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What would be the diameter and rotational speed of an O'Neill cylinder on the inside surface of which centrifugal and coriolis forces are equal to "gravity" and coriolis force on the surface of Earth at sea level?

Notes:

  • "Gravity on Earth" (in quotation marks) here means the sum of gravitation and centrifugal force. Assume that the sum of these forces results in an acceleration of 1 g.

  • Place on Earth for the coriolis force can be any latitude between the equator and 60°. Pick one you find represents living on Earth well, e.g. 45°.

  • On Earth, there is a slight difference in "gravity" between my feet and my head. This difference must be the same on the inside surface of the O'Neil cylinder.

  • You may ignore the gravitation of the cylinder's wall. If you want to estimate it, the wall of the cylinder is no thicker than 35 km and composed like the crust of the Earth.

  • Do not forget the coriolis force!

  • Your answer may prove that my requirements cannot be met! Just explain how far one or multiple parameters deviate from the requirement (e.g. "coriolis force would be x times larger than on Earth").

  • The dimensions of this O'Neill cylinder aren't fixed to that of O'Neill's or anyone else's original design. You may change the radius or any other measure to meet the requirements as closely as possible. The cylinder can be as big as the Earth or bigger, if it needs to be. 1 g of "gravity" and a head-foot-difference of gravity like that of Earth are the requirements that must be met.

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This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

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    $\begingroup$ The Coriolis force on Earth depends on the location. Where are you taking your reference? Also, please pick one between science based and hard science tag. They are mutually exclusive. $\endgroup$ – L.Dutch Nov 20 '18 at 20:30
  • $\begingroup$ if you don't want gravity to swing sideways, you may want to drop coriolis requirement. It would be fun to see size and rotation speed calculations, but once these two are set, coriolis force is also set and cannot be adjusted - and I bet it won't be anywhere near the one 45°. Whatever it'll be, it will not be possible to adjust it in any way without messing parameters that give "gravity" itself. $\endgroup$ – Mołot Nov 20 '18 at 23:56
  • $\begingroup$ You're just trying to get us to do math, aren't you? The corolis requirement, and gravitational gradient requirement, suggests that the cylinder would be nearly the diameter of earth. Maybe bigger. $\endgroup$ – FuzzyChef Nov 21 '18 at 1:52
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    $\begingroup$ OK, I've messed around with SpinCalc for a while, and you can't get what you want. In order to have the same gradient of "gravity" difference between head and feet as the Earth, you'd need to have a very large cylinder. And in order to have a cylinder that large, you'd have a very very fast rotation, like over 10x as fast as the earth rotates. You're going to have to eliminate one of your requirements. $\endgroup$ – FuzzyChef Nov 21 '18 at 2:02
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    $\begingroup$ Cool question. I'm not certain that you'll actually get the Coriolis force inside an Oneill cylinder, because you aren't changing the distance from the axis of rotation and thus the angular speed is always the same. On a globe, moving closer to the axis of rotation (toward the poles) decreases your radius and thus increases your rotational velocity to conserve angular momentum, which we observe as the Coriolis "force" $\endgroup$ – Dubukay Nov 21 '18 at 3:57
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TL;DR 3185501.499m radius, 0.00175460427 rotations per second, coriolis force will be different.

Let's take it point by point.

Gravity from walls would make people near the cylinder end feel like cylinder centre is downhill. Let's ignore it at the moment. So let's think about a thin slice thorough th middle for a moment:

enter image description here

Gravity increases or decreases with h3, where h is distance from certain point above or below. At the same time, amount of ground that pulls you in a slice decreases or increases with h2. So if you are 2 times closer to the middle depth of bottom than top, bottom will pull you 2 times stronger. 3 times closer? 3 times stronger. How important is gravity from the wall? Well, for 35km sphere of density similar to Earth crust, surface gravity is about 0.01g. You said You may ignore the gravitation of the cylinder's wall. and now I'm gonna do just that - ignore. Change in the order of magnitude of 1% change is not really worth it at the moment.


You want gravity experienced to be that on Earth, and difference from feet to head similar. OK. "Gravity" experienced by inhabitants is

$a = -\omega^2 r$ where $\omega$ is the angular velocity of the station, $r$ is its radius, and $a$ is linear acceleration at any point along its perimeter. source

Let's assume two meter tall human, a tiny bit higher than me. You want

$g = \omega^2 r$ and $a = \omega^2 (r-2m)$

where $g$ is surface gravity of Earth, and $a$ is "head gravity" of Earth for 2 meter tall human.

According to this calculator, $a = 0.9999993721553733g$

Thus, we have two equations with two variables:

$g = \omega^2 r$ and

$ 0.9999993721553733g = \omega^2 (r-2)$

Thus

$ \omega^2 r = \omega^2 (r-2) / 0.9999993721553733$

$ r = (r-2m) / 0.9999993721553733$

$ 0.9999993721553733r = r-2 $

$ 0.9999993721553733r-r =-2 $

$ (0.9999993721553733 - 1)r =-2 $

$ (1 - 0.9999993721553733)r =2 $

$ (1 - 0.9999993721553733)r =2 $

$ r = 2m / (1 - 0.9999993721553733) $

$ r = 3185501.499 $

With this radius, you will have the same proportion of perceived gravity between head and feet you have on Earth.

