Attacking Darius:

ThinksMarkedly wrote:And I bet you that grasers can be detected by the effect they have on the solar wind and interplanetary dust in their paths. Especially if there were multiple beams, and even more so if those were in multiple directions. Firing grasers thus means that any ship outside of energy range has now a VERY good locus for its missiles.

Loren Pechtel wrote:This. No matter how stealthy the LD is I don't believe a shipkiller graser shot can be hidden. So long as anything with eyes (including recon drones) is about the beam will be identified. Fire multiple beams and you have an origin. Given the resolution we have seen to date I think they have plenty of accuracy to ensure return fire hits the origin--the only escape would be if they can zig enough before the beams arrive. Note that the low power and lack of a wedge greatly limit the ability of a LD to zig.

People need to be careful, because there are two different types of grasers being talked about at this point in the thread: ship borne grasers and the graser on the torpedoes. Yes, in a dirty environment the graser beam should be visible; but that is only important if the graser is ship borne, because then the ionized path points to the ship. If it a graser torpedo, then you will know the source no matter what the environment, because at the end of the 3 second firing sequence the torpedo will spectacularly self-destruct. Moreover, if you could not detect the movement of the torpedoes, then you will have no clue for the location of the launch. Even if you can see the beam during the 3 seconds it is in existence; you will not have a clue for the launch, because the torpedo can twist and turn.

tlb wrote:People need to be careful, because there are two different types of grasers being talked about at this point in the thread: ship borne grasers and the graser on the torpedoes. Yes, in a dirty environment the graser beam should be visible; but that is only important if the graser is ship borne, because then the ionized path points to the ship. If it a graser torpedo, then you will know the source no matter what the environment, because at the end of the 3 second firing sequence the torpedo will spectacularly self-destruct. Moreover, if you could not detect the movement of the torpedoes, then you will have no clue for the location of the launch. Even if you can see the beam during the 3 seconds it is in existence; you will not have a clue for the launch, because the torpedo can twist and turn.

No doubt. That's why I said that the LD's standard tactic has to be to pre-deploy torpedoes and pods then get away from nearby, changing the velocity vector sufficiently so one can't backtrack those to their source.

Firing its own weapons would be a last-ditch defence, like CLACs. The LD needs to skedaddle after this, very quickly.

A 9-million-tonne SD takes over 3 minutes to transition to hyper after the button is pressed. An LD is bigger, so let's say 6 minutes. That means it's vulnerable to missile launch platforms within 30 million km. It'll depend on its own counter-missile systems, but with no wedge to obstruct the majority of the angles of attack, it's doubtful those can do much good against sufficiently large missile swarms. A single Nike-class BC(L) could produce multiple double-double broadsides of sufficient quantity to make the LD's day really bad.

Could be. Or it could deploy a fort class bubble and fight it out. Who wins in a fight between a BC and fort?

kzt wrote:Could be. Or it could deploy a fort class bubble and fight it out. Who wins in a fight between a BC and fort?

Between those two? The fort, of course. One would assume that it has a much bigger ammo depth than the battlecruiser.

But even then there are variables: the LD would be firing Ninurta missiles, whose performance we don't know (I assume they're at least equivalent to a true DDM like the Mk16). If they were equipped with only Cataphracts, they'd be outranged by the BC, which would place them at serious disadvantage. A bubblewall is tough but not impenetrable and it can be fired upon from all angles. The BC would be fighting with interposed wedges, so those Ninurta missiles need to be smart enough to actually fire through the openings and not depend on luck, like the Cataphracts would.

The equation changes if one of the sides gets reinforcements. A 4-drive system defence missile can reach 0.81c in 9 minutes, meaning it has a 15-minute range of 153 million km. That BC with Keyhole can assume control of the missiles and direct to target. This wouldn't happen, of course, because the LD would see the missiles fired with plenty of time to spare, at which point it would charge the hyper generators. It's also unlikely such 4DM emplacement would be within 8.5 light-minutes of a spider action that started at 1 or 2 light-hours away from the hyperlimit.

