Difference between revisions of "User:Cesrate/The New Future/Notes"

Revision as of 01:54, 4 June 2013

Some contents on this page are humorous. They should not be taken at face value.

This page includes some mathematics.

This is still a stub.

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 combined cargo ship 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

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

Where ${\displaystyle a}$ is the centripetal acceleration, ${\displaystyle v}$ is the tangetial velocity, and ${\displaystyle r}$ 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:

${\displaystyle F=-F}$

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

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

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

Where ${\displaystyle a}$ is acceleration and ${\displaystyle m}$ is the mass.

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

${\displaystyle v=r\omega }$

We have

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

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

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

By now, there are

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

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

So we will get

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

You have already known how to do.

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):

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

Where:

${\displaystyle g}$ is the gravity at sea level.

${\displaystyle R}$ is the radius of a planet (to sea level).

${\displaystyle h}$ 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 parachutes, lander lags and four liquid engines.

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

"Return to Kerbal Space Program"

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

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.

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, ${\displaystyle {\frac {a^{3}}{T^{2}}}}$ will be a constant according to Kepler's third law, where ${\displaystyle a}$ is the semi-major axis of orbit and ${\displaystyle T}$ is the orbit period. What's more, according to Newton's law, we have

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

Where:

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

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

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

Where:

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

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

And ${\displaystyle a}$ now should be 35,468,000 m.

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

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

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

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.