Rule of Thumb

https://books.google.com/books?id=0t7ZiYKV1KQC

Peter Nicholson’s Carpenter’s New Guide of 1803 contains an interesting technique:

To find a right line equal to any given Arch of a Circle. Divide the chord ab into four equal parts, set one part bc on the arch from a to d, and draw dc which will be nearly equal to half the arch.

Apparently this was an item of carpentry lore in 1803. In the figure above, if arc ad = bc, then cd is approximately half of arc length ab.

Nicholson warns that this works best for relatively short arcs: “This method should not be used above a quarter of a circle, so that if you would find the circumference of a whole circle by this method, the fourth part must only be used, which will give one eighth part of the whole exceedingly near.”

But with that proviso it works pretty well — in 1981 University of Essex mathematician Ian Cook found that for arcs up to a quadrant of a circle, the results show a maximum percentage error of 0.6 percent, “which I suppose can be said to be ‘exceedingly near.'” He adds, “[I]t would be of interest to know who discovered this construction.”

(Ian Cook, “Geometry for a Carpenter in 1800,” Mathematical Gazette 65:433 [October 1981], 193-195.)

Long Distance

https://galton.org/essays/1890-1899/galton-1893-diff-1up.pdf

Francis Galton was interested in communicating with Mars as early as 1892, when he wrote a letter to the Times suggesting that we try flashing sun signals at the red planet. At a lecture the following year he described more specifically a method by which pictures might be encoded using 26 alphabetical characters, which could then be transmitted over a distance in 5-character “words,” in effect creating a low-resolution visual telegraph. As a study he reduced this profile of a Greek girl to 271 coded dots, which he found yielded “a very creditable production.”

This had huge implications, he felt. In 1896 he imagined a whole correspondence with a civilization of intelligent ants on Mars; in three and a half hours they catch our attention; teach us their base-8 mathematical notation; demonstrate their shared understanding of certain celestial bodies and mathematical constants; and finally propose a specified 24-gon in which points can be situated by code, like stitches in a piece of embroidery.

That opens a limitless avenue for colloquy — the Martians send images of Saturn, Earth, the solar system, and domestic and sociological drawings, a new one every evening. Galton concludes that two astronomical bodies that are close enough to signal one another with flashes of light already have everything they need to establish “an efficient inter-stellar language.”

Band Practice

https://commons.wikimedia.org/wiki/File:Sphere_bands.svg

Drill a hole straight through the center of a sphere, leaving a band in the shape of a napkin ring. Suppose the height of this band is h. Now take a sphere of a different size and drill a hole through that, contriving the hole’s width so that its depth is again h. Remarkably, the two napkin rings will have the same volume.

As the sphere expands, the band must grow both wider and thinner, and it turns out that these two effects exactly cancel one another. It’s called the napkin ring problem.

A Pi Diet

https://commons.wikimedia.org/wiki/File:Academ_rosette.svg

Image: <a href="https://commons.wikimedia.org/wiki/File:Academ_rosette.svg">Wikimedia Commons</a>

Students beginning with the compass learn to draw this rosette, sometimes called the Flower of Life. If the arcs and the circle have the same radius, a, then the area of one petal is

\displaystyle  X = a^{2}\left ( \frac{\pi }{3} - sin \frac{\pi }{3}\right ) = \left ( \frac{\pi }{3} - \frac{\sqrt{3}}{2}\right ) a^{2}

and the unshaded area of the circle is

\displaystyle  Y = \pi a^{2} - 3X = \frac{3\sqrt{3}}{2}a^{2}.

Remarkably, though the area we sought is bounded entirely by the arcs of circles, the final expression is independent of π.

Related: When two cylinders of radius r meet at right angles, the volume of their intersection is 16r3/3 — again, no sign of π.

(J.V. Narlikar, “A Pi-Less Area,” Mathematical Gazette 65:431 [March 1981], 32-33.)

10/01/2024 UPDATE: Some deft rearranging shows that the unshaded area of the circle is just the area of an inscribed regular hexagon:

2024-09-28-a-pi-diet-2

So the absence of π isn’t that surprising. Thanks to readers Catalin Voinescu (who sent this diagram) and Gareth McCaughan for pointing this out.

A Belt Font

Suppose you have a collection of gears pinned to a wall (disks in the plane). When is it possible to wrap a conveyor belt around them so that the belt touches every gear, is taut, and does not touch itself? This problem was first posed by Manuel Abellanas in 2001. When all the gears are the same size, it appears that it’s always possible to find a suitable path for the belt, but the question remains open.

Erik Demaine, Martin Demaine, and Belén Palop have designed a font to illustrate the problem — each letter is a collection of equal-sized gears around which exactly one conveyor-belt wrapping outlines an English letter:

2024-09-26-a-belt-font-1.png

Apart from its mathematical interest, the font makes for intriguing puzzles — when the belts are removed, the letters are surprisingly hard to discern. What does this say?

2024-09-26-a-belt-font-2.png

Click for Answer

The Sussman Anomaly

MIT computer scientist Gerald Sussman offered this example to show the importance of sophisticated planning algorithms in artificial intelligence. Suppose an agent is told to stack these three blocks into a tower, with A at the top and C at the bottom, moving one block at a time:

https://commons.wikimedia.org/wiki/File:Sussman-anomaly-1.svg
Image: Wikimedia Commons

It might proceed by separating the goal into two subgoals:

  1. Get A onto B.
  2. Get B onto C.

But this leads immediately to trouble. If the agent starts with subgoal 1, it will move C off of A and then put A onto B:

https://commons.wikimedia.org/wiki/File:Sussman-anomaly-2.svg
Image: Wikimedia Commons

But that’s a dead end. Because it can move only one block at a time, the agent can’t now undertake subgoal 2 without first undoing subgoal 1.

If the agent starts with subgoal 2, it will move B onto C, which is another dead end:

https://commons.wikimedia.org/wiki/File:Sussman-anomaly-3.svg
Image: Wikimedia Commons

Now we have a tower, but the blocks are in the wrong order. Again, we’ll have to undo one subgoal before we can undertake the other.

Modern algorithms can handle this challenge, but still it illustrates why planning is not a trivial undertaking. Sussman discussed it as part of his 1973 doctoral dissertation, A Computational Model of Skill Acquisition.

The Sandwheel

https://commons.wikimedia.org/wiki/File:Sandwheel.gif

This is a variation on a perpetual motion machine proposed by the Indian mathematician Bhāskara II around 1150. Each of the wheel’s tilted spokes is filled with a quantity of sand. As the tubes descend on the right, the sand within them shifts outward, exerting greater torque in the clockwise direction and thus keeping the wheel turning forever.

Unfortunately the same design ensures that there’s always a greater quantity of sand on the left, so nothing happens.

Matriarch

https://commons.wikimedia.org/wiki/File:Laysan_albatross_fws.JPG

In 1956, ornithologist Chandler Robbins tagged a wild female Laysan albatross at the Midway Atoll National Wildlife Refuge in the North Pacific. The bird, dubbed Wisdom, went on to a stunning career, flying more than 3 million miles, equivalent to 120 trips around the Earth. She has been seen at the atoll as recently as last December, making her, at 72, the the oldest known wild bird in the world.

In that time she’s hatched as many as 36 chicks, a significant contribution to the struggling wild albatross population. The U.S. Fish and Wildlife Service wrote, “Her health and dedication have led to the birth of other healthy offspring which will help recover albatross populations on Laysan and other islands.” Bruce Peterjohn, chief of the North American Bird Banding Program, added, “To know that she can still successfully raise young at age 60-plus, that is beyond words.”