Not familiar with TNE or it's specific attractions, but you clearly have a handle on the problem set under that system. Anyone grappling with the issue and what play effect they want has to decide on this point.
I consider TNE to be the most "realistic" of the rule sets. You can see how Chadwick over time has tried to apply "the physics that we have" and extend them to "the future", and model this in to something "mostly" playable. From AHL to Striker to T2K to TNE, plus his 20th century armor work, you can see him refine his model of damage. If nothing else, he's reasonably consistent. Not saying it's perfect, of course. He has a minimum amount of handwavium in his work.
For example, I heard a presentation of his and he seems to believe the HePLAR model in TNE is far too conservative, and we'll be getting much better performance out of a drive, far sooner than expected. TNE uses orders of magnitude too much fuel. He didn't say this directly, but his explanation of his new views on such a drive talk of something much more efficient.
I'd say part of the calculus is also how much each costs to succeed at in terms of both tons/logistics to operate the blockade/raider force, and the war material/time value.
Early on in the Atlantic WWII U-boat campaign, destroying ships was several times more damaging as the Allies lost that tonnage-years capacity for the rest of the war and had to replace it to match and ultimately build up to an offensive level. Strategically too, without the US in the war the UK was much more vulnerable earlier on.
Destroying ships with finished products such as tanks and artillery was also destroying weapons at the end of a production chain, wasting all that effort to get the weapons to that point.
Ultimately however, if the planet cannot build, equip, and/or feed itself, a total successful blockade will reduce war potential and resistance in the long run. The age-old question of the siege, who will break first?
I consider the raiders disruptive, rather than sieges. I'm thinking small forces that can be shooed away from a defenseless system, but hinder traffic while they're there.
I note you are skipping over the fighter point. I would think that fighters are an excellent raider/blockade mechanism for both wide coverage and quickly bringing force to bear on breakthrough attempts without breaking the bank or risking big hulls. Possibly can risk their loss closer to PD forces then conventional ships.
The fighters can certainly broaden the space control aspect of the raider. The precept is that the PD forces are "impenetrable" to the Raider. The Raider simply will not engage the "coastal batteries". It's power is it's flexibility to engage targets of opportunity outside the shield of the PD.
Maybe, maybe not. Even if they do, there is the question of sensor range, autoengagement capability, and the rules surrounding 'minesweeping'.
Of course, but I think it's fair to say that mechanisms designed and deployed in space are reasonably hardened to the radiation hazards, and vacuum itself is not a particularly notorious "wear" mechanism. We have several examples of equipment that last year in space, I would think that level of engineering would be routine -- especially for something designed to be kept "on station".
I don't know anyone who is tracking on inherited system vee, nor wants to.
No, it's not necessary. While the differential exists, it's small enough to not be of any real impact. It can be "margin of error"d out of routine operations. With the combination of jump scatter due to timing, plus the error in actually "hitting" your plotted destination (it is said that you will arrive within "several thousand Km" of your plotted point), all ships have to do some compensation when they arrive in system.
For example, with Earths orbital velocity, and the 16.8 hr random arrival window from jump (10% of 168hr, 7 days), the Earth can be anywhere along a 60 hex (30Kkm hexes, 1.8Mkm) path on its orbit. So, if you "aimed" to arrive "right on it", the temporal jump shift could put you 60 hexes away from your target. Obviously, you would want to place your entry point on the leading path of the planet (well, the 100D leading edge of the planet), so you can take advantage of it vector approaching you.
A clever navigator would plot the back part of that temporal jump shift to put you in to the window of the 100D of the planet, then the planets 100D horizon would yank you out of jump.
My question was oriented towards the understanding that if you jumped at 100D having been under full burn the whole time, say 5 hours at 1G, that's 180,000 km per 1000 second CT turn, you dump out of jump at that vee and assuming you aimed square at 'where the planet is going to be in 5 hours' you are still going to be turning and burning the whole way in to slow down.
