How to control only the “second oscillations” of a curve

algebra-precalculuscurvesfunctionsgraphing-functions

I have the following "elastic out" curve:

$$f_b(t)=1+2^{-10t}\cdot \sin\left(\frac{(t-\frac{b}{4}) \cdot 2π}{b}\right)$$

I only need to use it in the interval $\left(0, 1\right)$ (values for $0$ and $1$ do not matter). It already describes the oscillation after "getting to" $y=1$ quite well, however, I am facing issues trying to design the dampening according to my needs and transforming this curve in general.

Problem

Above you can see what $b$ controls – I would say the dampening of the oscillation but also not really. What I want to have control over ideally is only the dampening about $1$ after the function comes back to $y=1$ for the first time, so I want to keep the initial bump past $y=1$ from say $b=0.4$, but I want more overshoot past $y=1$ afterwards.

This is a terrible sketch, but I hope that you see what I mean. The red graph would be what I have now with $b=0.4$ and the blue graph is what I am trying to achieve.
I tried a bunch of edits to the equation, but nothing I did got me close to what I want. The graph should keep oscillating about $y=1$ after some time.

Observations

Changing the factor before the $sin$ will obviously allow me to stretch and compress the whole graph, but that is not what I want.

Application

The point of this curve is to describe motion from one place to another. An object is supposed to move from say $A$ to $B$. $A$ is at $y=0$ and $B$ is at $y=1$. $y=1.1$ would mean that the object moved past the destination (past $B$), which is why this is "elastic". My goal is to keep the intial elasticity of the bounce, i.e. the overshoot past $B$, but I also want to have more bounciness when the object oscillates about point $B$ after going beyond it initially.

Best Answer

If I am understanding you correctly, you want an envelope function $g$ such that $g(0) = 0$, $g(x)\approx x$ for $x > 1$, but you want $g(x) < x$ in that range $(0, 1)$ so that $g(f_b(x))$ has lower troughs than you presently have.

If we were to use some sort of bell-shaped curve, say, a Gaussian $h(x) = e^{-x^2}$ then we could form such a function as $g(x) = x(1 - \alpha h(x/\beta)$ for two parameters $\alpha$ and $\beta$ that tell us about the strength and the spreadiness of the envelope. Some more shaping can be done by using $(h(x/\beta)/h(0))^n$ or so instead of just $h(x/\beta)$ in there. So for example with $g(x) = x(1 - 0.8 e^{-6 x^6})$ I can get a graph like

(blue is original, black is after the envelope transformation).

The only weird thing about this sort of approach is the way it messes with the slope near $x=0$.

Related Solutions

[Math] Bezier Curves and Acceleration

It depends what you're graphing.

The buttons on the left of your image say "xCurve, yCurve, zCurve", and this suggests that you have a 3D curve, and you are graphing one of the coordinates ($x$) versus a time parameter, $t$.

If so, the graph of the derivative is certainly wrong. It should have the value $0$ when the abscissa (the horizontal axis value, the $t$-value) is around 17 or 53.

On the other hand, your graphs don't look like "$x$ versus $t$" graphs, they look like "$(x,y)$ versus $t$" graphs. If this is the case, then your results might well be correct (though undesirable). See below for details.

Let's start from the beginning with a nice simple notation:

Suppose $P(t)$ is a cubic Bezier, with control points $A$, $B$, $C$, $D$. Then its equation is:

$$P(t) = (1-t)^3A + 3t(1-t)^2B + 3t^2(1-t)C + t^3D \quad (0 \le t \le 1) $$

Then the derivative curve is a quadratic (degree 2) curve, and its control points are $3(B-A)$, $3(C-B)$, $3(D-C)$, so it's equation is:

$$Q(t) = 3(1-t)^2(B-A) + 6t(1-t)(C-B) + 3t^2(D-C) \quad (0 \le t \le 1) $$

All of this applies regardless of whether $A$, $B$, $C$, $D$ are $x$ values or $(x,y)$ values.

If you want to draw "$x$ versus $t$" graphs, then drawing $P$ and $Q$ together on the same graph should be straightforward.

If you want to draw "$(x,y)$ versus $t$" graphs, then putting both $P$ and $Q$ on the same graph is more problematic. Suppose the control points $A$, $B$, $C$, $D$ were a great distance from the origin, but fairly close to each other. Then $B-A$, $C-B$, $D-C$ would be small, so the $Q$ curve would be close to the origin -- far away from the $P$ curve. In your case, it looks like (roughly) $A = (0,0)$ and $B=(32,12)$, so the first control point of the derivative curve $Q$ is $3(B-A) = (96,36)$, which is off the charts. Your derivative graph is clipped, so the end-points of the curve are not visible, which makes it harder to say whether or not it's correct. At least it looks like a parabola, though, which is correct (Bezier curves of degree 2 are parabolas).

These notes might help. Section 2.5 discusses derivatives, and there's a picture showing how the derivative curve relates to the original one (for the xy-vs-t case). Section 2.12 talks about the x-vs-t type of curve (which is variously referred to as a real-valued, explicit, or non-parametric Bezier curve).

[Math] How to plot a smooth curve function from given points

some options in R would be

t=seq(0,10,0.01)
y=sin(t)+rnorm(length(t))
plot(t,y,cex=0.1)

# puts a loess smoother through the points
lines(loess.smooth(t,y),col=2)

library(mgcv)
# puts a spline through the points - read Simon Woods work to learn more about this
g<-gam(y~s(t),data=data.frame(t=t,y=y))
lines(t,g$fitted.values,col=3)

and for your recently added data

t=1:400
y=rep(NA,length(t))
y[10]<-30
y[111]<-100
y[171]<-128
y[181]<-86
y[201]<-42
y[211]<-44
y[281]<-39
y[321]<-59
y[341]<-20
y[351]<-4
library(mgcv)

g<-gam(y~s(t),data=data.frame(t=t,y=y))
lm<-predict.gam(g,data.frame(t=t,y=y))

plot(t,y,cex=0.5,pch=16,ylim=c(min(lm),max(lm)))
lines(t,lm,col='red')