The winner of Marty and my book Putting Two and Two together is Alexander Svorre Jordan. Congratulations. :) Thank you again to everybody who submitted an implementation of the dance. Here are five particularly noteworthy submissions: (Kieran Clancy) kieranclancy.github.io/star-animation/ (this was the very first submission submitted in record time :) (Liam Applebe) tiusic.com/magic_star_anim.html (an early submission that automatically does the whole dance for any choice of parameters) (Pierre Lancien) lab.toxicode.fr/spirograph/ (with geared circles) (Christopher Gallegos) gallegosaudio.com/MathologerStars (very slick interface) (Matthew Arcus) www.shadertoy.com/view/7tKXWy (implements the fact that BOTH types of rotating polygons are parts of circles rolling around DIFFERENT large circles) Here is the complete list in the order that received them :) (Kieran Clancy) kieranclancy.github.io/star-animation/ (Richard Copley) bustercopley.github.io/star/ (Liam Applebe) tiusic.com/magic_star_anim.html (Nan Ma) observablehq.com/@nanma80/rotating-dots-along-star-polygons (Owen Bechtel) owenbechtel.com/mathologer-animation.html (Morten Barklund) spirograph.live (Chris du Plessis) www.desmos.com/calculator/1pf7ftgo8i (r57shell) www.desmos.com/calculator/h01p97im0o (Adam Zimny) github.com/AdamZimny349/StarAnimationManim (Gary Au) www.desmos.com/calculator/wcdzino9pa\ (Tamir Daniely) jsfiddle.net/tdaniely/ojcgydze/595 (Alexander Svorre Jordan) alex6480.github.io/MathologerCircles/ (Niyoko Yuliawan) 347-challenge.niyoko.id (Gijs Schröder) www.shadertoy.com/view/slKXRy (Azai) www.shadertoy.com/view/NtKXR3 (Hall Holden) editor.p5js.org/pi3point14/full/1aRnJUX9S (Eric Kaiser) misc.eric-kaiser.net/starograph (Nahuel Fouilleul) nahuelfouilleul.free.fr/programmation%20jeux/hypotrochoid/hypotrochoid.html (Pierre Lancien) lab.toxicode.fr/spirograph/ (Anton Älgmyr) www.algmyr.se/347/ (Lars Christensen) larsch.github.io/dancing-polygons/ (Ivan Sorokin) sorokin.github.io/littlemiracle/littlemiracle.html (Robin Lasne) twelveothirteen.itch.io/mathologer-spirograph-challenge (Christopher Gallegos) gallegosaudio.com/MathologerStars (Matthew Arcus) www.shadertoy.com/view/7tKXWy Other submissions without online implementations by (Joseph McGowan) github.com/Joetrahedron/star-animation (Octavia Togami) github.com/octylFractal/StellatedRoller (Jérémie Marquès) github.com/Joetrahedron/star-animation (Emmanuel Pinto) github.com/EmmanuelPinto/manim-3-4-7 Stefan Muntean, Qubix (Tim), Daniel Wilckens
These are incredible. The speed and shape options on the first link (Kieran Clancy) are really handy. I just spent over 15 minutes playing with that one. Marvelous!
My solution to the coin paradox: imagine the two coins are like gears with the centers fixed in space. Both turn once every time the other one does. Then imagine the camera rotates with one of them, so it looks fixed. This feels more intuitive since nothing "rolls away" or gets wound around.
There are many ways to skin a cat. And for this particular cat there are at least four simple ones I can think of to make sense of what is going on. Of course, for this video we really need a way that ends up giving us the number of times that the arrow/George Washington's head is pointing in a fixed direction which then also translates in our 7-3=4 equation for explaining the numbers in the animation :)
There is a special place in heaven for people like you who make math accessible and fun, who put the intermediate logical steps in their demonstrations, thank you.
If I had a math teacher like you instead of the monstrous bully in Gr 9 I'd probably be a retired engineer today having worked on super-projects. You're an inspiration.
Yes, on first seeing it I thought "projection of a hypercube". I don't *think* it actually is, though, and I don't have the first idea how to pursue the idea!
I'm pretty sure it is the rotation of some 4D object (I think the joining of two triangular prism?). really cool how in the second 7-pointed star you can actually see the rotation of a pentagonal prism! I think there's still much to explore here.
Yeah, with both the tringles and squares it's actually the skeleton graph of a 3-4 duoprism. I don't think the positions of the points line up perfectly with an orthographic projection, but the figure is definitely related.
I started playing with Meccano gears when I was 7 and rapidly discovered the anomalies that seem to happen when one gear rotates around another. It really bugged me til I got it figured out. Later, during my engineering degree, I couldn't understand why the lecturer made such a meal of it all, with complicated formulas, when to me it was (by now) all so obvious but then he lacked your clear insight and pretty animations. Many years ago Meccano issued a striking clock kit. A striking mechanism requires a wheel that goes round once in 12 hours in 1+2+3...+12=78 steps, one for each hour strike. Awkwardly 78=6*13. They had to use a 12:1 geared ratio with 13 given by mounting the 12:1 gears to go round with the big wheel epicyclically. For more puzzlement tie a loop of string into a trefoil knot and drop it over a post and discover that the string only goes around the post twice despite the three loops. Or count the number of times on a clock, the minute hand crosses the hour hand in 12 hours.
I get most of your comment. especially the thingy with the earth .. it's easy to see. it's similar to the moon, actually. that also rotates once in a 28 day period (equivalent to one moon year relative to the object it orbits), while there's exactly zero "earthrises or earthsets" going on. or in other words, it's one more rotation than observed from the moon. however, I can try as hard as I want, I fail to spot anything "unexpected" with your clock analogy. oh, I somehow lost count, lol. before actually hitting, "post" I recounted and now it's thirteen, all of a sudden. to anyone reading this: it's apparently very helpful not to just randomly start at let's say, "noon." start at any point when the hands don't overlap .. that makes things less confusing, lol.
Hypnotic! You obviously enjoy teaching people things. That is the only reason to go into teaching. I spent my entire adult life as an educator, so I know what I'm talking about.
This IS a crucial step in mathematics instruction: instructor preparatory analysis, modeling, planning and preparation! I have dozens (if not hundreds) of Geogebra, MATLAB, Maple, Scientific Workbook, TI-8x and other technology models that did NOT work out. Students love the ones that are beautiful, but I only used the ones in the full class that clearly and accessibly illustrate the relevant math. The rest I keep as curiosities and jumping off points for future work.
The off-by-one counter-intuitive logic reminded me of (and is clearly related to) how there are only _11_ places on the face of a 12-hour clock where the hands point the same direction. One of my favourite weird facts is that if either one of the two hands on that clock went backwards (what a strange clock that would be!) the hands would point in the same direction in _13_ places instead.
What an interesting weird fact! Yet Intuitively I guess that when the two hands start off together in the same direction the faster minute hand takes a bit longer than the circuit of the clock to catch up with the hour hand again since the latter has advanced too (like Achilles and the tortoise), resulting in fewer than n catch-ups per n circuits. But if it was going round in the opposite direction then they would for the very same reason meet up in less than a circuit, meaning more than n catch-ups per n circuits. Also, the "off-by-one" rule fully stated could be "Let F/M be the ratio of the perimeter of the fixed circle to that of the moving circle. When the moving circle rolls once outside then it makes F/M + 1 rotations, when inside F/M - 1 rotations". When the ratio is 1, that is they are identical, then the number of rotations inside is 0. This is kind of pro-intuitive, since how can a circle roll round inside another of exactly the same dimensions?
I am so glad that I subscribed to your channel many years ago, as this wonderful video popped-up on my thumbnails this morning - The content is truly inspirational - thank you for the unexpected pleasure.
The sum of angles: Imagine a point with a nose walking along the edges of the star. In every corner, it will rotate 180°-a, where a is any angle of the star. If you carefully watch its nosr, it does three rotations, so 360°×3 or 180°×6. We get (180°-a)×7=180°×6, so the sum of angles is equal to 7a, which from the equation is 180°
Fascinating and entertaining... and it completely took my mind off of the troubles in the world for the duration. That fact in itself made it worth the 23 minutes spent watching. thank you.
