Mathematical Representation of Curves

(a) Parametric equation

$$\mathbf P(u)=\mathbf P_1+u(\mathbf P_2-\mathbf P_1),\qquad 0\le u\le1$$

$\mathbf P_2-\mathbf P_1=(4,\,-4,\,6)$, so

$$x=1+4u,\qquad y=2-4u,\qquad z=3+6u$$

(b) Length and direction cosines

$$L=\sqrt{4^2+(-4)^2+6^2}=\sqrt{68}=8.246$$ $$l=\frac{4}{8.246}=0.485,\quad m=\frac{-4}{8.246}=-0.485,\quad n=\frac{6}{8.246}=0.728$$

Check: $l^2+m^2+n^2=0.235+0.235+0.529=1.0$.

(c) Point at u = 0.25

$$x=1+1=2,\quad y=2-1=1,\quad z=3+1.5=4.5$$

Point $(2,\ 1,\ 4.5)$.

(d) Intersection with z = 6

$3+6u=6 \Rightarrow u=0.5$, which lies in $0\le u\le1$, so the segment does cut the plane.

$$x=1+4(0.5)=3,\quad y=2-4(0.5)=0$$

Answer: L = 8.246; direction cosines (0.485, -0.485, 0.728); P(0.25) = (2, 1, 4.5); intersection with z = 6 at (3, 0, 6).

(a) Circle

$$x=x_c+R\cos\theta,\qquad y=y_c+R\sin\theta,\qquad 0\le\theta\le2\pi$$

With $(x_c,y_c)=(2,3)$, $R=5$: $x=2+5\cos\theta$, $y=3+5\sin\theta$.

Tangent: $\dfrac{dx}{d\theta}=-5\sin\theta,\ \dfrac{dy}{d\theta}=5\cos\theta$. At $60^\circ$: $(-4.330,\ 2.500)$. Its length is 5, and it is perpendicular to the radius vector $(2.5,\ 4.330)$ since the dot product is $-10.83+10.83=0$.

(b) Ellipse

$$x=x_c+a\cos\theta,\qquad y=y_c+b\sin\theta$$

With $a=6$, $b=3$, centre $(1,2)$: $x=1+6\cos\theta$, $y=2+3\sin\theta$.

Check at 45 deg: $\dfrac{(5.243-1)^2}{36}+\dfrac{(4.121-2)^2}{9}=0.5+0.5=1$.

The circle is the special case $a=b=R$. Note that $\theta$ here is a parameter (eccentric angle), not the polar angle of the point of the ellipse.

Answer: circle points (7, 3), (4.5, 7.33), (-0.5, 7.33), (-3, 3); tangent at 60 deg = (-4.33, 2.5); ellipse points (7, 2), (5.243, 4.121), (1, 5).

Conic sections (circle, ellipse, parabola, hyperbola) are the analytic curves used in CAD. Their parametric forms with centre or vertex at the origin are:

• Parabola $y^2=4ax$: $x=at^2$, $y=2at$, $-\infty<t<\infty$.
• Hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$: $x=a\sec\theta$, $y=b\tan\theta$ (or $x=a\cosh t$, $y=b\sinh t$).

Parabola $y^2=12x$

$4a=12\Rightarrow a=3$, so $x=3t^2$, $y=6t$.

Check at $t=2$: $y^2=144=12\times12$.

Tangent slope: $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{6}{6t}=\dfrac1t$. At $t=1$ the slope is 1 (45 deg).

Hyperbola with $a=3$, $b=4$

$x=3\sec\theta$, $y=4\tan\theta$.

Check at 45 deg: $18/9-16/16=2-1=1$. The parameter $\theta$ must stay away from $90^\circ$, where $\sec\theta$ is infinite; this gives only the right branch.

Answer: parabola points (0,0), (0.75,3), (3,6), (12,12), slope at t = 1 is 1; hyperbola points (3,0), (3.464,2.309), (4.243,4.0).

A Hermite cubic segment is defined by two end points $\mathbf P_0,\mathbf P_1$ and two end tangent vectors $\mathbf P_0',\mathbf P_1'$.

