Aristotle's wheel paradox

[Peter]

Ba’al wrote: Models a slipping wheel on a track. 

Well, when I saw no one was talking about this paradox I barked and dug up an old boner from the group, “Right Said Fred.”

I'm too sexy for my shirt too sexy for my shirt
So sexy it hurts
And I'm too sexy for Milan too sexy for Milan
New York and Japan
And I'm too sexy for your party
Too sexy for your party
No way I'm disco dancing.

[merjet]

On 8/3/2018 at 12:07 PM, Jonathan said:

do you understand why a tire or roll of duct tape is not an appropriate physical example? Those items are not appropriate because the inner circle is not a wheel that can trace out a path on a surface, but rather is a void. So, the idea would be to choose an appropriate physical setup in which the smaller wheel CAN trace out a path as it moves along attached to the larger wheel which is rolling on a surface and tracing out its own path. Understand yet?

Yes, I understand it well enough to identify your fallacies. First, the inner circle for a roll of tape or tire depicts the boundary between the roll of tape or tire and the hole, not merely the hole.  Second, your argument assumes that a moving object implies a traced path. By modus tollens no traced path implies no object. Apply those to a bullet or arrow that's shot upwards in the air and then falls to the ground. By your faulty premise the bullet or arrow doesn't have a trajectory/path because it doesn't leave a trace for you to see afterwards.

[merjet]

On 8/5/2018 at 8:15 AM, Brant Gaede said:

What rot.

Brant

I second that. Jonathan's remark was rot.

[merjet]

Here's another challenge. Aristotle's Wheel Paradox says the smaller circle moves horizontally 2*pi*R, R being the larger circle's radius. Given that and rolling = rotation & translation, use the following framework to decompose 2*pi*R quantitatively.

Rolling = Rotation + Translation + Slippage
2*pi*R =  Rotation + Translation + Slippage
If you wish, add another free variable to define and use: 
2*pi*R =  Rotation + Translation + Slippage + X

How much is each variable on the right side of the equation so that they add to 2*pi*R? Zeroes and negative amounts are okay. 

[Jonathan]

5 hours ago, merjet said:

Yes, I understand it well enough to identify your fallacies. First, the inner circle for a roll of tape or tire depicts the boundary between the roll of tape or tire and the hole, not merely the hole.  Second, your argument assumes that a moving object implies a traced path. By modus tollens no traced path implies no object. Apply those to a bullet or arrow that's shot upwards in the air and then falls to the ground. By your faulty premise the bullet or arrow doesn't have a trajectory/path because it doesn't leave a trace for you to see afterwards.

You're an idiot. You're arguing against positions no one has taken.

[Jonathan]

5 hours ago, merjet said:

Here's another challenge. Aristotle's Wheel Paradox says the smaller circle moves horizontally 2*pi*R, R being the larger circle's radius. Given that and rolling = rotation & translation, use the following framework to decompose 2*pi*R quantitatively.

Rolling = Rotation + Translation + Slippage
2*pi*R =  Rotation + Translation + Slippage
If you wish, add another free variable to define and use: 
2*pi*R =  Rotation + Translation + Slippage + X

How much is each variable on the right side of the equation so that they add to 2*pi*R? Zeroes and negative amounts are okay. 

Rolling does not include any slippage. Rolling means rolling.

J

[Jonathan]

Earlier, I wrote:

Quote

 

I have another challenge. Model the following setup: Two concentric wheels, one large, one small, affixed to each other, with each in contact with a horizontal line or surface, exactly the same as in the Aristotle's Wheel "Paradox." Make the large wheel's circumference 15 inches, and the smaller one's circumference 5 inches. Place marks on the circumferences of the wheels at 45 degree intervals, and make each mark exactly 0.25 inches wide. Roll the set of wheels so that the larger one rolls freely and without slippage or skidding. As the wheels move, mark on the surface or line that each contacts the precise point at which each wheel's 0.25 inch mark begins to make contact with its surface or line. Then, as the wheels continue to move, mark the surface or line that each contacts the precise point at which each 0.25 inch mark ceases to make contact with its surface or line. Do this for all of the marks on the wheels' circumferences.

Now, here's the challenge: Identify the length that each of the larger wheel's 0.25 inch marks make contact with the surface or line that the large wheel rolls on, and also identify the length that each of the smaller wheel's 0.25 inch marks make contact with the surface or line that the smaller wheel moves along.

 

No takers?!!! C'mon, this is super easy stuff. The math is simple.

Of course, the solution can't be found online, copied and pasted, but there shouldn't be any need for that.

Here's an illustration of challenge that I issued above:

The large wheel is 15 inches in circumference. The small one is 5 inches in circumference. Each of the orange and yellow marks on the wheels' edges are a quarter of an inch (0.25"). The black horizontal lines contact the bottoms of the wheels. The yellow and orange segments of the lines are the lengths that the yellow and orange marks on the wheels are in contact with the lines as the large wheel rolls freely and without slippage on the lower line.

Identify those lengths. What is the length of each orange segment on the bottom line, and what is the length of each yellow segment on the top line.

