[ Freelance Traveller Home Page | Search Freelance Traveller | Site Index ]

*Freelance Traveller

The Electronic Fan-Supported Traveller® Resource

A Starship Design Tutorial: Advanced Design 

Editor's Note: This article assumes that you have read and understood Peter's Basic design article.

Now let us do something a lot more complicated. TNE starship design requires a bit of practice and organisation. But once you have all of the basic rules and procedures down, it only takes a week or two to get in gear. So I'll do a bigger design as a means of incentive. Since it is no fun to have one person doing all of the design work - and this is as much true for me as anyone who reads this stuff, as I don't want this to become a chore - I'll do the Plankwell Battleship as an example of how straightforward converting older, larger designs can be.

Design Matrix #1: Establish the parameters of the design.

Plankwell Battleship
Displacement: 200,000 tons/2,800,000 m3
Tech Level:15
Configuration: Slab SL
Material Volume Multiplier (MVM): 1410
Length: 467.5 meters
Surface Area: 141,000 m2
Displacement Multiplier (DM): 2000
Acceleration: 5G
Jump: 3

Design Matrix #2: Volumetric systems.

Manuever Fuel (MF)
To obtain the volume of a single g-turn of fuel, divide the ships displacement in tons by 16. (200,000/16= 12,500 m3). Now one percent of a ships displacement, if given over to fuel, is equal to 2.24 g-turns at 1G. 25% of displacement gives 56 g-turns. The Plankwell doesn't need that much fuel since she is primarily for siegework. 50 G-Turns is enough. 50x12,500= 625,000 m3 of reaction mass. a cubic meter of fuel masses 0.07 tons. 825x0.07 equals 43,750 metric tons. Record fuel mass in parentheses to indicate that it adds its mass to a loaded ship only.
Jump Fuel (JF)
Jump fuel is always a basic 5% of displacement, plus 5% times the jump number. 5%+(5%x3)= 20% of displacement in KILOLITERS (do not confuse with the maneuver fuel use of tons). 2,800,000m3x0.2= 560,000 m3. 560,000/12,500= 44.8 additional g-turns of reaction mass. 560,000x0.07 equals 39,200 tons mass
Jump Drive (JD)
The jump drive is always the displacement of the jump fuel divided by 5. 560,000/5= 112,000 m3. At tech level 15 a jump drive masses two tons a kiloliter (112,000x2) or 224,000 tons. A kiloliter of drive equals 0.3 Mcr (112x0.3) or 33,600 Mcr
Armor (MV)
Choosing armor is one of the most important steps. The Plankwell is a battleship, so I need good armor. 70 centimeters (av=1960) might be plenty. 70x1410= 98,700 m3 of Bonded Superdense (BSD). 1 kiloliter of BSD= 15 tons of mass (98,700x15) or 1,480,500 tons. 1 kl of BSD costs 0.028 Mcr (98,700x0.028) times the configuration multiplier on pg.12 for a slab streamline (0.7) gives a final price of 1934.52 Mcr.
Internal Structure (IS)
The formula is the product of the material volume of the vessel (1410) and the acceleration (5G), divided by the toughness of the armor (28). (1410x5)/28= 251.79 m3 (x15), 3776.85 tons mass, (251.79x0.028) and 7.05 Mcr.

The top of the design matrix should look as follows:

System Volume (cu. m.) Mass (tons) Price (MCr) Power
MF 625,000 (43,750)   -
JF 560,000 (39,200)   -
JD 112,000 224,000 33,600 -
MV 98,700 1,480,500 1,934.52 -
IS 251.79 3,776.85 7.05 -
Total 1,395,951.79 1,674,500 35,541.57 -

Design Matrix #3: Volumetric and Power based systems

These include the ship's life support and artificial gravity and discretionary electronics.

