What is the formula for winding and unwinding a spool with changing radius due to the strap thickness being wound. I have a servo motor and I'm trying to come up with a formula that relates the angle of the motor to the length of a strap that is wound on it. Since the radius at <360 degrees is the radius of the motor shaft (r1) and at 360<theta<720 the radius is now r1 + Ct (cable thickness) and after 720 degrees, the new radius r2 = r1 + 2*Ct. Consider the thickness to be uniform. So the circumference is growing every 360 degrees when a new layer of cable is wound. Any ideas? The cable goes only on top of its previous layer, it does not move to the side.
The "spool" is cylindrical ie. servo motor shaft. I want to treat each layer as a concentric circle, not a spiral.
Thanks!
rough work of what I've tried
 A: The  solution depends on the shape of the spool, which you did not specify. A fairly complete class of spiral solutions can be constructed using Fourier series and a bit of calculus.
The unit normal $\vec N$  to the spiral is needed in order to describe how the spiral grows when one adds another layer:   at each point on the spiral, the thickness of the wrapped strap is equal to the distance to the next layer of the spiral as measured in the direction $\vec N$ that is  perpendicular to the current point on the spiral.
Caution. In what follows $\theta$   represents the angle of inclination of the unit normal vector (which is 90 degrees out of phase with the tangent vector of direction of travel).  This angle should not be confused with the "usual" symbol for a polar coordinate system that has a fixed origin. If a ship is traveling on a curvilinear path around a fixed buoy that is taken as the origin, $\theta$ described the instantaneous compass bearing of the moving ship rather than the angle that the position vector of the ship makes measured from that  fixed buoy.
The general strategy is to parametrize  the  spiral curve $\vec R$ as a function of $\theta$.
Notation.
The unit normal vector to the spiral is $\vec N=<\cos \theta, \sin \theta>$.
The unit tangent vector $\vec T$ to the spiral is perpendicular to the vector $\vec N$. In what follows $\vec T=<-\sin \theta, \cos \theta> =\frac{ d \vec N}{d\theta}$.
The position vector of the parametrized spiral is  $\vec R (\theta) $.
Thus,
$ \frac{d\vec R}{d\theta} = \vec T \frac{ds}{d\theta}$ where $\vec T =<-\sin \theta, \cos \theta >$ and $\frac{ds}{d\theta} = c(\theta) $ for some non-negative  function $c(\theta)$ whose detailed properties will now be determined.
If the thickness of the strap is $a$, then we want each revolution of the wrapped strap to be displaced in the normal direction by exactly this thickness. Thus we desire $\vec R(\theta+2\pi) -\vec R(\theta) = a \vec N(\theta)$.
Differentiation with respect to $\theta$ gives
$ \vec T(\theta +2\pi)  c(\theta+ 2\pi) - \vec T(\theta)c(\theta) = a \vec T(\theta)$. But $\vec T(\theta+2\pi)= \vec T(\theta)$, so
after cancelling $\vec T$ from both sides, deduce that
$c(\theta+2 \pi) -c(\theta)= a$. The general solution to this equation is $$c(\theta)= \frac{a}{2\pi} \theta+ p(\theta)$$ where $p(\theta)$ is periodic with period $2\pi$. We can call this function $c(\theta)$ a quasi-periodic function. (It is the sum of a linear function of $\theta$ and a periodic function.)
Thus working backwards from this, we see that starting with any periodic function $p(\theta)$ expanded as a Fourier series $p(\theta)= \sum_k A_k \cos (k\theta) + B_k \sin (k \theta)$
one can construct a  function $c(\theta)$ that has the required quasi-periodicity.  Note also that you want $c(\theta)$ to be non-negative. (This can be accomplished by trial and error, tinkering with the choice of the Fourier coefficients. There are many solutions, corresponding to different shapes of the spool! A good starting point is to use  a small number of such Fourier coefficients, e.g., just $A_0, A_1, B_1$.)Then integrating the equation $d\vec R= \vec T(\theta) c(\theta) d\theta$ gives the full parametric description of the spiral associated to this choice of quasi-periodic function.
P.S. In physical terms $c(\theta)$ is the radius of curvature and its reciprocal $\kappa$ is the curvature of the curve.
Note that any irregularities in the shape of the first wrap  (which is related somehow to the shape of the spool) is determined by $p(\theta)$.
Below is a plot of an illustrative example in which $c(\theta) =.5 \theta+1$  is especially simple.  In this case the total arc length along the spiral can be found by integrating $ds= c(\theta) d\theta$ to obtain $s= \frac{\theta^2}{4}+\theta$. This is a quadratic relationship between the angle $\theta$ and the length $s$ of the strap. (As mentioned above however, this $\theta$ should not be confused with the polar coordinate angle $\phi$ of the position vector $\vec R(\theta)$, although in practice for large values it is true that $\theta \approx \phi$.)
$c(t)=.5 t+1$ " />
A: This is similar to the problem of Calculating the length of the paper on a toilet paper roll.
In particular,
my answer to that question
assumes that the toilet paper is a material of uniform thickness (like your strap) that is rolled onto a cylinder so that each new layer of toilet paper is cylindrical.
