Get $t$ of ascending Bézier curve from $x$

Question

I have an ascending cubic Bézier curve. ($x_0 \leq x_1 \leq x_2 \leq x_3$) Considering this property, there is always one and only one $y$ value per $x$ value.

The point ($x, y$) along the curve is determined by the following equation, from Wikipedia:

$B(t) = (1-t)^3 P_0 + 3t (1-t)^2 P_1 + 3t^2 (1-t) P_2 + t^3 P_3$

However, this equation takes $t$ as a parameter, which is the progression along the curve, or, as Wikipedia explains,

The $t$ in the function for a linear Bézier curve can be thought of as describing how far $B(t)$ is from $P_0$ to $P_n$. For example when $t=0.25$, $B(t)$ is one quarter of the way from point $P_0$ to $P_n$.

I would like to use the $B(t)$ function with a $x$ parameter instead. Something like $B(x)$, or anything that can convert a given $x$ to a $t$.

Edit: I realized that I could solve $B(t)$ for $t$:

$B(t) = (1-t)^3 P_0 + 3t (1-t)^2 P_1 + 3t^2 (1-t) P_2 + t^3 P_3$

$x(t) = (1-t)^3 x_0 + 3t (1-t)^2 x_1 + 3t^2 (1-t) x_2 + t^3 x_3$

$x(t) = (-x_0+3x_1-3x_2+x_3)t^3 + (3x_0-6x_1+3x_2)t^2 + (-3x_0+3x_1)t + x_0$

$t(x) = ...?$

Edit 2: With the Newton-Raphson method, I managed to reach my goal. I'm attaching a screenshot.

The red dots represent my interpolation (and surprisingly, extrapolation as well. This was an unexpected but great surprise!).

The gray dots represent a small optimization I made to the method: the number of iterations based on the difficulty to approximate. I increase the precision in an invertly proportional relation to the absolute value of the derivative: $$p\ \alpha\ \frac{1}{|B'(t_n)|}$$

I have to tweak the parameters, but so far, this method is efficient and pretty safe, even at a nearly straight slope. Thanks for the help!

P.S.: The more explicit solution, without the precision adjustment (which is unfortunately only well represented in procedural code):

$$t\approx x-\sum_{n=0}^p (\frac{B(t)_x-x}{B'(t)_x})$$

(This takes advantage of the fact that $-x$ is nullified in the derivative Also note that $t$ is reassigned at each step of the sum).

P.P.S.: The accompanying code:

p = 5;
p_max = 100;

t = x;

for (n = 0; n < min(p, p_max); n++)
{
    d = dB(t).x;

    if (p < p_max) p += min(0.02 / d^2), 1);

    t -= (B(t).x - x) / d;
}

return t;

Answer 1

There are closed formulas for solving general cubic equations, but as Wolfram Alpha showed you, they are so horrible (involving complex cube roots even if your coefficients are real) that you might as well not bother, and instead find your $t$ numerically -- such as with bisection or Newton-Raphson.

Answer 2

Lazlo Bonin. Well, it's an equation, solved only with parameter t. In this equation P0, P1, P2, P3 are the points with (X,Y) (for 2D) parametrs. So you need functions that multiply Point and double (f.e. (t^3)*P3 ) Also you need function to sum points. So, after you make all calculations you get point. t is a parameter, in space [0,1]. So you need to make an accuracy or step of increasing parametr t. For example, step 0.01, you get 100 points between your points. Here is how a function of calaculating cube splain can be made:

private List<Point> cube(Point p0, Point p1, Point p2, Point p3)
    {
        List<Point> result = new List<Point>(accuracy + 1);
        double t = 0;
        while (t <= 1)
        {
            Point p = p3.Multiply(t * t * t);
            p = p.Add(p2.Multiply((1 - t) * 3 * t * t));
            p = p.Add(p1.Multiply(3 * t * (1 - t) * (1 - t)));
            p = p.Add(p0.Multiply(Math.Pow(1 - t, 3)));
            result.Add(p);
            t += 1.0 / accuracy;
        }
        return result;
    }