Example is the school of mankind, and they will learn at no other. Edmund Burke ( )

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1 Examples Example is the school of mankind, they will learn at no other. Edmund Burke (79 797) This section, like the previous section, is organised into 9 groups:. Trigonometry. Circles. Triangles.4 Quadrilaterals.5 Polygons.6 Three-dimensional objects.7 Coordinate systems.8 Vectors.9 Quaternions. Transformations. Two-dimensional straight lines. Lines circles. Second degree curves.4 Three-dimensional straight lines.5 Planes.6 Lines, planes spheres.7 Three-dimensional triangles.8 Parametric curves patches.9 Second degree surfaces in stard form The following examples illustrate how geometric formulas are used in practice. Hopefully, the reader will see the advantages of using unit vectors, the difference between using parametric equations the general form of line equations plane equations. There is no one strategy that overall is superior to another much will depend upon the context. 7

2 74 Geometry for computer graphics Vectors Vector notation provides a very compact way of expressing the solution to a geometric problem. For example, the formula for calculating the intersection of a line plane is given by p t lv where ( n i t d ) n i v The position vector p identifies a point P where the line intersects the plane. Therefore, the coordinates of P are given by x p x t lx v y p y t ly v z p z t lz v This sort of coordinate unpacking is used throughout the examples in this section.

3 Examples 75. Trigonometry Examples of cofunction identities sin a cos p a cos b tan a cot p a cot csc a sec p a sec sin cos 6.5 tan 45 tan 45 sin cos 6 Examples of even odd identities sin(a) sin a sin( ) sin.5 cos(a) cos a cos(6 ) cos 6.5 tan(a) tan a tan(45 ) tan 45 Examples of Pythagorean identities sin a cos a tan a sec a cot a csc a sin 4 cos tan 45 cos 45 cot 45 sin 45 4 Examples of compound angle identities sin(a b) sin a cos b cos a sin b sin( ) sin cos cos sin.5 cos(a b) cos a cos b sin a sin b cos( 5 ) cos cos 5 sin sin 5.5 tan a tan b tan( ab) tan a tan b tan tan 5 tan( 5 ) tan tan 5

4 76 Geometry for computer graphics Examples of double-angle identities sin a sin a cos a sin sin 5 cos 5.5 cos a sin a cos 6 sin.5 cos a cos a sin a cos 6 cos sin.5 tan a tan a tan b tan 5. tan 45 tan 5. Examples of multiple-angle identities sin a sin a 4 sin a sin sin 4 sin.5 cos a 4 cos a cos a cos 6 4 cos cos.5 tan a tan a tan a tan a sin 4a 4 sin a cos a 8 sin a cos a sin 4 sin 7.5 cos sin 7.5 cos cos 4a 8 cos 4 a 8 cos a cos 6 8 cos cos 5.5 4tan a 4tan a tan 4a 4 6tan atan a tan5 tan 5 tan 45 tan 5 4tan5 4tan 5 tan tan 5 tan 5 sin 5a 6 sin 5 a sin a 5 sin a sin 6 sin 5 6 sin 6 5 sin 6.5 cos 5a 6 cos 5 a cos a 5 cos a cos 6 6 cos 5 cos 5 cos.5 5 5tan atan atan a tan 5a 4 tan a5tan a 5 5tan 9 tan 9 tan 9 tan 45 tan 9 5tan 4 9 Functions of the half-angle a sin a cos cos a cos a sin cos 6 5. cos 6 cos 5. a tan cos a cos a tan 45 cos 9 cos 9

5 Examples 77 Functions converting to the half-angle tangent form a tan tan5 sin a sin 5. a tan tan 5 tan cos a tan a a tan cos 6 5. tan a tan tan 5. tan a tan 45 a tan tan 5. Relationships between sums of functions ab ab sin asin b sin cos ab ab sin asin b cos sin ab ab cos acos b cos cos ab ab cos acos b sin sin sin( a b) tan atan b cos acos b sin( a b) tan atan b cos acos b sin sin sin cos sin 6 sin cos 45 sin 5.66 cos 6 cos 6 cos 6 cos cos 6 cos sin 45 sin 5.66 sin 9 tan 45 tan 45 cos 45 cos 45 sin 5 tan 6 tan cos 6 cos 45

6 78 Geometry for computer graphics. Circles Example: Properties of circles r u 6 d 4 a c s Circle Area of circle A pr A p.57 Perimeter C pd C p4.57 Length of arc Area of sector Area of segment s a pd 6 u pr 6 ( ) r [ rad] [ rad] a sin a s p p p.46 Length of chord c rsin a c 4 sin 6.46

7 Examples 79. Triangles.. Checking for similar triangles Triangles A B are similar because three corresponding sides are in the same ratio: A 8 7 B Triangles C D are similar because two corresponding sides are in the same ratio, the 6 included angles are equal: the included angles equal. 8 6 C 8 D Triangles E F are similar because two corresponding angles are equal. E 55 F 55.. Checking for congruent triangles Triangles A B are congruent because three corresponding sides are equal. 6 4 A 6 4 B

8 8 Geometry for computer graphics Triangles C D are congruent because two corresponding sides are equal, the included angles are equal. 6 C 6 D Triangles E F are congruent because one side the adjoining angles are equal. E 55 F 55.. Solving the angles sides of a triangle Use the sine rule to find angle a. 6 4 sin a sin 6 a 4 6 sin a sin 4 6 a = sin sin Use the cosine rule to find side a. a 6 6 cos 6 a a cos a 5.7 Use the tangent rule to find side b. a b tan ab ab a b tan a b a b a a 6.87 b 5.

9 Examples 8 b 45 7 tan b tan( 8. ). 485 b 7( b) b 4 Given a b use Mollweide s rule to find side c. a b sin ab c g cos a b a 4 b g a c b 4 sin c cos 55 c.6 Given a b use Newton s rule to find side c. a b cos ab c g sin a b 6 a 4 b g a c b 4 6 cos 5. 6 c sin 55 c Calculating the area of a triangle Use Heron s formula to calculate the area of a triangle. a 8 b c Semiperimeter s 8 Area ss ( a)( sb)( sc) b a Area ( )( 8) c

10 8 Geometry for computer graphics Use a determinant to calculate the area of a triangle. C(, ) Area ABC Area Reversing the vertex order: x x x A B C y y y A B C ( 9) Area ( 9) A(, ) B(, )..5 The center radius of the inscribed circumscribed circles for a triangle Calculate the center of the inscribed circle for triangle ABC. a 8 b c A (, ) B (, ) C (, ) C x y M M ax bx cx abc A B C ay by cy abc A B C b r (x M, y M ) a x M A c B 8 4 y M Position of the center x M.5858 y M.5858 Calculate the radius of the inscribed circle for triangle ABC. s r 8 ( sa)( sb)( sc) s

