[FT] Orbit and FT

[quoted original message omitted]

> Brian Bell Wrote:

> Actually I've used the gravity mechanic for vector movement that others

I like this method, but I do have a couple of questions. First off, I always
get Geostationary and Geosynchronous mixed up. Do they mean the same thing? If
not which one means an orbit that keeps you directly over a fixed
point on the planet surface (i know, it was mentioned 2-3 posts ago, but
I already deleted it)? That's the one I mean, and I'll refer to it as GS for
the rest of my comments.

How would you simulate GS orbit using those rules? Do you have each band
around the planet, instead of having one fixed velocity for orbit, have a
range of allowable velocities, with GS somewhere within that range, or do you
have one band at which the orbital velocity matches the velocity necessary to
maintain GS? Or is there some direct relationship I'm forgetting between a
planets rotational speed and the orbital speed required by it's gravity (being
as they are both affected by its mass)?

2B^2

Brian Bell said:
> Using cinematic, you need to need to plan a hexigon that is larger than

First, decide how long a turn is, how much acceleration gravity
provides, and the distance of a mu (eg 7.5 minutes, 1 gee = 1 mu/turn, 1
mu = 1000km). Then calculate gravity for various distances. Oops! Since
anywhere significantly above the surface of the Earth is going to have less
than 1gee, this leads us to fractional mu. Let's go back and make double the
turn length to 15 minutes, so now 1 gee = 4 mu. Calculating the gravity bands
is left as an exercise for the student. Determine orbital circumference at
various orbits, and divide by 12 turns. Let gravity give you a 1 point change
of direction every turn.

> Laserlight wrote:

> Determine orbital circumference at various orbits, and divide by 12

> turns. Let gravity give you a 1 point change of direction every turn.

Also, not only do you need to determine orbital velocities, don't you have to
take into account orbit decay and decide how often the ship has to expend
some thrust to keep up orbital velocity?  Or is it such a long-term
problem as to be negligible in game terms?

2B^2

Geo-synchronous means you are over the same spot of the planet all the
time. To achieve this you have to in the orbit radius that gives an orbit time
exactly the same as the length of day of the point on the planet below. So GS
will depend on the rotation time of the planet (this gives you the orbit time
required), the size of the planet (this will give the gravity factor to
determine what orbit to be in). Note
that there are certain situations where GS is impossible - large planets
with slow rotations could generate orbits within the planet. Also GS can only
easily be done across the equator; orbits towards the poles are more unstable
since the orbit is no longer in line with the gravitational pull. High tech
ships would be able to pull it off since
they can continously apply thrust to counter-act the offset pull of
gravity.

Mass of the ship doesn't matter much. Mass of the planet determines the amount
of pull gravity has and that determines how fast the orbital speed will be.

One example to help visualize gravity and orbits. Take a perfectly smooth
sphere the size and mass of earth with no atmosphere. If I fire
a bullet horizontally from an altitude of 2 meters at 100 m/s, it will
hit the ground about 60 meters away as gravity pulls it down. However,
if I fire it at about 8,000 m/s the ground falls away, due to curvature,
at the same rate as the bullet falls - thus "free fall."  Effectively
the bullet is falling at the same rate as the earth is curving away from it.

As you get farther from the center of mass, the pull of gravity becomes less.
This is offset slightly by the fact that the curvature of your orbit is less
as you get further out. But curvature increases linearly and gravity drops off
by the square of the distance so as you get farther and farther away from the
center of gravity of the planet the less "falling" you have to do to never hit
the ground.

To paraphrase the Hitchhiker's Guide to the Galaxy - the art of flying
is throwing yourself at the ground and missing.

--Binhan

> -----Original Message-----

It's only a problem at low orbits, where you still have wisps of atmosphere.
Sky Lab came down because it was low enough to have a slight drag and it ran
out of fuel for the thrusters. It was thought uneconomical to try to push that
much mass into a higher orbit. When you get a few thousand miles out, there
isn't much atmosphere to deal with so it would be negligible.

