Treegonometry solves Christmas decoration dilemma

December 6, 2012
Treegonometry solves Christmas decoration dilemma

Mathematics has provided an answer for those striving for the perfect Christmas tree, Britain's University of Sheffield says.

The university's Maths Society was set the challenge of decorating a tree so that greenery and glitz are in harmonious proportion, resolving the problem of a tree that is either too barren or gaudy.

Here's their formula:

- Number of baubles: Take the square root of 17, divide it by 20 and multiply it by the height of tree (in centimetres).

- Length of tinsel: 13 multiplied by (3.1415) divided by 8, then multiplied by tree height.

- Length of tree lights: Pi multiplied by tree height

- Height (in centimetres) of star or fairy on top of tree: Tree height divided by 10.

"For example, a 180cm (six-feet) would need 37 baubles, around 919 cms of tinsel (30 feet) and 565 cms (19 feet) of lights, and an 18cm (seven-inch) star or angel is required to achieve the perfect look," the University says.

For those seeking an easier way of figuring this out, its website ( ) has a simple-to-use calculator.

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not rated yet Dec 06, 2012
This seems overly simplistic. The length of lights is pi * tree height but the surface area of the cone enclosing the tree is going to be pi * r * sqrt(r^2 h^2); without including the variable r (radius of tree branches at bottom) it's not going to be a good formula.
3 / 5 (1) Dec 07, 2012
depends on whether I like a mathematics geek"s idea of the ideal tree - where is the picture !
3 / 5 (1) Dec 14, 2012
You think it seems overly simplistic? It seems kind of silly to make everyone divide the square root of 17 by 20 and then multiply by the tree height (for number of baubles). Wouldn't it just be easier to say multiply the tree height by 0.206? And for tinsel multiply tree height by 5.1 instead of making the world multiply pi by 13 and divide by 8. A constant is a constant is a constant. I mean really that's just making it more difficult for the sake of making it difficult.

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