What do the lengths reveal about what is happening?

Should we call this another "paradox," since the yellow marks are covering a greater distance than the length of their sides which contact the line? Or should we stop being retarded, and instead reject the false premise that the smaller wheel doesn't slip or skid in comparison to its line?

Oh, and here's a version of the illustration for retards:

The same is true as the above in this illustration: The segments that the smaller circle's yellow sections contact the upper line are much longer than the segments that the larger circle's orange segments contact the lower line.

Do the math. Precisely how many times greater are the yellow segments on the upper line than the orange segments on the lower line?

J

[BaalChatzaf]

56 minutes ago, Jonathan said:

Earlier, I wrote:

No takers?!!! C'mon, this is super easy stuff. The math is simple.

Of course, the solution can't be found online, copied and pasted, but there shouldn't be any need for that.

Here's an illustration of challenge that I issued above:

The large wheel is 15 inches in circumference. The small one is 5 inches in circumference. Each of the orange and yellow marks on the wheels' edges are a quarter of an inch (0.25"). The black horizontal lines contact the bottoms of the wheels. The yellow and orange segments of the lines are the lengths that the yellow and orange marks on the wheels are in contact with the lines as the large wheel rolls freely and without slippage on the lower line.

Identify those lengths. What is the length of each orange segment on the bottom line, and what is the length of each yellow segment on the top line.

What do the lengths reveal about what is happening?

Should we call this another "paradox," since the yellow marks are covering a greater distance than the length of their sides which contact the line? Or should we stop being retarded, and instead reject the false premise that the smaller wheel doesn't slip or skid in comparison to its line?

Oh, and here's a version of the illustration for retards:

The same is true as the above in this illustration: The segments that the smaller circle's yellow sections contact the upper line are much longer than the segments that the larger circle's orange segments contact the lower line.

Do the math. Precisely how many times greater are the yellow segments on the upper line than the orange segments on the lower line?

J

15/5 = 3/1

 

[Jonathan]

36 minutes ago, BaalChatzaf said:

15/5 = 3/1

 

So far, so good. What about the rest?

The challenge was to specifically identify the lengths that the 0.25 inch orange and yellow marks on the edges of the wheels are in contact with the lines.

J

[Brant Gaede]

11 hours ago, merjet said:

I second that. Jonathan's remark was rot.

I knew you would say this. 

--Brant

[BaalChatzaf]

3 hours ago, Jonathan said:

So far, so good. What about the rest?

The challenge was to specifically identify the lengths that the 0.25 inch orange and yellow marks on the edges of the wheels are in contact with the lines.

J

Yellow marks three times longer than the orange marks.  

 

[merjet]

14 hours ago, Jonathan said:

Rolling does not include any slippage.

What an idiot! Can you read this? It says "rolling while slipping."  Can you read this? It says "rolling and slipping." 

 

 

[Jonathan]

10 hours ago, BaalChatzaf said:

Yellow marks three times longer than the orange marks.  

 

Correct!

J

[Jonathan]

4 hours ago, merjet said:

What an idiot! Can you read this? It says "rolling while slipping."  Can you read this? It says "rolling and slipping." 

 

 

The phrases "rolling while slipping" and "rolling and slipping" separate the two concepts, thus recognizing that rolling qua rolling does not include slipping. Thus your initial retarded statement' "Rolling = Rotation + Translation + Slippage," is wrong. Rolling is rotation plus the right amount of translation in proportion to rotation resulting in no slipping.

J

[merjet]

On 8/7/2018 at 7:57 AM, Jonathan said:

The phrases "rolling while slipping" and "rolling and slipping" separate the two concepts, thus recognizing that rolling qua rolling does not include slipping. Thus your initial retarded statement' "Rolling = Rotation + Translation + Slippage," is wrong. Rolling is rotation plus the right amount of translation in proportion to rotation resulting in no slipping.

You are wrong again. It is pure rolling that does not include slipping (link).

You are welcome to use the formula and say the small circle's amount of slippage is zero. I would.

[Jonathan]

6 hours ago, merjet said:

You are wrong again. It is pure rolling that does not include slipping (link).

You are welcome to use the formula and say the small circle's amount of slippage is zero. I would.

You're a moron. That's a fact. Nothing is getting through to you. You are willfully, petulantly rejecting reality. Idiot.

J

[Brant Gaede]

After 33 thread pgs it's the last man posting--at least in his mind 

--BRANT

[merjet]

On 8/9/2018 at 11:52 AM, Jonathan said:

You're a moron. That's a fact. Nothing is getting through to you. You are willfully, petulantly rejecting reality. Idiot.

Pitiful.

Suppose for the wheel and axle pictured here: 1. The diameter of each wheel is 12 inches and the diameter of the axle is 3 inches. The wheel and axle roll one revolution, or 12*pi = 37.7 inches. According to self-deluded Jonathan: 1. The axle must slip 28.27 inches, or some other non-zero distance his diseased imagination severed from reality can fabricate. 2. There is an invisible, 2nd support tangent to the axle that the axle slips on!!