Derivation

A cubic parametric curve is

$$\mathbf P(u)=\mathbf a_3u^3+\mathbf a_2u^2+\mathbf a_1u+\mathbf a_0,\qquad 0\le u\le1$$

with tangent

$$\mathbf P'(u)=3\mathbf a_3u^2+2\mathbf a_2u+\mathbf a_1$$

Apply the four boundary conditions:

$$\begin{aligned} \mathbf P(0)&=\mathbf a_0=\mathbf P_0 \\ \mathbf P(1)&=\mathbf a_3+\mathbf a_2+\mathbf a_1+\mathbf a_0=\mathbf P_1 \\ \mathbf P'(0)&=\mathbf a_1=\mathbf P_0' \\ \mathbf P'(1)&=3\mathbf a_3+2\mathbf a_2+\mathbf a_1=\mathbf P_1' \end{aligned}$$

Solving:

$$\begin{aligned} \mathbf a_0&=\mathbf P_0,\qquad \mathbf a_1=\mathbf P_0' \\ \mathbf a_2&=3(\mathbf P_1-\mathbf P_0)-2\mathbf P_0'-\mathbf P_1' \\ \mathbf a_3&=2(\mathbf P_0-\mathbf P_1)+\mathbf P_0'+\mathbf P_1' \end{aligned}$$

Substituting and collecting terms in $\mathbf P_0,\mathbf P_1,\mathbf P_0',\mathbf P_1'$:

$$\mathbf P(u)=F_1\mathbf P_0+F_2\mathbf P_1+F_3\mathbf P_0'+F_4\mathbf P_1'$$

Blending (Hermite) functions

$$\begin{aligned} F_1&=2u^3-3u^2+1 \\ F_2&=-2u^3+3u^2 \\ F_3&=u^3-2u^2+u \\ F_4&=u^3-u^2 \end{aligned}$$

Matrix form

$$\mathbf P(u)=\begin{bmatrix}u^3&u^2&u&1\end{bmatrix} \begin{bmatrix}2&-2&1&1\\-3&3&-2&-1\\0&0&1&0\\1&0&0&0\end{bmatrix} \begin{bmatrix}\mathbf P_0\\\mathbf P_1\\\mathbf P_0'\\\mathbf P_1'\end{bmatrix}$$

Properties

• Passes through both end points (interpolates).
• Tangent direction and magnitude at the ends are controlled; a larger tangent magnitude makes the curve stay longer in that direction.
• $F_1+F_2=1$ for every $u$.
• Two segments are joined with $C^1$ continuity by taking the same tangent at the common point.
• Local control: changing one segment does not change the others.
• Drawback: tangent vectors are not intuitive to the designer, which led to the Bezier form.

Formulae

$$\mathbf P(u)=F_1\mathbf P_0+F_2\mathbf P_1+F_3\mathbf P_0'+F_4\mathbf P_1'$$

with $F_1=2u^3-3u^2+1$, $F_2=-2u^3+3u^2$, $F_3=u^3-2u^2+u$, $F_4=u^3-u^2$. The tangent uses the derivatives $F_1'=6u^2-6u$, $F_2'=-6u^2+6u$, $F_3'=3u^2-4u+1$, $F_4'=3u^2-2u$.

Since $\mathbf P_0=(0,0)$:

$$\mathbf P(u)=F_2(4,2)+F_3(4,4)+F_4(4,-4)$$

At u = 0.5

$F_2=0.5$, $F_3=0.125-0.5+0.5=0.125$, $F_4=0.125-0.25=-0.125$.

$$\begin{aligned} x&=0.5(4)+0.125(4)-0.125(4)=2.0 \\ y&=0.5(2)+0.125(4)-0.125(-4)=1+0.5+0.5=2.0 \end{aligned}$$

At u = 0.25

$F_2=-0.03125+0.1875=0.15625$; $F_3=0.015625-0.125+0.25=0.140625$; $F_4=0.015625-0.0625=-0.046875$.

$x=0.15625(4)+0.140625(4)-0.046875(4)=0.625+0.5625-0.1875=1.000$

$y=0.15625(2)+0.140625(4)+0.046875(4)=0.3125+0.5625+0.1875=1.0625$

At u = 0.75

$F_2=-0.84375+1.6875=0.84375$; $F_3=0.421875-1.125+0.75=0.046875$; $F_4=0.421875-0.5625=-0.140625$.

$x=0.84375(4)+0.046875(4)-0.140625(4)=3.000$

$y=0.84375(2)+0.046875(4)+0.140625(4)=1.6875+0.1875+0.5625=2.4375$

Tangent vectors

Using $\mathbf P'(u)=F_2'(4,2)+F_3'(4,4)+F_4'(4,-4)$:

Example at $u=0.5$: $F_2'=1.5$, $F_3'=-0.25$, $F_4'=-0.25$, so $P'_y=1.5(2)+(-0.25)(4)+(-0.25)(-4)=3-1+1=3$ and $P'_x=4(1.5-0.25-0.25)=4$.

