scan conversion of point , line and circle

Lecture Outline
• Scan Conversion
– Point
– Line
– Circle

Point Scan
• Point plotting is furnished by converting a single
coordinate position into appropriate operations for
the output device in use.
• Random scan uses stored point-plotting instructions
(coordinates) and convert them in deflection voltage
that positions the electron beam at the screen
locations to be plotted during each refresh cycle.

Point Scan…1
• Raster scan depends on setting the bit value
corresponding the specified screen position within
the frame buffer to 1. The electron beam when
sweeps across such position in a line emits burst of
electrons with intensities corresponding to colour
codes to be displayed at that location.

Point Scan…2
• Let as scan-convert a point (x, y) to a pixel location
(x’, y’)
• x’ = Floor(x), and y’ = Floor(y)
• Where Floor is a function that returns largest integer that
is less than or equal to the argument.
• In such a case, the origin is placed at the lowest left
corner of the pixel grid in the image space
• All points that satisfy x’  x < x’+1 and y’  y < y’ + 1
will be mapped to pixel location (x’, y’) (refer Fig a)

Point Scan…4
• If the origin of coordinate system for (x, y) is to
be shifted to new location say (x+0.5, y+0.5),
then
– x’ = Floor(x+0.5), and y’ = Floor(y+0.5)
– This will place the origin of the coordinate system
(x, y) at the centre of pixel (0, 0)
– Now, all points that satisfy x’-0.5  x < x’+0.5 and
y’-0.5  y < y’+0.5 will be mapped to pixel location
(x’, y’) (refer Fig b)

Line Scan-Conversion
• Line plotting is done by calculating intermediate
positions between the two specified endpoint positions.
• Linearly varying horizontal and vertical deflection
voltages are generated in proportion to the required
changes in x, y directions to produce the smooth line.
• A straight line segment is displayed by plotting the
discrete points between endpoint positions.
– These intermediate positions are calculated from the equation
of the line
– Colour intensity is loaded into the frame buffer at the
corresponding pixel coordinate

Line Scan-Conversion…1
• Screen locations are referenced with integer values. E.g.
(10.48, 20.51) will be converted to pixel position (10, 20)
• This rounding effect causes lines to be displayed with a
stairstep appearance . This is noticeable on low
resolution systems.
• Better smoothing can be achieved by adjusting light
intensities along the line paths.

• Pixel column number is basically pixel position across a
scan line. These are increasing from left to right.
• Scan line number are increasing from bottom to top.
• Loading a specified colour into frame
Scan Line No

buffer at a pixel position defined by
column ‘x’ along scan line ‘y’
setPixel (x, y)
• Current frame-buffer intensity for a
specified location can be accomplished
2
1
getPixel (x, y)
0
0 12
Pixel
column No

• Line is defined by two end points P1, P2 and the equation of
line as shown in the Figure
• These can be plotted or scan-converted using different line
plotting algorithms.
y • The slope-intercept equation is
P2(x2, y2) not suitable for vertical lines.
y = m.x + b Horizontal, vertical and diagonal
(|m| = 1) lines can be handled
P1 (x1, y1) as special cases and can be
mapped to image space
b
straightforward.
x

Line Drawing Algorithms
• Direct use of line equation
• DDA algorithm
• Bresenham’s Line Algorithm

Direct use of Line equation
• Direct approach
y2
– The Cartesian slope
intercept equation for a
straight line:
y1
y=m.x+b (1)
x1 x2
– ‘m’ represents slope of
the line and ‘b’ is
intercept on y-axis, such
that y2
y 2  y1 y1
m (2)
x2  x1 x1 x2
b  y1  m . x1 (3)

Direct use of Line equation..1
• For a small change in ‘x’ • For lines with slope
along the line, the magnitudes |m|<1, x
corresponding change is set proportional to
in ‘y’ can be given by small horizontal
y = m . x (4) deflection voltage and
or x = y/m (5) the corresponding
– These equations form vertical deflection is
the basis for determining then set proportional to
the deflection voltages y as calculated from
in analog devices. the equation (4).

