Point Plotting Techniques

 Most of the general interactive computer graphics
display use the Cartesian co-ordinate system.
 This system serves as a reference for plotting
points, lines and other figures on the screen.
 Each point on the screen is addressed by its (x,y)
co-ordinate.
 In this system x increases from left to right and y
increases from top to bottom and origin is placed at
the left hand corner of the screen

 Precision of the co-ordinate system is generally
made equal to the resolution of the display screen.
 Let co-ordinate for both x and y values is of 10 bits
i.e. x and y is specified as 10 bit binary number this
allows 210* 210 distant x and y co-ordinates on the
screen.
 If precision on the screen of the co-ordinate system
is less than resolution. Then there will be points on
screen at which it is impossible to display the dots

 If precision is greater than the resolution than
some of the points may not appear on the
screen. Therefore, the precision is made
generally made equal to the resolution of
display.
 A point plotting display construct picture by
combining several 100pts.
 A pixel is the smallest picture unit that can be
displayed lines and curves are drawn with
closely spaced pixels.

Criteria for Good Computer Generated
Line
 Line should appear straight: Point plotting
techniques are well suited to the generation of line
parallel to x and y axis or line at 45 degree to x and
y axis for the other lines the point to be plotted must
be approximated to the nearest addressable points
suitable approximation makes the line appear
straight.
 Line should terminate accurately: Unless lines
are plotted accurately they may terminate at the
wrong place. The effect is often seen as a small gap
between the end points of line and the starting point
of the next.

 Line should have constant density: line
density is directly proportional to the number
of dots displayed by the length of the line. To
maintain constant density dots should be
equally spaced. This is possible only in lines
parallel to or diagonal to the axis. In other
cases we must attempt to achieve as even
spacing as possible. Bunching of dots may
appear as a bright region.

 Line should be drawn rapidly: Interactive
computer graphics application requires the lines to
appear rapidly on the screen. This requires that
there should be minimum of computation to draw the
line. Ideally this computation should be performed
by a special purpose hardware.
 Line density should be independent of line
length and angle: This is the difficult requirement
to satisfy, a line generation algorithm should be able
to keep line density approximately constant.

Incremental Method
 Incremental methods are useful in generating lines
on point plotting displays. So it is also used in
shading the computer generated pictures of solid
objects.
 Incremental computing techniques are form of
iterative computation in which each iterative step is
simplified by maintaining a small amount of state
or memory about the progress of the computation.
 Example: A newcomer to a city finds it way using
incremental method . If he is looking for a particular
house no. then he requires 3 pieces of information

 The direction in which he is moving
 The no. of house he is looking for
 The no. house he is just passed.
 In incremental method not only the final results but
the intermediate results are of use.
 When we plot lines incrementally we start at one
end and finish by computing the co-ordinates of the
other end and in between the incremental technique
generate the co-ordinates of all the points that lie on
the line

Symmetrical DDA
 The DDA is a mechanical device that solves differential
equations by numerical methods.
 It traces out successive (x, y) values by simultaneously
incrementing x and y by small steps proportional to the
first derivative of x and y.
 Since real variables have limited precision, summing an
inexact m repetitively introduces cumulative error
buildup and eventually a drift away from a true Round
(y), for most short lines ,this will not present a problem.

Symmetrical DDA
 Symmetrical DDA
works on the principle
that we simultaneously
increment x & y by
small steps.
 i.e. we can generate
line by incrementing x &
y by є∆x & є∆y
respectively.
є∆x
є∆y
Next pixel to plot

Symmetrical DDA
 Suppose (x1 , y1) &
(x2, y2) are end points of
the line.
 ∆x=x2-x1
 ∆y=y2-y1
 An ‘є’ is small quantifier
which is equal to 2-n
 2n
is the line length
estimate
 i.e. ‘є’is the reciprocal of line
length estimate
2n-1
≤ max (|∆x| |∆y|) ≤ 2n
(x1 , y1)
(x2, y2)

Symmetrical DDA
 Advantage of having є= 2-n
.In the computation shifting is much
faster because each step in the line is computed with just two
additions for generating adjacent points rather than division.
 We can round off to the nearest integer value of each incremental
step.
 We can use arithmetic overflow. In case of arithmetic overflow
both x & y are kept in registers which consist of both integer and
fractional parts.
 Incremental value (which are very small i.e. less than unity )are
repeatedly added to the fractional part and whenever the result
overflows the integer part is incremented.
 So we initialized DDA with 0.5 in each fractional part to achieve
true rounding.

