Max

Ah, finally I got the 3rd graph belonging to this post uplodaded:

I just realized that I should have saved those graphs in GIF format, which in this case is much more efficient. I still had to cut a part with explanations of the symbols to get it small enough, but these are the same as on the previous graph. so that's no great loss. 

But I'd still like to now how long I must wait before my upload limit is reset again. Obviously more than a day; a week, a month, a year, forever...?

Hm, uploading the third graph doesn't work... In the second one some lines have disappeared when I made it smaller to allow it to upload. Do I have an  upload limit for today or so?

Fig. 1

I‘ ve made some calculations for the general case with arbitrary distance Z for the 3 legs of the trip, instead of 1 mile. The calculations are made for an idealized Earth as a perfect sphere, with radius R = 6400 km. See the first figure for the meaning of the different symbols, the drawing is not to scale!

The trip starts anywhere on the upper circle with radius r2, goes southwards along a meridian over a distance Z km, then westwards along a circle with circumference Z and radius r1 = Z / 2π, and then back northwards.

2π * r₁ = Z → r₁ = Z / 2π

a² + r₁² = R² → a = √ (R² – r₁²)

sin (β) = r₁ / R

sin (β + γ) = r₂ / R = sin (β) * cos (γ) + cos (β) * sin (γ)

= (r₁ / R) * cos (Z / R) + (a / R) * sin (Z / R) →

r₂ = r₁ * cos (Z / R) + (√ (R² – r₁²)) * sin (Z / R)

With these expressions we can now draw a graph of the different variables r1, r2, M and M + Z as a function of Z, see the second figure.*) At the left, for relatively low values of Z, these variables have a linear dependence on Z, but for larger values of Z especially r2 is going to deviate, reaches a maximum and than decreases again to zero, meaning that the radius of the upper circle can increase to the radius of the Earth and then diminishes as the circle becomes smaller when it moves to the North. Finally it shrinks to a point, the North Pole, larger values of Z therefore have no physical meaning.

In the next figure *) the first 20 kilometers for Z are given. Here all the variables are quite linear and M ≈ r1, and M + Z ≈ r2. For small Z/R the expression for r2 becomes

r2 ≈ r1 – r1* Z²/(2!*R²) + R (1 – r1²/R²) * (Z/R – Z³/(3!*R³)) =

= r1 + Z + O(Z²/R²)

as you’d expect when the curvature of the Earth may be neglected. In the graph the value of 1 mile = 1.609344 kilometers is indicated, with the corresponding values for r1 and r2. The large number of decimals seems to be overdone, but this is just to show how small the difference between the exact values and the approximated values in this case is.

*) Uh oh, too big to upload here, I'll try it in a separate message

5 hours ago, anthony said:

Concrete you want. But are you sure ~your~ calculations and proof are reality based? As you've seen, I've been repeatedly concerned with real world "content", like friction, velocities, mass, force, drag and torque. Those factors have been generally ignored, in the 'track hypothesis'. Unless your calculations account for these, plus a physical track, not imaginary, they amount to abstractions. A "line" which escapes having "concrete" attributes, i.e. friction, for the wheel to sorta glide upon ("slippage") isn't good enough proof. 

You seem to forget that Aristotle posed this problem as a problem about circles and lines, not as a dynamical problem, but as a kinematic problem. If you're asked to prove Pythagoras' theorem for a triangle, you do this not by looking at a lot of triangles in the real world. These may be good for illustrative purposes, but to prove the theorem, you use mathematics. In the same way, Aristotle's paradox can be illustrated in real life situations by rolling wheels. But you can mathematically prove that if in the given situation one wheel rolls without slipping, the other wheel must slip, that is as certain as 2 + 2 = 4. Just as you can derive mathematically that when you roll a wheel without slipping, after one revolution the wheel has traveled a distance of 2*pi *R, with R the radius of the wheel. The correspondence between kinematics in the real world and geometry in mathematics is very well known for many centuries. Knowledge of forces and friction (btw also mathematical abstractions) may be necessary for constructing real life systems, but not for kinematic problems. The fact that after one revolution without slipping of a wheel this has traveled a distance of 2*pi*R does not depend on forces or friction, these are only important for ensuring that the wheel turns without slipping. 

5 hours ago, anthony said:

Concrete, you want. A real world challenge for you: Two identical, rotating wheels A, B, on separate axes, both turning clockwise are brought firmly together, perimeter meeting perimeter. 

