10001110100110101
Geez, I tried playing Counter-Strike last night and I had horrible ping times. This was even late last night! I have no idea why my connection here is so bad, but I really have to start looking into it and complain (if need be).
It seems as if putting the computer into the DMZ didn't work. I know that port forwarding would be fruitless, and I was so sure that the DMZ trick would be able to lick the problem. Hmm. I'm at a loss right now. I'm going to check the Linksys site, but I'm not too hopeful at finding anything..
In other news, Laz and I came up with a solution to the problem I posed a couple days ago. The answer is one diameter, or two radii. The solution is fairly simple, if you know how to determine the surface area of a sphere by taking the integral of:
r2 sin q df dq
r is the radius of the sphere, (and since it's constant, there's no integration over r.) f is the angle about the axis that connects the centre of the sphere with the location of the observer and goes from 0 to 2p, while q is the angle from the same axis so that a vector with q=0 means that the vector is pointing towards the observer, while q=p points away from the observer.
Now, this will be difficult to explain without a diagram, especially since I'm talking, but since the observer has basically a circularly symmetric view of the planet's surface, we can integrate over the entire range of f, giving us:
2pr2 sin q dq
integrated from 0 to the angle of q such that the integral is equal to one third of the total surface area of the sphere (one third of 4pr2 or 4pr2/3). This is a simple task that gives us a solution for q:
cos q = 1/3
Which can be solved by throwing it into your calculator, but can be used directly if you connect the centre of the sphere, the observer, and the tangent point which is the farthest part of the planet's surface that the observer can see into a right angled triangle. The angle at the centre of the sphere is q, and using simple trig, we know that cos q is equal to the length of the side of the triangle adjacent to the angle divided by the length of the hypotenuse. Clearly, the length of the adjacent side of the triangle is equal to the radius of the planet, which means that the observer is situated at a distance of 3r from the centre of the planet. Thus, the observer is in fact 2r, or the diameter of the planet, from the surface. QED.
I should come up with a cleaner and clearer proof than this (like adding a diagram), but I'm not sure if I have the time, or the gumption to do so.
In any case, I might put up a new problem later tonight.
While I was cleaning out some of my books that were lying on the floor, I came across this "Micro Adventure" book that I must've picked up ages ago. It's a story where you're a secret agent by the code name Orion working for ACT. The story is aimed at adolescents/early teens (from the language) and is littered with snippets of code which you actually program into a computer (in BASIC). The code is an integral part of the story, although you can figure out what's going on without it. It's clearly dated by the fact that the code's written for the Apple II+ or Apple IIe (although you can modify the program to work for the Atari 400/800, IBM PC/PCjr, Commodore 64, VIC-20, and Radio ShackNews Color Computer). Geez, what a geekbook. *picks up book and starts reading*
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Monday, September 11, 2000 at 18:50:56 (UTC)
Try some traceroutes to these servers. You may find that the servers are physically close to you, but network wise, are quite removed. At one point, all traffic on the @Home cable modems was going down into the states somewhere, as that was their only peering point. You might be in a similar situation.
me