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 $L(t)=P+{\vec {v}}\,t$ that starts at a point $P$ and extends to infinity in the direction of ${\vec {v}}$, 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 $F(x,y,z)=x^{2}+y^{2}+z^{2}-1=0$

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

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 $t$:

${\begin{aligned}{\frac {\partial F}{\partial x}}\,{\frac {\mathrm {d} x}{\mathrm {d} t}}+{\frac {\partial F}{\partial y}}\,{\frac {\mathrm {d} y}{\mathrm {d} t}}+{\frac {\partial F}{\partial z}}\,{\frac {\mathrm {d} z}{\mathrm {d} t}}&=0\\\left\langle {\frac {\partial F}{\partial x}},{\frac {\partial F}{\partial y}},{\frac {\partial F}{\partial z}}\right\rangle \cdot \left\langle {\frac {\mathrm {d} x}{\mathrm {d} t}},{\frac {\mathrm {d} y}{\mathrm {d} t}},{\frac {\mathrm {d} z}{\mathrm {d} t}}\right\rangle &=0\\&=\nabla F\cdot {\vec {v}}\end{aligned}}$

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

$P(s,t)=\left\langle x(s,t),y(s,t),z(s,t)\right\rangle$

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: $P(u,v)=\left\langle u,v,u+v\right\rangle$ and $L(t)=(0,0,1)+(-1,0,0)\,t$

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

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: $D(x,y)=(2x,y)$

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

$D^{-1}(L(t))={\begin{bmatrix}{\frac {1}{2}}&0&0\\0&1&0\\0&0&1\end{bmatrix}}\,\left({\begin{bmatrix}-1\\-1\\1\end{bmatrix}}+{\begin{bmatrix}1\\1\\0\end{bmatrix}}\,t\right)$

Normals

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

${\begin{aligned}C'(t)&\approx {\frac {C(t+h)-C(t)}{h}}\\D(C)'(t)&\approx {\frac {D(C)(t+h)-D(C)(t)}{h}}\end{aligned}}$

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:

${\begin{aligned}(M\,N)^{T}\,D\,T&=0\\N^{T}\,M^{T}\,D\,T&=0\\N^{T}\,\left(M^{T}\,D\right)\,T&=0\\M^{T}\,D&=I\\M&=D^{-T}\end{aligned}}$

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