# Thread: Spinning space station and artificial gravity

1. Originally Posted by profloater
I am not sure again. I think the the Corioilis acceleration is always at right angles to the radial velocity, due to spin w, not at right angles to the rail.
No, always at right angles to the velocity vector in the plane of rotation. If my diagram were rotating anticlockwise, all objects moving in the plane would be deflected to the right of their velocity vector, irrespective of their direction of travel. So spinwards if moving radially upwards, radially downwards if moving spinwards, antispinwards if moving radial downwards, and radially upwards if moving antispinwards. This is why air and water masses moving east are deflected towards the equator, and those moving west towards the poles, just as those moving towards the poles are deflected east, and those moving towards the equator are deflected west. (Diagram here.)
The magnitude of the pseudoforce depends on the velocity as measured in the plane of the rotating frame, and the angular velocity of the rotating frame.

My own explanation of the effect is here: https://oikofuge.com/coriolis/
Application to trajectories in a rotating habitat here: https://oikofuge.com/coriolis-effect...space-habitat/
And a supplement here: https://oikofuge.com/coriolis-effect...at-supplement/

Grant Hutchison

2. Originally Posted by grant hutchison
However, pulling up the radial component of Coriolis leaves me with a residual circumferentially directed component, which will serve to retard the spinward bead experiencing enhanced centrifugal gravity, and hasten the antispinward bead enjoying reduced gravity. I'm now thinking that Coriolis doesn't affect anything, because it always points in a direction that the sliding bead is constrained not to follow. Hence, parsing it into radial and circumferential components just gives me two forces that cancel out.
In other words, the Coriolis pseudoforce on the sliding bead never does any work, because it doesn't induce any displacement, so Coriolis has no effect on the energy of the bead. The only pseudoforce doing work is the centrifugal force, which has a component that accelerates the bead along its tautochrone track. Which is how I should have thought about the problem originally.

Grant Hutchison

3. The use of “pseudo accelerations” might be confusing. If we are inside the rotating frame , as we call it, when we release a tennis ball it falls not to our feet like on Earth, but away from us as if thrown. The actual path is a straight line at the velocity it had when released, but we see it fall and we call the apparent centrifugal acceleration “pseudo.”

But if we constrain the ball, say, in a radial tube, it will fall to our feet but it has been forced. The force is Coriolis and is a real force that does work on the ball, accelerating it to a higher tangential speed. ( and lower tangential speed again when it bounces).

So free fall in a rotating frame appears to have centrifugal acceleration and sideways Coriolis acceleration, which are only due to rotating observation, but constrained movement in a rotating frame generates actual centripetal and Coriolis accelerations and forces.

The only constraint in a rotating frame that has no actual forces on the ball is a curved path that allows the ball to free fall at constant speed in a straight line until it touches the wall. A radial pole or complex spiral pole will apply real forces to a bead sliding along.

4. Here is another thought experiment for artificial gravity. I choose a 20 metre diameter cylinder rotating in space at 1 radian/s to give 1 g at the wall. A pendulum 15 m long on a light string is suspended from the distant wall and starts perfectly diametrically so the string passes through the centre of the cylinder. It is stable wrt the rotating frame with the string providing the centripetal acceleration.

What happens if you move the pendulum sideways in the plane of rotation? Or if you like, you hold the bob off sideways, stationary and let it go.?

5. Originally Posted by profloater
The only constraint in a rotating frame that has no actual forces on the ball is a curved path that allows the ball to free fall at constant speed in a straight line until it touches the wall. A radial pole or complex spiral pole will apply real forces to a bead sliding along.
The following image is an example of a 'no constraints' Tautochrone spiral traced on the side of a cone.

Note the plane of rotation is at 45 degrees to the relatively stationary observer/camera and the observer is 2*Pi*r away from each rotating source at the sources start points 1,0 and 3,0. In this simple model photons emitted at the sources start points would have to be traveling at c to complete one rotation and arrive at the observer/camera so the actual distance traveled would be 2*Pi*r*c/v (for both) where v is the velocity of the source. At relativistic rotational velocities there would be a length contraction (in SR at least) of the spoke length (refer OP of link below).

