Our $20$-year old artificial Christmas tree has built-in lights, but the top and bottom thirds are burnt out. So I decided to buy several (much cheaper) new "tree light" strings (during a sale after the end of last Christmas) the majority of which would continue to work even if a light were to burn out (wired in parallel, I assume).
The bottom tree boughs are $4$-feet in diameter, $r_0 = 2$-feet, the top tree boughs are $1$-foot in diameter, $r_h = 0.5$-foot, and the vertical height distance between the top and bottom boughs is $h = 5.5$-feet. The top center spike rises, but will not include lights.
If I want to wrap the tree from bottom to top, with $1$-foot vertical spacing between wraps, what length of light string is required?
This is how I attacked the problem.
I related all variables in terms of the number of turns wrapping the tree from bottom to top.
The height (of light string wrapping) is $5.5$-feet with $1$-foot vertical spacing, thus there are $5.5$-turns = $11 \cdot \pi$ radians, of rotation as you progress from bottom to top.
I let $\theta$ (the angle) be the variable of integration from $0$ to $11 \cdot \pi$.
I, imaginatively, affixed a sector of cylindrical shell element at the end of a bough at the middle height of the tree. The light strand would wrap diagonally through the middle of this element. The edges of the element represent: the differential arc length component, $ds$, the differential height component, $dh$, and the differential radius component, $dr$.
The differential diagonal length, $dl$, in terms of $d\theta$ would be $$\frac{dl}{d\theta} = \sqrt{(ds_\theta)^2 + (dh_\theta)^2 + (dr_\theta)^2}$$
$r$ as a function of theta: $$r_\theta = r_0 - \frac{( r_0 - r_h ) \cdot \theta}{11 \cdot \pi}$$ $\frac{dr_\theta}{d\theta}:$ $$dr_\theta = - \frac{ r_0 - r_h }{11 \cdot \pi} \cdot d\theta$$ $h$ as a function of theta:$$h_\theta = \frac{5.5 \cdot \theta}{11 \cdot \pi}$$ $\frac{dh_\theta}{d\theta}:$ $$dh_\theta = \frac{5.5}{11 \cdot \pi} \cdot d\theta$$ $s$, arc length, as a function of theta:$$s = r \cdot \theta$$ $\frac{ds_\theta}{d\theta}:$ $$ds_\theta = r_\theta \cdot d\theta$$ $$ds_\theta = (r_0 - \frac{( r_0 - r_h ) \cdot \theta}{11 \cdot \pi}) \cdot d\theta$$ Plugging the above into an HP-$42$S simulator on my phone and integrating $dl$ as $\theta$ revolves from $0$ to $11 \cdot \pi$: $$L=\int_0^{11 \cdot \pi} \frac{dl}{d\theta} \cdot d\theta$$yields, $L\approx43.6$-feet. By varying the vertical spacing this allows one to evenly accommodate a modularly assembled light string length. (Note this also alters the total number of turns.)
Before I had a chance to test the modification, my wife said, "No.", and we returned the new "tree light" sets and got a new Christmas Tree (last year's model) on sale after last Christmas instead.$$\prod_0^{n=2}(Ho_n) = Ho \cdot Ho \cdot Ho$$