Direct use of Line equation…2
• For lines with slope • For lines with slope
magnitudes |m|>1, y magnitudes m=1, x =
is set proportional to a y and the horizontal
small vertical deflection and vertical deflection
voltage and the voltage are equal
corresponding
horizontal deflection is
then set proportional to
x as calculated from
the equation (5).

• This means:
– First scan convert end points to pixel locations (x1’, y1’) and
(x2’, y2’)
– Set ‘m’ and ‘b’ using equations
– For |m| 1, for integer value of x between and excluding
x1’, x2’, calculate the corresponding value of y and scan-
convert (x, y)
– For |m|> 1, for integer value of y between and excluding
y1’, y2’, calculate the corresponding value of x and scan-
convert (x, y)

• Example:
– End points are (0, 0) and (6, 18). Compute each value of x
as y steps from 0 to 18 and plot.
• For equation y = m.x + b, compute slope as
– m = y/x
• Using value of ‘m’ compute ‘b’
• Place y = 1 to 17 and compute corresponding value of ‘x’
• Plot

• Example:
– M = (18 – 0) /(6 – 0) = 3 m>1 b=0
– Change y: compute for y = 1 the value of x = ∆y/m = 1/3
• Cartesian coordinate point (1, 0.3)
• Pixel coordinate point (Case-I): (1, 0)
• Pixel coordinate point (Case-II): (1, 0)
– Compute for y = 2 the value of x = 2/3
• Cartesian coordinate point (1, 0.67)
• Pixel coordinate point (Case-I): (1, 0)
• Pixel coordinate point (Case-II): (1, 1)
– Plot the three graphs
– Continue …..

Digital Differential Analyzer (DDA)
• DDA is an incremental scan-conversion line algorithm
based on calculating either x or y as mentioned in
line primitives.
• Suppose at step ‘i’ the point is (xi, yi) on the line.
• For next point (xi+1, yi+1)
– yi+1 = yi + y, where y = m x
– xi+1 = xi + x, where x = y/m
– The process continues until x reaches x2’ (for |m| 1 case)
or y reaches y2’ (for |m|> 1 case)

Digital Differential Analyzer…1
• Consider a line with positive slope. Let the slope be
less than or equal to 1. For unit interval x=1,
yk+1 = yk + m, where k = 1, 2, …., end point (6)
– The value of yk+1 should be rounded to the nearest integer
• For slope greater than 1, increasing y by 1
xk+1 = xk + 1/m, where k = 1, 2, …., end point (7)
– The value of xk+1 should be rounded to the nearest integer

• The assumption here is that lines are processed from
left to right. If it is processed from right to left then
yk+1 = yk – m (8)
y
xk+1 = xk – 1/m (9)

x

• For negative slope of line as shown in figure:
• If moving from left to right:
– Set x = +1 and use equation (8) for m<1 to calculate y values
– Set  y = -1 and use equation (7) for m>1 to calculate x values
• If moving from right to left:
y
– If m < 1 then set x = -1 and obtain y
values from equation (6)
– If m > 1 then set y = +1 and obtain x
values from equation (9)

x

• This algorithm is faster in calculating pixel positions
• It eliminates multiplication by making use of raster
characteristics
• But rounding off in successive additions and its
accumulation may cause drifting of the pixel position
away from the true line
• This rounding off procedure is also time consuming
• By converting m or 1/m into integer or fractional
part, the time consumed can be reduced.

Bresenham’s Line Algorithm
• This raster-line generating algorithm is more accurate
and efficient
• It uses only incremental integer calculations
• It can be adapted to display circles and other curves.
• The method is based on identifying the nearest pixel
position closer to the line path at each sample step.

