CSCE 441 Lecture 24

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Surfaces

Implicit

Defined by a function; points are not directly defined.

$F(x,y,z)=0$

For example, a sphere is defined by the function

$x^2+y^2+z^2-r^2=0$

Shapes that are easy to define implicitly are:

• spheres
• planes
• cylinders
• cones
• tori

These types of shapes are easy for raytracers to handle.

Intersections

Given a ray $L(t)=P+\overrightarrow{v}t$ that starts at a point $P$ and extends to infinity in the direction of $\overrightarrow{v}$, find the intersection of $L(t)$ with $F(x,y,z)=0$:

Substitute $L(t)$ for $x,y,z$ in the function and solve for $t$ in $F(L(t))=0$.

Example: $L(t)=(0,0,-2)+(0,0,1)t$ and $F(x,y,z)=x^2+y^2+z^2-1=0$

$\begin{array}{rl}F(L(t))=0& =(0+0t{)}^2+(0+0t{)}^2+(-2+1t{)}^2-1\\ & =4-4t+t^2\\ t& =1,3\end{array}$

The solution to the intersection is $L(t)$, so $L(1)=(0,0,-1)$ and $L(3)=(0,0,1)$.

Normals

Given $F(x,y,z)=0$, find the normal at a point $(x,y,z)$

Assume we have a parametric curve $(x(t),y(t),z(t))$ on the surface of $F(x,y,z)$, we set $F(x(t),y(t),z(t))=0$ and differentiate with respect to $t$:

$\begin{array}{rl}\frac{\partial F}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial F}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t}+\frac{\partial F}{\partial z}\frac{\mathrm{d}z}{\mathrm{d}t}& =0\\ ⟨\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z}⟩\cdot ⟨\frac{\mathrm{d}x}{\mathrm{d}t},\frac{\mathrm{d}y}{\mathrm{d}t},\frac{\mathrm{d}z}{\mathrm{d}t}⟩& =0\\ & =\mathrm{\nabla }F\cdot \overrightarrow{v}\end{array}$

This represents conceptually the dot product between what must be the normal of the surface and the slope of a line on the surface (i.e. must be tangent to the surface)

Summary

Advantages

• easy to calculate intersections and normals

Disadvantages:

• hard to calculate points on the surface

Parametric

$P(s,t)=⟨x(s,t),y(s,t),z(s,t)⟩$

Intersections

Set $L(t)=P(u,v)$, and solve a system of three equations (each of $x$, $y$, and $z$) for the parameters $t$, $u$, and $v$.

Plug parameters back into equation to find solution.

Example: $P(u,v)=⟨u,v,u+v⟩$ and $L(t)=(0,0,1)+(-1,0,0)t$

$\{\begin{array}{ll}u& =-t\\ v& =0\\ u+v& =1\end{array}$

Normals

Assume $t$ is fixed. Set $P(s,t)=F(s)$ and differentiate with respect to $s$:

$\frac{\partial P(s,t)}{\partial s}=\frac{\partial F(s)}{\partial s}$

The RHS represents the tangent at $s$

Perform a similar operation with $t$:

$\frac{\partial P(s,t)}{\partial t}=\frac{\partial F(t)}{\partial t}$

The RHS represents the tangent at $t$.

To find the normal, take cross product of tangents:

$\frac{\partial P(s,t)}{\partial s}\times \frac{\partial P(s,t)}{\partial t}$

Summary

Advantages:

• easy to generate points on surface

Disadvantages:

• hard to determine if point is inside or outside
• hard to determine if point is on the surface

Deformed

Given a surface $S$ and a deformation function $D(x,y,z)$, $D(S)$ is a new surface representing the deformed surface.

This is useful for creating complicated shapes from simple objects.

Intersections

1. Assume $D(x,y,z)$ is a simple matrix (e.g. affine transformation)
2. First deform $L(t)$ by $D^{-1}$
3. Calculate intersection with undeformed surface $S$
4. Transform intersection point and normal by $D$.

Example: deformation of a circle that stretches by factor of two in the $x$ direction: $D(x,y)=(2x,y)$

$L(t)=(-1,-1,1)+(1,1,0)t$

$D^{-1}(L(t))=\left[\begin{array}{ccc}\frac{1}{2}& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right](\left[\begin{array}{c}-1\\ -1\\ 1\end{array}\right]+\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]t)$

Normals

Define how tangents transform first. Assume $C(t)$ is a curve on the surface:

$\begin{array}{rl}{C}^{\prime }(t)& \approx \frac{C(t+h)-C(t)}{h}\\ D(C{)}^{\prime }(t)& \approx \frac{D(C)(t+h)-D(C)(t)}{h}\end{array}$

Tangents deform by applying transformation $D$ (multiply by matrix)

Normals and tangents are orthogonal both before and after transformation.

Let $N$ be the normal and $T$ be the tangent:

$\begin{array}{rl}(MN{)}^{T}DT& =0\\ N^T{M}^{T}DT& =0\\ N^T\left(M^{T}D\right)T& =0\\ M^{T}D& =I\\ M& =D^{-T}\end{array}$

Hence

Normals transform by the inverse transpose of the deformation matrix, NOT by the deformation matrix.

(why? normal vectors are covectors, not vectors)

Summary

Advantages

• simple surfaces can represent complex shapes
• affine transformations yield simple calculations
• If we are given $D^{-1}$, we never have to compute any matrix inverse.

Disadvantages