10001110100110101
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.
Monday, September 11, 2000 at 23:04:17 (UTC)
Sheesh. "...I'm not sure I have the time...". Judging by the fact that you solve such pointless problems as a hobby - I'd say not having 'enough time' on your hands is a total farce, I'd say you have TOO much time on your hands.
In other news, Homeworld: Cataclysm is inbound.
In more other news, Deux Ex is imbound.
That is all.
*shakes head*
Tuesday, September 12, 2000 at 05:05:12 (UTC)
Hmm.. it seems as if Werdna is also getting moderately large ping times from his connection, although I think he's using a dialup, and if he's getting ping times like me, something is definitely wrong with my connection! (Especially since Laz tends to get a fraction of the ping I get.)
Ah yes, with this whole time business. It seems I may have erred in my "not having enough time" comment. However, I'll stick with my gumption defence.
QYV
Tuesday, September 12, 2000 at 05:13:31 (UTC)
Re: Inbound
Eggcellant. I will also be inbound.. *brings a couple of blank CD-Rs*
What's the ETA?
QYV