The size of the KF drive is established by the construction rules. So, if the sub-compact drive is 50% of the total mass of the Argos, it would be 47,500 tons (well over the 25k limit), but would have been enough to send the Argos by itself into hyperspace. But the restrictions would make little sense of the multi docking collars that the Argos does already have and render them even further uselessness by the inability to take dropships into hyperspace with it. Has to be a normal compact, or designate it entirely as a jumpship rather than a warship. It would have to drop all the habs and that would, once again, imo, make the entire design no different than just using a normal jumpship. Best bet is to redesign, add that pesky 5k to bump it into the 100k range for compact drive eligibility and bring everything else with it to make it play by the rules.
It's not the drive size that's limited to 25,000 tons, it's the size of the ship it's mounted in. But it does further illustrate the size issue:
- Standard Drive: for use with JumpShips (50,000 to 500,000 tons), mass is 95% of JS mass, so drive units will range from 47,500 to 475,000 tons, KF cost multiplier is x1
- Compact Drive: for use with WarShips (100,000 to 2,500,000 tons), mass is 45.25% of WS mass, so drive units will range from 45,250 to 1,131,250 tons, KF cost multiplier is x5
- Sub-Compact Drive: for use on "ultra-small" WarShips (5,000 to 25,000 tons), mass is 50% of WS mass, so drive units will range from 2,500 to 12,500 tons, KF cost multiplier is x16
JS construction rules show that it's possible to build a K-F drive for a vessel as small as 50,000 tons. WS construction rules show that it's possible for a Compact K-F drive core to be lighter than a Standard K-F drive core. So
theoretically, it should be possible to build a heavier Sub-Compact K-F Drive (& therefore build a larger "Sub-Compact" WarShip that's under 100k tons in mass).
Now, one of the things I noticed based on the construction rules in
SO on p. 149 is that the minimum K-F Drive Integrity for a Compact K-F Drive is 4 (2 base + drive weight / 25,000, round up to next whole number; 45,250 / 25,000 = 1.81; 2 + 1.81 = 3.81, rounded up to 4). For the Sub-Compact, the minimum and maximum are both 3, as it uses the same formula (2,500 / 25,000 = 0.1; 12,500 / 25,000 = 0.5; 2.1 rounds up to 3, 2.5 rounds up to 3). So, if we set the maximum K-F Drive integrity for a Sub-Compact Drive to be 3, then the maximum drive mass would be 25,000 tons...which means you could install them in a ship as large as 50,000 tons (or the minimum size of a JS).
Alternately, since the Compact Cores can be lighter than the Standard Cores on the lower end, we could say that the 100,000 ton standard WS size cut-off is due to that requirement, & set the upper size limit for a Sub-Compact Drive unit at a close number. If we set it at 40,000 tons, then you could go up to 80,000 tons for ship size; if we set it to 45,000 tons, then our Sub-Compact WS limit is 90,000 tons.
Either way, I feel like 25,000 tons is an artificially small cut-off. Even with the much more expensive base drive cost, you're somewhat offset by the lack of ability to carry DS collars, which have a much larger effect on the cost of the K-F Drive than any other factor.