Themistocles Cruiser

[Jame Rowe]

Themistocles-class Battleship
10,000 ton hull TL 13
Hull 222, Structure 222
15 points Crystaliron Armor
Jump 4 (500 tons)
Maneuver 6 (700 tons)
Power plant 6 (500 tons)
Emergency Power 1 (50 tons)
4753 tons total Fuel (4670 Jump, 83 tons Maneuver) OR 283 tons total if (Jump Drive*JN)*0.1
Fuel processors (250 tons)
Very Advanced sensors w/distributed arrays (45 tons)
Countermeasures (7 tons)
Enhanced Signal Processing (2 tons)
Hardened Bridge with holographic controls (60 tons)
Armored Bulkheads (200 tons)
286 staterooms (1144 tons)
286 escape pods (143 tons)
20 missile bays (200 tons)—I may have miscalculated on the bays but that’s my story for now
10 particle beam bays (100 tons)—again, miscalculation possible (if anyone wants to figure out how to get these in with the rest of the stuff at their real value let me know).
70 triple turrets (70 tons)
30 sandcasters in ten turrets
45 pulse lasers in fifteen turrets
45 beam lasers in fifteen turrets
45 beam lasers in fifteen turrets
45 particle beams in fifteen turrets
2 nuclear dampers (54 tons)
2 meson screens (108 tons)
1 command sections (10 tons)
2 briefing rooms (16 tons)
1 library (8 tons)
10 armories (20 tons)
3 pinnaces (120 tons)
4 ship’s boats (120 tons)
7 GCarriers which are TL-13 (42 tons)
Hangar space (299 tons)
373 tons cargo (245 tons missile and sand-canister storage, 148 tons general cargo)

7000 Megacredits (standard design); 92 weeks build time
247 crew, 125 marines

The Themistocles-class battleship is a intended as a high-end heavy cruiser in a smaller-ship universe (maximum vessel size being 20,000 tons), or a light cruiser for independent operations in a large-vessel universe.


I think that I calculated something wrongly, and that the tonnages are off; if anyone notices anything or has suggestions please point out.

 

[BP]

Themistocles-class Battleship
...
20 missile bays (200 tons)—I may have miscalculated on the bays but that’s my story for now
10 particle beam bays (100 tons)—again, miscalculation possible (if anyone wants to figure out how to get these in with the rest of the stuff at their real value let me know).
...
I think that I calculated something wrongly, and that the tonnages are off; if anyone notices anything or has suggestions please point out.

Use the TL modifiers (also the upgrades) from HG pg 53 (noticed you used for hull) - so bays = 50 tons x Mod% + 1 ton fire control (Carrrier example and Bomber)

So your bays are TL6 & 8 respectively so at TL11+ (you mention TL13 so there is room for upgrades as well) you get the bays for 31 tons each...

620 tons and 310 tons = you need to drop the bay counts or come up with 630 tons

Don't forget 75% reduction for M-Drive base TL-9 = 3.25% *.75 = 243.75 tons! Since you had 700 tons (which was incorrect BTW) you have 456.25 of your tons there...

So you only need to come up with 173.75 more tons or drop 6 bays!

You might also consider dropping some of the fuel processor tonnage - 115 tons would allow full processing in ~ 2 days rather than 1 - recovering 135 tons.

Leaves you needing only 38.75 tons (if the rest of your work is correct)...

If TL mods apply to turrets that would be 28 of those tons. (there is separate post about this...)

 

[snrdg121408]

Afternoon James Rowe,

I may have found a small miscalculation for armor which is listed as
15 points Crystaliron Armor if I am understanding the rules correctly anyway.

Is this really armor points or protection % per ton of hull?

HG p. 53 Armor table shows that crystaliron has a max armor point rating
of TL or 13, whichever is less.

If the 15 is protection % per ton of hull then the ship has

15/5 = 3 x 4 = 12 armor points and uses up 15% x 10,000 = 1,500 dtons.

If the 15 is armor points they need to be pared down to 13 armor points
and uses up 13/4 = 3.25% x 10,000 = 325 dtons.

 

[snrdg121408]

Hello James Rowe and BP,

If the 15 is armor points they need to be pared
down to 13 armor points and uses up 13/4 = 3.25% x 10,000 = 325 dtons.

Oops, I goofed on the math :shock: The correct formula is:

13/4 = 3.25 x 5% =.1625 x 10,000 = 1,625 dtons not 325 dtons.

 

[BP]

And for reduced tonnage he could go with Bonded Superdense at 6 points per 5% (13/6 *.05 *10,000 = 1083 1/3 tons - whole armour points, but...)

And how does one read the Cost column of that table = +50% per 5% - yikes, may want to stick with the Crystaliron!

 

[snrdg121408]

Evening BP,

And for reduced tonnage he could go with Bonded Superdense at 6 points per 5% (13/6 *.05 *10,000 = 1083 1/3 tons - whole armour points, but...)

The Themistocles is a TL13 design Bonded Superdesne is TL14.

And how does one read the Cost column of that table = +50% per 5% - yikes, may want to stick with the Crystaliron!

Looking at the example at the end of column 1 on HG p. 63 I'm not
sure how to figure armor cost for crystaliron armor.

Per p. 63 "This extra armor uses 2,000 tons of volume (10% of the hull
volume) and costs 20,000 x (MCr0.1 x 90%) x 20% for a total of MCr360."

