Monday, March 27, 2017

An Efficient Connector



Airport authorities would like to build a connector road (dashed line) so that maintenance vehicles can drive from one runway to the other without having to go all the way back to the airport terminal.

Given that the connector will be vertical, how far from the terminal should it be?

In my approach to the problem, I neglect the startup cost of building the connector road. My goal instead is to minimize the average distance that a maintenance vehicle would have to travel in order to get from the top runway to the bottom runway.

For example, with reference to the next figure, suppose that the vehicle is initially on the top runway, at the point indicated by the gray circle. To get to the indicated point on the bottom runway, the vehicle must drive along the magenta path to the other gray circle.


One way to begin the problem is to create a formula for the distance between a given initial location and a given final location. Then, one can use integration to average the distances over all initial and final locations. Because the average will depend on the location of the connector, the possibility arises of choosing the location of the connector that minimizes the average.

I did the problem for the dimensions shown:


In case anyone wants to solve it for themselves, I'll post my answer at a later time.

If you don't know calculus, you can try using your intuition to place the connector in an efficient location. Do you think that the most efficient connector will be halfway across, or less than halfway, or more than halfway?

Follow-up questions: Was it safe to assume that the most efficient connector is vertical? What if the runways are asymmetrical, as in this configuration?


Sunday, March 26, 2017

On Having A Favorite

To enroll in a frequent-flyer program online, I had to answer half a dozen security questions, including "What is your favorite kind of vacation?" and "What is your favorite cold-weather activity?" Who has answers to these kinds of questions?

"I like red the best!" says the toddler, as if his outfit didn't already make that clear. For grownups as well, having a favorite is for people who are at the toddler stage in their appreciation of something. I have a favorite bourbon, and that should tell you that I don't know much about bourbon. A good way to know that somebody isn't much of a reader is if they have a favorite book.

Now it is true that when I eat in a familiar restaurant, I almost always order the same thing. Always the pork curry at the Thai place in my neighborhood, always the chicken tikka at the Indian place. There's more to Indian food than chicken tikka, of course, but that's why God created other Indian restaurants. And if I didn't want a Shackburger and a strawberry shake, then I wouldn't be at Shake Shack, now would I?

This is not to say that Shake Shack hamburgers are the only hamburgers I like. I also like an occasional double quarter-pounder, or a "Mexican style" hamburger with Jack cheese, salsa, and avocado. Can't go wrong with a barbecue burger either—that bacon and zippy sauce!

Also, crumbled blue cheese is excellent on a thick hamburger.

Yet there are people out there who say things like, "My favorite hamburger is In-n-Out." Hearing this always makes me sad, not because I object to In-n-Out burgers in themselves, but because having a favorite hamburger just seems like a sad way to live.

To have a favorite X is to care too little about X's to arrange your life in such a way as to be surrounded only by wonderful X's. As few ties as I own, I can't say I have a favorite one. I like them all, or else why would I have them? Do you really want to put on a tie and think, "Eh, not my favorite"?

On the other hand, there are costs to not having favorites. I can spend a long time in the morning choosing a tie to wear. A trip to the bookstore can take me hours and lead to no firm decisions; likewise for a trip to my Netflix queue. If nothing else, having favorites is efficient: a fact well known to parents of toddlers.

Saturday, March 25, 2017

Piece Out

Back for a few more notes on grammar and spelling:


1. Please know that the past tense of lead is led.

Wrong:
LeBron James lead the Cavaliers through a relatively stress-free fourth quarter on the way to the win. (Sports Daily, December 2016)
Fixed:
LeBron James led the Cavaliers through a relatively stress-free fourth quarter on the way to the win.


2. The preferred past tense of plead is pleaded.

Bad:
When arraigned Friday morning Goff plead not guilty and was assigned a public defender. Shortly before 1:30, through a second public defender, he plead guilty and was sentenced to 12 months in the Arkansas Department of Correction. (Booneville Democrat, March 2017)
Best:
When arraigned Friday morning Goff pleaded not guilty and was assigned a public defender. Shortly before 1:30, through a second public defender, he pleaded guilty and was sentenced to 12 months in the Arkansas Department of Correction.
My American Heritage Fourth Edition (2000) also lists pled as an acceptable spelling of the past tense.

Unfortunately, I see from the online version that the Fifth Edition now lists plead is an acceptable spelling of the past tense. My advice though is not to use that spelling. For one thing, it's inconsiderate writing—you'll trip up some fraction of your readers when they mistakenly read plead as present-tense the first time through.

I do think pled is fine, though according to the American Heritage entry linked above, the usage panel for the Fifth Edition prefers pleaded to pled by a wide margin.