11 Examples 8 r ( 8)( )( ) r r.5858 x M.5858 y M.5858 Calculate the radius of the circumscribed circle for triangle ABC. C R b a (x P, y P ) A c B a 8 b c A (, ) B (, ) C (, ) abc R 4 Area ABC 8 4 Calculate the center of the circumscribed circle for triangle ABC. x y P P x A y A R abc R abc y y b c AC AB x x b c AC AB x P y P R x p y P

12 84 Geometry for computer graphics.4 Quadrilaterals Example: Calculate the area of a quadrilateral. a b c d AC d 4 BD d 8 a b c d s 5.54 By inspection A a a B d O 45 d d u b D c b C ABO BCO CDO DAO therefore Area ABCD 6. Here are four ways of computing the area: 4 8 Area dd sin u sin Area ( b d a c )tan u ( 4)tan Area 4 dd ( b d a c) ( 4) Area ( sa)( sb)( sc)( sd) abcdcos e e Area cos 9 6 It just so happens that the quadrilateral is a cyclic quadrilateral.

13 Examples 85 Example: Calculate the center radius of the circumscribed circle for a rectangle. C(, 4) D(, ) P A(, ) R B(, ) P A (, ) P B (, ) P C (, 4) P D (, ) The center of the circumscribed circle is x y y y P ( x A x C ) P ( A C ) x P ( ) = 5. y P ( 4) =.5 The radius of the circumscribed circle is R ( x x ) ( y y ) ( x x ) ( y y ) R ( ) ( ) ( ) ( 4) B A B A B C B C The circle has a radius of with a center at (.5,.5).

14 86 Geometry for computer graphics.5 Polygons Example: Determine the internal angles of a polygon The internal angles of an n-sided polygon sum to (n ) 8. Triangle (n ) Quadrilateral (n 4) a a a a a i i= 8 a 4 i i= 6 a4 Pentagon (n 5) Hexagon (n 6) a a a a a a 5 a 4 a a 6 a 5 a 4 5 a i i 54 6 a i i 7 Example: Determine the alternate internal angles of a cyclic polygon The alternate internal angles of an n-sided cyclic polygon sum to (n ) 9 [n 4 is even]. Cyclic quadrilateral (n 4) Cyclic hexagon (n 6) a a a 6 a 5 a 4 a a 4 a a a a a a a 4 8 a a a 5 a a 4 a 6 6

15 Examples 87 Example: Calculate the area of regular polygon Area ns 4 where n number of sides s length of side Let s n Area cot p n Example: Calculate the area of a polygon The figure shows a polygon with the following vertices in counter-clockwise sequence x 5 5 y By inspection, the area is.5 The area of a polygon is given by 4 5 Area i Area ( 5555) Area ( 6 5).5 n ( xy yx ) i i(mod n) i i(mod n)

16 88 Geometry for computer graphics.6 Three-dimensional objects.6. Cone, cylinder sphere Example: Area volume of a cone, cylinder sphere Area ( h r) ( s 5 r) (r ) Cone prr ( s) ( 5) pr ( 5) p Sphere 4pr 4p Cylinder prr ( h) 6pr 6p r Volume Cone Sphere 4 pr prh pr p 4 p Cylinder pr h pr p s = 5r h = r.6. Conical frustum, spherical segment torus Example: Area volume of a conical frustum, spherical segment torus Circular, conical frustum If r r h s Sp( r r s( r r )) S p(4 ( )) 9. h r s r V p h ( r r rr ) V p(4 ) 7. Spherical segment S prh If r h S V p h( r r h ) If r r h V.9 (half the volume) r r h r

17 Examples 89 Torus S 4p rr If r R S 9.48 V p r R If r R V 9.74 r R.6. Tetrahedron Example: Volume of a tetrahedron Tetrahedron Let A (,, ) B (,, ) C (,, ) x y z V a a a x y z b b b 6 6 x y z c c c Note: If the vertices are reversed the volume is negative. x y z b b b V x y z 6 a a a 6 x y z c c c 6 6 B O C A

18 9 Geometry for computer graphics.7 Coordinate systems.7. Cartesian coordinates in Example: Distance in Find the distance between the points (, 6) (9, ). Given therefore d ( 9 ) (6 ) 9 6 d 5 d ( x x ) ( y y ).7. Cartesian coordinates in Example: Distance in Find the distance between the points (, 6, ) (9,, ). Given d ( x x ) ( y y ) ( z z ) d ( 9) ( 6 ) ( ) therefore d Polar coordinates Example: Conversion between Cartesian polar coordinates Find the polar coordinates (r, u) for the points (4, ), (4, ), (4, ) (4, ). Given r x y u tan y (st 4th quadrants only) x For (4, ) r u tan (4, ) (5, 6.87 ) For (4, ) r 5 u (4, ) (5, 4. )

19 Examples 9 For (4, ) r 5 u (4, ) (5, 6.87 ) For (4, ) r 5 u 6.87 or. (4, ) (5,. ) Find the Cartesian coordinates (x, y) for the point (5, 6.87 ). Given x r cos y r sin u For (5, 6.87 ) x 5 cos y 5 sin 6.87 (5, 6.87 ) (4, ).7.4 Cylindrical coordinates Example: Conversion between Cartesian cylindrical coordinates Find the cylindrical coordinates (r, u, z) for the points (4,, 4), (4,, 4), (4,, 4) (4,, 4). Given r x y u tan y x z z (st 4th quadrants only) For (4,, 4) r u tan z 4 (4,, 4) (5, 6.87, 4) For (4,, 4) r 5 u z 4 (4,, 4) (5, 4., 4) For (4,, 4) r 5 u z 4 (4,, 4) (5, 6.87, 4)

20 9 Geometry for computer graphics For (4,, 4) r 5 u 6.87 or. z 4 (4,, 4) (5, 6.87, 4) Find the Cartesian coordinates (x, y, z) for the point (5, 6.87, 4). Given x r cos u y r sin u z z For (5, 6.87, 4) x 5 cos y 5 sin 6.87 z 4 (5, 6.87, 4) (4,, 4).7.5 Spherical coordinates Example: Conversion between Cartesian spherical coordinates Find the spherical coordinates (r, u, f) for the points (4,, 4), (4,, 4), (4,, 4) (4,, 4). Given r x y z u tan y x (st 4th quadrants only) f cos z x y z For (4,, 4) r u tan f cos (4,, 4) (6.4, 6.87, 5.4 ) For (4,, 4) r 6.4 u f 5.4 (4,, 4) (6.4, 4., 5.4 )

