5 ms·

How large would a ring have to be to have 1g at 1 rotation per day? (Edit: earth day)

by 83457 3y ago

Damogran6 3y ago
Shooting from the hip, approx the circumference of the earth.
- messe 3y ago
  > Shooting from the hip, approx the circumference of the earth. You're quite a bit off. It's actually a little under 4 million km; several times the diameter of the sun.
  - Damogran6 3y ago
    There's a reason why shooting from the hip isn't very accurate. :)
    - SideburnsOfDoom 3y ago
      It's not that it's an "not very accurate" loose relation; there is no relation at all. The two numbers concern different forces - gravity inward due to mass vs. centrifugal force outwards due to spin.
- gwbas1c 3y ago
  If that were true, we'd all fling off of the surface of the planet!
SideburnsOfDoom 3y ago
> A ring with 1g at 1 rotation per day That's the parameters of a "Banks Orbital" https://theculture.fandom.com/wiki/Orbital_(Wikipedia_version) https://theculture.fandom.com/wiki/Orbital_(Wikipedia_versio... Which is: > For such an orbital to reproduce the equivalent to the Earth's gravity, whilst maintaining Earth's 24-hour period of rotation, it would need to have a diameter of approximately 3.71 million kilometres, and spinning at 486,000 km/hr.
- hwc 3y ago
  And no known material has enough tensile strength to make it work!
  - SideburnsOfDoom 3y ago
    > no known material has enough tensile strength I know that's true of Niven's ringworld, which is just unholy scale and parameters - 1 rotation like the earth does in a year, every 9 days! (1) And so it cannot be made from atoms, something with "tensile strength similar to the strong nuclear force" is needed (2) But is it true of Banks's more practical Orbital as well? This reference says yes: "No form of ordinary matter will support the tensions of a Banks Orbital's spin, so exotic matter is required." (3) 1) http://www.alcyone.com/max/reference/scifi/ringworld.html http://www.alcyone.com/max/reference/scifi/ringworld.html 2) https://larryniven.fandom.com/wiki/Scrith https://larryniven.fandom.com/wiki/Scrith 3) https://www.orionsarm.com/eg-article/4845ef5c4ca7c https://www.orionsarm.com/eg-article/4845ef5c4ca7c
    - m4rtink 3y ago
      What about mass stream technology? Why everyone forgets mass stream technology? https://www.orionsarm.com/eg-article/47e1bb1fc898c https://www.orionsarm.com/eg-article/47e1bb1fc898c "Using accelerated streams of projectiles or particles to either transfer momentum or support a large structure." This is basically active super strong matter that requires uninterrupted continuous control, power supply and real time adjustments - or bad things happen very very quickly. What is there not to love?! ;-)
      - SideburnsOfDoom 3y ago
        Firing a machinegun up at something, from a planetary surface seems like a possible * way to to keep it from falling down (i.e. to push against gravity) but not so simply as a way to keep it from flying apart - the stresses on a ring rotating fast in zero G space, are Centrifugal. You would have to shoot inwards at it, from nowhere. * Possible, not necessarily practical.
gwbas1c 3y ago
It doesn't need to be that slow to avoid nausea. I've been to the top of the space needle with a rotating floor, and didn't feel any nausea.
- SideburnsOfDoom 3y ago
  The "1 day" rotational period is to give a natural-seeming day-night cycle, using natural sunlight from the star that the whole structure orbits. It would be oriented so that the sun appears to rise just to the side of the ring - When it's overhead the arc of the ring would appear to be near the sun, but not eclipsing it. This is nearer to vertical than the sun over much of Earth much of the time.
dredmorbius 3y ago
About 2 million km if GNU Units is serving me correctly: You have: (1 gravity) / (1/1440 * rpm)^2 You want: million km * 1.854336 / 0.5392766 By comparison: You have: (0.5 gravity) / (1 rpm)^2 You want: m * 447.12962 (That's the scenario in the film here.) For 1 g at 1 RPM: You have: (1 gravity) / (1 rpm)^2 You want: m * 894.25925 And presuming 3 RPM is tolerable (a common assumption in early space station / space colony proposals): You have: (1 gravity) / (3 rpm)^2 You want: m * 99.362139 (Almost exactly 100m or 330 ft.)
- dredmorbius 3y ago
  If the Earth revolved at a rate of ~17 times per day, or once every 1:24:42, the centrifugal force at the equator should just match the pull of gravity. Also via GNU Units: You have: sqrt(1 gravity / earthradius) You want: revolution/day * 17.060434 You have: 1 day / 17 You want: hour; minute; second 1 hour + 24 minute + 42.352941 second Stack Exchange confirms this: <https://physics.stackexchange.com/questions/136486/how-fast-would-the-earth-need-to-spin-for-us-to-feel-weightless#136495 https://physics.stackexchange.com/questions/136486/how-fast-...>) That's about once every 84 minutes. Which, I recognise is also roughly the orbital period for low-Earth orbit, that is, an object falling around the Earth without ever striking the surface. Though I think that may be a coincidence. Compare geosyncronous orbit (~42,000 km from Earth's centre), with the radius at which there is 1g of centrifugal acceleration at 1 revolution per 24 hours, ~2 million km. Fun to play with however.
- dredmorbius 3y ago
  Just to clarify, I was solving for radius. Diameter would be 2x, and circumference would be 2pi larger. So 11.6 million km circumference for the 24-hour rotating ring. If you had a trans-ring commute, at 100 kph, that would take you about 6 years 7 months one way. A jet aircraft operating at nearer 1,000 kph would cut that to a far more manageable 7 months. Though I'd still use it as a reasonable argument for return-to-office. Ping times would still be about 38 seconds even for free-vacuum direct transmission, around the ring itself. Given lightspeed is reduced to about 66% of its in vacuo* value in fibre optics, you'd have almost exactly a 1 minute delay. Something to keep in mind during those Ring Zoom sessions.

