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I made a small GIF animation showing the same thing as the
illustration in French. A non-rotating planet and non-rotating
moon revolve about their common barycenter:

http://www.freemars.org/jeff2/revo1.htm

Although neither planet nor moon rotates, the planet-moon
system does rotate.

-- Jeff, in Minneapolis

2. Originally Posted by Jeff Root
Ken,

If Richard really isn't sure what revolution is-- even after discussing it
for several days-- then he needs more basic info...

-- Jeff, in Minneapolis
Jeff and Ken, I'm not saying that I am 'not sure' what it means, I'm saying that it has to have mathematicaly defined properties. Same as translation, reflection and rotation. Don't keep telling me what it means, point me to where its mathematical properties are defined.

3. Originally Posted by RichardMB
Don't keep telling me what it means, point me to where its mathematical properties are defined.
I already did, insofar as the word is useful in this thread. If you are searching for anything different in "revolution" than simply "motion in a closed trajectory", that you need to understand in order to understand tides, then you are chasing windmills.

4. Originally Posted by Jeff Root
Also, revolution and orbit are often not synonymous, even in this
thread. I am revolving around the Earth, but I am certainly not
in orbit around the Earth. A penny revolving on a phonograph
turntable is not in orbit around the spindle.
I am using the word "revolution" the way it is used in astronomy, because this thread is about tides. I have no interest in all the other possible ways one might use the term, only how it relates to tides. In reality, there is a deep confusion about the difference between revolution and rotation, I am trying to rescue us all from that confusion by presenting the only definition of revolution that is needed for this thread.

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Originally Posted by Ken G
In reality, there is a deep confusion about the difference between revolution and rotation, I am trying to rescue us all from that confusion by presenting the only definition of revolution that is needed for this thread.
Over here, I'm still cranking the handle on a turntable loaded with accelerometers ...
RichardMB seems to have an impression that centripetal acceleration of itself will inevitably lead to tension developing in an extended object. So I quite liked the idea of hanging on to a more general concept of circular motion, in order to see why internal compression and tension are symptoms of the distribution of the applied force, which then links through French's diagram to the tidal effects of gravity.
I'm not sure you'll persuade RichardMB by coming at it through gravity in the first instance, but I'm certainly happy to step back and watch.

Grant Hutchison

6. Originally Posted by grant hutchison
Over here, I'm still cranking the handle on a turntable loaded with accelerometers ...
RichardMB seems to have an impression that centripetal acceleration of itself will inevitably lead to tension developing in an extended object. So I quite liked the idea of hanging on to a more general concept of circular motion, in order to see why internal compression and tension are symptoms of the distribution of the applied force, which then links through French's diagram to the tidal effects of gravity.
I'm not sure you'll persuade RichardMB by coming at it through gravity in the first instance, but I'm certainly happy to step back and watch.

Grant Hutchison
We seem to have reached something of an impasse here Grant, I've got more stuff to discuss but I'm reluctant to make my next move until this thing is resolved.

I guess you're still hanging in there with the 'revolutionaries' - that right?

Tell me what you think, how do we proceed - anybody?

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Originally Posted by RichardMB
I guess you're still hanging in there with the 'revolutionaries' - that right?
I don't know what this means, sorry.

I'm aware that Ken and Jeff are coming at the same problem from a different direction than the one I'm using. It seems like it just complicates matters to have two explanatory threads braided together, as has been happening. You presumably have a view on the four-way discussion that's going on. Is it helping or hindering progress?

Grant Hutchison

8. Originally Posted by grant hutchison
I don't know what this means, sorry.

I'm aware that Ken and Jeff are coming at the same problem from a different direction than the one I'm using. It seems like it just complicates matters to have two explanatory threads braided together, as has been happening. You presumably have a view on the four-way discussion that's going on. Is it helping or hindering progress?

Grant Hutchison
I have my next post prepared for you, and I'm hanging back because of this 'undefinable' revolution thing, but now I think maybe I'll just get on with it.

Originally Posted by grant hutchison
The acceleration is angular-velocity-squared times radius. Angular velocity is the same for all points of the penny, so acceleration in this scenario is varying linearly with radius of gyration.