I am sure about 3 of the items shown in the example which are listed
below:

Base hull = 20,000
Base hull cost per ton = MCr0.1
Where did 90% come from?
Armor Cost modifier = 20%

 

[BP]

The Themistocles is a TL13 design Bonded Superdesne is TL14.

Doh! Quite right - I kept thinking his TL was 15 and had to go back and change from my original calcs back down to 13 - then didn't even check to see if it was still valid (so much for just double checking my maths)!

Looking at the example at the end of column 1 on HG p. 63 ...
I am sure about 3 of the items shown in the example which are listed below:

Base hull = 20,000
Base hull cost per ton = MCr0.1
Where did 90% come from?
Armor Cost modifier = 20%

Well, at first I thought the 'x 90%' comes from the fact that 10% of the hull volume is armour (100% - 10%). But if this were the case it wouldn't make sense (since more armour will decrease the cost).
Then I re-read the paragraph above the example, which states 'Planetoids... armoured as if they were a close structure vessel...' and that explains it = see HG pg 62 where Cost for Close Structure configuration is -10%! (And this still doesn't look kosher to me since 2x the armour doesn't change the cost - but read on...)

And now - look at the Planetoid Monitor on pg 113 - it is +12 pts so 3 upgrades at 5% or 15% of 30,000 = 4,500 tons. Cost = 30,000 x (0.1 MCr x 90%) x 20% x 3 = 1,620 MCr. Note the 'x 3' for the 3 upgrades. The example did not properly do this.

Now that I answered yours - maybe you can help me out here - where did the 0.1 MCr per ton come from, since on the same page it is stated that there is a hull cost of 4,000 Cr per ton for planetoids. I can see this coming from cost of reinforced hull as planetoids have too low a cost for this metric (another reason I had a problem with the % cost on the table - it just didn't look right - somebody got too creative and now we have to apply numerous exceptions to avoid inconsistencies - argh!).


[I think its time to start my own personal errata list - for the last few months I've worked under the assumption that my misinterpretations are generally to blame for inconsistencies and errors - now I'm to the point were I believe I am identifying actual errata!]

 

[snrdg121408]

Evening again BP,

Thank you for the quick reply, solving the 90% mystery, and figuring
out the cost of the planetoid monitor on HG. 113.

Now that I answered yours - maybe you can help me out here - where did the 0.1 MCr per ton come from, since on the same page it is stated that there is a hull cost of 4,000 Cr per ton for planetoids. I can see this coming from cost of reinforced hull as planetoids have too low a cost for this metric (another reason I had a problem with the % cost on the table - it just didn't look right - somebody got too creative and now we have to apply numerous exceptions to avoid inconsistencies - argh!).

The MCr0.1 is, in my opinion, the base cost a standard hull per the text
above Configuration on HG p. 62. With that said the formula for figuring
planetoid and buffered planetoid hulls should be based on Cr4,000 or
MCr0.004 and not MCr0.1.

None of the other designs appear to use the armor upgrade factor in
calculating cost. Which means I'm so confused :shock:

I think its time to start my own personal errata list - for the last few months I've worked under the assumption that my misinterpretations are generally to blame for inconsistencies and errors - now I'm to the point were I believe I am identifying actual errata!

I agree that there appears to be a need for erratum, since I try to
reverse engineer the published designs.

Thanks again, my bunk is calling for me so I'll shutdown for the night.
Thanks again for figuring out the example.

 

[BP]

The MCr0.1 is, in my opinion, the base cost a standard hull per the text above Configuration on HG p. 62. With that said the formula for figuring planetoid and buffered planetoid hulls should be based on Cr4,000 or MCr0.004 and not MCr0.1.

None of the other designs appear to use the armor upgrade factor in calculating cost. Which means I'm so confused :shock:

Ahha! - I looked for that and then lost track of what I was looking for when I found the hull reinforcement cost! Thanks - that has to be it - in fact, it is even somewhat consistently named 'base cost' to the 'base hull'! (And my interpretation is going to be that the base hull TL cost modifiers don't apply - they obviously shouldn't apply to the planetoid type hulls - and they would just increase cost without gain for others - hence the 'base' terminology).

As for the other examples:

Sylea Class pg 101 -
Armour= 15pts /(6pts/5%) = 2.5 x 5% = 12.5% of hull tons (25,000 tons)
Standard Hull = no base hull cost modifier (=> 100% cost)
200,000 x(0.1 MCr x 100%) * 50% x 15/6 = 25,000 MCr

Cruiser pg 119 -
Armour = 10pts /(6pts/5%) = 8 1/3% of hull (6,250 tons)
Cone Hull = +10% base modifier (=> 110% cost)
75,000 x(0.1 MCr x 110%) x 50% x 10/6 = 6,875 MCr

Carrier pg 127 -
Armour = 2pts /(4pts/5%) = 2.5% of hull (750 tons)
Close Structure Hull = -10% base modifier (=> 90% cost)
30,000 x(0.1 MCr x 90%) x 20% x 2/4 = 270 MCr

Escort pg 136 -
Armour = 6pts /(4pts/5%) = 7.5% of hull (375 tons)
Cylinder Hull = no base modifier (=> 100% cost)
5,000 x(0.1 MCr x 100%) x 20% x 6/4 = 150 MCr

So these all appear correct - hope the above is clear enough...