That is NOT a casual vee, and that's just what Beowulf can do.
As such, that's minimal distance using the slowest speed in the game. If you have a faster M-drive and want minimal vulnerability at least outbound, you will have a faster vee to deal with at the end journey.
The closer the vee is to your maximum, the less options you have to slow it down, or take a LOT of time to correct it, giving an enemy time to maneuver optimally asw ell.
Whether you have safe system time or not, the maneuvers will cost total ship use time, depending on the time value of the cargo being carried that may be nothing or the loss of 5 hours here and 10 hours there may end up being an entire transport fleet's worth of effective carrying capacity.
Yes, absolutely. Whether that time involved is that impactful vs the week long jump is a different question, and, inevitably, I'd rather burn 10 more hours if I had to than a) not go at all or, b) get blown to smithereens.
That depends upon how you define what a jump lane is, and what the parameters of that jump vee business is, particularly if the vee HAS to be in a straight line from point of jump to exit of jumpspace.
If we simplify the Universe, to the point that we can virtually overlay all systems on top of each other, with the system primary as the origin, and all vectors relative to that origin. So, if my ship is at "system north" (i.e. the primary is "due south" of me), and I burn for 10GTurns, that gives me a vector, that's 10G long, pointing south.
Jump allows us to place that vector "anyplace" within the destination system, but vector remains the same. The position changes, but not the vector. So, I can position my jump entry to anywhere in that system, but when I arrive, my vector will still be 10G long pointing south.
So, to me, a "jump lane" is designed to take advantage of any acceleration that I need to do anyway just to get to the 100D. So for a contrived example, consider that my destination planet is orbiting counter clockwise, and is currently "due west" of the primary (or at least will be in 1 week), and heading "south".
When I leave my current planet, I will accelerate due north to the 100D horizon, giving me a due north vector. I will plot my arrival "due south" of the destination world. So, when I arrive, I will get to leverage my retained velocity to arrive at the destination world.
If the destination planet was at "due north", traveling "due west", then you can see ships leaving their worlds by going "due east".
So, that's what a jump lane is to me. It's an ever changing "most efficient" vector used by ships going from system to system. Any path will do, it's just a matter of correction. A ship may well not be going "full blast" on departure if they don't want a huge vector on arrival.
If you don't use the jump lane, you simply have to do some more correction when you arrive. You can see a "worst case" using the Earth example above. With the 1/2 hr turn, Earth has a velocity of 2 (hexes/turn). So, if the Earth is due north, and you have a due east vector, you can see arriving 60 hexes behind Earth, with your retained velocity, having to make up the 60 hexes plus overcome the Earth 2 hex/turn velocity. A full burn at 1G hits 100D (27 hexes) at about 9GTurns (1/2hr turn). So, on arrival, the ship needs to decelerate for 9 turns, traveling 27 hexes, now 60 + 18 (9 turns * 2) + 27 = 105 hexes away from Earth, and then reaccelerate back. The total trip is about 15-20 hours of correction (if anyone has the math to calc the arrival time on a moving object, I'd love to see it).
If so, there darn well are costs, time tactical or otherwise, ESPECIALLY jumping into a system where your information about where the enemy is deployed is at least two weeks out of date.
If not, the blockade runner has the advantage of coming out wherever they like and punching through the 100D limit at a maximum vee, maybe even faster if they can overshoot the planet a bit and still stay under PD guns.
Yea, the enemy intelligence is really old. You don't know when you're arriving, you don't know what will be nearby when you do arrive. If traffic is using "jump lanes", you can try and arrive near a highly trafficked jump lane to another system, hope it will be target rich on arrival. Get lucky, pop out of jump, "Hey, Targets!" and start blasting away. Otherwise, you arrive wide of the planet, ideally out of the reaction range of any picketing forces, assess the situation, then start moving in.
Simply, the village suddenly sees a lion on the edge of the bush -- looking for prey, and can't do much about it as long as it stays out of the clearing and out of rifle range.