13 53

12 52

11 51

10 50
Fig-1 Fig-2
10 11 12 13 50 51 52 53

Bresenham’s Line Algorithm…1
• In Fig 1 the line segment 13
starts from scan line 11 12
and pixel position
11
(column) 10. The next
10
point along the line may Fig-1
10 11 12 13
be at pixel position (11,
11) or (11, 12) 53
• Similarly, in Fig 2 the line 52
segment starts from (50,
51
52) and the next pixel
50 Fig-2
position may be (51, 52)
50 51 52 53
or (51, 51)

• Bresenham’s line algorithm tests the sign of an
integer parameter and the one whose value is
proportional to the difference between the
separations of the two pixel positions from the actual
line path is selected.
• Let the line starts from left-end k+3y

point (xo, yo) of a given line. Now, y
k+2

the pixel whose scan-line y value y k+1

is closest to the line path is plotted. y
k

xk xk+1 xk+2 xk+3

Fig-3

• Let the pixel displayed is (xk, yk). The next pixel to
plot will be in column xk+1 and may be represented as
(xk+1, yk) or (xk+1, yk+1) y k+3

• The difference between estimated y k+2
y = m.x + b
value of y and new y value along the y k+1 d 2
line is to be computed. y
d 1
k

For first point yest =m . (xk+1) + b (10) x k xk+1 xk+2 xk+3

Difference d1 = yest – yk (11) Fig-3
For second point d2 = (yk+1) – yest (12)
Difference between the two separations = d1 – d2

• Difference (d1 – d2) = 2m.(xk+1) – 2 yk + 2b – 1 (13)
• Let m = y/x denoting vertical and horizontal
separations. Let px is defined as y k+3

px =  x (d1 – d2) yk+2
y = m.x + b

px = 2 y . xk - 2 x . yk + c (14) yk+1 d2
Where, c = 2 y + x (2b – 1) (15) yk
d1

• Decision parameter px is –ve i.e. d1<d2 xk xk+1 xk+2 xk+3

– Pixel at yk is closer to line than that at yk+1 Fig-3

– Lower point will be plotted
– Vice-Versa

• The change in coordinate along the line takes place
in unit steps. The decision parameter at next step will
be:
pk+1 = 2y . xk+1 - 2 x . yk+1 + c
And pk+1 – pk = 2 y (xk+1 – xk) - 2 x (yk+1 – yk) (16)
But xk+1 = xk+1, so that
pk+1 = pk + 2 y - 2 x (yk+1 – yk) (17)
(yk+1 – yk) is either 0 or 1 depending upon the sign of pk.
The initial value of parameter will be
po = 2 y - x (18)

• Algorithm for |m|<1: (steps)
– Input and store (xo, yo). Plot it (load on buffer).
– Calculate x, y, 2 y - 2 x
– Compute decision parameter, po (= 2 y - x)
– Assuming pk < 0, the next point will be (xk+1, yk) and
pk+1 = pk + 2y
Otherwise, plot point (xk +1, yk +1) and
pk+1 = pk + 2 y - 2 x
– Repeat step x times

• E.g. Plot a line with end points (20, 10) and (30, 18).
The slope of line is 0.8.
– Calculate x (=10), y (=8)
– Compute decision parameter, po (= 2 y - x = 6)
– Compute 2 y (=16), 2 y - 2 x (= -4)
– Assuming pk < 0, the next point will be (xk+1, yk) and
pk+1 = pk + 2y
Otherwise, plot point (xk +1, yk +1) and
pk+1 = pk + 2 y - 2 x
– Repeat step x times

• It is generalized to lines with arbitrary slopes.
• Considers symmetry between octants and quadrants of
the xy plane.
• Algorithm allows moving along pixel position from either
end of the line.
• For positive slope and start from left end point, both the
increments are positive. It is vice-versa for start from
right end point.
• For negative slope, one increment will be positive and
other negative depending upon the start point.

• When the two vertical separations from the line path are
equal (d1 = d2), we choose upper (or lower) of the two
candidate pixels.
• Special cases like y = 0 (horizontal line); x = 0 (vertical
line) and |x| = |y| (diagonal line) can be loaded
directly into the frame buffer without processing them
through the line –plotting algorithm.