Simple solution for the sum of 7-star angles: 1. Without loss of generality, assume the 7 points lies on a circle 2. Those 7 angles are produced by 7 unique arcs that form a whole circle 3. For a certain arc, the angle at centre produce by that arc is always twice of the angle at circumference (Theorem "Angle at centre is twice angle at circumference") 4. Therefore, twice of the sum of all angles is 360 degrees. 5. Sum of required angles = 180 degrees
3,4,7 is actually really useful in music. Intervals of 3 and 4 are periodic complements in mod 12, because they also function as the smallest interval in basic trichords, intervals of 7 are the most common. I often visualize the structure of chords in a way similar to the graphic representation used in this video. The triangles and squares form a continuum that you can rotate along to find relatively valent harmonic structures in a set of closely related keys.
Your video is a treasure. I hope it is incorporated in early STEM learning. My experience watching, caused me dynamic insights in three separate domains. In my opinion, Mathologer is a "miracle" math educator.
I don't agree. I love these videos, but the content is just intimidating for people without an interest in it. Not everyone is cut from the same cloth.
Mind blown. I love it. And you explain it so well I even get the feeling I understand it a bit. Great vid, thanks. Now I'm off to watch the others you mentioned.
The discussion around the 2:37 mark sets up a lovely proof that the vertex-to-vertex segments radiating from a single vertex in a regular polygon divide the associated interior angle evenly.
God, those spirograph animations are incredibly beautiful, the dancing cubes makes me think of hypercubes and the potential of seeing higher dimensions in this kids toy.
I have always thought about the coin rotation “paradox” in a rather simple manner. The distance travelled by a coin with radius r around the outside of another circle with radius R is simply 2*pi*(R + r), or 2*pi*(R - r) if it travels on the inside. Given the constraint of no-slip and the correct ratios, the result is hardly surprising.
If it wouldn't rotate but was just slipping along, like a triangle for example, it would do exactly one rotation. On top of that the coin is rotating on its own, so it has to do more than one rotation.
@@bur2000 Don’t get me wrong, the extra rotation is still “hidden” in a sense, by the very act of rotating the coin around the circle. Imagine sliding the coin around the circle without moving the contact point of the coin. The coin will do a complete rotation separate to the rolling action. The coin is not slipping along the circle though! That’s an important part of the setup. However, by rolling the coin along the circle with no slip, you’re moving the coin by (R + r)/R more than you might think. By the simple fact that the coin center is radially further out than the rolling interface. The coin has to move along the circle with radius (R + r) in our setup. In fact, if you add together the contributions of just the part between the circle and coin center (the radius r of the coin) as the coin rolls around the circle; you will get another circle with radius r.
Fascinating and super simple explanations. I wish I had you as an instructor when I was in high school... I'd have paid more attention, Thank you for what you do.
You can't just throw Matt Parker's name on random geometric figures. Here it makes particularly little sense, since Matt Parker had nothing to do with any of these figures, animations, or calculations, and second, nothing here demonstrates the defining feature of a "Parker" figure, that of being, "close, but not quite right." Would I like to see a collab between Matt Parker and Mathologer? Absolutely. But just randomly bringing Matt up here is silly, indeed it's bizarre.
@@LeoStaley it's funny how you're wrong by the very definition of Parker something you gave. The curve is literally close to being a heptagram but not quite right
@18:46: Answer is {7;2}. We start from one corner and count to reach the next edge of straight line. This gives us 2. Now we know this is a 5 point star or flower pedal. Using a-b=c, we can solve for a where b=2 & c=5. 2+5=7. We get {7;2}. This means small circle does 7 rotations on straight line perimeter of larger circle. The 2 counterclockwise rotations from coiling back. This gives the 5 total rotations.
The challenge at 15:33, solved in a different way: The number of rotations of the smaller coin per each time it rolls around the bigger coin is equal to the ratio between the smaller coin's radius, and the distance between its center and the center of the larger coin. In the case of equal-sized coins, that ratio is 2, which explains why the outer coin rotates twice. For a smaller coin with diameter 2/3 of that of the bigger coin, its center is at a radius of 1/3+1/2=5/6, while its own radius is 1/3. (5/6)/(1/3)=5/2, so it should make two and a half rotations around its center for each time it rolls around the bigger coin.
Not only interesting to listen (why not all of my school teachers wasn't such), but also very nice visualization. It is a brain pleasure to watch your stuff. Too bad i don't understand all. But it makes me search and think. Thank you!
The coin rotation: 5. The stationary sircle is 3/2 times bigger than the rolling one, so it will be 3/2+1=5/2 reotations of a rolling sircle per a roll, so 5 for two rolls
@15:33 Answer is 5 rotations. 3 clockwise rotations for the 2 times the straight line perimeter length of large circle. Then coil the large circle back 2 times, which rotates small coin 2 clockwise rotations. 3+2 = 5 clockwise rotations.
I first encountered the coin problem studying engineering over 60 years ago. The context was designing what we referred to as epicyclic gear trains. The rule of thumb is : if the small gear [ coin ] rolls round the outer of the centre gear [ coin ] its motion relative to an outside observer is the ratio of diameters [ in gearing parlance the pitch diameters ] PLUS 1 rotation. If the smaller rolls round the inner surface of the larger then the number of rotations of the smaller is the ratio of the pitch diameters MINUS 1 rotation. If you imagine the problem of an interconnected train of epicylic stages it's essential to have simple rules of thumb to work with. It's clear that the coin problem is actually only a simple system that one is almost never going to encounter in the real world of gearing. The basic real-world system would be THREE gears interconnected and referred to as a sun, planet and ring gear combination. Any one of the three elements of this system can be held stationary with respect to an outside observer. As a final glimpse into the world of gearing, it's important to remember that if the basic function of a gear train is as a speed reducer there is a concomitant increase in the torque [ and vice versa ]. It can happen then that if the train is driven in reverse this can lead to excessive torque or tooth forces which can cause failure of the gear teeth at the former input end.
4:30 to 4:45 - Mathologer demonstrating his acceptance of less than precision. How human! I love watching and listening to you, Mathologer! P.S. - That animation jumps out to me in 3D when the shapes containing more than 3 points are being shown!
in some way, this is also tied to one of the most famous pieces in literacy, "around the world in eighty days." remember when they thought they gonna lose the bet but didn't adjust for the fact, they went around the earth one time? which also means that one must either add or subtract one day in regard weather you're going "with" the sun or "against" it. to me there's a very similar, underlying principle.
Is it me? How is the rotation at 14:27 to 14:32 "just one again"? The top tip of the smaller coin rolls 3 times before it gets upright... How does Mathologer end up with one again? Sleight of hand ✋! In the first two cases, he counts the rotations of the rotating coin. In the ⅔ coin scenario, he miscounts the rotation of the static coin.
@15:33 Answer is 5 rotations. 2/3, so 3 rotations clockwise of small circle over straight line of 2x perimeter of larger circle. Then plus 2 clockwise coiling of large circle back gives 5 rotations.
You can perform the solution for the 7-star angle problem without any math: 1. Draw the star on a piece of paper (it doesn't have to be exact). 2. Grab a pencil and place it on any line segment of the star. 3. Keeping one end of the pencil anchored, rotate it until it rests along an adjacent line segment. 4. Alternating which end of the pencil to use as an anchor point, continue this process until your trip returns to the original line segment. This means you have now traversed every angle of the star. 5. Your pencil will now be facing in the opposite direction of how it started, which is to say it is 180 degrees from its original position.
*Observations:* If one circle has a radius of A and the other has a radius of B, and the radius-B circle revolves around the radius-A circle, then there are two equations depending on if the revolution is internal or external. The tilt of the radius-B circle per revolution is *(A - B) ÷ B × 360°* if the revolutions are internal. If the number of revolutions is external, then that value is *(A + B) ÷ B × 360°.* The easiest explanation for these equations is that the tilt of a circle is always 360° times the distance the origin of the circle travels divided by its circumference. The radius can also be considered negative for internal revolutions.
Dear Burkard recently i stumbled on this video ua-cam.com/video/0dwJ-bkJwDI/v-deo.html and is closely related to the rolling coin paradox. That spirograph draws amazing fractals shapes and it must be the application you are looking for as a challenge for developers. I hope you will like it. Thank you for your videos. Are real food for my brain cells! Regards, Alessandro Cavalli
For "coin paradox" - Road of rolling circle's center shoul be calculated and divided with perimetar of rolling circle (RC) In first case, center of RC is rotating around center of static circle (SC) for radius of 2r, where r is radius same for both circle. It's trivial that ratio between perimetar of circle that center of RC made is 2 times bigger than perimetar of RC. For second case: Rottating radius for center of RC is 3r (r + 2r). The rest is trivial. Third case: Rottating radius is r (2r - r). So, ratio is 1.
