User:Cesrate/The New Future/Notes

This is the note page of story "The New Future", showing some references and background details. As for some terminologies, you can consult Terminology or http://www.wikipedia.org.

Albatross 23a

This mining ship is somewhat like the freighter Shuttle-A in Orbiter 2010 the space flight simulator.

Artificial Gravity

Using centrifugal effect to simulate gravitational force is the easiest and most possible way for future space flights.

Some large structures in scientific fictions may give people a wrong understanding that only very large ships can use centrifugal effect to simulate gravity. However, think about the centripetal acceleration, it's determined by

$a={\frac {v^{2}}{r}}$

Where $a$ is the centripetal acceleration, $v$ is the tangetial velocity, and $r$ is the radius of rotation.

The only thing you need is velocity and radius. That is to say, you just need two parts of ships and a rope.

When two parts are attached together and rotating, things inside the parts will receive acceleration - it's not because things receive some kind of force, but because your reference of frame receives acceleration. In orbits such an acceleration has little difference with the actual gravitational acceleration as long as the rope is long enough so there's no significant change of acceleration from a person's feet to head.

What if the masses of two parts are different? Well, the center of rotation will move closer to the heavier ships. If you want to put something to the center, the way to find it is given as follow:

Assume the parts are heavy enough and the rope is light enough so the mass of rope can be ignored.

According to Newton's third law, the forces of parts given by each other is equal in amount, negative in direction:

$F=-F$

And from Newton's second law, the acceleration is:

$a_{1}={\frac {F}{m_{1}}}$

$a_{2}={\frac {F}{m_{2}}}$

Where $a$ is acceleration and $m$ is the mass.

And as we are doing circular motion, the angular velocity $\omega$ is same, so according to

$v=r\omega$

We have

$a_{1}=\omega ^{2}r_{1}$

$a_{2}=\omega ^{2}r_{2}$

Where $r_{1}$ and $r_{2}$ are the radius of each part's rotation.

By now, there are

${\frac {F}{m_{1}}}=\omega ^{2}r_{1}$

${\frac {F}{m_{2}}}=\omega ^{2}r_{2}$

So we will get

${\frac {r_{1}}{r_{2}}}={\frac {m_{2}}{m_{1}}}$

You have already known what to do next.

"Auto RCS Adjust"

This means let computer adjust the ship to correct attitude before reentry by using reaction control system.

EV-2

You could think it as some kind of lunar vehicle.

Laythe Spaceport

Laythe Spaceport is the space center in Quark City, supporting this mission.

No. Laythe Spaceport is NOT related to Kerbal Spaceport in each aspect.

Low Laythe Orbit

Abbreviated as "LLO". It doesn't have a clear definition yet, but an orbit at 60km above Laythe sea level is surely LLO.

Nightingale Satellite

An ordinary surface scanning probe, now orbiting Vall on a polar orbit.

Nuclear Thermal Rocket

Abbreviated as "NTR". See http://en.wikipedia.org/wiki/Nuclear_Thermal_Rocket for more information.

Orbiting Vall

Orbiting just at 0 m (although impossible) will be 832.47 m/s and take 2264.29 s (0.6290 h, 37.7 min). For any altitude, there's a formular in Tutorial: Basic Orbiting (Math):

$v=R{\sqrt {\frac {g}{R+h}}}$

Where:

$g$ is the gravity at sea level.

$R$ is the radius of a planet (to sea level).

$h$ is the altitude of the orbit.

Picker Base

Picker Bases are similar to the Mars bases in Constellation Program.

Pegasus

Pegasus is a relatively small manned ships with docking port, parachutes, four landing legs and four liquid engines.

No. It's NOT related to Poseidon, Medusa, Marvel Comics, One Piece or My Little Pony: Friendship is Magic.

Reentry Plasma and Smoke

We can see plasma during reentry, then as the ship slows down followed by smoke from burnt heat shield. This experience is from Space Shuttle Mission 2007.

"Return to Kerbal Space Program"

The idea of "Return to Kerbal Space Program" was from Showchen on Deeptimes.org.

Spoiler: This is the truth! XD

This is Cesrate's translation in KSP Forum.

The Developers of KSP are actually Kerbals trapped on Earth, the navigation system of their ship was broken, so they had to use Earth's computing technology. While, unfortunately personal computers they can find on Earth were quiet awful, so they decided to set up a game-like distributed computation system, using many players' computer to draw the cosmic map.

And this evil plan was called "Return to Kerbal Space Program".

Rockomax 1X

Rockomax 1X is a fictional liquid fuel engine in this story. It's small and capable for ships with complex jobs because of its relatively high performance.

Temperature of Laythe

The temperature of Laythe is between -40.19℃ (232.96 K) to 6.21℃ (279.36 K). Kerbals are advised to wear thermal clothes in the open air.

Time Units

The "Day", "Month", "Year" and "Century" in this story are Kerbin's units shown as follow, not Earth's.

	Earth	Kerbin
Day	23 h 56 m 4.100 s	6 h 0 m 0.000 s
Month	27 d 7 h 43 min 11.5 s	1 d 14 h 36 m 24.4 s
Year	365.2564 d	106.5225 d
Century	36525.64 d	10652.25 d

Transfer from Laythe to Vall

How to estimate time costing of a transfer orbit roughly? Well, here's a simple method, but it could only be used for rough estimation because the actual cases are usually more complex.

Think if we are orbiting Jool at 27,184,000 m (to Jool's barycenter), Laythe's orbit, and after a buring our apoapsis reaches 43,152,000 m (to Jool's barycenter), Vall's orbit, it's easy to know the orbit period because if only the central body isn't changed, ${\frac {a^{3}}{T^{2}}}$ will be a constant according to Kepler's third law, where $a$ is the semi-major axis of orbit and $T$ is the orbit period. What's more, according to Newton's law, we have

${\frac {a^{3}}{T^{2}}}={\frac {\mu }{4\pi ^{2}}}$

Where:

$\mu$ is the standard gravitational parameter, for Jool it's 2.8252800×10¹⁴ m³/s².

So, now we can calculate the semi-major axis by

$a={\frac {Ap+Pe}{2}}$

Where:

$Ap$ is the distance of apoapsis from barycenter, now is 43,152,000 m.

$Pe$ is the distance of periapsis from barycenter, now is 27,184,000 m.

And $a$ now should be 35,468,000 m.

So, now we can put the numbers in and get $T=7.7959\times 10^{4}s$ .

Next, as we want to know the time of going to there, the time should be half of $T$ as the ellipse is axisymmetric. So we get

${\frac {T}{2}}=3.8980\times 10^{4}s$ .

This time is about 10.8 h. But remember this is just a rough approximation, in real cases your trajectory won't just like a perfect ellipse when transfering between celestial bodies.

Variable Specific Impulse Magnetoplasma Rocket

Abbreviated as "VASIMR". See http://en.wikipedia.org/wiki/Variable_Specific_Impulse_Magnetoplasma_Rocket for more information.