Deriving the best interpolation function

I wanted to play around with a simple VR app with fun affects around the hand detections from the Quest 3 headset.

I had plans for all kinds of effects I wanted to do, and I started with this simple sphere following my finger. But it didn’t look nice enough at first!

So I became interested in deriving a smoothstep function by myself.

Detour - deriving smoothstep ourselves

Idea - smoothness at the edges, constant speed inbetween. Because what if we want to travel large distances

Constraints:

• $f(0)=0$
• $f'(0) = 0$
• $D = f(p) + f'(p) * (T - 2p) + f(p)$
• Where $0 \leq D, 0\leq T, 0\leq p \leq T/2$
• Given $D$, $T$, and $P$ as input constraints, we then have 1 less degree of freedom for whatever parameters define $f(x)$
• Controlling for a, we get a set of curves, which all satisfy our needs
• Another thing: $f''(p) \geq 0$, so that our function on “warmup” is always curved in

Using a cubic function:

The function is nice! Desmos here: https://www.desmos.com/calculator/rlfikf4yzl

Also as a fun side-thing I just made a desmos visual of all functions from Inigo Quilez’s post on smoothstep functions : https://www.desmos.com/calculator/uo1b8hmmgy

But then I’ve got a problem - this function is complicated to compute the inverse of, which I’ll need later. There are formulas for computing the inverse for the cubic from above, and Inigo Quilez even did a post about it. But I don’t want to integrate it because it’s not something I want to derive right now

So I tried a couple more families of functions:

• $b(1 - \cos(ax))$
• $ax^b$

This is all fun & stuff, but the section is called a detour because this function specification isn’t very useful. The speed of the middle section is derived, not an input parameter. and the “warmup distance” isn’t an input parameter.

Deriving a “warmup/warmdown” function

The purpose is to get a function with specific warmup / warmdown properties. I.e.:

• The warmup/warmdown must take time $T$
• During the warmup/warmdown, distance $D$ should be traveled
• By the end of warmup we should have speed $V$

Or:

Didn’t manage to derive this for a cubic $ax^3 + bx^2$ or a trigonometrical $b(1 - \cos(ax))$

But it was very doable for $ax^b$ !

Here’s the Desmos: https://www.desmos.com/calculator/liqymuycjg

Assume current position is $x$ , target position is $t$, and current speed is $s$

$d = ||x-t||$ – the distance to target

• If $d \geq D$:

  • if $v = V$ , then we’re up to speed and don’t need to do anything, just update position

  • i.e. we’re not close enough to do “warmdown” yet, we should do “warmup”:

    $$\begin{equation} \begin{aligned} x_{\text{new}} \coloneqq & x + v \Delta t \\ v_{\text{new}} \coloneqq & f'(f'^{-1}(v) + \Delta t) \end{aligned} \end{equation}$$

• Otherwise we need to do “warmdown”, strictly according to the function:

  $$\begin{equation} \begin{aligned} x_{\text{new}} \coloneqq & f(f^{-1}(d) - \Delta t) \\ v_{\text{new}} \coloneqq & f'(f^{-1}(d) - \Delta t) \end{aligned} \end{equation}$$

Voila! It works