Bullets, wind and assassins
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Recently, the news media had the occasion to speculate about the effect of wind on the trajectory of a bullet. A gust of wind, they said, might have spared Trump’s life on July 13th when the would-be assassin Thomas Crooks shot at him. Fortunately, Crooks missed Trump’s head – though not by much more than an inch.
It is true that given how narrow the miss was, almost any small change in circumstances could’ve changed the outcome dramatically. The question I’m interested in is: had Crooks been a competent shooter, could wind make him miss at 140 yards (130 meters)? That is, if he had been aiming dead center, would a gust of wind really have caused him to miss?
My own instinctive reply to this question is "no". I’ve fired an AK-style assault rifle at 150 meters, and windy weather seemed to have little impact on accuracy. The greatest loss of accuracy was caused by the dope pulling the trigger. Still, we can actually try to calculate how much a gust of wind would cause a bullet to deviate.
First, what is wind? For an accurate simulation of wind, you would have to do comprehensive fluid dynamics simulations, much like e.g. Formula 1 teams do to optimize their aerodynamic parts. That’s way too difficult, though. My claim is that wind is approximately air resistance – drag, in other words.
How come? Well, when you fire a bullet, it slows down due to going through air. Bullet is fired, air slows it, eventually it would stop completely (though it’s way more likely to hit something like the ground first). Imagine the unfortunate circumstance in which an evil wizard has shrunken you so that you can sit on a bullet as it is fired. Close your eyes – what do you feel?
Obviously, you feel the air rushing in to your face, probably quite painfully. However, from your frame of reference, the bullet is not moving at all: you’re sitting on it, so it’s stationary relative to you. So you’re feeling "wind", wind therefore must be the same as air resistance.
This model is, of course, imperfect. There are things like turbulence, Magnus effect (what happens to spinning objects in air – bullets spin!) and who knows what else which could affect the trajectory of the bullet. My guesstimate is that treating wind as air resistance captures the most important properties, though. If calculating a cow is too difficult, assume a spherical cow of the same volume.
| Caliber | Mass (g) | Muzzle velocity (m/s) |
|---|---|---|
| 5.56x45mm XM193 | 3.56 | 993 |
| 5.56x45mm GP 90 FMJBT | 4.1 | 851 |
| 5.6×15mmR (.22LR) 40gr | 2.5 | 370 |
| 5.6x15mmR (.22LR) 30gr | 1.9 | 500 |
| 7.62x51mm NATO 147gr | 10 | 850 |
| 7.62x51mm NATO 175gr | 11 | 790 |
| .38 Special 130gr | 8.4 | 240 |
We also need some wind speeds. A very strong gust of wind would be something like $v_{\text{wind,strong}}=20\ \text{m/s}$, which would be 40+ mph – according to this site, this would "make walking against the wind very difficult". Around wind speed 10 m/s, "small trees sway, waves break on inland waters". This type of gust was probably the highest present at the rally – based on video footage anyway.
The bullet’s trajectory is primarily influenced by its mass and velocity when it leaves the barrel. I’ve tabulated these values for some common calibers above.
Let’s get to calculating!
A simple model
First, we make the simplest assumption: the bullet doesn’t slow down at all during its flight and the wind is precisely perpendicular. For a short distance under 150 meters, that’s a pretty good assumption; it doesn’t have the time to slow down very much. It also means we can simply solve the flight time first and solve the sideways motion separately (assuming the sideways movement doesn’t add much to the flight time).
The flight time then is
\[\begin{aligned} t_{flight} = \frac{d}{v_0}\end{aligned}\]where $d$ is the distance and $v_0$ the velocity of the bullet. For example, taking $d=130$ m and $v_0 = 993$ m/s (5.56 mm) yields $t_{flight} = 0.13$ s.
The drag force – thus our wind force – is
\(\begin{aligned} F_{drag} = \frac{1}{2}\rho cv^2A\end{aligned}\) where:
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$\rho$ is the density of the medium (in our case air),
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$c$ is a constant that depends on the shape of the object,
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$v^2$ is the speed of the wind,
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$A$ is the cross sectional area of the object.
Now, $\rho _{air} = 1.2$ kg/m$^3$. $A$ can be calculated roughly using the assumption that the bullet’s side profile is rectangular and lowering that estimate by, say, 10% because the tip is sharp. That’s not precisely true, but good enough. For example, for a 5.56x45 mm bullet we would have a cross section of roughly $1.15\cdot 10^{-4}$ m$^2$ (remember that the length of the bullet is not 45 mm – that’s the cartridge). So,
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$\rho = 1.2$ kg/m$^3$,
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$c=1.2$,
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$A$ depends on the bullet, for 5.56x45 mm about $A=1.15\cdot 10^{-4}$ m$^2$,
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$m$ depends on the bullet, for 5.56x45 mm $m=3.56\cdot 10^{-3}$ kg,
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wind speed $v=10$ m/s.
The coefficient c is more difficult to determine for a bullet, but since it’s sort of cylindrical, we’ll guess it’s around $c=1.2$.
Let’s calculate the effect of wind on the bullet. Newton’s equations are easily solved if we assume that the sideways speed of the bullet doesn’t depend on time. This is false, of course – if the bullet didn’t actually pick up sideways speed, it wouldn’t move sideways! The assumption amounts to saying that the final speed is so slow as to be insignificant. We’ll later see why this was a good guess.
\(\begin{aligned} F=m\ddot{x}=\frac{1}{2}\rho _{air} c v_0^2 A \implies \Delta x = \frac{\frac{1}{4}\rho _{air} c v_{wind}^2 A t_{flight}^2}{m}\end{aligned}\) where we have assumed that initially the bullet is not moving sideways and where $\Delta x$ is the amount of distance the bullet has moved due to wind. We know all the numbers on the right hand side. For the 5.56x45 mm bullet with 10 m/s wind, we get \(\begin{aligned} \Delta x = 1.95\ \text{cm} \approx 0.8\ \text{inches}.\end{aligned}\) Certainly not enough to cause a miss! You can plug in values for the other bullets too, but the result doesn’t really change significantly.
Slightly more complicated model
We just assumed that the bullet’s sideways speed doesn’t affect the force. A more accurate force would be:
\[\begin{aligned} F = \frac{1}{2}\rho _{air} c (v_0-\dot{x}(t))^2A\end{aligned}\]so that the relative wind speed takes in to account the increasing speed of the bullet. As you can observe for yourself, this makes essentially no difference after solving the equations:
As long as the flight time is below 0.4 seconds, the displacement is pretty much indistinguishable.
What’s in a model?
I couldn’t come up with a model better than this that could be solved with comparable effort. I think the assumptions going in to it are fairly reasonable – probably it gives an answer of the right magnitude. A better model would be, as I said, some sort of full fluid dynamics simulation, but that would be outside of my area of expertise and in any case way too much effort.
The end result is: the wind won’t mess up your shot at that distance. Bad aim does.
It would be amusing to test this model in the real life at a range – it wouldn’t be possible to make the wind perfectly perpendicular, of course, but the calculation could be fixed for different wind directions. For different wind angles, the cross section would change and so would the drag coefficient.
Unfortunately, empirical testing would require me to be a better shot – I don’t think I’d get a tight enough grouping to measure a few centimeters’ difference.