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I am looking into torus worlds for fiction, and I know how to calculate the necessary speed to rotate, and necessary radius, for such a world to get a specific force of gravity.

But what I am wondering, is that how would I find what rate of rotation would be safe for inhabitants given a specific radius. As the plans I have now call for 100 Second rotation period and 2.484 km radius, which I feel would exert a lot of force on the inhabitants. Am I right to assume this? And how would I be able to find safe and comfortable rotation periods?

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    $\begingroup$ You need a small enough $\omega$ from a biological perspective, so this SE alone can't cover everything, but space.se will be accustomed to these questions. The O'Neill cylinder design has a slightly smaller radius; the McKendree cylinder has a much larger one, but requires stronger construction materials due to the greater tangential velocity. $\endgroup$
    – J.G.
    Commented Mar 1, 2022 at 22:26

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Usually for inhabitants on the surface of the world, we'd include the centrifugal force as a part of the local gravity. If your formula to determine the local gravitational field already accounts for centrifugal force, then you don't need to worry. If it doesn't account for it, then you can easily add it in. The formula is:

$$ \vec a_\text{centrifugal} = \frac{4\pi^2 \vec r}{T^2} $$

where $T$ is the period of rotation and $\vec r$ is a vector that shoots out perpendicular to the axis of rotation and lands at the place where we want to know the gravitational field. Once you have a formula that includes both the local gravitational field, and the centrifugal force, you can just check that it gives a value that's about equal to 10m/s$^2$. Too much lower means your inhabitants will be walking in slow motion and there won't be any atmosphere, too much higher means they'll be unable to move at all in the strong gravity. (Assuming they're standard humans, at least.)

One thing that could be true in this situation, and I'd guess it probably is true, since a spinning torus world is a pretty complicated situation, is that the gravitational field may not be the same strength everywhere on the surface of the torus world. It might be stronger at some latitudes and weaker at others. I don't know if this would be a slight difference or a large one, probably it depends on the exact geometry of your world. Anyway, the formula for local gravity should probably depend on where in the torus world you're standing.

Last thing to consider is the Coriolis effect. This is an acceleration that isn't present when you're just standing still on the world's surface, but shows up when you move towards or away from the axis of rotation. The faster you're moving towards or away from the axis of rotation, the stronger the effect. The formula is:

$$ a_\text{Coriolis} = \frac{4\pi v_r}{T} $$

Where $v_r$ is the velocity with which you're moving towards or away from the axis of rotation. In your case, with a rotation period of 100s, someone standing on top of the torus and walking directly away from the axis of rotation at 1m/s would feel a Coriolis acceleration of about 0.1 m/s$^2$, which would probably be noticeable enough to be disorienting. Driving a car would be even weirder. Freeway speeds of 60 miles per hour would produce a Coriolis acceleration of 3.4 m/s$^2$, which is huge. You could work this into the worldbuilding, or choose a larger rotation period, so that the Coriolis effect is less noticeable.

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  • $\begingroup$ The one I used was the one shown on Wikipedia, R = a(T/2pi)^2, with R being Radius in metres, a is the gravity in m/s2, and T is time in seconds. $\endgroup$
    – Zoey
    Commented Feb 28, 2022 at 0:04
  • $\begingroup$ And on the exact situation, it is artificial, and atmosphere is artificially kept by a very strong glass overhang of sorts which surrounds it all. The surface is only the inside part, think you cut off the inner half of a torus, and have the surface be on the exposed inside. and thanks for the formulae. $\endgroup$
    – Zoey
    Commented Feb 28, 2022 at 0:07
  • $\begingroup$ Will have to increase rotation period, as I don't want the Coriolis acceleration to be too extreme. Just gonna have to make it be more resource intensive. With a larger radius. $\endgroup$
    – Zoey
    Commented Feb 28, 2022 at 0:10
  • $\begingroup$ Okay, cool, thanks for the additional detail, I was imagining something slightly different, where the torus has to be held together under its own gravity. Sounds like in your situation it's held together by the strength of the material. If the gravity from the torus itself isn't too strong, then that makes things much easier to analyze. Then the centrifugal force is the only source of gravity, and so you can just use the formula you already have. (And the force would be the same everywhere on the surface, since it's all the same distance from the axis of rotation.) $\endgroup$
    – pb1729
    Commented Feb 28, 2022 at 0:14
  • $\begingroup$ I see what you were thinking, no, it isn't like that, sorry for not being clear. It is all basically the same distance to the axis of rotation. Though there would of course be small differences due to a completely perfect circle being basically impossible. $\endgroup$
    – Zoey
    Commented Feb 28, 2022 at 0:18
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enter image description here

your position on a torus surface is described by this vector

$$\vec R=\left[ \begin {array}{c} \cos \left( \vartheta \right) \left( c+a \cos \left( \varphi \right) \right) \\ a\sin \left( \varphi \right) \\ -\sin \left( \vartheta \right) \left( c+a\cos \left( \varphi \right) \right) \end {array} \right] $$

where c is the radius to the center of the torus and a is the radius of the tube $~\varphi~$ and $~\vartheta~$ are the generalized coordinate $~\varphi~,\vartheta\in[0\,2\pi]$

generate the equations of motion in the rotate system (the rotation is about the y-axes) and put all time derivatives equal zero , this is the steady state solution.

the generalized steady state forces are:

$$F_\phi=\left( \left( -\sin \left( \varphi \right) a\,c-\frac 12\,{a}^{2}\sin \left( 2\,\varphi \right) \right) {\omega}^{2} \right) \,m \\F_\vartheta=0$$

with those data

$$ c=2484\,10^3~,a=30\,10^3,~\omega=10^{-6}~,m=50$$

you obtain that max. magnitude of $~|F_\phi|~$ is 40 [N]

$$\omega=\frac{2\,\pi}{T}$$

simulation Result $~\vec R(~\varphi(t)~,\vartheta(t)~)$

enter image description here

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