I believe the function $f:\mathbb R\to\mathbb R$ with
$$f(x)=(-1)^{\lfloor x/2\rfloor} \sqrt{1-(x-2\lfloor x/2\rfloor -1)^2}$$
will do the trick. Here $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$.
Explanation:
(1) The semicircle: An equation for the circle of radius $r$ centered at $(a,b)$ is $(x-a)^2+(y-b)^2=r^2$, so the graph of the function $s:[0,2]\to\mathbb R$ with $$s(x) = \sqrt{1-(x-1)^2}$$ is the upper semicircle of radius $1$ centered at $(1,0)$ (to see this, solve the first equation for $y$ with $y\geq 0$ and put $a=1,b=0,r=1$).
(2) The periodicity: If $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$, then $\lfloor x\rfloor \leq x <\lfloor x\rfloor + 1 $, so $0\leq x- \lfloor x\rfloor <1$ (the expression $x- \lfloor x\rfloor$ is often called the fractional part of $x$). The graph of the function $p:\mathbb R\to\mathbb R$ with $p(x) = x- \lfloor x\rfloor$ looks like a sawtooth wave. It coincides with the graph of $y=x$ on $[0,1)$ and is periodic with period $1$. So the graph of the equation $y/2 = p(x/2)$, which can be written $y=2p(x/2)$ looks similar but is scaled out by a factor of $2$ both horizontally and vertically; it coincides with the line $y=x$ on $[0,2)$ and is periodic with period $2$. Note that $$2p(x/2)=2(x/2-\lfloor x/2\rfloor) = x-2\lfloor x/2\rfloor$$
(3) Combining them: Now $2p(x/2)$ has the effect of mapping each interval $[2n, 2n+2)$ monotonically onto $[0,2)$, so we see that plugging this into $s$ gives a function $g:\mathbb R\to\mathbb R$ whose graph is a collection of semicircles over each interval $[2n,2n+2)$; we have
$$g(x) = s(2p(x/2)) = \sqrt{1-(2p(x/2)-1)^2}=\sqrt{1-(x-2\lfloor x/2\rfloor-1)^2}$$
(4) Flipping: Finally, note that the expression $(-1)^{\lfloor x/2\rfloor}$ alternates between $1$ and $-1$ on successive intervals $[2n,2n+2)$, so the function $f(x) = (-1)^{\lfloor x/2\rfloor}g(x)$ is what you want: $$f(x)=(-1)^{\lfloor x/2\rfloor}\sqrt{1-(x-2\lfloor x/2\rfloor-1)^2}$$
The factor $(-1)^{\lfloor x/2\rfloor}$ has the effect of flipping every other upper semicircle on the graph of $g$ to a "lower" semicircle for the graph of $f$, as desired.
Interpretation of your question:
You want to plot the graph
$$r = 14+0.69 \theta^{1.8} \qquad \theta \text{ from }4\pi \text{ to }0$$
relative to a moving origin that travels at constant speed in a straight line from $(0,0)$ to $(-15,-3)$.
Let the position of the moving origin be $\mathbf p$, with respect to a fixed origin at $(0,0)$. So $\mathbf p = \mathbf p(\theta)$ moves in a straight line at constant speed from $(0,0)$ to $(-15,-3)$ as $\theta$ varies from $4\pi$ to $0$. It follows that
$$\mathbf p(\theta) = \bigg(\frac {\theta}{4\pi}-1\bigg) (15,3)$$
Next, with respect to this moving origin (which you might think of as a "moving frame relative to an inertial frame"), the Cartesian coordinates of your spiral is given by
$$\mathbf r = (r\cos \theta, r\sin\theta)$$
where $(r,\theta)$ are related by $r = 14+0.69 \theta^{1.8}$. Plug this in to get
$$\mathbf r(\theta) = \big((14+0.69 \theta^{1.8})\cos \theta, (14+0.69 \theta^{1.8})\sin\theta \big)$$
Finally, the $(x,y)$ coordinates of your spiral relative to the moving origin is given by
$$\mathbf p(\theta) + \mathbf r(\theta) = \bigg(15\bigg(\frac {\theta}{4\pi}-1\bigg)+(14+0.69 \theta^{1.8})\cos \theta \, , \, 3\bigg(\frac {\theta}{4\pi}-1\bigg)+(14+0.69 \theta^{1.8})\sin\theta \bigg)$$
for $\theta$ going from $4\pi$ to $0$.
Best Answer
According to your picture, it looks like you want $f(0)=1.5$. So $\sin(a)+b=1.5$. If $b=0.5$, this implies $a=\pi/2$ (or that plus some multiple of $2\pi$). Since $\pi/2$ is close to $1.5$, that’s why your picture looks “almost right” but not exactly right.