A Lesson Involving the Social Use of Color

Posted on 4 September 2015 by RobertLovesPi

RobertLovesPi’s social-interaction lesson of the day: different colors of fabric can actually mean something else, besides simply reflecting different wavelengths of light, and these meanings can shift quickly. (I already knew this could happen once per day, but was only just taught that this is also possible for n = 2, allowing me to extrapolate that, for the general case, n > -1, presumably with an upper limit set by the individual’s speed at changing clothes.)

As far as I can tell, n = 0 on weekends and legal holidays, in most cases, and n = 1 on most workdays (but not today, when the needed reflection-wavelength shifts from ~475 nm to ~550 nm after I leave the city of Sherwood, Arkansas, bound for a spot approximately 20 km South of there, in Little Rock, which is still in the same county).

Apparently my key to understanding this stuff is finding a way to analyze it mathematically. Also, posting such “new” discoveries to my blog increases the odds of me remembering them. However, unlike my last such finding (it involved chocolate chips not being a sandwich topping at Subway), I did NOT figure these things out “all by myself.” In fact, without help from two very important people, I doubt I ever would have figured them out at all!

An Ethical Dilemma Involving a Polyhedral Crystal

Posted on 1 September 2015 by RobertLovesPi

I just ordered a crystal rhombcuboctahedron on eBay because I like its geometrical properties, despite the mystical claims in the item listing. I did so with the full knowledge and expectation that these claims are almost certainly false, because, well, they’re mystical claims.

Here’s my ethical dilemma: would it be ethical to lab-test those claims, then post the results in the feedback I leave?

[Image created using Stella 4d, available at http://www.software3d.com/Stella.php. This isn’t a picture of the crystal on eBay; it is made of quartz, and not rainbow-colored. It is of the same shape, however.]

Geometry Problem Involving Two Circles (See Comments for Solution)

Posted on 30 August 2015 by RobertLovesPi

This is a puzzle I made up not long ago. After trying to solve it for a bit (no success yet, but I haven’t given up), I decided to share the fun.

A small circle of radius r is centered on a large circle of radius R. It is a given that 0 < r < R. In terms of r and R, what fraction of the smaller circle’s circumference lies outside the larger circle?

I am 90% certain there is an extremely simple way to do this, using only things I already know. It’s frustrating that the answer isn’t simply leaping out of the computer screen, at me. For simple math problems, that’s what usually happens . . . so either this is merely deceptively simple, or I am missing something.

Two Views of the Truncated Tesseract

Posted on 30 August 2015 by RobertLovesPi

The figure above, rotating in hyperspace, is an orthogonal projection of a four-dimensional polychoron known as the truncated tesseract. It is analogous to the truncated cube, one of the Archimedean solids. The image below is of the same figure, but is shown as a perspective projection.

Both images were created using Stella 4d, software you can buy (with a free trial download available, first) at http://www.software3d.com/Stella.php. It’s great software, and a friend of mine wrote it — but no, he doesn’t pay me to give his program free advertising, as some have wondered.

Elementary School Mathematics Education Mysteries

Posted on 8 August 2015 by RobertLovesPi

Since these two problems are really the exact same problem, in two different forms, why not just use “x” to teach it, from the beginning, in elementary school, instead of using the little box? The two symbols have the exact same meaning!

To the possible answer, “We use an ‘x’ for multiplication, instead, so doing this would be confusing,” I have a response: why? Using “x” for multiplication is a bad idea, because then students have to unlearn it later. In algebra, it’s better to write (7)(5) = 35, instead of 7×5 = 35, for obvious reasons — we use “x” as a variable, instead, almost constantly. This wouldn’t be as much trouble for students taking algebra if they had never been taught, in the first place, that “x” means “multiply.” It’s already a letter of the alphabet and a variable, plus it marks spots. It doesn’t need to also mean “multiply.”

Why are we doing things in a way that causes more confusion than is necessary? Should we, as teachers, not try to minimize confusion? We certainly shouldn’t create it, without a good reason for doing so, and these current practices do create it.

