This seems to imply that the transit drive is not only very bright, but also spewing detectable charged particles in every direction, because... well.. you can see it from all directions. You can't see it if it's not sending something for you to see towards you.
(this also means the fantastically powerful transit drives of large ships are even worse than typical estimates, because typical guesses of power output don't factor in that an unknown amount of that energy isn't even going to accelerating the ship)
They are efficient, it's just that there's an unbelievably insane amount of energy involved that even very high efficiencies produce an insane amount of waste energy (heat, presumably).
Battletech warships weirdly get
more efficient as the ship gets bigger, with the exception of sharp drops after hitting certain tonnage milestones which reduces the thrust points you get per ton.
Let's take a really beefy bugger. 2,500,000 ton boat. Every ton yields 2 'thrust' (really acceleration) points. IIRC, an Aerospace turn is something like 1 whole minute.
Thrust is easy to calculate: Just multiply your acceleration times your mass, that gives you the amount of thrust you need to actually accelerate your massive hulk.
M*A=F
2,500,000,000 kg * 9.81 m/s^2 = 24,525,000,000 N = 24.5 GN
Delta-V is basically "fuel efficiency" for spacecraft: It's basically how much a unit can change their velocity vector before running out of fuel. Chemical rockets have absolutely terrible Delta-V, but there's not much other choice to counteract gravity and get off the Earth at the moment. Electric station-keeping drives ain't too bad with Delta-V
T(t) is thrust, a force. BT's engines are already set for acceleration though, so it's the same as the acceleration you get times the mass of the ship, or A(t)m(t)
m(t) again is the mass of the ship, which leaves you with the following (May ludd forgive me for this notation)
dV = integral
t0t1A(t)dt
For a test case of our 2,500,000 ton warship accelerating at a constant 1G (2 thrust points) for one aerospace turn (1 minute?):
dV = 9.81t m/s
2 [60s, 0]
dV = 589 m/s.
Thus, this warship can change its velocity vector by a bit over half a km/s (almost 2x the speed of sound if the Warship could enter Earth's atmosphere without disintegrating) by just one ton of fuel. With BT's simplified fuel calculations, every extra ton of fuel added increases your delta-V by the same amount: 10 tons yields 5,890 m/s dV, 1,000 tons yields 589,000 m/s. If you were feeling really ornery and filled that puppy up to 1,000,000 tons (almost all of its free weight), you can,
on paper, hit a delta-V of 589,000,000 m/s. (For reference, the speed of light is only about 300,000,000 m/s)
That's the delta-V of the
warship though. In order to move at all, that ship's gotta eject reaction mass out the back in the opposite direction. Due to the conservation of momentum we can actually calculate the exhaust velocity fairly accurately.
m
1v
1=m
2v
2m
1= warship mass=2,500,000 tons
m
2= exhaust mass=1 ton
v
1= warship velocity change=589 m/s
v
2= exhaust velocity=V
(2,500,000 ton) * (589 m/s) = (1 ton) * V
V = 1,470,000,000 m/s
Which is, umm, almost 5x that of the speed of light. Whoops, did I say we could estimate it
accurately? Things get spooky when you get within a couple percentage points from the speed of light, let alone several times over. Well, guess I'll pretend this value makes sense and continue with the Newtonian value for entertainment purposes until Cray smites me for high crimes against physics.
Going forward, we can also determine the ideal propulsive power, that is, the power required if there's exactly 0 waste anything: all of the energy to make the hydrogen exhaust go fast is solely devoted to make it go fast and not just make it hot (which is waste heat). In general, that looks something like this:
P = Propulsive power (thrust)
T = Thrust
v = Exhaust Velocity
Hey would you look at that, we already have the latter two numbers. Let's just plug in the thrust and the totally-not-meaninglessly-high exhaust values in to see what we get:
(24.5 GN) * (1,470,000,000 m/s) = 3.6e
19 Watts = 36ish Exawatts?
Assuming I somehow did these calculations right while I'm half-asleep, that means one super-sized warship with a totally, 100% efficient drive, consumes more power than the current electrical output of the Earth (350 W per capita * population of 7.8 billion = 2.7 TW power output). Like more than 10 million times more. Honestly, I almost find this Battletech technology more impressive than the literal physics-breaking K-F drives.
Usually you have to balance between the two attributes, ridiculously high delta-V and ridiculously high thrust, specifically
because the required propulsive power would just be insurmountable to get both. Battletech warships don't really have that problem, and I can only assume that it is only thanks to the pure anger of the spirit of McKenna inside each transit drive system that provides the necessary power to actually reach the required power output. Honestly, with this sort of energy behind them you could probably use these drives as weapons...
Even if that drive if 99% efficient, a 36 Exawatt level of effective propulsive power is going to result in several Petawatts of waste energy per second, and that's going to be distributed between not that much exhaust mass per second (16.7 kg/s). That's going to be really
really hot. Please don't ask me how hot; I'm much too tired for that and it might make the exhaust hydrogen hotter than the center of the sun. Plus neither my engineering training nor my two hours of dabbling in
Terra Invicta have adequately prepared me for Battletech physics.
Yeah, guess I better circle back to the topic at hand...
Well, for a stealth mode, maybe add some equipment, "Reduced Exhaust Signature Module" or something that costs a fraction of the transit drive's weight (thus scaling to be heavier for larger and more powerful transit drives). You could limit the exhaust velocity to station-keeping thrust, or alternatively to "just" safe thrust, but dramatically increase fuel consumption: For example, 1 "effective" thrust point costs 10 "real" fuel points. The excuse being that you're pumping way more mass at a lower velocity to get your thrust, but in return that mass is far cooler than normal and harder to detect (Hence it has a reduced exhaust signature).
I didn't
need to post a massive text wall to get to this paragraph-sized point, but there you go.