#or reuleaux triangles (for a three-circle)

[nullarysources]

Mathematicians Discover New Shapes to Solve Decades-Old Geometry Problem

Gregory Barber for Quanta Magazine:

Arguably the best known noncircular body of constant width is the Reuleaux triangle, which you can construct by taking the central region of overlap in a three-circle Venn diagram. For a given width in two dimensions, a Reuleaux triangle is the constant-width shape with the smallest possible area. A circle has the largest.

In three dimensions, the largest body of constant width is a ball. In higher dimensions, it's simply a higher-dimensional ball — the shape swept out if you hold a needle at a point and let it rotate freely in every direction.

But mathematicians have long wondered if it's always possible to find smaller constant-width shapes in higher dimensions. Such shapes exist in three dimensions: Though these Reuleaux-like blobs might look a bit pointy, sandwich them between two parallel planes and they will roll smoothly, like a ball. But it's much harder to tell whether this is true in general. It could be that in higher dimensions, the ball is optimal. And so in 1988, Oded Schramm, then a graduate student at Princeton University, asked a simple-sounding question: Can you construct a constant-width body in any dimension that is exponentially smaller than the ball?

Now, in a paper posted online in May, five researchers — four of whom grew up in Ukraine and have known each other since their high school or college days — have reported that the answer is yes.

New shapes just dropped

[goldislops]

Shapes That Roll

Mathematicians have created a new shape that can roll, but it’s not a circle or sphere. They devised a multi-dimensional "guitar pick" shape that rolls in ways beyond our three-dimensional comprehension. This breakthrough solves a decades-old geometry problem.

How it works: Circles and spheres have so-called "constant width," which means there’s always the same distance between the edge of the shape and two parallel planes, such as the ground, as it rolls along. The researchers figured out how to create new shapes in any dimension by examining the intersection of an infinite number of n-dimensional balls–shapes where all edge points are the same distance from the center in n-dimensions. The result is the shape of a guitar pick in three dimensions that can “roll” by maintaining a constant width between two parallel planes in any dimension.

What the experts say: “It’s really difficult to estimate volume in high dimensions, yet this whole proof is fairly simple and so elegant,” says Gil Kalai, a professor at the Einstein Institute of Mathematics. He suggests this could mark the beginning of a new era in studying constant-width shapes from a new perspective, and that in the future mathematicians might construct even smaller ones. --Max Springer, news intern

Animated diagram shows a circle moving along a wire track shaped into an equilateral triangle with sides that match the radius of the circle in length. As the circle completes its trip around the track, the area of common overlap among all its positions over time forms a Reuleaux triangle.

The infinite intersection of n-dimensional balls in two dimensions Credit: Amanda Montañez

[aeide-thea]

greek poetry nerd mutual

(possibly i am conflating posts you have written about poetry and translation with tags you have posted mentioning Greek but this is How My Brain Knows You, apparently)

hey, i’ll take it! :)

ultimately i feel like you’re correct that i’m interested in greek poetry insofar as it’s the, like, venn-diagram overlap of a bunch of my other interests, rather than because the thing itself has a totally singular appeal for me, but like. you could stick greek poetry in a summoning circle (or, you know, tumblr post) and i’d show up: it counts.

(truly wild that i made this sappho post less than five years ago. that really was a whole other K talking. but also i guess i did some homerposting in march so like. an Ongoing Interest for sure!)

thx for playing! :)

[come tell me what kind of mutual you think i am!]

[bubbloquacious]

Generalized Sides

This was inspired by a conversation in the new Mathblr Discord server, come join if it sounds like fun!

So a polygon is a 2-dimensional shape that has a certain, finite number of sides, right? These are the parts of the boundary of the shape that are straight line segments in the plane. It seems like intuitively there are more kinds of shapes that can be said to have a certain number of sides, though. The boundary of a Reuleaux triangle, for example, contains no straight line segments but it looks like it has three ‘sides’, for some definition thereof. You might even ask how many sides a circle has, or a teardrop shape. Let’s try to generalize.

The rest of this post is written in the online math typesetting format KaTeX, so viewing it on the Tumblr dashboard is not ideal. View the post in a browser on my Tumblr page to render the math properly.

A (planar) curve \(C\) is a subset of \(\R^2\) such that there is at least one continuous function \(\gamma \colon S^1 \to \R^2\) such that \(C\) is the image of \(\gamma\), where \(S^1\) is the unit circle in \(\R^2\). We wish to define a notion of the amount of ‘sides’ and ‘vertices’ that this curve has.

