how do I plot the curve of a conical spiral on an unwrapped (2D) conic surface? Example images
I need a real world solution to a reconstructive surgery problem!
A fairly commonly encountered clinical scenario results in the need to create a tube (from skin) in the shape of an inverted truncated right cone to reconstruct a defect in the pharynx after cancer resection.
The lengths of the defects are somewhat variable (8-12cm) as are the radii of the base of the truncated cones (upper pharynx, 2-3cm) and the radius where the cone has been truncated (lower pharyngeal defect, 1-2cm).
Obviously cutting the truncated cone along its perpendicular slant length would give a very simple template. However this is unusable in the real world as the width of the template means that the edges of the skin defect cannot be sutured back together after harvesting the skin. The maximum width of any template would have to be approximately 8cm.
I have created paper templates of the tubed reconstructions and then cut them in an approximate spiral fashion with a helical angle of between 30 and 40 degrees. This results in a template that can be transferred to the donor area (usually the thigh) from which the skin can then be harvested and the resulting defect closed with sutures without undue tension in the remaining skin. The harvested skin is then sutured along the spiral to recreate the truncated conic tube which is then transferred to the pharynx, sutured in at the upper and lower openings and the blood supply to the tube restored with microvascular techniques.
To my mind, this is an untidy method of planning and I would like to look for a general solution where I can plan a template based on the 2 radii and the length of the defect, and based on a conical spiral which gives a relatively long and narrow template pattern.
I have trawled the internet and have found the attached images (for calculating the length of Christmas tree lights) but no explanation of how the curves were drawn. There are many sites giving the formulae for conical helices but none that I can find showing how these can be constructed on an unfolded two-dimensional conical plan.
I am well aware that this is a verysimple problem for this site but I would be grateful if you would bear with me. I am unfamiliar with coordinate systems other than Cartesian but am prepared to put in some groundwork if necessary.
 A: This can be solved by unwrapping the helix from the truncated cone to a single flat strip. The strip can then be cut to the same pieces as the ones obtained from an unwrapped cone. So we first solve this problem: given a truncated cone of height $h$ and bottom and top radii $r_1,r_2$ respectively cut the cone into a helix strip of width $w$. The last paragraph of this post shows how to cut the strip so that the pieces can fit into the unrolled cone.
When unrolled into plane the strip's border is made of four curves: two arcs and two Archimedian spirals.
The equation of the curves uses several derived values:
$$L=\sqrt{h^2+(r_1-r_2)^2}$$
$$\phi = \frac{2\pi (r_1-r_2)}{L}$$
$$a =\frac{Lr_2}{(r_1-r_2)}$$
Note: $L$ is the slanted length of the cone, and $\phi$ is the angle of the unrolled (flattened) cone. Now we can describe all 4 curves:
1st arc has radius $a+L$ and angle range $[0, \phi]$
2nd arc has radius $a$ and angle range
$\left[\frac{L\phi}{w}, \frac{L\phi}{w} + \phi\right]$.
1st spiral is described by parametric polar equation (with $0\leq t\leq 1$):
$$r = a + L -Lt$$
$$\rho = \frac{L\phi}{w}t$$
2nd spiral is the rotation of the first by $\phi$, and the corresponding polar parameteric equation (again, with $0\leq t\leq 1$) is:
$$ r = a + L -Lt$$
$$ \rho =  \phi + \frac{L\phi}{w}t$$
The attached drawing shows the unrolled helix, with radial lines showing the borders of unrolled cone. These radial lines would have to match when the strip is rolled back into the cone. Also, if the spiral strip is cut along those radial lines the resulting pieces can be put together to form the unrolled cone. But note also that there can be other, potentially more practical, ways to cut the spiral strip into smaller pieces.

The equivalent spiral can be obtained from the unrolled cone, where for $0 \leq k \leq \left\lfloor\frac{L}{w}\right\rfloor$ the equation for the $k$-th segment of the spiral is:
$$\rho=\phi t$$
$$ r = a+L-kw - wt$$
As before $0\leq t \leq 1$ with possible exception for the last segment of the spiral, which can be shorter.

A: Assuming throughout that the angle at the vertex is $90^{\circ}$:
A natural mathematical strategy is to unroll the cone by making four copies of the quadrant and describing the cut in polar coordinates. Archimedean spirals wrapping $n = 1$, $2$, $3$, and $4$ times about the origin are shown in light gray.
At the center, it looks easier to "transfer" the small hump to the "neighboring strip" as shown in blue.
Any curves can be used, in the sense that if we draw one curve and rotate it about the center by multiples of a right angle, the strip to be cut out is bounded by "neighboring" curves.

A: If I interpret your problem correctly, your goal is essentially equivalent to this geometry problem: you want to spiral-wrap a flat paper ribbon onto a given conical surface  so that the edges of the ribbon fit neatly together on the cone. In practice, you probably will construct the ribbon in patches that are joined together to tile the cone.
Note that in order to construct a perfectly right-angled cone in space by rolling up a circular sector cut from a planar piece of  paper, the "notch" that must be removed  from the circular disk is not 90 degrees, but rather $(1- \frac{1}{\sqrt 2})$ of a full revolution (about 105.5 degrees removed).
The portion of the flat circular disk that you want to keep therefore has an angle of $\alpha=254 $ degrees =$4.42 =\frac{ 2\pi}{\sqrt{2}}$ radians. I point this out as the main caution you should keep in mind if you contemplate applying the techniques mentioned in other answers.
In general,  the cylindrical equation of a 3D cone is $z= c r$  where $c= tan \beta$ and $ 2\beta$ is the angular aperture. In this general case  $\alpha = 2\pi/ \sqrt{ c^2+1}$ is the angular width of the flat circular sector that will cover that cone. (For the right-angled cone, $c=1, \beta = $ 45 degrees. and $2\beta=90$ degrees.)
