Estimating the Chinese Spy Balloon's Power Budget

“The Balloon” has been a hot topic the past few days. The United States and Canada say it was a spy balloon that intruded into NORAD airspace. China claims it was a “meteorological airship” that drifted accidentally over the United States due to unforeseen force majeure. (Being good Communists, they refrained from using the phrase “act of God.”)

I don’t buy the Chinese alibi. When China first learned the ship would stray into our airspace, why didn’t they make it known to us immediately, rather than waiting until after we announced we were aware of its presence before they made any comment? Furthermore, how is it that an “errant weather balloon” just happened to loiter around Malmstrom AFB? Did the jet stream stop blowing?

I thought it would be interesting to examine the size of the ship’s solar panels and estimate its power budget from that. Knowing how much power it used could shed some light on what type of vehicle it was.

As seen in the schematic above, the length of the balloon’s payload bus has been estimated at 30 meters. I will use this as a reference length to obtain dimensions for the solar panels. I’ll analyze the below photo, taken on 4 February when the balloon was over the Atlantic:

According to my drawing program, the red line in the below picture is 84.1 mm long. (At least it was before I cropped and resized the picture for this post.) That makes this a 356:1 scale photo.

By that scale, the 5.63 mm red line in the below picture corresponds to a real length of 2004 mm. Given that my estimation method is coarse and that the Chinese engineers likely selected a nice round number, I will consider the solar panel to have sides of exactly 2000 mm. That means each solar panel is 4 m², and the sixteen solar panels have a total area of 64 m².

The solar constant is 1399 W/m². At sea level, it’s closer to 1000 W/m², but the balloon was discovered at an altitude of 18 km. When I consult the US Standard Atmosphere, I find that the density at that altitude is 0.12 kg/m³, or about 10% of the density at sea level. I therefore find no reason to alter the solar constant figure.

Using the 1399 W/m² value, then, the total solar power falling on 64 m² of solar panels is 89.53 kW. Assuming a typical efficiency of 18% for these solar panels, this corresponds to roughly 16 kW of actual electric power generation when those panels are in full sunlight.

16 kW is quite a lot of juice. That amount of power could supply 3 to 5 American households (~3 kW each), or charge two Teslas at once (~7 kW each.) To me, that seems like gross overkill for a “meteorological airship.” Reasonable power users on a weather ship would be an onboard computer, a satellite modem for transmitting data, and power for a few instruments. Maybe something like this:

I intentionally chose much larger than reasonable values for the above requirements (server-scale onboard computer, heavy satellite communications, a ridiculous margin, etc.) and I still arrived at a figure that was less than 10% of the real ship’s power budget.

I think it’s safe to say that, whatever this thing was doing, it was a lot more than a “meteorological airship.”