logarithmic spiral around cone stump Based on the answer on my previous question I managed to come up with the following equations:
$$\begin{eqnarray}
k &=& 1 \\
r_\Delta &=& r_b - r_t \\
r(\theta) &=& r_t * e^{\theta/360*k} \\
x(\theta) &=& r(\theta) \cos(\theta) \\
y(\theta) &=& r(\theta) \sin(\theta) \\
z(\theta) &=& (\frac{r_b}{r_\Delta} - \frac{r(\theta)}{r_\Delta}) * h_s \\
d(\theta) &=& diameter * \frac{r(\theta)}{r_b} \\
\end{eqnarray}$$
that give almost the desired spiral.

The problem is that with $k = 1$ the spiral goes beyond the stump (defined by the top radius $r_t$, the bottom radius $r_b$ and the height $h_s$).
The question is what does $k$ need to be so that the spiral ends at the bottom of the stump (at $z = 0$ and $r(\theta) = r_b$)
 A: With $k=1$ and with $\theta$ ranging from $0$ to $360$, measured in degrees
(as we appear to see in the figure),
we get $r(0) = r_t$ and $r(360) = r_t\cdot e.$
And then the largest sphere in the figure is at radius $r_t \cdot e$,
which is much larger than $r_b$ in this figure.
As the sphere is still on the surface of the cone,
it is not only much further from the axis than desired
but also at a much lower $z$ coordinate than desired.
If that is an accurate description of what we see in the figure in 
the question, and if the desired result is that we should
see $r(360) = r_b$, we can work backwards to figure out what $k$ should be:
$$r_b = r(360) = r_t \cdot e^{(360/360)\cdot k} = r_t \cdot e^{k}.$$
$$e^{k} = \frac{r_b}{r_t}.$$
The logarithms of these two equal quantities are equal:
$$\ln\left(e^{k}\right) = \ln\left(\frac{r_b}{r_t}\right).$$
$$k = \ln r_b - \ln r_t.$$
So if you measure angles in degrees and want a full
$360$-degree "wrap" around the cone to start at $r(0) = r_t$
and end at $r(360) = r_b$,
then you should set
$$r(\theta) = r_t \cdot e^{\theta \cdot (\ln r_b - \ln r_t) / 360}.$$
Note that if you measure angles in radians, then since $360$ degrees is
$2\pi$ radians, the formula would be
$$r(\theta) = r_t \cdot e^{\theta \cdot (\ln r_b - \ln r_t) / (2\pi)}$$
where $\theta$ starts at $0$ and grows to $2\pi$ to accomplish one full
turn around the axis.
