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.$$\prod_0^{n=2}(Ho_n) = Ho \cdot Ho \cdot Ho$$

  • $\begingroup$ Ah, it's that time of year. Lovely question :) do try to use the latex format please. $\endgroup$
    – user459879
    Dec 9, 2020 at 22:57

1 Answer 1


Known as involutes,their width $W$ along the generator is constant. But the strips are curved.

By using differential calculus we get a relation

$$ \text{ arc-length }=\dfrac{\pi( r_{top}^2- r_{bottom}^2)}{W \sin \alpha}, $$

$\alpha$ is semi-cone angle of Christmas tree, light string a has vertical climb rate W $ \cos \alpha$.

To find the total length directly, consider the packed situation when each upper strip lower edge touches / snugly fits the lower strip upper edge. The spacing is uniform either vertical, horizontal or along slant generator.

Projected area in horizontal plane$= \pi( r_{top}^2- r_{bottom}^2)$

Projected width in the horizontal plane $= W\sin \alpha$.

Divide to get the total arc-length.

The Involute design here requires multiple radial starts (eight in the shown example) at top minimum radius as shown:

$$ \text{ Number of strips = }\frac{2 \pi r_{top}}{W }$$

It can be truncated between any two desired levels.

enter image description here

  • $\begingroup$ This is interesting. W must somehow relate to the distance between the wraps (possibly is even related to alpha). Because, the total arc length is being modulated by W. If W is large the arc length is short. If W is infinitesimal the arc length is looong. But, thank you, I will seek more information about the above formula. $\endgroup$
    – Jim Clark
    May 9, 2021 at 14:12
  • $\begingroup$ Yes, W and total arc length are inversely proportional, as their product is constant, As you can expect in a flat rectangle also... Covered Area= W * (narrow width ) with ribbon arc lengths. In a full cone it is area before projection in the base plane. $\endgroup$
    – Narasimham
    May 9, 2021 at 15:33

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