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,