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Sanity checking my understanding of ships' weapons

TheDark

SOC-12
I'll lead this off with the disclaimer that none of my MegaTraveller experience has involved naval campaigns, so this is probably a bit basic of a question.
I'm trying to figure out some ship weapon layouts for personal conversion to other editions of the game. If I'm understanding correctly, ship weapons are in batteries that need to be backed through the UCP tables in the Referee's Manual to find the total number of emitters/tubes/whatever, correct? So using the Planet-class from page 83 of the Rebellion Sourcebook as an example:
Missiles are x93 with 20/10 batteries. At TL 15, only 50 ton bays are available, and they have a UCP of 9, so that's 20 50-ton batteries. Turrets are 3 tubes for 3 UCP (TL-13 turrets, since TL-21 isn't available), so it's 30 tubes split among various turrets.
Fusion guns are x04 with 5 batteries. With FG-14, you need 10 emitters for UCP 4, so there are 50 emitters.
Beam Lasers are xx7 with 3 batteries. That high of a UCP takes 15 emitters, so there are 45 total.
Particle Accelerators are 090, so also bays, but 100-ton bays this time, and there are 2 of them.
Meson guns are J0x, a single Type J spinal mount.
Sandcasters are xx7 with 15 batteries, so 150 'casters since TL-10 sandcasters are 10 launchers for UCP 7.
Repulsors are x9x. At TL-15, a UCP of 9 means 100 ton bays, of which there are 10.

So the total weapon layout of a Planet-class is probably 1 Meson spinal mount, 10 100-ton repulsor bays, 2 100-ton particle accelerators, 20 50-ton missile bays, 10 triple missile turrets, 15 triple blaser turrets, 50 triple sandcaster turrets, and 25 twin fusion turrets. Does that sound right to people with greater understanding of the system?
 
Be carefull using cannon ships to understand rules, be sure to check the errata pages to see if the ship has been changed via errata. Next is to look at the tons of displacement that the installed weapons and defenses require.
Spinal mounts consume thier listed displacement.
Bays consume 1000TD per bay regardless if 50 or 100 TD bay. 32 bays = 32,000TD needed
turrets use 100 TD per turret. 100 turrets = 10,000TD needed.
I am assumming the type J meson gun needs 8,000DT
8k+32k+10K= ship must be 50KDT or larger.
Note: You picked the best selections of turrets the sandcaster and fusion batteries could have been. batteries of 10 single turrets as well for 150 single sand caster turrets, and the fusion guns could have been 50 single fusion gun turrets. That is why you also have to look at less efficient layouts they can help explain why the ship displaces 62,500DT instead of 50,000DT. (No, I did not look at the ship, just giving you some pointers.)
 
So the total weapon layout of a Planet-class is probably 1 Meson spinal mount, 10 100-ton repulsor bays, 2 100-ton particle accelerators, 20 50-ton missile bays, 10 triple missile turrets, 15 triple blaser turrets, 50 triple sandcaster turrets, and 25 twin fusion turrets. Does that sound right to people with greater understanding of the system?

Yes, more or less. Turrets, not single weapons, are grouped into batteries (RM, p88), so 60 sandcaster turrets are needed. Note that a single Fusion Gun has factor 5 at TL-15, so 5 single fusion turrets are enough.

Something like this:
Skärmavbild 2024-09-12 kl. 09.13 1.png

15 batteries of 4 turrets (60 turrets) each with 3 sandcasters are 15 batteries of factor 7.
 
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Yes, more or less. Turrets, not single weapons, are grouped into batteries (RM, p88), so 60 sandcaster turrets are needed. Note that a single Fusion Gun has factor 5 at TL-15, so 5 single fusion turrets are enough.
Ah, yes, I was looking at my dead tree book instead of the consolidated errata, which has a 3-point difference in UCP for all the fusion guns.


15 batteries of 4 turrets (60 turrets) each with 3 sandcasters are 15 batteries of factor 7.

I'm guessing this is an example of a case where you have to look at displacement so you know it's not 15 batteries of 5 twin turrets instead of 4 triple turrets, since both would provide factor 7 batteries? (Which basically requires rebuilding the entire ship, since system tonnages aren't provided?)
 
I'm guessing this is an example of a case where you have to look at displacement so you know it's not 15 batteries of 5 twin turrets instead of 4 triple turrets, since both would provide factor 7 batteries? (Which basically requires rebuilding the entire ship, since system tonnages aren't provided?)
You don't know from the short description. I would generally assume the fewest turrets, hence the least tonnage used. Energy consuming weapons might minimise power consumption.

This ship in particular has over 200 hardpoints unallocated, so no clue there.

Five twin turrets would consume slightly more tonnage, but cost less, hence require fewer gunners, saving some tonnage. In this case it would seem that twin turrets are better, as the ship has hardpoints to spare.
 
To see what is really going on you have to make a full reverse engineering of the ship:
Skärmavbild 2024-09-12 kl. 11.02.png

Skärmavbild 2024-09-12 kl. 11.02 1.png
Skärmavbild 2024-09-12 kl. 11.03.png

The ship seems to be 35 000 kl overtonnage (even with the smaller crew of the errata), has a powerplant that is ~90 000 MW too large, and the mass is 50% too large. 290 hardpoints are unused.

About par for a MT ship...
 