Grant Hutchison
Yes I agree, that's fine. In the present case I suggest w^2r is more useful - same thing but saves having to work out v every time. That ok?

Originally Posted by grant hutchison

To be honest, I don't find the physics at all confusing.

Grant Hutchison
No harm intended Grant I thought you were conveying confusion by your choice of words. Sorry if I misunderstood.

Originally Posted by grant hutchison

"Revolution" is motion in a circle. As such, it requires acceleration directed towards the centre of revolution. In astronomy, the usage is restricted to motion around a distant centre: a planet revolves around the sun, and rotates on its own axis.

Grant Hutchison
Now you're misunderstanding me. I didn't ask what you thought it meant, I asked to pointed to a definition of its mathematical properties. This is not a ploy to faze you - it is an absolute essential if you want to use it to do mathematics.

Originally Posted by grant hutchison
And yet, the motorbike and rider, the objects doing the revolving, are in compression. We could achieve the same effect without any tension anywhere. Imagine flying in a circle in a rocket, with the engine firing outwards from the centre all the time, and the tip of the rocket facing inwards.
So you can't dismiss compression as "misleading", I'm afraid.

Grant Hutchison
Discussing how the 'wheel of death' works won't clarify the point I'm making - unless it is the case that you find it confusing and would like me to elucidate, but I don't think that is the case. It would therefore lead us away from our current path - in that sense it's misleading.

Anyway, here's the post I was hanging on to:

Before I start I must say that my maths is not good. So if I boob here and there don’t hold it against me – I’m doing my best.

O.K. We have a penny ‘going round’ on a platter, there is a radial delta-a, which we can calculate (if need be) using a = w^2r, and a transverse delta-a of, essentially, zero.

These accelerations are applied to the penny in some fashion but they appear as tension in the platter. The acceleration can therefore be any magnitude up to the tensile limit of the platter – depending on w.

Now we need to consider a more realistic force, one with an inverse square law. I can’t think how to do that so we will just have to imagine it.

To make it easy let’s make the penny 2 cm diam. And place it so the centre is 16 cm from the spindle, at our 9 o’clock start, this will put west at 17 cm and east at 15 cm. The centripetal force at the centre of the penny (16 cm from the spindle) we’ll set at 16 (16 anythings, milliNewtons maybe, doesn’t matter).

At 16 cm the centripetal force is 16 so our ‘gravity’ is also 16 – (otherwise there’d be an unbalanced force and it would go off somewhere else).

Sorry about all the numbers but it’s easy to make daft mistakes with algebra (for me anyway).

Now, we can find ‘gravity’ using g = k/r^2 where k is a constant and r is the distance, so k = g*r^2. Both g and r are 16 so k = 16^3 = 4096.

At the west point using g = k/r^2, we have g = 4096/17^2 = 14.17 and at the east point we have g = 4096/15^2 = 18.20.

This gives a total gravity difference west-east of 18.2 – 14.17 = 4.03 – but it isn’t linear, west-centre it is 1.83 and centre-east it is 2.2.

The centripetal force is proportional to distance so, by inspection, it is 17 west and 15 east. That is to say the west point requires a centripetal acceleration of 17 – but we only have 14.17 available from gravity, we need another 2.83 and the only candidate is the penny itself, so it must provide this extra force as tension. A similar condition applies at the east, here we need 15 but ‘gravity’ provides 18.2, an excess of 3.2 again this must be provided by tensile force from the penny.

There’s your tension Grant. You’d no doubt do it algebraically but I think it has more punch done like this.

The result is that gravity differential alone contributes a tension of 4.03, that’s when it is stationary. When it is ‘going round’ or ‘revolving’ or whatever, there is an additional tension of 6.03. This centripetal ‘tidal’ force is fully 50% greater than that due to ‘gravity’.

Far from the tidal force being due to gravity alone, gravity is actually very much the minor contributor. The major source is centripetal acceleration.