These things may not be mysteries to others, but they certainly are to me.

[Note: for those who do not already know, I am a teacher of mathematics. However, I do not have any experience teaching anything at the elementary level. For this particular post, that’s certainly relevant information.]

One Dozen Precious Metal Cubes: A Problem Involving Geometry, Chemistry, and Finance (Solution Provided, with Pictures)

Posted on 7 August 2015 by RobertLovesPi

The troy ounce is a unit of mass, not weight, and is used exclusively for four precious metals. At this time, the prices per troy ounce, according to this source for current precious metal prices, for these four elements, are:

Gold, $1,094
Palladium, $600
Platinum, $965
Silver, $14.82

(As a side note, it is rare for platinum to have a lower price per troy ounce than gold, as is now the case. I would explain the reasons this is happening, except for one problem: I don’t understand the reasons, myself, well enough to do so. Yet.)

A troy ounce equals 31.1034768 grams, but, for most purposes, 31.103 g, or even 31.1 g, works just fine.

Also, as you can see here, these “troy elements” are all in one part of the periodic table. This is related to the numerous similarities in these elements’ physical and chemical properties, which is itself related, of course, to the suitability of these four elements for such things as jewelry, coinage, and bullion.

To determine the volume of a given mass of one of these metals, it is also necessary to know their densities, so I looked them up, using Google (they are not listed on the periodic table above):

Gold, 19.3 g/cm³
Palladium, 11.9 g/cm³
Platinum, 21.46 g/cm³
Silver, 10.49 g/cm³

In chemistry, of course, one must often deal with elements (as well as other chemicals) in terms of the numbers of units (such as atoms or molecules), except for one problem: this is absurdly impractical, due to the outrageously small size of atoms. Despite this, though, it is necessary to count such things as atoms in order to do much chemistry at all, so chemists have devised a “workaround” for this problem: when counting units of pure chemicals, they don’t count such things as atoms or molecules directly, but count them a mole at a time. A mole is defined as a number of things equal to the number of atoms in exactly 12 grams of pure carbon-12. To three significant figures, this number is 6.02 x 10²³. To deal with moles, since atoms have differing masses, we need to know the molar mass (mass of one mole) of whatever we are dealing with to convert, both directions, between moles and grams. Here are the molar masses of the four troy-measured elements, as seen on the periodic table above, below each element’s symbol.

Gold, 196.97 g
Palladium, 106.42 g
Platinum, 195.08 g
Silver, 107.87 g

I’ve given these numbers as the information needed to solve the following problem: rank one dozen precious metal cubes (descriptions follow) by ascending order of volume. There are three cubes each of gold, palladium, platinum, and silver. Four of the twelve (one of each element) have a mass of one troy ounce each. Another four each have a value, at the time of this writing, of $1,000. The last set of four each contain one mole of the element which composes the cube, and, again, there is one of each of these same four elements in the set.

If you would like to do this problem for yourself, the time to stop reading is now. Otherwise (or to check your answers against mine), just scroll down.

In the solutions which follow, a rearrangement of the formula for density (d=m/v) is used; solved for v, this equation becomes v = m/d. In order, then, by both volume and edge length, from smallest to largest, here are the twelve cubes:

Smallest cube: one troy ounce of platinum

One tr oz, or 31.103 g, of platinum would have a volume of v = m/d = 31.103 g / (21.46 g/cm³) = 1.449 cm³. A cube with this volume would have an edge length equal to the its volume’s cube root, or 1.132 cm. (This explanation for the calculation of the edge length, given the cube’s volume, is omitted in the items below, since the mathematical procedure is the same each time.)

Second-smallest cube: $1000 worth of gold

Gold worth $1000, at the time of this posting, would have a troy mass, and then a mass in grams, of $1000.00/($1,094.00/tr oz) = (0.914077 tr oz)(31.103 g/tr oz) = 28.431 g. This mass of gold would have a volume of v = m/d = 28.431 g / (19.3 g/cm³) = 1.47 cm³. A cube with this volume would have an edge length of 1.14 cm.

