LBB2 M-Drives in LBB5: When does it help?

[Grav_Moped]

ut there are no abstract "potential-tons" in LBB2, there is only a table that forces some random hull sizes to be even remotely efficient.

A drive J in a 600 Dt hull is potential 3, so in a 300 Dt hull it's of course potential 6? No, it's potential 4 because the table isn't granular enough... Hence 300 Dt hulls are strongly discouraged for no good reason.

Why is my ship penalised just because I wanted to make it 300 Dt?

Why does it need to be 300Td specifically, anyhow? LBB2 arbitrarily penalizes a lot of things it probably shouldn't.

I'd be perfectly fine with a house rule allowing that sort of use of the underlying formulae though -- it's the other side of the intermediate-sized-drive house rule, and with less inference needed. The only problem is when you start doing this with the drives that don't follow the formula (J at 2000Td, and the W-Z drives).

 

[Grav_Moped]

As for small craft they generally use either a A class Power and drive. Note I have postulated subA and subB sized drives to allow for a wider range of small craft as well as a J1 100 ton hull.

The only problem with using drives derived from LBB2 for small craft is the fuel requirement (a 10Td 6G fighter needing 60Td fuel!?) The "negative-tonnage maneuver drive" problem can be mitigated -- or at least handwaved -- by bundling the power plant and maneuver drive tonnage together before scaling down. (Yeah, the math says a 100G*ton ("half-A") "small craft drive" is just a power plant, and at 50G*tons it's an oversize power plant at that, but we'll ignore that detail in favor of playability.)

Side-note, I have pondered using Book5 to generate the standard drives as match the relative preportion sizes between the two sources.

I think we're kind of on the same page here if I'm reading this right (something like flipping the percentage requirements of the jump and maneuver drives).

The thing is, at larger sizes the +1Td (power) or -1Td (maneuver) or +5Td (jump) components of the formulae become such a small part of the total that they can be disregarded.

 

[Spinward Flow]

The point of too large drives would be damage capacity, or tug work.

External towing capacity ... where have I heard that advocated before ...?

 

[whartung]

If you look at the LBB2 table, it's clear that each drive provides some amount of power, and the various drive ratings correspond to that power being divided by some factor (i.e. perhaps mass) and rounded up.

The standard drives could just be generating XXX tons of thrust and everything else falls out from that.

The issue is that you only get a single drive in the ship, rather than being able to gang drives. 2 100 tons thrust drives instead of 1 200 ton.

The benefit of standard drives is that they're standard. Which matches much of the "real world". As I've mentioned before, even US Navy ships uses "standard" drives, drives installed in different hulls, in different navies. It doesn't seem many are engineered specifically for the particular vessel. Rather the vessel is designed around the drive. Obviously this gives benefits of maintenance, parts, scales, etc. that you don't get with one off drives.

 

[Spinward Flow]

If you look at the LBB2 table, it's clear that each drive provides some amount of power, and the various drive ratings correspond to that power being divided by some factor (i.e. perhaps mass) and rounded up.

The best way to think of it (until you get to the TL=15 W+ drives that is) when it comes to standard drives and their performances is ... they're all Code: 1 designed for specific hull sizes.

A-Drives are Code: 1 in 200 ton hulls.
B-Drives are Code: 1 in 400 ton hulls.
C-Drives are Code: 1 in 600 ton hulls.
D-Drives are Code: 1 in 800 ton hulls.
... and so on and so forth.

Basically, each "drive letter" is a +200 ton displacement bump up the scale to yield a Code: 1 performance output.
That's why A/A/A standard drives yield J1, M1, PP1 in a 200 ton form factor (hello, A1 Free Trader!).

Once you have that "basic insight" ... everything else rather simply flows from there.

Put an A-Drive into a 100 ton hull and you get a Code: 2 performance output.
That's why A/A/A standard drives yield J2, M2, PP2 in a 100 ton form factor (hello, Type-S Scout/Courier!).
It's also why A/A standard drives yield M4, PP4 in a 50 ton form factor (hello, Modular Cutter!)

In other words, in order to get Code: 4 performance in a 400 ton hull ... you need a standard drive that yields Code: 1 performance in a 1600 ton hull. It's just straight multiplication.