The curve \(C\) is called differentiable in the point \(x \in C\) if there is some parametrisation of \(C\) such that this parametrizing function can be linearly approximated near \(x\). EDIT: if there is some homeomorphism from a real open interval to an open \(C\)-neighbourhood of \(x\) that can be linearly approximated near \(x\). The exact specifics are a little technical, and ultimately unnecessary; as long as you have an intuitive idea of what it means for a curve to be smooth at certain points it’s fine.

We will call a connected open subset \(A \subset C\) a smooth segment of \(C\) if for every \(a \in A\) we have that \(C\) is differentiable in \(a\). \(A\) is additionally called a side of \(C\) if for every smooth segment \(B\) we have that \(A \subset B\) implies \(A = B\). A vertex of \(C\) is then a point not contained in the union of all sides of \(C\).

The following three facts are true about sides and vertices:

1. If some \(x \in C\) has a \(C\)-neighbourhood that is a smooth segment, then there is some side \(E\) with \(x \in E\).

2. Any two distinct sides are disjoint.

3. If the set of vertices is finite and non-empty, and there is a parametrisation \(\gamma\) of \(C\) that is injective, then the number of sides equals the number of vertices.

Proof. This is the longest proof I’ve written on this blog so far. If you’re ever lost, I’d recommend drawing your favourite \(n\)-sided polygon, marking its vertices, then drawing a circle with \(n\) distinct points marked on it, and finally drawing a big arrow from the circle to the polygon labelled \(\gamma\).

Let \(C\) and \(x \in C\) be such that \(U\) is a smooth segment that contains \(x\). Define \(E(x) := \bigcup_{V \in \mathcal{V}} V\), where \(\mathcal{V}\) is the collection of all smooth segments \(V\) of \(C\) such that \(x \in V\).

We know that \(U \subset E(x)\), so it is not empty. \(E(x)\) is a union of open sets, so it is itself open. It is also a union of connected sets that each contain \(x\), so \(E(x)\) is connected. For every \(a \in E(x)\) it holds that \(a\) is contained in some smooth segment, so \(C\) is differentiable in \(a\). It follows that \(E(x)\) is a smooth segment. Assume that \(B\) is a smooth segment such that \(E(x) \subset B\). Then \(x \in B\), so \(B \subset E(x)\). We conclude that \(E(x)\) is a side of \(C\). Let \(A,B\) be sides of \(C\) such that there is some \(x \in A \cap B\). Then \(A,B\) are smooth segments that contain \(x\), so \(A,B \subset E(x)\). Because \(A\) and \(B\) are sides, it follows that \(A = E(x) = B\). We conclude that any two distinct sides of \(C\) are disjoint. Let \(V\) be the set of vertices, and \(\mathcal{E}\) the set of sides. Assume that \(C\) is such that \(V\) is finite and non-empty, and let \(\gamma \colon S^1 \to \R^2\) be an injective parametrisation of \(C\). \(C\) is precisely the image of \(\gamma\), so there is an inverse mapping \(\gamma^{-1} \colon C \to S^1\). Note that any continuous bijection from \(S^1\) into a Hausdorff space is a homeomorphism, so \(\gamma\) and \(\gamma^{-1}\) are additionally open maps between \(S^1\) and \(C\). For the sake of notational brevity, whenever \(x \in C\) or \(A \subset C\), define \(x_\circ := \gamma^{-1}(x)\) and \(A_\circ := \gamma^{-1}(A)\) respectively. Any finite subset of \(S^1\) is discrete, for every vertex \(v \in V\) we have that \(\{v_\circ\}\) is an isolated point. For every \(v \in V\) there is some angle \(\varepsilon(v) > 0\) such that \(B_{\varepsilon(v)}(v_\circ) \cap V_\circ = \{v_\circ\}\), where the metric on \(S^1\) is taken to be the smallest angle between two points. Let \(f \colon V \to \mathcal{E}\) be given by

\[ f(v) = E \left( \gamma \left( v_\circ + \frac{\varepsilon(v)}{2} \right) \right), \]

