Max 91 Posted January 18, 2019 Share Posted January 18, 2019 1 hour ago, Darrell Hougen said: Now, assume that the interval, [0, 2] contains N points. That is of course the fallacy. The interval [0,2] contains infinitely many points, and infinity is not a natural number, therefore the notion of density doesn't work, as the density is also infinite, and 2 * ∞ = ∞. Cantor, cardinality, continuum and all that. It isn't surprising that people like Aristotle and Galileo didn't understand such things well. Therefore those helpless attempts to consider circles "jumping" or "waiting" to make up for differences in traveled distance in Aristotle's paradox. 1 Link to post Share on other sites

Darrell Hougen 19 Posted January 21, 2019 Share Posted January 21, 2019 I promised Merlin that I would analyze his "solutions" to see if they were correct. Of course, Jonathan, Jon Letendre, and Max have already analyzed his "solutions," so my analysis won't really add anything new. Still, I need to make good on my promise. Here are Merlin's solutions as taken from the relevant Wikipedia page: Quote Analysis and solutions[edit] The paradox is that the smaller inner circle moves 2πR, the circumference of the larger outer circle with radius R, rather than its own circumference. If the inner circle were rolled separately, it would move 2πr, its own circumference with radius r. The inner circle is not separate but rigidly connected to the larger. So 2πr is a red herring. First solution If the smaller circle depends on the larger one (Case I), then the larger circle forces the smaller one to traverse the larger circle’s circumference. If the larger circle depends on the smaller one (Case II), then the smaller circle forces the larger one to traverse the smaller circle’s circumference. This is the simplest solution. Second solution This solution considers the transition from starting to ending positions. Let Pb be a point on the bigger circle and Ps be a point on the smaller circle, both on the same radius. For convenience, assume they are both directly below the center, analogous to both hands of a clock pointing towards six. Both Pb and Ps travel a cycloid path as they roll together one revolution. The two paths are pictured here: http://mathworld.wolfram.com/Cycloid.html and http://mathworld.wolfram.com/CurtateCycloid.html While each travels 2πR horizontally from start to end, Ps's cycloid path is shorter and more efficient than Pb's. Pb travels farther above and farther below the center's path -- the only straight one -- than does Ps. The nearby image shows the circles before and after rolling one revolution. It shows the motions of the center, Pb, and Ps, with Pb and Ps starting and ending at the top of their circles. The green dash line is the center's motion. The blue dash curve shows Pb's motion. The red dash curve shows Ps's motion. Ps's path is clearly shorter than Pb's. The closer Ps is to the center, the shorter, more direct, and closer to the green line its path is. If Pb and Ps were anywhere else on their respective circles, the curved paths would be the same length. Summarizing, the smaller circle moves horizontally 2πR because any point on the smaller circle travels a shorter, more direct path than any point on the larger circle. Third solution This solution only compares the starting and ending positions. The larger circle and the smaller circle have the same center. If said center is moved, both circles move the same distance, which is a necessary property of translation (geometry) and equals 2πR in the experiment. QED. Also, every other point on both circles has the same position relative to the center before and after rolling one revolution (or any other integer count of revolutions). Aristotle's paradox is related to the fact that it is possible to find a one-to-one mapping of all the points in an interval of a particular length to all of the points in an interval of a different length. Since none of the "solutions" above has anything to do with the actual paradox, they are not solutions to the paradox. In fact, the paradox is misstated in the quote above. The fact that the smaller circle moves a distance that is different from its circumference is a simple mechanical observation, not a paradox. Max gives the solution to the paradox above: On 1/18/2019 at 2:34 PM, Max said: On 1/18/2019 at 12:17 PM, Darrell Hougen said: Now, assume that the interval, [0, 2] contains N points. That is of course the fallacy. The interval [0,2] contains infinitely many points, and infinity is not a natural number, therefore the notion of density doesn't work, as the density is also infinite, and 2 * ∞ = ∞. Cantor, cardinality, continuum and all that. It isn't surprising that people like Aristotle and Galileo didn't understand such things well. Therefore those helpless attempts to consider circles "jumping" or "waiting" to make up for differences in traveled distance in Aristotle's paradox. Because the number and density of points in an interval are infinite, it makes no sense to compare the number or density of points in a interval to the length of the interval. 2 * ∞ = ∞. One of the "solutions" given by Merlin, involves the use of cycloids. Although cycloids don't help with the solution of Aristotle's paradox, they can be used to help solve another problem that some people seem to be having, namely comprehending the fact that a wheel may be rotating and translating at the same time. First, I will note the fact (pointed out by Max in an earlier post) that if a wheel rolls without slipping, the point of the wheel in contact with the ground must be stationary at the moment of contact. Since the quote above makes reference to Mathworld, I will use the equations listed there except that I will use "r" or "R" for the radius of the circle. If R is the radius of the large circle and if it rolls on its line, then the motion of a point on its circumference is given by the parametric equations: x = R * (t - sin(t)) y = R * (1 - cos(t)) Those are the equations of the point that starts in the 6 o'clock position. Let (u, v) be the velocity of the point. Then (u, v) = (dx/dt, dy/dt) or u = R * (1 - cos(t)) v = R * sin(t) Now, we can observe that (u, v) = (0, 0) whenever t = 2 * π * k, for k = 0, 1, 2, ... In other words, the point is stationary whenever it returns to the 6 o'clock position. Now, consider a point on an inner circle of radius r < R that starts in the 6 o'clock position. As claimed above, the point does indeed describe a curtate cycloid given by the equations: x = R * t - r * sin(t) y = R - r * cos(t) Again, we can calculate the velocity of the point by taking derivatives: u = R - r * cos(t) v = r * sin(t) Since -1 <= cos(t) <= 1, we have R - r <= u <= R + r. Therefore, the horizontal component of the velocity is never equal to zero. In fact, it is always strictly greater than zero. Therefore, the inner circle does not roll on its line. When the point is in the 6 o'clock position, its speed is equal to R - r which shows that the wheel is skidding or slipping in the +x direction. We can also consider the case in which the inner circle rolls on its line and the outer circle is along for the ride. In this case, we have a prolate cycloid given by: x = r * t - R * sin(t) y = r - R * cos(t) Again: u = r - R * cos(t) v = R * sin(t) In this case, the horizontal component of the velocity is zero whenever r - R * cos(t) = 0. That happens when cos(t) = r / R. Now, consider a right triangle with leg a = r and hypotenuse c = R. Then the length of the other side, b = √(R^{2} - r^{2}) so that sin(t) = √(R^{2} - r^{2}) / R. But, that implies that sin(t) =/= 0 so that the vertical component of the velocity is not zero at the same time as the horizontal component. Therefore, the outer circle (or wheel) never has a point in stationary contact with its line. It is always skidding or slipping on its surface. The prolate case can also be examined by first setting the y-component of the velocity equal to zero. That happens whenever the point is in either the 6 o'clock or 12 o'clock position. In this case, we are only interested in the case in which the point is at the bottom which happens whenever t = 2 * π * k, for k = 0, 1, 2, ... In that position, the horizontal component of the velocity, u = r - R which shows that the large wheel is slipping backward --- opposite the direction of motion of the center point. As I said at the outset, this demonstration doesn't show anything beyond what was already shown by numerous graphical and mathematical methods. It merely serves to illustrate the point that using cycloids results in the same conclusion as other methods. Darrell 1 Link to post Share on other sites

