I'm not a big fan of flying for a number of reasons. Before 9/11 and the tighter security restrictions, though, I'd finally gotten to being able to pack One True Bag for trips of up to a week. My One True Bag contained a week's worth of clothes and a toiletry kit. After 9/11, I started checking my bag and going on the plane with nothing but a book and a jacket. Of course this isn't true of ski trips or other gear-requiring travel.
The airlines have screwed it up, though. They've started charging to check even the first bag. If I don't have to spend money on something, I won't, so now I try to limit myself to one checked bag, containing my toiletries. It occurred to me yesterday that that's actually kind of dumb. I typically only travel with tooth paste and shaving gel. Those are $5 items, so if I bought a new tube of toothpaste and a new can of shaving gel every time I flew, I'd come out $5 ahead on a $15 bag fee. Of course, I usually fly round trips. If I leave my toothpaste and shaving gel at home, rather than getting them confiscated or something, I only buy them once per two flights, and come out $20 ahead. My mother made a "tooth paste in every port" joke that points out that a lot of the time, we don't fly to a new place each time. So realistically, I don't even have to re-buy those products on every trip.
The point of all that being that I can save $15/flight if I stop checking my bag. The carryon policy is open to a lot of abuse - you can easily roll up to the jetway with a bag that's a little bigger than the specifications, and there's no specific size given for the "personal" item. I sometimes carry my messenger bag, which has a capacity of over 2000in3. Between my ginormous messenger bag or a large backpack and my rolling bag, I'm well over the amount of volume that each passenger can carry on the plane and fit into the overhead bins. And the airlines are essentially paying people $15/head to do this on every flight.
I know that planes are sometimes delayed because people have trouble fitting all their luggage, have to check things from the jetway at the last minute, etc. And airports have some pretty restrictive rules about when planes can be on the runways to keep them from getting to close to each other, so those delays can snowball into bigger delays if the planes miss their spots or the departure times get bunched. Schedules are pretty tight, so those delays can send ripples throughout the system and cause more delays, or force the airlines to program more slack into their schedules and get less travel out of each plane. I think that all of that has to be costing money. I want to know (actually I don't care, it's more rhetorical) if the $15 that some rules-following people are paying for their first bag is enough to cover the cost of that stupidity. I suspect it would make more sense to encourage people to check their bags rather than carrying them on and having a melee in the cabin about who gets bin space.
Anyway, now that I've thought more about the issue of re-buying toiletries on every trip, or finding some 3oz containers to keep for travel only, I'm going to be one more person making it harder for the plane to take off on time with excess carryon baggage. And the airlines are going to be paying me to do that.
Tuesday, November 24, 2009
In my last post, I alluded to the trigonometric functions related to a hexagon being something one could derive easily, but I didn't actually do it. I also alluded to being able to keep slicing up the triangles forever, but didn't say why.
I'm not going to prove the Pythagorean theorem here. If you don't remember it, here it is:
A2 + B2 = C2
So why did I choose a hexagon for my shape? There are a number of triangles with relationships that are relatively easy to describe mathematically. Two of them are the equilateral triangle and the right isosceles triangle.
The first triangle is a right isosceles triangle. By definition, the angle united the two equal sides is a right angle. If the length of the equal sides is 1, applying the Pythagorean theorem tells me that the length of the hypotenuse is √2. I considered using a square as my initial shape, but I suspected it would make it harder for me to cut up the angles later. I think it would actually be almost as easy, but it doesn't really matter.
The second triangle is an equilateral triangle. Equilateral triangles are cool because if I draw a line from the middle of one face to the opposite corner, I know that I've cut the opposite corner in half. I can then use the Pythagorean theorem to find the length of the new line. Now I know several important things about the triangle.
These numbers are important because the angle θ/2, created by splitting an equilateral triangle, is the smallest one I know of that has trigonometric functions that can be worked out geometrically. After that, as far as I'm concerned, the trigonometric functions are buttons on my calculator. However, the calculator had to get its table somewhere, and creating such tables was the life's work of a few ancient mathematicians, to varying degrees of accuracy. They were able to do it because many of the formulas related to adding and subtracting and slicing up angles can be proven geometrically, and then applied to whatever number and measurement system seems like a good idea at the time. The formulas I'm interested in are the half-angle formulas.
sin2 x = (1 - cos 2x)/2
cos2 x = (1 + cos 2x)/2
What these formulas mean is that if I know the cosine of an angle, I can calculate the sine and cosine of half of that angle. If I can calculate the sine and cosine, I can also calculate the tangent. And I can do it again and again. That's important because the way my last post achieves higher degrees of accuracy in calculating pi is by inscribing polygons with higher and higher numbers of faces into the circle. The addition and subtraction formulas mean that it could be done with an arbitrary number of faces, but I think that the most reasonable choice is 6*2n with n being an integer. The half-angle formula has to be applied n times to give the correct cosine.
