# Could cars drive normally on the inside of an o'neil cylinder?

Say we had some type of four-wheeled vehicle on the inside of an O'Neil cylinder with roads. I imagine if it drove in the same direction as the spin, there would not be many problems. However, I am not sure what would happen if it were to drive down the length of the cylinder or in the opposite direction of the habitat's spin.

If the vehicle doesn't behave like a normal vehicle on earth in these situations, would there be some sort of device that could make it? For instance, a smart speed limiter? Im thinking if cars would not work well, then vehicles locked into tracks would have to be used instead.

• When travelling space in a gigantic, rotating structure, the most important questions revolve around engineering, life-support, self-sustainment, and… "can I bring my own car?". :) Jun 7, 2021 at 15:00
• Also consider the issues on the rotating parts of an engine! After all, kinds in an O'Neil Cylinder can't play with tops. Jun 7, 2021 at 21:25
• Other than scale, why would cars have more problem driving than people might in walking inside those cylinders? Jun 8, 2021 at 0:49
• It would appear that the correct spelling is "O'Neill". Jun 8, 2021 at 0:50

## 4 Answers

The Wikipedia page for an O'Neill Cylinder indicates that they would revolve at a rate of around 2.8 degrees per second – at a diameter of 5 miles, that would be a surface speed of about 0.12 miles per second, or ~440 mph.

If the velocity of the surface is 440 mph, then ±40 mph would be about 90–110% of Earth gravity. Not enough to float, but probably enough to require a specialized suspension.

(Edit: Note that the 1g ±10% is a rough estimate, see @TannerSwett's answer for more precise values)

However, are cars on roads actually the most likely transportation method? At 20 miles long and 5 miles in diameter a typical O'Neill cylinder has 300 square miles of area (~775 km$$^2$$), which is the size of a small county in the USA (or about the size of Bahrain). Roads are mostly wasted space, if you are trying to grow food or house humans. Public transportation is most likely, and trains or trams, connected to a central electric grid, would not pollute the air. And, if you use vehicles on rails, then designing them like roller coasters would 100% resolve any issues with "falling off".

• That's confusing phrasing; +/- 10% of Earth gravity is 1/10*g* not 90-110%.
– rek
Jun 6, 2021 at 22:10
• I threw numbers at my calculator, it says 82% and 121% actually. Still not at all significant. JUst dont ascend or descend relative to the axis, without some serious precautions. Jun 6, 2021 at 23:22
• O'Neill Cylinders would also be designed so people lived near where they worked so they could walk everywhere. Jun 7, 2021 at 0:53
• @rek: Yes, exactly. 1.0g +- 10% spans a range of 90% to 110% of Earth standard gravity. I think the phrasing is pretty clear that it's talking about deviations from an existing apparent gravity, for a range of horizontal speeds. (Possibly I just got lucky and interpreted it the right way by chance when I first read it, but on re-reading I disagree that it's confusing. If you skimmed too fast, then maybe you'd miss something, but if so you can read that one sentence. And clearly nothing is going to create -10% negative total absolute apparent gravitation, i.e. inward pull from moving.) Jun 7, 2021 at 11:01
• The gravity experienced varies with the square of the angular velocity. So as PcMan indicates, changing your own angular velocity by $\pm10\%$ of the station's would result in the gee-force on you either decreasing to $81\%$ or increasing to $121\%$ of normal (I assume PcMan actually used $440 \pm 40$ mph, not $\pm 10\%$). Jun 7, 2021 at 17:49

As long as your car stays on the same radius from the axis, it can do what it wants with very little effect.

Driving north/south along the axis is exactly just driving on flat road.
Driving spin/antispin will encounter an apparent upwards curve in the road, that the car somehow never reaches. Spinward will increase gravity a bit, Antispin will decrease it a bit. The amount is small though, driving antispin at 70km/h will reduce your gravity by only 18%, driving the same speed spinward will only increase your gravity by 21%. That is about the same as the acceleration felt when you are in grandma's car going shopping, and she pulls away at a green light. MAybe a third the acceleration of when she slams the brakes to avoid that nice kittycat crossing the road two city blocks ahead.

The only time when driving a car in an O'Neill cylinder will get weird, is when the car drives up a ramp towards or away from the axis. Then coriolis effect will rear its ugly head and try very hard to throw you off towards the spin or antispin direction, depending on whether you heading towards or away from the axis.

• It's worth noting that the change in gravity when you travel spinward or antispinward can also be viewed as a consequence of the Coriolis force. Jun 7, 2021 at 13:56
• @MichaelSeifert nope. Coriolis acts only if the distance from the axis of rotation changes. Jun 7, 2021 at 14:13
• I don't think that's true. If you work in a rotating reference frame and you take the cross product of a vector along the axis and a vector moving tangentially to the surface, you get a vector pointing either in towards the axis or out away from the axis. This would be in the same direction (or the opposite direction) as the centrifugal force that acts as gravity in such a space station. See the last two bullet points in the "Formula" section of the Wikipedia article on the Coriolis force. Jun 7, 2021 at 14:18
• (Of course, in the situation we're talking about, the amount of inward force exerted by the "ground" changes to compensate for the Coriolis force, so the trajectory of the object doesn't change in the same way it does for a projectile. But it's not accurate to say that it doesn't act.) Jun 7, 2021 at 14:26
• I think @MichaelSeifert is right. The CF is only zero if you drive along the length of the cylinder. Then v and omega are parallel and the cross-product is zero. If you drive spinward, the CF is a pitching force. Jun 7, 2021 at 14:41

IronEagle's answer explains that the cylinder would have a linear speed of about 440 mph (708 km/h). Driving spinward or "east" (which is to say, in the direction of rotation) would cause the occupants of the vehicle to feel heavier, and driving antispinward or "west" would make them feel lighter.

I've calculated just how much heavier or lighter they would feel at various speeds:

Speed relative to floor Speed in inertial frame Apparent gravity
75 mph (121 km/h) antispinward 365 mph (587 km/h) 0.67 g
50 mph (80 km/h) antispinward 390 mph (628 km/h) 0.77 g
25 mph (40 km/h) antispinward 415 mph (668 km/h) 0.87 g
Stopped 440 mph (708 km/h) 0.98 g
25 mph (40 km/h) spinward 465 mph (748 km/h) 1.10 g
50 mph (80 km/h) spinward 490 mph (789 km/h) 1.22 g
75 mph (121 km/h) spinward 515 mph (829 km/h) 1.34 g

So, the occupants would definitely notice a difference in how heavy they feel, but not a huge difference. Cars may be designed to be somewhat sturdier than cars here on Earth, and there would probably be a note in the owner's manual saying not to drive spinward at high speeds while heavily loaded.

Driving along the length of the cylinder would have no unusual effects.

• It's very likely that in an environment this compact, speed would be limited to much lower than 75 mph for safety. Among other considerations, the faster you're going antispinward, the less effective traction you have for braking or sharp swerves. It'd be like driving on wet pavement, or worse. Jun 7, 2021 at 17:11

The surface and car only move in reference to the axis they're jointly spinning around so as long as the car maintains surface contact it will behave as a car on Earth would.

O'Neill cylinders are small enough that people could detect the difference in spinward/antispinward direction by turning their heads, without feeling sick or other I'll effects, so the influence on driving would be negligible.