Using the diagram above, calculating a point on a quadratic Bezier curve at time $t=0.6$ using the De Casteljau Algorithm is done like this:
- Linearly interpolate between $P_0$ and $P_1$ for time $t=0.6$ to get a point $Q_0$.
- Linearly interpolate between $P_1$ and $P_2$ for time $t=0.6$ to get a point $Q_1$.
- Lastly, linearly interpolate between $Q_0$ and $Q_1$ for time $t=0.6$ to get the point on the curve $R$, where the black dot on the blue line is.
Is there any value or meaning in using different $t$ values for each linear interpolation instead of using $t=0.6$ for each of them?
For instance, one modification could be to square the weight when interpolating between $Q_0$ and $Q_1$. It would still have the property of going from 0 to 1, but would do so non linearly.
Other modifications may include using $1-t$, using a constant, or maybe using $t*0.5+0.5$.
Is there any value to doing this which a regular bezier curve couldn't give you? Is there a more generalized term which includes this type of a curve or modified version of De Casteljau's Algorithm?
In short, I'm just wondering if this sort of thing is a known thing with a known name, and what sort of functionality it can get you.
As an alternative to this algorithm, you can also write an equation to evaluate Bezier curve points in Bernstein form like the below, where $s = 1-t$, and $A,B,C$ are the control points of the curve:
$P = As^2 + 2Bt + Ct^2$
Which can also be seen mathematically as a linear interpolation between two linear interpolations, just like the De Casteljeau algorithm, of course.
If using $(1-t)$ instead of $t$ for the lerp between $Q_0$ and $Q_1$, and arrange it back into Bernstein form I get the below:
$P = Bs^2 + (A+C)st + Bt^2$
That indicates that at least for this modification, the result is still a Bezier curve of the same degree, but the starting and ending control point both have the value of B, while the middle control point has a value of (A+C).
On the other hand, if i use a constant value of $t = 0.7$ for the last interpolation I get this:
$P = 0.3As - (0.4B - 0.7C)t + 0.7B$
Which puts us at a linear interpolation (a linear Bezier curve), plus a constant.
If i use $s^2$ and $t^2$ instead of $s$ and $t$ for the last interpolation, then put it back in Bernstein form, I get the below, which is a cubic Bezier curve with control points: $A$, $B/3$, $B/3$, $C$.
$P = As^3 + B/3 * 3s^2t + B/3 * 3st^2 + Ct^3$