Symmetrical DDA
X: Integer Fractional
+
є∆x
Y: Integer Fractional
+
є∆y
Initially initialized at 0.5
Q.1 Plot a line between (0, 0) & (10, 5) using Symmetric DDA

Plot a line whose end points are given below using Symmetric
DDA
 (2,3) & (10,15)
 (5,10) & (20 ,15)
 (10,2) & (21, 8)
 (4, 5) & (20,11)
 (0,1) & (9, 5)
 (20,10) & (30, 18)

Simple DDA
 Principle of DDA is ,we may use any line length
estimate and any corresponding value of є such that
neither є∆x nor є∆y exceeds unit magnitude (i.e.
increment ≤1).
 In symmetric DDA we use power of 2(2n
) as a line
length estimate.
 In simple DDA, we choose a line length estimate equal
to larger of magnitude of ∆x & ∆y so that either є∆x &
є∆y is of unit magnitude.
 So, we can replace one of the address with simple
counter.
 The simple DDA generate unit steps in the direction of
greatest motion.

Simple DDA
 Suppose (x1 , y1) &
(x2, y2) are end points of
the line.
 ∆x=x2-x1
 ∆y=y2-y1
procedure DDA(x1,y1,x2,y2:integer)
var i, length : integer
var xincr, yincr, x,y : real
begin
length = abs(x2-x1);
if abs(y2-y1) > length then
length= abs(y2-y1)
xincr= (x2-x1)/length;
yincr= (y2-y1)/length;
x=x1;
y=y1;
plot( x, y);
x=x+0.5;
y=y+0.5;
for i=1 to length do
begin
x=x+xincr;
y=y+yincr;
plot( trun(x), trun(y));
end;
end;
Q.1 Plot a line between (0, 0) & (10, 5) using Simple DDA

Plot a line whose end points are given below using Simple DDA
 (2,3) & (10,15)
 (5,10) & (20 ,15)
 (10,2) & (21, 8)
 (4, 5) & (20,11)
 (0,1) & (9, 5)
 (20,10) & (30, 18)

Q:1 (2,3)&(10,15)
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
1 2 3 4 5 6 7 8 9 10
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
1 2 3 4 5 6 7 8 9 10
Symmetric DDA Simple DDA

3. PLOTTING OF (10,2) & (21,8)

4. PLOTTING OF (4,5) & (20,11)
BY SYMMETRIC DDA BY SIMPLE DDA

5. PLOTTING OF (0,1) & (9,5)
BY SYMMETRIC DDA BY SIMPLE DDA

Mid-point Algorithm
It uses the concept of incremental method.
It is based on integer arithmetic.
Assuming the constraint that we are
drawing line.
slope<1 i.e. in first quadrant.
1
1
1
2
3
4
5
6 7
8

Given the choice of current pixel, which
one do we choose next : E or NE ?
Equations:
1. y = dy / dx * x + B
If we rewrite the equation as
2. F (x ,y) = a * x + b * y + c = 0
This is also equation of line.
Given by: F (x, y)= dy * x –dx * y+ B* dx =0
a = dy , b = -dx , c=B*dx
Parametric line eq.
dy / dx is slope of line
dy / dx =y2-y1/ x2-x1

What is the advantage of this equation ?
You can use this equation to find out
where is the midpoint of the next choice of
the pixel.
Criteria:
Evaluate the mid- point , M w.r.t the equation of line.
Choice: E or NE ?
M
NE
E

 So, when we draw a line from NE to E, the
bisector will give the mid-point on the line w.r.t
the line equation
 Now we have to find wether the M is above the
line or below the line.
 If M is below the line NE pixel is closer to the
line.
 If line is below M then we will be choosing E
pixel.
M
NE
E

How do you use the equation to find
out whether the mid point is below the
line or above the line ?
Equation:
F (x, y)= dy * x –dx * y + B* dx =0
If the point (x, y) lies on the line then F (x, y) = 0
If we substitute any value of (x, y)
i.e. above the line or below the line
we get F (x, y) ≠ 0
Not equal to zero will give you two conditions.
So, what are the conditions?

