Shooting at a hurricane

A couple of years ago, Florida found itself staring down Hurricane Irma, a powerful category-5 storm which had already wrought havoc in the Leeward Islands and Greater Antilles. In true Florida Man fashion, a Facebook event sprung up that encouraged people to shoot at Irma – something which apparent thousands of Floridians agreed was a good idea.

Of course this was all tongue in cheek; nobody sane would suggest that a display of martial power could have any effect on a hurricane.

Nonetheless, it did make me wonder how much gunfire it would take to stop a hurricane, and how it could actually work. I came up with three possible scenarios:

( I should probably say at this point that I have an amateur knowledge of hurricanes, and even less knowledge about firearms. I accept no responsibility for facepalms induced by the following text – you have only yourself to blame )

1. A steady stream of bullets will entrain air, producing wind in the same direction as the gunfire. If the wind is strong enough, and well aimed, this could modify the hurricane’s airflow – perhaps even enough to disrupt it.
2. Hurricanes are heat engines, and fired rounds are warm. A huge number of bullets could be used to deliver huge quantities of thermal energy. The problem here is that adding extra energy will make the hurricane stronger, but there are ways around this.
3. Hurricanes can strengthen over water, generally weaken over land, and don’t like mountains at all. If effects 1 and 2 are weak enough, you’re going to need a literal mountain of ammunition. In this case, you’re better off constructing a wall of bullets to shield you from the hurricane.

Now, before we actually start working things out, we need to pick a gun. I’m going to go with the highest-selling firearm in the USA, the Smith & Wesson M&P shield 9mm pistol. This pistol can fire a 7.5 g bullet at a velocity of 360 m/s. Getting the temperature of the bullet is a bit trickier – you can find a surprising amount of discussion on this topic; the consensus for 9mm rounds seems to be around 150 °C. I’m going to assume that the entire bullet is this hot (even though this is likely an overestimate – the jacket gets heated due to friction with the barrel).

Finally, I’m going to make one further assumption. People who are willing to shoot at a hurricane will also be willing to take a boat out into the storm and deliver the momentum/heat/pile of bullets where it’s most effective. This avoids the problem that these bullets will get, at best, a couple of kilometres out to sea, and if the hurricane’s that close, you’re probably too late.

1: Disruption by wind

As bullets travel through the air, they slow due to air resistance. Momentum is still conserved, though, so the momentum lost from the bullet is transferred to the surrounding air. Each bullet carries a maximum of 2.7 N s of momentum; encouragingly, this is enough to slow 1 m³ of air by 2 m/s (about 5 mph). So now we need to estimate how much momentum it takes to disrupt a hurricane.

Hurricanes are roughly cylindrical, and extend most of the height of the troposphere. Since air density drops roughly exponentially with height, the total momentum should be relatively easy to estimate; starting from the integral

\( \mathrm{momentum} = \int_0^{\infty} \int_0^{\infty} \int_0^{2\pi} \rho(z) v(r) r ~ \mathrm{d}\phi~\mathrm{d}r~\mathrm{d}z\)

we can approximate the vertical (\(z\)) integration like so:

\(\int_0^{\infty} \rho(z)~\mathrm{d}z \approx \rho_0 H \)

I’ve added the variables \(\rho_0\) and \(H\), which are the surface air density and the scale height of the atmosphere – 1.2 kg/m³ and 8 km respectively.

The velocity integral is storm-specific, and a little trickier. Immediately before landfall in Florida, the National Hurricane Center predicted that hurricane-force winds extended approximately 100 km from the centre, while storm-forced winds reached 300 km. From these two points, we can fit a power law:

\( v(r) \approx 23 000~\left(r\,/\,\mathrm{m}\right)^{-0.57}~\mathrm{m/s} \)

This is obviously going to create problems when \(r\) is very small, so I’ll truncate the integral at 30 km, which was roughly the radius of Irma’s eyewall. This is less obviously going to create problems when \(r\) is large, so I’ll limit the outer radius to 300 km, the extent of the storm-force winds. Finally, putting all of this together, we get a total momentum of around \(7 \times 10^{16}\) m/s.