Warning! Earth radius is 6371km, and your cylinder has radius of 3185.5km! It is truly a planetary scale construction!


Now to omega:

$g = \omega^2 r$

Using Wikipedia $ g = 9.807 m/s^2$

$9.807 = \omega^2*3185501.499$

$9.807/3185501.499 = omega^2$

$omega = \sqrt 9.807/3185501.499$

$omega = 0.001754604$

0.00175460427 rotations per second.


Coriolis force experienced when moving on the surface in parallel to axis will be 0, because it only appears when motion is not parallel to rotation axis. Earth is roughly a sphere, so any motion not exactly on equator and perpendicular to it will experience Coriolis force - 0 is possible only for infinitesimally small distance and infinitesimally small time, when in cylinder you can have 0 for the whole cylinder length.

For up and down motion on the equator, effect will be opposite: in your cylinder, going up will mean smaller radius. On Earth, going up means bigger one. These effects will also probably have different magnitudes, but since I already proved that having identical is impossible, I'm going to leave calculation of absolute values as an exercise to the reader.

Coriolis on earth illustration

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    $\begingroup$ “Coriolis force experienced when moving on the surface will be 0, because it only appears when radius of rotation changes.” — Wrong. Coriolis force happens whenever your movement is not parallel to the axis of rotation. Yes, this means that on earth you'll experience Coriolis force when moving along the equator. However the Coriolis force is directed perpendicular to both the rotation axis and the direction of movement, so when moving along the equator (or generally, along a circle of latitude), it amounts to just a slightly higher/lower weight depending on the direction. $\endgroup$ – celtschk Nov 21 '18 at 18:02
  • $\begingroup$ @celtschk oook, I'll need to revise that part. Or maybe I should ask to convert to community wiki so you can do it yourself? $\endgroup$ – Mołot Nov 21 '18 at 18:32
  • $\begingroup$ Eötvös effect :en.wikipedia.org/wiki/E%C3%B6tv%C3%B6s_effect Cylinder is rotating fast, so no need to care too much. $\endgroup$ – Artemijs Danilovs Nov 22 '18 at 0:30
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    $\begingroup$ If my calculations are correct, an object falling from a height of 2 m above the inside wall of this O'Neill cylinder should be deflected by about 1.5 mm. Formula (in R): (sqrt(1 / ((r - 2) / r)^2 - 1) - acos((r - 2) / r)) * r using r = 3185501 m as given in this answer. $\endgroup$ – user57423 Nov 22 '18 at 10:44
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    $\begingroup$ @user57423 If you run in vacuum, yes. But don't forget that your foot will be dragged by air, and air will be dragged by floor, so this will compensate for a huge part of that millimeter. Assuming that you have air there, of course. $\endgroup$ – Mołot Nov 22 '18 at 10:56
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It is possible to find a radius and angular velocity that satisfies the conditions very closely. As stated in another answer, the head-gravity and feet-gravity for a 2m human comes to 3185501.499 m radius, 0.00175460427 rot/s. I have not rigorously verified these numbers, but at a glance they seem correct. HOWEVER, where the difficulty arises is that in an O'Neill cylinder (or any rotating system) the centrifugal force in the rotating reference frame is directly proportional to the radius, whereas gravity above a planet falls off with the square of the distance. This means that if you were to plot the felt force at any height off the ground, you would see the following two curves: Graph of straight line curve intersecting with inverse square curve at two points

As you can see, the straight line curve of the rotating habitat's "gravitational" force as one moves from the floor to the center can be made to intersect with the inverse square curve mapping the gravity field of a massive body at only up to two points. Between these two points, the gravity in the habitat will be greater, and above (to the right on the graph) the higher elevation point (also below the lower elevation point, although "below the floor" in a rotating habitat is not a fun place to be) the gravity on the massive body will be greater. So in effect, the numbers above hold for the force at any person's feet, at ground level of the habitat. And they hold for the 2m height. However, any other person of different height will feel a different force at their head level, and even the 2m person will (if having a sufficiently sophisticated sense of weight around their midsection) experience more downward force toward their middle in the cylinder than on Earth, even while experiencing equivalent force at their head and feet.

For a cylinder of this magnitude, or any where the radius is this significantly greater than the height of the people involved, the difference between an inverse square curve and a straight line is imperceptible on the human scale.

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  • $\begingroup$ Very useful supplement! About my numbers, I'm pretty sure they are OK for assumptions and simplifications I made, but for the sake of completeness I also included the way I got them, so anyone can adjust them to their needs or point out mistakes, if there are any. To be honest, I believe process is more important and useful than raw numbers. $\endgroup$ – Mołot Nov 22 '18 at 10:54
  • $\begingroup$ @Mołot I checked your formula and your calculations and they seem correct to me also. Thank you for your answer, and thank you, Christyn, for your additional information. $\endgroup$ – user57423 Nov 22 '18 at 10:59

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