End conditions

At $u=0$: $F_1=1$, others 0, so $\mathbf P=(0,0)$ and $\mathbf P'=(4,4)$. At $u=1$: $F_2=1$, so $\mathbf P=(4,2)$ and $\mathbf P'=(4,-4)$. Both are satisfied.

Answer: P(0.25) = (1.00, 1.0625), P(0.5) = (2.00, 2.00), P(0.75) = (3.00, 2.4375); tangents (4, 4.25), (4, 3), (4, 0.25).

A Bezier curve is a parametric curve defined by a set of control points forming a control polygon. The curve approximates the polygon: it starts at the first and ends at the last control point, and is pulled towards the other points. It was developed by Pierre Bezier at Renault.

Equation

For $n+1$ control points $\mathbf B_0,\dots,\mathbf B_n$ (degree $n$):

$$\mathbf P(u)=\sum_{i=0}^{n}\mathbf B_i\,J_{n,i}(u),\qquad 0\le u\le1$$

with the Bernstein basis

$$J_{n,i}(u)=\binom{n}{i}u^i(1-u)^{n-i}=\frac{n!}{i!(n-i)!}u^i(1-u)^{n-i}$$

For a cubic ($n=3$): $J_{3,0}=(1-u)^3$, $J_{3,1}=3u(1-u)^2$, $J_{3,2}=3u^2(1-u)$, $J_{3,3}=u^3$.

Properties

1. The curve is of degree one less than the number of control points.
2. It passes through the first and last control points: $\mathbf P(0)=\mathbf B_0$, $\mathbf P(1)=\mathbf B_n$.
3. End tangents lie along the first and last polygon sides: $\mathbf P'(0)=n(\mathbf B_1-\mathbf B_0)$, $\mathbf P'(1)=n(\mathbf B_n-\mathbf B_{n-1})$.
4. Convex hull property: the curve lies inside the convex hull of its control points, because the basis functions are non-negative and sum to 1.
5. Variation diminishing: the curve does not cross any straight line more times than the control polygon does.
6. Symmetry: reversing the order of the points gives the same curve.
7. Invariant under affine transformation: transform the control points and redraw.
8. Global control: moving one control point changes the whole curve.

Limitations

• Degree rises with the number of points, so many points give a high-degree, costly curve.
• No local control.
• Cannot represent exact conics such as a circle (rational Bezier is required).
• Joining segments with $C^1$ continuity needs the three points around the joint to be collinear.

Equation

$$\mathbf P(u)=(1-u)^3\mathbf B_0+3u(1-u)^2\mathbf B_1+3u^2(1-u)\mathbf B_2+u^3\mathbf B_3$$

At u = 0.5

Weights: $0.125,\ 0.375,\ 0.375,\ 0.125$.

$$\begin{aligned} x&=0.125(0)+0.375(1)+0.375(4)+0.125(5)=0.375+1.5+0.625=2.5 \\ y&=0.125(0)+0.375(3)+0.375(4)+0.125(0)=1.125+1.5=2.625 \end{aligned}$$

At u = 0.25

Weights: $(0.75)^3=0.421875$; $3(0.25)(0.75)^2=0.421875$; $3(0.25)^2(0.75)=0.140625$; $(0.25)^3=0.015625$.

$$\begin{aligned} x&=0.421875(1)+0.140625(4)+0.015625(5)=0.421875+0.5625+0.078125=1.0625 \\ y&=0.421875(3)+0.140625(4)=1.265625+0.5625=1.828125 \end{aligned}$$

End tangents

$$\mathbf P'(0)=3(\mathbf B_1-\mathbf B_0)=3(1,3)=(3,\ 9),\qquad \mathbf P'(1)=3(\mathbf B_3-\mathbf B_2)=3(1,-4)=(3,\ -12)$$

De Casteljau check at u = 0.5

Repeated mid-points:

The result $(2.5,\ 2.625)$ equals the formula value.

Answer: P(0.25) = (1.0625, 1.828); P(0.5) = (2.5, 2.625); P'(0) = (3, 9); P'(1) = (3, -12).

A B-spline curve is a piecewise polynomial parametric curve defined by control points and B-spline basis functions. It is a generalisation of the Bezier curve and removes its two main weaknesses: global control and the link between the number of points and the degree.

Equation

$$\mathbf P(u)=\sum_{i=0}^{n}\mathbf B_i\,N_{i,k}(u)$$

$N_{i,k}$ are basis functions of order $k$ (degree $k-1$) given by the Cox-de Boor recursion over the knot vector $\{t_0,t_1,\dots\}$.