Conditions:
1. F (x, y) > 0;
If point below the line
2. F (x, y) < 0;
If point above the line
If M is below the line we get F (x, y) to be positive
If M is above the line we get F (x, y) to be negative.
So, just looking the sign of the F (x, y) ,it itself will tell you
whether M is below or above the line
That would be sufficient to help you choose E or NE pixel

M
Q
NE
E
(xp+1,yp+1)
(xp+1,yp+1/2)
(xp+1,yp)
M
Q
NE
E
(xp+1,yp+1)
(xp+1,yp+1/2)
(xp+1,yp)
Q is below M, so
E is selected
Q is above M, so
NE is selected

ALGO-for next choice
If F(M)>0 /* Q is above M */
then select NE
/* M is below the line */
else select E;
/* also with F(M)=0 */

Evaluate Mid-point M using a
decision variable d
d=F (x, y)
d=F(xp+1,yp+1/2)
=a(xp+1)+b(yp+1/2)+c at M
set
d old = d
Based on the sign of d , you choose NE or E

CASE I : Chosen E
d new = F(xp+2,yp+1/2)
=a(xp+2) + b(yp+1/2) + c
(∆d) E = d new - d old = a /* = dy */
CASE II : Chosen NE
d new = F(xp+2,yp+3/2)
=a(xp+2) + b(yp+3/2) + c
(∆d) NE = d new - d old = a + b /* = dy- dx */
UPDATE USING d new= d old+ ∆d
∆d has two choices o f (∆d) E or (∆d) NE

Mid-point Criteria
d=F(M)=F(xp+1,yp+1/2);
if d>0 choose NE
else /* if d<=0 */ choose E
CASE EAST:
Increment M by 1 in x
d new=F (M new) = F (xp+2,yp+1/2)
(∆d) E = d new - d old = a /* = dy */
(∆d) E= dy

CASE NORTH EAST:
Increment M by 1in both x and y
d new = F( M new)=F(xp+2,yp+3/2)
(∆d) NE = d new - d old = a + b /* = dy- dx */
(∆d) NE= = dy - dx
What is d start ?
d start = F (x0+1,y0+1/2) = ax0 + a + by0 + b/2 + c
= F(x0,y0) + a + b/2
= dy – dx/2
Let’s get rid of the fraction and see what we end up with
for all the variables

Solution: Multiply all the variables by 2
and we get
d start = 2dy – dx
(∆d) NE = 2(dy - dx)
(∆d) E= 2dy
Now we have
integer arithmetic

Bresenham’s Algorithm: (|m|<1)
1. Input the two line end points and store the left end
point in (x0,y0)
2.Load (x0,y0) into the frame buffer, that is plot the first point.
3.Calculate constants dx,dy,2dy and 2dy-2dx and
obtain the starting value for the decision parameter as
d start = 2dy-dx.
4.At each xp ,along the line starting p=0,perform the following
Test If d<=0(-ve) the next point to plot is (xp+1,yp)
and d = d + 2dy or d = d + (∆d) E
Otherwise ,the next plot is (xp+1,yp+1)
and d =d + 2dy - 2dx or d =d +(∆d) NE
5.Repeat step 4. dx times

ALGORITHM:
while( x<x1)
if (d<=0)/* choose E */
d = d + (∆d) E
else
d =d + (∆d) NE
y = y+1
end if
x = x+1
Plot_point (x, y)
end while

Example:
SP- (5,8)
EP- (9,11)
SUCCESSIVE STEPS:
Initialization : dy =3 , dx =4 ,dstart =2 ,
(∆d) E=6 , (∆d) NE = -2
4 5 6 7 8 9 10 11
6
7
8
9
10
11
12
13
14
d=2, (6,9)
d=0, (7,9)
d=6, (8,10)
d=4, (9,11)

For a line with positive slope greater than 1,we
interchange the roles of the x and y directions.
That is ,we step along the y direction in unit steps
and calculate successive x values nearest the line
path.
Ex: Plot a line (1,6) and (3,16)
Special Cases :
Horizontal lines (∆y=0),
Vertical lines (∆x=0) and diagonal lines with (
∆x=∆y) each can be loaded directly into the frame
buffer without processing them through the line
plotting algorithm.