I hoped, when I figured that out, that there'd be some totally cool way to speed up the calculation. I don't think there is. If n=3, the polygon has 6*23=48 faces, the cosine of the angle at the apex of the 96 skinny right triangles is
cos(360°/96) = √(1+√(2 + √(2 + √(2 + √3)))/2)
Multiply that be twice the number of faces and you have your perimeter. For each iteration, the half-angle formula is applied again to that increasingly ugly number. No wonder Archimedes only went to 96 faces in his calculation. To get a little better than the level of accuracy of the 10,000-sided figure I suggest at the end of my post, it would take n=11, which is a lot of finding roots for anyone but a computer. To make it additionally difficult, the roots aren't rational numbers, so the computer would have to decide just how accurate an approximation he really needed; 11 approximations later, the number will have drifted pretty far from being exact.
To return to the point of my last post, if this were being applied to calculating pi, a person could get a little more accuracy each time he sliced the angle in half, approximated the trigonometric functions for the new angle, and re-calculated. The formula could even be restated like this:
6*2nsin(30/2n) < π < 6*2ntan(30/2n)
Monday, November 23, 2009
I'm planning to go back to school in January to take prerequisite classes enabling me to apply to engineering programs. In order to save time and money, I'm trying to avoid taking first-year calculus again, so I've been re-teaching myself math. This time around, I've been finding myself fascinated with certain things - why is the Pythagorean theorem true? How did people figure out pi?
In reading about pi, I read about a formula to calculate it used in some of the earlier computer approximations. Pi is the ratio of a circle's circumference to its diameter.
A circle can be thought of as a decayed polygon. As the number of sides of the polygon approaches infinity, the polygon approaches a circle and its perimeter approaches C. So the problem of calculating pi can be solved by calculating the perimeter of a polygon with lots and lots of sides, until either pi stops changing (this never happens - it's an irrational number) or one runs out of computing resources, gets bored, or decides that it's accurate enough. This is, essentially, Archimedes' approach - he calculated the perimeter of a 96-sided polygon and decided that was close enough. It's accurate to three places, which isn't a ton by today's standards, but it was the best anyone in Europe did for a very long time.
The approach that really got me thinking, though, was to place the value of pi between a known larger and known smaller value, then compare digits yielded by the two formulae. The value of pi is accurate as long as the corresponding digits agree.
"too small" formula < π < "too big" formula
That leaves deriving the formulae. It's relatively easy to find out the perimeter of a regular hexagon...
I know the angle a is 60 degrees because there are six triangles in the circle and 360 degrees is the total measurement they have to take up if they use all available real estate. Since it's a regular hexagon, they have to match each other. I can set the radius of the circle to be anything I want, so I'm going to make it 1. That means Rs, the radius of the smaller hexagon, is also 1. The ancients actually used a trigonometric function no longer in use, "chord" abbreviated "crd" quite frequently. Chord refers to the measurement of the chord of an arc with a given angle, so
Ps = crd(a)
Now all I have to do is multiply by six and I have the perimeter of the internal hexagon. It's not very close to pi, but it doesn't need to be yet. If I infinity was a number, I could set the number of triangles to infinity, calculate
∞Ps/d = π
and call it a day. Unfortunately, infinity's not a number. The other big problem I had was that I assumed that the trigonometric functions relied on knowledge of pi in their calculation, so I was having a chicken-and-egg issue in calculating the perimeters of my polygons.
I found this article on the chord function, though, that describes the calculation of a chord table to an arbitrary degree of accuracy based on special triangles and formulae for cutting angles into smaller and smaller pieces. Interestingly, Ptolemy uses a circle with radius 120 and measures the angle in 360 degrees, likely a throwback to the belief that the circumference of a circle is three times the diameter. It just goes to show that units of measurement are pretty arbitrary.