So you could install B/B standard drives which yield M6, PP6 into a 66 ton form factor if you wanted (sounds like a heavy fighter to me!).
Between 66-100 tons, you're going to have to resort to C/C standard drives in order to achieve a M6, PP6 yield throughput.

As soon as you derive a "standard drives every 200 tons" step parameter system for Code: 1 performance and then base everything else associated with those steps on a straight multiplier basis ... it becomes a LOT EASIER to convert LBB2 standard drives into a more cleanly coherent "formula based" approach to the entire problem. That then lets you sidestep the "maneuver drives get a 2c-1 formula" conundrum entirely ... because the smallest standard drive is intended for a 200 ton form factor as a Code: 1.

A natural consequence of this method of extrapolation is that an A-Drive in a 10 ton form factor ought to yield a Code: 20 performance output (as in ... M20, PP20). Of course, those A/A standard drives would cost 1 and 4 tons respectively for Maneuver and Power Plant, leaving only 5 tons remaining for everything else to be installed ... but I'm just wanting to illustrate the point here of what happens when you take the "training wheels off" the chart based approach to the presentation of information used in LBB2 and instead move into the much more powerful realm (mathematically speaking) of using formulas instead. Simply stipulate that when it comes to the computation of USP style performance profile coding, whenever you multiply and divide to determine drive performance, you're always looking at a "drop fractions" condition.

C-Drive that is Code: 1 in a 600 ton hull is still Code: 1 in a 400 ton hull (because 1.5 drop fractions is still 1), but becomes Code: 2 in a 300 ton hull ... and Code: 3 in a 200 ton hull.

Such a system can theoretically be extended upwards all the way to 1,000,000 tons of displacement, in +200 tons increments.
You'll need 50,000 discrete standard drives in order to do so in +200 ton increments ... but that's the idea.

 

[Grav_Moped]

So you could install B/B standard drives which yield M6, PP6 into a 66 ton form factor if you wanted (sounds like a heavy fighter to me!).
Between 66-100 tons, you're going to have to resort to C/C standard drives in order to achieve a M6, PP6 yield throughput.

You certainly could.

You'd also need 60Td of fuel for that 66Td fighter, unless you change the formula for that (and in all fairness, you probably should).

The change would either be to cut the requirement (not the full 4 weeks) or cut the consumption rate (not 10Pn -> 4weeks).

 

[mike wightman]

I noticed a couple of transcription errors, and since we have proletariat editing rights here is the corrected* table:

Spoiler:

Drive	A	B	C	D	E	F	G	H	J	K	L	M	N	P	Q	R	S	T	U	V	W	X	Y	Z
Output	200	400	600	800	1000	1200	1400	1600	1800	2000	2200	2400	2600	2800	3000	3200	3400	3600	3800	4000	5000	6000	8000	12000

Output/hull tonnage = performance number (round down)
*with entries for J, W-Z

 

[mike wightman]

You certainly could.

You'd also need 60Td of fuel for that 66Td fighter, unless you change the formula for that (and in all fairness, you probably should).

The change would either be to cut the requirement (not the full 4 weeks) or cut the consumption rate (not 10Pn -> 4weeks).

In Mayday and LBB:2 '77 smallcraft have a limited number of burns. 60tons of fuel is enough for that 6g fighter for 1000 turns, so 6t gets you 100 turns (which is 16 hours, the Mayday fighter has 5 hours of continuous thrust)

 

[Spinward Flow]

The change would either be to cut the requirement (not the full 4 weeks) or cut the consumption rate (not 10Pn -> 4weeks).

The 4 weeks minimum fuel requirement for STARSHIPS is there to provide sufficient margin in the event of a misjump.
If a misjump duration is 1D6 weeks, then 4 weeks of power plant endurance will suffice at least half of the time (3 week misjump leaves 1 week of maneuvering after breakout).

This requirement is NOT needed for small craft (which by definition lack jump drives).
The absolute minimum fuel requirement is 1 ton. (LBB5.80, p34) (CT Errata, p15)
CT Errata p15 further stipulates that the minimum fuel requirement is 24 hours of endurance, not 28 days/4 weeks of endurance.

