# Falling in an O'Neill Cylinder

• I
Summary:
Could a small fall cause serious injury?
So I'm working on a project that involves the design of an O'Neill Cylinder, and there was a consideration that I had never made before. Say you are in a cylinder that is generating enough force for 1G in its spin. This means that while you are spinning, the motion means that your body is being pinned to whatever you are standing on. If that were to fall, or if you took a bad step, would that mean you would be immediately falling at 9.8 m/s? For a point of reference, you don't actually hit 9.8 m/s speed until 1 full second. During this time, you would have fallen about 4.9 meters.

So if you took a tumble, since there's no acceleration and you're right at that speed, would even a small fall translate to a pretty gnarly impact? If so, does this mean the cylinder is safer to design with a lower gravity to prevent these injuries, and perhaps have secondary rings that people can spend short bursts in for workouts on their body?

## Answers and Replies

PeroK
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Are you confusing speed with acceleration?

The Earth's surface is rotating at about 1,000 miles per hour, but you don't hit the ground at that (relative) speed if you fall over.

vanhees71
Are you confusing speed with acceleration?

The Earth's surface is rotating at about 1,000 miles per hour, but you don't hit the ground at that (relative) speed if you fall over.
I understand that, because everything is moving relative to one another. However, in a rotating cylinder, you are not accelerating. If you were to push off gently from an object at the center, you wouldn’t accelerate towards the outside because gravity is not a substantial force moving you towards the outside. When you are spinning on a merry-go-round and are flung off, you are flung off at the speed of rotation at the circumference.

An O’Neill cylinder is functionally similar, just on a massive scale. You are still moving relative to the vessel, but some of those forces would still act upon you. There would be no acceleration, and I’m trying to decide whether simulated 1g in this case would be dangerous.

PeroK
PeroK
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I understand that, because everything is moving relative to one another. However, in a rotating cylinder, you are not accelerating. If you were to push off gently from an object at the center, you wouldn’t accelerate towards the outside because gravity is not a substantial force moving you towards the outside. When you are spinning on a merry-go-round and are flung off, you are flung off at the speed of rotation at the circumference.

An O’Neill cylinder is functionally similar, just on a massive scale. You are still moving relative to the vessel, but some of those forces would still act upon you. There would be no acceleration, and I’m trying to decide whether simulated 1g in this case would be dangerous.
Gravity on the Earth's surface and centrifugal force on a large rotating object are essentially indistinguishable. In both, there is a real force from the ground/exterior of the cylinder. But, there is no relative motion between the person and the ground in either case.

An O'Neill cylinder in space and a merry-go-round on Earth are not the same physically!

Janus
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I understand that, because everything is moving relative to one another. However, in a rotating cylinder, you are not accelerating. If you were to push off gently from an object at the center, you wouldn’t accelerate towards the outside because gravity is not a substantial force moving you towards the outside. When you are spinning on a merry-go-round and are flung off, you are flung off at the speed of rotation at the circumference.

An O’Neill cylinder is functionally similar, just on a massive scale. You are still moving relative to the vessel, but some of those forces would still act upon you. There would be no acceleration, and I’m trying to decide whether simulated 1g in this case would be dangerous.
While standing on the inside of a rotating cylinder, you are accelerating. Acceleration is a change in velocity, and velocity includes both speed and direction. Since the cylinder fall is forcing you to travel in a circle, and this causes you to constantly change direction, you are accelerating.

With the merry-go-round, you are not flung-off as you are now free to travel in a straight-line rather than being forced to travel in a circle. So you continue in a straight line in the direction you were moving when you let go.

In an O'Neill cylinder, while standing on the inside surface, both you and "ground" are traveling at the same speed. If you jump, you will begin to travel in a straight line trajectory which is a combination of speed of your jump and the speed of the "ground" But remember, the cylinder also keeps rotating. You will intersect with the ground at almost the same point as you left it and with a difference in velocity that better much matches that of your initial jump.

The caveat here is that the cylinder diameter needs to be large compared to your jump. With a small cylinder, the curvature will be such that there can be a significant difference, and you would land awkwardly.

Or, for example, if it were possible for you to jump hard enough to travel across the diameter of the cylinder, the ground on the far side has a velocity in the opposite direction of the ground you left, and there will be a significant velocity difference between you and it when you contact.

But with a large cylinder, and normal activity, it would be hard to casually distinguish between standing on the Earth and being in a O'Neill cylinder using just the way things "fall".

SolarisOne, Aerdan and PeroK
Wonderful, thank you both! I wasn’t thinking about the velocity changing because of the direction. So if you were to step out of a building on the 5th floor, assuming the cylinder is large enough that the difference in gravity is negligible, once you are no longer touching the building would your momentum towards the ground be a static speed? Because at that point there are no other notable forces acting on your body, and this is where my confusion lies.