--Binhan

> -----Original Message-----

> B Lin wrote:

> Geo-synchronous means you are over the same spot of the planet all the

That answers my first question.

To achieve this you have to in the orbit radius that gives an orbit time

exactly the same as the length of day of the point on the planet below.

And that answers the second question.

*SNIP*

> Mass of the ship doesn't matter much. Mass of the planet determines

This I knew, notice I never mention ship mass.

> One example to help visualize gravity and orbits. *SNIP*

All of this I was aware of as well. I was merely wondering how
Geosynchronous was achieved - apparently from what you said it is
dependent on what diostance you orbit at. That's all I really needed. Thanks.

2B^2

> From: "B Lin" <lin@rxkinetix.com>

> It's only a problem at low orbits, where you still have wisps of

So maybe the lowest orbital band in the game would have this facotered in, but
no others.

2B^2

From: B Lin lin@rxkinetix.com
> Geo-synchronous means you are over the same spot of the planet all the

To be picky, geosynchronous is orbiting with the same time period as
spot below, but may vary north/south if in a different plane than the
equator.  Geostationary is geosynch with no north/south movement, and
IIRC is only possible with points on the equator (assuming an unpowered
orbiter).

Of course, if you have enough thrust, you can be geostationary anywhere you
like.

laserlight

> To be picky, geosynchronous is orbiting with the same time period as

THIS is why I love this list. Thanks, just what the doctor ordered.

2B^2

> laserlight@quixnet.net wrote:

> First, decide how long a turn is, how much acceleration gravity

I'll be playing FT this weekend, and the scenario I'm working on will most
likely involve a planet big enough to matter gravitationally speaking. I had
done range band calculations a while back, but never used them in play yet...

So I'm interested in how you apply the gravitational force in this type
of gravity-handling setup.  Do you base the force on where the ship is
at the end of its movement? This seems very simple and definitely in keeping
with the spirit of the game.

The downside is that the result will be very different for two ships, moving
the same high speed, narrowly missing the planet's surface, where

one ship just happens to end its turn 1 MU from the surface but the other say
10 MU away, because of staggered starting positions.

Alternatively, you could move the ship as normal, then determine the band
closest to the planet that the ship passed through, and apply that amount of
grav. accel. A little more work, but really not much. After all, you have to
measure from start to finish anyway with your tape measure to determine
velocity. Easy enough to find out what the closest

band the ship passed through was.

Or use the midpoint of the vector, and use whatever band that is in... etc.
etc. Or you could get really nutty and do a pull averaged over the

path taken... (Well, maybe *insane* is a better word than "nutty"...)

Anyone have experience playing in a game like this? Comments on how to handle
it?

> laserlight Wrote:

> Geostationary is geosynch with no north/south movement, and IIRC is

Ok, again, how to simulate this in a scenario that calls for Geostationary
orbit over a specific point NOT on the equator? Could you just require the
ship to expend X amount of thrust each turn to maintain that spot? Am I

right in assuming that the farther north/south of the equator, the more
thrust would be required and the shorter the orbit? Am I also right in
extrapolating from this that at some point, in concept directly over the

poles, you would cease to actually orbit, and instead be "Hovering" over the
planet and expending much thrust? It also seems that this thrust would be
equal to the necessary velocity to maintain orbit at that range (ie, if at X
range, Y velocity is required to orbit, you could hover by expending Y thrust
every turn), and that for any point between the equator and the pole, there
should be some fraction of this relationship at work. Sounds pretty daunting
with modern thrust technology, but with the right PSB........

2B^2

[quoted original message omitted]

[quoted original message omitted]

[quoted original message omitted]

> From: KH.Ranitzsch@t-online.de (K.H.Ranitzsch)

> Over the pole, the Thrust will have to exactly counterbalance the pull

I thought so.

I'm not sure of this scales closer to teh equator,

Intuitively I'd think it does, but that's just a hunch.

> but the Sine of the latitude * the gravity would be my first guess.

Sounds plausible to this math challenged boy.

> It's perfectly feasible with present-day thrust technology.

Feasible, but not very practical.