The danger is if the reinforcements come from hyperspace. One would assume that the BC is out there in the first place because there was a hyper footprint detection, so reinforcements could have scrambled and gone to hyper to wait the call. If the call goes out the moment the LD raises bubblewalls or fires missiles, the reinforcements could drop in within 4 minutes (30 seconds for the destroyer to go to alpha, 3 minutes for SDs to come back to n-space). A single SD with rolled pods would more than outmatch the LD in that situation.

ThinksMarkedly wrote:A single SD with rolled pods would more than outmatch the LD in that situation.

According to David, a fort can beat a SD squadron like a rug. I have no idea how that is supposed to work, but I offer it as a data point.

kzt wrote:Could be. Or it could deploy a fort class bubble and fight it out. Who wins in a fight between a BC and fort?

Depends on if they're at equivalent tech levels.

(Pretty sure a single Agamemnon class BC(P) could crush one of the pre-Buttercup Junction forts. It can throw and control more missiles than a pre-pod SD squadron; and its Mk16s are moving faster and with more powerful ECM, jamming, and decoys -- and it need not come into range of the fort's missiles)


But yeah, normally the fort is the obvious winner.
However a fort class bubble sidewall doesn't necessarily give you the full protections of a fort. The hull shape the Spider Drive imposes would make it harder for an LD to carry the same extent and coverage of point defense as a fort of similar tonnage would.

kzt wrote:According to David, a fort can beat a SD squadron like a rug. I have no idea how that is supposed to work, but I offer it as a data point.

As Jonathan says, it depends on the tech level. I doubt even a modern fort can just shrug off a several thousand Mk23E+F.

A Manticore fort would be firing the same type of missile back at the SD, though. So if it does survive the alpha launch from the SD, it will be putting more missiles into space per salvo than the SD will. That means the advantage swings back to the fort.

An LD would be firing Ninurtas back. However advanced those are, Apollo should still come out ahead. Plus, we don't know how many launchers an LD will carry, but I doubt it'll be as many as a fort.

Jonathan_S wrote:But yeah, normally the fort is the obvious winner.
However a fort class bubble sidewall doesn't necessarily give you the full protections of a fort. The hull shape the Spider Drive imposes would make it harder for an LD to carry the same extent and coverage of point defense as a fort of similar tonnage would.

And as we've seen at Galton, forts can use wedges for defence. Nothing can penetrate a wedge.

LDs -- as far as we know -- can't.

This is talking about the dozen fortresses at the Lynx Terminus. So no, one SDP is not going to take a fort. You'll need on the order of a squadron. And that assumes nobody is trying to drill a hole in the back of the SDPs with a cruiser graser. Oh, and you want a fun time? Punch a hole in one of the pods with a graser as it is ejecting.

"There seems to be something wrong with our bloody ships today," he said as he watched the rear of the second SD(P) explode.

"Your Majesty, at the risk of sounding immodest, the only real question would be how long it took us to blow all seventy-one of them out of space. Those forts were designed to hold that terminus without any outside support against the attack of two hundred and fifty of our own pre-Apollo podnoughts. Now that they have Apollo, their defensive capability's been multiplied many times. We still aren't sure by exactly how much, but it's got to be at least a factor of four."

kzt wrote:This is talking about the dozen fortresses at the Lynx Terminus.[cut]
"Your Majesty, at the risk of sounding immodest, the only real question would be how long it took us to blow all seventy-one of them out of space. Those forts were designed to hold that terminus without any outside support against the attack of two hundred and fifty of our own pre-Apollo podnoughts. Now that they have Apollo, their defensive capability's been multiplied many times. We still aren't sure by exactly how much, but it's got to be at least a factor of four."

A dozen pre-Apollo forts against pre-Apollo 250 SD(P)s means a single fort equals probably under 20 SD(P)s, due to synergistic effects for synchronised defence. Against an Apollo SD(P), I wouldn't divide by 4, but certainly by two. That means a lonely pre-Apollo fort might be equal to 5-7 Apollo SD(P)s.

