Can a bullet travel all the way around a centrifuge? I'm not planning to shoot any astronauts, but researching a problem for a story.
Picture a rotating centrifuge in space, spun to generate artificial gravity, like so: https://media.wired.com/photos/5b33cf8f8d9ebd40cd3edeee/master/w_1600%2Cc_limit/rotatingspacecraft1.png
What I want to happen is to have this spinning at the right rate so that a bullet fired from a typical handgun will travel in a path maintaining constant height above the floor (where a person would be standing),completing at least one orbit before atmospheric affects cause it to fall. For this to happen, the bullet must be travelling in the direction opposite to the spin direction--the coriolis effect will cause it to drift away from the "floor", so that it drifts upwards.
My initial assumption is that this will work if the bullet's velocity matches the rim velocity of the centrifuge. A quick calculation (http://www.artificial-gravity.com/sw/SpinCalc/SpinCalc.htm) tells me that this can work just fine--with a bullet velocity of ~200 m/s for a typical pistol, this would work in a centrifuge of a radius if 4 km or so. Big, of course, but not outside the realms of possibility or comfort, and it would give it a reasonable speed of half a rotation per minute.
But what I'm not sure about is the atmospheric affects. How much will drag on a bullet move it off its initial course? Does this need to happen in a vacuum to work? Are there any errors in my assumption that make this impossible? Any help would be appreciated.
Edit:

This image illustrates what I'm talking about. (Source: https://oikofuge.com/coriolis-effect-rotating-space-habitat/) The projectile to an outside observer would appear to be stationary. Which is bad news for anyone firing bullets in this thing.
This and other sources seem to confirm my thinking. Am I reading them right?
 A: The bullet will move straight from the point of view of an inertial observer, there is no force on the air that provides it with a centripetal force, such as the friction on the foot for humans and furniture. Thus in the frame of the rotating centrifuge the bullet will "fall" towards the floor, if you point the gun horizontally.
A: I decided to simulate the flight of the bullet, including air resistance, in Mathematica.  In a rotating reference frame, the equation of motion for the bullet is
$$
m \vec{a} = -\frac{1}{2} \rho C_d A |\vec{v}|\vec{v} + 2 m \vec{\omega} \times \vec{v} + m \vec{\omega} \times (\vec{\omega} \times \vec{r}),
$$
where:

*

*$m$ is the mass of the bullet (I assumed 5 grams);

*$\rho$ is the density of the air (about 1.225 $\text{kg/m}^3$ on Earth);

*$C_d$ is the drag coefficient of the bullet, which I took to be 0.295 according to this random website;

*$A$ is the cross-sectional area of the bullet (I assumed a 10 mm diameter); and

*$\vec{\omega}$ is the angular velocity vector of the station.  This will point along the station's axis and have a magnitude of $\sqrt{g/R}$, where $R$ is the radius of the station and $g$ is the strength of "gravity" on its inner surface.  I took $g = 10 \text{ m/s}^2$ and $R = 4000$ m (at least initially).

A word about my assumptions concerning air resistance:  I assumed that the quadratic drag assumption was valid at all times in the bullet's flight.  This is probably a valid assumption for firearms with subsonic ammunition, but would not be a valid assumption for firearms with supersonic ammunition.  To avoid this complication, I assumed a muzzle velocity of $v_0 = 200$ m/s, well below the speed of sound.
I also assumed that the air density was constant throughout the station and was rotating with the station.  The bullets were "fired" from a height of 1.5 m "above" the inner surface of the station.  The trajectory of the bullets was assumed to be perpendicular to the axis of the space station's rotation.
Based on this, here are my results.  The blue line is the trajectory;  the black curve is the outer wall of the station.  Code is available at the bottom of this answer.
Horizontal firing against rotation, large station

The bullet lands about 270 m away after a time of flight of about 2 seconds.  Note that this time of flight is substantially longer than a bullet dropped from a similar height on Earth (that would be about 0.55 s), since the Coriolis force is pushing "up" on the bullet.  But unfortunately, since the bullet is constantly losing speed to the drag force, there is insufficient Coriolis force to maintain the circular trajectory, and it eventually hits the inner surface of the cylinder.
As pointed out in the comments, a space station could use a low-pressure pure oxygen atmosphere instead of an Earth-like atmosphere.  Since the density of the air would be lower, the bullet would go further.  Wikipedia says that the lowest pure-oxygen spacesuits use 20.7 kPa of oxygen, which according to the ideal gas law reduces $\rho$ to about $0.272 \text{ kg/m}^3$.  Running the code with this gas density leads to a range of about 615 meters rather than 270 — farther, but still nowhere near all the way around.
Can we reduce the density sufficiently to get the bullet all the way around?  From playing with the code it appears that a density of about $0.2 \text{ g/m}^3$ would do the trick.  For comparison, the atmosphere of Mars has a density of about $20 \text{ g/m}^3$;  a density of $0.2 \text{ g/m}^3$ is (I think) comparable to the density of Earth's atmosphere at about 60–70 km, which is getting pretty close to what some people call outer space.
Near-vertical firing, large station