Incidentally, Cycloid arc's have some other interesting properties you may not be aware of:-

(1) In a 2D plot of 1 rotation the total area = 2*Pi*r*2*r = 4*Pi*r^2 = the surface area of a sphere of radius r.
(2) The area under a 2D plotted Cycloid arc = the area of the circle that created the arc.
(3) The area above a 2D plotted Cycloid arc = 3 x the area of the circle that created the arc.
(4) The length of the 2D plotted Cycloid arc = 4*r.
(5) You cannot create a circle of radius r with 1/4 of the surface area on a sphere of radius r (unless you 'measure' the curve as flat).

The above image is not an ATM claim. https://forum.cosmoquest.org/showthr...27#post2495527

6. Wow, synchonicity of posts! While I absorb that cone diagram, I can add that if you displace my pendulum axially and release it, it has a different motion again. Not at all like an earthly pendulum in gravity. But of course for large scales Earth is also a rotating frame for air and ocean currents.

7. Originally Posted by grant hutchison
No, always at right angles to the velocity vector in the plane of rotation. …
Grant Hutchison
I want to clarify this in my mind. Staying in the rotation plane but analysing from the static reference frame:
The instantaneous velocity seen must be along the constraint rail.
That velocity can be resolved to radial and tangential components at right angles. The Coriolis tangential acceleration is at that moment 2.dr/dt.w
The centripetal acceleration at that moment is radial at rw^2

The force from the rail must be a normal at right angles to the rail at that point because it is friction free.
That normal force can also be resolved into radial and tangential components or can be calculated from the acceleration components just mentioned.

8. Originally Posted by LaurieAG
The following image is an example of a 'no constraints' Tautochrone spiral traced on the side of a cone.
But if they're light paths, they're not going to be tautochrones.

Grant Hutchison
Last edited by grant hutchison; 2021-Nov-24 at 05:17 PM.

9. Originally Posted by profloater
The use of “pseudo accelerations” might be confusing. If we are inside the rotating frame , as we call it, when we release a tennis ball it falls not to our feet like on Earth, but away from us as if thrown. The actual path is a straight line at the velocity it had when released, but we see it fall and we call the apparent centrifugal acceleration “pseudo.”

But if we constrain the ball, say, in a radial tube, it will fall to our feet but it has been forced. The force is Coriolis and is a real force that does work on the ball, accelerating it to a higher tangential speed. ( and lower tangential speed again when it bounces).

So free fall in a rotating frame appears to have centrifugal acceleration and sideways Coriolis acceleration, which are only due to rotating observation, but constrained movement in a rotating frame generates actual centripetal and Coriolis accelerations and forces.

The only constraint in a rotating frame that has no actual forces on the ball is a curved path that allows the ball to free fall at constant speed in a straight line until it touches the wall. A radial pole or complex spiral pole will apply real forces to a bead sliding along.
The pseudo-accelerations and pseudo-forces evident in a rotating reference frame are as real (or unreal) as the accelerations and forces produced by Einsteinian gravity, but no-one ever seems to get into a frothing state of fury over mention of "gravitational force", in the same way one encounters with "centrifugal force".
My old physics teacher used to hammer on his desk while declaiming "There's. No. Such. Thing. As. Centri. Fugal. Force." But he never got around to denouncing gravity and claiming that the only real force was the one generated by the Earth pushing on the soles of your feet to accelerate you off a spacetime geodesic.

Grant Hutchison

10. Originally Posted by grant hutchison
The pseudo-accelerations and pseudo-forces evident in a rotating reference frame are as real (or unreal) as the accelerations and forces produced by Einsteinian gravity, but no-one ever seems to get into a frothing state of fury over mention of "gravitational force", in the same way one encounters with "centrifugal force".
My old physics teacher used to hammer on his desk while declaiming "There's. No. Such. Thing. As. Centri. Fugal. Force." But he never got around to denouncing gravity and claiming that the only real force was the one generated by the Earth pushing on the soles of your feet to accelerate you off a spacetime geodesic.

