Something's Off About This Slow-Motion Bullet Video

March 20, 2019 | Story | No Comments

Who doesn't love a good slow motion video? The Slow Mo Guys—Gav and Dan—sure do! In this video of theirs, they use a high speed camera to capture the motion of four different bullets. And lucky for me, the motion looks to be perfect for a video analysis: They give both a reference scale (the black and white markers in the back) as well as the frame rate (100,000 frames per second).

Let's just jump right into an analysis. I will be using Tracker Video Analysis to get position and time data for each bullet after it leaves the weapon. The bullets are so small that it can be difficult to always see them—for all but the largest bullets, I can only mark the bullets when they are passing in front of the white backgrounds. Still, this should be enough for an analysis.

Now for the data. I marked all the bullet positions so you don't have to. Here's a plot of position vs. time for each one (you can also view the plotly version).

I'm pretty happy with this—however, there is a problem. During the video, Gav and Dan switched from the slow-mo view back to a commentary view because the 45 caliber bullet was taking too long. When they switched back to the slow motion view, their timing was off. You can see this in a graph of position vs. time for that bullet. Oh, you can also notice the missing data when the bullet passed in front of the black parts of the background.

But how off is it? Let me first make the assumption that the bullet has a constant velocity in the horizontal direction. If this is the case, then a linear fit to the first part of the data gives a speed of 287.6 m/s. I should add that this speed would convert to 642 mph, which is faster than the speed listed on the video at 577 mph. Perhaps the displayed frame rate is different than the recorded frame rate? Maybe Gav and Dan could give me the answer here.

Anyway, back to the data. From the linear fit, I get the following equation of motion for the bullet.

This equation of motion should give the position of the bullet for any time. The "jump" time is at 0.00825 seconds. The constant velocity equation says that the bullet should have a position of 1.795 meters but the data from the video puts it at 1.886 meters. What about the reverse of this problem? If I know that the position is 1.886, what time should it be? That's a pretty straightforward problem to solve (algebraically). You can do that for yourself as a homework assignment, but I get a correct time of 0.008567 seconds. So, they were "late" by 0.000317 seconds. But wait! That's how far they were off in "real" time, but the video was played back in slow motion. It was recorded at 100,000 fps—but I assume it was displayed at 30 fps. That means this short time interval was actually off by 1 second. That's the mistake.

But that's just a cosmetic error, not really what I wanted to look at. Instead, I want to know if it's possible to estimate the amount of air resistance on these bullets as they leave the muzzle. I have to admit that air resistance on bullets can be pretty tricky. When these suckers are moving super fast, the simpler models for air resistance don't always work. But no matter what, an air drag force on a bullet should push in the opposite direction of the motion of the bullet and slow it down. So I will see if I can estimate the acceleration of the bullet during this short flight.

In one dimension, the acceleration is defined as the change in velocity divided by the change in time. That can be written as the following equation.

I just need to find the velocity at the beginning of the trajectory and then at the end. This will just be the slope of the position-time graph at these two points. Then I can divide by the time of flight for a rough approximation of the acceleration. Here's what I get:

Barrett: v₁ = 934 m/s, v₂ = 854 m/s, Δt = 0.0051 sec, acceleration = 15,686 m/s². This seems very high.
AK-47: v₁ = 752 m/s, v₂ = 698 m/s, Δt = 0.0062 sec, acceleration = 8710 m/s².
45 cal: v₁ = 246 m/s, v₂ = 242 m/s, Δt = 0.012 sec, acceleraiton = 333 m/s².
9 mm: v₁ = 351 m/s, v₂ = 330 m/s, Δt = 0.0105 sec, acceleration = 2000 m/s².

Since these values for the acceleration seem super high, I am going to roughly estimate the acceleration using a basic model for air drag. Here is the equation I will use:

In this expression, ρ is the density of air (about 1.2 kg/m³), A is the cross sectional area of the bullet and C is the drag coefficient. I can approximate the bullet size and mass from this wikipedia page and I will just use a drag coefficient of 0.295. With these values and the velocity right out of the barrel, I get an acceleration of 624 m/s. OK, that is high—but not quite as high as the measured acceleration. Still, I think the values from the video aren't super crazy. That bullet is moving really fast and interacting with the air that will make it slow down quite a bit—especially at first.

Of course ballistics physics can get pretty complicated, but that will never stop me from making some rough estimates.