Terms

• Control points $\mathbf B_i$: vertices of the control polygon; the curve is attracted to them but usually does not pass through them.
• Knot vector: non-decreasing parameter values where the polynomial pieces join. Number of knots $=n+k+1$ for $n+1$ points. Uniform, open-uniform (end knots repeated $k$ times) and non-uniform vectors are used.
• Degree: polynomial degree of each segment (cubic is common); order $=$ degree $+1$.

Difference

Advantages

1. Local control: moving one point changes only $k$ segments.
2. Degree stays low (cubic) for any number of points.
3. High continuity ($C^2$ for cubic) at joints.
4. Convex hull and variation diminishing properties.
5. NURBS (rational B-splines) can represent exact conics and free-form curves in one form.

Matrix form

For one uniform cubic segment ($0\le u\le1$):

$$\mathbf P(u)=\frac16\begin{bmatrix}u^3&u^2&u&1\end{bmatrix} \begin{bmatrix}-1&3&-3&1\\3&-6&3&0\\-3&0&3&0\\1&4&1&0\end{bmatrix} \begin{bmatrix}\mathbf P_0\\\mathbf P_1\\\mathbf P_2\\\mathbf P_3\end{bmatrix}$$

This is equivalent to the blending functions

$$\begin{aligned} N_0&=\tfrac16(1-u)^3 \\ N_1&=\tfrac16(3u^3-6u^2+4) \\ N_2&=\tfrac16(-3u^3+3u^2+3u+1) \\ N_3&=\tfrac16u^3 \end{aligned}$$

Weights

Points

At $u=0$:

$$x=\frac{1+4(2)+5}{6}=\frac{14}{6}=2.333,\qquad y=\frac{1+4(4)+5}{6}=\frac{22}{6}=3.667$$

At $u=0.5$:

$$\begin{aligned} x&=0.0208(1)+0.4792(2)+0.4792(5)+0.0208(7)=3.521 \\ y&=0.0208(1)+0.4792(4)+0.4792(5)+0.0208(2)=4.375 \end{aligned}$$

At $u=1$:

$$x=\frac{2+4(5)+7}{6}=4.833,\qquad y=\frac{4+4(5)+2}{6}=4.333$$

Tangents

$\mathbf P'(0)=\tfrac12(\mathbf P_2-\mathbf P_0)=\tfrac12(4,4)=(2,\ 2)$

$\mathbf P'(1)=\tfrac12(\mathbf P_3-\mathbf P_1)=\tfrac12(5,-2)=(2.5,\ -1)$

The curve does not touch the control points; it starts at $(2.333, 3.667)$, a weighted average of $\mathbf P_0,\mathbf P_1,\mathbf P_2$.

Answer: P(0) = (2.333, 3.667); P(0.5) = (3.521, 4.375); P(1) = (4.833, 4.333); tangents (2, 2) at u = 0 and (2.5, -1) at u = 1.

Relation between the forms

A cubic Bezier curve with control points $\mathbf B_0..\mathbf B_3$ has end points $\mathbf P(0)=\mathbf B_0$, $\mathbf P(1)=\mathbf B_3$ and end tangents

$$\mathbf P'(0)=3(\mathbf B_1-\mathbf B_0),\qquad \mathbf P'(1)=3(\mathbf B_3-\mathbf B_2)$$

If this curve is to equal the Hermite curve, the end conditions must be identical:

$$\mathbf B_0=\mathbf P_0,\quad \mathbf B_3=\mathbf P_1,\quad \mathbf B_1=\mathbf P_0+\tfrac13\mathbf P_0',\quad \mathbf B_2=\mathbf P_1-\tfrac13\mathbf P_1'$$

Both are cubics with the same four end conditions, so they are the same curve.

Control points

$$\begin{aligned} \mathbf B_0&=(0,0) \\ \mathbf B_1&=(0,0)+\tfrac13(3,6)=(1,\ 2) \\ \mathbf B_2&=(6,0)-\tfrac13(3,-6)=(5,\ 2) \\ \mathbf B_3&=(6,0) \end{aligned}$$

Point at u = 0.5

Weights $0.125, 0.375, 0.375, 0.125$:

$$x=0.375(1)+0.375(5)+0.125(6)=3.0,\qquad y=0.375(2)+0.375(2)=1.5$$

Point at u = 0.25

Weights $0.421875,\ 0.421875,\ 0.140625,\ 0.015625$:

$$\begin{aligned} x&=0.421875(1)+0.140625(5)+0.015625(6)=1.2188 \\ y&=0.421875(2)+0.140625(2)=1.125 \end{aligned}$$

Answer: Bezier control points (0,0), (1,2), (5,2), (6,0); P(0.5) = (3.0, 1.5); P(0.25) = (1.219, 1.125).