You could also push off from the very center. You accelerate as you push off, but once you are done pushing, your speed won’t change minus the minor effects of air resistance from atmosphere you move through. As you move towards the ground, that speed never changes, but the ground is moving rapidly relative to you. Your impact with the ground wouldn’t be hard from your momentum, but it would be from the spinning of the cylinder.

If you stepped off a platform at half the height, you would have a similar effect once you are no longer touching the platform. You will move at a higher velocity towards the ground (4.9 m/s). Even with the motion, you would still land in a different spot relative to your starting position: not directly below. As this distance towards the cylinder is decreased, you will eventually land closer to the spot vertically from where you started.

My question then is if you fell from a small height close to the edge, you would still land in the same spot, but wouldn’t your velocity be immediately 1g?

I hope I’m being clear in what I’m asking, and no doubt there is some detail that is not clicking. Thank you everyone for weighing in :)

Last edited:
PeroK
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You could also push off from the very center. You accelerate as you push off, but once you are done pushing, your speed won’t change minus the minor effects of air resistance from atmosphere you move through. As you move towards the ground, that speed never changes, but the ground is moving rapidly relative to you. Your impact with the ground wouldn’t be hard from your momentum, but it would be from the spinning of the cylinder.
Jumping from the centre of the cylinder is not really the intended function! That would be dangerous, of course. It's similar to jumping onto a moving train. As opposed to jumping up inside a moving train, which is safe - as long as you avoid the ceiling.

My question then is if you fell from a small height close to the edge, you would still land in the same spot, but wouldn’t your velocity be immediately 1g?
##1g## is still an acceleration, not a speed.

I hope I’m being clear in what I’m asking, and no doubt there is some detail that is not clicking. Thank you everyone for weighing in :)
Picturing motion in a rotating reference frame is not easy. One approach is to trust the mathemetics of fictitious forces, like the centrifugal force, and then it's simple. It's no different from gravity on Earth (for a sufficiently large cylinder) - if you analyse things from the rotating reference frame.

If you imagine yourself outside the cylinder, then it's more difficult to picture the inertial motion of the jumper being equivalent to the perceived non-inertial motion of someone jumping on Earth.

Aerdan
Jumping from the centre of the cylinder is not really the intended function! That would be dangerous, of course. It's similar to jumping onto a moving train. As opposed to jumping up inside a moving train, which is safe - as long as you avoid the ceiling.

##1g## is still an acceleration, not a speed.

Picturing motion in a rotating reference frame is not easy. One approach is to trust the mathemetics of fictitious forces, like the centrifugal force, and then it's simple. It's no different from gravity on Earth (for a sufficiently large cylinder) - if you analyse things from the rotating reference frame.

If you imagine yourself outside the cylinder, then it's more difficult to picture the inertial motion of the jumper being equivalent to the perceived non-inertial motion of someone jumping on Earth.
You’re right about 1g: I’m framing things poorly. Thanks for being flexible with your responses! I think it’s clicking better now :)

And thank you Bob, I’ll review that thread now!

bob012345
bob012345
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Wonderful, thank you both! I wasn’t thinking about the velocity changing because of the direction. So if you were to step out of a building on the 5th floor, assuming the cylinder is large enough that the difference in gravity is negligible, once you are no longer touching the building would your momentum towards the ground be a static speed? Because at that point there are no other notable forces acting on your body, and th

You could also push off from the very center. You accelerate as you push off, but once you are done pushing, your speed won’t change minus the minor effects of air resistance from atmosphere you move through. As you move towards the ground, that speed never changes, but the ground is moving rapidly relative to you. Your impact with the ground wouldn’t be hard from your momentum, but it would be from the spinning of the cylinder.

If you stepped off a platform at half the height, you would have a similar effect once you are no longer touching the platform. You will move at a higher velocity towards the ground (4.9 m/s). Even with the motion, you would still land in a different spot relative to your starting position: not directly below. As this distance towards the cylinder is decreased, you will eventually land closer to the spot vertically from where you started.

My question then is if you fell from a small height close to the edge, you would still land in the same spot, but wouldn’t your velocity be immediately 1g?

I hope I’m being clear in what I’m asking, and no doubt there is some detail that is not clicking. Thank you everyone for weighing in :)
Here is a scenario where you step off a building in an O'Neill colony assuming at first the building does not interfere with your path (you step to the side a bit). The cylinder is spinning clockwise and ##V_b## is the tangential velocity at the height of the building while ##V_c## is at the ground level and ##V_{rel}## is the relative velocity wrt to the ground. This assumes you just step off the building with no other velocity. I have not shown where the building is when you reach the ground.

The larger perspective;

DaveE