For my purposes, we'll say that the radius of the circle is 1, and we don't know the circumference. I'm going to stick with degrees as a unit of measurement because while I like radians better, they use pi in their definition and if I already knew pi, this whole exercise would be even more pointless. Then the perimeter of an inscribed polygon with n sides is
π ≈ n*crd(360/n)/2
It gets more equal as n approaches infinity:
π = limn→∞n*crd(360/n)/2
The problem is that infinity's not a number, which comes back to why I'm also interested in the perimeter of the outside polygon. Since I know the angle - it's whatever I want it to be - and I know the altitude of the triangle, which is the radius of the circle and which I've set to equal one, I can calculate the length of the base of the triangle, which is the length of one face of the polygon, as
Pl = 2tan(a/2)
In a polygon with n sides, the perimeter is
n*2tan(360/2n) = n*2tan(180/n)
So here's another way to calculate pi - the numbers will start too big, and approach pi as n approaches infinity.
π = limn→∞n*tan(180/n)
The twos cancel each other, so the equation's actually rather pretty.
What this all means in more practical terms, in a world in which we can't just throw infinity around and act like it's a calculable number is that
n*crd(360/n)/2 < π < n*tan(180/n)
n can be set to whatever number your calculator, computer, or patience with slicing angles in a trigonometric table can put up with, and you get a pretty good approximation of pi. For those whose calculators don't happen to have a chord function,
crd(θ) = 2sin(θ/2)
So the above formula for pi can be restated as
n*sin(180/n) < π < n*tan(180/n)
On my calculator, n must be set to 10,000 to get an accuracy of six places for either formula. If you really want to make your graphing device chug, try graphing these and watching them converge.
Here's the results using some different values of n
For reference, pi to seven places, from Wikipedia, is 3.1415927 - so the value of n of 10,000 is accurate to all the places I went to which should be adequate to prove my point.
Saturday, November 21, 2009
I learned a couple of things about training last season. One is that it's important. Another was that ultimately, all tracking methods are somewhat arbitrary. However, the information yielded by a cycle computer is pretty useless, especially when training on the road for off-road riding, or trying to combine on- and off-road training numbers. A mile of mountain biking is a lot harder than a mile of road riding, and road riding can be very different depending on whether I'm riding to commute or somewhere on the East Side on a long stretch of uninterrupted road. An hour on the bike at an easy pace is an hour on the bike at an easy pace no matter which one I happen to be using that day, where I am, or why. A cynic might point out that I also lost my cycle computer early last season, so in a sense I didn't have a choice about going to a watch and my own rate of perceived exertion as my only training tools. Oh well.
Anyway, I think riding my bike a lot last winter was a big part of my success in the early part of last season. So I'm going to try to do it again. I'm also going to try to follow a schedule, because I think that ramping up over time was a big part of how I did the season before last, when I went from having to severely limit my ride time to being able to do my first Century. I'm also trying to talk my friends into riding bikes with me because while these rides may not be catatonically slow, they're not going to be speed work either. Also because it's always more fun to ride with friends and when the weather sucks, that's extra-important.
So without further ado, here are my time goals for the next few months. I'm planning to do my first race on February 21st, after which I'll decide what I need to work on between then and my first 'A' race, probably in mid-April.
|Week Ending||Goal (Hours)|
I'm going to be giving myself credit for time spent commuting - basically, any time spent on a bike I'll count as time spent on a bike (go figure.) I don't have a great sense of the times I spent last winter. My records weren't great and I was counting miles, not hours, which started to annoy me pretty quickly when it came to riding around town. I think that this schedule is going to be a little more demanding at the end, but then I'm racing in a harder category than I expected to when I was training last year.
Saturday, November 14, 2009
Tomorrow is the last race I'm planning to enter this season. After that, I don't race again until the BuDu Racing series starts near the end of next February. I'd like to keep riding my bike between now and then, but it gets a lot harder when it's dark and nasty out. So if I can rope my friends into it, I want to do some regular rides between now and then. If you're interested in riding bikes with me (and are near enough to me for it to be convenient to meet somewhere in Seattle, like Green Lake or Gasworks Park) let me know, and also what kind of format you'd be interested in.
My thought is probably 20+ mile road rides, starting at 11am on Sundays. For those who are into structured training, this would be a long or recovery ride. However, I'm not set on anything yet. Mostly I just want to figure out a format and schedule that works for at least a couple of my friends so we can all keep active through the winter when it's tempting to hide inside. Blog comments would be an excellent way to give me some feedback since then if anyone else is interested, they can see them too.