21 Examples 9 For (4,, 4) r 6.4 u f 5.4 (4,, 4) (6.4, 6.87, 5.4 ) For (4,, 4) r 6.4 u tan f 5.4 (4,, 4) (6.4,., 5.4 )

22 94 Geometry for computer graphics.8 Vectors.8. Vector between two points Given P (,, ) P (4, 6, 8) x x PP a y y 4 z z 5 a i 4j 5k.8. Scaling a vector Given scale by a i 4j 5k a 9i j 5k.8. Reversing a vector Given a i 4j 5k a i 4j 5k.8.4 Magnitude of a vector Given a i 4j 5k a Normalizing a vector to a unit length Given check a i 4j 5k 4 5 aˆ i j k. 44i. 566j. 77k aˆ Vector addition/subtraction Given a i 4j 5k b i 4j 6k a b 5i 8j k

23 Examples Position vector Given a point (, 4, 5) its position vector is i 4j 5k..8.8 Scalar (dot) product Given a i 4j 5k b i 4j 6k a b Angle between two vectors Given a i 4j 5k b i 4j 6k Let a be the angle between a b. a b a cos xx yy zz a b a b a b a b a cos cos Vector (cross) product Given a i j 5k b i j 8k i j k ab 5 i9jk 8 Remember that Proof i 9j k is orthogonal to a b. a b b a i j k ba 8 i9jk 5 i 9j k is still orthogonal to a b but is in the opposite direction to i 9j k.

24 96 Geometry for computer graphics.8. Scalar triple product Given a j k b k c 5i ai( bc) x y z x y z x y z a a a b b b c c c Volume ai ( bc) 5.8. Vector normal to a triangle Given P (5,, ) P (,, 5) P (,, 5) x x a y y z z a 5i 5k x x b y y z z b 5i 5k i j k nab 5 5 5j 5 5 Surface normal n 5j.8. Area of a triangle Given P (5,, ) P (,, 5) P (,, 5) x x x x a y y b y y z z z z a 5i 5k b 5i 5k i j k Area ab 5 5 5j 5 5 Area 5

25 Examples 97.9 Quaternions.9. Quaternion addition subtraction q q [(s s ) (x x )i (y y )j (z z )k] Given q [ i j 4k] q [ i j 5k] then q q [ i 5j 9k].9. Quaternion multiplication q q [(s s v v ), s v s v v v ] Given q [ i] q [ j] then q q [ i j k].9. Magnitude of a quaternion s x y z q Given q [ i j 4k] then 4 q.9.4 The inverse quaternion Given q [ i j 4k] q [ sxi yjzk ] q then q [ ij4k] [ i j k] Rotating a vector Rotate p using pqpq where q [cos( ),sin( ) v u u ˆ] Let p be the quaternion for (,, ) i.e. p [ i]

26 98 Geometry for computer graphics Let q be a unit quaternion aligned with the z-axis which rotates p 8 i.e. q [cos 9, sin 9 (k)] [ k] then q [ k] but q therefore p[ k] [ i] [ k] [ j] [ k] [ i] [ i] points to the rotated point: (,, ), which is correct..9.6 Quaternion as a matrix Let s express the previous rotation quaternion as a matrix: Given [ k] then s, x, y, z therefore then R(u) which confirms the previous result. s x y z ( xy sz) ( xy sz) R( u) ( xy sz) s y x z ( yzsx) ( xz sy) ( yz sx) s z x y

27 Examples 99. Transformations In the following examples the coordinates of the original shape A are shown on the righth side of the transform enclosed in brackets, whilst the coordinates of the transformed shape A are shown on the left-h side... Scaling relative to the origin in Scale shape A by a factor of in the x-direction in the y-direction relative to the origin. x S x x y S y y A Transform A 4 4 A A 4.. Scaling relative to a point in Scale shape A by a factor of in the x-direction in the y-direction relative to the point (, ). x S x ( S ) x P x x y S y ( S ) y P y y A Transform A A A.. Translation in Translate shape A by in the x-direction in the y-direction. x T x x y T y y A Transform A A A 4

28 Geometry for computer graphics..4 Rotation about the origin in Rotate shape A 9 about the origin. x cos a sin a x y sin a cos a y A Transform A A A..5 Rotation about a point in Rotate shape A 9 about the point (, ). x cos a sin a x ( cos a) y sin a P P x y sin a cos a y ( cos a) x sin a P P y A Transform A A A..6 Shearing along the x-axis in Shear shape A 45 along the x-axis. x tan a x y y A Transform A 4 A A 4

29 Examples..7 Shearing along the y-axis in Shear shape A 45 along the y-axis. x x y tan a y A Transform A A A 4..8 Reflection about the x-axis in Reflect shape A about the x-axis. x x y y A Transform A A A..9 Reflection about the y-axis in Reflect shape A about the y-axis. x x y y A Transform A A A

30 Geometry for computer graphics.. Reflection about a line parallel with the x-axis in Reflect shape A about the line y p. x x y y P y A Transform A y P A A 4.. Reflection about a line parallel with the y-axis in Reflect shape A about the line x p. x x P x y y A A A Transform A x P 4.. Translated change of axes in The axes are subjected to a translation of (, ). x x T x y y T y A Transform A A A 4

31 Examples.. Rotated change of axes in Rotate the axes 9. x cos a sin a x y sin a cos a y A Transform A A A..4 The identity matrix in x x y y A Transform A = A A 4..5 Scaling relative to the origin in Scale shape A.5 in the x-direction, in the y-direction in the z-direction. x S x x yz S y yz S z A Transform A 5. 4 A A

32 4 Geometry for computer graphics..6 Scaling relative to a point in Scale shape A.5 in the x-direction, in the y-direction in the z-direction relative to the point (,, ). x S x ( S ) x P x x yz S y ( S ) y P y yz S z z ( P S ) z A Transform A 5. A A..7 Translation in Translate shape A by (,, ). x T x x yz T y yz T z A Transform A A A..8 Rotation about the x-axis in Rotate shape A about the x-axis 9. x x yz cos a sin a sin a cos a yz A Transform A A A

33 Examples 5..9 Rotation about the y-axis in Rotate shape A about the y-axis 9. x cos a sin a x yz sin a cos a yz A Transform A A A.. Rotation about the z-axis in Rotate shape A about the z-axis 9. x cos a sin a x yz sin a cos a yz A Transform A A A.. Rotation about an arbitrary axis in x a K cos a abk c sin a ack b sin a x yz abk c sin a b K cos a bck a sin a yz ack b sin a bck asin a c K cos a K cos a Axis v ai bj ck v Given v k a 9 then K A Transform A A A