Example:
SP- (1,6)
EP- (3,16)
SUCCESSIVE STEPS:
Initialization : dx =2 , dy =10
dstart = 2dx-dy=4-10=-6
(∆d) E=2dx=2x2=4
(∆d) NE = 2dx-2dy=2x2-2x10=-16
1.d=-6, (1,7)
2.d=-2, (1,8)
3.d=2, (2,9)
4.d=-14, (2,10)
5.d=-10, (2,11)
6.d=-6, (2,12)
7.d=-2, (2,13)
8.d=2, (3,14)
9.d=-14, (3,15)
10.d=-10, (3,16)

Lines with Negative Slopes:
•If the slope of the line we are using is between 0 and -
1,then ∆y=-∆y and we step in x using exactly the same
tests but decrement the dependent variable rather than
increment it.
Ex: Plot a line (5,18) and (15,10)
•If slope is more negative than -1 ∆x=-∆x and interchange
the roles of x and y as in case where the slope is greater
than 1
Ex: Plot a line (8,-12) and (2,3)

Example:
SP- (5,18)
EP- (15,10)
SUCCESSIVE STEPS:
Initialization : dy =-8 8, dx =10 ,dstart =6 ,
(∆d) E=16 , (∆d) NE = -4
5 6 7 8 9 10 11 12 13 14 15
10
11
12
13
14
15
16
17
18
d=6, (6,17)
d=2, (7,16)
d=-2, (8,16)
d=14,(9,15)
d=10,(10,14)
d=6,(11,13)
d=2,(12,12)
d=-2,(13,12)
d=14,(14,11)
d=10,(15,10)
CONDITION : - When the slope is between 0 & -1

CONDITION : - When the slope is greater than -1
S P=(8,-12)
E P=(2,3)
dy = (y2-y1) = 3-(-12)= 15
dx = (x2-x1) = 2-8 = -6 6
dstart = 2*dx-dy = 2*6-15 = -3
(∆d) E= 2*dx = 2*6 = 12
(∆d)NE =2*(dx-dy) = 2*(6-15) = -18

d=(8,-11)
d=(7,-10)
d=(7.-9)
d=(6,-8)
d=(6,-7)
d=(6,-6)
d=(5,-5)
d=(5,-4)
d=(4,-3)
d=(4,-2)
d=(4,-1)
d=(3,0)
d=(3,1)
d=(2,2)
d=(2,3)
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
3
2
1
0
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
-11
-12

Mid-point Algorithm( Circle)
Here we assume that we start from the top of the
circle.
It uses the concept of incremental method.
Unlike the line drawing where we were drawing
line between 0 & 1 ( means 1 st octant).
We plot points in second octant and last point in
the second octant will have special property i.e (
x= y ) x is equal to y
1
1

1
2
3
4
5
6 7
8 (xp,yp-1)
(xp,yp-2)
(xp+1,yp)
M
E
SE
ME
MSE
(xp, yp)
(xp+1,yp-1)
(xp+1,yp-1/2)
Suppose (xp,yp) is selected at previous iteration
Evaluate the mid-point w.r.t the arc or line and decide the pixel E or SE
Choice is E or SE ? After the previous iteration.
Only considers circles centered at origin with integer radii.
Can apply translations to get non-origin centered circles.