9. Originally Posted by Ken G
Right-- "rotation" implies rigid body motion.
Not in astronomy, but I was just trying to make it clear.
This tidal contrast is in 2D, not 1D, and if you just look at 1D, then rotation can be an issue, but only if the particles are acting like a solid body. More to the point, the fact that you are only using 1D means that rotation is being used as a kind of replacement for having to consider the other direction-- you set up rotation to give no pinching along the orbit, so anything you see radially will explain the football shape.
I don't understand how considering the side to side effects as well as the radial effects is "only using 1D". I mentioned it because I thought it was 2D, and you seem to be an advocate for that sort of analysis.

10. Originally Posted by RichardMB
The result is that gravity differential alone contributes a tension of 4.03, that’s when it is stationary. When it is ‘going round’ or ‘revolving’ or whatever, there is an additional tension of 6.03. This centripetal ‘tidal’ force is fully 50% greater than that due to ‘gravity’.
What you mean is that the difference in the "centripetal" (17-15) adds another 2 units to the total (4.03 + 2 = 6.03). In other words, your calculation is about 50% larger than what someone who used only gravity would calculate. Hence my comments earlier, that your figure was 50% larger than the observed values.

Why?

Your penny is revolving around the spindle, and as we pointed out, it is also rotating--if you follow its motion, it is rotating once per revolution. In other words, the "centripetal" force produced in the penny's reference frame uses the same angular velocity, and so the same "anything" units produce a value that is linearly proportional to radius, of the penny.

So, it's 16 at the penny, and 1 outwards all along the edge of the penny! That's where the 16-1=15 arises for the east edge, and 16+1 for the west edge, but there is also a 1 at the north and south edges. That is the "equatorial bulge" due to the rotation of the penny, and it does not show up in the tides--because it is a uniformly raised amount, all around the edge. The "1 force" produces the oblateness of the earth, but does not contribute to the tides.

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Originally Posted by RichardMB
In the present case I suggest w^2r is more useful - same thing but saves having to work out v every time. That ok?
We're saying the same thing: angular velocity is angle/time, symbolized by ω, which people sometimes write as w.

Originally Posted by RichardMB
I didn't ask what you thought it meant, I asked to pointed to a definition of its mathematical properties. This is not a ploy to faze you - it is an absolute essential if you want to use it to do mathematics.
So tell me what particular mathematical properties of "movement in a circle around a remote centre" you require. I honestly can't see what properties you want that aren't already implicit in that description. Do you want an equation for the coordinate change with time?

Originally Posted by RichardMB
The result is that gravity differential alone contributes a tension of 4.03, that’s when it is stationary. When it is ‘going round’ or ‘revolving’ or whatever, there is an additional tension of 6.03. This centripetal ‘tidal’ force is fully 50% greater than that due to ‘gravity’.

Far from the tidal force being due to gravity alone, gravity is actually very much the minor contributor. The major source is centripetal acceleration.
You're double-counting. If gravity gives us 4.03, and gravity plus rotation gives us 6.03, then rotation has provided 2. That's the difference between 17 and 15, the centripetal accelerations you worked out to start with.

Let's look again at the centripetal acceleration relative to the centre of the coin, since tidal effects are raised relative to the centre of gravity. The centripetal acceleration relative to the centre of the coin is 1 unit westwards for the east side of the coin, and one unit eastwards for the west side of the coin (from your figures). If you do some vector algebra for the north and south sides of the coin, you'll find the north side has an acceleration southwards of 1, and the south side has an acceleration northwards of 1. So each of the four quadrant points has an acceleration relative to the coin's centre of 1, in your units. It's ones all the way round!
Hmmm. Since these forces are uniform around the coin's equator, they're certainly not associated with a tidal bulge, despite what you say: any distortion associated with them must be symmetrical right around the equator. In fact, they represent the equatorial bulge due to the coin's rotation, as it revolves synchronously around the centre of the platter; and since the coin rotates at the same rate as the platter, the forces are perfectly consistent with your rate of 1 unit force per centimetre radius of gyration, because each of these points is 1cm from the coin's centre.
You've simply fooled yourself by looking at the equatorial bulge where it assists tidal gravity (nearest to and farthest from the centre of the platter), and not looking (north and south) to where the equatorial bulge opposes tidal gravity.

