Hoverboard Physics

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.

[1] As you can see, I’m trying to impress you with my use of technical jargon.

[2] 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.

Sneeze Breeze

I often catch myself thinking about the great things that humanity could do if we all worked together.  And while it’s nice to think about things like stopping hunger or working for world peace, it’s more fun to think about sneezing.  If all of humanity faced the same direction and sneezed at once, how fast could we push a sailboat?

From physics we know that momentum has to be conserved.  As an upper bound, let’s assume all of the momentum from the sneezed air gets transferred to the boat.  Our lungs can hold about 500 cm³ of air and air has a density of 1.2 kg/m³, meaning that the total mass of a sneeze is given by,

mass = (density) · (volume)

= (1.2 kg/m³) · (500 cm³)

= 0.6 g.

According to Wikipedia, sneeze speeds lie between 75 km/hr and 1045 km/hr, so I’ll assume an average sneeze speed of 200 km/hr.  The world population is 6.7×10⁹ people sneezing.  From this info, we can compute the total momentum of sneezed air,

momentum = (number of sneezes) · (mass per sneeze) · (speed)

= (6.7×10⁹ sneezes) · (0.6 g) · (200 km/hr)

= 2.2×10⁸ kg m/s.

We’ll assume that all this momentum is transferred to the ship.  

For our model ship, we’ll consider the fully rigged tall ship Bounty II, which has been featured in numerous films including Mutiny on the Bounty, Treasure Island, and Pirates of the Caribbean.  It is listed at 412 tons = 3.7×10⁵ kg.  From this we can estimate an upper bound for the total speed it would gain by the entire world sneezing on it,

speed = (momentum) / (mass)

= (2.2×10⁸ kg m/s) / (3.7×10⁵ kg)

= 600 m/s.

In contrast, the sound barrier is only 340 m/s, meaning the boat would be traveling at close to Mach 2 or twice the speed of sound..  Remember, however, that this is only an upper bound since a vast majority of the air’s momentum will not be transferred to the ship.

Brian Pothier on Ulam's Conjecture

Today’s question comes from this week’s special guest, Brian Pothier.  In addition to being an NHL defenseman for the Washington Capitals, Brian may also be an amateur physicist.  Perhaps without realizing it, Brian suggested a question that contains some fairly deep physics:

“How many pucks can you fit in a hockey net?”

The packing of three-dimensional geometric objects has been an on-going topic of research from antiquity to the present day¹.  Since we’re only concerned with an order of magnitude estimate, we won’t be treating this problem with the rigor that a mathematician might, but the odd shape of the goal makes finding a rigorous solution an interesting—or at the very least challenging—problem.

An official NHL hockey goal is 4 ft tall and 6 ft wide.  As mentioned above, the back of the net has a funny shape, but it looks about 2 ft deep.  This gives a total volume of ~1.4 m³ (42 ft³).

According to Wikipedia, a standard hockey puck is 1 in thick with a 3 in diameter giving a total volume of 116 cm³ (~7.06 in³).  If we stack them on top of each other in a hexagonal pattern, they’ll consume about 90.7% of the volume².  From this and the volume above, we can compute the number of pucks that can fit in a net,

# of pucks = (packing fraction) · (vol. per net) · (vol. per puck)

= (0.907) · (1.4 m³) · (116 cm³)

= 9,300 pucks per net

That’s a lot of pucks, roughly equal to the number of goals that Brian’s team the Capitals have scored in their entire history.  Alex Ovechkin would have to keep up his current scoring pace for 150 years to score that many goals. Thanks for the question, Brian!

[1] For more info on packing problems, you can check out this article or this one.

[2] This is equivalent to a packing circles in two-dimensions. 

You’re messing up my seasonal affective disorder

According to NASA scientists, our days are 1.26 microseconds shorter because of the Chilean earthquake.  Apparently, this is fairly common.  The magnitude 9.1 earthquake that caused the 2004 tsunami shortened the day by 6.8 microseconds.  The length of the day decreases because the distribution of Earth’s mass changed and angular momentum must be conserved.  The effect is similar to an ice skater rotating in a circle.  As she pulls her arms in, she rotates faster.  How much skinnier would the Earth have to get for the day to last only one second?

The angular momentum of a rotating body can be computed using the equation,

angular momentum = (moment of inertia) · (angular velocity).

Using the Earth’s mass (~5.97×10²⁴ kg) and radius (6371 km), we can estimate its moment of inertia as being equal to that of a sphere,

moment of inertia = 2 · (mass) · (radius)² / 5

= 2 · (5.97×10²⁴ kg) · (6371 km)² / 5

= 9.7×10³⁷ kg m².

