I am working on reverse engineering the animation system of a video game. I initially assumed it used a Bezier curve to interpolate between keyframes, but I've since worked out the curve it actually uses is given by
$$
(2t+1)(t-1)^2\cdot{P_1}+t(t-1)^2\cdot{P_2}+(t-1)t^2\cdot{P_3}+(3-2t)t^2\cdot{P_4}
$$
where $P_1$ and $P_4$ are the starting and ending keyframe values, respectively, and $P_2$ and $P_3$ are control points. This feels awfully close to a cubic Bezier curve, but as I understand it, that would be given by
$$
(1-t)^3\cdot{P_1}+t(1-t)^2\cdot{P_2}+(1-t)t^2\cdot{P_3}+t^3\cdot{P_4}
$$
Does the first curve have a name? Is it related to Bezier curves in any way? Is it a Bezier curve in disguise and I just can't see it?
What kind of animation interpolation curve is this
bezier-curveinterpolationpolynomials
Related Solutions
Another approach. Let
$$
\mathbf{x} = (x_1,\ x_2,\ x_3) = (0.1,\ 0.5,\ 0.9)
$$
and define the following polynomials:
\begin{align}
\theta_1(k) &= c_{11}k^4 + c_{12}k^3 + c_{13}k^2 + c_{14}k + 1,\\
\theta_2(k) &= c_{21}k^4 + c_{22}k^3 + c_{23}k^2 + c_{24}k + 1,\\
\theta_3(k) &= c_{31}k^4 + c_{32}k^3 + c_{33}k^2 + c_{34}k + 1,
\end{align}
where $c$ is the following matrix:
$$
\begin{bmatrix}
0.16321393821380 & -1.14859165394532 & 2.51140325233548 & -2.52602553660396\\
0.26949508735386 & 0.09905839089292 & -0.89714007482260 & -0.47141340342418\\
-2.08424569212321 & 1.94150558704136 & -0.87990952487339 & 0.02264962995523
\end{bmatrix}.
$$
Now, given $k\in[0,1]$, fit a parabola to the points $(x_i, \theta_i(k)),\ i=1,2,3$. That is, let
$$
\Theta(x) = \frac{(x-x_2)(x-x_3)}{(x_1-x_2)(x_1-x_3)}\theta_1(k) +
\frac{(x-x_1)(x-x_3)}{(x_2-x_1)(x_2-x_3)}\theta_2(k) +
\frac{(x-x_1)(x-x_2)}{(x_3-x_1)(x_3-x_2)}\theta_3(k).
$$
Then $y$ as a function of $x$ is approximated by
$$
y = \Theta(x) (-2x + 3x^{2/3}) + (1-\Theta(x)) x.
$$
Below is the fitted function when $k=0.25$ (blue = true y, red = approx.).
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.
Best Answer
This is an interpolation that specifies the endpoints and the derivatives at the endpoints. P1 and P4 are the initial and final values, while P2 and P3 are the derivatives.
Also, which video game? As it happens, I also came across this interpolation while reverse engineering the animation system of a video game.