I think that if the surface the paper/strap/cable is wound onto is initially cylindrical, this is a better model of the shape of the wound paper/strap/cable than any of the usual mathematical spirals.
There are some ways in which my toilet-paper answer does not perfectly match your question. I assumed we want to account for the transition from one cylindrical layer to the next; this transition occurs during the last few degrees of rotation before the start of the next cylindrical layer, and during that part of the rotation the paper/strap follows a non-cylindrical path (as it would have to in reality).
Unlike the earlier question I was answering, you want to use the total angle of rotation rather than the final radius of the spool as your independent variable.
Also, you are interested in the behavior in between full rotations.
One thing to consider is that as you wrap the strap/cable around the spool, either the outer surface will stretch or the inner surface will compress. Possibly both will happen. Presumably somewhere in the strap (at the inner surface, the outer surface, or some layer in between) the strap neither stretches nor compresses.  If you want an accurate answer, you want to determine the layer within the strap where the length of that layer is preserved. In the toilet paper problem, as a simple guess I assumed this occurred exactly halfway between the inner and outer surface of the paper, but you could determine a more accurate estimate experimentally.
When you discover (or guess) which layer of the strap preserves its length, you use that layer as the cylinder that measures the length of strap rolled onto the spool.
If you simply assume that the strap is wrapped all the way around the spindle ($360$ degrees) in a complete cylindrical layer and then somehow starts a new layer offset by the thickness of the strap,
you can take the integer part of the total number of revolutions of the spindle and assume that the strap is wrapped in that many cylindrical layers around the spool, in addition to which there is a partial layer wrapped by rotating the spindle through the remaining fractional part of the total number of revolutions.
Since this problem involves lengths of circular arcs, I highly recommend converting your rotation angle to radians before proceeding.
Then if you rotate the spindle by $\theta$ radians,
the number of full revolutions is
$$\left\lfloor \frac{\theta}{2\pi} \right\rfloor$$
and the number of fractional revolutions is
$$\left\{\frac{\theta}{2\pi} \right\}
 = \frac{\theta}{2\pi} - \left\lfloor \frac{\theta}{2\pi} \right\rfloor.$$
The total length of the complete cylindrical layers is a function of
$\left\lfloor \frac{\theta}{2\pi} \right\rfloor.$
The length of the partial layer is
$2\pi \left\{\frac{\theta}{2\pi} \right\} R,$
where $R$ is the of the cylinder to use for the last partial layer.
Let $r_1$ be the spindle/shaft radius,
$C_t$ the thickness of the strap/cable,
and $\delta$ the distance from the inner surface of each cylindrical wrap of the strap/cable to the place inside the strap/cable where the length is preserved ($\delta = 0$ if the inner surface does not change length,
$\delta = 1$ if the outer surface does not change length).
So the cylindrical surface representing the true length of the strap/cable during its first wrap around the spindle is $r_1 + \delta,$ and
the length of the first full wrap (assuming it goes a full $360$ degrees around) is $2\pi(r_1 + \delta).$
The length of the second full wrap is $2\pi(r_1 + \delta + C_t),$
the third full wrap is $2\pi(r_1 + \delta + 2C_t),$ and so forth.
For $n = \left\lfloor \frac{\theta}{2\pi} \right\rfloor$
full wraps we have length
$$ 2\pi(n(r_1 + \delta) + (1 + 2 + 3 + \cdots + (n-1)) C_t). $$
Using the formula for the sum of consecutive integers, this comes out to
$$ 2\pi\left(n(r_1 + \delta) + \frac{n(n-1)}{2} C_t\right). $$
The cylindrical radius to use when computing the length of the final partial layer is $r_1 + \delta + nC_t,$ leading to a length of
$$ 2\pi \left\{\frac{\theta}{2\pi} \right\} (r_1 + \delta + nC_t). $$
Putting it all together, and using the fact that
$n + \left\{\frac{\theta}{2\pi} \right\} = \frac{\theta}{2\pi},$ we have
\begin{align}
2\pi &\left(n(r_1 + \delta) + \frac{n(n-1)}{2} C_t\right)
+ 2\pi \left\{\frac{\theta}{2\pi} \right\} (r_1 + \delta + nC_t)\\
&= (r_1 + \delta)\theta
+ 2\pi C_t n\left(\frac{n-1}{2} + \left\{\frac{\theta}{2\pi}\right\}\right)\\
&= (r_1 + \delta)\theta
+ 2\pi C_t \left\lfloor \frac{\theta}{2\pi} \right\rfloor\left(\frac12 
\left(\left\lfloor \frac{\theta}{2\pi} \right\rfloor-1\right)
 + \left(\frac{\theta}{2\pi}
     - \left\lfloor \frac{\theta}{2\pi} \right\rfloor\right)\right) \\
&= (r_1 + \delta)\theta
+ 2\pi C_t \left\lfloor \frac{\theta}{2\pi} \right\rfloor
   \left(\frac{\theta}{2\pi}
         - \frac12\left\lfloor \frac{\theta}{2\pi} \right\rfloor
         - \frac12\right).
\end{align}
