HomeWork 2 Solution Chen Zhanqing.

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HomeWork 2 Solution Chen Zhanqing

1.What are the two main applications of a surface modeling system in a product cycle? Answer : Generation of visual model for aesthetic evaluation. Generation of NC tool paths for machining.

2. Give at least two types of primitives in solid modeling that can also be defined as sweeps. Answer : Cylinder,Torus

3. Suppose a solid is stored in computer with CSG data structure 3.Suppose a solid is stored in computer with CSG data structure. When it is displayed on the screen, computer needs its faces to triangulate them and render those triangles on the screen. Does CSG have those faces? Why? If not, how does the computer get those faces? What is the main disadvantage of this? What is the main advantage of CSG? Answer : CSG does not have those faces because it only stores the history of applying Boolean operations on the primitives . It is necessary to transform the CSG into other types of representation.

Answer Advantage: The CSG tree data structure has the following advantages. 􀁺 Data structure is simple and stores compact data. Accordingly, the management of data is easy. 􀁺 The solid stored in a CSG tree is always a valid solid. A valid solid implies a solid of which inside and outside region can be clearly identified. An example of an invalid solid would be a solid with a strut edge. In this case, the notion of inside and outside is not clear around the vertex where the strut edge is attached. 􀁺 CSG representation of a solid can be always converted to the corresponding B-rep. Thus CSG tree representation can be interfaced to the application programs written for B-Rep. 􀁺 Parametric modeling can be realized very easily by changing the parameter values of the associated primitives.

Disadvantage: However, the CSG tree structure has the following disadvantages. 􀁺 Since CSG tree structure stores the history of applying the Boolean operations, only the Boolean operations are allowed in the modeling process. With Boolean operations alone, the range of shapes to be modeled is restricted severely. Furthermore, the convenient local modification functions such as lifting and rounding can not be used. 􀁺 It requires a large amount of computation to derive the information on the boundary surfaces, their boundary edges, and the connectivity between these boundary entities from CSG tree representation. Unfortunately, there are many applications which require these boundary information. One example would be a display of solids. Whether a shaded image or a line drawing of a solid is to be displayed as explained in Chapter 3, the information on the faces and/or edges is required. Thus CSG tree representation has been assumed to be inappropriate for the interactive display and manipulation of solids. Another example would be the calculation of NC tool paths to machine the surfaces of a solid by a milling machine. In this application, the information on the surface to be machined and its boundary edges is necessary. Furthermore, the adjacent surfaces need to be derived for the gouging detection. It is not a simple task to derive all these boundary information from a CSG tree representation of a solid.

4.(1)During building the Oct-tree representation of a solid, when deciding the color of an octant (i.e., whether it is inside, outside, or intersecting the boundary of the object), you can not simply decide it by merely looking at the center point of the octant, why? (2)Then, how do you decide the color of the octant? What’s the major disadvantage of it? Answer: (1)Because only decide the center point, the area that inside or outside can not be known clearly. This cause large error. To decide the color of the Octant in the following way : Step 1.Make a hexahedron H that completely encloses the solid Step 2.Divide H into 8 octants Step 3.For each octant obtained in step 2, mark it by one of three colors White: completely outside the solid Black: completely inside the solid Gray : partially inside the solid Step 4:For each gray octant obtained in step 3, go to step 2, until either no gray octants are left or the size of the octant reaches a minimum threshold. (2)The major disadvantage is time consuming.

5. Euler-Poincare formula of solids with voids (1) What is the relationship between V, E, F, H and P of the solid below? (The small cube is a hollow part, also called a void.) (2) Let W be the number of voids in the solid, establish the relationship between V, E, F, H, P, and W. (Hint: first exam the case of two voids and compare with that of a single void.)

Answer : V =24, E = 36, F = 16, H = 2, P = 1, By Euler-Poincare formula ,that is V-E+F-H =2(1-P) However 24-36+16-2= 2 not equal to 2(1-1) Therefore this situation does not satisfy the formula, we add an unknown x to balance the formula, that is V-E+F-H = 2(1-P)+x , here x=2 for one void Assume there are two voids in the solid V=32, E=48, F=22, H=2, P=1,in to above formula x=4, Similarly, for solid with 3 x=6, solid with 4, x=8 We could get the equation: V-E+F-H = 2(1-P)+2W W is the number of the voids