How large would a ring have to be to have 1g at 1 rotation per day? (Edit: earth day)

by 83457 3y ago

How large would a ring have to be to have 1g at 1 rotation per day? (Edit: earth day)

Damogran6 3y ago

Shooting from the hip, approx the circumference of the earth.

messe 3y ago
> Shooting from the hip, approx the circumference of the earth. You're quite a bit off. It's actually a little under 4 million km; several times the diameter of the sun.
- Damogran6 3y ago
  There's a reason why shooting from the hip isn't very accurate. :)
  - SideburnsOfDoom 3y ago
    It's not that it's an "not very accurate" loose relation; there is no relation at all. The two numbers concern different forces - gravity inward due to mass vs. centrifugal force outwards due to spin.
gwbas1c 3y ago
If that were true, we'd all fling off of the surface of the planet!

SideburnsOfDoom 3y ago

> A ring with 1g at 1 rotation per day That's the parameters of a "Banks Orbital" https://theculture.fandom.com/wiki/Orbital_(Wikipedia_version) https://theculture.fandom.com/wiki/Orbital_(Wikipedia_versio... Which is: > For such an orbital to reproduce the equivalent to the Earth's gravity, whilst maintaining Earth's 24-hour period of rotation, it would need to have a diameter of approximately 3.71 million kilometres, and spinning at 486,000 km/hr.

