SV: FTFB Turn Arcs

1 posts ยท Nov 27 1998

From: Oerjan Ohlson <oerjan.ohlson@t...>

Date: Fri, 27 Nov 1998 19:55:24 +0100

Subject: SV: FTFB Turn Arcs

Chris Lowrey asked:

> >Because the timescales are different. One game turn in FT is

The FT time scale was derived from the MT orbital mechanics a long time ago
(...Thomas Barclay, wasn't it? Or have I mixed you up again? Can't find the
post now, but it is somewhere in the archives.) I put the
"commonly agreed upon" within quotation marks - they're a rough
guideline rather than written in stone, and there are many (Laserlight, for
example) who have derived other scales as well.

The EFSB time scale is derived from comparing how fast ships are
destroyed in the show with how fast they die in the game :-)

> Laserlight wrote:

> I still think 450 seconds (7.5 minutes) is the way to go.

Not in EFSB - or, rather, not if you compare EFSB battles with battles
from the show - especially the amount of damage they inflict on one
another :-/ For FT, your scale works as well as anyones as long as the
size of an MU and the g rating of a thrust point matches :-)

> > "Realistically", a change of facing for a starship takes on the

The longer axis doesn't matter that much - or, rather, it also increases
the lever (and therefore the torque) which helps compensate. The main
importance of the length of the ship is that the structural integrity needs to
be better the bigger the ship is, since the shear forces increase. You need to
assume that the thruster strength is proportional to the size of the ship, but
the FT and FB design systems do that already.

Example:

Assume that you want to turn a ship around 180 degrees in one minute. How fast
must the bow and stern be accelerated sideways to achieve this?

Let:

Phi be the angle turned by the ship a the angular acceleration ( = the 2nd
time derivative of Phi) t the time in which the ship turns

If you start turning with constant a from a standstill, you get the relation

Phi = 0.5 * a * t^2

a is what we're looking for here, so for a given Phi and t we get a = 2
*
Phi / t^2.

We accelerate the rotation half the way (90 degrees) and then reverse the
thrust. a is constant throughout the acceleration and through the retardation,
but is reversed inbetween the two, so I'll just calculate how large a needs to
be in order to turn the ship 90 degrees in 30 seconds (leaving it with a
respectable angular velocity). This gives

a = 2 * 90 / 30^2 = 0.2 degrees/second^2

which isn't very much. If your ship is 1000 meters long and has its
center of mass roughly half-way along its length, an angular
acceleration
of 0.2 degrees/s^2 means that the bow and stern must accelerate sideways
by about 0.9 m/s^2, or less than 0.1 g.

Using the "usual" game scale, 1 thrust point is IIRC about 0.125 g. A
thrust-2 ship has maneuvering thrust 1, but this is thrust 1 sideways
for
the *entire* ship - you usually need more force to accelerate the ship
sideways (or forward, or whatever direction you like) than to set it spinning,
depending on the shape and mass distribution of the ship. You never need
*more* force to set it spinning than to push it sideways
though, so even a 1-km long thrust-2 battleship is capable of turning
180 degrees in a minute or less.

If your background assumes that ships are more than 1 km long (Renegade
Legion, for example), you may have to adjust the assumptions. However,
even a Shiva-class BB (2.7 km long IIRC) wouldn't need more than a
maneuvering thrust of 2 to flip around in a minute using the "usual" game
scale.

So yes, unless you assume *very* big ships they won't have any problems
turning to any new heading in a minute or less. Depending on what assumptions
you make for the time scale of one game turn, this may or may not be too fast
to reduce the amount of main thrust available; and EFSB as written doesn't use
the same assumptions as the FB.

Later,