Forget the flying cars, the coolest thing in Back to the Future II was the hoverboard. I’m not sure what Marty and Doc Brown did to screw up the space-time continuum, but they better fix it because we’ve only got five years to invent to the necessary hover technology. How much power would a hoverboard need to stay afloat?
I’m assuming that the hoverboard works like a conventional hovercraft, which floats by creating a cushion of highly pressurized air underneath it. Marty’s hoverboard is about 30 cm (~1 ft) by 90 cm (~3 ft) and appears to have two circular propulsion doodads.1 I’ll assume the circles act like fans that push pressurized air down. From the picture, it appears they each have about a 13 cm (~5 in) radius, leading to a total area of about 1100 cm2.
In order for the board to hover, the pressure underneath must be large enough to lift the weight of the board plus Marty. If the total mass of Marty and the board is 80 kg (~180 lbs), then the total force pushing down is,
force = (mass) · (gravitational acceleration)
= (80 kg) · (9.8 m/s2)
= 780 N.
This force is applied evenly over the area of the board, which is about 2700 cm2. From this, we can calculate the extra pressure under the board that’s needed to make the craft hover,
pressure = (force) / (area)
= (780 N) / (2700 cm2)
= 2900 Pa.
In order to generate this extra pressure, air has to be pumped under the board through the circles. We can estimate the speed of air flowing through the circles using Bernoulli’s equation2,
velocity = [ 2 · (pressure) / (air density) ]1/2
= [ 2 · (2900 Pa) / (1.2 kg/m3) ]1/2
= 70 m/s.
In a previous problem, we estimated the power absorbed by a wind turbine. The hoverboard is basically the same process in reverse, so we can use the same formula to estimate the power it takes to run the hoverboard,
power = (air density) · (area swept by the rotor) · (air speed)3 / 2
= (1.2 kg/m3) · (1100 cm2) · (70 m/s)3 / 2
= 23,000 W.
That’s roughly the amount of power needed to run 20 air conditioners. If we were going to run this on 100% efficient solar power, we’d need a solar panel with a total area of about 18 m2.
 As you can see, I’m trying to impress you with my use of technical jargon. Rigorously speaking, Bernoulli’s equation doesn’t work here since air is compressible. There are also some subtleties about the system not being enclosed, but this should be a reasonably good order of magnitude estimate.