So if you retrofit that understanding to the 10Pn fuel formula for LBB2 standard drives, you get the following:

• PP1 = 10 tons fuel/28 days = 0.358 tons fuel/day ... 1 ton minimum fuel = 67.0 hours endurance
• PP2 = 20 tons fuel/28 days = 0.715 tons fuel/day ... 1 ton minimum fuel = 33.5 hours endurance
• PP3 = 30 tons fuel/28 days = 1.072 tons fuel/day ... 1.1 tons minimum fuel = 24.6 hours endurance
• PP4 = 40 tons fuel/28 days = 1.429 tons fuel/day ... 1.5 tons minimum fuel = 25.1 hours endurance
• PP5 = 50 tons fuel/28 days = 1.786 tons fuel/day ... 1.8 tons minimum fuel = 24.1 hours endurance
• PP6 = 60 tons fuel/28 days = 2.143 tons fuel/day ... 2.2 tons minimum fuel = 24.6 hours endurance

Now, given that standard life support endurance for an acceleration couch is ... 24 hours (12 hours in combat), as per LBB5.80, p35 ... the obvious "smart play" for small craft fuel using the LBB2 (77 or 81) standard drives is to assume a minimum fuel requirement of 24 hours endurance (which the 6G Ship's Boat unfortunately misses, having only 1.8 tons of fuel standard), things start looking a lot more reasonable and coherent and intentional (despite the occasional error here and there).

This is then where Interplanetary Travel Distance by Time and Acceleration becomes relevant.

	1G	2G	3G	4G	5G	6G
12 hours (0.5d)	4,665,600 km (0.031 AU)	9,331,200 km (0.062 AU)	13,996,800 km (0.093 AU)	18,662,400 km (0.124 AU)	23,328,000 km (0.156 AU)	27,993,600 km (0.187 AU)
24 hours (1.0d)	18,662,400 km (0.124 AU)	37,324,800 km (0.249 AU)	55,987,200 km (0.374 AU)	74,649,600 km (0.499 AU)	93,312,000 km (0.623 AU)	111,974,400 km (0.748 AU)

Under continuous acceleration (coasting on inertial changes this calculus), you're only going to get but so far in 12-24 hours. This will often times be plenty sufficient for planetary satellite maneuvers (planet to moon, moon to moon or moon to planet) but will usually be entirely inadequate for interplanetary maneuvering burns.

For longer duration continuous acceleration (over longer distances), you're going to need more fuel ... and small craft cabins to extend the life support capacity beyond 24 hours per person.

Interplanetary maneuvering is "faster" than microjumping until you get to 192 hours/8 days of maneuvering required at continuous acceleration.

• PP1 = 10 tons fuel/28 days*8days = 2.9 tons fuel = 8.12 days endurance
• PP2 = 20 tons fuel/28 days*8days = 5.8 tons fuel = 8.12 days endurance
• PP3 = 30 tons fuel/28 days*8days = 8.6 tons fuel = 8.02 days endurance
• PP4 = 40 tons fuel/28 days*8days = 11.5 tons fuel = 8.05 days endurance
• PP5 = 50 tons fuel/28 days*8days = 14.3 tons fuel = 8.008 days endurance
• PP6 = 60 tons fuel/28 days*8days = 17.2 tons fuel = 8.02 days endurance

For properly "long haul interplanetary maneuvering" of up to 8 days in duration, as a Standards Regulator involved in writing the laws, I would require an additional +1 day of endurance for every 4 days of designed endurance over 36 hours, rounding up.

• 1 day acceleration requires 1 day of endurance
• 2 days acceleration requires 3 days of endurance
• 3 days acceleration requires 4 days of endurance
• 4 days acceleration requires 5 days of endurance
• 5 days acceleration requires 7 days of endurance
• 6 days acceleration requires 8 days of endurance
• 7 days acceleration requires 9 days of endurance
• 8 days acceleration requires 10 days of endurance

That way, if the craft (small or big) is going to be continuously accelerating for more than 24 hours (the life support limits of acceleration couches) then it will require an additional fuel reserve margin to account for errors in course plotting as well as mishap avoidance and/or the need to divert around hazards, response to distress calls ... etc. etc.