hwc 3y ago
And no known material has enough tensile strength to make it work!
- SideburnsOfDoom 3y ago
  > no known material has enough tensile strength I know that's true of Niven's ringworld, which is just unholy scale and parameters - 1 rotation like the earth does in a year, every 9 days! (1) And so it cannot be made from atoms, something with "tensile strength similar to the strong nuclear force" is needed (2) But is it true of Banks's more practical Orbital as well? This reference says yes: "No form of ordinary matter will support the tensions of a Banks Orbital's spin, so exotic matter is required." (3) 1) http://www.alcyone.com/max/reference/scifi/ringworld.html http://www.alcyone.com/max/reference/scifi/ringworld.html 2) https://larryniven.fandom.com/wiki/Scrith https://larryniven.fandom.com/wiki/Scrith 3) https://www.orionsarm.com/eg-article/4845ef5c4ca7c https://www.orionsarm.com/eg-article/4845ef5c4ca7c
  - m4rtink 3y ago
    What about mass stream technology? Why everyone forgets mass stream technology? https://www.orionsarm.com/eg-article/47e1bb1fc898c https://www.orionsarm.com/eg-article/47e1bb1fc898c "Using accelerated streams of projectiles or particles to either transfer momentum or support a large structure." This is basically active super strong matter that requires uninterrupted continuous control, power supply and real time adjustments - or bad things happen very very quickly. What is there not to love?! ;-)
    - SideburnsOfDoom 3y ago
      Firing a machinegun up at something, from a planetary surface seems like a possible * way to to keep it from falling down (i.e. to push against gravity) but not so simply as a way to keep it from flying apart - the stresses on a ring rotating fast in zero G space, are Centrifugal. You would have to shoot inwards at it, from nowhere. * Possible, not necessarily practical.

gwbas1c 3y ago

It doesn't need to be that slow to avoid nausea. I've been to the top of the space needle with a rotating floor, and didn't feel any nausea.

SideburnsOfDoom 3y ago
The "1 day" rotational period is to give a natural-seeming day-night cycle, using natural sunlight from the star that the whole structure orbits. It would be oriented so that the sun appears to rise just to the side of the ring - When it's overhead the arc of the ring would appear to be near the sun, but not eclipsing it. This is nearer to vertical than the sun over much of Earth much of the time.

dredmorbius 3y ago

About 2 million km if GNU Units is serving me correctly: You have: (1 gravity) / (1/1440 * rpm)^2 You want: million km * 1.854336 / 0.5392766 By comparison: You have: (0.5 gravity) / (1 rpm)^2 You want: m * 447.12962 (That's the scenario in the film here.) For 1 g at 1 RPM: You have: (1 gravity) / (1 rpm)^2 You want: m * 894.25925 And presuming 3 RPM is tolerable (a common assumption in early space station / space colony proposals): You have: (1 gravity) / (3 rpm)^2 You want: m * 99.362139 (Almost exactly 100m or 330 ft.)

dredmorbius 3y ago
If the Earth revolved at a rate of ~17 times per day, or once every 1:24:42, the centrifugal force at the equator should just match the pull of gravity. Also via GNU Units: You have: sqrt(1 gravity / earthradius) You want: revolution/day * 17.060434 You have: 1 day / 17 You want: hour; minute; second 1 hour + 24 minute + 42.352941 second Stack Exchange confirms this: <https://physics.stackexchange.com/questions/136486/how-fast-would-the-earth-need-to-spin-for-us-to-feel-weightless#136495 https://physics.stackexchange.com/questions/136486/how-fast-...>) That's about once every 84 minutes. Which, I recognise is also roughly the orbital period for low-Earth orbit, that is, an object falling around the Earth without ever striking the surface. Though I think that may be a coincidence. Compare geosyncronous orbit (~42,000 km from Earth's centre), with the radius at which there is 1g of centrifugal acceleration at 1 revolution per 24 hours, ~2 million km. Fun to play with however.
dredmorbius 3y ago
Just to clarify, I was solving for radius. Diameter would be 2x, and circumference would be 2pi larger. So 11.6 million km circumference for the 24-hour rotating ring. If you had a trans-ring commute, at 100 kph, that would take you about 6 years 7 months one way. A jet aircraft operating at nearer 1,000 kph would cut that to a far more manageable 7 months. Though I'd still use it as a reasonable argument for return-to-office. Ping times would still be about 38 seconds even for free-vacuum direct transmission, around the ring itself. Given lightspeed is reduced to about 66% of its in vacuo* value in fibre optics, you'd have almost exactly a 1 minute delay. Something to keep in mind during those Ring Zoom sessions.