A has a Vt of 6m/s, B has the Vt of 4m/s.

What will be the effect? Both A and B will turn at 10 m/s? Both will turn at 2m/s? Both at 0m/s? Think about it.

You're evading my question: what is wrong with my proof? Where is your proof, with concrete calculations, that the smaller wheel doesn't slip? Introducing a new question is no answer, it's just an attempt to evade an answer that you can't give. Much talk about tangential velocities, but you don't do anything with them, no formulas to show how there can be no slipping, nothing! They are not some mantra with magical effects, you should use them constructively, if you can.

5 hours ago, anthony said:

No, you have it in reverse, which answers itself if you'd paid any dues to Aristotelian metaphysics. What pre-exists all the above? A wheel. What are its attributes? Just for one, different tangential velocities of different radii/points within it. [An entity acts according to its nature]. Only then, for the Paradox, is there an add-on feature (an accessory) -- an extra wheel, "with a common center..." etc. This extra wheel acts according to the nature and actions of the first wheel: Revolution (1 : 1), tangential velocity (proportionally to where it is positioned - its radius), angular velocity, translational velocity, direction.

Artistotle's paradox could very well have been stated and solved even if there hadn't been any wheel yet in the world at his time. "Slipping" can also be described mathematically  "An entity acts according to its nature" is of course an empty tautology. What is the definition of "its nature"? How do we know what its "nature" is? Only by observing what that entity "does", how it "acts", don't you see the circularity in that statement? What if an entity suddenly "acts" differently? Well it's of course in it's nature to "act" suddenly differently! You just observed it! It is simply meaningless blah blah, bad philosophy.

20 minutes ago, anthony said:

I'll trust my recall. I began remarking on this t-velocity (whatever I named it) several months ago and have tried to explain it and its ramifications several times, before, in the last few days, it's now gained purchase. Of all of you, only Darrell, Merlin and yourself even alluded to Vt. The v-s experts were silent. Now, everyone acts as if they knew all about it. (But won't admit to the effects this must have on the group theory).

If someone doesn't mention it, that doesn't imply that he disagrees with it, it is after all a very well known fact. And the fact that the small wheel is slipping can easily be proved without any reference to speeds, as I've done in my very first post, comparing distances is sufficient. Later, in February 2018, I showed in a detailed calculation the relation between the different velocities, how these fit into the picture of the slipping disk, so that it would be easier to understand how it works. Oh, and group theory is an important and fascinating subject, but not in your meaning...

20 minutes ago, anthony said:

However, one inconvenient fact - different Vt's:  A) blows away the 'track and slippage' idea - B. explains why and how an inner, smaller wheel maintains its 1: 1 integrity with the larger, outer (and travels laterally as identically far and as fast as the latter).

A) "Blowing away" and "exploding" may be satisfying activities to some people and in the metaphoric sense they may suggest an argument of great force, but I'd like to see a proof that "different Vt's falsify the 'track and slippage idea". I have proved, that, using your 'different Vt's', slippage must occur. According to you, this proof must then be erroneous. I challenge you to show the error in my proof. Not with some vague "floating abstractions" as the "identity of the wheel", but quite concrete please. If you are concrete enough to use the "different Vt's" as an argument, you should also be able show quantitatively how these different Vt's result in a solution without slipping. Otherwise it is just gratuitous talk, meant to impress people but without content.

B. It is of course the other way around: it is the integrity of smaller and larger wheel as a rigid body with a common center that causes the Vt's to differ.

18 minutes ago, merjet said:

Well, whoop-de-do! That's about as useful as 2 + 2 = 4 for what's needed. 

Well, at least I could do it without cycloids.

15 minutes ago, anthony said:

The best visual depiction of the different Vt's of inner circles. For the visuo-spatial experts who doubted that.

Which visuo-spatial experts doubted that? Names please and links to the posts where they said that.

6 minutes ago, 9thdoctor said:

Did you mean to say "it's hard to conclude it won't be good", or "it's hard not to conclude it will be good"?  The way it stands is odd; if you expect the book to be a stinker, then instead of "although" you'd probably use "and". 

Always tricky, those double negations...

45 minutes ago, merjet said:

Try proving that the circumference of the disk equals the distance its center travels from one end to the other end.

