I'm ok with using LBB2 drives in small craft, though you need to break the 4-week powerplant fuel requirement to do it (house rule). The TCS/JTAS #14 power-down rule might be a way around it, if you break it down to affect periods shorter than one month (also a house rule).
You end up with something analogous to the LBB2:'77 small craft fuel burn rates.
I'd suggest that any maneuver drive with less than 200 "tons of thrust" (that is, a Size A maneuver drive from LBB2) should be larger and more expensive per "ton of thrust" because starship maneuver drives are just better. If they were worse, then the better small-craft drives would replace them in at least the smaller starships.
Probably won't be using a Size C drive in anything under 95Td since 6G is the best you can get. So, it's Size B and A then.
Benchmark fuel use rates on what's needed in a 100Td hull just to keep from breaking LBB2 any further than necessary. (If I were starting from scratch, I'd benchmark all LBB2 power plant use on the Pn it would have if it were in a 400Td hull or so -- basically, 5Td per month per (ordinal value of) drive letter. Size A always takes 5Td, B takes 10Td, etc. regardless of what ship it's in. But I'm trying not to break too much here.)
Consider benchmarking drive cost at the cost of the drive in a 100Td hull, because the price formula would otherwise heavily discount these drives when used in smaller craft.
[Deleted: Redundant. This only matters when trying to calculate the cost of fractional/intermediate drive sizes, and this exercise only uses ones specified by LBB2.]
Size A maneuver drives yield the following performance:
34Td hull: 6G (rounded from 33.33)
40Td hull: 5G
50Td hull: 4G
67Td hull: 3G (rounded from 66.67)
99Td hull: 2G
Size B maneuver drives yield the following performance:
67Td hull: 6G (size is rounded)
80Td hull: 5G
99Td hull: 4G
This is where it's necessary to go off the reservation.
Rationale behind spoiler so the stats are up front:
Spoiler:
There are two limit conditions for fuel consumption.
The first is the LBB2 rate of 10Td per Pn. This gets unworkable in small craft VERY quickly, a 50Td craft with a Size A drive (Pn-4) needs 40Td fuel for its 5Td of drives. A bridge at 20% of tonnage isn't going to fit, let alone anything else.
For the purposes of this exercise (house rule alert!), I'm capping that at the worst-case requirement in LBB2: 20Td per (ordinal value of the) drive letter, from use in a 100Td hull; that is, A=1, B=2, C=3. Thus, a Size A powerplant uses 20Td fuel in 4 weeks regardless of how small a hull it's in; Size B, 40Td in 4 weeks, Size C, 60Td.
The second limit is based on the TCS/JTAS#14 power down rule, pro-rated to shorter periods than 4 weeks. From that, the minimum fuel consumption rate is the Pn=1 rate.
Based on that, the minimum possible fuel consumption is: 10Td in 4 weeks, based on Pn=1 in a 100Td hull.
If the 4 week requirement can be bent (house rule!) so that a small craft must have 4 weeks of Pn-1 fuel instead of 4 weeks at its max burn rate, then the minimum fuel requirement is 10Td for any LBB2 drive.
This kind of works for the Size A drives: Using it at full power full-time only cuts that 4 week reserve to 2 weeks at Pn-2*, and 3 weeks is possible at 3/4 power (50Td hull, 3G instead of rated 4G; pro-rate appropriately for operating at reduced power in other hulls).
It gets problematic for the Size B drives: Full power, full-time cuts the 4-week Pn-1 reserve to 1 week at Pn-4*. (2 weeks at half power).
It's a real issue for the Size C drives: Full power, full-time cuts the 4-week Pn-1 reserve to 4.7 days at Pn-6* (1.33 weeks at half power).
Using the Size A Drives as a baseline, the 4 week Pn-1* allocation is 2 weeks at constant full power. So, set the minimum fuel requirement as 2 weeks at full power so the bigger drives aren't stuck with too-short endurance if they really use their drives. (This is half the starship/non-starship requirement, and obviously it's a house rule.)
