Found the radius of a 3D spiral (corkscrew type) I am trying to find the radius of a 3D spiral with the following constraint:


*

*Center of spheres of 3mm radius are on the spiral

*Each sphere is tangent to other


Basically, it's a spiral of 6mm sphere, and I want to know the radius of the spiral as a function of the number of spheres per tour.
With picture :
(16 sphere per tour)

(11 sphere per tour)

I know the parametric equation for a spiral is
\begin{align}
x &= r * \cos(a)\\
y &= r * \sin(a)\\
z &= r * a * \tan (b)
\end{align}
with
$r$ = radius
$a$ = angle (degre)
$b$ = constant
But there begin my problem, because I don't know what to do now.
If I put a sphere $S_1$ of center 'A' at the beginning of the spiral ($a = 0$), I have
\begin{align}
Ax &= r \\
Ay &= 0 \\
Az &= 0
\end{align}
But I don't know how to calculate the intersection of the spiral and the sphere $S_1$.
Anybody can give me a hint ?
(ps : this is a personal project, not school homework).
Thank for reading me.
 A: This isn't an exact answer, but a close approximation when the radius is reasonably large (as in the two models pictured).
Suppose the individual spheres have radius $r$, and that a string of $n$ spheres make "one full turn" on a cylinder of radius $R$ in the sense of the diagram:

If the centers of all the spheres lie on a cylinder of radius $R$, each sphere cuts the cylinder (very nearly) in a disk of radius $r$. The diagram comes from cutting this cylinder open and rolling it flat. The shaded disk represents the initial ball in the chain.
The straight-line distance between the centers of the shaded disks is $2\pi R$, the circumference of the cylinder.
If the left-hand center lies at the origin, then the right-hand center has coordinates $\bigl((2n - 1)r, -\sqrt{3}r\bigr)$, so the Pythagorean theorem gives
$$
2\pi R \approx r\sqrt{(2n-1)^{2} + 3} = 2r\sqrt{n^{2} - n + 1},
$$
of $R \approx \frac{1}{\pi} r\sqrt{n^{2} - n + 1}$. For instance:


*

*If $n = 11$, then $R \approx 3.354r$, or just over $10$mm for beads of diameter $6$mm;

*If $n = 16$, then $R \approx 4.94r$, or about $14.8$mm for beads of diameter $6$mm.
This calculation is approximate because spheres (as opposed to disks) occupy space, and so will intersect slightly (inside the cylinder) when the flat diagram is rolled into a cylinder.
A: Here's an exact (but non-explicit) treatment, which I'm posting separately since the original approximation may also be of interest to posterity.
Let $r$ denote the radius of the "beads", $R$ the radius of the cylinder on which the centers lie, and $n$ the number of beads in "one turn". The centers of the beads lie along a helix of the form
$$
(x, y, z) = (R\cos t, R\sin t, Rmt).
\tag{1}
$$

If $\theta$ denotes the angular separation (measured around the axis of the cylinder) between the centers of successive beads, then without loss of generality, an "initial" bead has position $(R, 0, 0)$, while the $j$th bead has center
$$
\bigl(R\cos(j\theta), R\sin(j\theta), Rmj\theta\bigr).
\tag{2}
$$
The three unknowns $\theta$, $m$, and $R$ satisfy three constraints (shown in blue):


*

*The centers $p_{0}$ and $p_{1}$ of adjacent beads have a straight-line separation of $2r$:
\begin{align*}
  4r^{2}
  &= R^{2}\bigl[(1 - \cos\theta)^{2} + \sin^{2}\theta + (m\theta)^{2}\bigr] \\
  &= R^{2}\bigl[2[1 - \cos\theta] + (m\theta)^{2}\bigr].
\tag{3a}
\end{align*}

*The centers $p_{0}$ and $p_{n}$ of the initial bead and the $n$th bead have a straight-line separation of $2r$. From (2) with $j = n$:
$$
4r^{2} = R^{2}\bigl[2[1 - \cos(n\theta)] + (m\theta)^{2} n^{2}\bigr].
\tag{3b}
$$

*The centers $p_{0}$ and $p_{n+1}$ of the initial bead and the $(n + 1)$th bead have a straight-line separation of $2r$:
$$
4r^{2} = R^{2}\bigl[2[1 - \cos((n + 1)\theta)] + (m\theta)^{2}(n + 1)^{2}\bigr].
\tag{3c}
$$
Subtracting (3c) from (3b), simplifying, and using the cosine addition formula for $\cos((n + 1)\theta) = \cos(n\theta + \theta)$ gives
\begin{align*}
0 &= 2\cos((n + 1)\theta) - 2\cos(n\theta) - (m\theta)^{2}(2n + 1) \\
&= 2\cos(n\theta)(\cos\theta - 1) - 2\sin(n\theta)\sin\theta - (m\theta)^{2}(2n + 1),
\end{align*}
or
$$
m^{2} \theta^{2} = \frac{2}{2n + 1} \bigl[\cos(n\theta)(\cos\theta - 1) - \sin(n\theta)\sin\theta\bigr].
\tag{4a}
$$
Subtracting (3b) from (3a) and rearranging gives, similarly,
$$
m^{2} \theta^{2} = \frac{2}{n^{2} - 1}\bigl[\cos(n\theta) - \cos\theta\bigr].
\tag{4b}
$$
Subtracting (4a) from (4b) and cancelling the factor of $2$ gives
$$
0 = \frac{1}{n^{2} - 1}\bigl[\cos(n\theta) - \cos\theta\bigr] - \frac{1}{2n + 1} \bigl[\cos(n\theta)(\cos\theta - 1) - \sin(n\theta)\sin\theta\bigr],
\tag{5}
$$
an equation for $\theta$ alone. If $\theta_{n}$ denotes the smallest positive solution of (5), which is a bit larger than $4\pi/(2n + 1)$, substitution in (4b) gives $m$, and then substitution in (3a) gives $R$.