 Mid-point function uses implicit equations
given by
 Use of Symmetry:
 Only calculate one octant.
 Can get points in other seven octants
F (x, y)= x2 + y2 - r2 =0
1
2
3
4
5
6 7
8
(x, y)
(y, x)
(y, -x)
(-y, x)
(-y, -x)
(-x, y)
(-x, -y) (x, -y)
Shows Symmetry

 Criteria: Choice is between pixels E or SE
 Conditions:
1. F (x, y) > 0; If point is outside the circle
2. F (x, y) < 0; If point is inside the circle
 Directly starting with Bresenham’s Mid-point
concept..
F (x, y)= x2 + y2 - r2 =0

Evaluate Mid-point M using a
decision variable d
F(M)=F(xp+1,yp-1/2)
=(xp+1)2+(yp-1/2)2-r2 at M
set
d old = d
Based on the sign of d , you choose E or SE
d old = F(M)

 If d>=0 choose SE
 Next mid point increment +1 in x and -1 in y
which gives
 If d<0 choose E.
 Next mid point increment +1 in x
d new = ????
d new = ????

(∆d) SE = d new - d old
= F(xp+2,yp-3/2)-F(xp+1,yp-1/2)
= 2xp-2yp+5
(∆d) E = d new - d old
UPDATE USING d new= d old+ ∆d
∆d has two choices o f (∆d) E or (∆d) SE
= F(xp+2,yp-1/2)-F(xp+1,yp-1/2)
= 2xp+3

What is d start ?
d start = F (x0+1,y0-1/2) = F(1, r-1/2)
= 1+ (r-1/2)2 - r2
= 1+ r2 –r + 1/4 -r2
= 5/4-r = 1-r
hstart = 1-r

Mid-point Circle Algorithm( Version 1)
 x=0 y=r h=1-r
 DrawCircle (x,y)
 while (y>x)
 if h<0 /* select E */
 h=h+2x+3;
 else /* select SE */
 h=h+2(x-y)+5;
 y=y+1;
 endif
 x=x+1;
 DrawCircle(x,y);
 end-_while

Example : r=10 r = 10
h= 1-r = -9
x=0 y=10
2x=0 2y=20
(∆d) E=2x+3 (-ve)
(∆d)SE=2x-2y+5 (+ve)
K h 2x 2y (x,y)
0 -9 2 20 (1,10)
1 -6 4 20 (2,10)
2 -1 6 20 (3,10)
3 6 8 18 (4,9)
4 -3 10 18 (5,9)
5 8 12 16 (6,8)
6 5 14 14 (7,7)
7 6
0,10
1 2 3 4 5 6 7 8
9
8
7
6
5

example r=15
r = 15
h= 1-r = -14
x=0 y=15
2x=0 2y=30
(∆d)SE=2x-2y+5 (+ve)
K h 2x 2y (x,y)
0 -14 2 30 (1,15)
1 -11 4 30 (2,15)
2 -6 6 30 (3,15)
3 1 8 28 (4,14)
4 -18 10 28 (5,14)
5 -7 12 28 (6,14)
6 6 14 26 (7,13)
7 -5 16 26 (8,13)
8 12 18 24 (9,12)
9 7 20 22 (10,11)
10 6 22 20 (11,10)
0,15 1 2 3 4 5 6 7 8 9 10 11
14
13
12
11
10
(∆d)E=2X+3 (-ve)

Ques. Draw a circle centered at (2,3) and whose radius is
15 units?
Sol:- Using the mid point algorithm firstly we will draw a circle
centered at origin i.e. at (0,0) .
where H=1-r
(∆d)E = 2x+3
(∆d)SE =2x-2y+5
MID POINT ALGORITHM (EXAMPLE)

K h 2x 2y
1. -14 2 30
2. -11 4 30
3. -6 6 30
4. 1 8 28
5. -18 10 28
6. -7 12 28
7. 6 14 26
8. -5 16 26
9. 12 18 24
10. 7 20 22
11. 6 22 20
H=1-r
(∆d)E = 2x+3(-ve)
(∆d)SE =2x-2y+5 (+ve)
if d>= 0 then we choose SE and next point will be (xc+1) and (yc-1).
if d<0 then we choose E and next point will be (xc +1) and yc .

