I just saw the Veritasium video, The Fatal Physics of Falling Objects. At around 16:30, it's said that a falling bullet would start tumbling and likely land on it's side. Adam Savage says that the bullet chooses an orientation such that it feels maximum resistance.

  1. Why is this true?

Immediately afterwards, Derek says that if you fire a bullet upwards at some angle with the horizontal, it maintains it's horizontal velocity at it's peak, and "that combined with the spin imparted to the bullet by the grooves inside the gun barrel, keep it moving pointy end forwards", in his words.

  1. In this case, why doesn't the bullet tumble? I didn't seem to get a satisfactory explanation from the video for these two questions.

3 Answers 3


The video is frustrating because while some of his content is great, he really needs to stay out of the “probably” business. Nearly all bullets have their CP (center of pressure) forward of their CG (center of gravity). Thus if you drop one from a height, it will tend to fall tail-first. Instead of resorting to “probably” he could have just tested this. A falling object seeks force equilibrium, not minimum drag. The minimum drag configuration is nose low. The forces-in-equilibrium configuration is the CG leading into the relative wind, the CP trailing directly behind it, and the CG-CP arm aligned with the relative wind. This is the equilibrium attitude. Keeping the bullet in its lowest-drag configuration means keeping the “pointy end” forward. This is done with spin in a CP-forward projectile, and with fins in a CP-rearward projectile.

A bullet spinning in flight is experiencing extraordinary gyroscopic forces. A typical bullet is spinning in the range of 150,000 to 300,000 RPM. Like all objects under gyroscopic influence, that means a bullet in flight has two properties:

  • Rigidity in space. The bullet resists perturbation along its longitudinal access. This is good, it keeps the bullet from flipping around and flying tail-first. But it also means the bullet’s orientation is called “intractable”… It tends to stay in the orientation that it left the barrel, and as the bullet’s path starts to descend, the bullet is starting to fly more and more “nose high” compared to the relative wind. This creates upward pressure on the CP of the bullet, forward of the CG. So the nose should go up, right? Actually, no. This is because of the other property of gyroscopes…

  • Precession 90 degrees in the direction of spin. The gyroscopic force translates that nose-up, tail-down torque on the fast-spinning bullet into a nose-right yaw on the bullet because most rifling is to the right (clockwise ) from the shooter’s perspective. There will be a tiny Magnus force pulling the bullet to the left and a much larger gyroscopic force pulling the bullet to the right, called spin drift.

People try to simplify ballistics by applying intuition. But that intuition is almost always formed by a lifetime of observing things with “high-Magnus, low-spin drift” like a ball, or things with their CP behind their CG, like model rockets or arrows, in an effect called “weathervaning.” But that intuition is incorrect because most of us will never visibly observe an object in flight that has a CP forward of the CG, nor one that is traveling a long enough distance or time of flight while remaining in our view. I cannot hit a golf ball far enough for Coriolis to really matter. I can’t spiral a football with nearly enough spin for me to be able to observe spin intractability. And so our intuition about how a bullet flies will almost always end up wrong.

  • $\begingroup$ "The minimum drag configuration is nose low." Nose-down is also an equilibrium position. It's just not a stable equilibrium. $\endgroup$ Oct 4, 2022 at 1:17
  • $\begingroup$ @Acccumulation Absolutely agree. Thanks for “pointing” that out, pun intended. $\endgroup$
    – Max R
    Oct 4, 2022 at 1:28

If the bullet was fired straight up then at the peak it is at some point of unstable equilibrium perhaps with the point up. As it starts to fall there is some instability in air flow so it tilts to one side or the other, and the drag at the ends has more torque than the drag at the middle so it will likely tumble. The pointy end down has less drag, but the torques are larger when they are acting on the ends of the bullet, and the bullet shape is not symmetric so probably there is a little more torque at one end than the other, so the bullet ends up tumbling, or depending might end up falling at some angle. It is not just drag, but also if there are forces from the airflow over the surface.

If you think about a sheet of paper or a leaf falling, even though there is less drag if the edge was pointed down, Even if the edge is pointed down, it is not a stable equilibrium and the leaf or paper will then rotate so the broader area is parallel to the ground. The air flows and motion can be complicated. A parachute on the other hand would be designed to fall in a way where it would be near a stable equilibrium.

For the second question. I think they are assuming the bullet is also spinning on its axis after being shot from a rifled barrel. In that case the spinning is helping to keep the bullet from tumbling, and the airflow is a lot more predictable. Since you have a high spin rate around the axis it has substantial angular momentum to counter the torques that would cause it to tumble. If you google gyroscopic stability factor, you will see bullet manufactures claim that if the value is bigger than some number it will keep the bullet from tumbling.


Basically after a shot, bullet fights against turbulence with high speed and spinning, but when it reaches vertical highest point and turns back,- turbulence and air flow produces some torque around bullet COM, so it starts rotating. Due to rotation cross-section increases, which in turn increases air drag force, which amplifies this instability.

If bullet would have some sort of stabilizing wings (like rockets has),- then probably air flow could set it back to a stable alignment.


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