It’s that time of year again. Time to break out the ol’ DVD of National Lampoon’s Christmas Vacation, the best holiday movie ever filmed.

If you haven’t seen the movie, there’s a part where Clark Griswold decides to put up 25,000 Christmas lights on his house. After the predictable shenanigans caused by having them plugged into a switched outlet, they are finally lit, causing a brownout in the rest of the city (Chicago), which is rectified by someone switching on the “auxilliary nuclear” power plant.

We also get a short clip of his power meter spinning incredibly quickly, for comedic effect.

In the 41 frames, or 1.7 seconds, that this is shown, the tens dial spins around two and a half times, meaning that 250 kilowatt-hours were used in this amount of time. That’s a lot of power! How much? Well, if you divide by 1.7, and then multiply by the conversion factor for kWh/s to watts (or the number of seconds in an hour times 1000), it turns out that these imported Italian twinkle lights are using five hundred and twenty-nine Megawatts.

No wonder the rest of the city lost power. In 1989, the year of the movie’s release, Chicago had a population of 2.7 million, using perhaps 2.25 GW of electrical power residentially. That means the local electrical grid saw a sudden load increase of almost 25%. I’m impressed it handled it as well as it did.

Also, each bulb of those 25,000 was using more than 20 kilowatts. Aren’t you glad we have LED Christmas lights these days? I knew incandescents were inefficient, but wow.

When the house is lit up, the snooty neighbours get an eyeful of light streaming in their window, causing all sorts of chaos.

How much light? Well, let’s assume the Griswold house is radiating equally in all directions, and Todd and Margo are 50 meters from the epicenter. Incandescent light bulbs typically had an efficiency of about 2%, meaning that of those 529 Megawatts, only about 10.6 MW became light. At a distance of 50 meters, this would only be a radiative flux of 340 W/m^2, or about a third as bright as that old standby, the sun.

The sun is really, really bright.

But Mr. Griswold certainly gets an A for effort.

“Is your house on fire, Clark?”

If 2% of the power is turning into light, that leaves 98% that is becoming heat. A rough estimate for the size of the house gives a surface area of 5000 square feet, or 460 square meters. That’s about 1.13 Megawatts/m^2. Assuming we can model the house as a blackbody radiator (and with that amount of power being pumped into it, it would rapidly become one), then we can use the Stefan-Boltzmann law to determine its temperature. Dividing the radiative flux by the Stefan-Boltzmann constant and then taking the fourth root, we find that the temperature at the surface of the house must be about 2100 K, or 3300 degrees Fahrenheit. Aunt Bethany was right all along.