34 6 Geometry for computer graphics.. Reflection about the yz-plane in Reflect shape A in the yz-plane. x x yz yz A Transform A A A.. Reflection about the zx-plane in Reflect shape A in the zx-plane. x x yz yz A Transform A A A..4 Reflection about the xy-plane in Reflect shape A in the xy-plane. x x yz yz A Transform A A A

35 Examples 7..5 Reflection about a plane parallel with the yz-plane in Reflect shape A in the yz-plane x p. x x P x yz yz A Transform A A A x P..6 Reflection about a plane parallel with the zx-plane in Reflect shape A in the zx-plane y p. x x yz y P yz A Transform A 4 4 y P A A..7 Reflection about a plane parallel with the xy-plane in Reflect shape A in the xy-plane z p. x x yz zp yz A Transform A z P A A

36 8 Geometry for computer graphics..8 Translated axes in The axes are subjected to a translation of (,, ). x x T x yz y T z yz T A Transform A A A..9 Rotated axes in The axes are subjected to a rotation as illustrated. x r r r x yz r r r yz r r r A Transform A A A.. The identity matrix in A Transform A A A

37 Examples 9. Two-dimensional straight lines.. Convert the normal form of the line equation to its general form the Hessian normal form Given the normal form of the line equation 5 y x 4 4 The general form of the line equation is obtained by rearranging the equation to x 4y 5 The Hessian normal form is obtained be dividing throughout by the magnitude of the line s normal vector: x4y5 4 x y The line intersects the x-axis at x the y-axis at y. The unit normal vector to the 4 line nˆ = 6. i 8. j the perpendicular from the origin to the line is... Derive the unit normal vector perpendicular from the origin to the line for the line equation x 4y 6 The normal vector is n i 4j The unit normal vector is nˆ ( i 4j) 4 n d 6. i8. j The distance is d c

38 Geometry for computer graphics.. Derive the straight-line equation from two points Normal form of the line equation Given P (x, y ) P (x, y ) then y mx c m y y x x P y c y x x y x P If the two points are P (, ) P (, 4) then 4 y x 4 y x General form of the line equation Given P l (x, y ) P (x, y ) Ax By C then A y y B x x C (x y x y ) If the two points are P (, ) P (, 4) then (4 )x ( )y ( 4 ) 4x y 4 or x y Determinant form of the line equation Given y y x x x y x x y y If the two points are P (, ) P (, 4) then 4 x y 4 4x y 4 or x y

39 Examples Hessian normal form of the line equation Given 4x y 4 The normalizing factor is a b ˆn then 4 4 x y x y The normal unit vector to the line is nˆ ( i j) 5 The perpendicular from the origin to the line 5 Parametric form of the line equation Given P (x, y ) P (x, y ) p p lv v p p If the two points are P (, ) P (, 4) v i 4j Therefore x l y 4l For example, when l x y when l.5 x y..4 Point of intersection of two straight lines General form of the line equation Given a x b y c a x b y c They intersect at x P cbcb ab ab y P ac ab ac ab P

40 Geometry for computer graphics Let the straight lines be x y 4 x 4y 4 Therefore 44 x P y P 4 4 The point of intersection is (, ) as confirmed by the diagram. Parametric form of the line equation Given p r la q s b where r x R i y R j s x S i y S j a x a i y a j b x b i y b j x ( y y ) y ( x x ) then b S R b S R l x y x y b a a b Point of intersection x P x R lx a y P y R ly a Given r j a i j s j b i j s r P b a ( ) ( ) l ( ) ( ) x P y P () The point of intersection is (, ) as confirmed by the diagram...5 Calculate the angle between two straight lines General form of the line equation Given a x b y c a x b y c where n a i b j m a i b j Angle n i m cos n m Let the line equations be x y 4 x 4y 4 Therefore a cos a

41 Examples Normal form of the line equation Given y m x c y m x c Angle Let the line equations be y x where Therefore a m m mm cos m m x y a cos ( )( ) ( ) ( ) Parametric form of the line equation Given p r la q s b Angle a i b a cos a b Let the line equations be p r la q s b where r j a i j sj b i j Therefore ( )( ) a cos Test if three points lie on a straight line Given P (x, y ), P (x, y ) P (x, y ) r PP s PP P The three points lie on a straight line when s lr. Let the points be P (, ) P (, ) P (4, ) P P Therefore r i j s 4i 4j s 4r Therefore the points lie on a straight line as confirmed by the diagram.

42 4 Geometry for computer graphics..7 Test for parallel perpendicular lines General form of the line equation Given a x b y c a x b y c where n a i b j m a i b j The lines are parallel if n lm. The lines are mutually perpendicular if n m. Given three lines L : x y L : x y L : x y L L are parallel because the normal vectors to the lines are n i j n i j n ln (l ) L L are perpendicular because n m () L L L Normal form of the line equation Given y m x c y m x c The lines are parallel if m m. The lines are mutually perpendicular if m m Given three lines L : y x L : y x L : y x L L are parallel because m m L L are perpendicular because m m () Parametric form of the line equation Given p r la q s b where a x a i y a j b x b i y b j The lines are parallel if a kb. The lines are mutually perpendicular if a b. Given three lines p r la q s b u t bc

43 Examples 5 where L : a i j L : b i j L : c i j L L are parallel because a b L L are perpendicular because x a x c y a y c ()..8 Find the position distance of the nearest point on a line to the origin General form of the line equation Given ax by c where n ai bj q ln where l c n i n q n Q Distance OQ q ln O Given the line equation x y where a b c Therefore l The nearest point is x lx y ly Q n Q n (, ) Q Distance OQ n. 77 Parametric form of the line equation Given q t lv where l v i t v i v O T t q Q v

44 6 Geometry for computer graphics Distance OQ q Given the direction vectors t j v i j l x x lx Q T v The nearest point is Distance y y ly ( ) Q T v (, ) Q OQ tlv i j Find the position distance of the nearest point on a line to a point General form of the line equation Given ax by c n P where n ai bj q p ln Q where n i pc l n i n Distance PQ ln Given P(, ) x y then a b c l Therefore The nearest point is x x lx Q P n y y ly Q P n (, ) Q Distance PQ ln i j. 77

45 Examples 7 Parametric form of the line equation Given where Distance q t lv v i ( p t) l v i v PQ ptlv T t q Q v P Given the direction vectors t j v i j p i j l x x lx Q T v The nearest point is Distance y y ly Q T v (, ) Q PQ ptv i j Find the reflection of a point in a line passing through the origin General form of the line equation Given ax by c where n ai bj q p ln P ( n i pc) l n i n Q q n Given the line equation x y where a b c P(, ) Therefore l x Q x P lx n y Q y P ly n The reflection point is Q(, )