That's rather trivial. If a disk rolls without slipping, by definition the distance between the contact points between disk and support before and after one revolution equals the circumference of the disk. The center has always the same position and distance relative to the contact point (perpendicular, distance R), so that has also traveled a distance of the circumference of the disk.

6 hours ago, Brant Gaede said:

In the fall of 1970 Ayn gave a talk to a fairly crowded auditorium at a NYC community college and during the following Q and A session was flat out asked if she had ever read a book by Kant. She went blah blah blah but did not answer the question though with something--if I remember correctly--of a rebuke.

Well, Milgram didn't mention this...(she probably would have found Rand's anger justified).

In the past I was dumb enough to buy those ARI books about Rand, her journals, marginalia, whatever, thinking that those would give me interesting information about her. It turned out that these were not just edited, but altered to hide less desirable statements by Rand. They may not have been literally biographies, but the hagiography was there. Oh yes, I also bought then that slimy sense of life DVD. So I'm skeptical about anything that comes from ARI.

I've read the interview with Milgram, and my conclusion is that her Rand biography will be another hagiography. For anything that might be seen as a criticism of Rand she has a ready excuse. If Rand was often angry, there was according to Milgram always a good reason, it was just the consequence of her sense of justice. And if that didn't seem to be so, there must have been somewhere some misunderstanding, the idea that Rand might in some cases also have been unreasonably angry is simply not considered.

When people object that Rand didn't read Kant herself, Milgram's answer is "you cannot prove that Rand didn't read Kant! (just like the religionist who says "you can't prove that God doesn't exist!"). "If she hadn't read Kant, she surely would have told us, just as she told us that she hadn't read Rawls' Theory of Justice" when she criticized that book (of course pretending that she'd read that book would have been a bit too obvious then). If there is any doubt about some particular aspect of Rand, we can be sure that this always will be interpreted by Milgram to Rand's advantage, she just can't be wrong. Not an interesting biography, I'm afraid.

1 hour ago, anthony said:

Forget my occasionally lax terminology. I use a word loosely at times for explication. "Rotational" instead of "tangential".

Geez, the nit-picking here. 

Nitpicking? This is as misleading as confusing voltage and current, or force and energy, entropy and enthalpy.

Quote

In your words, you have just conceded the differing 'tangential" velocity (= m/s).

Conceded?!   I used this fact already on February 4  2018 for calculating the translation speed for different points on the wheels:

Quote

Suppose the large wheel/circle rolls without slipping. After 1 period in time T the center of the circle is translated over a distance 2*pi*R, with a uniform translation speed of its center v= 2*pi*R/T. The point at the top of the circle is translated with speed 2*v and the bottom (that touches the line (=support) has translation speed zero. The translation speed of the point at the top of the smaller circle ≡ v2 = 2*pi*(R+r)/T. This can be checked by substituting r=R and r=0. Similarly, the translation speed of the point at the bottom of the smaller circle ≡v3= 2*pi*(R-r)/T > 0 for r < R. So we see that for the smaller circle and its tangent (support) the condition for tracing out the circumference is not met. That the bottom point of the smaller circle has a translation speed > 0 is the mathematical equivalent of saying that the smaller wheel is rotating and slipping.

How do you think these formulas were derived? Where are your calculations? Your proof that my calculation has errors?

Quote

So. What do YOU think is the only cause and driver of the small wheel's and large wheel's equal rotation - i.e. 1: 1 ?

They form a rigid body and rotate about a common center (axis in 3 dimensions).

.

Quote

By trial and error, consider and remove both the other, equal, velocities of a circle and what are you left with? 

Right. The varying tangential velocity! Which is, commonsensically, how a wheel can function! And why the inner one turns 1 : 1 with the outer!  And why everyone has consistently got this wrong! "The rotational speeds are equal..." No, the "tangential" speeds cannot be equal. A wheel would disintegrate if they were so.

I cannot make head or tail of this. But one sentence is revealing: "And why everyone has consistently got this wrong!" 

You'd think this is satire, but I'm afraid it is not...

32 minutes ago, anthony said:

No. You misrepresent me. The "small wheel after one revolution moves a larger distance than its own circumference." -- is "a given", I have kept saying. A fundamental. The large wheel exceeds the size of the small one; and the small one - or any point or circle in the large one - is merely along for the ride.

Straw man. No one denies that the small wheel moves a larger distance that its own circumference, that is part of the paradox.

32 minutes ago, anthony said:

"Where is the proof?"

