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08 February 2006

Astrogation of the Joltcola System in Galaxy Dwingeloo-2


1. Introduction

Congratulations, Space Cadet! You have been accepted for training to become a Space-o-naut in the Vleeptron Space Academy!

If you successfully complete the Academy's rigorous training and pass all your examinations, you will soon be cruising through the Dwingeloo-2 Galaxy in a variety of remarkably fast, sexy spacecraft, and will also become a Frequent Flier on the Zeta Beam, hurtling back and forth between Earth and Vleeptron whenever the Zeta Beam isn't on the fritz. (Currently it's on the fritz.)

2. The Vleeptron Space Academy Honor Code

Do your own work, don't ask a smarter Cadet, don't ask your Mom or Dad, don't telephone Klaas in Rotterdam.

3. Our Sun

The name of our Sun is Joltcola. It is round, very hot and bright, and pretty large.

4. Planets which orbit Joltcola

The Planets which orbit Joltcola, from nearest to farthest from the Sun, are:

* Hoon

* Vleeptron

* Yobbo

* Linguica

* Bratwurst

* Cheddarwurst

* Banger

* Mollyringwald

* Nyu

5. The Frongo: Unit of distance measurement

Distance between things in the Joltcola System is measured in Frongoz, abbreviated Fz. It's still Frongoz whether you go in a straight line or a curved line or a wiggly line.

6. Sznqrüûüûp's First Law

The orbit of every planet is an Ellipse, with Joltcola at one Focus, and Nothing at the other. Don't go to the Empty Focus. There's nothing there.

EXAMINATION 1

The orbit of Planet Yobbo
is shown in Figure 1.

How far does Planet Yobbo
travel around Joltcola
in exactly one Yobboyear?

If you get the Right Answer, you win 3 slices of Pizza with your choice of toppings.

9 Comments:

Blogger Abbas Halai said...

LOL at Mollyringwald

10:40  
Blogger Vleeptron Dude said...

I didn't discover these planets, but the first time I took the Zeta Beam to Vleeptron, I was really surprised to find out they just didn't have names for their planets (except Vleeptron, Yobbo and Hoon). They saw that this seemed to upset me, so they let me name all the other planets.

There really is a new 10th planet in our Solar System discovered last year, and although it might not really be The Permanant Official Name, the astronomers who discovered it named it Xena. "I always wanted to name a planet Xena," the guy said, "I really love that show."

15:13  
Blogger Vleeptron Dude said...

guess where i went to college

15:17  
Anonymous Anonymous said...

What ?? Xena ?? Why not Serling ? Or Rupert ?
Or Kepler . Like in Kepler's Third Law. Or the Second. Can't remember now which one it was, but it was one of dem buggers allright. Right ?

18:46  
Blogger Vleeptron Dude said...

Kepler? Who's that bloke? Here on Vleeptron we know Sznqrüûüûp's Laws (see Section 6. above) but I never heard of any Kepler guy.

Here's the deal. You find a planet or a comet or a big-ass meteor, you get to give it any name you like. This guy out in California I think, he just loved watching Xena. Vleeptron had a post about the discovery with the usual photo of Xena in her usual outfit.

You got those Alps, so there must be some famous optical observatories in CH on the tops of the Alps, what are some famous Swiss optical observatories? France has got a famous big-ass one in Pic du Midi.

Oh wait! I think the very first "exoplanet" -- a planet circling some other faraway sun -- was discovered and verified by a Swiss optical telescope team. (I'm too lazy to Google right now.) That was about 8 years ago, and since then, astronomers have found a dozen more exoplanets. But the Helvetians were first!

19:45  
Blogger Vleeptron Dude said...

Discovery of Xena the new planet, also photo of Xena Warrior Princess:

http://vleeptron.blogspot.com/2005/07/10th-planet-found-in-our-solar-system.html

19:49  
Blogger Vleeptron Dude said...

All this interglobal chit-chat is just delightful and everything, but Bob notices nobody has answered the Yobbo Length of Elliptical Orbit Exam Problem yet. What will you all do when you wash out of the Vleeptron Space Academy?

btw regarding recent communique -- you better believe I am working overtime to fix the damn Zeta Beam! We all have to die sometime and somehow, but I absolutely refuse to die in a Cartoon Riot. When I get the Urge to Perish, I go visit my Polar Bear friends in Churchill, or I lean over the crater of the Montserrat volcano. I ain't dyin' over no Dansk cartoons.

Or, in the words of The Animals:

We got to get out of this place
If it's the last thing we ever do
We got to get out of this place
Girl there's a better place
For me and you

19:56  
Anonymous Anonymous said...

9999719.26522921 Fz.
hoo boy.
~ben n.

used allercalc and mostly openoffice sheet.
plz let me know if I got it right.

a = semimajor axis
149597887.500000000000

b = semiminor axis
149576999.826000000000

dist btwn ctr and focus
2499813.496133650000

e = eccentricity
0.016710219228

Phi = arcsin(e) = 0.016710997
.016710997000

E(e) = (I used allercalc here)
.016710996780

L
9,999,719.265229210000
Units are Fz. Or Frongoz


FORMULAS ONLY BELOW

L=4aE(e)
e=sqrt(1-(b^2)/(a^2))
e*a=dist btwn ctr and focus

E(e) is the complete elliptic integral of the second kind for the eccentricity e = sin φ (tables are usually given in terms of φ instead of e). E(0) = π/2, and E(1) = 1, corresponding to the limits of a circle, b = a, and a straight line, b = 0.



Modular angle = alpha,
arcsin(eccentricity)=alpha=phi
sin(alpha)=eccentricity=.016710997000
Eccentricity ^ 2 = (a2*a2-b2*b2)/(a2*a2)
Alpha = arcsin k
k = .016710219230

20:14  
Anonymous Anonymous said...

let me just come right out and say I'm not definitively solving this problem this time either, just sharing where I'm at.

how about Ramanujan's formula approximation of 939,885,629.4 Fz?

hmmm. solving the complete elliptic interval of the 2nd kind (ellipticE function) seems too tricky, in light of my current inability to comprehend the concept of the elliptic modulus or the modular angle.

939,885,629.2 Fz. is the eerily similar alternate answer I come up with when I use pi/2 instead of whatever phi really is in allercalc's ellipticE function. I've figured out that phi is supposed to refer to the angle at which a hypothetical circle is positioned relative to the x-y plane to make the present ellipse, but exactly calculating it is something my searches have not answered. It's not just the golden ratio, 1.61803399, right?

I can proudly say I didn't consult with anyone to get these answers, except of course google, wikipedia, answers.com, and probably half a dozen other sites.

22:51  

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