I was inspired. Inspired by this awesome estimation of the cost to build a Death Star. I won't tell you their estimate; you will have to click the link to find out yourselves.

Then there was a discussion I had with some friends. We came to the conclusion that the Lego Millennium Falcon was so cool because it was to the proper scale for Lego mini figs. No, we are not talking about the Millennium Falcon on the Lego store; this older one that they don't sell anymore (the Ultimate version).

Just like the Lego Falcon they made a Lego version of the Death Star, but it wasn't to scale. Yes, they do make another version of the Death Star, but they don't even try to pretend it is to scale.

How Big? ——–

First, how big is the REAL Death Star? Well, there were two (Episode IV and Episode VI). Apparently, these two Death Stars were not the same size. According to Wookiepedia, the first Death Star had a diameter of 160 km.

Need the dimensions of a mini fig? BOOM. The internet is here for you. That site appears to say the height of a mini fig is 38.6 mm tall. If I assume an average human height of 1.77 meters, this would mean the scale of the mini fig is:

So, a to-scale Lego Death Star (first version) would be 0.022 times the diameter of the REAL Death Star. This would put the diameter of the Lego Death Star at 3.52 km. That's a pretty big Lego model. This is what it would look like next to the world's tallest buildings. (The tallest one is around 600 meters.)

I told you it was huge.

How Many Lego Pieces? ———————

If the scale version of the Death Star came in a set, how many pieces would it have? This is tough. The first question we need to answer (we will answer it together) is: what will be on the inside of the Lego Death Star? There will have to be some things in there to make it support the outside. Probably if you want a scale model of the Death Star, you want everything. Garbage compactor and all.

So, assuming the inside of the model has structure I need to get an estimate for the density. Let's go back to the Ultimate Millennium Falcon model. According to brickset.com, the model has 5,195 pieces. It has dimensions of 84 cm x 56 cm x 21 cm. If I assume this is rectangular-ish, I can determine the Lego piece-density:

This is just an estimate, but one I am fairly happy with. Sure there are some large pieces in the Millennium Falcon model but there are also some small ones. I guess it is possible the Death Star would have a lower piece density (if it has more larger pieces).

Using this density and the volume of the Ultimate Death Star model, I can get the number of pieces in the set.

Maybe the Ultimate Death Star has more large pieces in the set. Let estimate there would be 1014 pieces in the set. Just to be safe.

How Much Would it Weigh? ————————

Really, I mean mass – but I like "weigh" in the title better. So, for this, I need the mass density of a Lego set. The Ultimate Millennium Falcon is listed at a shipping weight of 24.2 pounds. Of course this must include the box and the instructions, so maybe the pieces would weigh around 21 pounds (9.5 kg). This would give a mass density of 96.2 kg/m3. Just a quick check on the Lego Death Star II, it has a mass density of about 85 kg/m3 – and it isn't even complete.

Let me just go with a density of 90 kg/m3. With this density (mass density) my Super Ultimate Lego Death Star will have a mass of:

I don't know what to say about the mass.

How Much Would it Cost? ———————–

This is going to be a stretch. But here is a graph of the price of different Lego sets as a function of the number of pieces (from a very old post):

If I assume the function stays linear for up to 1015 pieces (which would be odd to not give some sort of large set discount), then I get about $0.098 USD per piece. So, for all the pieces this would cost about 9.8 x 1012 US dollars, yes almost 10 trillion dollars.

Put This Thing in Orbit ———————–

Really, this is your only option if you want to build something like this. The biggest problem on the surface of the Earth would be supporting the thing. Suppose I build a base to hold it up that is about 0.3 km across. All of the weight of the Death Star would have to be supported on top of this. This would be a compressive pressure of about 2.4 x 108 N/m2. Just for a comparison, granite (Engineering Toolbox) has a maximum compressive strength of 1.3 x 108 N/m2. So, we are talking about some structural failures here.

If you put it in orbit, you don't have to worry about this compressive strength problem. Also, you could move around to different parts of the model to build it.

Homework Questions ——————

Here are some other questions that I didn't get around to answering:

If you wanted to build it in orbit, how much would it cost to get all the material up there?

Suppose you can have as many people as you like working on the model; how long would it take to build?

If the model were in orbit, would the gravitational force from the mass in the model be enough so you could walk on it?

Really, I cheated with some of my data. I just looked at a couple of models to get the piece-density and the mass-density. Look at many more sets and make a plot of number of pieces versus model volume to get an estimate of the piece-density. Do the same for the mass density.

That should keep you busy for a while. Oh, I noticed that a few more of the Lego Star Wars models were not to scale. This needs to be fixed.

Bonus Picture ————-

Just when you thought everything was over, it keeps going. What if this Lego set were indeed in orbit around the Earth? Low Earth orbit (with an altitude of 300 km). What would it look like? Well, first let me say that the angular size of the the moon is about 0.53 degrees. If this 3.52 km diameter radius object was in orbit, it would have an angular size of:

So, it would appear bigger than the actual real moon. You know I am going to make a diagram showing that. Here it is.

How cool would that be? People would mistake it for a moon, just like Han Solo did. Well, it look just like the moon except that it would just take a couple of minutes to pass across the sky where the moon doesn't really seem to change its position.

Top photo: bdesham/Flickr