Life Support (LS)
Being a long term use vessel, extended life support is a must. This is where the displacement multiplier (dm=2) recorded at the top first comes into play. 100 displacement tons of ship requires 11.2 kiloliters of extended life support. 2000x11.2= 22,400 kiloliters. The mass and volume of life support are the same: 22,400 tons. 11.2 kiloliters of life support costs 0.7 Mcr and consumes 0.28 Megawatts, or 1400 Mcr and 560 MW.
Artificial Gravity (AG)
Always a must unless you're playing with low tech vessels. A ship requires 14 kiloliters of AG per 100 tons displacement. 2000x14= 28,000 kiloliters. Its mass is two tons per kiloliter, price 0.7 Mcr and power 7 MW per 100 tons displacement= 56,000 tons, 1400 Mcr, and 14,000 MW.
Controls (Ctrl)
Multiply AG volume by 0.1 to obtain controls volume. Multiply control volume by 0.1 to obtain mass. Multiply life support (ls) power by 0.1 to obtain controls price (multiply this number by 0.75 for vessels of tech level 12 and less). Controls consume 0.1 mw for every 100 tons of ship, or 0.2 MW.
At her 467.5 meters length, the Plankwell can carry the largest fixed array PEMS sensors. Four 240,000 km Fixed Arrays couple with six 480,000 km AEMS units and ten 300,000 km Ladars. From the integrated PEMS table, a TL-15 240,000 km PEMS is 65.5 kl, 68.5 tons, 68.5 Mcr, and 0.3 MW. Each AEMS is 5 kl, 10 tons, 10 Mcr and 25 MW. The Ladar are 0.6 kl, 1.2 tons, 6 Mcr, and 0.6 MW
Communications (Comm)
I choose six 1000 AU Radio units from pg.49: each is 0.05 kl, 0.1 tons, 0.15 Mcr and requires 20 MW. And twelve 1000 AU Masers (0.03 kl, 0.06 tons, 0.18 Mcr, 0.6 MW each).
System Volume (cu. m.) Mass (tons) Price (MCr) Power
LS 22,400 22,400 1,400 560
AG 28,000 56,000 1,400 14,000
Ctrl 2,800 280 56 200
Snsr 298 346 394 157.2
Comm 0.66 1.32 3.06 127.2
Totals 53,498.66 79,027.32 3,253.06 15,044.4

Matrix #4: Power intensive systems

These include main engineering, armaments and defensive systems.

Maneuver Drive (MD)
We're using Heplar here. The trick is that you need 5 kiloliters of Heplar for every 100 tons of displacement, times the acceleration number. 5x2000x5= 50,000 kiloliters of drive. Mass is the same as volume. Price is drive volume x0.01, or 500 Mcr. Every kiloliter requires 10 MW, or 500,000 MW total.
Conta-Grav (CG)
High Efficiency CG lifters are used at TL-15. A ship requires 30 kiloliters, at 20 tons mass, at 3 Mcr, and 10 MW per 100 tons displacement. Mutilplying by our trusty DM gives us 60,000 kiloliters, 40,000 tons, 6000 Mcr and 20,000 MW.
The old Plankwell in CT Supplement 9: Fighting Ships has a spinal meson gun combined with myriad secondary armament. However the depiction of the ship in FS and on the cover of GT:Rim of Fire both depict it with a meson gun (MG) well short of spinal length. I choose a 280 meter meson gun of 150,000 MJ discharge. The gun tunnel is 280 meters times the (discharge energy times 0.01) for volume of 420,000 kl. The gun tunnel masses 252,000 tons, costs 42,000 Mcr, and at normal rate-of-fire requires [(150,000 MJ times 5 times 10)/1800] 4166.67 MW, or 20,833.33 MW at -1 DMs. The gun requires input energy of (150,000/0.2) 750,000 MJ, times 0.035 to derive an Homopolar Generator stack of 26,250 kl (52,500 tons, 262.5 Mcr). Using CT rules would give me 2000 hardpoints for turret and barbette weaponry. I only need half that number for weaponry. 250 huge 454 MJ remote triple laser barbettes (each 252 m3, 491.026 tons, 4.289 Mcr, 37.83 MW), 500 smaller 140 MJ remote triple laser barbettes (each 84 m3, 156.74 tons, 1.783 Mcr, 11.67 MW), and 250 Missle Barbettes(MB) (each 84 m3, 70.4 tons, 0.11 Mcr, 0.15 MW)
I allocate another 250 turrets as sandcasters(SC) (each 42 m3, 50 tons, 1 Mcr, 1 MW), On top of that I want a meson screen of PV=1200. Dividing 1200 by 80 (PV value for tech level 15 Msn Scrn), square the result, multiply times 8 (size factor of large sized ship) to obtain a value of 1800 MW. Multiplying by 20 gives a 36,000 kl screen, 27,000 tons mass, and 3600 Mcr.
System Volume (cu. m.) Mass (tons) Price (MCr) Power
MD 50,000 50,000 500 500,000
CG 60,000 40,000 6,000 20,000
Meson Gun:
Tunnel 420,000 252,000 42,000 20833.33
HPG 26,250 52,500 2,625 -
LB(454MJ) 63,000 122,756.5 445.75 9,457.5
LB(140MJ) 42,000 78,370 891.5 5,835
MB 21,000 17,600 27.5 37.5
SC 10,500 12,500 250 250
MScr 36,000 27,000 3,600 1,800
Total 728,750 652,726.5 56,339.75 558,213.33

