bezier-curve
I have a cubic Bezier curve, and I need to divide it and create same curve between point on the original curve and the end point of the original curve.
From my research, I found the DeCasteljau algorithm helps in doing this. But, to use this algorithm, I need to know the '$t$' value at the dividing point.
Can anyone please help me out to find '$t$' at a given point on a cubic Bezier curve
I need to find the value of '$t$' at a given point B.
I think that I understand your confusion. Let me know if this works.
For a Bezier curve defined as you did, you have a parametric form:
$$x(t)=(1-t)^3\cdot 0+3t(1-t)^2\cdot .2+3t^2(1-t)\cdot.5+t^3\cdot 1$$
and
$$y(t)=(1-t)^3\cdot 0+3t(1-t)^2\cdot .5+3t^2(1-t)\cdot.9+t^3\cdot 1$$
When you plug in $t=1/2$, you get the point $(0.3875,0.65)$, which is not the point that you're looking for.
I think that you're interested in the case where $x=\frac{1}{2}$. Using Maple, I solved $x(t)=\frac{1}{2}$, and found that this happens when $t\approx0.60969549401666900$.
Plugging this $t$ value into $y(t)$ results in $\approx0.7576964125$, which, I think, is the value you're looking for.
Given 2D cubic Bezier segment $\mathcal{B}(t,A,B,C,D)=A\,(1-t)^3+3B\,(1-t)^2t+3C\,(1-t)t^2+D\,t^3$, $t\in[0,1]$ and line segment $\mathcal{L}(t,A,D)=A\,(1-s)+D\,s$, $s\in[0,1]$, the value of $t:\mathcal{B}(t,A,B,C,D)=\mathcal{L}(s,A,D),\ t\ne0, t\ne1$ is the same as the value of $t:\mathcal{B}(t,0,B-A,C-A,D-A)=\mathcal{L}(s,0,D-A)$. So, with a substitution $b=B-A,\ c=C-A,\ d=D-A$ we can solve a system of two equations with two unknowns $t,s$:
\begin{align} \mathcal{B}(t,0,b_x,c_x,d_x) &= \mathcal{L}(s,0,d_x) \\ \mathcal{B}(t,0,b_y,c_y,d_y) &= \mathcal{L}(s,0,d_y) \end{align}
which gives the value of parameter $t$ as
\begin{align} t &= \frac{b_x\,d_y-d_x\,b_y}{b_x\,d_y-c_x\,d_y-d_x\,b_y+d_x\,c_y} \end{align}
If $0<t<1$ than the intersection point of $\mathcal{B}$ and $\mathcal{L}$ is $X=\mathcal{B}(t,A,B,C,D)$.
Best Answer
Suppose we let $\mathbf{P}(t)$ be the Bezier curve, and let the given point be $\mathbf{B}$.
In theory, we want to find a value of $t$ such that $\mathbf{P}(t) = \mathbf{B}$. If you write out the $x$ and $y$ components separately, this will give you two cubic equations, which should have a common root, $t$.
But, in floating point arithmetic, $\mathbf{B}$ will be some small distance away from the curve, and we need to accomodate this. So, a better approach is to find the point on the curve that is closest to $\mathbf{B}$. In other words, find the value of $t$ that minimises the distance from $\mathbf{B}$ to $\mathbf{P}(t)$. At the point where this happens, the vector $\mathbf{P}(t) - \mathbf{B}$ will be perpendicular to the curve tangent, so $$ (\mathbf{P}(t) - \mathbf{B}) \cdot \mathbf{P}'(t) = 0 $$ Since $\mathbf{P}(t)$ is cubic and $\mathbf{P}'(t)$ is quadratic, this is an equation of degree 5 in $t$, which you can solve using your favorite numerical root-finder. This is the brain-dead brute force approach.
Now a smarter way ...
If you look in these notes by Tom Sederberg, you will find some techniques for "inversion" of polynomial and rational curves, based on the use of resultants. Specifically, on page 200, you will find a formula for doing "inversion" of a cubic curve written in Bezier form, and there's a nice worked example on page 201. This is exactly what you need.
As I mentioned above, you will often meet cases where the given point $\mathbf{B}$ is not exactly on the curve. You can apply the inversion formula, anyway, and it will give you some value of $t$. I'm not 100% sure what point this value of $t$ represents. I don't think it's the point on the curve that's closest to $\mathbf{B}$, but it should be good enough if $\mathbf{B}$ is already very close to the curve.