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Figure 5-13. Finding the normal vector of a triangle Note that Vector3.Cross(Vec1, Vec2) is not the same as Vector3.Cross(Vec2,Vec1). Both resulting vectors will be on the same line, but they will have opposite directions. This is because a plane, and thus the previous triangle, has two perpendicular directions: one pointing out of this page and the other pointing into this page.

For example, the following code shows the use and results of the operators: int x = 10; int y = -x; int z = -y; // x = 10 // y = -10 // z = 10

When defining a large object, you ll mostly want to define your structure using indices, because this is the only way a vertex can be used by multiple triangles, which allows smooth lighting on your object (see recipe 6-2). Therefore, this recipe will base its calculations on two arrays containing the vertices and indices of your structure:

n Note You probably won t need to edit your image on your hard drive or with an external editor if you just

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private VertexPositionNormalTexture[] GenerateNormalsForTriangleList(VertexPositionNormalTexture[] vertices, int[] indices) { } This method receives the array of vertices, which do not yet contain any normal data. It will store the correct normal information inside each vertex and will return the array. Based on the indices, it can determine which vertices are used to create triangles. However, depending on whether you are drawing the triangles as a TriangleList or TriangleStrip, the contents of the indices array will be different, so some lines in the method are different for both cases.

User-defined conversions will be discussed in greater detail in 18, but I will mention them here as well because they are operators. You can define both implicit and explicit conversions for your own classes and structs. This allows you to convert an object of your user-defined type to some other type, and vice versa. C# provides implicit and explicit conversions. With an implicit conversion, the compiler will automatically make the conversion, if necessary, when it is resolving what types to use in a particular context. With an explicit conversion, the compiler will only make the conversion when an explicit cast operator is used. The syntax for declaring an implicit conversion is the following. The syntax for the explicit conversion is the same, except that explicit is substituted for implicit. The modifiers public and static are required for all user-defined conversions. Required Target Source public static implicit operator TargetType ( SourceType Identifier ) { ... return ObjectOfTargetType; } The following code shows an example of declarations for conversion operators that will convert an object of type LimitedInt to type int, and vice versa. class LimitedInt { Target Source public static implicit operator int (LimitedInt li) { return li.TheValue; } Target Source public static implicit operator LimitedInt (int x) { LimitedInt li = new LimitedInt(); li.TheValue = x; return li; } private int _TheValue = 0; public int TheValue{ ... } }

If the vertices already contain any data for their normals, reset them all to 0: for (int i = 0; i < vertices.Length; i++) vertices[i].Normal = new Vector3(0, 0, 0); Next, as indicated in the previous pseudocode, you re going to scroll through all triangles of the structure and calculate their normals. In a TriangleList, each triangle is defined by three consecutive indices. This means the number of triangles in your structure equals indices.Length/3. This for loop scrolls through each triangle defined in the indices array: for (int i = 0; i < indices.Length/3; i++) { Vector3 firstVec = vertices[indices[i*3 + 1]].Position - vertices[indices[i*3]].Position; Vector3 secondVec = vertices[indices[i*3 + 2]].Position - vertices[indices[i*3]].Position; Vector3 normal = Vector3.Cross(secondVec, firstVec); normal.Normalize(); vertices[indices[i*3]].Normal += normal; vertices[indices[i*3 + 1]].Normal += normal; vertices[indices[i*3 + 2]].Normal += normal; } For each triangle, you define the two vectors that are shown in Figure 5-13. As for any vector between point P0 and P1, you find it by subtracting P0 from P1. This is done in the first line, and the second line calculates the vector from P0 to P2. Next, you find the vector perpendicular to these two vectors by taking their cross product. Don t forget to normalize this result so its length becomes exactly 1.

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