Celebratory Gunfire in Tripoli

There was a letter this morning in our local newpaper (the Calgary Herald) from a reader concerned about all of the celebratory rounds fired off in Libya after the death of Gadhafi and if anyone was killed by those rounds falling back to earth [1]. I was curious if this was a realistic worry, so here’s my incomplete, extremely rough analysis.

The population density of Tripoli is given by Wonderful Wikipedia as 4500 people/km2 [2]. If we assume that the cross-sectional area of an average human being from above is roughly 0.5 m2 and knowing that one square kilometer is composed of 1,000,000 m2, then:

\frac{0.5\ \mathrm{m^2}}{\mathrm{person}} \times \frac{4500\ \mathrm{persons}}{1\ \mathrm{km^2}} \times \frac{1\ \mathrm{km^2}}{1,000,000\ \mathrm{m^2}} = 0.225\%

This means that looking straight down on a square kilometer of Tripoli, we would expect 0.225% of the total area to be occupied by bits of “people”. This also means that a random bullet falling down on our imagined piece of Tripoli has a 0.225% chance of hitting part of a person. That is all well and good but a far more interesting question is how many bullets do we have to shoot to have a 50% chance of hitting somebody? or a 90% chance of hitting somebody?

A Bit of Probability
A person needs to roll a six (on a six-sided die) to win a game. How many times would they have to roll the die in order to have a 50% chance of getting a 6? This is analogous to our bullet scenario, but a bit easier to visualize. It turns out that the best way to calculate this is the combine the probabilities of not rolling a 6 on each throw. If P(A) is the probability of rolling a 6 in n throws, and P(A)’ is the probability of not rolling a 6 in n throws, then we know that P(A) = 1 – P(A)’. We are interested in the number of throws of the dice at which P(A)’ becomes less than 0.50, where P(A)’ is the probability of not throwing a 6 on each toss (an independent event):

P(A)' = \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \cdots \\ \\  P(A)' = (\frac{5}{6})^n

Then the probability “break-even” point can be calculated as follows:

\log{P(A)'} = n \cdot \log{\frac{5}{6}} \\ \\  n = \frac{\log{P(A)'}}{\log{\frac{5}{6}}} \\ \\  n = \frac{\log{0.50}}{-0.079} \\ \\  n = 3.8

So this means that 4 rolls are required to have a > 50% chance of rolling one six. We can apply the same reasoning to our bullet problem. The chance (P(A)) of a bullet hitting someone is 0.00225, so the individual, independent chance of a bullet not hitting someone is 0.99775:

n = \frac{\log{0.5}}{\log{0.99775}} \\ \\  n = 308\ \mathrm{bullets}

To have a 90% chance of hitting someone (by accident of course!) we would have to shoot off:
n = \frac{\log{0.1}}{\log{0.99775}} \\ \\  n = 1,022\ \mathrm{bullets}

While 1,022 is a lot of bullets, if you had a sizeable crowd celebrating, that wouldn’t take long at all! Of course with all this shooting going on, it is quite likely that most people are indoors, but even if 75% of people were hiding indoors (and presumably the bullets were stopped by their roofs), that’s still only 1,232 bullets to have a 50% chance of a hit, or 4,093 bullets for a 90% chance. There is still a lot of room for error in this approximation, but it seems reasonable that there were some accidental hits – I would be very surprised if it came anywhere near the death toll of the conflict though!

(For those interested, Libya has an average population density of 3.6 per km2, again from our trusty Wikipedia [3]. This means that if the gunfire and people were spread out over the entire country, the 50% level would be 385,082 bullets and the 90% level would take an astonishing 1,279,213 bullets. The moral of the story is, of course: don’t live in cities!)

The Quick Physics/Medicine Digression
So now that we’ve established that some unlucky souls have a reasonable chance of having a thoughtlessly fired bullet coming down on them: will it be worse than a bump on the head? The unfortunate answer is yes. Basic physics tells us that a bullet shot up will return to earth at the same speed it left, in the absence of any air resistance. Air resistance however means that the bullet will come down slower, but how much slower? Amazingly, this research has been undertaken and the result ends up being a roughly 90 m/s terminal velocity for a .30 caliber bullet (the size used in an AK-47) [4]. This is more than fast enough to penetrate not only human skin, but also the human skull [5]: a target that is rather unfortunately positioned to intercept bullets falling from the sky.

In fact a case report in The Annals of Thoracic Surgery published in 2007 describes a case report of a patient presenting to a Michigan ED with severe injuries from celebratory gunfire [6]. The bullet entered the patient’s chest at the 5th intercostal space about 2 cm left of the sternal margin (see figure at left). The bullet then passed downwards through the right ventricle of the heart, through the diaphragm, and through the lesser curvature of the stomach. The bullet then came to rest inside the stomach but not before impacting the greater curvature of the stomach and injuring the splenic hilum (requiring a splenectomy). The bullet is clearly visible in the left upper quadrant in the pyelogram shown below.

So What!?
So there you have it! While it is unlikely that celebratory gunfire associated with the death of Colonel Gadhafi caused deaths to rival the death toll of the conflict (estimates in excess of 10,000 people are being bandied about), it is absolutely likely that some unfortunate innocent bystanders were hit and seriously injured or killed. While the terminal velocity of a bullet falling from the sky is much lower than the muzzle velocity of a typical assault rifle, the speed is more than enough to maim or kill.

[1] Martin Kennedy, “Letter: Careless shooting may have marred Libyan celebration“, The Calgary Herald. October 22, 2011.
[2] “Tripoli“, Wikipedia. Retrieved October 22, 2011.
[3] “Libya“, Wikipedia. Retrieved October 22, 2011.
[4] Julian Hatcher, “Hatcher’s Notebook”, p. 154, Stackpole Books. 1962.
[5] Michael J Stewart, “Head, Face and Neck Trauma: Comprehensive Management”, p. 189, Thieme Medical Publishers. 2005.
[6] Angelo Incorvaia MD, Despina Poulos MPH, Robert Jones MD, James Tschirhart MD, “Can a Falling Bullet Be Lethal at Terminal Velocity? Cardiac Injury Caused by a Celebratory Bullet“, The Annals of Thoracic Surgery, Vol. 83 No. 1, p. 283-284. January 2007.


One thought on “Celebratory Gunfire in Tripoli

  1. Seems to me to sound rather like a chapter from a biomedical engineering thesis… Did you check the maths of the terminal velocity? And did it consider turbulent flow and the likelihood that the bullet will tumble since its spin will probably have decayed.

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