Road construction; geometric construction
There's construction on the road that I usually take to the highway. It's been sort of off-and-on for a few weeks now, and today it is on. The road is down to one lane open, so a pair of cops at either end keep the traffic moving here in alternating directions. Note that I use the word "moving" here very loosely, since while one side is moving, the other side clearly can't.
So as I approached that road from my side road, I saw the cop directing traffic, and it was going exactly the direction that I wasn't. No problem, he should reverse it any time now. Right? Right, Mr. Officer? Apparently not. Five minutes later, I gave up and turned around. It wasn't a major problem: there's another exit from my subdivision that should take me out to the main road just past the construction.
A twisty maze of roads, all alike. That's my subdivision. I've just about mastered this maze, though, so I know a handful of exits. I easily found my way to my usual backup -- the one that should be just past the construction. Unfortunately, it seems today it is in fact not past the construction. Instead, it was blocked by cones and behind them a large bulldozer. So much for my valiant escape.
Fortunately there's yet another exit, just this side of the cop and construction. I made my way around to that one and discovered traffic actually flowing my way! I was ecstatic. I was at a red light, but as soon as that turned green, I turned with the traffic and headed toward the cop. The cop stopped traffic. Right in front of me. I was the first person not allowed to continue. I was rather annoyed.
Fortunately, this twisty maze of roads has yet more exits. I turned around to pursue another. Unfortunately the next exit required me going way out of my way. Much, much annoyance.
Anyway, I did finally get out of that maze and into the gas station for some much-needed carfood. And while I was there, I saw a Best Buy truck. The truck in and of itself wasn't very interesting. Honestly, the Best Buy Bright Yellow Tag logo isn't very interesting either, but it did get me thinking. The roundy end is in the shape of some sides of an octagon. I found myself wondering, given a squared-off Bright Yellow Tag of a give width w, how could we construct the familiar angly logo? What would the side length be? Yes, this is the kind of thing I randomly think about.
So imagining a correctly cropped tag, I realized that the end could be decomposed into two right isosceles triangles -- one on each side -- with a rectangle in the middle. If the final octagon side length is s, the triangles would have a hypotenuse length s with legs we'll call s'. The middle rectangle, then, would have sides s and s'. Handy. The three shaped together -- one leg of each triangle plus the long end of the rectangle -- add up to the width of the tag, so 2s' + s = w.
Because the triangles are right isosceles, and since we've named the leg and hypotenuse lengths s and s' respectively, Pythagoras tells us that s'^2 + s'^2 = s^2. That simplifies to sqrt(2)*s' = s or s' = s * sqrt(2) / 2. Plugging this s' into the width equation, we have 2 * (s * sqrt(2) / 2) + s = w, which simplifies to s * (sqrt(2) + 1) = w or s = w * (sqrt(2) - 1). Plugging back into Pythagoras, s' = s * sqrt(2) / 2 = w * (sqrt(2) - 1) * sqrt(2) / 2, which simplifies to s' = w * (1 - sqrt(2)/ 2). That's pretty easy to measure out.
But we don't need to measure it out, do we? It seems a simple enough number. We should be able to construct it the way the ancient geometers did, with a compass and straightedge! Constructing s from w should be pretty simple -- it's just a matter of constructing sqrt(2) - 1 from 1. Easy. sqrt(2) is the hypotenuse of a right isosceles triangle with leg length 1, and 1 is... well... 1. With a little mental fiddling in traffic, I realized this is even easier than I first thought.
Here's how you construct it: You've got a tag of width w, squared at the end, and for our purposes infinitely long. We'll call the endpoints A and B. Use the distance between A and B to set your compass to width w. Draw an arc centered on A, beginning at B, and reaching down across the tape to the side. This arc reaches the tape at distance w along its length. Call this point C. Turn the compass around and draw a similar arc centered on B, beginning at A, and reaching down across the tape to the side at a point we'll call D. Our four points, A, B, C, and D form a square. Construct a line from B down across the square to C. Since ABCD is a square, ABC and BCD are right isosceles triangles. Pythagoras tells us that since AB and AC have length w, BC mush have length w * sqrt(2). Interestingly, it also crosses arc AD. Call their meeting point E. Now, all points on arc AD have distance w from B, so BE must have length w. CE, then, must have length w * sqrt(2) - w, or w * (sqrt(2) - 1). You'll recognize this as our s above. Construct line AD similarly to line BC. Find the point where line AD crosses arc BC and label it F. DF, like CE, has length s = w * (1 - sqrt(2)). It can be shown geometrically that line EF also has length s. If anyone's still reading, consider it an exercise. And thus we have our nice, pretty, even tag end in lines CE, EF, FD, and down the tag.
That pretty much got me to work. For some reason, just after that, my brain launched into a decision to dig into computational linguistics and particularly phone/phoneme/word recognition in audio. There's more behind that thought, but that'll have to wait for another post.