Grant Hutchison
It is Newton’s fault with his force equals mass times acceleration. If I spin my head around I can accelerate the whole world! As an engineer I encountered Coriolis in mechanisms and centrifugal pumps, as we carelessly call those things. And the Coriolis of ocean currents conflated with the hairy sphere of topology. My friends in the physics department insisted Coriolis to be virtual. So I learned it depends which building you are in. There are engineers who should have studied physics and vice versa. And then there were gyroscopes.

11. Originally Posted by grant hutchison
But if they're light paths, they're not going to be tautochrones.
A Tautochrone curve is only a straight line part of a Cycloid arc so there must be some part of some spiral curve that produces the same result wrt gravity.

12. Originally Posted by LaurieAG
A Tautochrone curve is only a straight line part of a Cycloid arc so there must be some part of some spiral curve that produces the same result wrt gravity.
I don't know what you mean by "straight line part of a cycloid".
A tautochrone involves an object starting from rest and accelerating under gravity along an inclined plane (by definition). A photon does not start from rest, does not accelerate, and is not constrained by an inclined plane. Therefore, not a tautochrone.

Grant Hutchison

13. Originally Posted by grant hutchison
I don't know what you mean by "straight line part of a cycloid".
Viewed from above both the Cycloid arc and the Tautochrone curve are in a straight line. The light path model is just an example of a cycloid traced on a cone, the photons themselves are irrelevant apart from the original derivation of the curve. I have actually seen a funnel referred to as an inclined plane, "The curved part of the cone serves as an inclined plane to the fluid or the liquid poured into it"

A tautochrone involves an object starting from rest and accelerating under gravity along an inclined plane (by definition).
The Tautochrone curve you describe is a 'straight' one and not a 'spiral' one although both are allowed by definition.

14. Originally Posted by grant hutchison
The pseudo-accelerations and pseudo-forces evident in a rotating reference frame are as real (or unreal) as the accelerations and forces produced by Einsteinian gravity, but no-one ever seems to get into a frothing state of fury over mention of "gravitational force", in the same way one encounters with "centrifugal force".
My old physics teacher used to hammer on his desk while declaiming "There's. No. Such. Thing. As. Centri. Fugal. Force." But he never got around to denouncing gravity and claiming that the only real force was the one generated by the Earth pushing on the soles of your feet to accelerate you off a spacetime geodesic.

Grant Hutchison
One of my college physics professors -- I think it was Porter Johnson (https://www.iit.edu/directory/people/porter-johnson) -- said something to the effect of centrifugal force acts like a force, so it's a force. What I tell my students (as a high school physics teacher) is that centripetal force is what you see when you're on the outside of a rotating system, looking in, but centripetal force is what you see when you're inside the system.

While the centripetal force/centrifugal force "controversy" is really just being overly pedantic, high school physics teachers have been known to present false explanations, most when explaining how wings work.
Last edited by swampyankee; 2021-Nov-28 at 04:16 AM.

15. Originally Posted by swampyankee
One of my college physics professors -- I think it was Porter Johnson (https://www.iit.edu/directory/people/porter-johnson) -- said something to the effect of centrifugal force acts like a force, so it's a force. What I tell my students (as a high school physics teacher) is that centripetal force is what you see when you're on the outside of a rotating system, looking in, but centripetal force is what you see when you're inside the system.

While the centripetal force/centrifugal force "controversy" is really just being overly pedantic, high school physics teachers have been known to present false explanations, most when explaining how wings work.
It is an old chestnut. When the centripetal acceleration is gravity for a moon, the planet is not an immovable object, it gets tugged around by its moon, what force is that? It is the constraint that reveals these forces. Maybe just the name causes the problems? In statics it is quite easy to see that there are equal and opposite forces, ties and struts and so on, forces without accelerations.