For VIth
octant
For IIIrd
octant
For Vth
octant
For VIth
octant
For VIIth
octant
For VIIIth
octant
For Ist
octant
For IInd
octant
For center at origin(0,0)
IInd(x,y) Ist(x,y) IIIrd(x,y) IVth(x,y) Vth(x,y) VIth (x,y) VIIth(x,y) VIIIth(x,y)
(0,15) (11,10) (0,15) (11,-10) (-15,0) (-10,-11) (0,-15) (-11,10)
(1,15) (12,9) (-1,15) (12,-9) (-15,-1) (-9,-12) (1,-15) (-12,9)
(2,15) (13,8) (-2,15) (13,-8) (-15,-2) (-8,-13) (2,-15) (-13,8)
(3,15) (13,7) (-3,15) (13,-7) (-15,-3) (-7,-13) (3,-15) (-13,7)
(4,14) (14,6) (-4,14) (14,-6) (-14,-4) (-6,-14) (4,-14) (-14,6)
(5,14) (14,5) (-5,14) (14,-5) (-14,-5) (-5,-14) (5,-14) (-14,5)
(6,14) (14,4) (-6,14) (14,-4) (-14,-6) (-4,-14) (6,-14) (-14,4)
(7,13) (15,3) (-7,13) (15,-3) (-13,-7) (-3,-15) (7,-13) (-15,3)
(8,13) (15,2) (-8,13) (15,-2) (-13,-8) (-2,-15) (8,-13) (-15,2)
(9,12) (15,1) (-9,12) (15,-1) (-12,-9) (-1,-15) (9,-12) (-15,1)
(10,11) (15,0) (-10,11) (15,0) (-11,-10) (0,-15) (10,-11) (-15,0)

Ist
octant
2nd
octant
3rd
octant
4th
octant
5th
octant
6th
octant
7th
octant
8th
octant
(18,2) (12,14) (-2,18) (14,-12) (-18,-2) (-12,-
14)
(2,-18) (-14,12)
(18,3) (11,15) (-3,18) (15,-11) (-18,-3) (-11,-
15)
(3,-18) (-15,11)
(18,4) (10,16) (-4,18) (16,-10) (-18,-4) (-10,-
16)
(4,-18) (-16,10)
(18,5) (9,16) (-5,18) (16,-9) (-18,-5) (-9,-16) (5,-18) (-16,9)
(17,6) (8,17) (-6,17) (17,-8) (-17,-6) (-8,-17) (6,-17) (-17,8)
(17,7) (7,17) (-7,17) (17,-7) (-17,-7) (-7,-17) (7,-17) (-17,7)
(17,8) (6,17) (-8,17) (17,-6) (-17,-8) (-6,-17) (8,-17) (-17,6)
(16,9) (5,18) (-9,16) (18,-5) (-16,-9) (-5,-18) (9,-16) (-18,5)
(16,10) (4,18) (-10,16) (18,-4) (-16,-10) (-4,-18) (10,-16) (-18,4)
(15,11) (3,18) (-11,15) (18,-3) (-15,-11) (-3,-18) (11,-15) (-18,3)
As given in the question,circle is centered at (2,3).Therefore we will add 2 in
x and 3 in y.
X=x+2 Y=y+3

1 2 3 4 5 6 7 8 9 10 1112 13 14 15 16 1718 19 20
-1
-2
-3
-4
-5
-6
-7
-8
-9
-10
-11
-12
-13
-14
-15
-16
-17
-18
-19
-20
0
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
-20-19-18-17-16 -15-14-13-12-11-10-9 -8 -7 -6 -5 -4 -3 -2 -1
Center (2,3)

 Is this a good mid point algorithm- NO!!!
 It is costlier and complex.
 Requires atleast 1 multiplication and three
addition per pixel
 Because update functions are linear functions
not constants.
 Solution VERSION 2….

Curves
 Various curve functions are useful in object
modeling, animation path specifications and
other graphic functions.
 A spline is a flexible strip used to produce a
smooth curve through a designated set of points.
 We can easily describe such a curve
mathematically with a piecewise polynomial
function.