It is possible for you to hit yourself with a handgun bullet on a large space station if you fire it nearly (but not quite) vertically, with a small component of the initial velocity against the rotation direction.  With the given initial height and muzzle velocity I assumed, an elevation angle of about 79.5° seems to do the trick.  The bullet returns after about 19.4 seconds with a speed of about 55 m/s.  I don't know whether this would be fast enough to be lethal, but it would definitely leave a mark.
More baroque trajectories

It's possible to get a bullet a large fraction of the way around the space station if the space station is smaller and you fire it with a significant vertical component.  Above is a a bullet fired at an elevation angle of 35° on a 400-m space station.  Roughly, if the station is smaller, it is easier to fire the bullet so it gets close to the axis;  and if it gets close to the axis, then the centrifugal force it feels is smaller, giving it more "hang time".  However, it does not seem to be possible to get a bullet to travel around the entire space station without using this effect;  and this does not fulfill the brief of a bullet that stays near the inner wall of the space station for its time of flight.

Code:
(* All numbers are in SI units *)
r = 4000.; (* station radius *)
g = 10 ; (* "gravity" at inner surface of cylinder *)
omega = Sqrt[g/r]; (* angular velocity of station *)
vst = omega r (* tangential velocity of station rim *)
period = 2 Pi/omega (* rotational period *)

m = 0.005;(* bullet mass *)
rho = 1.225; (* air density *)
cD = 0.295; (* drag coefficient for bullet *)
a = Pi*(0.010/2)^2; (* cross-sectional area of bullet *)
v0 = 200.; (* muzzle velocity *)
theta0 = 0 Degree; (* "elevation angle" for shot;  0 = "horizontal" & against rotation, 90 Degree = "vertical" *) 

eoms = Thread[D[{x[t], y[t]}, {t, 2}] == 
        -(1/2) (rho cD a)/m Sqrt[x'[t]^2 + y'[t]^2] {x'[t], y'[t]} 
        + 2 Drop[Cross[{0, 0, omega}, {x'[t], y'[t], 0}], -1] 
        + omega^2 {x[t], y[t]}]
ics = {x[0] == 0, y[0] == -(r - 1.5), x'[0] == v0 Cos[theta0], y'[0] == v0 Sin[theta0], WhenEvent[x[t]^2 + y[t]^2 == r^2, "StopIntegration"]}

solution = First[NDSolve[Join[eoms, ics], {x, y}, {t, 0, 2 period}]];
tend = (x["Domain"] /. solution)[[1, 2]] (* time of flight *)
{x[tend], y[tend]} /. solution (*landing coordinates *)
Norm[{x'[tend], y'[tend]}] /. solution (*landing velocity*)

ParametricPlot[{x[t], y[t]} /. solution, {t, 0, tend}, Epilog -> Circle[{0, 0}, r]]

A: 
What I want to happen is to have this spinning at the right rate so that a bullet fired from a typical handgun will travel in a path maintaining constant height above the floor (where a person would be standing)


[...]for a typical pistol, this would work in a centrifuge of a radius of 4 km or so.

Summerizing the setup:
A 4 kilometer diameter rotating space station will have the rim going at a velocity of 200m/s

Let me first go over the case without atmospheric drag, subsequently I will bring in atmosphere.
The initial state is that the handgun is co-rotating with the hull of the space station, moving at 200 m/s.
When the handgun is fired, in the opposite direction of the direction of motion of the hull, the bullet is brought to a standstill; the bullet is now stationary with respect to the center of mass of the space station.
In the absence of air friction there is now no physical connection between the bullet and the space station. The person who fired the handgun is still co-rotating with the space station, so if after a full revolution the bullet hasn't been hit by something the person who originally fired the bullet will intercept that bullet, with a relative velocity to the bullet of 200 m/s.
Air friction
Firing the bullet brought it to a standstill with respect to the center of mass of the space station. The air of the space station will give that bullet velocity again, in the direction of co-rotating with the space station. This velocity will be (to a first approximation) in a straight line. That straight line will intersect the hull of the space station.
As seen from a point fixed to the space station the bullet will start to drop.
So what you need to obtain is an estimate for how much velocity the bullet will gain (velocity relative to the center of mass of the space station) That will give you an estimate of how much time until the bullet has dropped so much that it is of no use for the plot of the story.
There was an episode of Mythbusters where they moved to a remote location to set up a rig that would fire a bullet from a handgun, attempting to fire straight up so accurately that the bullet would hit the ground again very nearby. Those bullets had very long flight times, so it may well be that in the space station story after a full revolution of the station the bullet hasn't dropped all that much.
So I think the follow-up question for this one is:
"In the absence of gravity causing drop, how much is a bullet from a handgun slowed down after half a minute?"
A: I think you're right that atmospheric effects are an issue. This is easiest to see if you picture the situation in a non-rotating reference frame.
In that case, just after the bullet is fired it is stationary in space. However, it is inside a 4km radius rotating habitat, and the rim of the habitat is moving relative to it at exactly the speed of a speeding bullet. That means the air is moving past it at that speed as well.
So the situation is analogous to a single bullet in a wind tunnel, exposed to 200 m/s wind speeds. The bullet is stationary initially but it's not going to stay that way for long. The bullet has aerodynamics on its side (it's pointing into the wind and designed to travel far), but given the size and speed of the habitat it will take a couple of minutes to complete a full revolution, and that's really more than enough time for the wind to start the bullet moving, which will cause it to crash into the side due to its inertia. (For someone inside the habitat, in the rotating reference frame, that corresponds to the bullet slowing down and falling.)
On the other hand, if the shooter aims high it might be possible for this effect to be canceled out. The bullet wouldn't be stationary but would still end up back at its starting position. I don't know anything about guns, but in the rotating reference frame it has to travel about 25 km around the whole rim, so you would need one that's capable of that kind of range.
A: Coriolis does not help - it would affect a bullet that's flying parallel to the rotation axis but not one that's perpendicular to the axis.
A shot perpendicular to the rotation axis will behave just like a shot on the equator of Earth: fly straight but losing speed due to friction, and height due to gravity (centrifugal force in this case). The shot will, however, hit the ground much more quickly than on Earth because of the curved cylinder ground.
A: Trying to build this up one step at a time, and intuitively.
Forget the atmosphere. Assume a perfectly aligned shot, "retrograde" if you will, which exactly cancels out the spinning velocity of the shooter (and gun). Assume an instant bullet launch, i.e. the spin of the barrel (from the centrifuge) is assumed zero, so the bullet travels in an exactly straight path.
I am also ignoring the fact that the "wind" (from the spinning atmosphere) doesn't move in a straight line, because I assume that your centrifuge will be rather large and I can therefore approximate a near-straight trajectory along a 1 cm length (which is what I assume your bullet length will be).
MythBusters have provided us with an excellent example:

