# All hail the dome

How much did the dome covering Springfield weigh, and what was it made out of, in The Simpsons Movie? Another math intensive post, but this time, it wasn’t me who did this. This was done by user phengineer on Reddit’s “They Did The Math“. I’ve copy and pasted it here, and a source link is at the very bottom 🙂

Wikipedia says that Springfield has 30,720 residents. Let us assume 3500 people per square mile as the population density (corresponds to a nice suburb with a fair amount of open spaces as well) Then sqrt(30720/3500/pi) = 1.67 mile radius. But of course the city isn’t perfectly round, and we know the dome is next to the simpson house, which is not in the middle of the city. Fuck it, 2 mile radius.

Now, the sound made when the dome is contacted suggests a plastic. The structure of glass disallows phonons that would produce the sounds used in the movie. We also know that it is bulletproof. When the dome is broken however, the remnants are glass. While such a material as we see in the movie doesn’t exist, it is a reasonable guess that is would be a thick acrylic. Glass would simply be impractical, as we will see in the stress analysis.

We know that the dome must be able to support itself, as well as external environmental forces. For domes, typically, a fairly low safety factor will suffice, because by far the highest stresses are in its self-support. The thing we are most concerned about when we’re talking about dome strength is the compressive yield strength. Acrylic has a yield strength of about 100 MPa, with an elastic modulus of 3 GPa (the elastic modulus is just for fun), and the greatest stress occurs at the base of the dome.

At this point, we consider an operations research problem. Assuming the government wants to minimize its own material and lifting cost, the dome should have a tapered thickness so that the entire dome experiences a constant stress. Let’s go with a safety factor of 2.

We want to solve for 50MPa = gmassAbove(theta)/sin(theta)/(2piradius(theta)thickness(theta)) Where massAbove(theta) = int_0phi 2 pir(phi)*thickness(phi) d phi (assuming thickness<<radius) For a perfect hemisphere, r(phi) = sin(phi)

The solution requires a least action integral and solving a functional equation to find the function thickness(t), at which point we can integrate this over the dome to find the volume, and then the mass using the density. In fact, we must use variable thickness, because if we let the thickness be constant, the stress is approximately linear in t, and t drops out of the equation. The stress at the base is 2* pi* r* rho* g = 217MPa, exceeding our strength. “But phengineer, what about the nonlinearities in t?” I crunched the numbers taking into account the exact dependence of t, and we get just over 11 kilometers as the thickness, much greater than the radius. That’s why weak linearity is essentially linearity.

With that, we’ve established an upper bound of about 20 billion tons. As a drastic lower bound, if we assume 1 meter thick, we’re merely at 2.5 million tons (weight of the great pyramid). But the answer is certainly much higher than this.