Third-smallest cube: $1000 worth of platinum

Platinum worth $1000, at the time of this posting, would have a troy mass, and then a mass in grams, of $1000.00/($965.00/tr oz) = (1.0363 tr oz)(31.103 g/tr oz) = 32.231 g. This mass of platinum would have a volume of v = m/d = 32.231 g / (21.46 g/cm³) = 1.502 cm³. A cube with this volume would have an edge length of 1.145 cm.

Fourth-smallest cube: one troy ounce of gold

One tr oz, or 31.1 g, of gold would have a volume of v = m/d = 31.1 g / (19.3 g/cm³) = 1.61 cm³. A cube with this volume would have an edge length of 1.17 cm.

Fifth-smallest cube: one troy ounce of palladium

One tr oz, or 31.1 g, of palladium would have a volume of v = m/d = 31.1 g / (11.9 g/cm³) = 2.61 cm³. A cube with this volume would have an edge length of 1.38 cm.

Sixth-smallest cube: one troy ounce of silver

One tr oz, or 31.103 g, of silver would have a volume of v = m/d = 31.103 g / (10.49 g/cm³) = 2.965 cm³. A cube with this volume would have an edge length of 1.437 cm.

Sixth-largest cube: $1000 worth of palladium

Palladium worth $1000, at the time of this posting, would have a troy mass, and then a mass in grams, of $1000.00/($600.00/tr oz) = (1.6667 tr oz)(31.103 g/tr oz) = 51.838 g. This mass of palladium would have a volume of v = m/d = 51.838 g / (11.9 g/cm³) = 4.36 cm³. A cube with this volume would have an edge length of 1.63 cm.

Fifth-largest cube: one mole of palladium

A mole of palladium, or 106.42 g of it, would have a volume of v = m/d = 106.42 g / (11.9 g/cm³) = 8.94 cm³. A cube with this volume would have an edge length of 2.07 cm.

Fourth-largest cube: one mole of platinum

A mole of platinum, or 195.08 g of it, would have a volume of v = m/d = 195.08 g / (21.46 g/cm³) = 9.090 cm³. A cube with this volume would have an edge length of 2.087 cm.

Third-largest cube: one mole of gold

A mole of gold, or 196.97 g of it, would have a volume of v = m/d = 196.97 g / (19.3 g/cm³) = 10.2 cm³. A cube with this volume would have an edge length of 2.17 cm.

Second-largest cube: one mole of silver

A mole of silver, or 107.87 g of it, would have a volume of v = m/d = 107.87 g / (10.49 g/cm³) = 10.28 cm³. A cube with this volume would have an edge length of 2.175 cm.

Largest cube: $1000 worth of silver

Silver worth $1000, at the time of this posting, would have a troy mass, and then a mass in grams, of $1000.00/($14.82/tr oz) = (67.48 tr oz)(31.103 g/tr oz) = 2099 g. This mass of gold would have a volume of v = m/d = 2099 g / (10.49 g/cm³) = 200.1 cm³. A cube with this volume would have an edge length of 5.849 cm.

Finally, here are pictures of all 12 cubes, with 1 cm³ reference cubes for comparison, all shown to scale, relative to one another.

A third of these cubes change size from day-to-day, and sometimes even moment-to-moment during the trading day, if their value is held constant at $1000 — which reveals, of course, which four cubes they are. The other eight cubes, by contrast, do not change size — no precious metal prices were used in the calculation of those cubes’ volumes and edge lengths, precisely because the size of those cubes is independent of such prices, due to the way those cubes were defined in the wording of the original problem.

Silly U.S. Map Puzzles #4a and 4b

Posted on 30 July 2015 by RobertLovesPi

First, for puzzle #4a, what are the meanings of the colors on this map?

For puzzle #4b, what do the colors mean on this second, similar map?

To find the answers, simply scroll down.

Keep scrolling….

Solution:

In the first map, consider the number of letters in the name of each state. Is this number prime or composite?