Bays consume 1000TD per bay regardless if 50 or 100 TD bay. 32 bays = 32,000TD needed
turrets use 100 TD per turret. 100 turrets = 10,000TD needed.
This is badly worded.

Bays do not "consume" 1000TD, rather there is a limit of 1 bay per 1000TD. A 50 Ton bay consumes 50 tons in the ship. So, if you have a 10,000 ton ship you are limited to 10 bays. If you have 10 50 ton bays, you've only used up 500 tons of the total 10,000 available in the hull.

It's also a but more nuanced, the limits are based on tonnage available that not already allocated to other weapon systems (at least in HG).

So, if you have a 10,000 ton ship, and place a 1000 ton spinal in it, that leave 9000 tons left, which mean you can have 9 bays. If you add 9, 100 ton bays, you are left with 8100 tons of empty hull. That's enough room for 81 turrets (8100/100 = 81).

With Turrets, you can have 1 hardpoint per 100 tons, and 1 turret per hardpoint. and all turret are the same size, whether single, double, or triple, but, at least in High Guard, different weapon types require different size turrets. Energy weapons require 2 ton turrets, for example.

So, going with the above contrived example, a 1000 ton spinal, 9 100 ton bays, 81 energy turrets = 162 tons. So, 2062 tons allocated to weapons.
 
So, if you have a 10,000 ton ship, and place a 1000 ton spinal in it, that leave 9000 tons left, which mean you can have 9 bays. If you add 9, 100 ton bays, you are left with 8100 tons of empty hull. That's enough room for 81 turrets (8100/100 = 81).
No, all weapons require hardpoints in MT. You are describing how LBB5'79 worked.

MT RM, p72:
Skärmavbild 2024-09-12 kl. 18.37.png
 
This is badly worded.
Agreed.

The wording in MT, where a ship has 1 HP per 100 dtons, and weapons consume them (1 HP/Turret, 10 HP/Bay, variable on spinals) is quite better and clearer.
 
It's me again, asking really basic questions again because apparently I'm easily confused by how MT is organized. Can someone walk me through a couple examples of firing weapons at a ship? I'm confused at how to interpret the Combat Tables at the back of the Referee's Manual. I know these tables are heavily errata'd in the 2013 Consolidated document, but I'm still confused on how they function mechanically.

(also, are antimatter missiles just the same +6 DM on surface explosions as nuclear missiles, but affected by proton screens instead of nuclear dampers?)
 
Can someone walk me through a couple examples of firing weapons at a ship? I'm confused at how to interpret the Combat Tables at the back of the Referee's Manual.

I’ll try... All page references are from MT:RM

Let’s imagine a ship with computer 3 is firing its triple missile turret to another ship with computer 3 that has one triple sandcaster turret and a triple blaser one acting as defenses. Both are TL14, 400 dton size and (as most ships in MT) agility 0. They fight at far orbit (50-500 kkm)
  • The TL 14 triple missile battery means factor 3
  • The TL 14 triple sandcaster batery means factor 4
  • The TL 14 triple BLaser batery means factor 4
  • Defensive DM is (page 89) 3 for computer size + 1 for size and 0 for agility, so 4

So, the firer’s task to hit is (page 97):
  • Difficult, off= computer size (3) + weapon table DM (5 for a factor 3 missile) + range DM (+1 for missiles beyond planetary range), Def is 4 (Defensive DM), confrotnation
So, he must roll 11+ wit ha total DM of +5, so 6+ to hit.

Should he hit, he must penetrate both, the Blaser defensive fire and the Sandcasters. In both cases, the task is:

Difficult, off= computer size+ penetration table DM Def= computer size, confrontation (see consolidated errata, page 38)

The penetration table DM, the DM is found in the to pen: disintegrator, sand, beam (page 100). By looking at the factor 3 missile battery against factor 4 defenses, we find a 5). Offensive DM would be 3 (computer size) + 5 (to pen table DM), so 8, while defensive DM would be 3 (computer size)

So, he needs to roll 11+ with a +5 DM, so 6+ to penetrate each defensive layer (BLasers and Sand)

If both are penetrated, then he will roll in the damage table.

Does that help?

NOTE: I left skill effects aside because it is quite confusing in MT. See just this very old post about it. It was discussed in the MT errata thread, and I'll look for it, but I'll need a little time
 
And note that, being a TL 15, the x04 fusion batteries are not possible, as even if single turrets, they would be x05...

Yeah, that mistake was because I had just gotten the book and hadn't read through all the errata yet. Page 74 of MT:RM has a single TL-14 to TL-16 fusion turret as UCP 2. Page 31 of the V2.21 errata adds 3 to all the UCPs on that table.

Does that help?

Yes, that does help. I'm assuming that also means pen is automatic if a target doesn't have the proper defenses? e.g. if you're firing a Disintegrator at a ship without a nuclear damper, you skip the pen test? After going through the examples, that seems obvious, but (unless I missed it) it's not stated in the process.
 
I'm assuming that also means pen is automatic if a target doesn't have the proper defenses?
Yes. That's one of the advantages of the PAs: there are no defenses against them, if they hit, they roll directly for damage (both, in surface explosion and radiation tables)
 
As I promised to look for the discussion about skill use, this was the discussion, on this side thread derivated from the errata one.
 
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