But again, those are Manticoran forts, which can presumably switch to full-on wedge, have shoals of missiles (even pre-Apollo ones), and are designed for a single purpose. They probably also mass in the range of 25-30 million tonnes, so they can dedicate a LOT to armour, especially if they can block a significant portion of their aspects with wedges.

An LD is not a fort. It probably tops at 18 million tonnes, probably less. It's designed with multiple roles in mind, and while it is designed to survive an ambush, that's not going to be its main role. The hull surface dedicated to the spider is going to be considerable too, which limits the number of CM launchers and PDLCs it can mount. The hull geometry in the form of a rounded triangular pyramid is also going to limit the amount of armour such a ship could be built with.

So if an LD is the fighting equivalent to a third of a pre-Apollo fort, it could hold its own against 2 Apollo-equipped SD(P)s or less.

That goes back to the original point: such a ship would not try to slug it out, even with a BC. Not because it can't win against the BC, but because once it became known, it has to charge its hyper generator and leave. Its last-ditch defensive capabilities are probably designed to survive the 6 minutes it takes to transit out.

ThinksMarkedly wrote:

kzt wrote:This is talking about the dozen fortresses at the Lynx Terminus.[cut]
"Your Majesty, at the risk of sounding immodest, the only real question would be how long it took us to blow all seventy-one of them out of space. Those forts were designed to hold that terminus without any outside support against the attack of two hundred and fifty of our own pre-Apollo podnoughts. Now that they have Apollo, their defensive capability's been multiplied many times. We still aren't sure by exactly how much, but it's got to be at least a factor of four."

A dozen pre-Apollo forts against pre-Apollo 250 SD(P)s means a single fort equals probably under 20 SD(P)s, due to synergistic effects for synchronised defence. Against an Apollo SD(P), I wouldn't divide by 4, but certainly by two. That means a lonely pre-Apollo fort might be equal to 5-7 Apollo SD(P)s.

But again, those are Manticoran forts, which can presumably switch to full-on wedge, have shoals of missiles (even pre-Apollo ones), and are designed for a single purpose. They probably also mass in the range of 25-30 million tonnes, so they can dedicate a LOT to armour, especially if they can block a significant portion of their aspects with wedges.

An LD is not a fort. It probably tops at 18 million tonnes, probably less. It's designed with multiple roles in mind, and while it is designed to survive an ambush, that's not going to be its main role. The hull surface dedicated to the spider is going to be considerable too, which limits the number of CM launchers and PDLCs it can mount. The hull geometry in the form of a rounded triangular pyramid is also going to limit the amount of armour such a ship could be built with.

So if an LD is the fighting equivalent to a third of a pre-Apollo fort, it could hold its own against 2 Apollo-equipped SD(P)s or less.

That goes back to the original point: such a ship would not try to slug it out, even with a BC. Not because it can't win against the BC, but because once it became known, it has to charge its hyper generator and leave. Its last-ditch defensive capabilities are probably designed to survive the 6 minutes it takes to transit out.

Difference between a ship and a fort

1) Shape - Forts are designed to oppose ships from every direction
2) Hyper drive - Forts do not have hyper drives, hyper sensors, and alpha nodes to make sails.
3) Comp - too large for a compensator (LD also lack comps)
4) Drive power - Forts only go 50-100 gs - they only need 20% of the drive capability a warship would, so as a matter of Size % their drive spaces are much smaller.
5) Endurance - Forts are placed near to a home base, they don't need months worth of consumables and spare parts. They probably also have more spartan crew accommodations as the crews can rotate to a location specialized for crew comfort.
6) Pre positioned missile pods - modern forts have shoals of external missile pods. Older forts had IEWPs and minefields surrounding them to augment their internal firepower. ships only rely on internal resources.
7) Bubble side wall. We're told a SD could also carry one of these massive defenses, but it would severely cut into it's weaponry for something it would rarely use. LDs may carry one of these, but could not use the spider in congress with the bubble.