And that's why French uses a diagram in which the "coin" does not rotate as it revolves: to stop you fooling yourself in this way.

Grant Hutchison

Edit: To acknowledge that hhEb09'1 got in to say exactly the same thing while I was still typing.

12. Originally Posted by grant hutchison
We're saying the same thing: angular velocity is angle/time, symbolized by ω, which people sometimes write as w.
So we are, reading quickly I thought I read v^2/r - my mistake.

Originally Posted by grant hutchison
So tell me what particular mathematical properties of "movement in a circle around a remote centre" you require. I honestly can't see what properties you want that aren't already implicit in that description. Do you want an equation for the coordinate change with time?
Just WIKI for 'transformation' (mathematical) you'll see what we have for translation, reflection and rotation - that's the sort of thing we need.

Originally Posted by grant hutchison
The centripetal acceleration relative to the centre of the coin is 1 unit westwards for the east side of the coin, and one unit eastwards for the west side of the coin (from your figures). If you do some vector algebra for the north and south sides of the coin, you'll find the north side has an acceleration southwards of 1, and the south side has an acceleration northwards of 1. So each of the four quadrant points has an acceleration relative to the coin's centre of 1, in your units.

Grant Hutchison
That was the purpose of the 4 accelerometers. You recall that the west & east read lateral accelerations, whilst the north & south accelerometers read zero. Vector algebra and reference frames won't change that will they? You are making statements about what you think ought to happen, what needs explaining is what does happen - as evidenced by the instruments.

You'd agree with that - yes?

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Originally Posted by RichardMB
Just WIKI for 'transformation' (mathematical) you'll see what we have for translation, reflection and rotation - that's the sort of thing we need.
I know what they are: the nature of the transformation relies only on the start and end coordinates. Revolution requires us to specify the path taken between the start and end coordinates. That's why I said that you might consider it to be an infinite series of infinitesimal translations. It's also why I offered to give you the general coordinates as a function of time.

Originally Posted by RichardMB
That was the purpose of the 4 accelerometers. You recall that the west & east read lateral accelerations, whilst the north & south accelerometers read zero. Vector algebra and reference frames won't change that will they? You are making statements about what you think ought to happen, what needs explaining is what does happen - as evidenced by the instruments.
Umm. Can you see how ridiculous that statement is?
You described the instruments and said, in a handwaving way, what you thought the instruments would read. The imaginary readings from imaginary instruments give you imaginary results: nature feels no obligation to match.
I have more confidence in the maths.

But let's follow through on your original description of the experiment (in which the coin rotates along with the platter):
Originally Posted by RichardMB
The north and south accelerometers both show zero (in reality there may be a very small vertical delta-a, analogous to the gravity pinch effect, but for the present case we can consider it zero).
You've now given us the measurements and units with which we can look at that negligible "pinch". The north accelerometer is 16 cm west of the platter centre and 1cm north: so that's a distance of sqrt(16&#178;+1&#178, forming the hypotenus of a right triangle with sides 16x1. Distance units and centripetal units match, thanks to your formulation, so we can immediately solve for the unknown "pinch" force directed towards the coin's centre ... presto, it's one force unit. Identical reasoning gives you the magnitude of the pinch on the south accelerometer. So the north and south forces pinch towards the centre of the coin with a combined magnitude of two force units; that's identical to the pinch arising from the difference between the western and eastern accelerometers, which we already know to be two units.

Grant Hutchison
Last edited by grant hutchison; 2007-Feb-02 at 12:26 AM. Reason: Fixed last sentence

14. Originally Posted by hhEb09'1
Not in astronomy, but I was just trying to make it clear.
Your link didn't work for me, so I'm not sure what you mean. I maintain that, even in astronomy, rotation implies a rigid rotation of a patch of coordinates, and we can say that if the identified locations on a body (say fluid parcels, or sites in a lattice) are following said coordinates, then we may say the body is "rotating". I am not aware of any other definition of that concept, in any application. In particular, there is no local meaning to the term, it is a relationship, which is the main contrast with revolution (when the latter word is reserved for an orbit, not a turntable).