In the second map, consider the number of characters, rather than letters, in each state’s name. This number is different for states with two-word names, due to the single character, a blank space, needed to separate the two words. Again: prime, or composite?

On Binary Planets, and Binary Polyhedra

Posted on 26 July 2015 by RobertLovesPi

This image of binary polyhedra of unequal size was, obviously, inspired by the double dwarf planet at the center of the Pluto / Charon system. The outer satellites also orbit Pluto and Charon’s common center of mass, or barycenter, which lies above Pluto’s surface. In the similar case of the Earth / Moon system, the barycenter stays within the interior of the larger body, the Earth.

I know of one other quasi-binary system in this solar system which involves a barycenter outside the larger body, but it isn’t one many would expect: it’s the Sun / Jupiter system. Both orbit their barycenter (or that of the whole solar system, more properly, but they are pretty much in the same place), Jupiter doing so at an average orbital radius of 5.2 AU — and the Sun doing so, staying opposite Jupiter, with an orbital radius which is slightly larger than the visible Sun itself. The Sun, therefore, orbits a point outside itself which is the gravitational center of the entire solar system.

Why don’t we notice this “wobble” in the Sun’s motion? Well, orbiting binary objects orbit their barycenters with equal orbital periods, as seen in the image above, where the orbital period of both the large, tightly-orbiting rhombicosidodecahedron, and the small, large-orbit icosahedron, is precisely eight seconds. In the case of the Sun / Jupiter system, the sun completes one complete Jupiter-induced wobble, in a tight ellipse, with their barycenter at one focus, but with an orbital period of one jovian year, which is just under twelve Earth years. If the Jovian-induced solar wobble were faster, it would be much more noticeable.

[Image credit: the picture of the orbiting polyhedra above was made with software called Stella 4d, available at this website.]

Thirty-Four Rotating, Convex, and Non-Chiral Polyhedra with Icosidodecahedral Symmetry

Posted on 22 July 2015 by RobertLovesPi

Most in the field call this type of symmetry “icosahedral,” but I prefer the term George Hart uses — along with “cuboctahedral” in place of “octahedral.”

Each polyhedral image here was created with Stella 4d: Polyhedron Navigator. At this linked page, you can try a trial version of that program for free.

By the way, when I described these polyhedra, in this post’s title, as non-chiral, I was not referring to the coloring-schemes used here, many of which obviously are chiral, but only the shapes themselves.

That 10 GB space-upgrade, which most bloggers don’t ever need, is really coming in handy right now. In other words, some of these .gif files are huge!

Why, yes, I am including some words after every fifth polyhedron. That will help, later, when I count them for the title of this post.

I’m not sure why that last one is spinning the opposite direction from the others. Perhaps this polyhedron is trying to start a trend. On the other hand, it could just simply upside-down.

That’s twenty-five so far.

Clearly, I should have checked the number of files in that file folder before deciding to simply post them all together, based on what they have in common. That’s thirty so far.

Order-Six Radial Tessellations of the Plane, Using Elongated and Equilateral Hexagons, Rendered with Twelve Different Coloring-Schemes

Posted on 20 July 2015 by RobertLovesPi

I explored radial tessellations of the plane, using only hexagons, in this earlier post. Order-three tessellations of this type are the familiar regular-hexagon tessellations of the plane. With higher-order all-hexagon radial tessellations, though, the hexagons must be elongated, although they can still remain equilateral, and all congruent, with bilateral symmetry. In that previous post, examples were shown of order 4, 5, and 8, in addition to the familiar order-3 regular-hexagon tessellation.

This left out order-6, of which I show many examples below. As it turns out, this particular radial tessellation lends itself particularly well to a variety of coloring-schemes. In the first picture, the construction-circles, -points, and -lines I used are shown; in the rest, they are hidden.

No upper limit exists to the order-number of these all-hexagon radial tessellations — although the larger that number gets, the thinner the hexagons become, relative to their edge length. At some point (which I expect would vary from person to person), as the order-number increases, the hexagons needed will become so thin that they will no longer be recognizable as hexagons.