Although the (ruthlessly capitalist) economics dictate allocating as little fuel tankage as possible for a specified application (so as to maximize revenue tonnage available, go figure eh? ) ... from a safety regulator of reliability that is fit for purpose standpoint, you REALLY DO NOT WANT to be arriving at your destination with "bone dry" fuel tanks every single time you complete an in-system maneuver. That kind of "no margin for error" thinking works absolutely GREAT ... except when it doesn't ... because when it doesn't, you lose the entire craft (and/or need to mount a salvage operation to retrieve it from wherever it went).

When you have NO MARGIN FOR ERROR ... the failure modes are ... not pretty ...

 

[kilemall]

The 4 weeks minimum fuel requirement for STARSHIPS is there to provide sufficient margin in the event of a misjump.
If a misjump duration is 1D6 weeks, then 4 weeks of power plant endurance will suffice at least half of the time (3 week misjump leaves 1 week of maneuvering after breakout).

This requirement is NOT needed for small craft (which by definition lack jump drives).
The absolute minimum fuel requirement is 1 ton. (LBB5.80, p34) (CT Errata, p15)
CT Errata p15 further stipulates that the minimum fuel requirement is 24 hours of endurance, not 28 days/4 weeks of endurance.

So if you retrofit that understanding to the 10Pn fuel formula for LBB2 standard drives, you get the following:

• PP1 = 10 tons fuel/28 days = 0.358 tons fuel/day ... 1 ton minimum fuel = 67.0 hours endurance
• PP2 = 20 tons fuel/28 days = 0.715 tons fuel/day ... 1 ton minimum fuel = 33.5 hours endurance
• PP3 = 30 tons fuel/28 days = 1.072 tons fuel/day ... 1.1 tons minimum fuel = 24.6 hours endurance
• PP4 = 40 tons fuel/28 days = 1.429 tons fuel/day ... 1.5 tons minimum fuel = 25.1 hours endurance
• PP5 = 50 tons fuel/28 days = 1.786 tons fuel/day ... 1.8 tons minimum fuel = 24.1 hours endurance
• PP6 = 60 tons fuel/28 days = 2.143 tons fuel/day ... 2.2 tons minimum fuel = 24.6 hours endurance

Now, given that standard life support endurance for an acceleration couch is ... 24 hours (12 hours in combat), as per LBB5.80, p35 ... the obvious "smart play" for small craft fuel using the LBB2 (77 or 81) standard drives is to assume a minimum fuel requirement of 24 hours endurance (which the 6G Ship's Boat unfortunately misses, having only 1.8 tons of fuel standard), things start looking a lot more reasonable and coherent and intentional (despite the occasional error here and there).

This is then where Interplanetary Travel Distance by Time and Acceleration becomes relevant.

	1G	2G	3G	4G	5G	6G
12 hours (0.5d)	4,665,600 km (0.031 AU)	9,331,200 km (0.062 AU)	13,996,800 km (0.093 AU)	18,662,400 km (0.124 AU)	23,328,000 km (0.156 AU)	27,993,600 km (0.187 AU)
24 hours (1.0d)	18,662,400 km (0.124 AU)	37,324,800 km (0.249 AU)	55,987,200 km (0.374 AU)	74,649,600 km (0.499 AU)	93,312,000 km (0.623 AU)	111,974,400 km (0.748 AU)

Under continuous acceleration (coasting on inertial changes this calculus), you're only going to get but so far in 12-24 hours. This will often times be plenty sufficient for planetary satellite maneuvers (planet to moon, moon to moon or moon to planet) but will usually be entirely inadequate for interplanetary maneuvering burns.

For longer duration continuous acceleration (over longer distances), you're going to need more fuel ... and small craft cabins to extend the life support capacity beyond 24 hours per person.

Interplanetary maneuvering is "faster" than microjumping until you get to 192 hours/8 days of maneuvering required at continuous acceleration.