And I see that the usage note doesn't even address plead as a past-tense spelling. I suspect that's because zero percent of the panel would find this spelling acceptable.

Although plead as a past-tense spelling is apparently widespread enough to be listed in the dictionary, I don't think it is used very often in high-status writing. Anecdotally, I find several thousand hits for a google search of "He pleaded guilty" on nytimes.com, but only about a hundred hits for "He pled guilty" or "He plead guilty." A quick scan suggests that the instances of "He plead guilty" tend to come from complicated constructions ("...it was required that he plead guilty...."), or from direct quotes of speech, or from internet comments.

Bottom line: use pleaded, or use pled if you prefer, but don't use plead because you might trip up your readers and/or come across like an internet commenter.


3. Don't hand-wave with around.

Lazy:
The committee developed a set of guidelines around ethics.
Better:
The committee developed a set of ethics guidelines for members.

The use of around, in the sense of the first sentence, is almost always a sign of vagueness. In the second sentence, we better serve the reader by stating more clearly what is going on.

Around, in the sense of the first sentence, is gaining currency. In addition to saving the writer the effort of being clear, it appeals to those writers who worry that more common prepositions just don't sound smart enough.


4. Know your own verbal tics. Is piece one of them?

With the tic:
We haven't figured out the professional development piece.
Without:
We haven't figured out professional development.

Normally I don't comment on speech, just writing, but piece is a prominent verbal tic among knowledge workers. Their jobs require them to analyze systems into parts; it's natural to think of the parts as "pieces," and to speak accordingly. But I have yet to hear a sentence that wouldn't be improved by just not doing that.

Because it's unpretentious, I would even prefer
We haven't figured out the professional development thing.


Sunday, March 19, 2017

The Sliding Ladder Problem

Yesterday a colleague mentioned a problem that professors often assign in their Calculus courses. The problem went something like this:
A 5m ladder is leaning against a wall. If the bottom of the ladder is pulled along the ground away from the wall at a constant rate of 0.40m/s, how fast will the top of the ladder be moving down the wall when its bottom is 3m away from the wall?
(This is a version of the problem I found today at math.stackexchange.)

The problem is straightforward as stated (the answer is 0.3 m/s), but my colleague wanted to know what I thought, as a physicist, about the fact that the speed of the top point of the ladder exceeds any finite value as this point nears the ground. I can see from a different math.stackexchange page that this is a frequently asked question, so I thought I'd post some thoughts about it here.

Before I get to that, however, I ought to substantiate the point my colleague was making about the ladder's speed. The following animation helps to visualize the motion:


I added a reversed phase of the motion just because I think it aids the eye to see the "bounce."

The equations I used to generate this figure were

x(t) = t
y(t) = (1 − t2)½
0 ≤ t ≤ 1

where point F = (x(t), 0) is the location of the ladder's point of attachment to the floor at time t, and point W = (0, y(t)) is the location of the ladder's point of attachment to the wall at time t. At time t = 0, the ladder is vertical; at time t = 1, the ladder is horizontal.

These equations are the non-dimensionalized versions of

X(T) = vT
Y(T) = (L2 - v2T2)½
0 ≤ TL/v

where v is the constant speed of the ladder's foot, and L is the length of the ladder. (That is, x = X/L, y = Y/L, and t = vT/L.)

The functions x(t) and y(t) are differentiable on the interval (0, 1), which means that points F and W both have well-defined speeds during this interval. However, y(t) has no derivative at t = 1, not even a left-sided one; this means that the speed of point W doesn't exist at the moment of time when the ladder is horizontal.

People usually say that the speed of point W "becomes infinite" at t = 1…but if "the speed of point W" means |dy/dt|, then saying that the speed is infinite is saying that dy/dt = ∞, and I used to discourage my beginning Calculus students from writing such things, because I wanted them to appreciate that dy/dt is defined by a limit, and limits are numbers that either exist or don't.

(It's certainly true that the speed of point W exceeds any finite value as t approaches 1, and that's enough to justify asking for a physicist's take.)


To think about the ladder's motion physically, concentrate on the motion of the ladder's center of mass point. (The motion of the center of mass point tells you about the net force acting on the ladder.) Here's an animation showing the motion of the center-of-mass point.


The coordinates of the CM point are ½(x(t), y(t)). As the ladder moves, the CM point traces a circle of radius ½. This isn't uniform circular motion, however, because the CM point isn't moving at constant speed.

The next animation shows the horizontal and vertical locations of the CM point.


The CM point has zero horizontal acceleration during the motion. But the CM point has a downward vertical acceleration that increases without bound throughout the motion.