I love to study Sacred Geometries and this video relates right into hidden tricks hidden in the hidden geometries of the world around us! Very awesome!
OMG mathloger is alive! Looks like triangles are spinning counter clockwise like in an invisible ferry with equilateral triangles 🔺 When I've seen the animation 2:40 I chuckled as I got to know something so deep in a trivially easy manner Looks like a star approximation algorithm so chilling..
If my math teachers in high school had been like this guy then I may have been more interested in mathematics or at least would've been less frustrated.
Same energy: why does a sidereal year last an extra day compared to a solar year? You have to count the extra rotation of earth that results from its revolution around the sun. Or, the earth doesn't have to spin quite one full turn around its axis per day, since it will move around its orbit and make up the rest of the rotation (about 4 minutes per day)
Yes! I thought of the similar relationships that lead to the final crisis in Jules Verne's "Around the World in 80 Days" ... [SPOILERS] Phileas Fogg returns to London thinking he is on the 81st day, having lost his bet, but by accident finds out the date is one day earlier than he was expecting. By traveling east around the world, he experienced one more sunrise that those who stayed in London, ... each day, on average, being 18 mins shorter than a stationary person would experience.
A question to think about: If the earth instead rotated on its axis in the opposite direction, retrograde to its rotation about the sun, but both rates of rotation were unchanged, which number would change, and what would it become?
I recall a different discussion about the paradox with one quarter rotating around the other relating to astronomy with the rotation of the earth around the sun, and the difference between a solar day and a sidereal day. Rather than a mathematical description of the paradox, the explanation points out that from the point of view of an observer at the center of the stationary quarter, the rotating quarter APPEARS to rotate only once. At the beginning, the observer in the center of the stationary quarter sees the bottom of the rotating quarter. As the rotating quarter rotates to the bottom of the stationary quarter, the stationary observer sees the top of the rotating quarter. As the rotating quarter continues its movement back to its origin, the observer again sees the bottom of the quarter -- each part of the rotating quarter was closest to the center of the stationary quarter only once, even though from a distance, the rotating quarter made two revolutions to produce that effect. Still confuses me!
I think I follow. I simplified things (for myself anyway) by having the sun as a square on a sheet of squared paper, with the orbiting rotating earth as another square at an arbitrary distance, say 5 squares, drawn at 8 different stages in its orbit, and in its rotation of 90 degrees at a time. The whole thing forms a big 3x3 arrangement of squares. Starting at the "south", a stationary observer on the earth's "north" face would see the sun full on. Only once in the subsequent orbit will the observer on our rotating square be again facing in the same direction, and that's when the earth is due "north" of the sun. But then of course s/he will be facing directly away from the sun. It's only when the earth returns to the starting position in the orbit that the sun becomes visible in the same way, so as far as the observer is concerned the earth has made only one full rotation, not two.
Wow, this needs a lot of calculation power to understand... I love math, and giving it a lot of thought, maybe, just maybe I'd get it, but thanks Mathologer, your videos are always the best
@21:01. Answer 180 degrees. Arrow makes 3x360 degree rotations or 6x180 degree rotations. 180- each of the angle is the rotation the arrow makes. It does it 7 times. 6x180 cancels the 6 of the 7 x 180 degree. Remaining is the angles are equal to 180 degrees.
I just had an epiphany that made the uncoiling explaination given for the coin paradox feel totally reasonable, whereas previously it just didn't sit right. Consider the coin's first little movement, just as it starts to roll. If the red perimeter has been uncoiled, then it just rolls straight ahead. But if the perimeter is still wrapped around the coin, then it has to rotate down a bit further, in order to contact the coin. This little extra bit of rotation, compared to what is needed to reach the straight line, is what makes up the extra rotation.
I didn't here a lot of the words you said, I was too busy trying to figure out which non-regular polychoron you generated by having both the squares and triangles overlayed on the points. It's a prism formed with triangular prisms on each end
Coin rotation paradox summary: If the ratio of the radius of the rotating coin to static coin is k/n then the number of rotations in a single trip is: Outside: (n + k)/k Inside: (n - k)/k So for 1/2 ratio the # of rotations is 3 outside and 1 inside. For 2/3 ratio the # of rotations is 5/2 outside (5 rotations per two trips) and 0.5 inside (1 rotation per two trips).
Very interesting video - thanks for putting this together! The {7/3} star and {7/4} stars are essentially the same stars, just traced in reverse. I noticed that the triangles in the {7/3} star are rotating in the opposite directions as the squares, which I think we can take as squares rotating in the {7/4} star. So perhaps what’s going on here is that if you have a {p/q} star (with p and q coprime) we expect to see q-gons rotating one way and (p - q)-gons rotating the other way? Perhaps that would make it easier to prove why this works.
"The {7/3} star and {7/4} stars are essentially the same stars" Great observation. Having said that there is some very interesting extra depth here. It is actually not true that "just traced in reverse". To get the {7,4} the point that is being traced actually has to be placed somewhere outside the rolling circle. Have a look at the Hypotrochoic wiki page it has an animation of a {5,3} which illustrates the same phenomenon :)
@@Mathologer I stand corrected! I had first learned about star polygons via a different construction - lay out p points on the perimeter of a circle. Jump from one point to the point q steps ahead of it, and repeat until you return home. In that context {p / q} and {p / (p - q)} do end up being identical but traced backwards. It’s interesting to hear that the nested circle construction treats these differently!
You are the best Math teacher. NOBODY does proofs better than you in my experience. The more you understand something the easier it is to explain to others, so, you really know your stuff. If you can make me understand, you are a true master. Yeah, I think you're better than 3b1b (No offense!) If you happen to read this I'd also like to state I've memorized most of your videos. Deep learning occurring here. I went to school for Math but you make it intuitive. The Pi formula video and the Summation masterclass are some of the best Math content ever produced by man.
@@dhoyt902 Well, if you end up programming this one please post what you come up with here and also let me know in an e-mail. burkard.polster@monash.edu
Each angle of star corresponds to arc of circle. In sum they correspond to full circle. But each angle of star is inscribed angle so it is one half of corresponding arc, so they sum to half of circle which is pi or 180 degrees.
@@Mathologer each arc between points is "inside" three different inscribed angles, so the total is 3 pi. Or, from another angle (pun only partially intended), in the pointy seven-point star, each point is opposite 1/7 of the circle and is therefore 1/14 of a circle, so all seven together is 7/14 of a circle. In the blunter seven-pointed star, each point is opposite 3/7 of a circle and is therefore 3/14 of a circle, and all together they are 21/14 of a circle.
@@Mathologer I did it the long way with angle theorems and got 25 5/7 degrees, which I see is 180/7. The reason for that is beyond my comprehension, but makes some kind of sense.
Wow, I really like that the star notation {7/3} refers to the sizes of circles in a spirograph you use to generate them. Also {8/4} is not legal as a star but it's definitely drawing the edges of a hypercube.
For the coin rolling paradox, it helps to think of where the center of the coin travels opposed to the edge. inside the large coin, the small coin only travels a very small circle while on the outside it must travel much farther.
Yes... they came with pins to fix them on thick cardboard but the cardboard got full of holes and dents to was no good for drawing on... so I used blue tack instead and then the 'fixed' ring kept slipping. Thin cardboard was no good with pins sticking out the other side. Then the pens that came with it ran out and most Biro tips would not fit through the little holes... and felt tip pens stained the holes so I knew which holes I had used the most!! I'm sure that amazing toy has helped a lot of people love maths and geometry, or understand it better over the years.
Yay Keynote! I've been using Keynote for years to make the math animations on my channel's math videos. These animations are a beautiful testament to what is possible.
15:30 If the larger coin is x times the size of the smaller coin, then the smaller coin does x+1 rotations on the outside and x-1 on the inside. Did I get that irght?
Good afternoon from Slovakia, thank You Mr. Polster and Your team for another great lecture. Being a visual solver, I see rotating (precessing, wobbling) 3D wireframe of pentaprism at 7:48 of the video and a more complex and a kind of precession of 4D 3,4 duoprism at 6:34. I am now thinking, if this approach can be useful in visualization of nets of geometric shapes of >2 dimensions or are those emerging shapes I mentioned a rare coincidence... just a thought.
My thoughts too! Thanks for mentioning the duoprism, I probably wouldn't have known the correct polytope otherwise. I guess an n-gon prism can be seen as a (n)-2 (degenerate) duoprism.