• PP1 = 10 tons fuel/28 days*8days = 2.9 tons fuel = 8.12 days endurance
• PP2 = 20 tons fuel/28 days*8days = 5.8 tons fuel = 8.12 days endurance
• PP3 = 30 tons fuel/28 days*8days = 8.6 tons fuel = 8.02 days endurance
• PP4 = 40 tons fuel/28 days*8days = 11.5 tons fuel = 8.05 days endurance
• PP5 = 50 tons fuel/28 days*8days = 14.3 tons fuel = 8.008 days endurance
• PP6 = 60 tons fuel/28 days*8days = 17.2 tons fuel = 8.02 days endurance

For properly "long haul interplanetary maneuvering" of up to 8 days in duration, as a Standards Regulator involved in writing the laws, I would require an additional +1 day of endurance for every 4 days of designed endurance over 36 hours, rounding up.

• 1 day acceleration requires 1 day of endurance
• 2 days acceleration requires 3 days of endurance
• 3 days acceleration requires 4 days of endurance
• 4 days acceleration requires 5 days of endurance
• 5 days acceleration requires 7 days of endurance
• 6 days acceleration requires 8 days of endurance
• 7 days acceleration requires 9 days of endurance
• 8 days acceleration requires 10 days of endurance

That way, if the craft (small or big) is going to be continuously accelerating for more than 24 hours (the life support limits of acceleration couches) then it will require an additional fuel reserve margin to account for errors in course plotting as well as mishap avoidance and/or the need to divert around hazards, response to distress calls ... etc. etc.

Although the (ruthlessly capitalist) economics dictate allocating as little fuel tankage as possible for a specified application (so as to maximize revenue tonnage available, go figure eh? ) ... from a safety regulator of reliability that is fit for purpose standpoint, you REALLY DO NOT WANT to be arriving at your destination with "bone dry" fuel tanks every single time you complete an in-system maneuver. That kind of "no margin for error" thinking works absolutely GREAT ... except when it doesn't ... because when it doesn't, you lose the entire craft (and/or need to mount a salvage operation to retrieve it from wherever it went).

When you have NO MARGIN FOR ERROR ... the failure modes are ... not pretty ...

The classic sci-fi story Cold Equations has such no margin fuel operations standards as the central conceit of the story. We can reasonably presume orders of magnitude costs driving that over Traveller economics.

 

[Grav_Moped]

In Mayday and LBB:2 '77 smallcraft have a limited number of burns. 60tons of fuel is enough for that 6g fighter for 1000 turns, so 6t gets you 100 turns (which is 16 hours, the Mayday fighter has 5 hours of continuous thrust)

Yes. This is one of the things that I figured out: the LBB2 '77 10Pn "at least 288 g-turns" starship fuel allocation is almost exactly (94% of) a week of full thrust at the small craft 10kg per g-turn rate -- which is all you need for a starship, since if the trip takes longer than that an in-system jump is faster. (At least it's close enough to handwave the matter...)

And smallcraft were generally presented as interface craft for starships, rather than intended for independent operation. That is, they might very well be operated outside the context of starships, but that's not how PCs generally interacted with them.

High Guard and LBB2'81 enabled System Defense Boats by shifting to the "allocation is good for one month" rate.

 

[kilemall]

Yes. This is one of the things that I figured out: the LBB2 '77 10Pn "at least 288 g-turns" starship fuel allocation is almost exactly (94% of) a week of full thrust at the small craft 10kg per g-turn rate -- which is all you need for a starship, since if the trip takes longer than that an in-system jump is faster. (At least it's close enough to handwave the matter...)

And smallcraft were generally presented as interface craft for starships, rather than intended for independent operation. That is, they might very well be operated outside the context of starships, but that's not how PCs generally interacted with them.

High Guard and LBB2'81 enabled System Defense Boats by shifting to the "allocation is good for one month" rate.

I’ve taken the interface metaphor one step further and have small craft bridges giving much reduced sensor and radio ranges. Most of the time they are little blind guppies without full computers.

As I recall the wording for fuel was one month of normal operation. With the above understanding, a full week of burn would not be normal.

 

[Grav_Moped]

I’ve taken the interface metaphor one step further and have small craft bridges giving much reduced sensor and radio ranges. Most of the time they are little blind guppies without full computers.

I don't go that far on small craft bridges, but I understand where you're coming from.

