In the article Sparta from the book What If?, the given problem is to know how many archers would be required for the scene of flying arrows blotting out the sun. His method is to calculate the number of arrows in a meter width of the archer battery. Each archer is assumed to have a fire rate of 150 mHz, or 0.15 arrow per second. The arrow flies 3 seconds on air. The number arrows in the air on 1 s for each archer is 0.45.

Attempt 1

  • 2 archers per meter in one row
  • A row every 1.5 m
  • 20 rows (30 m)

-> number of arrows in the air in 1 s in 1 m width: 2 x 20 x 0.45 = 18 (agrees with the book)

enter image description here

Attemp 2

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  • 6 archers per meter width (Since triple the number of archers in square foot essentially means triple the number per square meter, and since each row stands 1.5 m from each other, hence you just simply triple the number in one row.)
  • 40 rows (60 meters x 1.5 row/meter)

-> number of arrows in the air in 1 s in 1 m width: 6 x 40 x 0.45 = 6 x 18 = 108

This does not match with the number of 130 in the book. Why? 130 is even indivisible to 40. More importantly, it says "130 archers per meter". The correct number of archers per meter width column is 6 x 40 = 240.

In the end of the attempt 2, there is a bit of information: enter image description here

I assume that the number 339 is actually the number of arrows fired per meter width per second. But it is $240 \frac{archer}{m} \times \frac{7}{8} \frac{arrow}{archer.s} = 210 \frac{arrow}{m.s}$, not even close to 339. Why?

  • Only 0.15 arrows per second? Legolas can do much better than that if you believe the LOTR movies... (Joking aside, given the fantasy-style of the movie 300 with its improbably high fighting quality of various characters, a rate of 1 arrow every >6 seconds seems pretty pessimistic.) – TMM Mar 7 '17 at 11:52
  • The only thing that comes to my head is that in one meter wide column there are 240 archers, and this number was misspelled (2 lies next to 1, 4 lies next to 3). Is this number used in further text? By the way, 18*6=108. – Jaroslaw Matlak Mar 8 '17 at 8:52
  • @JaroslawMatlak This is a hilarious surprising correct answer. I have added more information in my edit – Ooker Mar 8 '17 at 12:03

It seems, that I've found out 'what author was thinking about'. - In second attempt density of archers from $2/m$ is increased to $1/ft$. Then, if we take the greek foot (as Spartans were Greek), ie. $1ft = 0.3083 m$, we have $ \frac{40}{0.3083} \approx 129.74 \approx 130$ archers in one meter firing column.

  • hmm, $2/m$ is increased to $6/m$, which equals $0.5/ft$. And using the greek foot seems overcomplicated to me – Ooker Mar 8 '17 at 15:11

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