How much will a cable sag? A simple approximation

Suppose you have a cable of length 2s suspended from two poles of equal height a distance 2x apart. Assuming the cable hangs in the shape of a catenary, how much does it sag in the middle?

If the cable were pulled perfectly taut, we would have s = x and there would be no sag. But in general, s > x and the cable is somewhat lower in the middle than on the two ends. The sag is the difference between the height at the end points and the height in the middle.

The equation of a catenary is

y = a cosh(t/a)

and the length of the catenary between t = −x and t = x is

2s = 2a sinh(x/a).

You could solve this equation for a and the sag will be

g = a cosh(x/a) − a.

Solving for a given s is not an insurmountable problem, but there is a simple approximation for g that has an error of less than 1%. This approximation, given in [1], is

g² = (s − x)(sx/2).

To visualize the error, I set x = 1 and let a vary to produce the plot below.

The relative error is always less than 1/117. It peaks somewhere around x = 1/3 and decreases from there, the error getting smaller as a increases (i.e. the cable is pulled tighter).

Related posts

[1] J. S. Frame. An approximation for the dip of a catenary. Pi Mu Epsilon Journal, Vol. 1, No. 2 (April 1950), pp. 44–46

Binary to text to binary

Gnu Privacy Guard includes a way to encode binary files as plain ASCII text files, and turn these text files back into binary. This is intended as a way to transmit encrypted data, but it can be used to convert any kind of file from binary to text and back to binary.

To illustrate this, I’ll use Albrecht Dürer’s Melencolia I as an example. (More on this image and its mathematical significance here.)

Albrecht Dürer’s engraving Melencolia I

This image is saved in the file Melancholia.jpg.

Binary to text

If we run

   gpg --enarmor Melencolia.jpg

at a command line, it produces a file Melancholia.jpg.asc, the asc suffix indicating an ASCII file.

We can look inside this file if we’d like.

-----BEGIN PGP ARMORED FILE-----
Comment: Use "gpg --dearmor" for unpacking

/9j/4AAQSkZJRgABAQEBLAEsAAD/4QBoRXhpZgAATU0AKgAAAAgABAEaAAUAAAAB
AAAAPgEbAAUAAAABAAAARgEoAAMAAAABAAIAAAExAAIAAAARAAAATgAAAAAAAAEs
AAAAAQAAASwAAAABUGFpbnQuTkVUIHYzLjUuOAAA/9sAQwACAQECAQECAgICAgIC
...
wJ/wkzRf8IN4LCCVVwNCtM8sDnPl5zkk565NbcH7Lnw01K0gtrn4feCriB4QfLl0
S2dV+bdgApwMknAwMk+tFFcktjqif//Z
=rpd1
-----END PGP ARMORED FILE-----

The cryptic text /9j/4A…//Z is a base 64 encoded representation of the binary file. Think of the file as one big binary number. Write that number in base 64, i.e. partition the bits into groups of six. Then represent the base 64 “digits” as alphanumeric characters, plus the symbols + and /. More on the details here.

The line =rpd1 is a 24-bit CRC checksum. The equal sign is a separator, and rpd1 is a base 64 encoding the checksum.

The JPG file is 91,272 bytes and the ASCII file is 123,712 bytes. The ASCII file is about 1/3 larger because every six bits in the binary file corresponds to an eight-bit ASCII character. The ASCII file is a little bit more than 1/3 larger because of the human-friendly text above and below the base 64 encoding, the newline characters, and the checksum.

Text to binary

If we run

    gpg --dearmor Melencolia.jpg.asc

at a command line, it produces a file Melancholia.jpg.asc.gpg. This file is bit-for-bit exactly the same as the original file, which we could confirm by running

    diff Melencolia.jpg Melencolia.jpg.asc.gpg

Related posts

A curious pattern in January exponential sums

The exponential sum page on this site draws a new image every day based on plugging the month, day, and year into a formula. Some of these images are visually appealing; I’ve had many people ask if they could use the images in publications or on coffee mugs etc.

The images generally look very different from one day to the next. One reason I include the date numbers in the order I do, using the American convention, is that this increases the variety from one day to the next.

If you first became aware of the page on New Year’s Day this year, you might think the page is broken because there was no apparent change between January 1 and January 2. Yesterday’s image was different, but then today, January 4, the image looks just like the images for January 1st and 2nd. They all look like the image below.

The plots produced on each day are distinct, but they are geometrically congruent.

The exponential sum page displays a plot connecting the partial sums of a certain series given here. The axes are turned off and so only the shape of the plot is displayed. If one plot is a translation or dilation of another, the images shown on the page will be the same.

Here’s a plot of the images for January 1, 2, and 4, plotted red, green, and blue respectively.

This shows that the images are not the same, but are apparently translations of each other.

There’s another difference between the images. Connecting consecutive partial sums draws an image clockwise on January 1, but counterclockwise on January 2 and 4. You can see this by clicking on the “animate” link on each page.