This is exactly what it will look like to an observer outside of the centrifuge, except the projectile won't start falling downwards due to gravity. It will just hover there. That is, if there is no atmosphere.
However, with an atmosphere, as it hangs there, the atmosphere passes by. This is effectively a wind tunnel:

Which will start accelerating the bullet leftwards over time. This indeed messes with the positioning of the bullet.
You could try to account for it. Let's say we shoot the bullet at slightly less than the spinning velocity of the shooter (and gun). Therefore, the bullet drifts a bit rightwards because it did not cancel out the spin entirely.
However, over time the wind will push the bullet leftwards, which is capable of "undoing" the rightward drift due to undershooting our shot.
Given you know the shot distance to your target, and that the station's spin is a constant, you can figure out how much travel time the bullet will have, which means that you can figure out exactly how much leftward drift from the wind there will be, which means that you can figure out how much to undershoot your shot (and thus cause the exact same amount of rightward drift as the wind will cause leftward drift).
This all assumes perfect knowledge and execution, but it is not technically impossible.

However, it seems considerably easier to just shoot upwards instead of trying to make the bullet travel along the surface, for many reasons.

*

*It only requires "simple" shot leading, rather than somehow controlling bullet speed.

*Less buildings/things in your way to block your shot

*You can shoot anyone in sight even if not at the same longitude (across the spinning axis) from you

*Due to the concave nature of a centrifuge's surface, you inherently have the high ground on everyone who is not near to you (but they also have the high ground on you, simultaneously). The high ground is always advantageous in combat.


As a last minute alternative idea, you could also shoot your bullet faster than the spinning surface speed and give your bullet little wings so that it generates lift, thus traveling along the curve of the centrifuge.
I guess this makes it more of a mini-missile than a bullet, but it should be viable given the right kind of engineering.
A: The answer is simple, and it is identical to the question:
Can you shoot something $2\pi\cdot 4km = 25km$ away from you with the gun in question?
The answer is obviously "no". Otherwise it wouldn't be possible to safely build out-door shooting ranges, or hunt birds in the vicinity of an airport (25km safety radius!). The truth is, that the ranges of firearms are limited by the atmospheric drag of the bullets they fire, and most rifles do not allow you to hit a target at even 1km distance.
I'm also sorry to inform you, that this means that this story idea won't work without resorting to handwavium: If you scaled down your centrifuge, the bullet would need to become slower due to the lower speed of the centrifuge rim, and by the time you've shrunk the centrifuge to 1km circumference, your bullet won't be deadly in any way (and still not be able to make the voyage).
A: Without atmosphere, the bullet relative to the hub of the centrifuge, would hang motionless, and the poor soul who shot it would run into their own bullet after completing one revolution.
If there were an atmosphere the rim of the centrifuge will give it some velocity too. From the shooters perspective the bullet flys away, slows, and drops to the ground.
From the hub perspective, the shooter moves away at 200 m/s, the bullet is motionless, but the air gradually speeds it up in the direction of the shooter.  The bullet "drops" as the (curved) floor rises to meet it.
Even harder would be to hit a target in a rotating cylinder with atmosphere.  Essentially, one would have to "time" where the bullet and the moving target intersected, and aim a little high.  As with earthly shooting, wind speed, direction, and density play crucial roles.