Baal: "The inner wheel [e.g. axle] is rigidly affixed to the outer wheel.  since it has a smaller radius its circumference is less than the circumference of the outer wheel so it slip on its rail by a distance equal to the difference of the circumferences." (Link, my bold).

Max: "When the larger wheel makes one rotation without slipping, it travels over a distance of 2 π R. So does the smaller wheel [e.g. axle], but if this wheel [e.g. axle] wouldn't slip, it would only travel over a distance 2 π r (r < R). However, it has to travel over a distance of 2 π R, so apart from its rotation it must also slip with respect to its support, to keep up with the larger wheel." (Link, my bold).

Its rail?? Its support?? Anyway, to both:
2*pi*R =  Rotation + Translation + Slippage
2*pi*R =  2*pi*r + 0 + 2*pi*(R-r)
Both imply that translation is irrelevant.

Alternatively, Translation = Rotation + Slippage, which is ad hoc and far short of comprehensive.

Me: "Translation fully accounts for [the smaller circle/axle] moving the horizontal distance 2πR, like it does for its center and the wheel with radius R and the same center." (Link).

 

 

[Jonathan]

4 hours ago, merjet said:

Pitiful.

Suppose for the wheel and axle pictured here: 1. The diameter of each wheel is 12 inches and the diameter of the axle is 3 inches. The wheel and axle roll one revolution, or 12*pi = 37.7 inches. According to self-deluded Jonathan: 1. The axle must slip 28.27 inches...

Yes, that is indeed what happens in reality. Nothing delusional about it. If the 12 inch wheels roll on a surface without slipping, then the 3 inch axle attached to the wheels must slip in comparison to the line/surface at its base (the line/surface that the Aristotle's Wheel "Paradox" setup includes and requires us to consider). The smaller wheel (axle) cannot roll without slipping on its surface if the large wheels are rolling without slipping on theirs.

This is really elementary stuff. 

 

4 hours ago, merjet said:

2. There is an invisible, 2nd support tangent to the axle that the axle slips on!!

No, it's not invisible. It doesn't become invisible and cease to exist just because you don't want to consider it. The initial scenario, the one that you provided and linked to, includes the second support tangent to the second wheel. Moron.

J

[merjet]

The Wikipedia page about Aristotle's wheel paradox (link) has been vastly improved!

By me. I included an image that shows (a) the circles before and after rolling one revolution and (b) paths of motion for three points. The image thus helps to show why the smaller circle moves 2*pi*R -- the path of motion of every point on the smaller circle is shorter and more direct than the path of motion of any point on the larger circle.

[Brant Gaede]

A snow ball just "improved" hell.

--Brant

[anthony]

It never was a "paradox", or else reality is really screwy. I wrote that this seeming puzzle is posed as a red herring (circumference of large wheel vs. small wheel) trapping those trying to 'solve' a non-existent problem of 'slippage' (with gears). Simple powers of observation confirm that any fixed inner circle conforms to the motion of its outer wheel/circle. The outer circle, alone, dictates distance. You roll a wine bottle on its side and its narrower neck rotates equally--and traverses the same distance as the bottle. Why and how? The neck must revolve slower. Inductive observation beats everything! 

[BaalChatzaf]

On 7/31/2018 at 6:24 PM, anthony said:

Unfortunately, this Ferris wheel broke loose from its mounts one night, luckily no one was killed. It rolled over the ground for 65 meters, exactly its own circumference, would you believe. The inner ring (of lights) rolled too (naturally) and when everything stopped, altogether, the wheel was at the identical position it began, as pictured here. No slippage between outer and inner rings was reported. "The path traced by" the inner ring which has a smaller circumference naturally, was equally 65m.

 

The inner circle of lights was not in contact with a flat or straight surface.

[BaalChatzaf]

On 7/30/2018 at 2:52 PM, Jonathan said:

Heh. The retard says that "slipping is irrelevant," and that it is a "metaphor," and then he says that he has not denied that the inner circle slips, but uses scare quotes on the word "slips," and then says that he has said that "the inner circle doesn't slip (literally)."

So, what's his retarded position? I can't decipher the above nonsense. He says that it does't slip, but he doesn't deny that it slips??? What in the fuck is a "metaphorical" slip versus a literal slip???

Slipping is the ratio of the circumference traversed to the horizontal motion of the center.  It is a well defined concept.  If 5 feet of circumference move the center 3 feet the slippage is 5:3

 

[Jonathan]

6 minutes ago, BaalChatzaf said:

Slipping is the ratio of the circumference traversed to the horizontal motion of the center.  It is a well defined concept.  If 5 feet of circumference move the center 3 feet the slippage is 5:3

 

I know what slipping is, thanks.

What I don't know is what Merlin thinks he means in calling slippage "metaphorical," or what he thinks that he means in saying that slippage is irrelevant to the "paradox," or what he means in saying that he doesn't deny that the inner wheel slips while also saying that the inner wheel does not slip.

That's what I was asking, not what does slippage mean.