46 8 Geometry for computer graphics Parametric form of the line equation Given where s t lv q t v p v i ( p t) e v i v Q q v P T Given x P y P t v i j e ( ) Therefore x Q x T x v x P y Q y T y v y P () The reflection point is Q(, ).. Find the reflection of a point in a line General form of the line equation Given ax by c where n ai bj P q p ln ( n i pc) l n i n Q Given the line equation x y where a b c x P y P l ( ) Therefore x Q x P lx n y Q y P ly n The reflection point is Q(, ) Parametric form of the line equation T P Given s t lv q t v p t v where v i ( p t) e v i v Q

47 Examples 9 Given x P y P t j v i j e Therefore x Q x T x v x P y Q y T y v y P () The reflection point is Q(, ).. Find the normal to a line through a point General form of the line equation If line m is ax by c line n is perpendicular to m passing through the point P (x P, y P ) n P The line equation for n is bx ay bx P ay P m Given m is x y then a b x P y P Line n is x y Parametric form of the line equation Given line m where q t lv a point P u p (t lv) v i ( p t) l v i v t v q n P u p m line n is p u where is a scalar. Given v i j p i j t j ( i j) i i l ( i j) ( i j) i u( i j) ( j ( i j)) i j Line n is n( i j) e( i j) ( e) i( e) j where is a scalar, which is equivalent to x y.

48 Geometry for computer graphics.. Find the line equidistant from two points General form of the line equation P m P n P Given ax by c Line n is x y Line m is given by ( x x ) x( y y ) y ( x x y y ) with P (, ) P (, ) Line m is ( ) x( ) y ( ) x y Parametric form of the line equation P P v q Q p u P Given where q p lv q ( ( x x ) l( y y )) i( ( y y ) l( x x ) j with P (, ) P (, ) Therefore q ( ( ) l( )) i( ( ) l( ) j q ( l) i( l) j ( ) e.g. when l we have P, when l the point is Q(, ) This is equivalent to y x or x y

49 Examples..4 Creating the parametric line equation for a line segment P q p v P P P (x, y ) P (x, y ) delimit the line segment the parametric line equation is given by p q l v where q x i y j v (x x )i (y y )j Therefore x P x l(x x ) y P y l(y y ) Given P (, ) P (, ). P is between P P for l [, ] i.e. x P l( ) l y P + l( ) l For example, when l.5 x y Intersecting two line segments P P 4 r a P i P P s b Given two line segments with equations r la s b where a x a i y a j b x b i y b j

50 Geometry for computer graphics The point of intersection is x i x r lx a y i y r ly a x ( y y ) y ( x x ) b b where l x y x y Let the two line segments be PP PP with P (, ), P (, ), P (, ), P 4 (, ) Therefore r i j a i j s i b i j 4 b a a b Therefore ( ) ( ) l ( ) As l there is a point of intersection x i y i ( ) The point of intersection is (, ), which is correct.

51 Examples. Lines circles.. Line intersecting a circle General form of the line equation L L L C r The diagram shows a circle radius r centered at C(x C, y C ) (, ) three lines: L, L L that miss, touch intersect the circle respectively. The line equation is ax by c Point(s) of intersection x x ac c ( a ) b r () C T T C T T y y bc c ( b ) a r where c T ax C by C c Miss condition Line L is x y () L normalized is where x y a b c then c T T c ( b a r ) ( ) The negative discriminant confirms the non-intersection.

52 4 Geometry for computer graphics Touch condition Line L is y (which is already normalized) () therefore a b c c T c T (b ) a r ( ) The zero discriminant confirms the touch condition: using () x () y Therefore the touching point is (, ) which is correct. Intersect condition Line L is x y (4) L normalized is x y where a b c c T T 4 c ( b ) a r The positive discriminant confirms the intersect condition. Using () x 4 (4) y The intersection points are (, ) (, ) which are correct. Parametric form of the line equation L T λv P T λv λv C L r L T

53 Examples 5 The diagram shows a circle radius r centered at C(x C, y C ) (, ) three lines: L, L L that miss, touch, intersect the circle respectively. The lines are p t lv p t lv p t lv where t j v i j t j v i t v i j c i j Let us substitute the lines into the following equations: Point(s) of intersection x p x T lx v y P y T ly v where L : l s v ( s v) s r s c t s i (s v) s r 4 The negative discriminant confirms a miss condition. L : s i j (s v) s r 4 5 The zero discriminant confirms a touch condition. Therefore l The touch point is x P y P which is correct. L : s i j ( s i v) s r The positive discriminant confirms an intersect condition. Therefore l The intersection points are l x P

54 6 Geometry for computer graphics y P l x P y P The intersection points are (, ) (, ) which are correct... Touching intersecting circles Touching circles The diagram shows two circles touching one another at a point P(x P, y P ). r r C P C One circle with radius r is centered at C (, ), the other with radius r.5 is centered at C (.5, ). Given C C C C d ( x x ) ( y y ) The touch condition is d r r The touch point is x r x ( d x x ) P C C C y r y ( d y y ) P C C C then d (. 5) ( ) 5. The touch condition is satisfied. x P (. 5 5 ). y P ( 5. ) Therefore the touch point is P(, ) which is correct.

55 Examples 7 Intersecting circles The diagram shows two circles intersecting one another at points P (x P, y P ) P (x P, y P ). One circle with radius r is centered at C (, ), the other with radius r is centered at C (.5, ). r P r The intersect condition d r r The points of intersection are x P x C lx d y d y P y C ly d x d x P x C lx d y d y P y C ly d x d C C P where l r r d d e d r d Therefore the intersect condition is satisfied e therefore x P 7 y 4 P x P 7 ( 4 4 ) 7 y 4 P 6 ( 4 ) 4 7 The intersection points are,, which are correct. 7 4

56 8 Geometry for computer graphics. Second degree curves.. Circle General equation Center (x C, y C ) (x x C ) (y y C ) r Given a radius r center (, ) then (x ) (y ) 4.. Ellipse General equation Center origin x a y b with a, b then x 4 y.. Parabola Parametric equation Vertex origin x t y t t [ 55, ] t 4 5 x y

57 Examples 9..4 Hyperbola General equation x a y b y 4 x Foci at (c,) c a b (5, ) (5, ) x y then with c y 4 x

58 Geometry for computer graphics.4 Three-dimensional straight lines.4. Derive the straight-line equation from two points Given P P v p p p p lv Given P (,, ) P (,, ) P j k P P p P p v v i j k p p p lv.4. Intersection of two straight lines Given two lines p t la q s b where t x t i y t j z t k s x s i y s j z s k a x a i y a j z a k b x b i y b j z b k Step : If a b the lines are parallel do not intersect. Step : If (t s) (a b) the lines do not touch. Step : Solving lx a x b x s x t ly a y b y s y t lz a z b z s z t T t b a s S provides values for l which, when substituted in the original line equations, reveal the intersection point. Given t j k s i j a i j k b i j k Step : Prove that the lines are not parallel. Although it is obvious that a b are not parallel, let s prove it by ensuring that a b. i j k a b a b Therefore the lines are not parallel. Step : Prove that the lines are touching. If (t s) (a b) the lines touch. Therefore (i k) (5i 5j 5k) so the lines touch.

59 Examples Step : Compute the intersection point. Create the three equations: l () l () l () From () l Substituting l in () l e 5 Substitute l in the original line equations p( j k) ( i jk) i j k The intersection point is (,, ) Calculate the angle between two straight lines Given p r la q s b angle a cos Given a i j k b i j ( i jk) i ( i j) a cos 6 cos a i b a b R r s S b a a.4.4 Test if three points lie on a straight line Given three points P, P, P. Let r = PP s = PP The points lie on a straight line when s lr where l is a scalar. P r P s P Given P (,, ) P (,, ) P (,, ) therefore r i k s i k s r Therefore the points lie on a straight line.

60 Geometry for computer graphics.4.5 Test for parallel perpendicular straight lines Given p r ma q s b The lines are parallel if a lb where l is a scalar. The lines are perpendicular if a b. Given three lines L : p ma L : p b L : p lc where a i k b i k c j L L are parallel because a b. L L are perpendicular because a c (i k) (j). L L P P b a c P L.4.6 Find the position distance of the nearest point on a line to the origin Given p t lv where l v i t v i v Distance OP p Given t j k v i k l (i k) i ( j k) 9 ( i k) i ( i k) 8 T t O P p v therefore x x lx P T v y y ly P T v z z lz ( ) P T v Distance OP p i j k Find the position distance of the nearest point on a line to a point Given where Distance Given q t lv l v i ( p t ) v i v PQ p (t lv) t j k v i j k p i j T t v Q q p P

61 Examples ( i j k) i (i k) 8 l ( i jk) i (i j k) 9 then 8 x x lx.84 Q T v 9 8 y y ly.947 Q T v 8 z z lz ( ).579 Q T v 9 9 Distance PQ ( i j) ( ( jk) 8 ( i j k) ) Find the reflection of a point in a line Given where Given s t lv a point P with reflection Q q t v p v i ( p t) e v i v t j k v i j k p i j ( i jk) i ( ik) e ( i j k) i ( i j k) T t v Q q p P then x x ex x.4545 Q T v P y y ey y.88 Q T v P z z ez z ( ).88 Q T v P The reflection point is Q(.45,.8,.8).4.9 Find the normal to a line through a point Given the normal is where Given therefore q t lv u p (t lv) v i ( p t) l v i v t j k v i j k p i j ( i jk) i ( ik) l ( ) ( ) i j k i i j k T t v q p Q u P

62 4 Geometry for computer graphics x x ( x lx ) ( ).77 u P T v y y ( y y ) ( ).99 u P T v z z ( z lz ) ( ( )). 99 u P T v therefore u.7i.99j.9k The line equation for the normal is n p u.4. Find the shortest distance between two skew lines Given Shortest distance Given Calculate v v p q tv p q tv ( qq) i ( vv) d v v q j k q k v i j k v k Q Q q q v v i j k v v v v ji ( i j) d i j

63 Examples 5.5 Planes.5. Cartesian form of the plane equation Given where the normal is ax by cz d n ai bj ck p x i y j z k n P d n p If the normal is n j k the point P (,, ) then x y z The plane equation is y z O.5. General form of the plane equation n P O Given ax by cz (ax by cz ) where the normal is n ai bj ck a point is p x i y j z k If the normal is n j k the point P (,, ) then x y z ( ) The plane equation is y z.5. Hessian normal form of the plane equation To convert the previous equation into Hessian normal form, rearrange the formula divide throughout by n. Given y z where the normal is n j k n

64 6 Geometry for computer graphics therefore or y z y z.5.4 Parametric form of the plane equation Given vectors a b that are parallel to the plane point T is on the plane where c la b p t c then x P x T lx a x b y P y T ly a y b z P z T lz a z b b t p T P λa The plane is parallel with the xz-plane intersects the y-axis at y. Let a b be unit vectors parallel with the plane i.e. a i b k T(,, ) is a point on the plane therefore t i j k p t la b As a b are unit vectors, l measure Euclidean distances. Therefore if l x P y P z P.5.5 Converting a plane equation from parametric form to general form Given p t la b for P to be perpendicular to O ( aib)( bit) ( ait) b l a b ( ai b) ( aib)( ait) ( bit) a e a b ( ai b) λa P p O t b

65 Examples 7 then x y z P x P P y z p p p p We know in advance that the general equation of this plane is intersects the y-axis z-axis at y z respectively. The vectors for the parametric equation are a j k b i t k therefore y z ()() ( ) l () therefore ()( ) () e () x P y P z P ( ) p The plane equation is x y z or y z y z.5.6 Plane equation from three points T R S

66 8 Geometry for computer graphics Given three points R(x R, y R,z R ), S(x S, y S, z S ), T(x T, y T, z T,) the plane equation is ax by cz d where y z z x R R R R a y z b z x c S S S S y z z x T T T T x x x R S T y y y R S T d( ax by cz ) R R R If the three points are R(,, ), S(,, ), T(,, ) a b c d() then the plane equation is x y z.5.7 Plane through a point normal to a line n Q Given n ai bj ck Q(x Q, y Q, z Q ) the plane equation is ax by cz (ax Q by Q cz Q ) If the line is n i j k Q(,, ) the plane is x y z.5.8 Plane through two points parallel to a line M b a N Given a line s direction vector a two points M(x M, y M, z M ) N(x N, y N, z N ) where a x a i y a j z a k

67 Examples 9 b (x N x M )i (y N y M )j (z N z M )k the plane equation is ax by cz (ax M by M cz M ) where a y a z b y b z a b z a x b z b x a c x a y b x b y a Given M (,, ) N (,, ) a i j therefore a b n ai bj ck i j k x y z ( ) The plane equation is x y z or x y z.5.9 Intersection of two planes Given two planes a x b y c z d a x b y c z d where n a i b j c k n a i b j c k The direction vector of the intersection line is given by n n n the point P on the intersection line is given by DET y a b c a b c a b c d a c d a c a c a c DET d b c d b c b c b c x DET z d a b d a b a b a b DET Example Let the two intersecting planes be the xy-plane the xz-plane, which means that the line of intersection will be the y-axis. P n n n P The plane equations are z x where n k n i d d i j k n j

68 4 Geometry for computer graphics Therefore DET x y z therefore the line p ln equation is where n j Example Let the two intersecting planes be the xy-plane the plane x, which means that the line of intersection will be parallel with the y-axis passing through the point (,, ) P n n P n The plane equations are z x where n k n i d d i j k n j DET x y z Therefore the line equation is p p ln where p i n j

69 Examples 4 Example Let the two intersecting planes be x y x y. n n P n P Therefore n i j n i j d d i j k n k DET 4 x 4 y 4 z 4 Therefore the line equation is p p ln where p i j n k.5. Intersection of three planes Given three planes a x b y c z d a x b y c z d a x b y c z d the intersection point (x, y, z) is x d b c d b c d b c DET y a d c a d c a d c DET z a b d a b d a b d DET where DET a b c a b c a b c

70 4 Geometry for computer graphics Example i k j Given the planes x y z which are the three orthogonal planes intersecting at the origin. DET x y z The intersection point is the origin, which is correct. Example i j k i j k Given the planes x y z z y DET x y The intersection point is (,, ) which is correct. z

71 Examples 4.5. Angle between two planes a n n Given two planes a x b y c z d a x b y c z d where n a i b j c k n a i b j c k the angle between the normals is cos Given the planes x y z z where n i j k n k n n n n n n cos Angle between a line a plane Given the plane ax by cz d where n ai bj ck the line p r la n a the angle between the line the plane s normal is cos n a n a Given the plane x y z then n i j k a i j n a cos 5.6 6

72 44 Geometry for computer graphics.5. Intersection of a line a plane Given a plane ax by cz d where n ai bj ck a line p t lv for the intersection point P l ( n t d ) n v n T v P(x, y, z) Example Given the plane x y z the line p t lv where t v i j then The point of intersection is ( ) P l (),,. Example With the same plane x y z but t i j k v i j k l () The point of intersection is P p t lv ( ),,..5.4 Position distance of the nearest point on a plane to a point P Given the plane ax by cz d where n ai bj ck Q a point P with position vector p.

73 Examples 45 The position vector of the nearest point Q is given by q p ln where l ( n p d ) n n The distance PQ is PQ ln Given the plane x y where n i j a point P(,, ) where p i j l () The nearest point is Q(,, ) the origin. The distance is ( ) PQ i j.5.5 Reflection of a point in a plane Given the plane ax by cz d where n ai bj ck P is a point with position vector p P Q P s reflection Q is given by q p ln where l ( n p d ) n n Given the plane x y P(,, ) n i j ( ) l The reflection point is (,, )..5.6 Plane equidistant from two points P P

74 46 Geometry for computer graphics Given two points P (x, y, z ) P (x, y, z ) the plane equation is Given P (,, ) P (,, ) the plane equation is xy ( 44) or x y ( ( x x ) x( y y ) y( z z ) z x x y y z z ).5.7 Reflected ray on a surface Given then where Given the surface normal n the incident ray s the reflected ray r r s ln l n i s ni n n i j k s i j k 4 4 s n r then l x r 7 y r 4 7 z r 4 with 7 7 r i j k Let s check this vector out. Its magnitude should equal the magnitude of the incident vector s, the reflection angle should equal the incident angle. 8 s r 4

75 Examples 47 The reflection angle equals The incident angle equals For u a cos a cos ni r n r nis n s nir nis n r n s but s r therefore n r n s 7 7 ni r( i jk) i j k nis( i jk) i j k 4 4 which confirms that the angle of reflection equals the angle of incidence.

76 48 Geometry for computer graphics.6 Lines, planes spheres.6. Line intersecting a sphere P Given a sphere with radius r located at C with position vector c λv C r λv λv a line equation p t lv where v a touch, miss or intersect condition is determined by l c t P P T L L L where l siv ( siv) s r s c t The diagram shows a sphere with radius r centered at C with position vector c i j three lines L, L L that miss, touch intersect the sphere respectively. The lines are of the form p t l v therefore p t lv p t lv p t lv where t i v i j t i v j t i v i j c i j Let us substitute the lines in the original equations: L : s i j (s v) s r The negative discriminant confirms a miss condition. L : s i j (s v) s r The zero discriminant confirms a touch condition, therefore l. The touch point is P (,, ) which is correct. L : s i j (s v) s r The positive discriminant confirms an intersect condition

77 Examples 49 therefore l or The intersection points are: if l x y z P P P ( ) ( ) if l x y z P P P ( ) ( ) The intersection points are P which are correct.,, P,,.6. Sphere touching a plane Given a plane ax by cz d where n ai bj ck Q n the nearest point Q on the plane to a point P is given by q p ln ni p d where l ni n P r The distance is given by for a plane a sphere ln ln r The diagram shows a sphere radius r centered at P(,, ) The plane equation is y therefore n j p i j k therefore l ( )

78 5 Geometry for computer graphics which equals the sphere s radius therefore the sphere the plane touch. The touch point is x Q y Q z Q therefore the touch point is Q(,, ) which is correct..6. Touching spheres r r P C C Given C C C C C C d ( x x ) ( y y ) ( z z ) the touch condition is d r r the touch point is x y z r x ( d x x ) r y ( d y y ) r z C d ( z C z C ) P C C C P C C C P Given that one sphere with radius r is centered at C (,, ) the other with radius r.5 is centered at C (.5,, ) then d ( 5. ) ( ) ( ) 5. The touch condition is satisfied x y z P P P (. 5 5 ). ( 5 ). ( ) 5. therefore the touch point is P(,, ) which is correct.

