Page 87 - DCAP313_LAB_ON_COMPUTER_GRAPHICS

Unit 5: Implementing Ellipse Algorithm

area 2 = ellipse_line_overlap (PHI, A, B, H, K, X 1 , Y 1 , X 2 , Y 2 , &rtn); Notes
sprintf (msg, “Fig 5.3 reverse: area = %15.8f, return_value = %d\n”,
area 2 , rtn);
printf (msg);
sprintf (msg, “sum of ellipse segments = %15.8f\n”, area 1 + area 2 );
printf (msg);
sprintf (msg, “total ellipse area by pi*a*b = %15.8f\n”, pi*A*B);
printf (msg);
return rtn;
}
Determining Intersection Points

To find intersection points, the two ellipse equations are solved simultaneously, using the implicit
polynomial forms. The first ellipse equation, in its translated + rotated position, is written using
the appropriate semi-axis lengths:
x 2 y 2
+ = 1 (11)
A 1 2 B 2 1
In a general form of this problem, the translation + rotation that positions the first ellipse
centered at the origin and oriented with the coordinate axes will typically leave the second ellipse
displaced and rotated, as in Figure 5.4. The implicit polynomial form for a more general ellipse
that is rotated and/or translated away from the origin is written in the conventional way as:
2
2
AA · x + BB · x · y + CC · y + DD · x + EE · y + FF = 0 (12)
Any points of intersection for the two ellipses will satisfy Equation (11) and Equation (12)
simultaneously. An intermediate goal is to find the implicit polynomial coefficients in Equation
(12) that describes the second ellipse after the translation + rotation that positions the first ellipse
centred at the origin and oriented with the coordinate axes. The second ellipse centre (h 2 , k 2 ) is
translated through an amount (–h 1 , –k 1 ), then centre-point is rotated through –φ 1 to give a new
TR
centre point (h 2 , k 2 ):
TR
h = cos(–φ 1 ) · (h 2 – h 1 ) – sin(–φ 1 ) · (k 2 – k 1 ) (13a)
2 TR
k 2 TR = sin(–φ 1 ) · (h 2 – h 1 ) + cos(–φ 1 ) · (k 2 – k 1 ) (13b)
The coordinates for any point (x TR , y TR ) from Equation (12) on the second ellipse in its new
translated + rotated position can be found from the following parametric equations, based on an
ellipse with semi-axis lengths A 2 and B 2 that is centred at the origin, then rotated and translated
to the desired position:

x = A ◊ cos( t ◊)cos(j - j ) - B ◊ sin( t ◊)sin(j - j ) + h 2 ¸
TR Ô
TR 2 2 1 2 2 1 ˝ 0 ≤ t ≤ 2 (14)
t ◊
y TR = A ◊ coos() sin(t ◊ j 2 - j 1 ) - B ◊ sin() cos(j 2 - j 1 ) + k 2 TR ˛ Ô
2
2
To find the implicit polynomial coefficients from the parametric form, transform the locus of
points (x TR , y TR ) so that they lie on the ellipse (A 2 , B 2 , 0, 0, 0), which is accomplished by first
translating through (–(h 1 – h 2 ),(k 1 – k 2 )) and then rotating the point through the angle –(φ 1 – φ 2 ):
x = cos(φ 2 – φ 1 )  (x TR – (h 1 – h 2 )) – sin(φ 2 – φ 1 )  (y TR – (k 1 – k 2 )) (15a)
y = sin(φ 2 – φ 1 )  (x TR – (h 1 – h 2 )) – cos(φ 2 – φ 1 )  (y TR – (k 1 – k 2 )) (15b)
The locus of points (x, y) should satisfy the standard ellipse equation with the appropriate
semi-axis lengths, A 2 and B 2 . Finally, the implicit polynomial coefficients for the second ellipse

LOVELY PROFESSIONAL UNIVERSITY 81