That doesn't need proof or argument or math, anyone with sight can observe what it does.

Read better, I didn't ask proof of the triviality you mention here, I asked for a proof of your opinion that the fact that the tangential velocity of the small wheel is smaller than that of the large wheel explains the paradox without slipping of the small wheel, something you've repeatedly asserted. You cannot give such a proof, as this proposition is false, I've given proof of the opposite. 

32 minutes ago, anthony said:

Then separately, is an explanation for the above, NOW, citing the relative tangential velocities of the inner, to the outer wheels. But with the same translational (forward) velocity and same angular speed.

That the inner wheel rotates exactly once, to the outer's once, is evidently a property of tangential velocity. I.E., it turns less quickly than the larger one.

Therefore -- it does not "slip".

There, now you say it again!

Further, your terminology is wrong: the tangential velocities differ, but the wheels turn equally fast. Tangential velocity is a linear velocity (m/s), "turning" of "rotation" is an angular velocity (rad/s)'

32 minutes ago, anthony said:

But IF - you wrongly accept the premise of equal rotational speeds, then and only then could it be imagined to slip.

The rotational speeds are equal, in contrast to the tangential speeds.

6 hours ago, Ellen Stuttle said:

What is there for the inner wheel to have tangential velocity in relation to?  You won't accept the idea of an imaginary support, but then you turn around and rely on the idea of a tangent which doesn't physically exist.

He seems to have some vague notion that the fact that the tangential velocity of the smaller wheel is smaller than that of the large wheel "somehow" compensates for the fact that the small wheel after one revolution has to move over a larger distance than its own circumference. He always comes back with that argument. Where is the proof? Well, I've shown with quite simple math that this is not the case, but, on the contrary, that the math shows that the small wheel must slip to keep up with the large one. But aww, mathematics, that is just all about floating abstractions, you should of course consider the causal identity of the wheel! I give up...

26 minutes ago, BaalChatzaf said:

if he starts out from a pole he will end up back at the same place. Look at case 2 where he starts at at a non polar point.  Unless step 2 of his journey ends at exactly the same point step  1 ended he does not get back because terminates step 2 and a different meridian of longitude  form the meridian he was on at step 1. 

And why shouldn't step 2 end exactly at the same point step 1 ended, if that gives a valid solution to the puzzle? It may be a surprising move, but it satisfies all the conditions of the original question.

2 hours ago, BaalChatzaf said:

The original problem was walk a mile south walk a mile west (was it east -- no matter)  walk a mile north.  The east-west walk  is less than the length of line of latitude  reached by the southword leg so that the return trip is along a  different line of longitude. 

Nowhere in the puzzle is stated that the east-west walk must be less than the lenght of line of latitude, that is a condition you yourself arbitrarily impose. In fact, valid solutions to the original puzzle include versions in which the walk is many times the length of line of latitude. This becomes then a relatively very small circle, but why shouldn't that be allowed?

5 minutes ago, Brant Gaede said:

According to the original problem Bob is right except for the one mile walk near the South Pole. Everybody is right about the North Pole.

Some busybody moved the goal posts.

--Brant

to make it more interesting

The "South Pole solution" is perfectly in accordance with the original problem, nobody moved the goal posts. Most people see only the first solution, the North Pole solution, but the South Pole solutions are just as good, only a bit less obvious.

2 minutes ago, BaalChatzaf said:

That last leg is NOT on the same line of longitude as the first leg.  Why? Because the second leg  is a traverse along a line of latitude  which changes the longitude.  

The last leg IS on the same line of longitude as the first leg, because following a line of latitude over its total length brings you back to the point were you started, lines of latitude are circles! If you follow the equator westwards for about 40000 kilometers you'll be back again at where you started. However, the green circle in the picture is very close to the South Pole, so in this case you have only to walk one mile, but the principle is the same.

17 minutes ago, Jon Letendre said:

There is still a dominant wheel, the large one rolling on the road.

Simply make the friction of the small wheel less than the friction of the big wheel, then the big wheel will remain dominant one.