@@godfreypigott before the end of 2021 i meant, was cleaning up my watch later queue from this channel in the last couple of days, but gladly carve out time for this one. Happy new year =)
8:50 Another way to understand the coin paradox: Think of the first coin as rolling half of its circumfence on a flat floor. Then it will stand upside down. But, rolling on the edge of a second coin, it will "stand" upside down on it as well, just like a human on the south pole. So, upside down of an upside-down coin is an upright coin! Thus it has completed a full rotation. Then repeat the process a second time to get to the top (north pole) again. In total two rotations.
In fact, it's not only that one. The other ones look like rotating duoprisms (a kind of 4D shape). The {7/3} looks like the triangle-square duoprism for example.
The animations and coin rollings can also be understood in terms of group theory. The 12 pointed star animation is 12 points tracing the Cayley graph Cay(Z_12, z) where z is the generator of Z_12. If you label each tracing point with the group element of the vertex where it starts, then the subsets of Z_12 which are cosets of Z_3 and Z_4 form the triangles and the squares respectively. As for the coins, the differences between the different cases becomes clear when you a pick a point on the circumference of the rolling coin and construct the rotatory group for that coin. The number of times it turns around is equal to the quotient of the subgroup of the rotary group which is isomorphic to the rolling coin just turning on its axis. I know this is a much more involve way of seeing this, but once you do it lets you see the group theory which is hidden inside geometry. It also explains why the two numbers mentioned at the end must be relatively prime for the star animation to work.
My brain REALLY wants to see a cuboctahedron in the figure where the triangles and squares are connected, but the closer I look, the less it even looks like a possible polyhedron.
A very fun visual approach to solving the coin paradox! The approach that was my initial thought was a bit less visual, but it gives the same conclusion (and it works for irrational radii ratios). The condition of no slipping means that the "speed" of the contact point is 0 (Any relative speed between the surfaces of the two circles at the contact point would imply slipping). The speed of the contact can be written as a sum of 2 parts: 1) The speed of the moving circle's center relative to the immobile circle's center. This has a simple expression, as it is just a point rotating around another point. We have v_1 = Ω*(R + r), where Ω is the angular velocity of the moving circle's center, and R and r are the radii of the stationary and moving circles respectively. 2) The speed of the contact point relative to the moving circle's center. This part is only due to the moving circle rotating around its axis. We have v_2 = -ω*r, where ω is the angular velocity of the moving circle's rotation around its axis. Note the minus sign. Using the no slipping condition (v_1 + v_2 = 0), we get ω = Ω * (1 + R/r). This means that after one revolution around the immobile circle, the moving circle will rotate around its axis 1 + R/r times. The result works for all cases, but in the case of the moving circle being inside the immobile one, the quantity "r" is negative.
2:30 This explains an amusement park ride I used to like as a kid! It was a spinning "scrambler" style ride, but it felt like you were just being launched back and forth across the room.
MATH QUESTION A |\ | \ |. \ B |.___\. C ABC is right angled triangle, AB = 1728 BC = 1050 AC = ? (Write you answer in reply section) HAPPY NEW YEAR EVERYONE
@@Mathologer Yeah - it was a bit weird feeling to see Mathographer here, I thought something was wrong and my brain misfired. But once I understood it isn't Mathologer with different avatar - it became hilarious. Congrats to both of you for sense of humor.
@@jannegrey actually I came to know of this channel through 3B1B, my google account was named before I knew of Burkard so this is a happy coincedence, as is almost customary to all things mathematical :)
@@Nothingtonnobodson No problem - it was just that my mind first read "Mathologer" probably because I was watching his video and have seen the name many times. Only after like 2 seconds I realized my mistake and I of course believe you that it wasn't copy-cating or anything like that. As you say - just a happy coincidence. I'm really glad I got to see it though - your comment doesn't have many likes, so many people will not know about this coincidence soon :( Stay Safe!
The coin paradox can be illustrated and new versions developed with squares on printed squared paper. Just outline squares of relative sizes you're interested in, and make them "roll" and "orbit" round each other, and "rotate" by one right angle at a time, in successively drawn stages. The same rules seem to apply but with new insights.
This lends itself to turtle graphics, via the The Total Turtle Trip Theorem, which states that the turtle will draw a closed figure with n sides when the sum of the angles turned is a multiple of 360. (From Turtle Geometry by by Andrea diSessa and Hal Abelson)
18:50 The label for the pentagram is: {5/2}; and we’ll get rotating line segments (2), and equilateral triangles (5-2). Also; showing the triangles and squares together, with the corresponding vertices connected, in the original animation, really reminds me of the animations of the 3D-projection of a 4D-equilateral triangular prismatic prism rotating in 4D-space 😅.
when I saw the original animation with the squares and triangles I thought it was some sort of rotation of a polyhedron, but I'm glad that it turned out to be even cooler!
The winner of Marty and my book Putting Two and Two together is Alexander Svorre Jordan. Congratulations. :) Thank you again to everybody who submitted an implementation of the dance. Here are five particularly noteworthy submissions:
(Kieran Clancy) kieranclancy.github.io/star-animation/ (this was the very first submission submitted in record time :)
(Liam Applebe) tiusic.com/magic_star_anim.html (an early submission that automatically does the whole dance for any choice of parameters)
(Pierre Lancien) lab.toxicode.fr/spirograph/ (with geared circles)
(Christopher Gallegos) gallegosaudio.com/MathologerStars (very slick interface)
(Matthew Arcus) www.shadertoy.com/view/7tKXWy (implements the fact that BOTH types of rotating polygons are parts of circles rolling around DIFFERENT large circles)
Here is the complete list in the order that received them :)
(Kieran Clancy) kieranclancy.github.io/star-animation/
(Richard Copley) bustercopley.github.io/star/
(Liam Applebe) tiusic.com/magic_star_anim.html
(Nan Ma) observablehq.com/@nanma80/rotating-dots-along-star-polygons
(Owen Bechtel) owenbechtel.com/mathologer-animation.html
(Morten Barklund) spirograph.live
(Chris du Plessis) www.desmos.com/calculator/1pf7ftgo8i
(r57shell) www.desmos.com/calculator/h01p97im0o
(Adam Zimny) github.com/AdamZimny349/StarAnimationManim
(Gary Au) www.desmos.com/calculator/wcdzino9pa\
(Tamir Daniely) jsfiddle.net/tdaniely/ojcgydze/595
(Alexander Svorre Jordan) alex6480.github.io/MathologerCircles/
(Niyoko Yuliawan) 347-challenge.niyoko.id
(Gijs Schröder) www.shadertoy.com/view/slKXRy
(Azai) www.shadertoy.com/view/NtKXR3
(Hall Holden) editor.p5js.org/pi3point14/full/1aRnJUX9S
(Eric Kaiser) misc.eric-kaiser.net/starograph
(Nahuel Fouilleul) nahuelfouilleul.free.fr/programmation%20jeux/hypotrochoid/hypotrochoid.html
(Pierre Lancien) lab.toxicode.fr/spirograph/
(Anton Älgmyr) www.algmyr.se/347/
(Lars Christensen) larsch.github.io/dancing-polygons/
(Ivan Sorokin) sorokin.github.io/littlemiracle/littlemiracle.html
(Robin Lasne) twelveothirteen.itch.io/mathologer-spirograph-challenge
(Christopher Gallegos) gallegosaudio.com/MathologerStars
(Matthew Arcus) www.shadertoy.com/view/7tKXWy
Other submissions without online implementations by
(Joseph McGowan) github.com/Joetrahedron/star-animation
(Octavia Togami) github.com/octylFractal/StellatedRoller
(Jérémie Marquès) github.com/Joetrahedron/star-animation
(Emmanuel Pinto) github.com/EmmanuelPinto/manim-3-4-7
Stefan Muntean, Qubix (Tim), Daniel Wilckens
These are incredible. The speed and shape options on the first link (Kieran Clancy) are really handy. I just spent over 15 minutes playing with that one. Marvelous!
10:43 The number of rotations equals the distance between the center points between both coins divided by the radius of the rotating coin.
Harvard MATH245
Try vertices: 23 Density 11... shape to the right, speed to the left Wow!
Can you discuss Riemann Hypothesis?
My solution to the coin paradox: imagine the two coins are like gears with the centers fixed in space. Both turn once every time the other one does. Then imagine the camera rotates with one of them, so it looks fixed. This feels more intuitive since nothing "rolls away" or gets wound around.
exactly how I thought about it
Thanku
i would guess that the upper coin turns +1+1 times but the bottom one makes -1+1 rotations, and that's more intuitive
I like this because it also includes frames of reference, which come in handy in other areas...