 

Square root factorial

What factorial is closest to the square root of 2024 factorial?

A good guess would be 1012, based on the idea that √(n!) might be near (n/2)!.

This isn’t correct—the actual answer is 1112—but it’s not wildly off.

Could it be that (2n)! is asymptotically (n!)²?

No, Gauss’ duplication formula

\Gamma(2z) = \frac{1}{\sqrt{2\pi}} 2^{2z - 1/2} \Gamma(z) \Gamma\left(z + \frac{1}{2}\right)

shows that the ratio of (2n)! to (n!)² grows exponentially as a function of n.

However, the ratio only grows exponentially, and factorials grow faster than exponentially. I believe that the value of m minimizing

| √(n!) – m! |

asymptotically approaches n/2. In other words, the inverse factorial of  √(n!) approaches n/2. And more generally the inverse factorial of (n!)1/k asymptotically approaches n/k. I haven’t written out a proof, but the plot below shows numerical evidence.

So √(n!) is not asymptotically (n/2)!, but the inverse factorial of √(n!) is asymptotically n/2.

See the next post for a way to compute inverse factorials.

222nd Carnival of Mathematics

A blog carnival is a round up of recent blog posts. The Carnival of Mathematics is a long-running carnival of blog posts on mathematical topics. This is the 222nd edition of the carnival.

Facts about 222

By longstanding tradition, the nth Carnival of Mathematics must begin with trivia about the number n, and so here are five facts about the number 222.

  • There are six groups of order 222, and 222 groups of order 912.
  • 222=9+87+6×5×4+3+2+1.
  • You can encode the number 222 in the Major mnemonic system as “unknown” or “one-on-one.”

The posts

Gil Kalai looks at what happens when you ask ChatGPT to solve Elchanan Mossel’s dice problem.

Γιώργος Πλούσος (@plousos2505 on the social media platform formerly know as Twitter) posted an image showing how to compute π via a sequence of approximations requiring only basic geometry.

David Eppstein’s blog 11011110 has a post on Pyramidology. It’s particularly fitting to have a post from 1101110 in this carnival. As the author explains in his About page,

This journal’s name comes from interpreting my initials as the hexadecimal number 0xDE (decimal 222) and then converting the result to binary.

The blog ThatsMaths has a post on Sharkovsky numbering, Oleksandr Sharkovsky (1936–2022), and Sharkovsky’s Theorem. This post includes the beautiful image from Wikimedia.

The post Patterns of reality by Timothy Williamson gives an overview of logic from basics to the work of Gödel and Turing.

Larissa Fedunik-Hofman posts an interview with Andrew Krause discussing dynamical system.

Finally, we have Matthew Scroggs’ Advent calendar of math puzzles.

To see previous editions of the carnival, or to submit a post to the next edition, see the Carnival of Mathematics page on Aperiodical.

Wrapping Christmas lights around a tree trunk

Suppose you want to wrap Christmas lights around a tree trunk that we can approximate by a cylinder of radius r.

You want to wrap lights around the tree in a helix, going up a distance h every time you go around the tree once. What length of lights do you need to make n turns around the tree?

You can model the lights as a parametric curve with equation

(r cos t, r sin t, ht/2π)

If t ranges from 0 to T, the corresponding curve length is

(r² + h²/4π²)1/2 T

and you can set T =2πn if you want to find the length of n turns.

How much does h matter? Not much if h is less than r, as is often the case when wrapping a tree trunk with Christmas lights. My daughter Allison discovered this while wrapping lights around our pine tree, and then I wrote this post adding the math details.

If we expand (r² + h²/4π²)1/2 as a function of h in a Taylor series centered at 0 we get

(r² + h²/4π²)1/2  = r + / 8π²r + O(h4).

For example, suppose a tree is r = 10 inches in diameter and move h = 4 inches vertically with each turn. To complete one turn we let T = 2π. The exact length of one turn is

(r² + h²/4π²)1/2  T = (10² + 4²/4π²)1/2 (2π) = 62.96 inches.

If we ignore the h term we get

rT = 10 (2π) = 62.83 inches.

In short, the length of n turns around the tree is 2πrn, the same as 10 circles around the tree. The difference in length between a helix with n turns and n circles is negligible, provided h is smaller than r. Even if h = r = 10 in the example above, we’d get a length of 63.6 inches and our approximation would still be off by less than an inch per turn.

The heart of this calculation pops up frequently in various contexts. For example, the same calculation appears in the post It doesn’t matter much if the tape is straight.

F# and G

I was looking at frequencies of pitches and saw something I hadn’t noticed before: F# and G have very nearly integer frequencies.

To back up a bit, we’re assuming the A above middle C has frequency 440 Hz. This is the most common convention now, but conventions have varied over time and place.