Jonathan 299 Posted January 30, 2019 Share Posted January 30, 2019 On 1/14/2019 at 5:55 AM, merjet said: ... Merlin? Hello? Merlin? Have you been spending your time trying to learn projective geometry? If so, it shouldn't be taking this long. J Link to post Share on other sites

Brant Gaede 198 Posted February 13, 2019 Share Posted February 13, 2019 A 76 page thread is what happens when you try to explain the obvious to the oblivious. ---Brant Link to post Share on other sites

Peter 195 Posted February 13, 2019 Share Posted February 13, 2019 6 hours ago, Brant Gaede said: A 76 page thread is what happens when you try to explain the obvious to the oblivious. ---Brant Are halo's mythical? Do they represent something in reality? Link to post Share on other sites

Jonathan 299 Posted May 11, 2019 Share Posted May 11, 2019 Helloooo? Just checking in. Has Merlin made any progress learning projective geometry? Many questions and challenges on this thread remain unanswered, unmet. And some probably remain ungrasped. Is Wikipedia still fucked up from Merlin's molestations? Heh. What a shitshow. J Link to post Share on other sites

Rodney 11 Posted May 11, 2019 Share Posted May 11, 2019 One does not even need any math to resolve the paradox (explain the apparent contradiction). It was resolved at the beginning. It only takes a little thought and a discovery of the fallacious assumptions in the statement of the "contradiction." One can state things in mathematical terms afterwards, but that is not necessary. The thread is useful only in showing how screwed-up a mind can be, and still seem brilliant (and perhaps be such, in certain quite limited aspects). 1 Link to post Share on other sites

Jonathan 299 Posted July 12, 2019 Share Posted July 12, 2019 On 1/16/2019 at 1:29 PM, Darrell Hougen said: For your amusement: Hey Darrell, I was curious if you've seen Merlin's statement about you over on a different thread: 5 hours ago, merjet said: Your "proof" is hogwash and con art. Nobody on OL endorsed your "proof." I bet Hougen knows way more about geometry than you do. Responding to your "proof", he wrote, "To be honest, I'm not sure what you're doing above myself." Also, your con art shows no numerical distances. That by itself proves your "proof" is incorrect. Idiot, that's what's in dispute -- numerical distances. Jonathan pretends to know something about projective geometry. Heh. "Projective geometry is an elementary non-metrical form of geometry, meaning that it is not based on a concept of distance" (link, my bold). Fascinating, no? J Link to post Share on other sites

Jon Letendre 521 Posted January 31 Share Posted January 31 On 12/19/2018 at 1:45 AM, Jon Letendre said: Now let's ride world-class auto racing track, High Plains Raceway, a 2.55 mile track developed and owned by a handful of Colorado car clubs. I am in the white helmet on the same INTERCEPTOR, up ahead. We are following me onboard a modern 1050cc Triumph Speed Triple. Unlike my 750cc VFR, this bike has major power for the straights, modern wide tires, fantastic suspension. The Triumph's sound is cleaner and a higher pitched whizzing. You can see the Triumph's rpm needle, her sound is the one going up and down with that needle. Mine is the foghorn sound. The gauge displays mph. The long straight is only 0.65 miles long and I can consistently reach about 130mph on the old VFR once the tires are warmed up and I get in the groove. Here I am pushing harder than on Squaw Pass, a public road. On the street, about 75-80% is my maximum. This is 90-95%. I want to keep my vintage bike, I don't want to crash, but it is acceptable to crash. There is an ambulance on site on the track days that I attend. Clockwise ... Here is the promised footage I was shooting on the same occasion as the recording above. I recorded this video with a GoPro camera on my 1986 VFR pointed rearward at the Triumph Speed Triple videoing from behind me. 1 Link to post Share on other sites

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