Dec 19

A Photonicly perfect Christmas tree

This week is our last week of work before breaking for the Christmas/New year holiday. As it's the last week, the office has taken on a decidedly festive air (someone had a christmas card on their desk) and I was thinking how we could make our work more seasonally appropriate. So after some group brain storming we decided that we can help everyone out this Christmas by calculating the best possible arrangement of christmas tree lights 🙂 I have written up the math in this post as a python script for anyone wanting to double check it.

Through extensive research, I have found that people <3 lights, therefore I assume that the best  arrangement is the most possible lights. So what I think is the best place to start is to calculate the optimal viewable density of christmas tree lights. What I mean by viewable is; if you are standing say 10 m away from the tree what density of lights can you have before they all just blend into one big glow. I chose 10 m as this is average width of a street (UK) and I assume that part of the point of a christmas tree is that your neighbours can see how much better it is than theirs.

Diagram may not be truly accurate

Project aim: make people walking past your house super jealous

With a little googling I discovered that people with 20/20 vision can resolve objects 1 arc minute apart. What is 1 arc minute I hear you ask, well it's complicated but basically it's 1/60 of a degree in your field of view. So if you had two objects on a table and drew an imaginary triangle (the best kind) between you and the two objects, providing your viewing angle is greater than 1/60 of a degree, your eye can resolve the difference between them. Simple.

Viewing angle

Viewing angle between two objects in your field of vision

So with a little trigonometry we can calculate that from 10 m a christmas tree with the hight 1.8 m (seems like a pretty average tree) from top to bottom is around 40 degrees which is equivalent to 1239 arc minutes. If you then do the same for the width, assuming the tree is 1 m wide which gives you a further value of 687 arc minutes. Multiplying these two together will then give us the number of resolvable points available on the tree: 85,1193. Pedants might point out that trees are rarely exactly 1.8 x 1 m square and come in many different shapes, to which my reply is "shut up, this is difficult enough".

If this makes anything clearer then there might be something wrong with you

See simple!

If we feed the number of visible points back into the measurements of the perfectly rectangular tree we can calculate that there are 47 viewable points every cm2; therefore if you covered your tree in lights at this density they would still look like nice twinkly individual lights. Job done, yet another triumph for practical science.

[ED: err 85,1193 lights doesn't sound very practical, care to try again?]

No, and anyway it's 1,702,386 lights, the tree has two sides. I suppose I could meet you half way and run the numbers on powering up this awesome spectacle.

Assuming that firstly you have found some lights that you can pack at a density of 47 per cm2; and secondly that you have found a tree capable of supporting their weight, one thing I assume you'll want to know is what electricity bill you might expect. There are two routes you can choose here, LED or Mini-bulbs. Mini-bulbs (~0.5 watts per light) are still more common and are essentially little incandescent lights, if you run the numbers these will draw around 0.8 Mega Watts which is around 90% of the output of the UKs largest solar energy park. If you are a bit more environmentally conscious then you might opt for LED lights (~0.07 watts per light) which work out far more efficient and only use 0.1 mega watts. If you then work this out in terms of actual cost (based on average UK prices) you can have your tree on 5 hours a day for the very reasonable sum of £70/day, not including the initial cost of installing specialist cabling to handle the power load. Considering you will have already spent around £140,000 on the lights in the first place this really seems like a bargain. Besides the cost of this will pale in comparison to the warm feeling you'll get from looking at a tree that is around x1500 brighter than a laser pointer (assuming only 7% of the energy is converted to light), although it will very likely be the last thing you see.

This picture reminds me of back when I had retinas

The lights may be a little brighter than other trees, torches, car headlights, stadium lights or IMAX projectors you may have lying around.

[ED: Only 7% of the energy is converted to light...where's the rest going?]

Well, you know that warm feeling I mentioned...? LEDs produce heat as well as light (around 75% of the energy, compared to 90% in mini-bulbs), and 1,702,386 lights produces a LOT of heat. Making the numbers for heat make any kind of sense is a bit tricky as there are several ways of measuring the heat output of any object. I think the easiest way is to work out what this heat might do to our poor christmas tree (i.e. how hot will the tree get). Using some basic chemistry math and assuming the tree is mostly water, the simplest way of explaining the heat generated is by calculating the increase in the temperature of the tree while the lights are on.

Teh burninator tree

Worryingly people watching your tree burn seem quite pleased with the idea

If you work this out (assuming the tree weighs around 25kg) when with mini-bulbs, the temperature of your tree will increase by about 200℃ per minute while you have them on. As wood ignites at around 400℃ you'll get a nice 2 minutes viewing followed by a nice evening round the bonfire that is where your house used to be. LEDs are a lot better and only raise the temperature of the tree by 26℃ per minute, meaning you get a nice 16 minutes to enjoy your display before running for your lives.

[ED: seriously, this is your idea of practical, burning houses?]

Fine, I suppose I could work the math backwards and work out what number of lights WON'T burn your house down. A quick estimate put this value at around 4.5 LEDs every cm2 (170,239 total), which is both practically achievable given the size of LEDs, and infinitely more boring. I think a better alternative is to stick to 47 LEDs per cm2 but set them up so that only 1/10 are on at any one time. That way you can have them twinkling nicely. If you replaced the tinsel (which is buried in lights anyway) with some kind of water cooling system then you might even be able to get this down to 1/5.

So I hope this has been a helpful little exercise in choosing how to decorate your festive tree. Once again I feel science has provided a clear and simple answer to a complex problem millions of people struggle with every year. MERRY CHRISTMAS.