# Is it possible to do an integration over a Bézier curve "area"?

I have a shape defined by a Bézier spline that has a width, and I want to give it an attractive force.

Is such a thing even doable without approximating it?

Edit:

To integrate the area, I think I will first have to express the answer for a line (crossection) of the path, then integrate it through the path.

Edit:

Moving into the teritory of actually answering the question, but I think I'll keep editing the question until I actually have a final answer.

So, now I want to rename the spline, it will new be defined by $$Q(t) = (1-t)^3\ Q_0+(1-t)^2t\ Q_1+(1-t)t^2\ Q_2+t^3\ Q_3$$ rearranging gives $$Q_3\ t^3+Q_2\ (t^2-t^3)+Q_1(t-2t^2+t^3)+Q_0(1-3t+3t^2-t^3)\\=\\(\underbrace{Q_3-Q_2+Q_1-Q_0}_{\mathcal{A}})\ t^3+(\underbrace{Q_2-2Q_1+3Q_0}_{\mathcal{B}})\ t^2+(\underbrace{Q_1-3Q_0}_{\mathcal{C}})\ t+Q_0$$

Finding the speed vector and its square $$Q′(t)$$:$$Q′(t)=3\mathcal{A}t^2+2\mathcal{B}t+\mathcal{C}$$ $$(Q′(t))^2=9\mathcal{A}^2t^4+4\mathcal{B}^2t^2+\mathcal{C}^2+12\mathcal{A}\mathcal{B}t^3+6\mathcal{A}\mathcal{C}t^2+4\mathcal{B}\mathcal{C}t$$

Now I'm going to find $$P_0(t)$$ and $$P_1(t)$$.

Then, I want to rotate the speed vector left and right, putting $$P_0$$ on the left side and $$P_1$$ on the right. Rotating a vector $$(x,\ y)$$ gives $$(-y,\ x)$$. Rotating right gives $$(y,\ -x)$$.

We also devide by the length of the speed vector to get a unit vector. Then multiplying with the spline width $$W$$.

We name the length of the speed vector $$\Gamma(t)=\sqrt{Q′_x(t)^2+Q′_y(t)^2}$$

We get $$P_0(t) = Q(t) + W{(-Q′_y(t),\ Q′_x(t))\over\Gamma(t)}\\[1.5em]P_1(t) = Q(t) + W{(Q′_y(t),\ -Q′_x(t))\over\Gamma(t)}$$

Can I now just do this? $$\iint{1 \over \lVert P(u,t)\rVert ^2}\Gamma(t)\ du\ dt$$

I'm trying to express: the force of attraction by a spline cross section times how far the cross section moves at time $$t$$.

• If you want to view the object as being "quite thin", then you should content yourself to integrate using a one-dimensional path integral along the bezier curve. On the other hand, if you want to give it some appreciable thickness, you will need to parameterize the region about the curve. As long as the thickness is not larger than the radius of curvature, this parameterization should not be difficult: let one parameter $s$ go "along the curve" and let another parameter $t$ go "perpendicular" to the curve. Commented Oct 26, 2020 at 14:29

## 2 Answers

Take a look at the following illustration featuring a curve similar to the one you have shown and the acceleration vectors that have the nice (ill-known) representation:

$$a=(1-t)(A-2B+C)+t(B-2C+D) \ \ \text{for} \ \ t \in [0,1]$$

Isn't it a tool for expressing what you call attraction ?

Matlab program for the figure:

clear all;close all,hold on;
P=[(4-2*i) (5-i) (5+i) (3+2i)]; % points A,B,C,D
plot(P,'b');
t=0:0.01:1;s=1-t;
m=P(1)*s.^3+P(2)*3*s.^2.*t+P(3)*3*t.^2.*s+P(4)*t.^3;
plot(m,'r');
a=(1-t)*(P(1)-2*P(2)+P(3))+t*(P(2)-2*P(3)+P(4));
quiver(real(m),imag(m),real(a),imag(a),1); % "acceleration field"

• I like your motto "all I need is a good idea"... Commented Oct 26, 2020 at 18:09
• See this. Commented Oct 26, 2020 at 18:14

Reposting an answer that I got on paper for posterity.

### Part 1

$$C(t)=P_0B_{0,3}+P_1B_{1,3}+P_2B_{2,3}+P_3B_{3,3}$$ where B is a Bernstein polynomial

$$t\in[0,1]$$ Force on the line: $$F_{line}=\int_{t=0}^1{1\over(Q(t))^2}\ dt\\[2em] =\int_{t=0}^1{1\over((1-t)^3\ Q_0+(1-t)^2t\ Q_1+(1-t)t^2\ Q_2+t^3\ Q_3))^2}\ dt$$
Assume: Roots $$\alpha$$, $$\beta$$, $$\gamma$$ of $$Q(t)$$ exist {1\over((t-\alpha)(t-\beta)(t-\gamma))^2}\\[2.5em]=\overbrace{{A\over(t-\alpha)}+{B\over(t-\beta)}+{C\over(t-\gamma)}+{D\over(t-\alpha)^2}+{E\over(t-\beta)^2}+{F\over(t-\gamma)^2}}^{constants}\\[3.5em] \begin{align}\int_0^1{A\over t-\alpha}\ dt &= A(log(1-\alpha)-log(-\alpha))\\[1.5em] \int_0^1{D\over (t-\alpha)^2}\ dt &= \left[\begin{array}{}u=t-\alpha\\du=dt\\t=0\implies u=-\alpha\\t=1\implies u=1-\alpha\end{array}\right]\\[1.5em]&=D\int_{-\alpha}^{1-\alpha}{1\over u^2}\ du=D\left[{-1\over u}\right]_{-\alpha}^{1-\alpha}\\[1.5em]&=D\left({-1\over 1-\alpha}+{1\over\alpha}\right)\end{align}

