[Spaceships] Is This Legal?

[Ulzgoroth]

Originally Posted by Varyon

I had initially done it that way, but I'm not certain how to calculate when the ship has gone through all the fuel in the tank, because that fuel doesn't account for exactly 3/4 of the overall delta-v, unless I'm horribly mistaken. If I am, then instead of overall delta-v of 2.88 you instead have initial delta-v of 2.52 mps, then once you've used up 3/4 of the delta-v (1.89 mps, for 0.63 mps remaining) you drop the drop tank and end up with 1.26 mps remaining, for total overall delta-v of 3.78 mps.

I'm not exactly sure what you're doing, but it sounds like you're working too hard.

The ship+tank combination has 9 system-equivalents of fuel in the drop tank. (18 systems divided by 2 since the combination is double sized.) That gives you 9*1.4*dV = 12.6*dV where dV is your drive's per-tank delta-V. (I don't know what engine you're assuming here.)

After you've burned that the external tank is empty and you jettison it. All the internal tanks are still full.

At that point you've got 6 systems internal fuel, for another 6*1.2*dV = 7.2*dV. (You also are no longer limited to half your nominal acceleration.)

Total delta-V is thus 19.8*dV, which falls into the large gap between 16.8 for 12 tanks and 20.8 for 13 tanks.