Matrix #5: Power plant and fuel

Add and total the power usage from modules 2-3. Usually the remaining unallocated systems, such as fuel processing plants and computers, do not add much to this total. Add + or -5% to this total as you see fit. So far the Plankwell requires 573,257.73MW. I'll add a few more MWs for the other systems I haven't done yet. I'll settle on 573,600 MWs as my final power plant. A TL-15 powerplant (PP) generates 6 MW per kiloliter, weighs 2 tons per kiloliter, and costs 0.2 MW. My final powerplant is 95,600 kiloliters, weighs 191,200 tons, and costs 19,120 Mcr. Powerplant fuel (PPF), for one year, at this tech level is equal to the power output times 0.1, so I need 57,360 kls for h2 (4015.2 tons mass)

System Volume (cu. m) Mass (tons) Price (MCr) Power
PP 95,600 191,200 19,120 <573,600>
PPF 57,360 (4,015.2) - -
Total 152,960 191,200 19,120 <573,600>

Matrix #6: Crew and auxiliary systems

Crew allocation
Using the formulas in FF&S, pg.13, (and bearing in mind the computer multiplier for a TL-15 computer, 0.2) design the ship crew. I usually add additional maneuver and electronics crewmen for larger vessels. I need 16 Maneuver and 16 Electronics. I need 840 gunnery (volume of meson gun tunnel times 0.01 times Computer multiplier) for the meson gun, and decide to allocated the other weapons in batteries of five apiece, for an additional 200 crew. Using the same means for the meson screen as the meson gun, I discover I need 72 crew, coupled with one hundred twenty five sandcaster batteries, to obtain 197 Screen crew. I also need 3824 Engineering, 30 Small craft (as I need 10 Fast Cutters) and 460 Maintenance. That is 5583 crew so far, requiring an additional 931 Command. They require 139 Stewards, and 55 Medics. For a grand total of 6708.
Master Fire Directors (MFD)
I need 200. At 6 kls apiece for a total of 1200 kls, 1200 tons, 1200 Mcr and 12 MW
Computers and Workstations (Cmp)
The bridge section holds the workstations for the 32 Maneuver/Electronics and 931 command crew(13,482 kls, 192.6 tons, 1.926 Mcr), and another 5061 normal workstations (35,427 kls, 1012.2 tons, 10.122 Mcr) and three TL-15 Fiber computers (42 kls, 8.4 tons, 36 Mcr, 3.3 MW total)
Staterooms (SSR)
I decide that each command personnel and the chief medic rates a single occupancy small stateroom, and the rest of the crew shares small staterooms in double occupancy. 3820 small staterooms times (28 kls/2tons mass/0.04 Mcr/0.0005 MW)=106,960 kls/7640 tons/152.8 Mcr/1.91 MW
Fuel Processing Plants (FPPs)
Things get a little tricky in this step. FPPs are allocated according to six hour intervals. At tech level 15, I need 0.2 kiloliters of FPP for every kiloliter of fuel. This covers maneuver and jump fuel only, as the powerplant requires only annual refuels. This totals 1,185,000 klsx0.2= 237,000 kiloliters for six hours of refining time. This number is too high for my purposes. Since every six hours (every DOUBLING) added of fuel processing time added halves the fuel plants displacement, I decide on an 24 hour fuel processing time. 4x6=24 hours. Then I take the inverse of 4 (0.25) and multiply that by original fuel plant's volume to obtain a value 59,250 kls. Multiply times 2 to obtain mass. To obtain the price and powerplant divide the fuel volume by 100 (11,850). First multiply this times the FPP price per 100 kls of fuel (0.015 Mcr) then by the inverse integer from before (x0.25) to obtain the price (44.438 Mcr). Secondly multiply 11850 times the power requirement for 100 kls of fuel (0.5 MW at TL-15) times the inverse integer (0.25) for a total 1481.25 MW.
Hangar (Hngr)
For the ten times 50 ton Lushina Cutters (7000 kls total), minimal hangars will suffice (14,000 kl, 2800 tons, 2.8 Mcr) plus 10 Launch Ports
Airlocks (AL)
I need one airlock for every 100 tons displacement. This gives the Plankwell 2000 airlocks (6000 kls, 400 tons, 10 Mcr, 2 MW)