I don't understand how considering the side to side effects as well as the radial effects is "only using 1D".
It isn't-- my point is that one can use tidally locked rotation as a special case, a device that allows you to focus entirely on the radial (1D) effects, because in that special case, you know there's nothing going on in the sideways direction (for just the reasons you mentioned).
I mentioned it because I thought it was 2D, and you seem to be an advocate for that sort of analysis.
Oh yes, indeed-- I am more than an advocate, I am pointing out that no other analysis of tide-causing deformations is remotely close to complete. That's just the truth, because a football shape is a contrast of two directions, inescapably.

15. Originally Posted by Ken G
Your link didn't work for me, so I'm not sure what you mean.
http://en.wikipedia.org/wiki/Galaxy_rotation_problem
I maintain that, even in astronomy, rotation implies a rigid rotation of a patch of coordinates, and we can say that if the identified locations on a body (say fluid parcels, or sites in a lattice) are following said coordinates, then we may say the body is "rotating". I am not aware of any other definition of that concept, in any application. In particular, there is no local meaning to the term, it is a relationship, which is the main contrast with revolution (when the latter word is reserved for an orbit, not a turntable).
I'm not sure how you'd apply that to the rotations of galaxies, but not to the example we were talking about.

16. Originally Posted by hhEb09'1
http://en.wikipedia.org/wiki/Galaxy_rotation_problemI'm not sure how you'd apply that to the rotations of galaxies, but not to the example we were talking about.
"Galaxy rotation" is an obvious misnomer, right up there with "planetary nebula". It doesn't change the definition of rotation, nor of planets. It should certainly be called "galactic revolution" to be consistent with established nomenclature, and all astronomers know this, but the traditions are what they are. If you ever find an astronomer who doesn't know this, ask him or her: what is the period of galaxy rotation?

17. Originally Posted by Ken G
If you ever find an astronomer who doesn't know this, ask him or her: what is the period of galaxy rotation?
What about solar rotation? also a misnomer?

18. Originally Posted by hhEb09'1
What about solar rotation? also a misnomer?
Now you are really stretching. The Sun does rotate very nearly as a solid body, you know that. We all know that nothing in the real world is exact, terms are always idealizations. But when one is trying to be more exact, one adds a layer of complexity, and for the Sun, you have "differential rotation". Notice that I did not say rotation had to apply to the entire body (indeed, what is the entire body of the Sun?), it applies to a patch of coordinates, or a collection of matter following those coordinates. "Differential rotation" is a term that indicates you must break the Sun into subpieces in order to apply the concept of rotation more precisely, but notice this is still quite distinct from the concept of revolution. Nevertheless, in many applications such precision is unnecessary, and you'll hear "the rotation period of the Sun" without apology. So the short answer is, "no", and the other short answer is, "yes".

19. Yes and besides, "solar revolution" is reserved for the upcoming color change.

20. Originally Posted by Ken G
Notice that I did not say rotation had to apply to the entire body (indeed, what is the entire body of the Sun?), it applies to a patch of coordinates, or a collection of matter following those coordinates.
Then you don't have a problem with the two points I mentioned before? Why three then? What is it about three that makes it wrong? Isn't it just as "nearly" right?

21. Originally Posted by hhEb09'1
Then you don't have a problem with the two points I mentioned before? Why three then? What is it about three that makes it wrong? Isn't it just as "nearly" right?
I think you are missing what I'm saying. I'm saying that "revolution" is a property of one point. Period, you don't need any more to analyze centripetal force. Tidal stress is a property of multiple points that are bound to each other, but it is a much more general relation than rotation. Rotation also involves multiple points, but with only one degree of freedom, so it's a tiny subset of all the possible types of motion. What I'm saying is that just because tidal stress involves multiple points, and so does rotation, in no way suggests these two are necessarily related to each other. Furthermore, when you restrict your attention to football-shaped tidal stresses, you find that this is completely independent of anything that has to do with rotation. That is because of the different types of constraints involved in defining these kinds of interrelations between the points. This is what I'm saying, and so that's why you need three points, not two, to see the difference-- otherwise you just don't have enough degrees of freedom to see what a football shape even is, or why it has nothing to do with rotation.