16. Originally Posted by LaurieAG
Viewed from above both the Cycloid arc and the Tautochrone curve are in a straight line.
A straight line is not a cycloid. There are tautochrone curves that are straight lines (one through the centre of a uniformly gravitating sphere, for instance), but they are not applicable to our rotating habitat. The geometry of the plane is specified in the plane. You can't just turn it on its side and announce that everything is a straight line.

Originally Posted by LaurieAG
The Tautochrone curve you describe is a 'straight' one and not a 'spiral' one although both are allowed by definition.
Please show me the definition to which you are referring.

Grant Hutchison

17. I am still worried about the kinetic energy differences. In gravity all the balls start with no kinetic energy but different potentials, and arrive at the same time with no potential energy but differing kinetic energy. In the rotating ship, we have the centre position which never starts but all other starts have tangential velocity, increasing toward the wall where the gravity potentials are decreasing toward the ground. So I guess to be the same travel time, the excess energy remains as increased tangential energy, provided by the coriolis acceleration. At all points along the rail the actual energy is all tangential kinetic energy, there is no equivalent to the gravity potential, as in “there is no centrifugal force” . So it worries me that the tautochronic tracks are actually concentric circles with infinite fall time.

18. Originally Posted by profloater
So it worries me that the tautochronic tracks are actually concentric circles with infinite fall time.
But they're not. They are (according to the paper I linked to) epicycloids. The diagram I drew earlier in this thread is just such an epicycloid.
Coriolis turns out to be irrelevant, because it's always at right angles to the tautochrone track and does no work. Centrifugal pseudoforce does work on our sliding bead all the way down its tautochrone track, because there is always a component of the radially directed force directed along the sloping line of the track (except at a dimensionless point at the very bottom of the track, where it is tangential to the habitat floor).
Each bead arrives at the bottom of the curve with a kinetic energy (measured in the rotating frame) that matches its centrifugal potential energy when it started from rest (in the rotating frame).

BTW: The fact that the epicycloid (or cycloid in the classic uniform-gravity case) is tangential to the floor at the arrival point is another thing that tells us these tautochrone tracks are not spirals, which always increase their radial coordinate as they turn.

Grant Hutchison

19. I am imagining , as visualisation, a bead starting stationary in the rotating frame on the spiral at half radius, it has a tangential velocity at that instant in the non rotating frame.
The spiral at that point allows the bead to move but constrains it. So in the rotating frame we see it move clockwise or anticlockwise depending on the left or right hand spiral. Now it has a speed along the rail so its tangential speed is either greater or less than the required velocity at that new radius. This must continue until it reaches the wall going either faster or slower than the wall at that point of impact, jerking positively or negatively to the ship w.
Fine, now we start another bead at 3/4 radius with a higher actual tangential velocity. It must accelerate along the rail at a slower rate so that the first one can catch up.
So my energy problem is resolved by the first bead having less or more KE when it reaches the 3/4 point. A bead starting nearly at the wall must find a very gentle slope, nearly a circle, in order that the others catch up and the centre part must be near radial like in the diagram.
OK, I think that works as you described. An axial slope would be more like the gravity field equivalent, but with an axial impact at the wall. So if the bead is massive you can devise a rail to change the spin rate or the axial velocity, in the static frame.

20. OK, Rana & Joag have a nice short chapter on tautochrones and brachistochrones, dealing with the three cases of a uniform field, a central attractive field with force proportional to distance from the centre (uniform gravitating sphere), and a central repulsive field with force proportional to distance (the "centrifugal gravity" case).

Setting up the tautochrone problem the way that Huygens treated it, in terms of oscillatory motion with constant period irrespective of amplitude, we can draw epicycloids from cusp to cusp in our rotating habitat.
In the plot below, the starting point (and the radial coordinate of the epicycloid cusps) is at 0.5 of the radius of the habitat floor. I've drawn epicycloids connecting a starting point at left to positions at R=0.5 that are 180, 90 and 45 degrees away from the starting point. So these are all tautochrones connecting the two endpoints. And, by symmetry, they're tautochrones connecting the start point at R=0.5 to the point on each curve which reaches maximum radius.