• For negative slope of a line (|m|<1)
– When moving from left to right
– Po = - 2∆y - ∆x
– If po is – ve point will be (xk+1, yk)
and pk+1 = po - 2 ∆y
– If po is + ve point will be (xk+1, yk-1)
and pk+1 = po - 2 ∆y - 2 ∆x

• For negative slope of a line (|m|<1)
– When moving from right to left
– Po = 2∆y + ∆x
– If po is – ve point will be (xk-1, yk)
and pk+1 = po - 2 ∆y
– If po is + ve point will be (xk-1, yk+1)
and pk+1 = po + 2 ∆y + 2 ∆x

y = m. x + b
• For negative slope of a line (|m|>1) yk+3
– Increment is ∆y yk+2
d1 d2
– When moving from left to right
yk+1
– ∆x = ∆y/m or (1/m) = ∆x/ ∆y
yk
– d1 = xest - xk
xk xk+1 xk+2 xk+3
– d1 = (1/m)(yk+1 – b) – xk
– d2 = xk+1 – xest
– d2 = xk+1 – (1/m)(yk+1 – b)
– d1 – d2 = (2/m)(yk + 1) – 2xk – (2b/m) – 1
– pk = ∆y . (d1 – d2)
– pk = 2 ∆x . yk - 2 ∆y . xk + 2 ∆x (1-b) - ∆y

y = m. x + b
– When moving from left to right yk+2
d1 d2
– pk = 2 ∆x . yk - 2 ∆y . xk + 2 ∆x (1-b) - ∆y yk+1
– pk = 2 ∆x . yk - 2 ∆y . xk + c yk
– Where c = 2 ∆x (1-b) - ∆y xk xk+1 xk+2 xk+3
– When pk ≤ 0, then d1 < d2 point (xk, yk+1)
– When pk > 0, then d1 > d2 point (xk+1, yk+1)
– Now, pk+1 = 2 ∆x . yk+1 - 2 ∆y . xk+1 + c
– pk+1 – pk = 2 ∆x . (yk+1 – yk) - 2 ∆y . (xk+1 –xk)
– pk+1 = pk+2 ∆x . (yk + 1 – yk) - 2 ∆y . (xk+1 –xk)

y = m. x + b
d1 d2
– pk+1 = pk + 2 ∆x . (yk + 1 – yk) - 2 ∆y . (xk+1 y k)
–x
k+1

– pk+1 = pk + 2 ∆x - 2 ∆y . (xk+1 –xk)
yk
– Now, if pk+1 ≤ 0, then d1 < d2 point (xk, yk+1) xk xk+1 xk+2 xk+3
– Decision parameter pk+1 = pk + 2 ∆x
– When pk+1 > 0, then d1 > d2 point (xk+1, yk+1)
– Decision parameter pk+1 = pk + 2 ∆x - 2∆y
– Initial decision parameter po = 2 ∆x - ∆y

y = m. x + b
d1 d2
– Steps:
yk+1
– Initial decision parameter po = 2 ∆x - ∆y
yk
– If ≤ 0, then point (xk, yk+1) and
xk xk+1 xk+2 xk+3
Decision parameter pk+1 = pk + 2 ∆x
– If > 0, then point (xk+1, yk+1) and
Decision parameter pk+1 = pk + 2 ∆x - 2∆y
– For other orientations or start from right to left, change the sign
of ∆x and ∆y based on change in value of respective x or y

• Example: Line 6x + 5y = 30
– Points are (0, 6) and (5, 0) (|m|>1) increment ∆y
– ∆x = 5; ∆y = -6; 2 ∆x = 10; 2 ∆y = -12
– po = 2 ∆x + ∆y = 10 - 6 = 4 point (1, 5)
– p > 0; pk+1 = pk + 2 ∆x + 2∆y = 4 + 10 – 12 = 2
point (2, 4)
– p2 = 2 + 10 -12 = 0 point (2, 3)
– p3 = pk + 2 ∆x = 0 + 10 = 10 point (3, 2)
– p4 = 10 + 10 - 12 = 8 point (4, 1)
– p5 = 8 + 10 - 12 = 6 point (5, 0)