I do not need any other basic equipment. This module should look like

System Volume (cu. m.) Mass (tons) Price (MCr) Power
MFD 1,200 1,200 1,200 12
Cmp 48,951 1,204.8 48.048 3.3
SSR 106,960 7,640 152.8 1.91
FPP 59,250 118,500 44,438 1,481.25
Hngr 14,000 2,800 2.8 -
AL 6,000 400 10 2
Total 236,361 131,744.8 45,851,648 1,500.71

Now we total the "totals" of each matrix. The vessel has allocated 2,567,520.66 kl/m3. Subtracting it from 2,800,000 kl, I obtain 232,479.34 kl/m3 of cargo. The craft has an empty mass of 2,729,198.62 tons tons, and a loaded mass (adding cargo and fuel, but not the air/raft) of 3,048,643.16 tons, which exceeds the density based constraints. To compensate I'll eliminate 100,000 kls of cargo and replace it with 8 more g-turns. This changes the loaded mass to 2,955,643.16 tons and the Fuel Processing Plant now takes 27.84 hours to refine a full tank. She has a total cost of 160,106.028 Mcr. Her systems consume 574,758.44 MW, leaving a 1158.44 MW shortfall.

Matrix #7: Damage Table Allocations

Divide the ships displacement (2,800,000 kl) by 400= 7000 kls. This is the volume allocated to every hit location on the ships damage chart. Add, using the guide on pg. 15 and 16 to lump the sum of each system in one of five categories: electronics, quarters, weaponry, hold and engineering. Electronics includes sensors, computers and workstations, communications, screens, EMM/ECM and controls. Qtrs include the material volume, LS and AG, staterooms, shops and sickbays, and airlocks. Weaponry is offensive systems and "point" defenses like repulsors, nuclear dampers and sandcasters. Hold is fuel, cargo and carried craft. Engineering is power plant, maneuver and jump drives, lifters and fuel processing plants. For the Plankwell these numbers are:

System Allocation (kl) # Hit Locs
Electronics 89,249.66 12.75
Quarters 262,311 37.47
Weaponry 582,750 83.25
Hold 1,488,839.34 212.69
Engineering 376,850 53.84
Total 2,800,000 400

Since these numbers must be rounded off to the nearest whole number, these become:

Elec:13 Qtrs:37 Wpn:83 Hold:213 Eng:54

Now to correct a little error under the basic tutorial: when allocating surface hits, one takes the surface area of that system, divides it by the total surface area, and MULTIPLIES it times 400 hit locations. Sensors and communicators are lumped under "antenna". The radios have 200 m2 each, the masers are 1 m2, the PEMS are 1250 m2, AEMS are 10 m2 each, and the Meson Screen is 18,000 m2 for a total of 23,846 m2. (23,846/141,000)x400 equals ~67.65 hit locations, or a total of 68. EMM radiators are allocated in m2=(displacment in tons)x0.14= 28,000 m2. (28,000/141,000)x400= ~79.43 hit locations, or 79. Other systems include the 100 cargo hatches (2000/141,000)x400 = ~5.67 or 6, and the airlocks, (6000/141,000)x400= ~17.02 or 17, and finally the launch ports, [(11.2^2x10)/141,000]x400= ~3.56 hit locations, or 4. Totalling these up gives 174 locations, which is more than enough for j-drive, cg lifters and fuel scoops.

This design is completed.