You can see for start and end points at R=0.5 and 180 degrees apart, the tautochrone is tangent to the floor of the habitat. If we make R>0.5, we can produce tangent tautochrones that span <180 degrees.

All these curves are also brachistochrones--that is, they represent the trajectory that minimizes time of travel.

Now, if we make R<0.5, then we can't reach the floor of the habitat with an epicycloid that spans <=180 degrees. But we can do it if we push beyond 180 degrees. In the case of my previous diagram, with start and end at R=0.1, the total span of the epicycloid was 1620 degrees! However, these curves aren't brachistochrones any more. For instance, if we had an epicycloid spanning, say, 270 degrees, we could shorten the journey time by linking its endpoints with a 90-degree epicycloid.

It was this business of brachistochrones not being able to reach the habitat floor from starting points high in the habitat that was getting me in a tangle earlier. Of course, tautochrones are less constrained, because they don't need to minimize time.

Grant Hutchison

21. Originally Posted by grant hutchison
The geometry of the plane is specified in the plane.
The path of the cycloid would be a 'straight' line when viewed from above.

If you consider a cylinder per Rama your curves are still 'straight' lines and their planes are perpendicular to the cylinders axis of rotation. A spiral Tautochrone curve would be in a plane at an angle to the axis of rotation and would therefore appear 'spiral' from above (end of cylinder) and not 'straight'.

22. Originally Posted by grant hutchison
OK, Rana & Joag have a nice short chapter on tautochrones and brachistochrones, dealing with the three cases of a uniform field, a central attractive field with force proportional to distance from the centre (uniform gravitating sphere), and a central repulsive field with force proportional to distance (the "centrifugal gravity" case).

Setting up the tautochrone problem the way that Huygens treated it, in terms of oscillatory motion with constant period irrespective of amplitude, we can draw epicycloids from cusp to cusp in our rotating habitat.
In the plot below, the starting point (and the radial coordinate of the epicycloid cusps) is at 0.5 of the radius of the habitat floor. I've drawn epicycloids connecting a starting point at left to positions at R=0.5 that are 180, 90 and 45 degrees away from the starting point. So these are all tautochrones connecting the two endpoints. And, by symmetry, they're tautochrones connecting the start point at R=0.5 to the point on each curve which reaches maximum radius.

You can see for start and end points at R=0.5 and 180 degrees apart, the tautochrone is tangent to the floor of the habitat. If we make R>0.5, we can produce tangent tautochrones that span <180 degrees.

All these curves are also brachistochrones--that is, they represent the trajectory that minimizes time of travel.

Now, if we make R<0.5, then we can't reach the floor of the habitat with an epicycloid that spans <=180 degrees. But we can do it if we push beyond 180 degrees. In the case of my previous diagram, with start and end at R=0.1, the total span of the epicycloid was 1620 degrees! However, these curves aren't brachistochrones any more. For instance, if we had an epicycloid spanning, say, 270 degrees, we could shorten the journey time by linking its endpoints with a 90-degree epicycloid.

It was this business of brachistochrones not being able to reach the habitat floor from starting points high in the habitat that was getting me in a tangle earlier. Of course, tautochrones are less constrained, because they don't need to minimize time.

Grant Hutchison
Thank you for clarifying all those cases. All aspects of artificial gravity that are not often considered.

23. Originally Posted by LaurieAG
The path of the cycloid would be a 'straight' line when viewed from above.

If you consider a cylinder per Rama your curves are still 'straight' lines and their planes are perpendicular to the cylinders axis of rotation. A spiral Tautochrone curve would be in a plane at an angle to the axis of rotation and would therefore appear 'spiral' from above (end of cylinder) and not 'straight'.
None of this is relevant to either plane geometry, classical mechanics, or the fact that your spiral is not a tautochrone, for all the reasons already given.