There are many ways to skin a cat. And for this particular cat there are at least four simple ones I can think of to make sense of what is going on. Of course, for this video we really need a way that ends up giving us the number of times that the arrow/George Washington's head is pointing in a fixed direction which then also translates in our 7-3=4 equation for explaining the numbers in the animation :)
There is a special place in heaven for people like you who make math accessible and fun, who put the intermediate logical steps in their demonstrations, thank you.
Math is awesome, but mediocre teachers are soul-killing.
If I had a math teacher like you instead of the monstrous bully in Gr 9 I'd probably be a retired engineer today having worked on super-projects. You're an inspiration.
Hmm maybe.
Once you add both the triangles and squares, it looks like a 4D projection.
Yes, on first seeing it I thought "projection of a hypercube". I don't *think* it actually is, though, and I don't have the first idea how to pursue the idea!
I thought the same thing!
@@bauerwesterentarot That makes three of us.
I'm pretty sure it is the rotation of some 4D object (I think the joining of two triangular prism?). really cool how in the second 7-pointed star you can actually see the rotation of a pentagonal prism! I think there's still much to explore here.
Yeah, with both the tringles and squares it's actually the skeleton graph of a 3-4 duoprism. I don't think the positions of the points line up perfectly with an orthographic projection, but the figure is definitely related.
I started playing with Meccano gears when I was 7 and rapidly discovered the anomalies that seem to happen when one gear rotates around another. It really bugged me til I got it figured out. Later, during my engineering degree, I couldn't understand why the lecturer made such a meal of it all, with complicated formulas, when to me it was (by now) all so obvious but then he lacked your clear insight and pretty animations.
Many years ago Meccano issued a striking clock kit. A striking mechanism requires a wheel that goes round once in 12 hours in 1+2+3...+12=78 steps, one for each hour strike. Awkwardly 78=6*13. They had to use a 12:1 geared ratio with 13 given by mounting the 12:1 gears to go round with the big wheel epicyclically.
For more puzzlement tie a loop of string into a trefoil knot and drop it over a post and discover that the string only goes around the post twice despite the three loops. Or count the number of times on a clock, the minute hand crosses the hour hand in 12 hours.
Was something like this? ua-cam.com/video/La6t59ol3Ng/v-deo.html
I get most of your comment. especially the thingy with the earth .. it's easy to see. it's similar to the moon, actually. that also rotates once in a 28 day period (equivalent to one moon year relative to the object it orbits), while there's exactly zero "earthrises or earthsets" going on. or in other words, it's one more rotation than observed from the moon. however, I can try as hard as I want, I fail to spot anything "unexpected" with your clock analogy. oh, I somehow lost count, lol. before actually hitting, "post" I recounted and now it's thirteen, all of a sudden. to anyone reading this: it's apparently very helpful not to just randomly start at let's say, "noon." start at any point when the hands don't overlap .. that makes things less confusing, lol.
Thats not awkward, thats n times n+1 by 2
Hypnotic! You obviously enjoy teaching people things. That is the only reason to go into teaching. I spent my entire adult life as an educator, so I know what I'm talking about.
The amount of effort you put into making these animations are truly astonishing.
This IS a crucial step in mathematics instruction: instructor preparatory analysis, modeling, planning and preparation! I have dozens (if not hundreds) of Geogebra, MATLAB, Maple, Scientific Workbook, TI-8x and other technology models that did NOT work out. Students love the ones that are beautiful, but I only used the ones in the full class that clearly and accessibly illustrate the relevant math. The rest I keep as curiosities and jumping off points for future work.
The off-by-one counter-intuitive logic reminded me of (and is clearly related to) how there are only _11_ places on the face of a 12-hour clock where the hands point the same direction. One of my favourite weird facts is that if either one of the two hands on that clock went backwards (what a strange clock that would be!) the hands would point in the same direction in _13_ places instead.
What an interesting weird fact! Yet Intuitively I guess that when the two hands start off together in the same direction the faster minute hand takes a bit longer than the circuit of the clock to catch up with the hour hand again since the latter has advanced too (like Achilles and the tortoise), resulting in fewer than n catch-ups per n circuits. But if it was going round in the opposite direction then they would for the very same reason meet up in less than a circuit, meaning more than n catch-ups per n circuits.
Also, the "off-by-one" rule fully stated could be "Let F/M be the ratio of the perimeter of the fixed circle to that of the moving circle. When the moving circle rolls once outside then it makes F/M + 1 rotations, when inside F/M - 1 rotations". When the ratio is 1, that is they are identical, then the number of rotations inside is 0. This is kind of pro-intuitive, since how can a circle roll round inside another of exactly the same dimensions?
@@chrisg3030 Very well put, sir 👏🏻👍🏻.
I am so glad that I subscribed to your channel many years ago, as this wonderful video popped-up on my thumbnails this morning - The content is truly inspirational - thank you for the unexpected pleasure.
The sum of angles: Imagine a point with a nose walking along the edges of the star. In every corner, it will rotate 180°-a, where a is any angle of the star. If you carefully watch its nosr, it does three rotations, so 360°×3 or 180°×6. We get (180°-a)×7=180°×6, so the sum of angles is equal to 7a, which from the equation is 180°
oh just realised that's what you show at the end
@@bot24032 :)
In general, for the polygon {p/q} with p≥3, p,q coprime, and 1≤q
The formula actually works fine for other values of p and q too, but the interpretation is slightly more involved.
Doesn't this also describe gearing ratios?
This is why I love your channel, a machine gun of “Aha!” moments makes me feel like Baldrick with a cunning idea. Love it!
what's a baldrick?
@@chalichaligha3234 ?
You will be pleased to know that I always show the black adder bean counting clip at the beginning of my calculus course at uni :)
Fascinating and entertaining... and it completely took my mind off of the troubles in the world for the duration. That fact in itself made it worth the 23 minutes spent watching. thank you.
Bruh chillax with the Russia talk
@@klembokable Dude, you team ukraine or team russia?
Simple solution for the sum of 7-star angles:
1. Without loss of generality, assume the 7 points lies on a circle
2. Those 7 angles are produced by 7 unique arcs that form a whole circle
3. For a certain arc, the angle at centre produce by that arc is always twice of the angle at circumference (Theorem "Angle at centre is twice angle at circumference")
4. Therefore, twice of the sum of all angles is 360 degrees.
5. Sum of required angles = 180 degrees
8:07 If you can make a 12 pointed star with this, you can design a CLOCK with this fancy movement!
What an EXCELLENT way to start 2022!!! Thanks, Mathologer!!!
Coin paradox blew my mind. Great video, great visual and explanations.
6:32 The beauty of the movement looks like a 4-dimensional hyper cube / Tesseract. So many great videos on this channel. Thank You!
@mathologer needs to do a whole video on that point.
I really enjoy watching your enthusiastic explantation of these concepts. The animations are beautiful.
My head hurts.
Thank you for your efforts. May you and yours stay well and prosper.
3,4,7 is actually really useful in music. Intervals of 3 and 4 are periodic complements in mod 12, because they also function as the smallest interval in basic trichords, intervals of 7 are the most common. I often visualize the structure of chords in a way similar to the graphic representation used in this video. The triangles and squares form a continuum that you can rotate along to find relatively valent harmonic structures in a set of closely related keys.
Nice. Would be nice to be able to see what you see :)
This has been the best, most dynamic after effects tutorial I've watched in my life. Thank you! 🤙🏻✨ I learned SO MUCH TODAY!
Your video is a treasure. I hope it is incorporated in early STEM learning. My experience watching, caused me dynamic insights in three separate domains. In my opinion, Mathologer is a "miracle" math educator.
Glad you think so and thank you very much for saying so :)
Strongly agree
I don't agree. I love these videos, but the content is just intimidating for people without an interest in it. Not everyone is cut from the same cloth.
@@garethb1961 if no interest, why are you even here?
@@wvwoman3193 Hi Shannon, how is your reading comprehension? Try reading my post again, and then yours.
Mind blown. I love it. And you explain it so well I even get the feeling I understand it a bit. Great vid, thanks. Now I'm off to watch the others you mentioned.
I love the beautiful sequence of adding elements starting at @6:40
The discussion around the 2:37 mark sets up a lovely proof that the vertex-to-vertex segments radiating from a single vertex in a regular polygon divide the associated interior angle evenly.