The angular velocity of the earth is given by,

angular velocity = 2 · π /(period)

= 2 π /(1.0 day)

= 7.3×10^-5 Hz.

From this, we can compute Earth’s angular momentum to be about 7.1×10³³ kg m²/s. 

If the day was only one second long, the new angular velocity would have to be 6.26 Hz.  Using the fact that the angular momentum must remain the same, we can compute the new moment of inertia,

moment of inertia = (angular momentum) / (angular velocity)

= (7.1×10³³ kg m²/s) / (6.26 Hz)

= 1.1×10³³ kg m².

For reasons that will be clear in a moment, we will set this equal to the moment of inertia for a cylinder (rather than a sphere) and solve for the cylinder’s radius,

cylinder radius = [ 2 · (moment of inertia) / (mass) ]^1/2

= [ 2 · (1.1×10³³ kg m²) / (5.97×10²⁴ kg) ]^1/2

= 19 km.

To have a one-second long day, that Earth would become so skinny that it would start to look like a long cylinder.  If it maintained the same density, it would now be 9.7×10⁸ km long, about 2.5 times the distance to the Moon.

If you’d like to help earthquake victims in Chile and Haiti, please visit the Red Cross.

Spot, Good Boy!

Two of my favorite themes to estimate are space travel and green energy.  I view the two as related.  With population growth, I think the only way we’re going to maintain the planet is if some of us get off of it.  (Some people should probably leave sooner than others.)  With that said, I really like Thomas’s question from the comments below: “If we built a farm of giant wind turbines in Jupiter's Great Red Spot, how much energy could they provide?”

According to Wikipedia, the Great Red Spot has lasted for at least 180 years and possibly as long as 345 years.  Its measures 24–40,000 km west–to–east and 12–14,000 km south–to–north.  To make life easy, I’ll assume the Spot is a circle 20,000 km in diameter.  Air currents flow around the edges with a maximum rate of 120 m/s (432 km/h).  Jupiter’s surface pressure lies within the range 20-200 kPa and its average temperature is 125 K = -148°C.  If we assume the atmosphere is an ideal gas¹, we can compute the number density using the equation,

number density  = (pressure) / [ (Boltzmann’s constant) · (temperature) ]

= (100 kPa) / [ (1.38×10^-23 J / K) · (125 K) ]

= 6.0×10²⁵ m^-3.

This is the number density, i.e. the number of molecules in a cubic meter.  Since most of Jupiter’s atmosphere is made of hydrogen gas, we can estimate the mass density by multiplying the number density by the mass of a single hydrogen molecule²,

mass density = (mass of H₂) · (number density)

= (3.3×10^-24 g) · (6.0×10²⁵ m^-3)

= 0.20 kg/m³.

Wind turbines convert the kinetic energy of the wind into electric energy.  I’ll assume the turbine’s blades are 45 m long, so that they sweep out an area of 6,400 m². We can estimate the energy produced by using the formula,

power = (density) · (area swept by the rotor) · (speed of the wind)³ / 2

= (0.20 kg/m³) · (6,400 m²) · (120 m/s)³ / 2

= 1.1×10⁹ W

This is the total energy absorbed by the turbine, but not all of this will be converted into useable energy.  If we assume the turbine is only 10% efficient, we’d end up with a total of 1.1×10⁸ W per turbine. As the phrase “per turbine” suggests, this is how much energy we’d get from only one wind turbine.  Wind farms usually have multiple turbines running simultaneously.  If we rigged one up every 100 m in a circle around the edge of the Spot, we’d have a total of 630,000 turbines.  We could then compute the total amount of useable power,

total power = (# of turbines) · (power per turbine)

= (630,000 turbines) · (1.1×10⁸ W per turbine)

= 6.9×10¹³ W.

That’s enough energy to power 21 USAs.  Although clearly an impractical solution to the energy problem, the sheer amount of energy makes it an intriguing option as an energy source.  At the very least, it will hopefully make government officials think twice before slashing NASA’s budget.   

Thanks, Thomas.  This was a fun problem.

[1] Jupiter’s atmosphere is probably not an ideal gas, but this is a reasonable first estimate.

[2] You can calculate the mass of an H₂ molecule by taking it molecular weight (~2.016 g/mol) and dividing by Avogadro’s number.

Contest Reminder...

There's still plenty of time to enter the estimation contest. The winner gets a signed copy of How Many Licks?. Email your entries to “aaron at aaronsantos period com.”

As extra motivation, I’m posting this picture I found on reddit today. What can I say?  It's a small Internet.