> (at least until the fuel runs out,

THIS is the tech I was referring to - we could do it now, but not for
very long. Future PSB regarding fuel efficiency was what I was referencing.

2B^2

Just look at the Harrier II...

--Binhan

> -----Original Message-----

Over the poles, you just have to use thrust to counteract the acceleration of
gravity, so if you want to maintain an altitude of 2*radius of earth
about pole, you need to be able to use thrust that would provide 1/4 G.
(gravity goes as 1/r^2).  Not over the poles gets a little more
complicated because you have to counteract gravity and provide a rotational
velocity that matches the rotation below. But with enough thrust you can be
place yourself anywhere in a powered orbit.

Randy Wolfmeyer

> On Tue, 12 Mar 2002, Brian Bilderback wrote:

> Ok, again, how to simulate this in a scenario that calls for
over the
> planet and expending much thrust? It also seems that this thrust

[quoted original message omitted]

> Randy W. Wolfmeyer wrote:

> Over the poles, you just have to use thrust to counteract the

That's what I suspected.

so if you want to maintain an altitude of 2*radius of earth
> about pole, you need to be able to use thrust that would provide 1/4 G.

How does this compare to the required velocity for an orbit at the same
altitude?

Not over the poles gets a little more
> complicated because you have to counteract gravity and provide a

Ostensibly, you should be able to give yourself a nudge for the rotation, and
only continue thrust for the lift. The added velocity may even reduce the
amount of lifting thrust necessary, but that's just a guess.

2B^2

> From: "B Lin" <lin@rxkinetix.com>

> Just look at the Harrier II...

Perfect example of what my point was MEANT to be - not that
thrust/guidance
technology is not yet capable (It IS, I agree), but that energy efficiency
technology is not yet capable of sustaining it for long. Take said Harrier,
and hold it in place. Now hold it there for a long time. How long will

it's tanks hold out? Not as long as it would if the then converted it into
forward flight.  It's a fuel issue, not  a thrust/precision issue.

2B^2

KH Ranitzsch

> Of course, the pull of gravity goes down the further up you go. - like

True...

> For a spaceship at the height of a geostationanry orbit, you need about

Depending on the desired orbital altitude for a given mission. It still

costs fuel. Mind you, if we assume the kind of future that the game allows
for, we should assume ships efficient enough that this is not a problem.

2B^2

Jerry A said:
> So I'm interested in how you apply the gravitational force in this

> Brian Bilderback wrote:

Just a point of real satellite ops, the Hubble Space Telescope's nominal
operating height is ~370 miles above sea level. Over the years the wisps
of atmosphere drag on it, dropping its orbit anywhere from 20-50 miles
in ~3 years. During each Servicing Missions (which we just finished up with a
successful 3b last week) the shuttle reboosts the Hubble back up to approx 370
miles (this time I think they only went to 360). The effects of atmospheric
drag are relatively negligible in the time frame of a
single game/scenario even up a few hundred miles [from a planet very
much like our own; obviously other planets with different atmospheric
composition, and stellar interactions, will vary this to some extent]. During
times of high solar activity, the Earth's atmosphere expands.
During low solar activity, it contracts (hence the 20-50 mile drop in
our orbit over a period of 3 years). If you are assuming 1 mu is equivalent to
1000 miles, "low Earth orbit" (which is where the Hubble and most other
satellites are) is well within 1 mu of the planet. Your lowest orbital band is
going to be pretty tight if you want to model
atmospheric effects.  ;-)

Just some stuff to chew on.

> Randy Wolfmeyer wrote:

> Over the poles, you just have to use thrust to counteract the

Err... yes, if you measure the altitude from the center of the planet rather
than from the surface. At an altitude of 2*radius of Earth above the *pole*
(which is 3*radius from the center of the planet), the required
thrust is only 1/9 G.

Later,

I think I meant radius instead of altitude there. Oops.

Randy Wolfmeyer Dept. of Physics Washington University

> On Thu, 14 Mar 2002, Oerjan Ohlson wrote:

> Randy Wolfmeyer wrote: