Geometric Primitives. Chapter 5

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1 Chapter 5 Geometric Primitives In this chapter, we discuss the basic geometric primitives we will use to represent the world in which our graphic objects live. As discussed at the beginning of this class, there are many different ways to describe and represent objects. The way or ways we select will determine the kind of geometric primitives we use. Because this is a beginner s class, and also a one semester class, we will simplify greatly the model we use to represent objects. First, we will represent objects by representing their outside surface and ignore the inside of the object. This outside surface of each object is made of too many points to represent all of them. We will use as many points as we need to capture the general shape but no more. We will further simplify our surface by approximating it with flat polygons, mostly triangles. In other words, once our surface has been approximated with a finite (small) number of points, we will join these points with triangles. This is called triangulating a surface. The way this is done is fairly complex and the subject of active research beyond the scope of this class. There is software available to do this, and we will use it. Therefore, for the purpose of this class, our objects will have already been triangulated. The world in which our graphic objects live will be made of points, edges and flat polygons, mostly triangles. These are the objects we will work with. Therefore, the kind of primitives we use will include lines, planes, rectangles, and triangles in both 2-D and 3-D. In some instances, we will also use the exact surface we are representing. For example, when trying to detect if a sphere is going to touch a plane, even if the sphere has been triangulated, it is not efficient to see if each triangle making up the sphere is touching the plane. It is much easier to see if the distance between the plane and the center of the sphere is less that its radius. Therefore, we will also mention some properties of curves and surfaces in 2-D and 3-D. For each primitive, we will talk about how to represent it as well as discuss some of its important properties and operations. Because this class is also applied, we will discuss and present some Java code for representing these primitives and perform operations on them. 51

2 52 CHAPTER 5. GEOMETRIC PRIMITIVES 5.1 Representation Techniques There are different ways to represent a given primitive. Often, the way we select depend on the situation. We discuss the different ways to represent primitives, and discuss when and where each way is best applicable. It is important to remember that all but one of the representation techniques we discuss will tell us in their own way what relation must exist between the coordinates (x, y) or (x, y, z) of a point so that this point is on the primitive Explicit Form This is the form mathematics students are the most familiar with. In this form, an explicit relation between the coordinates (x, y, z) of a point is given so that the point is on the primitive if and only if its coordinates satisfy the relation. This relation is usually of the form z = f (x, y) in 3-D or y = f (x) in 2-D. It is called explicit form because we can explicitely compute points on the primitive. Example 60 z = x 2 + y 2 describes a paraboloid along the z axis. This form makes it easy to test if a point belongs to a primitive. Because we have an explicit formula, it is also easy to derive any information about the primitive which is mathematical in nature such as the slope, the tangent plane, the normal, Implicit Form Similar to the explicit form, however, only an implicit relation exists or is given between the coordinates of a point. This relation is usually of the form f (x, y, z) =0in 3-D or f (x, y) in 2-D. It is called implicit because we cannot directly compute the coordinates of points on the primitive, though we can test if a point is on the primitive. Example 61 The equation of a sphere centered at the origin, of radius r is: Parametric Form x 2 + y 2 + z 2 = r 2 Instead of being given a relation between the coordinates of a point, each coordinate is given in terms of one or more parameters. If the coordinates of a point are given in terms of one parameter, as the parameter changes, the point will describe a curve. If the coordinates of a point are given in terms of two parameters, as the parameters change, the point will describe a surface. Example 62 describes a circle of radius r centered at the origin. x = cos 2πt y =sin2πt 0 t 1

3 5.2. SOME PROPERTIES OF CURVES AND SURFACES "Straightforward" Form We call it this, for a lack of a better term. There is not a recipe for this method. It tries to capture the most important information about the primitive directly. Therefore, it is the most intuitive form for human to work with. For example, to store a line segment, we would give two endpoints. To store a circle, we would give its center and its radius Degrees of Freedom Each primitive has an inherent number of degrees of freedom. This is the minimum number of pieces of information needed to represent the primitive unambiguously Practice Problems 1. What object is describes by x = r cos 2πt y = r sin 2πt 0 t 1 and 1 r 2 2. How many degrees of freedom does a line have in 2D, in 3D? 3. How many degrees of freedom does a plane have in 3D? 5.2 Some Properties of Curves and Surfaces We discuss 2-D curves and 3-D surfaces D Curves Recall from above that an equation of the form f (x, y) =0 determines a 2-D curve. Definition 63 The 2-D gradient, denoted f (x, y) is defined to be f f (x, y) = x, f y (5.1) Thus, it is a 2-D vector. The gradient is an important quantity. We summarize the information which can be obtained from it:

4 54 CHAPTER 5. GEOMETRIC PRIMITIVES 1. f (x, y) points in the direction of greatest increase of f. 2. If (a, b) is a point on the curve determined by f (x, y) =0,then f (a, b) points in the direction of the region where f (x, y) > If (a, b) is a point on the implicit curve determined by f (x, y) =0, then f (a, b) is a vector perpendicular to the tangent vector to the curve at that point. Thus, the gradient is an important tool to compute normals. Example 64 Consider the unit circle of radius 1 centered at the origin. Its ( ) 2 2 equation is f (x, y) =x 2 + y 2 1=0. The point 2, is a point on 2 ( ) 2 2 this circle. The gradient is f (x, y) = 2x, 2y. Therefore, f 2, = 2 ( ) 2 2 2, 2, is the normal to the circle at 2,. It points toward the outside 2 of the circle, which is the region where f (x, y) > 0 (outside the circle, the distance to the origin is larger than the radius) D Surfaces Recall from above that an equation of the form determines a 3-D surface. f (x, y, z) =0 Definition 65 The 3-D gradient, denoted f (x, y, z) is defined to be f f (x, y, z) = x, f y, f z (5.2) Thus, it is a 3-D vector. Like its 2-D counterpart, the 3-D gradient provides needed information. We summarize the information which can be obtained from it: 1. f (x, y, z) points in the direction of greatest increase of f. 2. If (a, b, c) is a point on the surface determined by f (x, y, z) =0, then f (a, b, c) points in the direction of the region where f (x, y, z) > If (a, b, c) is a point on the implicit curve determined by f (x, y, z) =0, then f (a, b, c) is a vector perpendicular to the tangent plane to the surface at that point. Thus, the gradient is an important tool to compute normals.

5 5.3. LINES AND RAYS Practice Problems 1. Explain why f points in the direction of greatest increase of f. 2. Explain why if (a, b) is a point on the curve determined by f (x, y) =0, then f (a, b) points toward the region where f (x, y) > Find the normal to the unit sphere centered at the origin at the points below: (a) (1, 0, 0) ( ) (b) 3, 3, Lines and Rays We need two quantities to fully determine a line. Either two points on the line, or a point on the line and the slope of the line. In mathematics, in classical geometry, we have the following definitions. Definition 66 A line extends infinitely in two directions. Definition 67 A line segment is a finite portion of a line. It has two endpoints. Definition 68 A ray has an origin and extends infinitely in one direction. However, in computer graphics, we alter the definition of a ray slightly. A ray is not infinite. It has an origin, an endpoint and a direction. In other words, a ray is a directed line segment. Rays are very important in computer graphics Two point Representation The most obvious way to represent a line, a line segment or a ray is to use two points. In the case of a line, they are just two points the line goes through. In the case of a line segment, the two points are the endpoints. In the case of a ray, we also need to know which point is the origin. For the other representations, we need to differentiate between 2-D and 3-D Lines in 2-D Standard or Implicit Representation The general equation of a line in 2-D is ax + by + d =0 (5.3) This is the implicit representation of a line. From this form, we can easily extract a vector parallel to the line and a vector perpendicular to the line. The following proposition can be proven:

6 56 CHAPTER 5. GEOMETRIC PRIMITIVES Proposition 69 Consider the line whose implicit equation is ax + by + d =0. The following is true: 1. The vector a, b is perpendicular to the direction of the line. 2. The vector b, a is parallel to the direction of the line. Proof. See problems at the end of this section, on page 58. Explicit Representation From equation 5.3 and when b 0, we can solve for y to obtain the explicit representation: y = a b x d (5.4) b which is also known as the slope-intercept form. In this form, the coefficient of x is the slope of the line and the constant is the y intercept.we usually write this form as y = mx + b (5.5) Given a point P 0 =(x 0,y 0 ) and the slope m, the equation of the line is y y 0 = m (x x 0 ) (5.6) This way is not the most convenient way to represent line segments or rays. Example 70 Find the equation (implicit and explicit) of the line through (1, 2) and (3, 8). Theslopeofthislineism = 8 2 =3. Using equation 5.6 and one of the 3 1 given point, we have y 2 = 3(x 1) y = 3x 1 Parametric Representation Given two Points Suppose we are given two points P 0 and P 1 on a line. A point P is on the line if for some real number t we have OP = OP 0 + tp 0 P 1 (5.7) Equation 5.7 is known as the vector equation of a line. Switching to coordinates, assuming that P =(x, y), P 0 =(x 0,y 0 ) and P 1 =(x 1,y 1 ), we obtain { x = x0 + t (x 1 x 0 ) y = y 0 + t (y 1 y 0 ) or { x =(1 t) x0 + tx 1 y =(1 t) y 0 + ty 1 (5.8)

7 5.3. LINES AND RAYS 57 Equation 5.8 is known as the parametric equation of a line. t, the parameter takes on every real number. If we limit the values of t between 0 and 1, then wehavetheequationofthelinesegmentbetweenthetwopointsp 0 and P 1.It is also the equation of the ray starting at P 0 and ending at P 1 when 0 t 1. Example 71 Find the parametric equation of the line through (1, 2) and (3, 8). Using equation 5.8, we get { x =(1 t)+2t y =2(1 t)+8t { x =1+t y =2+6t Lines in 3-D Lines in 3-D do not have a representation like the ones shown in equations 5.3 and 5.4. We can derive the vector and parametric equation of a line in a similar mannerthanin2-d.giventwopointsp 0 =(x 0,y 0,z 0 ) and P 1 =(x 1,y 1,z 1 ),a point P =(x, y, z) is on the line if OP = OP 0 + tp 0 P 1 (5.9) Equation 5.9 is called the vector equation of a line. If we switch to coordinates, we have x = x 0 + t (x 1 x 0 ) y = y 0 + t (y 1 y 0 ) (5.10) z = z 0 + t (z 1 z 0 ) If we solve for t, we get x x 0 x 1 x 0 = y y 0 y 1 y 0 = z z 0 z 1 z 0 (5.11) These are called the symmetric equations of the line. Example 72 Find the parametric and symmetric equations of the line through (1, 2, 3) and (2, 4, 1). Using equation 5.10, we get If we solve for t, weget x =1+t y =2+2t z =3 2t x 1= y 2 2 = z 3 2

8 58 CHAPTER 5. GEOMETRIC PRIMITIVES Practice Problems 1. Whatisavectornormaltothe2-Dlinegivenbyax + by + d =0? What is a vector parallel to the same line? 2. Using the implicit equation of a 2-D line, explain how one can test if two given points are on the same side of the line. 3. Derive all the forms of the equation of a 2-D line given a point P 0 =(x 0,y 0 ) on the line and a vector u = a, b parallel to the line. 4. Derivetheimplicitequationofa2-DlinegivenapointP 0 =(x 0,y 0 ) on thelineandavector u = a, b perpendicular to the line. 5. Sow that the distance in R 2 between a point P 1 =(x 1,y 1 ) and the line ax + by + d =0is ax 1 + by 1 + d a2 + b 2 Use this to find the distance between the line ax + by + d =0and the origin. 6. Derive the vector, parametric and symmetric forms of the equation of a line in R 3 given a point P 0 =(x 0,y 0,z 0 ) on the line and a vector u = a, b, c parallel to the line. 7. Let L be the line through the points Q and R and let P be a point not on L. Let a = QR and b = QP. (a) Show that the distance d from the point P to the line L is a b d = a (b) Use the formula you just derived to find the distance between the point P (1, 1, 1) and the line through Q (0, 6, 8) and R ( 1, 4, 7). 5.4 Planes A plane, in 3-D, is uniquely determined given three points. It is also uniquely determined given a point and a normal to the plane Implicit Equation Given a point P 0 =(x 0,y 0,z 0 ) and a normal n = a, b, c to a plane, a point P =(x, y, z) will be on the plane if it satisfies n P0 P =0 (5.12)

9 5.4. PLANES 59 This is known as the vector equation of a plane. Switching to coordinates, we get a, b, c x x 0,y y 0,z z 0 = 0 a (x x 0 )+b(y y 0 )+c(z z 0 ) = 0 We usually write this as ax + by + cz + d =0 (5.13) where d = ax 0 by 0 cz 0. This is known as the scalar equation of a plane. It is also the equation of a plane in implicit form. Example 73 The scalar equation of a plane through (1, 2, 3) with normal 2, 1, 4 is 2(x 1) + 1 (y 2) + 4 (z 3) = 0 2x + y +4z 16 = 0 If instead of being given a point and the normal, we are given three noncolinear points P 1, P 2,andP 3, we form the vectors P 1 P 2 and P 1 P 3. The cross product P 1 P 2 P 1 P 3 is a vector perpendicular to both P 1 P 2 and P 1 P 3 and therefore perpendicular to the plane. We can then use one of the point, the vector obtained from the cross product to derive the equation of the plane. See the problems at the end of the section on page Practice Problems 1. Show that the distance from a point P (x 1,y 1,z 1 ) to the plane ax + by + cz + d =0is ax 1 + by 1 + cz 1 + d a2 + b 2 + c 2 2. Use the previous problem to give an interpretation to the quantity d in the scalar equation of a plane, when a, b, c is a unit vector. 3. Using the scalar equation of a plane, explain how one can determine if two given points are on the same side of a plane. 4. Find the normal to the unit sphere centered at the origin at the point ( ) , 3, and find the equation of the tangent plane at this point Derive a general formula which gives the equation of the plane tangent to a surface given by f (x, y, z) =0atapoint(x 0,y 0,z 0 ).

10 60 CHAPTER 5. GEOMETRIC PRIMITIVES 5.5 Triangles and Barycentric Coordinates Triangles are fundamental to 2-D and 3-D graphics as most objects are approximated with triangles. A basic triangle consists of three vertices, three edges and one face. The easiest way to draw a ytriangle is to draw its edges. However, the face of a triangle is often colored, depending of the color of its vertices as well as other factors such as lighting. In addition, one often has to determine if a point is on one side of a triangle, or the other, if a ray intersects with a triangle or not. The easiest way to handle this is to use a new coordinate system, called barycentric coordinates. We first define them in 2-D, then estend their definition in 3-D Calculus Review Figure 5.1: Barycentric Coordinates The area of a 2D triangle whose vertices are a =(x a,y a ),b =(x b,y b ),c = (x c,y c ) (as shown in figure 5.1) is given by Area = 1 2 x b x a x c x a y b y a y c y a (5.14) Note that this area is a "signed" area. In other words, it has a sign. To obtain the area the way we are used to define it, we take the absolute value. It is easy to implement this computation using our vect3d class. If we convert the 2D points in 3D points by adding 0 for their z-coordinate, then x b x a x c x a y b y a y c y a

11 5.5. TRIANGLES AND BARYCENTRIC COORDINATES 61 will be the z-coordinate of (b a) (c a). If the vertices live in 3D space, the area of the corresponding triangle is This always gives a positive answer. Area = 1 (b a) (c a) (5.15) Barycentric Coordinates in 2D Consider a 2D triangle whose vertices are a =(x a,y a ),b=(x b,y b ),c=(x c,y c ) (as shown in figure 5.1) and a point p. Barycentric coordinates allow us to express the coordinates of p =(x, y) in terms of a, b, c. More specifically, the barycentric coordinates of p are the numbers β and γ such that If we regroup a, b and c, we obtain p = a + β (b a)+γ (c a) (5.16) p = a + βb βa + γc γa = (1 β γ) a + βb + γc Figure 5.1 shows the meaning of the numbers β and γ. At this point, it is customary to define a third variable, α, by α =1 β γ We then have p = αa + βb + γc Definition 74 (Barycentric Coordinates) The barycentric coordinates of the point p in terms of the points a, b, c are the numbers α, β, γ such that p = αa + βb + γc (5.17) with the constraint α + β + γ =1 (5.18) We first look at some of the properties of the barycentric coordinates. Barycentric coordinate are defined for all points in the plane. They have several nice features: 1. A point p is inside the triangle defined by a, b, c if and only if 0 < α < 1 0 < β < 1 0 < γ < 1 This is very important. It provides an easy way to test if a point is inside a triangle.

12 62 CHAPTER 5. GEOMETRIC PRIMITIVES Figure 5.2: Barycentric Coordinates as a Ratio of Areas 2. If one of the barycentric coordinates is 0 and the other two are between 0 and 1, the corresponding point p is on one of the edges of the triangle. 3. If two of the barycentric coordinates are zero and the third is 1, thepoint p is at one of the vertices of the triangle. 4. By changing the values of α, β, γ between 0 and 1, the point p will move smoothly inside the triangle. This can (and will) be applied to other properties of the vertices such as color. 5. The center of the triangle is obtained when α = β = γ = 1 3. If the triangle is made of a certain substance which is evenly distributed throughout the triangle, then these values of α, β, γ wouldgiveusthecenterofgravity. Finding α, β, γ Let us first notice that because of equation 5.18, it is enough to find two of the three barycentric coordinates. One way of finding α, β, γ is to write equation 5.16 in terms of the coordinates of the various points involved. This gives us the following system { x = xa + β (x b x a )+γ(x c x a ) (5.19) y = y a + β (y b y a )+γ(y c y a ) which can be solved using your favorite method. We can also use geometry to find α, β, γ. Let A a, A b and A c be as in figure 5.2 and let A denote the area of the triangle. Also note that the point inside the triangle on figure 5.2 is the point we called p. Consider the triangles in figure 5.3. These are different triangles drawn for a fixed value of β. They have the same area since they have the same base and height. This area was denoted A b on figure 5.2. Thus, we see that A b only depends on β. Therefore, we have A b = Cβ (5.20)

13 5.5. TRIANGLES AND BARYCENTRIC COORDINATES 63 Figure 5.3: A b Only Depends on β for some constant C. When p is on b that is when β =1,wehaveA b = A. Equation 5.20 gives us A = C. Therefore we see that Similarly, we have β = A b A γ = A c A α = A a A These various areas can then be computed using formula More specifically, x a x c x x c y a y c y y c β = x b x a x c x a y b y a y c y a x b x a x x a y a y a y y a γ = x b x a x c x a y b y a y c y a α =1 β γ Application of Barycentric Coordinates to Coloring Triangles Let us assume that we are using the RGB model. That is all colors can be obtained by mixing R (red), G (green) and B (blue). Usually, with such a

14 64 CHAPTER 5. GEOMETRIC PRIMITIVES model, the level of each color channel is a number between 0 and 255. To specify the color at any point, we must specify a triplet (R, G, B) where R, G, and B are integers between 0 and 255. They indicate how much of red, green and blue is used in the color. When using Java, there is a built-in class to handle colors. It is called Color. This class has some built-in predefined colors. Here are some examples: Color.black, Color.blue, Color.cyan, Color.gray, Color.green, Color.magenta, Color.orange, Color.pink, Color.red, Color,white, Color.yellow. To get any other color, one uses a statement such as new Color(R,G,B) where R,G,B are integers between 0 and 255. It is also possible to use a single integer to represent colors. Keeping in mind that an integer has 32 bits, bits 0 7 contain the R level, bits 8 15 contain the G level and bits contain the B level. The remaining bits are unused and set to 0. Let us see now how barycentric coordinates can be used to smoothly color a triangle, given the color of its vertices. Using the notation above, let us assume that C a is the color of a, C b is the color of b and C c is the color of c. Each color is in fact a triplet. We will use the notation C a =(R a,g a,b c ) and similar notation for the remaining points. We would like to color the triangle so that there is a smooth coloring throughout the triangle. We use the fact that by changing the values of α, β, γ between 0 and 1, thepointp = αa + βb + γc will move smoothly inside the triangle. In other words, small changes in α, β, γ will result in small changes in the location of p. We apply this to colors. We let C = αc a + βc b + γc c (5.21) (we really do this for every color channel). Small changes in α, β, γ will result in small changes in the color. Therefore, the color will change smoothly as we move within the triangle. To color smoothly a triangle given the color of its vertices, we can use the following algorithm: 1. ForeachpointP =(x, y) inside the triangle, find α, β, γ 2. Use α, β, γ to interpolate the color of the point from the color of the vertices using relation Plot the point with coordinates (x, y) and color computed above. Thereisanappletonthewebsitefortheclasstoillustratethis. Itallowsthe user to draw a rectangle and specify the color of the vertices. Clicking inside the triangle will result in the coordinates (Cartesian and barycentric) of the point where the user clicked to be displayed. If Coloring of the triangle is checked, thergbcolorofthepointwillalsobedisplayed Barycentric Coordinates in 3D We use the same notation as in the 2D case. The only difference is that now points have three coordinates. So, we have a =(x a,y a,z a ), b =(x b,y b,z b ) and

15 5.5. TRIANGLES AND BARYCENTRIC COORDINATES 65 c =(x c,y c,z c ). Barycentric coordinates extend naturally to 3D triangles and they have the same properties. In other words, we have: Definition 75 (Barycentric Coordinates) The barycentric coordinates of the point p in terms of the points a, b, c are the numbers α, β, γ such that p = αa + βb + γc (5.22) with the constraint α + β + γ =1 (5.23) Barycentric coordinate are defined for all points in the plane. They have several nice features: 1. A point p is inside the triangle defined by a, b, c if and only if 0 < α < 1 0 < β < 1 0 < γ < 1 This is very important. It provides an easy way to test if a point is inside a triangle. 2. If one of the barycentric coordinates is 0 and the other two are between 0 and 1, the corresponding point p is on one of the edges of the triangle. 3. If two of the barycentric coordinates are zero and the third is 1, thepoint p is at one of the vertices of the triangle. 4. By changing the values of α, β, γ between 0 and 1, the point p will move smoothly inside the triangle. This can (and will) be applied to other properties of the vertices such as color. 5. The center of the triangle is obtained when α = β = γ = 1 3. If the triangle is made of a certain substance which is evenly distributed throughout the triangle, then these values of α, β, γ would give us the center of gravity. What differs from the 2D case is how they are computed. α is still related to the ratio Aa A. There is a similar relation for β and γ. However, in the 2D case, the formula we used for the area gave us a "signed" area. It depended on the orientation of the orientation of the triangle. The formula for the 3D area of a triangle, as noticed above, is always positive. To also make it a sign area, we also need to take a dot product. More specifically, we define the following quantities: n is the normal to the triangle T with vertices (a, b, c) in counterclockwise order. In other words, n =(b a) (c a). n a is the normal to T a, the triangle with area A a as shown in figure 5.2. T a =(b, c, p) in counterclockwise order. Thus, n a =(c b) (p b).

16 66 CHAPTER 5. GEOMETRIC PRIMITIVES n b is the normal to T b, the triangle with area A b as shown in figure 5.2. T b =(c, a, p) in counterclockwise order. Thus, n b =(a c) (p c). n c is the normal to T c, the triangle with area A c as shown in figure 5.2. T c =(a, b, p) in counterclockwise order. Thus, n c =(b a) (p a). The quantity n.n b n.na n n a =1if p is inside T, 1 otherwise. The same is true for n n b and n.n c n n. Multiplying A a n.n c A by a n n a will then give us a signed area, depending on whether p is inside T or outside. Since A a = n a and A = n, we have: A a A n.n a n n a = n a n.n a n n n a = n.n a n 2 We obtain similar formulas for the other ratios. Thus, in the case of a 3D triangle, we can define the barycentric coordinates by: Practice Problems α = n.n a n 2 (5.24) β = n.n b n 2 γ = n.n c n 2 1. Solve the system in equation 5.19 and find α, β, γ. 2. Verify that the second method shown in the text gives us the same answer. 3. Prove that if R, G, andb are integers between 0 and 255, then αr + βg + γb is also between 0 and 255. This is important. It means we can mix RGB colors using barycentric coordinates, the values we obtain are still within the allowable range. 4. Explain how you would convert colors between the RGB format and the single integer format. 5. Explain how, without using barycentric coordinates, one can determine if and where a line intersects with a triangle.

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