the midpoint f(x,y) = x 2 + y 2 -r 2 = 0 (x -1,y +1) midpoint e (x,y +1) Then for next decision (midpoint decides row) f(x-1/2, y +1) = d = (x - 1/2) 2 + (y + 1) 2 -r 2 Starting at y = 0, x = r this becomes = (r - 1/2) 2 + (0 + 1) 2 -r 2 = 5/4 - r (x,y) 28 14

15 Fast Circles - via the midpoint All thats left is to find the new d for the next step If (go up and over) d* = f(x - 1-1/2, (y + 1) + 1) = (x -3/2) 2 + (y + 2) 2 -r 2... = d - 2(x -1) + 2 (y + 1) + 1 = d + 2(y - x) + 5 Else (just go up) d* = f(x -1/2, (y + 1) + 1)... = d + 2(y)