79 Examples 5.7 Three-dimensional triangles.7. Coordinates of a point inside a triangle To locate points inside outside the triangle P, P, P using barycentric coordinates. For any point P (x, y, z ) we can state P (,, ) x ex lx bx y ey ly by z ez lz bz where e l b P (,, 4) x y z P (,, ) The table below shows values of P for various values of e, l b. Let us check that the positions of P reside on the plane of the triangle. The vertices of the triangle are P (,, ), P (,, 4), P (,, ) therefore the Cartesian plane equation is ax by cz d (see plane equation from three points) where y z z x a y z b z x c y z z x x x x y y y d ax by cz a 4 4 b 4 c 6 d 46 4 therefore the plane equation is 4x y 6z 4 The table also confirms that the values of P satisfy the plane equation. e l b x y z 4x y 6z

80 5 Geometry for computer graphics.7. Unknown coordinate value inside a triangle The x z-coordinates of a point P are known it is required to determine its y-coordinate inside the triangle P, P, P. Using barycentric coordinates we have y ey ly ( e l)y where x x x e z z z x x x l z z z x x x z z z For P to be inside the triangle (e, l) [, ]. If P is positioned at P i.e. x z, y should be. Therefore e 4 l 4 e l which makes e l therefore y ( ) which is correct. The table below shows the values of e, l, e l y for different values of x z. Let us check that the interpolated values of P reside on the plane of the triangle. The vertices of the triangle are P (,, ), P (,, 4), P (,, ) therefore the Cartesian plane equation is ax by cz d (see plane equation from three points) where y z z x a y z b z x c y z z x x x x y y y d ax by cz a 4 4 b 4 c 6 d 46 4 therefore the plane equation is 4x y 6z 4

81 Examples 5 x y z e l e l 4x y 6z The table below also confirms that the above values of P satisfy the plane equation. Let us test a point outside the triangle s boundary, e.g. P (4,, ) e 4 4 l 4 4 e l 4 therefore e which confirms that P is outside the triangle s boundary. Similarly, for P (,, 5) e 5 4 l 5 4 e l 5 therefore e 4 l 4 which confirms that P is also outside the triangle s boundary.

82 54 Geometry for computer graphics.8 Parametric curves patches The following examples illustrate how various curves can be created by mixing together different parametric functions..8. Parametric curves in Sine curve t p a x t y asin t t [, t ] Cosine curve t p a x t y acost t [, t ] Sine curve with growing amplitude t p t a t t [, t ] x t y asin t

83 Examples 55 Cosine curve with growing amplitude t p t a t x t y acost t [, t ] Sine curve with decaying amplitude t p t a t x t y asin t t [, t ] Cosine curve with decaying amplitude t p t a t x t y acost t [, t ] Sine-squared curve t p a x t t [, t ] y asin t

84 56 Geometry for computer graphics Cosine-squared curve t p a x t y acos t t [, t ] Lissajous curve t p a x asin t y asin t t [, t ] Circle t p a x acost y asin t t [, t ] Ellipse t p a b x acost y bsin t t [, t ].5.5

85 Examples 57 Spiral t p t r t x rcost y rsin t. t [, tm ax ] Logarithmic spiral t p a 6. b 8. t x ae cosbt t y ae sin bt t [, ] t Parabola t 4 p x t y p t t t t,

86 58 Geometry for computer graphics Neil s parabola t 4 a x t y at t t t, Cardioid t p a xa( costcos t) ya( sin tsin t) t [, t ].8. Parametric curves in Circle t p a x acost y t [, tma x ] z aasin t

87 Examples 59 Ellipse t p a b x acost y bsin t t [, tm ax ] z Spiral t 4 a x acost y asin t t [, t ] z t 4π Spiral t 4p a b x acost y bsin t t [, tm ax ] z t 4π

88 6 Geometry for computer graphics Spiral t 4p a b x acost y bsin t t [, tm ax ] z t 4π Spiral 4 t 4p a xacost y asin t t [, t ] z t 4π Spiral 5 t 4p a x acost yasin t t [, t ] z t 4π

89 Examples 6 Spiral 6 t 4p t r t x rcost y rsin t z t t [, t ] 4π Spiral 7 t 4p t r t x rcost y rsin t z t t [, ] t 4π Sinusoid t = p a x asin t y t [, t ] z t π

90 6 Geometry for computer graphics Sinusoidal ring t p a b. n 8 x acost y bsin nt z asin t t [, t ] Coiled ring t p R ( major radius) r 5. ( minor radius) n 4 x( Rrcos nt) cost y rsin nt z ( Rrcos nt) sint t [, t ].8. Planar patch Given P, P, P, P in P P P uv u P P v Given P (, ), P (, ), P (4, ), P (4, ) P P P x 4 4 P P y

91 Examples Parametric surfaces in Modulated surface y sin(x z) T p a yasin( xz)} ( x, z) [ T, T].5.5 y cos(x z) T p a yacos ( xz)} ( x, z) [ T, T].5.5 y sin(xz) T p a yasin( xz)} ( x, z) [ T, T].5.5

92 64 Geometry for computer graphics y cos(xz) T p a yacos( xz)} ( x, z) [ T, T].5.5 y cosx sinz T p a yacos xasin z} ( x, z) [ T, T] y z cosx x sinz T 4p y zcos xxsin z} ( x, z) [ T, T]

93 Examples 65 sin x y T 9 sin x y x, y [ T, T] ( ) Quadratic Bézier curve Quadratic Bézier curve in A quadratic Bézier curve is given by p(t) ( t) p t( t) p C t p Given the points P (, ), P C (,.5), P (, ) the quadratic Bézier curve is shown with its control points. Quadratic Bézier curve in Given the points P (,, ), P C (,.5, ), P (,, ) the quadratic Bézier curve is shown with its control points Cubic Bézier curve Cubic Bézier curve in A cubic Bézier curve is given by p(t) ( t) p t( t) p C t ( t) p C t p Given the points P (, ), P C (, ), P C (., ), P (.5, ) the cubic Bézier curve is shown with its control points

94 66 Geometry for computer graphics Cubic Bézier curve in Given the points P (,, ), P C (,.5, ), P C (,, ), P (,, 4) the cubic Bézier curve is shown with its control points..8.7 Quadratic Bézier patch A quadratic surface patch is described by p p p p( uv, ) [( u) u( u) u ] p p p p p p Given p (,, ) p (,, ) p (,, ) p ( 5.,, ) p (,, ) p (,, ) p (,, ) p (,, ) p (,, ) The surface patch is shown in the diagram ( v) v( v) v P P P P P P P P P

95 Examples Cubic Bézier patch A cubic surface patch is described by p p p p p p p p p( uv, ) [ ( u) u( u) u( u) u] p p p p p p p p ( v) v( v) v ( v) v Given p (,, ) p (,, ) p (,, ) p (,, ) p (,, ) p (,, ) p (,, ) p (,, ) p (,, ) p (,, ) p (,, ) p (,,) p (,, ) p (,, ) p (,, ) p (,, ) The surface patch is shown in the diagram P P P P P P P P P P P P P P P P

96 68 Geometry for computer graphics.9 Second degree surfaces in stard form Sphere Ellipsoid 4 x y z x y z Elliptic cylinder Elliptic paraboloid y z Elliptic cone x z y Elliptic hyperboloid of one sheet x z y x z y Elliptic hyperboloid of two sheets x z y y x z

97