22. Originally Posted by Ken G
I think you are missing what I'm saying.
Yes, perhaps we aren't talking about the same thing. I'm asking about your comment "Fine, but now look at three motes! Where's the rotation now?"

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Ken,

I said that my animation...

http://www.freemars.org/jeff2/revo1.htm

... shows a non-rotating planet and non-rotating moon revolving
about their common barycenter. Although neither the planet nor
the moon rotates, the planet-moon system does rotate.

Do you disagree?

This is what I said in another thread last November 26:

Originally Posted by Jeff Root
Anything which can hold together can rotate. Solid bodies are
held together by electrochemical forces, so solid bodies can rotate.
All the parts of a rotating solid body revolve about the axis of
rotation. A spinning dancer's arms and legs, lungs and heart,
atoms and molecules are all revolving about her axis of rotation.
A galaxy can rotate because all of its parts are held together by
gravity. The rotation is more complex than that of a solid body
because the different parts revolve about the axis of rotation at
different speeds and in different directions, and constantly change
speed and direction as they interact.

The planets in the Solar System revolve about the Sun. Although
it is unusual to put it this way, the Solar System can be said to
rotate in exactly the same way that a disk galaxy rotates. The
parts revolve about the center in pretty much the same direction.

On the other hand, stars in a globular cluster revolve about the
center of mass in all different directions, so the cluster as a whole
cannot be said to rotate. That is why globulars are globular. The
paths of the stars in a globular may be even more complex than
those in a disk galaxy, on average. The paths can change wildly
as the stars interact, without changing the shape or overall lack
of rotation of the globular. In a disk galaxy, which is much larger
than a globular cluster, interactions between stars cause much
smaller changes in the stars' paths relative to the overall size of
the whole system, so in this case, too, the overall rotation of the
galaxy is not changed, unless two stars actually collide, or one is
thrown out of the galaxy entirely.

Also, when a dancer rotates, each part of her body rotates, in
addition to revolving about her axis of rotation. We can see that
she is rotating as a whole, because at one moment we see her
front, then we see her left side, then her back, then her right
side, and then her front again. Likewise, first we see the front
of her left knee, then the left side of her left knee, then the back
of her left knee, then the right side of her left knee, and then the
front again.

But the individual parts are not required to rotate along with
the rest of the body. Suppose that the dancer is holding a lazy
susan in each hand. The lazy susans have good bearings, so they
will rotate almost frictionlessly when given a torque. However, the
dancer and the lazy susans start out motionless. Then the dancer
starts rotating, being careful not to drop the lazy susans. As she
turns, the lazy susans, being lazy and having inertia, do not rotate
even though they do revolve about the dancer's axis of rotation.
We always see the same side of the objects on the lazy susans,
and never see the other sides.
Do you disagree?

-- Jeff, in Minneapolis

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Originally Posted by Jeff Root
... shows a non-rotating planet and non-rotating moon revolving
about their common barycenter. Although neither the planet nor
the moon rotates, the planet-moon system does rotate.

Do you disagree?
I disagree, but I left it alone because I thought you might come back and edit. If these things were in synchronous rotation, then I think the whole could be said to "rotate": they'd be in rigid rotation, as Ken describes, with all parts of the system moving as one. As it is, I see two objects moving in circles around each other, but no rotation anywhere.

Grant Hutchison

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Originally Posted by grant hutchison
I disagree, but I left it alone because I thought you might come back and edit.
Very funny.

Originally Posted by grant hutchison
If these things were in synchronous rotation, then I think the whole could
be said to "rotate": they'd be in rigid rotation, as Ken describes, with all
parts of the system moving as one. As it is, I see two objects moving in
circles around each other, but no rotation anywhere.
So the cocoa in my mug, swirling around in circles after I have stirred
it, is revolving but not rotating?

