Blog: 3. Module 01
  1. For a bi-linear surface P00(0,0,1), P01(1,1,1), P10(1,0,0), P11(0,1,0). calculate the surface at u=0.5 &v=0.5.

Sol. P00[0,0,1]; P01[1,1,1]; P01=[1,0,0]; P10=[1,0,0]; P11=[0,1,0]. Equation of bi linear surface is given by, P(u,v)=(1-u)(1-v)P00+u(1-v)P10+(1-u)vP10+uvP11 The values for x, y and z coordinates can be determined as, Px(u,v)=(1-u)(1-v)(0)+u(1-v)(1)+(1-u)v(1)+uv(0) =u+v-2uv Py(u,v)=(1-u)(1-v)(0)+u(1-v)(0)+(1-u)v(1)+uv(1) =v Pz(u,v)=(1-u)(1-v)(1)+u(1-v)(0)+(1-u)v(1)+uv(0) =1-u Therefore, P(u,v)=[u+v-2uv v 1-u] At u=0.5 &v=0.5, coordinate of point is, P(u,v)=[0.5 0.5 0.5]

  1. The four points P00=(0,0,1), P10=(1,0,0), P01=(0.5,1,0) and P11=(1,1,0). Find the equation of surface. Ans. P(u,v)=[0.5(1-u)v+u v (1-u)(1-v)]
    *Lofted surface (Ruled Surface):

In a lofted surface, two bounding entitles are defined to patch a surface between them . A lofted surface is a linear surface and is boundary surface. It is bounded by two arbitrary curves denoted by P(u,0) and P(u,1) and by two segments P(0,v) and P(1,v) connecting them. Surface line in the v direction are straight. Therefore surface is created by linearly interpreting between the boundary surface. P(u,v)=(1-v)P(u,0)+vP(u,1)

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modified 3 months ago  • written 3 months ago by gravatar for Nivetha Soundarrajan Nivetha Soundarrajan0
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