Size A drives need 10Td for 2 weeks. Idled down to 1G, they can run for 4 weeks.* This (house rule!) satisfies the minimum fuel requirement.
Size B drives need 20Td for 2 weeks. Idled down, 8 weeks.
Size C drives need 30Td for 2 weeks. Idled down, 12 weeks.
This does require some math, where the other design systems just say 4 weeks no matter what.
And it takes a lot more fuel than the LBB2 small craft use, even the '77 versions (but they had much less total delta-v).
EDIT: Not for the '77 rules, by a long shot! The Size A drives get 2 weeks of 200 tons thrust out of 10Td fuel, rules-as-written*.
In LBB2:'77 (or at least DA1) the 40Td pinnace only gets 100 minutes of the same output for the same amount of fuel (1Td fuel for 200 tons thrust for 10 minutes, with a 10Td tank).
At 5G, using the travel time table in LBB2, that's enough for a 400,000km trip -- about the distance from the Earth to the Moon. One way.**
'77 had a far more limited vision of what small craft are capable of, compared to the '81 rules (and HG). I don't think I really want to go to that extreme...
*calculated based on usage in a 100Td hull; i.e. 20Td for 4 weeks but half that if powered-down per the TCS/JTAS#14 rule.
**if you want to be able to come home without needing to refuel when you reach the Moon, throttle back to maybe 2G or just coast most of the way.
To make it simple I would use a few base drive packages:
A-drives × _10% = _20 u = 0.5 Dt, MCr _1.2, 0 EP(=lasers): for the poor Launch
A-drives × _50% = 100 u = 2.5 Dt, MCr _6.0, 1 EP(=lasers): fighter, slow boat & pinnace
A-drives × 100% = 200 u = 5.0 Dt, MCr 12.0, 2 EP(=lasers): ship's boat, pinnace, cutter
A-drives × 150% = 300 u = 7.5 Dt, MCr 18.0, 3 EP(=lasers): shuttle
Since the drive packages are linear with regards to size, cost, and performance we can trivially scale them as desired, thereby reinventing a LBB5 style percentage based system...
For fuel I would just drop the 4 week requirement and allow much shorter mission endurance. Unless we have staterooms installed we will not spend much more than 12 h in them anyway. If we start with 20 Dt/4 weeks for the A drive package, we get 5 Dt per week or ~1 Dt per day scaled as above.
If we are house-ruling anyway, I would allow capacitors from LBB5 to be used to power small craft as batteries, replacing the power plant and fuel. Good for a few hours, but fusion plants are better for days or weeks of endurance. A few hours are often enough for small craft. If carried aboard ship, the small craft can be recharged by the fusion plant on the ship.
For simplicity just halve the size and cost of the drive package to remove the fusion power plant, and add a battery at 36 EP×turns and MCr 4 per Dt. A 1 Dt battery (36 EP×turns) would power an A drive (2 EP) for 36 / 2 = 18 turns ≈ 6 h, saving a few Dt and MCr.
40 tonnes of thrust per cubic meter of drive, with a mass of 2 tonnes and requiring 1MW of power. It also requires thrust/200 sq meters of surface area, and costs 1MCr.
There's no claimed minimum size.
So, a 1 tonne thrust drive would be roughly 30cm on a side, weigh 50 kilos and consume 25Kw of power.
Clearly there must be some minimum size. "How much for a gram of thrust!?"
If you want a justification for even more burn outside of unworkable reaction mass, I worked up a missile as mini-craft paradigm, needed limited range fuel use, and figured the smaller missiles have to burn off fuel as a way to shed heat from the inefficient mini-fusion reactors. You could say that putting in full size reactors in small craft means not enough hull/maneuver drive 'bubble' to burn off the heat and they need to eject the stuff the same way.