As I recall the wording for fuel was one month of normal operation. With the above understanding, a full week of burn would not be normal.

The thing is, for an interstellar voyage between two Size 8 planets that are both outside the star's 100D limit, you're looking at about 2 days of full acceleration (inbound and outbound combined) at 1G, plus the Jump. However, LBB2 '77 called for expending the entire power plant fuel allocation on every trip. I see this as a handwave allowing avoidance of having to do the math every trip, and thus the possibility that the rules might show that there wasn't actually enough fuel to do it quite that way (easily worked around by coasting through part of each side of the trip, but that's unnecessary complexity).

 

[infojunky]

In Mayday and LBB:2 '77 smallcraft have a limited number of burns. 60tons of fuel is enough for that 6g fighter for 1000 turns, so 6t gets you 100 turns (which is 16 hours, the Mayday fighter has 5 hours of continuous thrust)

Ships have a simple limit as well of 288 burns...

Also note one could use the fuel use from Beltstrike as well...

 

[Spinward Flow]

Also note one could use the fuel use from Beltstrike as well...

I prefer to think of the Beltstrike fuel consumption rules as being "HEPlaR for CT" (kinda sorta) since it's really reaction drive maneuvering ... albeit incredibly efficient reaction drive maneuvering (0.0005MG/day).

 

[Spinward Flow]

The third is that if you're not intermixing drives, agility is almost invariably limited to a maximum of 5 since LBB2 power plants cannot have a rating higher than Pn-5, and computers larger than Mod/2 draw energy points.

A fourth is that TL affects LBB5 power plant size if you're intermixing LBB2 and LBB5 drives. I personally dislike the concept of intermixing them, as it's cherry-picking from systems with different underlying purposes.

My personal preference is ... no intermixing. Your options are:

1. All LBB2 drives
2. All LBB5 drives

Sure the rules don't "explicitly SAY" that explicitly, but it is definitely the way I would prefer to interpret them ... primarily as a matter of intellectual consistency and Good Taste™ as a naval architect.

However, I would not construe LBB2 drive EP outputs using the LBB5 methodology (code factor x 1% of hull tonnage = EP).
That works just fine for LBB5 drives, which are defined by formula (and integers, for the most part) ... but DOESN'T work just fine for LBB2 drives, which can easily (*easily*!) run into fractional code yields due to the structural granularity of the LBB2 drive system paradigm.

For example:
LBB2.81/LBB3.81 Power Plants have the following benchmark yields (@ 200 tons per letter baseline, M=12 and N=13, because I and O are skipped as letters for disambiguation purposes):

• M (@TL=12) = Code: 1 @ 2400 tons = 24EP for LBB5.80 purposes
• N (@TL=12) = Code: 1 @ 2600 tons = 26EP for LBB5.80 purposes

The nuance of interpretation then comes in when dealing with smaller hull sizes than the Code: 1 benchmarks.

• M (@TL=12) = Code: 1 @ 2400 tons = 24EP for LBB5.80 purposes
• M (@TL=12) = Code: 2 @ 1200 tons = 24EP for LBB5.80 purposes
• M (@TL=12) = Code: 3 @ 800 tons = 24EP for LBB5.80 purposes
• M (@TL=12) = Code: 4 @ 600 tons = 24EP for LBB5.80 purposes

However, if you used a Power Plant-N drive instead in 2400/1200/800/600 tons displacement hulls, you would get 26EPs out of it ... not 24EPs ... in a LBB5.80 rules crossover interpretation. Power Plant-N in a 600 ton hull would still yield Power Plant-4 (technically 4.333, which then rounds down to 4) as far as LBB5.80 is concerned (for USP and so on) but it would output 26EP ... not 24EP ... and it would consume 40 tons of fuel per 4 weeks (because, PP4), which is still more than the 26 tons of fuel per 4 weeks that a LBB5.80 custom power plant would require for the exact same output yield in a 600 ton hull.