I would say that each individual molecule or clump of molecules in my
cocoa is revolving around the center of the mug, and that the cocoa
as a whole is rotating. I can't tell anything about the rotations or
translations of the individual molecules or clumps of molecules, but
since the cocoa is hot, I presume that they are moving in random
directions at speeds on the order of a kilometer per second, and
changing direction typically several billion times per second.

I must admit, though, that it was a mistake to put accelerometers in
the cocoa. They are too crunchy, and not at all tasty. Maybe I'll try

-- Jeff, in Minneapolis

26. Originally Posted by grant hutchison
I know what they are: the nature of the transformation relies only on the start and end coordinates. Revolution requires us to specify the path taken between the start and end coordinates. That's why I said that you might consider it to be an infinite series of infinitesimal translations. It's also why I offered to give you the general coordinates as a function of time.

Grant Hutchison
You can’t resist telling me what you think it means – that isn’t of any interest at all. What is required is a mathematical definition of its properties.

Originally Posted by grant hutchison
Umm. Can you see how ridiculous that statement is?
You described the instruments and said, in a handwaving way, what you thought the instruments would read. The imaginary readings from imaginary instruments give you imaginary results: nature feels no obligation to match.
I have more confidence in the maths.

Grant Hutchison
It’s called a thought experiment. It is a well understood tool of physics and has, in the past, included apples and lifts.

Your confidence in maths is misplaced – see next.

Originally Posted by grant hutchison
But let's follow through on your original description of the experiment (in which the coin rotates along with the platter):You've now given us the measurements and units with which we can look at that negligible "pinch". The north accelerometer is 16 cm west of the platter centre and 1cm north: so that's a distance of sqrt(16˛+1˛), forming the hypotenus of a right triangle with sides 16x1. Distance units and centripetal units match, thanks to your formulation, so we can immediately solve for the unknown "pinch" force directed towards the coin's centre ... presto, it's one force unit. Identical reasoning gives you the magnitude of the pinch on the south accelerometer. So the north and south forces pinch towards the centre of the coin with a combined magnitude of two force units; that's identical to the pinch arising from the difference between the western and eastern accelerometers, which we already know to be two units.

Grant Hutchison
By all means use trig if you want – but let’s do it properly:

The distance to the north accelerometer is root(16^2 + 1^2) which is 16.031. The centripetal force is 16.031 along this radial and the gravitation is 4096/16.031^2 = 15.938 for a deficit of 16.031 –15.938 = 0.093, this is the additional stress force required from the penny. This force also is directed along the radial, and makes an angle to the horizontal of tan^-1 1/16 = 3.576 deg.

The force at the north point, required from the penny, is a vector pointing down at 3.576 deg towards the spindle. It has a length of 0.093. I’m not going to do the sums – it comes to a horizontal force of 0.09 and a vertical ‘pinch’ force of 0.06. Both, as I said, are essentially zero.

There is no ‘oblating’ force, for that you would need to rotate the penny on its axis – as French does – the force is all radial with the exception of a small pinch force, which I pointed out at the start.

Originally Posted by hhEb09'1
What you mean is that the difference in the "centripetal" (17-15) adds another 2 units to the total (4.03 + 2 = 6.03). In other words, your calculation is about 50% larger than what someone who used only gravity would calculate. Hence my comments earlier, that your figure was 50% larger than the observed values.
Correct. No excuses. It's 50% larger not 150% as I claimed. Gravity therefore is still the dominant cause of the tidal force. I did say that my maths wasn't good.

Don't accept rest of your comments, the centripetal force results in a tidal force not an oblating force - see response to Grant, above.

Disclaimer: there is a difference between properly and correct.

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Originally Posted by RichardMB
What is required is a mathematical definition of its properties.
So tell me what you want: no insight I've offered so far seems to suit you. It's just circular motion.
Here's the motion of a horizontal line segment that starts out on the x-axis, with initial coordinates (x,0)-(x+dx,0). If point (x,0) revolves around the origin and the line segment does not rotate, the line segment will translate through all coordinates that fit (x.cosθ, x.sinθ)-(x.cosθ+dx, x.sinθ), where θ is the angle of revolution. A vertical line segment (x,0)-(x, dy) will translate through all coordinates (x.cosθ, x.sinθ)-(x.cosθ, x.sinθ+dy). That should cover all you need for your accelerometers, and can be generalized without too much trouble, I think.