True, but this is all totally beside the point: Tony seems to think that the fact that in practice there are seldom such smaller wheels slipping against a second support would somehow be a refutation of the paradox. Now of course such slipping is in general undesirable, as it causes wear and costs extra energy. But that is with regard to Aristotle's paradox completely irrelevant. Aristotle describes a thought experiment about rolling circles, this was not an exercise in good wheel design or a discussion about the merits of different existing wheel designs. That is not to say that you cannot demonstrate it with a physical model, it is perfectly possible to realize such a wheel with a smaller wheel slipping while the outer wheel rolls without slipping and it is a good illustration of the principle of the paradox. But Aristotle was interested in the question how it is possible to map the circumference of the small circle 1-1 onto a longer track, without skipping or temporarily stopping of one of the circles. That was his problem, and the mechanism of slipping and modern insights about infinite sets are the solution that escaped him. Don't try to change his subject and turn it into you own hobby horse!

We now know that the small wheel cannot roll without slipping if the large wheel rolls without slipping. If you take away the support of the small wheel, it still makes exactly the same movement, which you could call slipping against an imaginary support. After all, circles and lines are also imaginary constructions, and that is what Aristotle was talking about. Talk about forces and friction is irrelevant, perhaps useful for practical applications, but we are not talking about a consumer test of car wheels, we're talking about mapping circles onto line segments.  

I'll repeat what I wrote earlier: this reminds me of Einstein's twin paradox: solving that is not arguing that in practice we don't send a twin with huge speeds through space, so that when he returns he's still a young man, while his brother in the meantime has become an old man. Anyway, that would be far beyond our current technical possibilities. Yet no one doubts that the twin paradox is about a very real effect.
 

6 minutes ago, BaalChatzaf said:

Since two legs of the walk, the first and the third are along lines of longitude one must end up where the lines of longitude intersect.  Under the conditions of the puzzle that would be the north geographic pole.  The three logs are along a line of longitude, a line of latitude and a line of longitude  the conditions of the puzzle require that the journey begin and end  at a pole. 

In this case the line of longitude "intersects" itself over its whole length, you travel namely twice the same line of longitude. Again: the same line of longitude. When it is the same line, it doesn't have to intersect another line of longitude to arrive at its starting point. That is the point!

3 minutes ago, anthony said:

And as I just explained, any slippage can and will cause the other to slip. . 

Wrong.

10 minutes ago, anthony said:

One can't remove the inner wheel from its context. 

It's that context that makes the inner wheel slip when the outer wheel doesn't slip.

1 hour ago, BaalChatzaf said:

North mean North a line of longitude.  Longitudes all intersect at exactly two points. The geographic north pole and the geographic south pole.   Walking north-west is NOT  walking north.

Bob, look at the picture in your own post. You start walking at the yellow point at the right, you walk along a meridian (the red line) southwards, to the South Pole (indicated by the barber pole). After 1 mile walking you arrive at the small, green latitude circle. There you start walking to the west, always keeping the South Pole to your left. After walking one mile westwards, you are back at the point where the red line meets the green circle, as the circumference of that latitude circle is exactly 1 mile (a very small latitude circle, you are really very close to the South Pole!). Now you walk back along the red meridian, and after another 1 mile you're back at your starting point. Voilà!

You have fulfilled all the conditions of the exercise: walked 1 mile southwards, then 1 mile westwards, then 1 mile northwards (along the same meridian as when you went to the south) and you are back again at your starting point. Any questions?

y₁² + x₂² = M²

x₃/x₂ = (M + L)/M    x₃ = ((M + L)/M) x₂

(y₃ - y₁)/y₃ = (M + L)/L     y₃ = (-L/M) * y₁

x₃²/((M+L)²/M²) + y₃²/(L²/M²) = y₁²  + x₂² = M²

which is indeed the equation of an ellipse. As the denominator in the x-term is larger than the denominator in the y-term, the long axis of the ellipse is in the x-direction.

8 hours ago, Jon Letendre said:

1 – (2*X*pi) = X divided by (1+X)

But I confess I have forgotten the algebra to solve it.

My estimate, X = 0.1396522

So now it's again:

1 - 2*pi*X = x/(1+x)

(X+1)*(1 - 2*pi*X) = X

X + 1 - 2*pi*X² - 2*pi*X - X = 0

X² + X - 1/(2*pi) = 0

X = - 1/2 + 1/2 SQRT (1 + 2/pi) (here only the positive root makes sense)

= 0.13965220479

Your estimate was not bad!

My attempt was more difficult because I used a general distance "walked" Z instead of 1 mile, and the fact that I made a correction for the difference between the radius of a circle in a plane and measured along a meridian. For circles with radius about 1 mile on Earth the difference is negligible, but not for larger circles. Perhaps I was just too ambitious...