God, those spirograph animations are incredibly beautiful, the dancing cubes makes me think of hypercubes and the potential of seeing higher dimensions in this kids toy.
14:02 "The universe makes sense again"...hahaha
This reminded me of the winding number and the argument principle.
Happy New Year!
I have always thought about the coin rotation “paradox” in a rather simple manner. The distance travelled by a coin with radius r around the outside of another circle with radius R is simply 2*pi*(R + r), or 2*pi*(R - r) if it travels on the inside. Given the constraint of no-slip and the correct ratios, the result is hardly surprising.
this is my favourite explanation :)
rather than looking at the circumference, we look at the distance the center of the circle travels
A very straitforward answer
🤯
If it wouldn't rotate but was just slipping along, like a triangle for example, it would do exactly one rotation. On top of that the coin is rotating on its own, so it has to do more than one rotation.
@@bur2000 Don’t get me wrong, the extra rotation is still “hidden” in a sense, by the very act of rotating the coin around the circle. Imagine sliding the coin around the circle without moving the contact point of the coin. The coin will do a complete rotation separate to the rolling action.
The coin is not slipping along the circle though! That’s an important part of the setup. However, by rolling the coin along the circle with no slip, you’re moving the coin by (R + r)/R more than you might think. By the simple fact that the coin center is radially further out than the rolling interface. The coin has to move along the circle with radius (R + r) in our setup. In fact, if you add together the contributions of just the part between the circle and coin center (the radius r of the coin) as the coin rolls around the circle; you will get another circle with radius r.
Fascinating and super simple explanations. I wish I had you as an instructor when I was in high school... I'd have paid more attention, Thank you for what you do.
A Parker heptagon!
You can't just throw Matt Parker's name on random geometric figures. Here it makes particularly little sense, since Matt Parker had nothing to do with any of these figures, animations, or calculations, and second, nothing here demonstrates the defining feature of a "Parker" figure, that of being, "close, but not quite right." Would I like to see a collab between Matt Parker and Mathologer? Absolutely. But just randomly bringing Matt up here is silly, indeed it's bizarre.
@@LeoStaley it's because the heptagram in the original animation has to cheat slightly for the points traveling along it.
A star is an Ntangle not an Ntogram surely?
@@LeoStaley nah, it's definitely a parker heptagram
@@LeoStaley it's funny how you're wrong by the very definition of Parker something you gave. The curve is literally close to being a heptagram but not quite right
@18:46: Answer is {7;2}.
We start from one corner and count to reach the next edge of straight line. This gives us 2. Now we know this is a 5 point star or flower pedal. Using a-b=c, we can solve for a where b=2 & c=5. 2+5=7. We get {7;2}.
This means small circle does 7 rotations on straight line perimeter of larger circle. The 2 counterclockwise rotations from coiling back. This gives the 5 total rotations.
The challenge at 15:33, solved in a different way: The number of rotations of the smaller coin per each time it rolls around the bigger coin is equal to the ratio between the smaller coin's radius, and the distance between its center and the center of the larger coin. In the case of equal-sized coins, that ratio is 2, which explains why the outer coin rotates twice.
For a smaller coin with diameter 2/3 of that of the bigger coin, its center is at a radius of 1/3+1/2=5/6, while its own radius is 1/3. (5/6)/(1/3)=5/2, so it should make two and a half rotations around its center for each time it rolls around the bigger coin.
Not only interesting to listen (why not all of my school teachers wasn't such), but also very nice visualization.
It is a brain pleasure to watch your stuff. Too bad i don't understand all. But it makes me search and think. Thank you!
The coin rotation: 5.
The stationary sircle is 3/2 times bigger than the rolling one, so it will be 3/2+1=5/2 reotations of a rolling sircle per a roll, so 5 for two rolls
or simply, it rolls 3 times as shown in the video but 2 in the same direction so 3+2=5
that’s what i think also
Or (R/r + 1)X = (3/2 +1)2
(x=number of rolls) (R/r = radius ratio)
Probably doesn't work in all examples given but it's close I assume.
@15:33 Answer is 5 rotations.
3 clockwise rotations for the 2 times the straight line perimeter length of large circle. Then coil the large circle back 2 times, which rotates small coin 2 clockwise rotations. 3+2 = 5 clockwise rotations.
I first encountered the coin problem studying engineering over 60 years ago. The context was designing what we referred to as epicyclic gear trains. The rule of thumb is : if the small gear [ coin ] rolls round the outer of the centre gear [ coin ] its motion relative to an outside observer is the ratio of diameters [ in gearing parlance the pitch diameters ] PLUS 1 rotation. If the smaller rolls round the inner surface of the larger then the number of rotations of the smaller is the ratio of the pitch diameters MINUS 1 rotation.
If you imagine the problem of an interconnected train of epicylic stages it's essential to have simple rules of thumb to work with.
It's clear that the coin problem is actually only a simple system that one is almost never going to encounter in the real world of gearing. The basic real-world system would be THREE gears interconnected and referred to as a sun, planet and ring gear combination. Any one of the three elements of this system can be held stationary with respect to an outside observer.
As a final glimpse into the world of gearing, it's important to remember that if the basic function of a gear train is as a speed reducer there is a concomitant increase in the torque [ and vice versa ]. It can happen then that if the train is driven in reverse this can lead to excessive torque or tooth forces which can cause failure of the gear teeth at the former input end.
4:30 to 4:45 - Mathologer demonstrating his acceptance of less than precision.
How human!
I love watching and listening to you, Mathologer!
P.S. - That animation jumps out to me in 3D when the shapes containing more than 3 points are being shown!
in some way, this is also tied to one of the most famous pieces in literacy, "around the world in eighty days." remember when they thought they gonna lose the bet but didn't adjust for the fact, they went around the earth one time? which also means that one must either add or subtract one day in regard weather you're going "with" the sun or "against" it. to me there's a very similar, underlying principle.
Is it me? How is the rotation at 14:27 to 14:32 "just one again"? The top tip of the smaller coin rolls 3 times before it gets upright... How does Mathologer end up with one again? Sleight of hand ✋! In the first two cases, he counts the rotations of the rotating coin. In the ⅔ coin scenario, he miscounts the rotation of the static coin.
Mr.Mathologer you are incredibly wonderful thanks for everything ❤
@15:33 Answer is 5 rotations. 2/3, so 3 rotations clockwise of small circle over straight line of 2x perimeter of larger circle. Then plus 2 clockwise coiling of large circle back gives 5 rotations.
You can perform the solution for the 7-star angle problem without any math:
1. Draw the star on a piece of paper (it doesn't have to be exact).
2. Grab a pencil and place it on any line segment of the star.
3. Keeping one end of the pencil anchored, rotate it until it rests along an adjacent line segment.
4. Alternating which end of the pencil to use as an anchor point, continue this process until your trip returns to the original line segment. This means you have now traversed every angle of the star.
5. Your pencil will now be facing in the opposite direction of how it started, which is to say it is 180 degrees from its original position.
Thank you very much for this amazing gift!! Happy new year professor, and keep up the good work!!🥳
*Observations:*
If one circle has a radius of A and the other has a radius of B, and the radius-B circle revolves around the radius-A circle, then there are two equations depending on if the revolution is internal or external. The tilt of the radius-B circle per revolution is *(A - B) ÷ B × 360°* if the revolutions are internal. If the number of revolutions is external, then that value is *(A + B) ÷ B × 360°.* The easiest explanation for these equations is that the tilt of a circle is always 360° times the distance the origin of the circle travels divided by its circumference. The radius can also be considered negative for internal revolutions.
Dear Burkard recently i stumbled on this video
ua-cam.com/video/0dwJ-bkJwDI/v-deo.html
and is closely related to the rolling coin paradox. That spirograph draws amazing fractals shapes and it must be the application you are looking for as a challenge for developers. I hope you will like it. Thank you for your videos. Are real food for my brain cells! Regards, Alessandro Cavalli
For "coin paradox" - Road of rolling circle's center shoul be calculated and divided with perimetar of rolling circle (RC)
In first case, center of RC is rotating around center of static circle (SC) for radius of 2r, where r is radius same for both circle.
It's trivial that ratio between perimetar of circle that center of RC made is 2 times bigger than perimetar of RC.
For second case: Rottating radius for center of RC is 3r (r + 2r). The rest is trivial.
Third case: Rottating radius is r (2r - r). So, ratio is 1.