Circle Generating Algorithms
• Properties of circle:
– A circle is defined as the set of points that are all at a given
distance r from a center position (xc,yc)
– This distance relationship is expressed by the Pythagorean
theorem in Cartesian coordinates as
(x – xc)2 + (y – yc)2 = r2 (22)
– For position of points on a circle
• Move from (xc – r) to (xc + r) in unit steps r
yc
• Calculate corresponding y values as:
y = yc ± √(r2 – (xc – x)2 (23) xc

Circle Generating Algorithms…1
• This is not the best method for generating a circle.
• One problem with this approach is that it involves
considerable computation at each step.
• Moreover, the spacing between plotted pixel
positions is not uniform
• The spacing can be adjusted by interchanging x and y
whenever the absolute value of
slope of the curve is greater than 1.
• This is computationally more
cumbersome.

• Another approach is to use polar coordinates r and θ,
yielding
x = xc + r Cos θ (24)
y = yc + r Sin θ (25)
• Larger angular separations along the circumference can
be connected with straight line segments to approximate
the circular path.
• For a more continues boundary on a raster display, the
step size is set at 1/r. This plots pixel positions that are
approximately one unit apart.

• Computation can be reduced either by
– Considering symmetry w.r.t. y-axis for segment in 1st and
2nd quadrant, and then considering symmetry w.r.t. x-axis
for segment in 3rd and 4th quadrant or
– Considering the symmetry between adjacent octants
within a quadrant w.r.t 45o line (-y, x) (y, x)

– Therefore calculations are required (-x, y)
(x, y)
for points between x=0 and x=y 450

and then these are mapped to rest (-x, -y) (x, -y)

eight sectors. (-y, -x) (y, -x)

Circle Generating Algorithm…4

• All these procedures require a good deal of
computation time
• More efficient circle algorithms are based on
incremental calculations as in the Bresenham line
algorithm
• Bresenham’s line algorithm is adapted to circle
generation by setting up decision parameters for
finding the closest pixel to the circumference at each
sampling step
• It avoids square-root calculations by comparing the
squares of the pixel separation distances

Midpoint Circle Algorithm
• The closest pixel position to the specified circle path
is determined by moving unit distance at each step
• Let radius is r and screen center position is (xc, yc)
• Set up algorithm to calculate pixel positions around a
circle path centered at the coordinate origin (0, 0)
• To move each calculated position (x, y) to its proper
screen position add 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

Midpoint Circle Algorithm…1
• To apply the midpoint method, circle function is
defined as:
fcircle (x, y) = x2 + y2 – r2 (26)
Such that
fcircle (x, y) < 0, if (x, y) is inside the circle boundary
= 0, if (x, y) is on the circle boundary
> 0, if (x, y) outside the circle boundary
• This becomes the decision parameter for midpoint
algorithm

• Let us consider the midpoint between the two
candidate pixels at sampling position xk+1.
• After plotting of the pixel at (xk, yk), next to
determine is whether the pixel at position (xk+1, yk)
along a circular path.
or the one at position (xk+1, yk-1) is closer to the
circle.
• Decision parameter will be: y x +y -r =0 2 2 2
k

 1
pk  f circle  xk  1, yk   yk-1
 2 Midpoint

2
 1
 x k  1 2
  yk    r 2 xk xk + 1 xk + 2
 2 (27)

• If pk<0, this midpoint is inside the circle and the pixel
on scan line yk is closer to the circle boundary.
• Otherwise, the mid position is outside or on the
circle boundary, and we select the pixel on scanline
yk-1.
• Successive decision parameters are obtained using
incremental calculations. For pixel location (xk+1+ 1)
p k 1  p k  2(x k  1)  ( y 2 1  y 2 )  ( y k 1  y k )  1
k k (28)
• yk+1 is either yk or yk-1 depending on the sign of pk

• Increments for obtaining pk+1 are either 2xk+1 +1 (if pk
is negative) or (2xk+1 + 1 - 2yk+1)
• Evaluation of the terms 2xk+1 and 2yk+1 can also be
done incrementally as
2xk+1 = 2xk + 2; 2yk+1 = 2yk - 2
• At the start position (0, r) these two terms have the
values 0 and 2r, respectively.
• Each successive value is obtained by adding 2 to the
previous value of 2x and subtracting 2 from the
previous value of 2y.