To make it simple I would use a few base drive packages:
A-drives × _10% = _20 u = 0.5 Dt, MCr _1.2, 0 EP(=lasers): for the poor Launch
A-drives × _50% = 100 u = 2.5 Dt, MCr _6.0, 1 EP(=lasers): fighter, slow boat & pinnace
A-drives × 100% = 200 u = 5.0 Dt, MCr 12.0, 2 EP(=lasers): ship's boat, pinnace, cutter
A-drives × 150% = 300 u = 7.5 Dt, MCr 18.0, 3 EP(=lasers): shuttle
Since the drive packages are linear with regards to size, cost, and performance we can trivially scale them as desired, thereby reinventing a LBB5 style percentage based system...
For fuel I would just drop the 4 week requirement and allow much shorter mission endurance. Unless we have staterooms installed we will not spend much more than 12 h in them anyway. If we start with 20 Dt/4 weeks for the A drive package, we get 5 Dt per week or ~1 Dt per day scaled as above.
If we are house-ruling anyway, I would allow capacitors from LBB5 to be used to power small craft as batteries, replacing the power plant and fuel. Good for a few hours, but fusion plants are better for days or weeks of endurance. A few hours are often enough for small craft. If carried aboard ship, the small craft can be recharged by the fusion plant on the ship.
For simplicity just halve the size and cost of the drive package to remove the fusion power plant, and add a battery at 36 EP×turns and MCr 4 per Dt. A 1 Dt battery (36 EP×turns) would power an A drive (2 EP) for 36 / 2 = 18 turns ≈ 6 h, saving a few Dt and MCr.
My objective here is to break as little as possible of the existing rules. Allowing capacitors to power drives allows smaller power plants across the board. Doing so in combination with LBB2 maneuver drives REALLY breaks things if it's applicable across the board.
Example:
200Td A2 Far Trader at 6G
PP-B, MD-F, 2.2Td capacitors at MCr4.0/Td
For MCr20.4 and 9.1Td extra compared to a base A2, it can be 6G for 5 combat turns before reverting to 2G (MCr24.8 and 10.2Td allows 10 combat turns at 6G before reversion to 2G).
6G is possible in a J2, 200Td ship in LBB2:'81. In the A2 it would need 20 more tons of drives (+MCr48) and 40 more tons of fuel.
You could. The problem, as I keep pointing out, is that under the formula for LBB2 maneuver drives that 1/2 ton module of maneuver drive puts out 150 tons of thrust, not 100. It's like the glitch in High Guard maneuver and jump drives, but beneficial rather than a penalty.
I think it's just artistic license, combined with the designer not overthinking it to this extent.
The glitch:
Spoiler:
Using jump drives as an example: a 100TD J4 ship has 5% of its tonnage as jump drive. If it has 100Td of drop tanks/cargo modules/whatever attached, the Jump Drive is 2.5% of the total, and the ship can now only do Jump-1 because it needed 3% to do Jump-2.
On the other hand, a canon LBB2:'77 Xboat (or a house-ruled LBB2'81 version with 50Td fuel) could carry 100Td of drop tanks or whatever at J2 because it's using a Size B drive that has a rating of 2 in a 200Td hull.
I ran into that problem going the other direction, trying to build a Scout version of the A2 Far Trader. I had it using the Jump Drive B and power plant B from the XBoat because they'd be in the warehouse already. However, there's no other canon IISS ship that needs a Size C maneuver drive, which means there'd be no spares in the logistics system. So, it used two Size A maneuver drives that they'd keep on hand for Scout/Couriers.
That left one ton extra in the drive bay. Same cost though, since the price formula is just based on rating and ship tonnage. I house-ruled that the two maneuver drives needed a 1Td adapter bracket or something. (Apparently T5 has a rules mechanic for this.)
If you want a justification for even more burn outside of unworkable reaction mass, I worked up a missile as mini-craft paradigm, needed limited range fuel use, and figured the smaller missiles have to burn off fuel as a way to shed heat from the inefficient mini-fusion reactors. You could say that putting in full size reactors in small craft means not enough hull/maneuver drive 'bubble' to burn off the heat and they need to eject the stuff the same way.
That might work. Make it like jump fuel burn rates?
