Well, I've finally got time to go through Symon's treatise on lasers,
How much armour are we assuming a ship has. This has been discussed on the list before with relation to current and old designs but I don't have the figures anymore. IIRC it was said that a current destroyer may have about a foot (say 25cm) of armour. (Please correct me if I'm wrong). I would expect that a fighter however would have approximately half a cm if they wanted a decent power to weight ratio.
This only applies to velocities perpendicular to the attacker. If the fury is flying at the attacker then the beam will not travel across the surface at all. I'd suggest from scenes in 'The fall of night' and 'Severed Dreams' that most fighters attacking ships are driving straight at them. If the fighters were going at say 5 degrees to the ships the relative velocity across the line of fire is now below 100m/s, the beam only travels 10cm across the surface and the 0.02Mj/cm^2 is up to 0.2Mj/cm^2 which is a bit more reasonable.
I would estimate from the show that a laser sweeps across a Fury's wing in approximately 1/5 of a second. If a Fury has a wing width of a couple of metres we would get 20 impulses (in .2s) across 2m or a pulse traversing 10cm. However if the Fury is heading towards the ship then the first part of the beam will strike the wing at an angle and its spot will be larger, but when it hits the leading edge it will take the full force of the beam
___________ ---------> / beam /--- /--- |--- | WING |__________________________
If the leading edge is half the visible width but only 10cm high then the laser will get 10 impulses (0.1s) in 10cm and the beam only travels 1cm per pulse. We thus get our 2Mj/cm^2 back.
Use this with the 5mm skin of the Fury and we get serious damage inside the wing, certainly enough to blow control wires to the engine.
In SD I'd estimate (the video is on loan) that the rake of the Roanoake is about 40m in 2s or 20m/s. Each pulse is therefore traverses 2cm long, enough to blow 1/2 cm of armour (assuming a linear relationship between energy transfer and cutting power).
The barrels of the HL are over 1m wide or 7500cm^2 Oh dear, that's suddenly a lot more power required. However it does mean that traversing at 2m/s the beam is on target for 1/2s, or 50 pulses and will therefore blow 50cm of armour, a reasonable figure (I think) for a Cap ship and enough to fry any fighter completely.
I'm not sure that the beams are actually focussed at all, they seem to me to be a collimated beam with no focal point. However not knowing enough about the optics I'll assume that it means that you can't make a parallel beam either.
You've assumed 500km, I assume 10km wide hexes and a range of 20 hexes is then 200km (less if it's not directly down a hex spline). This gives a 6 fold increase in the power/area and the UV with a 50cm dish becomes better than 1.2Mj/cm^2. that's close enough for me given all the unknowns. AT 100km (10 hexes) it rises to 12.5MJ/cm^2 more than enough to blow through any reasonable armour.
I thought that 25% was about the efficiency of a coal power plant, I'd hope/expect that a laser would be more like 50%. Again I may be talking out of me rear end here as I have no figures to back up this assumption.
The temperature in deep space (in a shadow) is not a lot. With a minimal cover to conduct away the direct radiation from the sun cooling should not be a problem, I'd expect superconductors to easily attain their working temperature
Improvements will be expected in 200 years This is where I go off into the realms of fiction and guesswork so don't get too rough with me.
Room temperature Superconductor. - Reaching 0 degrees Centigrade wouldn't seem preposterous
Laser efficiency - maybe 50%.
Electronically tuneable lasers - Marconi have just invented these so no problem there.
Armour - Unfortunately I'd expect armour effectiveness to increase so armour values may be too low.
Capacitors - A big problem as the power required is way beyond our current capabilities
With 1m beams and a hundred 1/1000s pulses at 50% efficiency we need 7500 x 2Mj x 10% (percentage beam is on) x2s (beam duration) x 2 (efficiency) = 6GJ. Powering up over 4 turns (which I take as 10s per turn or 40s) then we need 150Mw per HL. This I think is reasonable for a pretty small fusion reactor in 200 years.