• The initial decision parameter is obtained by
evaluating the circle function at the start position
(x0, y0) = (0, r) as:
 1
p0  f cricle 1, r  
 2
2
 1
 1  r    r 2
 2
5
p0   r (29)
4
• If the radius r is specified as an integer, we can simply
or
round to p0 = 1 – r (for r an integer).

Steps
• Input radius r, circle center (xc, yc), and obtain the first point
on the circumference of a circle centered at the origin as (x0,
y0) = (0, r)
• Calculate the initial value of the decision parameter as
p0 = 5/4 – r
• At each xk position, starting at k = 0, perform the following
test: if pk < 0, the next point along the circle centered on (0,0)
is (xk+1 , yk) and pk+1 = pk + 2xk+1 + 1
• Otherwise, the next point along the circle is (xk+1, yk-1) and
pk+1 = pk + 2xk+1 + 1 - 2yk+1
Where 2xk+1 = 2x k + 2 and 2yk+1 = 2y k - 2
• Determine symmetry points in the seven other octants
• Move pixel position (calculated) to screen position by adding

Example
• Plot the generated pixel positions for a circle with
radius r = 10 using midpoint circle algorithm.
– po = 1 – r = -9 y
y=x

– Initial point (0, 10) 10

– Decision parameters
9

8

2xo = 0; 2yo = 20 7

6

5

4

3

2

1

0
x
0 1 2 3 4 5 6 7 8 9 10

BRESENHAM’S ALGORITHM
• Assume that (xi, yi) are the coordinates of the last scan-
converted pixel upon entering step i.
• Let the distance from the origin to pixel T squared minus
the distance to the true circle squared = D(t).
• Then let the distance from the origin to pixel S squared
minus the distance to the true circle squared = D(s).
• As the coordinates of T are (xi + 1, yi) and those of S are
(xi + 1, yi -1), the following expression can be developed:

BRESENHAM’S ALGORITHM…1
(xi, yi) T (xi + 1, yi)
y

(xi, + 1, yi - 1)
S

2
i
y

)2
r
1
i
(x

xi  12   yi  12

x

Choosing Pixels in Bresenham’s Circle Algorithm

D(t) = (xi + 1)2 + yi2 - r2
D(s) = (xi+1)2 + (yi – 1)2 – r2

• This function D provides a relative measurement of the
distance from the center of a pixel to the true circle.
• Since D(t) will always be positive (T is outside the true
circle) and D(s) will always be negative (S is inside the
true circle), a decision variable, di may be defined as
follows:
di = D(t) + D(s)
Therefore
di = 2(xi + 1)2 + yi2 + (yi – 1)2 – 2r2

• When di<0, |D(t)|<|D(s)|, then select pixel T.
• When di ≥ 0, |D(t) ≥ |D(s)|, then select pixel S.
• The decision variable di+1 for the next step:
di+1 = 2(xi+1+1)2 + y2i+ 1 + (yi+1-1)2 – 2 r2
Hence
di+1- di = 2(xi+1 + 1)2 + y2i+1 + (yi+1 - 1)2
– 2(xi + 1)2 – y2i – (yi – 1)2
Since xi+1 = xi + 1 so,
di+1 = di + 4xi + 2(y2i+1 – y2i) – 2(yi+1 - yi) + 6