From the Second Law, Fnet, cm = Mtotacm, we conclude that during the time when the ladder is dropping, there is zero net horizontal force on the ladder, while the net vertical force on the ladder is downward, increasing without bound as the ladder reaches the ground.

So I think the main takeaway from physics is that no finite force is capable of making a ladder move this way.

To analyze the problem any further, I would want to specify the coupling between the ladder and the wall and floor. For instance, on the wall there could be a freely sliding ring, allowing point W to move vertically without resistance, but constraining point W so that it can't move horizontally. That version of the problem is considered in this article, again leading to the conclusion that the constraint forces cannot be finite. This had to happen, given the motion of the CM point during the stipulated motion.

***

The ladder problem reminds me of a class of problems treated by philosophers of science, called supertasks. A supertask is an infinite sequence of operations conducted in a finite amount of time. One question is whether supertasks ever happen, or are ever possible. Zeno's Paradox could be thought of as a question about whether motion itself is a supertask. You can read about supertasks on Wikipedia.

After Zeno, the most famous of a supertask is the example of Thomson's lamp; here is a paper about that paradox by John Earman and John D. Norton. I learned about some of Norton's work while writing my paper on the Law of Inertia and determinism in Newton's Laws.

Sunday, March 5, 2017

Points In A Circle Problem: Solution for N = 3

Let's take another look at the second-to-last figure from the previous post.


Given minimal obstacles X, Y and Z, we rotated and reflected the disk as necessary so that segment XY is horizontal and located in the upper half-disk. Then:

  • Obstacle Z can be located no higher than the lowest point of the small circle, otherwise there would exist a large feasible triangle in the part of the disk below Z.
  • Z cannot be located below the long chord ending at L, otherwise we could draw a large feasible triangle by sliding L downward.
  • Similarly, Z cannot be located below the long chord ending at M.

Altogether then, Z belongs to the (closed) grey triangular region, the bottom point of which is (0, −r0).

An interesting thing about this situation is that the location, shape, and size of the grey triangle is independent of the precise location of points X and Y—these points aren't even shown on the diagram. One could translate segment XY arbitrarily within the upper part of the disk, or lengthen or shorten segment XY, and it would remain the case that Z would have to belong to the grey triangle—because the grey triangle was defined by considerations about the lower half-disk. Of course, we almost certainly had to rotate and reflect the disk to make segment XY horizontal and up-top in the first place. So if you imagine not doing that reflection and rotation step, then when you construct the grey triangle based on the location of X and Y, it will end up being rotated or reflected from what is shown above.


Here's an animation showing all of the possible locations of the shaded triangle. (I'm shading the triangle green from here on out. Also, note that the dashed circle has radius r0.)


No matter how we orient the green triangle (i.e., no matter where X and Y are located), the green triangle can't escape the grey annulus, ½r0 ≤ r ≤ q, where q = 0.36606…. Therefore, no matter where X and Y are located, obstacle Z belongs to the grey annulus. 

But "Z" could be any of the three obstacles, so actually,

All three obstacles in a minimal triple must belong to the annulus ½r0rq.

This is a pretty severe constraint, especially when combined with the result of from the previous post (any two obstacles in a minimal triple must be at least r0Sqrt[3] apart).

***

Let's start again. Suppose {X, Y, Z} is a minimal set of obstacles, and rotate the disk if necessary so that segment XY is horizontal. Reflect if necessary so that segment XY is in the lower half-disk. (The switch from upper to lower is just so that I can use my existing codebase for generating diagrams.) And finally, for definiteness, reflect once more if necessary so that X is to the left of Y. Then with reference to the figure below, Z belongs to the green triangular region, and horizontal segment XY belongs to the part of the grey annulus located below chord DE.


Segment XY has to be at least r0Sqrt[3] long, which constrains its position in the annulus in several ways:

  • The segment can't be too close to the bottom of the annulus, because the width of the annulus decreases as you move towards the bottom.
  • The leftmost point can't be located very far to the right—there has to be a distance at least r0Sqrt[3] to the right-hand boundary of the annulus.
  • The rightmost point can't be located very far to the left—there has to be a distance at least r0Sqrt[3] to the left-hand boundary of the annulus.

Here's the real estate available for X and Y once we take into account these implications of the minimum separation requirement. Point X must belong to the blue region, and Point Y must belong to the red region.


On the same figure, lay down Segment PQ as shown:


Segment PQ represents an extreme case in the following sense:

(*) If a segment has one endpoint in the green triangle to the right of PQ, and passes tangent to the circle, then its other endpoint cannot be located in the blue region.

To see this, imagine choosing any point in the green triangle to the right of PQ, and from this point project a ray tangent to the circle: pretty clearly, this ray "flies too high" and misses the blue region entirely. This is because of the way the tangent to the circle "rocks" against the circle like a rocker-arm.