Since each bullet is capable of delivering 2.6 N s of momentum, we’ll need about \(\mathbf{2.7 \times 10^{16}}\) bullets. That’s quite a lot – let’s see if the heat is any better.

2: Disruption by heat

The problem with adding heat to a hurricane is that it’s quite likely to intensify. Hurricanes work by exchanging heat from the warm ocean and cold upper atmosphere, so adding heat near sea level only feeds the hurricane. What if, instead, we used gunfire to heat the upper atmosphere? No temperature gradient means no convection; no convection means no hurricane.

Well, there’s an immediate problem – I’m happy to let our gunners use a boat, but I think a plane is a bit too far, and since our 9mm bullets can only reach an altitude of 1 km or so, we don’t stand much chance of heating the upper atmosphere. It should be enough, however, to heat the lower kilometre: this will prevent convection of water vapour, which is what carries most of the heat.

So, to work out how much heat we need, we must know three things: how temparature changes with height, how air density changes with height, and the heat capacity of (pretty humid) air.

This graph gives us the first piece of information – assuming a surface temperature of 30 °C, the temperature of moist air at 1 km is 27 °C. Also changing the surface temperature doesn’t have much effect on the temperature gradient; it’s always around 3° cooler at an altitude of 1 km. We can do a quick bit of interpolation to get

\( T(z) = (30 – 3 z\,/\,\mathrm{km}) °C \)

For the density, I can use an exponential formula: \( \rho(z) \, = \, \rho_0 \, \mathrm{exp}\left(\frac{z}{8~\mathrm{km}}\right) \). The heat capacity is more complex than it initially appears, since it depends on the humidity of the air. If we assume it’s saturated with water, we’ll get an upper bound for the heat capacity of \(C_v\) = 754 J / kg °C.

Combining these three, we get another integral for the total heat needed (where \(z\) is measured in km, and \(R\) is the nominal radius of the hurricane):

\( \mathrm{heat} = \pi R^2 C_v \rho_0 \int_0^1 (30 – 3 z)~\mathrm{exp}(z / 8)~\mathrm{d}z \)

Plugging in the numbers again, we get a total heat requirement of \(6.9 \times 10^{15}\) J, or in bullets, \(\mathbf{4.8 \times 10^{16}}\). Bizarrely, this is almost identical to the previous answer; if you found yourself in possession of around 10¹⁷ bullets, you could try both and see what works better.

Alternatively, you could try…

3: Disruption by giant pile of ammo

Both the previous estimates gave numbers of bullets in the tens of quadrillions – for comparison, the total number of bullets fired during the Second World War is estimated at 100 billion (10¹¹) or so. If we armed every Floridian with two pistols, and they fired each 10 times per second, it would take just under four years to deliver the required number of bullets. That’s actually a lot less than I’d have thought, but still far too slow.

Since we’ve got all these bullets, why not just dump them in a huge pile? If our rounds are roughly 1x1x2 cm, they take up a volume of around 100 cubic kilometres. Surely this is enough to disrupt any weather system that tries to pass over.

Oh – that’s actually quite a lot smaller than I expected.

Between Florida and the Bahamas, the sea is about 500 metres deep, meaning the rounds could form an artificial island with an area of over 200 km². Again, that’s big, but probably not big enough – for comparison, Irma’s eye had an area of around 3000 km².

So, how about instead of disrupting the hurricane, we just try to defend against it? 100 km³ of material would allow us to build a flood barrier 185 metres tall (including borders with Alabama and Georgia), which would protect it even if all the ice on Earth melted.

Unsurprisingly, this giant flood barrier does create one problem (aside from decreasing Florida’s tourism income): hurricanes cause a lot of damage from rainfall. This wall would do a pretty good job of retaining water, which slightly misses the point of avoiding hurricane damage.

Luckily, the wall has a convenient built-in solution: hold a match to it, and it will provide enough energy to evaporate the entire rainfall produced by Irma… in a few hours. Then you’ll need another wall.

This whole exercise was obviously ridiculous, but I think it really shows the power of slightly-organised hand-waving (a.k.a Fermi estimation). Two very different methods involving very different physics gave basically the same answer. Taking one of these answers and calculating something different (but related), we get basically the same thing. That, and the knowledge that you can’t stop a hurricane by shooting at it.