 There are several different kinds of spline specifications
that are used in graphics applications.
 Each individual specification simply refers to a particular
type of polynomial with certain specified boundary
conditions.
 Splines are used in graphics applications to design curve
and surface shapes , to digitize drawings for computer
storage and to specify animation path for objects.
 CAD applications for splines include the design of
automobile bodies, aircraft and spacecraft surfaces and
ship structures.

Interpolation and Approximation Splines
A set of six control points interpolated with piecewise
continuous polynomial sections
A set of six control points approximated with piecewise
continuous polynomial sections

 We specify a spline curve by giving a set of co-ordinate
positions called control points, which indicate the general
shape of the curve.
 These control points are then fitted with piecewise
continuous parametric polynomial functions in one of the
two ways:
 Interpolate
 Approximate
 When the polynomial sections are fitted so that the curve
passes through each control point, the resulting curve is
said to interpolate the set of control points

 When the polynomial sections are fitted to the general
control point path without necessarily passing through
any control point, the resulting curve is said to
approximate the set of control points.
 Interpolation curves are commonly used to specify
animation paths.
 Approximation curves are used as design tools to
structure object surfaces.
 A spline curve is defined , modified and manipulated with
operations on the control points.
 By interactively selecting spatial positions for the control
points, a designer can setup an initial curve.

 After the polynomial fit is displayed for a
given set of control points, the designer can
then reposition some or all of the control
points to restructure the shape of the curve.
 In addition curve can be rotated, translated ,
or scaled with transformation applied to the
control points.
 Packages can also insert extra control points
to aid a designer in adjusting the curve
shapes

Polynomial and spline curves
 A polynomial function of nth degree in x is defined as
 When n is a non negative integer and ak are constants and
an!=0.
 We get a quadratic when n=2, a cubic polynomial when n=3 , a
quartic when n=4 and so forth and a straight line when n=1
 So polynomials are useful in number of graphics applications for
design of objects.
n
n
n
n
n
k
k
x
a
x
a
x
a
a
x
a
y k 



 


 1
1
1
0
0
.........

Convex Hull
 The convex polygon boundary that encloses
a set of control points is called the convex
hull.
 Convex hull provide a measure for the
deviation of a curve or surface from the
region bounding the control points
 Will always remain within bounding region
(convex hull) defined by control points

Important properties for designing curves
 Control points: A common way to control the shape of a curve
interactively is to locate points through which the curve must
pass or points that control the curve’s shape in a predictable
way .These points are called control points, or sometimes
knots when they lie on the curve.
 The Bezier curve is predictably related to the locations of the control
points. In bezier curve not all the control points lie on the curve.
 Properties: curve does pass through the two end points(p0 and pn)
 The curve is tangent at the end points to the corresponding edge of the
polygon of the control points(e.g the curve at p0 is tangent to the vector
joining p0 and p1 )
 Multiple Values: In general is not a graph of single-valued
function of a coordinate system.
 The parametric formulation of bezier curve allows it to represent multi-
valued shapes. If the first and last control coincide, the curve is closed

Important properties for designing curves
contd..
 Axis Independence: The shape of an object must not change
when the control points are measured in different co-ordinate
system. If , for example, the control points are rotated 90
degrees , the curve should rotate 90 degrees but not change
shape.
 Bezier curve is independent of the co-ordinate system used
to measure the locations of control points.
 Global or local control: As a designer manipulates a control
point, a curve may change shape only in the region near the
control point, or it may change shape throughout . Means
global control , may be annoying to the designer trying to
make fine adjustments to just one portion of the curve
 Bezier curve doesn’t provide the local control moving any control point
will change shape of every part of the curve because all blending
functions are non zero almost every where (the two values u=0 and u=1
are exceptions ), consequently the location of each control point will
influence the curve location almost everywhere.

 Variation diminishing property always smooth the designer’s
control points.
 Bezier curve are variation diminishing , a curve is guaranteed to lie
within the convex hull of the control points that define it. Thus the bezier
curve doesn’t oscillate wildly away from its defining control points.
 Versatility: More flexible techniques allow the designer to control
the versatility of a curve representation, often by adding or removing
control points .e.g. a curve specified by two control points might be a
straight line connecting the points, introducing the third control point
allows the curve to take on a large number of additional shapes,
depending on the location of control point.