(LJ Spellchecker Genius of the Day: carfood -> carroty)
So as I approached that road from my side road, I saw the cop directing traffic, and it was going exactly the direction that I wasn't. No problem, he should reverse it any time now. Right? Right, Mr. Officer? Apparently not. Five minutes later, I gave up and turned around. It wasn't a major problem: there's another exit from my subdivision that should take me out to the main road just past the construction.
A twisty maze of roads, all alike. That's my subdivision. I've just about mastered this maze, though, so I know a handful of exits. I easily found my way to my usual backup -- the one that should be just past the construction. Unfortunately, it seems today it is in fact not past the construction. Instead, it was blocked by cones and behind them a large bulldozer. So much for my valiant escape.
Fortunately there's yet another exit, just this side of the cop and construction. I made my way around to that one and discovered traffic actually flowing my way! I was ecstatic. I was at a red light, but as soon as that turned green, I turned with the traffic and headed toward the cop. The cop stopped traffic. Right in front of me. I was the first person not allowed to continue. I was rather annoyed.
Fortunately, this twisty maze of roads has yet more exits. I turned around to pursue another. Unfortunately the next exit required me going way out of my way. Much, much annoyance.
Anyway, I did finally get out of that maze and into the gas station for some much-needed carfood. And while I was there, I saw a Best Buy truck. The truck in and of itself wasn't very interesting. Honestly, the Best Buy Bright Yellow Tag logo isn't very interesting either, but it did get me thinking. The roundy end is in the shape of some sides of an octagon. I found myself wondering, given a squared-off Bright Yellow Tag of a give width w, how could we construct the familiar angly logo? What would the side length be? Yes, this is the kind of thing I randomly think about.
So imagining a correctly cropped tag, I realized that the end could be decomposed into two right isosceles triangles -- one on each side -- with a rectangle in the middle. If the final octagon side length is s, the triangles would have a hypotenuse length s with legs we'll call s'. The middle rectangle, then, would have sides s and s'. Handy. The three shaped together -- one leg of each triangle plus the long end of the rectangle -- add up to the width of the tag, so 2s' + s = w.
Because the triangles are right isosceles, and since we've named the leg and hypotenuse lengths s and s' respectively, Pythagoras tells us that s'^2 + s'^2 = s^2. That simplifies to sqrt(2)*s' = s or s' = s * sqrt(2) / 2. Plugging this s' into the width equation, we have 2 * (s * sqrt(2) / 2) + s = w, which simplifies to s * (sqrt(2) + 1) = w or s = w * (sqrt(2) - 1). Plugging back into Pythagoras, s' = s * sqrt(2) / 2 = w * (sqrt(2) - 1) * sqrt(2) / 2, which simplifies to s' = w * (1 - sqrt(2)/ 2). That's pretty easy to measure out.
But we don't need to measure it out, do we? It seems a simple enough number. We should be able to construct it the way the ancient geometers did, with a compass and straightedge! Constructing s from w should be pretty simple -- it's just a matter of constructing sqrt(2) - 1 from 1. Easy. sqrt(2) is the hypotenuse of a right isosceles triangle with leg length 1, and 1 is... well... 1. With a little mental fiddling in traffic, I realized this is even easier than I first thought.
Here's how you construct it: You've got a tag of width w, squared at the end, and for our purposes infinitely long. We'll call the endpoints A and B. Use the distance between A and B to set your compass to width w. Draw an arc centered on A, beginning at B, and reaching down across the tape to the side. This arc reaches the tape at distance w along its length. Call this point C. Turn the compass around and draw a similar arc centered on B, beginning at A, and reaching down across the tape to the side at a point we'll call D. Our four points, A, B, C, and D form a square. Construct a line from B down across the square to C. Since ABCD is a square, ABC and BCD are right isosceles triangles. Pythagoras tells us that since AB and AC have length w, BC mush have length w * sqrt(2). Interestingly, it also crosses arc AD. Call their meeting point E. Now, all points on arc AD have distance w from B, so BE must have length w. CE, then, must have length w * sqrt(2) - w, or w * (sqrt(2) - 1). You'll recognize this as our s above. Construct line AD similarly to line BC. Find the point where line AD crosses arc BC and label it F. DF, like CE, has length s = w * (1 - sqrt(2)). It can be shown geometrically that line EF also has length s. If anyone's still reading, consider it an exercise. And thus we have our nice, pretty, even tag end in lines CE, EF, FD, and down the tag.
That pretty much got me to work. For some reason, just after that, my brain launched into a decision to dig into computational linguistics and particularly phone/phoneme/word recognition in audio. There's more behind that thought, but that'll have to wait for another post.
(LJ Spellchecker Genius of the Day: carfood -> carroty)