Thus, if a segment has one endpoint in the green region to the right of PQ, and another endpoint anywhere in the blue region, then the segment can't be tangent to the circle. Indeed such a segment enters the circle, i.e., it passes closer than ½r0 to the center of the disk.


This has implications for the location of point Z. If Z were located to the right of segment PQ, then segment ZX would necessarily pass closer than ½r0 to the center of the disk, and there would exist a large feasible edge-type triangle on the chord through ZX. (Feasible, because obstacle Y is safely away from this action.)

Conclusion: Z can't be located to the right of segment PQ. And by considering the red region analogously, Z can't be located to the left of segment RS either.


The allowable regions for X (blue), Y (red), and Z (green) are therefore as shown here:



***

Let's forget about the red and blue regions now. I was really only using them to help me clip off the annoying "wide lapels" of our initial green triangle. Having done that, we can now re-run the same annular arguments that we used for the initial green triangle.


Since the outermost point of the clipped green region is a distance r0 from the disk center, we can replace q by r0 in the reasoning and conclude this time that

All three obstacles in a minimal triple must belong to the annulus ½r0 ≤ r ≤ r0.

And with that, we're in the home stretch. In an annulus ½r0 ≤ r ≤ r0, the longest possible segment that doesn't pass closer than ½r0 to the center is a segment of length r0Sqrt[3] tangent to the inner circle.


This fact is probably standard, but at any rate I think it's not hard to see that the longest segment has to be a chord on the outer circle. (If either endpoint were in the interior, the segment could be lengthened; and if an endpoint were on the small circle, rotating the segment a bit about the other endpoint would free up some space to lengthen it again). And given that it's a chord, it's not hard to see that its length is maximized by pushing the chord toward the center until it's tangent to the inner circle.

What this means is that if all three pairs of obstacles must be at least r0Sqrt[3] apart, then all three pairs of obstacles must be exactly r0Sqrt[3] apart, i.e., triangle XYZ is isosceles, with its sides tangent to the circle of radius ½r0.

That is, {X, Y, Z} = {O1, O2, O3}.

***

Readers, please let me know if you see mistakes or lapses in the above! But assuming for now that it is basically right, we can add to our results in the Points In A Circle Problem.

Given N obstacle points X1, …, XN in the closed unit disk, let
aN(X1, …, XN)
denote the greatest possible area of a triangle in the disk that doesn't strictly contain any of the Xi. Let
aN* = minimum possible value of aN(X1, …, XN).
Then for each given value of N, our problem is to calculate aN* and find a minimal set of obstacles, that is, N points Xi* for which aN(X1*, …, XN*) = aN*. 

Known results so far:
  • a1* = 1 and a1(X) = 1 if and only if X = O, the center of the disk.
  • a2* = 1 and a2(X1, X2) = 1 if X1 = O or X2 = O.
  • a3* = 0.8286369… and a3(X1, X2, X3) = 0.8286369… if and only if {X1, X2, X3} = {O1, O2, O3} in some orientation.
Here, the points Oi are equally spaced around a circle of radius r0 = 0.320963…. As noted in an earlier post, the value of r0 is given exactly by the real root of the cubic polynomial 5x3 − 4x2 + 7x − 2, namely


and the value of a3* is given exactly by positive root of the polynomial 160000x6 − 265824x4 +  412857x2 − 209952, namely



***

Not all of the partial results from the past few months figured in the eventual solution to N = 3. Mostly, the argument relies on the tension between edge triangles and spotlight triangles. Calculating a3(O1, O2, O3) to begin with was actually harder, I thought, than showing a3(X, Y, Z) ≥ a3(O1, O2, O3) for all X, Y, Z. Needless to say, there could be much easier ways of solving the problem than the ways I developed, and perhaps there are theorems known to experts that would make quick work of the problem.

Part of me is disappointed that a careful investigation of N = 3 failed to turn up any better obstacles, or even any additional minimizers beyond the Oi originally conjectured. But I guess I'm even more surprised that an initial, intuitive choice of obstacles eventually held up to scrutiny.

Speaking of scrutiny, I have listed our "known results"—but in mathematics, nothing is really known until it's proved to professional standards of rigor. I'm not under the illusion that this series of blog posts meets that standard. I've been approaching the Points In A Circle problem recreationally, not professionally. Some of the arguments I've made are solid, but many others are just sketches that would need filling in. Journal peer review would certainly reveal gaps, if not errors, along the way. Nevertheless, I don't have any plans right now to clean up the analysis or organize it more linearly.

Also, I plan to keep my promise and leave the N = 4 problem to others!