 Order of continuity: A complex shape is usually not modeled by a
single curve , but by several curves piece together end-to-end. Joint
is introduced to increase versatility, a shape that cannot be
described by a single curve can often be described by a several
curves joined together . When creating joints designer often wants to
control the order of continuity at joints.
 Zero order continuity means simply that two curves meet
 First order continuity requires the curve to be tangent at the point of
intersection.
 Second order continuity requires that curvatures be the same.

Bezier Curve
 Bezier curve can be fitted to any no. of control points and control
points are nothing but the co-ordinate positions that are given.
 No. of control points and their relative position determine the degree
of Bezier polynomial and their bezier curve must pass through first
and last control points.
 Given n+1 control points, the parametric equation of the Bezier curve
is:
P(u) = p0BEZ0,n(u) + p1BEZ1,n(u) + … + pnBEZn,n(u)
 Once again, u ranges from 0 to 1. Given a particular value of u, a
particular point on the curve is generated.
 Thus, each point on the curve is a linear combination of the control
points.

 If there are n+1 control points then
 pk = (xk , yk , zk) where 0 ≤ k ≤ n
 And these control points can be blended to produce the
control vector P(u) which describes the path of Bezier
polynomial functions between Po and Pn and the path
vector.
Control point
Point on the curve Coefficient



n
k
n
k
k u
BEZ
p
u
p
0
, )
(
)
(

Bezier Curve
 The Bezier blending functions BEZk,n(u) are
the Bernstein polynomials
BEZk,n(u) = C(n,k)uk(1-u)n-k
where C(n.k) are the binomial coefficients:
n!
C(n,k) =
k!(n-k)!
Where, BEZk,k(u) = uk
BEZ0,k(u) = (1-u)k

Bezier Curve Example
 Example with four control points, p0, p1, p2
and p3 (that is, n=3).
 Then,
P(u) = p0 x 1 x u0 x (1-u)3 + p1 x 3 x u1 x (1-u)2 +
p2 x 3 x u2 x (1-u)1 + p3 x 1 x u3 x (1-u)0
C(3,2) = 3
Control point
BEZ0,3(u)

Blending functions with 4 control points
(n=3) k=0,1,2,3
u
u u
u
BEZ0,3(u)
BEZ2,3(u) BEZ3,3(u)
BEZ1,3(u)
0.2 0.4 0.6 0.8 1
When
K=0
When
K=1
When
K=2
When
k=3
0.4
0.6

Properties of Bezier Curve
 It always passes through the first and last
control points i.e. the path vector .
 p(0) = p0
 p(1) = pn
 It lies with in the convex polygon boundary of
the control points. This follows from the
properties of the Bezier blending functions
that is they all are +ve and their sum is
always equal to 1

 Individual curve co-ordinates are defined as



n
k
n
k
k u
BEZ
x
u
x
0
, )
(
)
(



n
k
n
k
k u
BEZ
y
u
y
0
, )
(
)
(



n
k
n
k
k u
BEZ
z
u
z
0
, )
(
)
(

Example: if the control points are given as
po(1,1,1), p1(2,4,1), p2(6,2,1)
 n=2 ,k=0,1,2
 Individual co-ordinates
 For u=0 x(0)=1×1+0+0=1 y(0)=? Z(0)=?
 For u=1 x(1)=1×0+0×0+1×6=6 , y(1)=? , Z(1)=?
2
2
,
2
2
,
1
2
2
,
0
2
,
2
2
2
,
1
1
2
,
0
0
)
(
)
1
(
2
)
(
)
1
(
)
(
)
(
)
(
)
(
)
(
u
u
BEZ
u
u
u
BEZ
u
u
BEZ
u
BEZ
x
u
BEZ
x
u
BEZ
x
u
x









Question:
Draw bezier curve for the given control
points p0(1,4) ,p1(4,8), p2(6,1), p3(9,6).