I love to study Sacred Geometries and this video relates right into hidden tricks hidden in the hidden geometries of the world around us! Very awesome!
OMG mathloger is alive!
Looks like triangles are spinning counter clockwise like in an invisible ferry with equilateral triangles 🔺
When I've seen the animation 2:40 I chuckled as I got to know something so deep in a trivially easy manner
Looks like a star approximation algorithm so chilling..
If my math teachers in high school had been like this guy then I may have been more interested in mathematics or at least would've been less frustrated.
This tickles just the right part of my brain. Thank you.
this animation looks like a 4d shape of some kind
i know it probably isn't one but it's still cool
It is, a 3,4-duoprism 🙂.
what a nice christmas present you give us... so funny and revealing.. thank you very much. Happy new year!
Same energy: why does a sidereal year last an extra day compared to a solar year? You have to count the extra rotation of earth that results from its revolution around the sun. Or, the earth doesn't have to spin quite one full turn around its axis per day, since it will move around its orbit and make up the rest of the rotation (about 4 minutes per day)
Yes! I thought of the similar relationships that lead to the final crisis in Jules Verne's "Around the World in 80 Days" ... [SPOILERS]
Phileas Fogg returns to London thinking he is on the 81st day, having lost his bet, but by accident finds out the date is one day earlier than he was expecting. By traveling east around the world, he experienced one more sunrise that those who stayed in London, ... each day, on average, being 18 mins shorter than a stationary person would experience.
my wording here is imprecise... sidereal year and solar year are the same year; it's the sidereal day and solar day that are different
Thanks for your videos, I just love watching them and trying not to wreck my brain. I wasn't a fan of Math but you are making a difference.
Related to the coin rolling paradox: Despite there being 365.25 days in a year, the earth actually rotates 366.25 times on its own axis in a year.
A question to think about:
If the earth instead rotated on its axis in the opposite direction, retrograde to its rotation about the sun, but both rates of rotation were unchanged, which number would change, and what would it become?
@@godfreypigott 364.25 rotations?
@@LukeSumIpsePatremTe Sorry ...
I guess 364.26. j just like linear things.
@@Noam_.Menashe Sorry - same answer as the other guy - still not correct. The relationship is still linear though.
Great video as always!
There's a gear-rotation argument which is equivalent to the coil-uncoil one.
Happy holidays!
I recall a different discussion about the paradox with one quarter rotating around the other relating to astronomy with the rotation of the earth around the sun, and the difference between a solar day and a sidereal day. Rather than a mathematical description of the paradox, the explanation points out that from the point of view of an observer at the center of the stationary quarter, the rotating quarter APPEARS to rotate only once. At the beginning, the observer in the center of the stationary quarter sees the bottom of the rotating quarter. As the rotating quarter rotates to the bottom of the stationary quarter, the stationary observer sees the top of the rotating quarter. As the rotating quarter continues its movement back to its origin, the observer again sees the bottom of the quarter -- each part of the rotating quarter was closest to the center of the stationary quarter only once, even though from a distance, the rotating quarter made two revolutions to produce that effect. Still confuses me!
I think I follow. I simplified things (for myself anyway) by having the sun as a square on a sheet of squared paper, with the orbiting rotating earth as another square at an arbitrary distance, say 5 squares, drawn at 8 different stages in its orbit, and in its rotation of 90 degrees at a time. The whole thing forms a big 3x3 arrangement of squares.
Starting at the "south", a stationary observer on the earth's "north" face would see the sun full on. Only once in the subsequent orbit will the observer on our rotating square be again facing in the same direction, and that's when the earth is due "north" of the sun. But then of course s/he will be facing directly away from the sun. It's only when the earth returns to the starting position in the orbit that the sun becomes visible in the same way, so as far as the observer is concerned the earth has made only one full rotation, not two.
Wow, this needs a lot of calculation power to understand... I love math, and giving it a lot of thought, maybe, just maybe I'd get it, but thanks Mathologer, your videos are always the best
Mathologer never ever disappoints ;) 😄❤❤❤🤗
@21:01. Answer 180 degrees.
Arrow makes 3x360 degree rotations or 6x180 degree rotations. 180- each of the angle is the rotation the arrow makes. It does it 7 times. 6x180 cancels the 6 of the 7 x 180 degree. Remaining is the angles are equal to 180 degrees.
Correct :)
As a mid-level PowerPoint wizard, just be glad you didn’t have to use google slides. Those animations looked difficult enough as is.
I just had an epiphany that made the uncoiling explaination given for the coin paradox feel totally reasonable, whereas previously it just didn't sit right.
Consider the coin's first little movement, just as it starts to roll. If the red perimeter has been uncoiled, then it just rolls straight ahead. But if the perimeter is still wrapped around the coin, then it has to rotate down a bit further, in order to contact the coin. This little extra bit of rotation, compared to what is needed to reach the straight line, is what makes up the extra rotation.
I didn't here a lot of the words you said, I was too busy trying to figure out which non-regular polychoron you generated by having both the squares and triangles overlayed on the points. It's a prism formed with triangular prisms on each end
Coin rotation paradox summary:
If the ratio of the radius of the rotating coin to static coin is k/n then the number of rotations in a single trip is:
Outside: (n + k)/k
Inside: (n - k)/k
So for 1/2 ratio the # of rotations is 3 outside and 1 inside.
For 2/3 ratio the # of rotations is 5/2 outside (5 rotations per two trips) and 0.5 inside (1 rotation per two trips).
Very interesting video - thanks for putting this together! The {7/3} star and {7/4} stars are essentially the same stars, just traced in reverse. I noticed that the triangles in the {7/3} star are rotating in the opposite directions as the squares, which I think we can take as squares rotating in the {7/4} star. So perhaps what’s going on here is that if you have a {p/q} star (with p and q coprime) we expect to see q-gons rotating one way and (p - q)-gons rotating the other way? Perhaps that would make it easier to prove why this works.
"The {7/3} star and {7/4} stars are essentially the same stars" Great observation. Having said that there is some very interesting extra depth here. It is actually not true that "just traced in reverse". To get the {7,4} the point that is being traced actually has to be placed somewhere outside the rolling circle. Have a look at the Hypotrochoic wiki page it has an animation of a {5,3} which illustrates the same phenomenon :)
@@Mathologer I stand corrected! I had first learned about star polygons via a different construction - lay out p points on the perimeter of a circle. Jump from one point to the point q steps ahead of it, and repeat until you return home. In that context {p / q} and {p / (p - q)} do end up being identical but traced backwards. It’s interesting to hear that the nested circle construction treats these differently!
Fantastic. Your obvious love of this material is infectious.
I had a spyrograph 30 years ago.
It's the kind of "analog" toy that children are missing these days.
get a cheapy simple spirograph in a dollar store
Your videos are always interesting and fun to watch, thank you for making them!
Looking forward to the hardcore video!
Three quarters finished :)
You are the best Math teacher. NOBODY does proofs better than you in my experience. The more you understand something the easier it is to explain to others, so, you really know your stuff. If you can make me understand, you are a true master. Yeah, I think you're better than 3b1b (No offense!)
If you happen to read this I'd also like to state I've memorized most of your videos. Deep learning occurring here. I went to school for Math but you make it intuitive. The Pi formula video and the Summation masterclass are some of the best Math content ever produced by man.
I write so much code based on your videos.
Glad you think so and thank you very much for saying so :)
@@dhoyt902 Well, if you end up programming this one please post what you come up with here and also let me know in an e-mail. burkard.polster@monash.edu
Each angle of star corresponds to arc of circle. In sum they correspond to full circle. But each angle of star is inscribed angle so it is one half of corresponding arc, so they sum to half of circle which is pi or 180 degrees.
How about the angle sum of the second type of 7-pointed star I show in this vides. What is its angle sum?
Brilliant!
@@Mathologer each arc between points is "inside" three different inscribed angles, so the total is 3 pi. Or, from another angle (pun only partially intended), in the pointy seven-point star, each point is opposite 1/7 of the circle and is therefore 1/14 of a circle, so all seven together is 7/14 of a circle. In the blunter seven-pointed star, each point is opposite 3/7 of a circle and is therefore 3/14 of a circle, and all together they are 21/14 of a circle.
@@Mathologer 180(7-2x3)/7
@@Mathologer I did it the long way with angle theorems and got 25 5/7 degrees, which I see is 180/7. The reason for that is beyond my comprehension, but makes some kind of sense.
Wow, I really like that the star notation {7/3} refers to the sizes of circles in a spirograph you use to generate them. Also {8/4} is not legal as a star but it's definitely drawing the edges of a hypercube.