Originally Posted by RichardMB
It’s called a thought experiment. It is a well understood tool of physics and has, in the past, included apples and lifts.
But neither Newton nor Einstein handwaved their way past any awkward bits that didn't fit their theory; so if you're going to compare yourself to them, you have to do the maths.

Originally Posted by RichardMB
By all means use trig if you want – but let’s do it properly
Well, that wasn't trig, it was Pythagoras; all that's required for the problem at hand.
So now, having demonstrated that gravity and the equatorial bulge "cooperate" at the tidal bulges, you've gone on to demonstrated how gravity and the equatorial bulge oppose each other at the tidal pinches. This is good. But you have rather unexpectedly skipped past calculating what those north and south accelerometers are going to show in the absence of gravity, in the nice rotating turntable setup we started with. Let's just focus on that.
We've established that, because centripetal acceleration is proportional to distance, the west accelerometer will measure 17, and the east 15: so a difference of 2 units, east-west.
The north accelerometer is 16 centimetres west and one centimetre north of the coin centre. You've already worked out its radial acceleration, 16.031. A component of that is parallel to the acceleration of the coin centre, and a component is at right angles to the coin centre (the "pinch"). Please complete your thought experiment by calculating that resultant acceleration, in a simple coin glued to a rotating turntable, with no gravity.

Grant Hutchison

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Originally Posted by Jeff Root
So the cocoa in my mug, swirling around in circles after I have stirred it, is revolving but not rotating?
Well, like Ken, I have trouble thinking of that as rotation, since it's a liquid flow: it wouldn't as a whole meet RichardMB's mathematical criteria for the transformation "rotation", for instance, since not all the coordinates in the cocoa transform in the same way in the same time period.
Some identifiable marker in the cocoa (a marshmallow, for instance) would almost certainly be seen to revolve around the centre of the cup, though.

This coordinate transformation idea is what's underlying the divergence between hhEb09'1 and Ken, above. Ken's definition of "rotation", if I understand it correctly, requires the whole rotating body to undergo the same rotational coordinate transformation (="rigid rotation"). Galaxies and stirred cocoa fail this test. The Sun essentially passes it, barring a bit of minor flow business (the same is true of the Earth, since ocean currents and winds diverge from the rigid rotation of the solid body on a short time scale; as plate tectonics, mantle plumes and the core "dynamo" do on a long time scale).

Grant Hutchison

29. Originally Posted by hhEb09'1
Yes, perhaps we aren't talking about the same thing. I'm asking about your comment "Fine, but now look at three motes! Where's the rotation now?"
Yes I know, that's what I explained in my last post. Three particles can rotate, but they don't have to-- they can do all kinds of things and still stay in a group. I was explaining why I made them dust, so they would not have to be a rigid system, yet would still make a football shape. The point is, they'd all have their own orbit, rattling around in the potential well. It would be all about centripetal accelerations, and nothing about rotation. It is the potential well that causes the football shape, not any rotation, which might not even exist until one pictures the body as a fluid.

30. Originally Posted by Jeff Root
This is what I said in another thread last November 26:

All the parts of a rotating solid body revolve about the axis of
rotation.

-- Jeff, in Minneapolis
Jeff, I have not come across this as a property of rotation. I'm guessing that it is derived from the use of 'revolution' common in astronomy (a reasonable surmise from recent correspondence). I think the way out of this dilemma is to restrict the use of 'revolution' to it's astronomic utility, and to realise that this use does not extend to the mathematics of mechanics, the calculation of force being a mechanics problem - not an astronomy problem.

I'd be interested to hear your comments on the coin (accelerometers) on platter thing - are there any issues using accelerometers in this way that you can see?

Originally Posted by grant hutchison
neither Newton nor Einstein handwaved their way past any awkward bits that didn't fit their theory

Grant Hutchison