• If T is the chosen pixel (di < 0) then yi+1=yi and so
di+1 = di + 4xi + 6
• If S is the chosen pixel (di ≥ 0) then yi+1 = yi-1and so
di+1 = di + 4(xi–yi) + 10
• Set (0, r) to be the starting pixel coordinates and
compute the base case value di :

d1 = 2 (0 + 1)2 + r2 + (r – 1)2 – 2r2
= 3- 2r

Example
• Given a radius of 10 units.
• Compute di = 2(xi+1)2 + yi2+(yi-1)2-2r2.
• Set the value as (0,r) as the start point

d0 = 2(0+1)2 + 102 + 10-1)2 - 2x102
= 2+100+81-200
= 183-200
= -17. Or = 3 – 2r = 3 – 20 = -17

Example…1
• As d0 < 0, yi+1=yi.
• So next point (1,10)
d1 = d0 + 4xi + 6.
= - 17+ 4 + 6
= -7
• So next point is (2, 10).
d2 = d1 + 4x2+6
= -7+8+6.
=+7.

Example…2
• So next point (3, 9).
d3 = di+4(xi-yi)+10
= 7+4(3-9)+10
= 7-24+10
= -7.
• So next point (4, 9)
d4 = -7+16+6 = 15.
• So next point (5, 8).
d5 = 15+4(5-8)+10 = 13.

Example…3
• So next point (6,7).
d6 = 13+4x(6-7)+10
= 19.
So next point is (7,6)

Comparison

Iteration Mid point Bresenham
0 (0,10) (0,10)
1 (1,10) (1,10)
2 (2,10) (2,10)
3 (3,10) (3,9)
4 (4,9) (4,9)
5 (5,9) (5,8)
6 (6,8) (6,7)
7 (7,7) (7,6)

Coordinate Representations
• General graphics packages are designed to be used with Cartesian
Coordinate specifications.
• If coordinate values for a picture are specified in some other
reference frame, they must be converted to Cartesian system.
• Several different Cartesian reference frames are used to construct and
display a scene:
– Modeling or local coordinates
– World coordinates
– Device or screen coordinates
• Graphics system first converts world coordinates to normalized
coordinates before final conversion to device coordinates.

64

Coordinate Representations

Modeling
Transformations

Modeling Coordinates

65

FRAME BUFFER

A frame buffer is a video output device that drives a video display from a
memory buffer containing a complete frame of data. The information in the
memory buffer typically consists of color values for every pixel (point that can be
displayed) on the screen.

Color values are commonly stored in 1-bit binary (monochrome), 4-bit palettized,
8-bit palettized, 16-bit highcolor and 24-bit truecolor formats.

The total amount of the memory required to drive the frame buffer depends on
the resolution of the output signal, and on the color depth and palette size.

66

Loading Frame Buffer
• When objects are scan converted for display with
raster system, frame buffer positions must be
calculated.
• Scan conversion algorithms generate pixel positions
at successive unit intervals, this allows the use of
incremental methods to calculate frame buffer
addresses.
• This is accomplished with setPixel procedure, which
stores intensity values for the pixels at the
corresponding addresses within the frame-buffer
array.

67

Frame-buffer array is addressed in row-major order

Pixel screen positions stored linearly in row-
major order within the frame buffer

For 1 bit per pixel, the frame buffer address
for pixel position (x,y) is calculated as

addr(x,y)=addr(0,0)+y(xmax+1)+x

addr(x+1,y)=addr(x,y)+1

addr(x+1,y+1)=addr(x,y)+xmax+2 68

SOFTWARE STANDARDS

 The primary goal is portability among the graphics softwares.
 Without standards, programs designed for one hardware system can not be
transferred to another system without extensive rewriting of programs.
 Several Int. & national standards planning organizations have cooperated
and developed Graphical Kernel System (GKS). The is adopted by Int.
Standards organisation (ISO) and American National Standards Institute
(ANSI).
 Other system is PHIGS (Programmer’s Hierarchial Interactive Graphics
Standard), which is extension of GKS.
 PHIGS+ is developed to provide 3D surface-shading.
 Standard graphics functions are defined as a set of specifications that is in-
dependent to any programming language.

PHIGS polyline(n,x,y)
Fortran call gpl(n,x,y)
C ppolyline (n,pts)
69

scan conversion of point , line and circle

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