With Mathologer you can always count that the subject will be deeply explored :)
These animations leave me hypnotized....and cross-eyed!
18:41 should be {5/2}, and that would be either 3 line segments ("2-gons?"), or 2 triangles.
Those “line segments” are digons
For the coin rolling paradox, it helps to think of where the center of the coin travels opposed to the edge. inside the large coin, the small coin only travels a very small circle while on the outside it must travel much farther.
Fascinating stuff indeed! I remember spirograph as a kid, but got quickly frustrated with it - cogs kept slipping lol
Yes... they came with pins to fix them on thick cardboard but the cardboard got full of holes and dents to was no good for drawing on... so I used blue tack instead and then the 'fixed' ring kept slipping. Thin cardboard was no good with pins sticking out the other side. Then the pens that came with it ran out and most Biro tips would not fit through the little holes... and felt tip pens stained the holes so I knew which holes I had used the most!! I'm sure that amazing toy has helped a lot of people love maths and geometry, or understand it better over the years.
Yay Keynote! I've been using Keynote for years to make the math animations on my channel's math videos. These animations are a beautiful testament to what is possible.
15:30 If the larger coin is x times the size of the smaller coin, then the smaller coin does x+1 rotations on the outside and x-1 on the inside.
Did I get that irght?
That's correct :)
Good afternoon from Slovakia, thank You Mr. Polster and Your team for another great lecture. Being a visual solver, I see rotating (precessing, wobbling) 3D wireframe of pentaprism at 7:48 of the video and a more complex and a kind of precession of 4D 3,4 duoprism at 6:34. I am now thinking, if this approach can be useful in visualization of nets of geometric shapes of >2 dimensions or are those emerging shapes I mentioned a rare coincidence... just a thought.
My thoughts too! Thanks for mentioning the duoprism, I probably wouldn't have known the correct polytope otherwise. I guess an n-gon prism can be seen as a (n)-2 (degenerate) duoprism.
Lets see if I can finish this one before the end of 2021 =) Thanks Burkard
You mean 2022
This one should be an easy watch :)
@@godfreypigott before the end of 2021 i meant, was cleaning up my watch later queue from this channel in the last couple of days, but gladly carve out time for this one. Happy new year =)
@@morkovija Happy new year :)
@@morkovija nice name bro! Thank you!! And happy new year :)
8:50 Another way to understand the coin paradox: Think of the first coin as rolling half of its circumfence on a flat floor. Then it will stand upside down. But, rolling on the edge of a second coin, it will "stand" upside down on it as well, just like a human on the south pole. So, upside down of an upside-down coin is an upright coin! Thus it has completed a full rotation. Then repeat the process a second time to get to the top (north pole) again. In total two rotations.
The {7/2} animation looks like a rotating pentagonal prism to me.
It will look like that for any shape {n/2}
In fact, it's not only that one. The other ones look like rotating duoprisms (a kind of 4D shape). The {7/3} looks like the triangle-square duoprism for example.
Yes! I wonder if there exists a "3-dimensional explanation" for all these animations?
@@galoomba5559 Nice! I was going to ask in another comment.
The animations and coin rollings can also be understood in terms of group theory. The 12 pointed star animation is 12 points tracing the Cayley graph Cay(Z_12, z) where z is the generator of Z_12. If you label each tracing point with the group element of the vertex where it starts, then the subsets of Z_12 which are cosets of Z_3 and Z_4 form the triangles and the squares respectively.
As for the coins, the differences between the different cases becomes clear when you a pick a point on the circumference of the rolling coin and construct the rotatory group for that coin. The number of times it turns around is equal to the quotient of the subgroup of the rotary group which is isomorphic to the rolling coin just turning on its axis.
I know this is a much more involve way of seeing this, but once you do it lets you see the group theory which is hidden inside geometry. It also explains why the two numbers mentioned at the end must be relatively prime for the star animation to work.
My brain REALLY wants to see a cuboctahedron in the figure where the triangles and squares are connected, but the closer I look, the less it even looks like a possible polyhedron.
I know right? It seems to be a (4D) 4-3 duoprism actually.
Always nice to see a new Mathologer video :D
Happy new mathologer 🎥
A very fun visual approach to solving the coin paradox!
The approach that was my initial thought was a bit less visual, but it gives the same conclusion (and it works for irrational radii ratios). The condition of no slipping means that the "speed" of the contact point is 0 (Any relative speed between the surfaces of the two circles at the contact point would imply slipping).
The speed of the contact can be written as a sum of 2 parts:
1) The speed of the moving circle's center relative to the immobile circle's center. This has a simple expression, as it is just a point rotating around another point. We have v_1 = Ω*(R + r), where Ω is the angular velocity of the moving circle's center, and R and r are the radii of the stationary and moving circles respectively.
2) The speed of the contact point relative to the moving circle's center. This part is only due to the moving circle rotating around its axis. We have v_2 = -ω*r, where ω is the angular velocity of the moving circle's rotation around its axis. Note the minus sign.
Using the no slipping condition (v_1 + v_2 = 0), we get ω = Ω * (1 + R/r). This means that after one revolution around the immobile circle, the moving circle will rotate around its axis 1 + R/r times. The result works for all cases, but in the case of the moving circle being inside the immobile one, the quantity "r" is negative.
MY NEW YEAR'S PRESENT!!
Still meant to be a Christmas present, just running a bit late :(
@@Mathologer it's a wonderful gift either way.
Your choice and visualization of proofs is always beautiful.
@@sebastianjost :)
Let's give the under-rated genius Schlaefli some credit for the {7/3} notation.
Our turn: sum of all 7 angles is also 180°.. awesome maths magic.
I got the same answer ^^
Red + Orange + Yellow + Green + Blue + Purple + Pink = 180°
2:30 This explains an amusement park ride I used to like as a kid! It was a spinning "scrambler" style ride, but it felt like you were just being launched back and forth across the room.
MATH QUESTION
A
|\
| \
|. \
B |.___\. C
ABC is right angled triangle,
AB = 1728
BC = 1050
AC = ? (Write you answer in reply section)
HAPPY NEW YEAR EVERYONE
And a happy 2022 to you as well.
1728^2=12^6=2.985.984
1050^2=1102500
Sum(both)=4.088.484 (=h)
Squar.root(h)= 2022
Oh. (Checks date)
I hope that you will never stop making videos!
awesome
Mathographer, I like the name :)
@@Mathologer Yeah - it was a bit weird feeling to see Mathographer here, I thought something was wrong and my brain misfired. But once I understood it isn't Mathologer with different avatar - it became hilarious. Congrats to both of you for sense of humor.
@@Mathologer Thanks :) actually I will also start a math youtube channel soon once I learn MANIM
@@jannegrey actually I came to know of this channel through 3B1B, my google account was named before I knew of Burkard so this is a happy coincedence, as is almost customary to all things mathematical :)
@@Nothingtonnobodson No problem - it was just that my mind first read "Mathologer" probably because I was watching his video and have seen the name many times. Only after like 2 seconds I realized my mistake and I of course believe you that it wasn't copy-cating or anything like that. As you say - just a happy coincidence. I'm really glad I got to see it though - your comment doesn't have many likes, so many people will not know about this coincidence soon :(
Stay Safe!
The coin paradox can be illustrated and new versions developed with squares on printed squared paper. Just outline squares of relative sizes you're interested in, and make them "roll" and "orbit" round each other, and "rotate" by one right angle at a time, in successively drawn stages. The same rules seem to apply but with new insights.
This lends itself to turtle graphics, via the The Total Turtle Trip Theorem, which states that the turtle will draw a closed figure with n sides when the sum of the angles turned is a multiple of 360. (From Turtle Geometry by by Andrea diSessa and Hal Abelson)
18:50 The label for the pentagram is: {5/2}; and we’ll get rotating line segments (2), and equilateral triangles (5-2). Also; showing the triangles and squares together, with the corresponding vertices connected, in the original animation, really reminds me of the animations of the 3D-projection of a 4D-equilateral triangular prismatic prism rotating in 4D-space 😅.
this blows my mind
It's 23 minutes long and only been out 8 minutes. You can't have watched it yet.
@@godfreypigott Well, there is something mindblowing in the first 8 minutes :)
@@Mathologer Right😙😄😅
when I saw the original animation with the squares and triangles I thought it was some sort of rotation of a polyhedron, but I'm glad that it turned out to be even cooler!