Right now, the results look like '77 fuel tank size with '81 endurance.
To get '77 endurance and '81 fuel tank size, just lower the idle fuel consumption. Let it run on batteries for a week at a time if it's not under acceleration, recharging by powering up for an hour once a week. Justify it by using the TCS power-down rule, but (unlike for all of my other calculations) at the rate for Pn-1 in the small craft hull rather than Pn-1 in a 100Td hull (10Td/month). Best case is in a 30 (33.3)Td Ship's Boat, which can idle down to 1/6 power (1G), needing only 3.3 tons per month. It works out to about twice the HG fuel requirement (1.8 tons).
20 tons fuel in 4 weeks (full power burn rate) is about 0.03 tons per hour (this is for a Size A power plant supporting 200 G-tons thrust).
A 1Td fuel tank (HG minimum) provides 33.6 hours of operation at full power (Ship's Boat at 6G, Pinnace at 5G, Cutter at 4G).
The LBB2:'81 versions of each of these small craft carry 2 tons of fuel, which would be good for 67.2 hours (2.79 days) at full power.
At idle, that's 16.74 days.
This is much closer to the '77 concept, but not so much so that you need to count every G-turn in a battle.
Just as a reference, lets compare a 50Td, 4G Cutter with a Size A drive with one build from High Guard.
Size A:
Drives: 5Td, MCr 12.
Fuel: 3.3Td.
Total: 8.3Td, MCr 12 (2.79 days at full power, 4 weeks at idle.)
High Guard (TL9-12):
Drives: 11.5Td, MCr20.75
Fuel: 2.0Td
Total: 13.5Td, MCr20.75* (4 weeks at full power)
Add fuel to the LBB2 drive package to match the tonnage of the TL9 HG package: 8.5Td fuel, total 13.5Td.
That's 1.7 weeks of full power operation, where the HG package runs for four weeks.
To match it for endurance (rules as written), it needs 20Td fuel for a total of 25Td with the drives.
To sum up, if you're willing to trade endurance for cost savings, Size A drives in a cutter save MCr8.75. (That's a 31% cost reduction! Maybe 28% with class discount.)
If you're matching endurance though, it's giving up 11.5Td of payload for that MCr8.75.
Sacrificing endurance for both cases cuts the payload penalty of using Size A drives significantly.
*At TL13-14: 9.5Td, MCr14.75; at TL15: 9.5Td, MCr8.75
The problem is trying to make sense of the drive formulations, except in that for Tee Five and Mongoose, you're back to five tonnes overhead plus two and a half percent per jump factor, minimum ten tonnes (otherwise you could have one at seven and a half tonnes).
With manoeuvre drives, classic High Guard is three percent per acceleration factor, minus one.
With the alphabet drives, the best interpretation is that they're commercial off the shelf models.
On the other hand, model Zed jump drive on a five kay tonne hull, with a jump factor of two, comes to two and a half percent net, two million bux per tonne, while model Why is five tonnes and ten million bux less, and can only achieve jump factor one, with no other drives qualifying.
Manoeuvre drives are two tonnes per ascending letter, minus one, at four million bux per ascending letter.
Fo manoeuvre drives you could come to the conclusion of diminishing returns per ascending letter.
So yeah, if you took a line from hundred to a thousand tonnes, it's one tonne to fifteen, you'd end up with 67 kilogrammes at ten tonne hull if you drew the line in the other direction, for factor one acceleration.
Which means for a ten kay hull, it should be 225 tonnes for factor one acceleration; yet, you could presume that a forty seven tonne Zed model which can accelerate a five kay tonne hull at two gees, should be able to do one gee for a ten kay tonne hull.
The biggest problem with all this is that maneuver drive isn't linear.
The power equation is P = k·a²
When you look at HP/ton in quarter-mile drag racing, it fits almost perfectly from your 18 s street vehicle to a sub-4 s top fuel beast. 350 HP/ton at 1 G (9 s), iirc, with top fuel at about 5 G.