where \(+\) on \(S^1\) is taken as counterclockwise addition of angles. First we need to show that \(f\) is well-defined: we must show that \(C\) is differentiable at \(p := v_\circ + \varepsilon(v)/2\). Define \(A := B_{\varepsilon(v)/2}(p)\). We know that \(A \cap V_\circ = \emptyset\), and \(A\) is connected, so \(A \cap (S^1 \setminus V_\circ) = A\) is a connected open set containing \(p\) that is itself contained in the inverse image of some smooth segment of \(C\), hence \(\gamma(A)\) is a connected open subset of \(C\) containing \(\gamma(p)\) on which \(C\) is differentiable. We conclude that \(f\) is well-defined. Now we prove that \(f\) is bijective. \(V\) is not empty, so let \(v,w \in V\) be such that \(f(v) = f(w)\). By definition of \(f\) there is some connected subset \(K \subset S^1 \setminus V_\circ\) such that \(v_\circ + \varepsilon(v)/2\) and \(w_\circ + \varepsilon(w)/2\) are both in \(K\). We have that the counterclockwise open arcs \((v_\circ,v_\circ+\varepsilon(v)/2)\) and \((w_\circ,w_\circ+\varepsilon(w)/2)\) are subsets of \(S^1 \setminus V_\circ\), so we have \(w_\circ \notin (v_\circ, v_\circ + \varepsilon(v)/2)\) and \(v_\circ \notin (w_\circ, w_\circ + \varepsilon(w)/2)\). Hence the two arcs are either disjoint or equal. If the two arcs are equal, then \(v = w\) and we are done. If the arcs are disjoint, then there is no connected subset of \(S^1\) that contains \(v_\circ + \varepsilon(v)/2\) and \(w_\circ + \varepsilon(w)/2\) but not one of \(v_\circ\) and \(w_\circ\). Because \(K\) must satisfy this, it follows that \(v = w\), so \(f\) is injective. Now let \(E \in \mathcal{E}\) be some side of \(\gamma\). Because \(E_\circ\) is a connected open subset of \(S^1 \setminus V_\circ\), and \(V_\circ\) is non-empty, it follows that \(E_\circ = (a,b)\) for some (not necessarily distinct) \(a,b \in S^1\), where \((a,b)\) is, again, the counterclockwise open arc from \(a\) to \(b\). We prove that \(a,b \in V_\circ\). Assume that \(a \notin V_\circ\). Then there is some \(A \subset S^1\) containing \(a\) such that \(\gamma(A)\) is a smooth segment of \(C\). Then \(E \cup \gamma(A) = \gamma((a,b)) \cup \gamma(A)\) is a smooth segment strictly larger (because it contains \(\gamma(a)\)) than \(E\), which is a contradiction. We conclude that \(a \in V_\circ\). The proof for \(b\) is exactly analogous. Then, because \(a + \varepsilon(\gamma(a))/2 \in E\), it follows that \(E = f(\gamma(a))\), and we conclude that \(f\) is surjective, hence bijective. We have constructed a bijection between \(V\) and \(\mathcal{E}\), so the number of vertices equals the number of sides, QED.

Note that nowhere in this proof we actually used the definition of what it means for \(C\) to be differentiable in some point \(x\). In fact, if we were to replace differentiability with an arbitrary property \(\Phi(x)\), these three facts we’ve just proven would still hold true. The only thing we need is that a ‘\(\Phi\)-side’ is connected, open, and is a maximal such set on which \(\Phi(x)\) holds for all \(x\).

However, if we wanted to determine what kinds of planar transformations preserve the number of sides and vertices, we would need to use the proper definition. It is my hope that a set being a side is preserved under a diffeomorphism of \(C\), but this post is already way too long so I leave that to any curious readers :)

[y1s1-20182019]

Rolling Object - Final Model

Side By Side - This model is the model that move in the uncommon way, it’s moving side by side in the slow speed. The shape of the model is compose of 2 circle join in the perpendicular plane, which is call “Oloid” shape. The strips that cover the object are for support the object to move smoothly. The pattern for this object rolling on the ramp is moving slowly side by side left and right, which is the movement that I want it to be. 

Rolling Object - Study Model I

For the first prototype, it’s kind of like a blade rolling object. The object composed of 8 blades, which half of it will load with weighted and heavier than the normal one. It will be place swap one by one to make it balance. The movement that I want this object to be is the heavier blade make a momentum for the object, to push it further while it’s moving down the ramp. The performance of this is quite good but it stuck sometimes and can’t go further when it touch the ground.

Rolling Object - Study Model II

This 2nd prototype is the object that made from a shape call “reuleaux triangle”. It’s a shape with a curve of constant width. I use the idea of interlocking two pieces of this triangle and have more part to interlocking to make it fully three-dimensional and for support when it’s rolling.

Rolling Object - Study Model III

The 3rd prototype, this is the one that use for develop into the final model. The shape of this is the same was final, which is oloid shape. But I want to use the triangle shape from 2nd prototype to use it instead of circle. And I put the weight at the side of triangle to make it have momentum to push the object further. But the movement for this object is fast and not so much interesting.