We’re assuming 12-tone equal temperament (12-TET), and so each semitone is a ratio of 21/12. So the nth note in the chromatic scale from A below middle C to A above middle C has frequency

220 × 2n/12.

I expected the pitch with frequency closest to integer would be an E because a perfect fifth above 220 Hz would be exactly 330 Hz. In equal temperament the frequency of the E above middle C is 329.6 Hz.

The frequency of F# is

220 × 29/12 Hz = 369.9944 Hz.

The difference between this frequency and 370 Hz is much less than the difference between equal temperament and other tuning systems.

The frequency of G is

220 × 25/6 Hz = 391.9954 Hz

which is even closer to being an integer.

In more mathematical terms, stripped of musical significance, we’ve discovered that

23/4 ≈ 37/22

and

25/6 ≈ 98/55.

 

Straight on a map or straight on a globe?

Straight lines on a globe are not straight on a map, and straight lines on a map are not straight on a globe.

A straight line on a globe is an arc of a great circle, the shortest path between two points. When projected onto a map, a straight path looks curved. Here’s an image I made for a post back in August.

Quito to Nairobi to Jerusalem and back to Quito

The red lines form a spherical triangle with vertices at Quito, Nairobi, and Jerusalem. The leg from Quito to Nairobi is straight because it follows the equator. And the leg from Nairobi to Jerusalem is straight because it follows a meridian. But the leg from Quito to Jerusalem looks wrong.

If you were flying from Quito to Jerusalem and saw this flight plan, you might ask “Why aren’t we flying straight there, cutting across Africa rather than making a big arc around it? Are we trying to avoid flying over the Sahara?”

But the path from Quito to Jerusalem is straight, on a globe. It’s just not straight on the map. The map is not the territory.

Now let’s look at things from the opposite direction. What do straight lines on a map look like on a globe? By map I mean a Mercator projection. You could take a map and draw a straight line from Quito to Jerusalem, and it would cross every meridian at the same angle. A pilot could fly from Quito to Jerusalem along such a path without ever changing bearing. But the plane would have to turn continuously to stay on such a bearing, because this is not a straight line.

A straight line on a Mercator projection is a spiral on a globe, known as a loxodrome or a rhumb line. If a plane flew on a constant bearing from Quito but few over Jerusalem and kept going, it would spiral toward the North Pole. It would keep circling the earth, crossing the meridian through Jerusalem over and over, each time at a higher latitude. On a polar projection map, the plane’s course would be approximately a logarithmic spiral. The next post goes into this in more detail.

loxodrome spiral

I made the image above using the Mathematica code found here.

Although straight lines the globe are surprising on a map, straight lines on a map are even more surprising on a globe.

Related posts

Latus rectum of an ellipse

Ellipses have been studied for over two thousand years, and so some of the terminology is ancient and sounds odd to modern ears. One such term is latus rectum. What is the latus rectum and does have it anything to do with anatomy?

This post defines the latus rectum for an ellipse. See the next post for a more general definition that extends to parabolas and hyperbolas.

Background

An ellipse has two foci, two points such that the distance to one plus the distance to the other remains constant for every point on the ellipse. For a circle, the two foci coincide with the center.

Given an ellipse in the plane, we can pick a coordinate system so that our ellipse has equation

x²/a² + y²/b² = 1

where ab > 0. Here a is called the semi-major axis and b the semi-minor axis. The two foci are located at ±c where

c² = a² – b².

The ratio c/a is the eccentricity. As discussed here, eccentricity can be moderately large and yet the ellipse still be close to a circle.

Definition

Pick one of the foci, say the one located at (c, 0). Then the line segment through the focus, parallel to the minor axis and connecting two points on the ellipse, is called the latus rectum. Its length is 2b² / a.

Example

Let the semi-major axis be a = 5 and the semi-minor axis be b = 3. Then c = 4 and so the foci are located at (-4, 0) and (4, 0).

When x = 4, the equation of the ellipse tells us

16/25 + y²/9 = 1

and so y = ±9/5. So the latus rectum is the line connecting (4, -9/5) and (4, 9/5), the red vertical line below.

Why the name?

The first part latus mean side and the second part rectum means straight. According to Etymonline, rectum was

the name given to the lowest part of the large intestine by Galen, who so called it because he dissected only animals whose rectum (in contradistinction to that of man) is really straight.

So latus rectum means straight side. If you wanted to draw straight sides on an ellipse, where would you put them? Through the foci seems like as good a choice as any.

Radius of curvature

An interesting fact about the latus rectum is that its length is the diameter of a kissing circle tangent to the vertex of the ellipse. In other words, the semi-latus rectum, half the length of the latus rectum, is the radius of curvature at the vertex.

The dashed orange circle below has radius 9/5, equal to the semi-latus rectum. So the radius of curvature at the right end of the ellipse is 9/5 and the curvature is 5/9.

More on ellipses