Tornado Gun

Tornadoes are a big problem in much of the United States. In addition to ruining homes, cleaning up after them can cost the government millions of dollars each year. Perhaps there’s a way to shoot them down before they do any damage. A tornado is basically a giant pocket of air rotating really fast. Any mass moving in a circle must be accelerating. An accelerating electric charge releases power in the form of electromagnetic radiation, i.e. light.¹ In principle, if you dumped a bunch of electric charges into the tornado and let them spin around, the tornado should lose energy. How many charged pith balls would you need to fire at a tornado to sap all its energy out in one second?

To solve this problem, it helps to know some physics. In 1897, Joseph Larmor derived an equation² for the power P emitted by a charge as it accelerates,

P = e² a² / 6 π ε₀ c³.

Here, e is the electric charge, a is the acceleration, π = 3.1415926... , ε₀ = 8.85×10^-12 F/m is the permittivity of free space³, and c = 3.00×10⁸ m/s is the speed of light. Pith balls were one of the first materials used in an electroscope. Much like a balloon that you rub on your head, they can pick up a static charge. I’m imagining that we drop a bunch of pith balls onto a van de Graff generator to charge them and then fire them into the tornado. Pith balls can hold anywhere from 10^-9-10^-6 coulombs (C) of charge, so I’ll assume each one holds e = 10^-8 C. To calculate how much power the charges will drain out, we need to know what their acceleration will be once they’re inside the tornado.

For simplicity, I’m going to assume the pith balls are light enough that they’ll be caught in the air stream and get flung around like Dorothy’s house in The Wizard of Oz. If that’s the case, then we can estimate the acceleration of the charges from the wind speed of the tornado. According to Wikipedia, tornados have winds between 64 and 177 km/h (~40-110 mph) and are approximately 75 m (~250 feet) across. They maintain contact with both the ground and a cumulonimbus cloud, meaning they can be anywhere from 150 to 3,960 m tall. To make things easy, I’ll assume the tornado is shaped like a 2000 m tall cylinder with a 75 m diameter and an outer edge that’s moving at a speed off 177 km/h. Since most of the pith balls will land somewhere in the middle of the tornado, they’ll be moving slower than the edges. I’ll assume that on average they’re traveling 100 km/h at a distance 20 m away from the center. From these numbers, we can estimate the acceleration of particles inside the tornado using the formula,

a = v² / r

= (100 km/h)² / (20 m)

= 39 m/s².

By plugging this result into the Larmor formula, we can compute the power radiated by a single pith ball each second,

P = (10^-8 C) ² · (39 m/s²)² / [6 π (8.85×10^-12 F/m) · (3.00×10⁸ m/s)³ ]

= 3.4×10^-12 W per pith ball.

To calculate how many pith balls we would need to kill the tornado in one second, we’d need to know how much kinetic energy is in the tornado. The rotational kinetic energy of a cylinder is given by,

KE = I ω² /2,

where I is the moment of inertia for a solid cylinder and w is the angular velocity given by,

I = m r² /2,

and,

ω = v / r,

respectively. The mass of the tornado can be calculated using the density of air (1.2 kg/m³) and the dimensions of the tornado to get a total mass of 1.1×10⁷ kg. From this and the equations above, we can solve for the rotational kinetic energy of a tornado,

KE = m v² / 4,

= (1.1×10⁷ kg) · (177 km/h)² / 4

= 6.6×10⁹ J.

From this number and the total power emitted by a single charged pith ball, we can compute how many pith balls we would need to stop the tornado in one second,⁴

# of balls = (energy) / [ (power per ball) · (time)]

= (6.6×10⁹ J) / [ (3.4×10^-12 W per ball) · (1 s)]

= 1.9×10²¹ pith balls

This number is huge. If each pith ball were a centimeter in diameter, they would fill up a cube 100 km on a side.⁵

[1] Since a charged tornado would emit light, you might wonder if actually doing this would look cool. In truth, the frequency of the light emitted will be about the same as the rotational frequency of the tornado. This would produce invisible radio waves.

[2] Rigorously speaking, this formula only works in vacuum. The main result won’t change, but in principle you should use the permittivity of air.

[3] The “F” here is short for “Farad”, a unit of electrical capacitance.

[4] To solve this correctly, you need more advanced mathematics since the acceleration of the charges will not be constant, (i.e. the charges will slow down as the tornado loses energy). It should be possible to get a better estimate using calculus.

[5] There’s another fundamental problem we haven’t considered. Namely, once you have a few pith balls in the tornado, it will be harder to get more in there since the charges will repel.