Grant Hutchison

24. Originally Posted by grant hutchison
None of this is relevant to either plane geometry, classical mechanics, or the fact that your spiral is not a tautochrone, for all the reasons already given.
Considering that a funnel/cone is an inclined plane " A tautochrone involves an object starting from rest and accelerating under gravity along an inclined plane (by definition)." a spiral non tautochrone curve could produce the same results.

I could understand you claiming it is not relevant if you exclude 3D geometry but that doesn't mean that there is not a 3D solution when you don't restrict your 'solutions' to 2D tauhtochrones.

25. Originally Posted by LaurieAG
Considering that a funnel/cone is an inclined plane " A tautochrone involves an object starting from rest and accelerating under gravity along an inclined plane (by definition)." a spiral non tautochrone curve could produce the same results.
Firstly, a tautochrone travels along a curve, not a plane. (Google denies the existence of your quotation on the internet.)
Secondly, a spiral cannot be a tautochrone, since the radial coordinate of a spiral increases without limit, while that of a tautochrone (in a uniform field or a Hooke field) does not.

Grant Hutchison

26. Originally Posted by grant hutchison
Firstly, a tautochrone travels along a curve, not a plane. (Google denies the existence of your quotation on the internet.)

Originally Posted by grant hutchison
A tautochrone involves an object starting from rest and accelerating under gravity along an inclined plane (by definition).
Originally Posted by grant hutchison
Secondly, a spiral cannot be a tautochrone, since the radial coordinate of a spiral increases without limit, while that of a tautochrone (in a uniform field or a Hooke field) does not.
A tautocrone is a 2D geometric curve while a spiral is a 3D geometric curve.

Abels Solution requires a distance measure along a curve and a height drop that, when combined, allow for the calculation of the gravitational potential energy of the frictionless ball. If you scale a spiral up into circles and experiment you will find a solution soon enough.

https://en.wikipedia.org/wiki/Tautoc...#39;s_solution
The kinetic energy is 1/2 mv^2, and since the particle is constrained to move along a curve, its velocity is simply dℓ/dt, where ℓ is the distance measured along the curve. Likewise, the gravitational potential energy gained in falling from an initial height y0 to a height y is mg(y0 − y) thus: 1/2 m(dℓ/dt)^2 = mg(y0 − y).

27. Originally Posted by LaurieAG
Apologies. I made an error in my post by careless use of language. I've now corrected myself. The object does not accelerate down an inclined plane, but a curve of continuously varying gradient, relative to the local "gravity" vector. The instantaneous acceleration is calculated by treating each point on the curve as an infinitesimal inclined plane, but the whole curve is of course not an inclined plane. Sometimes what's in my head doesn't come out of my fingers--I'm sorry if I misled you in that regard.

Originally Posted by LaurieAG
A tautocrone is a 2D geometric curve while a spiral is a 3D geometric curve.

Abels Solution requires a distance measure along a curve and a height drop that, when combined, allow for the calculation of the gravitational potential energy of the frictionless ball. If you scale a spiral up into circles and experiment you will find a solution soon enough.
Given that "scale a spiral up into circles" makes no sense to me, I doubt it.
One last attempt. The restoring force along a tautochrone curve must tend steadily to zero as it approaches the base of the curve. The Archimedean conical spiral in your diagram moves steadily into regions of increasing centrifugal gravity while maintaining a constant gradient relative to the radial centrifugal pseudoforce. The force component directed tangent to its curve therefore increases steadily and without limit. Therefore, it cannot be a tautochrone. That's all there is to it, that's the point I was making, and I don't propose getting into further discussion about it.

Grant Hutchison
Last edited by grant hutchison; Yesterday at 05:22 PM.

28. I suppose it isn't surprising that the 2D tautocrone curve is as far as they got with this avenue of Newtonian mechanics as 3D spin/artificial gravity made no sense to them either.

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