So $$\int_0^1{1\over(Q(t))^2}\ dt$$ turns into six solvable integrals.

As an approximation of your desired force over the area, you could do $$F_{area}\approx wF_{line}$$. $$w$$ is width of band.

### Part 2

More serious attempt. No approximations.

Say we have two equations $$Q_l(t)$$ and $$Q_r(t)$$ that fulfill $${Q_l(t)+Q_r(t)\over2}=Q(t)$$

$$Q_l(t) = Q(t) + \delta\ \vec{v}(t)$$

$$Q_r(t) = Q(t) - \delta\ \vec{v}(t)$$

$$\vec{v}(t)$$ is always orthogonal to $$Q(t)$$ and has length 1

$$\vec{v}(t)={\vec{T}′(t)\over\lVert\vec{T}′(t)\rVert},\ \text{where}\ \vec{T}(t)={\vec{Q}′(t)\over\lVert\vec{Q}′(t)\rVert}$$

So $$Q_l(t)$$ and $$Q_r(t)$$ exist and should be not too hard to find. We could also construct them by moving $$Q$$'s control points.

Now introduce a new variable $$\omega$$ that goes from 0 to 1, and swap out $$Q(t)$$ in the original problem with $$(1-\omega)\ Q_l(t)+\omega\ Q_r(t)$$. We can set up the integral $$\int_{\omega=0}^1\int_{t=0}^1{1\over((1-\omega)\ Q_l(t)+\omega\ Q_r(t))^2}\ dt\ d\omega$$

Solve for the $$\omega$$ integral first \begin{align}\int_{\omega=0}^1{1\over((Q_r-Q_l)\ \omega+Q_l)^2}\ d\omega&=\int_0^1{1\over((Q_r-Q_l)(\omega+{Q_l\over{Q_r-Q_l}}))^2}\ d\omega\\[2.5em]&={1\over(Q_r-Q_l)^2}\int_0^1{1\over(\omega+{Q_l\over Q_r-Q_l})^2}\ d\omega\\[2.5em]&= {1\over(Q_r-Q_l)^2}\left[{-1\over(1+{Q_l\over Q_r-Q_l}}+{Q_r-Q_l\over Q_l}\right]\\[2.5em]&={-1\over((Q_r-Q_l)^2+Q_l(Q_r-Q_l))}+{1\over Q_l(Q_r-Q_l)}\\[2.5em]&={-1\over Q_r^2-Q_rQ_l}+{1\over Q_lQ_r-Q_l^2}\\[2.5em]&={1\over Q_rQ_l-Q_r^2}+{1\over Q_lQ_r-Q_l^2}\end{align}

Two very similar terms. Let's solve the integral for one of them, the other will be analogous. $$\require{cancel}\int_{t=0}^1{1\over Q_lQ_r-Q_l^2}\ dt\\[2.5em] {1\over Q_lQ_r-Q_l^2}={1\over(Q(t)+\delta\ v(t))\ (Q(t)-\delta\ v(t))-(Q(t)+\delta\ v(t))^2}\\[2.5em]{1\over\cancel{Q^2}-(\delta\ v)^2-\cancel{Q^2}-2Q\delta\ v- (\delta\ v)^2}\\[2.5em]={-1\over2(Q\delta\ v+(\delta\ v)^2)}$$

So you get $$-{1\over2\delta}\int_0^1{1\over Q(t)v(t)+\delta\ v(t)^2}\ dt$$

$$v(t)$$ should be 1st order polynomial. Denominator is then 4th order polynomial. Its roots can be found with some closed form equation. $$-{1\over2\delta}\int_0^1{1\over(t-\epsilon)(t-\zeta)(t-\eta)(t-\theta)}\ dt$$

The integrand can be split into partial fractions, and we get $$-{1\over2\delta}\left(\int_0^1\overbrace{{G\over(t-\epsilon)}\ dt+\int_0^1{H\over(t-\zeta)}\ dt+\int_0^1{I\over(t-\eta)}\ dt+\int_0^1{J\over(t-\theta)}}^{constants}\ dt\right)$$

Each integral is solvable with variable substitution like we did before, we get $$F_{Area}={-1\over2\delta}\left(G(log(\ldots)-log(\ldots))+H(log(\ldots)\cdots\right)$$

• I'm struggling to find $v(t)$. I get a difficult 4th degree polynomial under the square root already in $T(t)$. Commented Nov 10, 2020 at 11:19