So the moral of the story is that if you have a Maneuver-X drive and a Power Plant-Z drive (both TL=15) in an 800 ton hull ... you CAN have Agility=6 with some additional EP surplus for weapons, screens and computers (although it won't be but so much!), even though the code factors for both on the USP yield Maneuver-6 and Power Plant-6 due to the "drop fractions" nature of USP integers only coding. However, due to the limitations of the "200 ton step Code: 1 drives" baseline formulation for LBB2 standard drives, there's only so "far" you can go with standard drives in terms of technology and size (see: Small Starship Universe).

Note also that having a Power Plant letter that is higher than the Maneuver letter enables use of the Double Fire computer program in LBB2 combat (LBB5.80 substantially ignores computer programming as a concern in combat, greatly simplifying the abstract system).



I'm currently working on yet another revision of my Five Sisters Clipper design that upgrades the core fundamentals from a(n honestly too tight for comfort) 400 tons @ TL=10 design into a much more capable and versatile 600 tons @ TL=12 design that includes a lot of conceptual framework refinements over my previously published notions and interpolations/extensions of various CT rules and constraints. Among those considerations is the fact that M-N standard drives are both TL=12 (as per LBB3.81, p15), but you only need Drive-M to yield Drive-4 performance in a 600 ton form factor.

That then means that Jump-M(4), Maneuver-M(4) and Power Plant-N(4) is possible @ TL=12 ... which then enables Double Fire programming in the model/4fib computer with equipped lasers (if you like to play with computer programming rules in your ship to ship combat encounters).

• A Power Plant-N(4) producing a 26EP output (see above)
• Cross-pollinating with LBB5.80 for EP and Agility rules ... Agility=2 in a 600 ton hull requires 12EP.
• Computer model/4 (needed for J4) requires 2EP, whether fib or not.
• That then leave 12EP remaining for weapons ... which is a perfect fit for 4x Triple Laser Turrets (beam or pulse) ... which can then benefit GREATLY from use of Double Fire programming in the main computer (assuming you even want to bother with such details as a Referee). In LBB5.80 "batteries bearing" terms, this gives some variability in the organization of lasers (2x Code: 4 Beam Lasers ... or 4x Code: 2 Pulse Lasers) depending on loadout, with the pulse lasers being the cheaper option (obviously).
• Add 2x Triple Sandcaster Turrets (2x Code: 4) requiring 0EP for defense to round out the ship's weaponry.

This combination of (slightly) overdesigned power plant, minimum computer requirement, decent Agility=2 (and Emergency Agility=4) along with "lots of lasers" and a couple sandcasters means that the ship ought to be relatively safe in a Small Ship Universe as outlined by LBB2 against most threats of piracy. It also has a crew of 10(!) which makes it more difficult to capture via boarding action (LBB5.80, p43) and of whom only the Navigator receives the minimum salary of Cr5000 per 4 weeks (I like my crews to be ... skilled).



However, this particular exercise (with my revision) once again highlights that there are going to be particular tech levels and tonnages where you can get a performance profile out of a LBB2.81 standard drive that simply IS NOT AVAILABLE to you from a LBB5.80 custom drive ... and more specifically in jump drives.

• Jump-H @ TL=10 yields J4 in a 400 ton hull ... while LBB5.80 requires TL=13 for J4.
• Jump-J @ TL=11 yields J4 in a 450 ton hull ... while LBB5.80 requires TL=13 for J4.
• Jump-K @ TL=11 yields J4 in a 500 ton hull ... while LBB5.80 requires TL=13 for J4.
• Jump-L @ TL=12 yields J4 in a 550 ton hull ... while LBB5.80 requires TL=13 for J4.
• Jump-M @ TL=12 yields J4 in a 600 ton hull ... while LBB5.80 requires TL=13 for J4.
• Jump-N @ TL=12 yields J4 in a 650 ton hull ... while LBB5.80 requires TL=13 for J4.
• Jump-P @ TL=13 yields J4 in a 700 ton hull ... while LBB5.80 requires TL=13 for J4.

There are other breakpoint examples of this phenomenon of course (@Grav_Moped's Shugushaag design for a J5/2G @ TL=13 starship being one of the more notable ones), but it does highlight the fact that there are some very notable differences in the underlying paradigms between LBB2.81 and LBB5.80 when it comes to standard vs custom drives. In many cases, "pushing the envelope" like this will yield a design that isn't economically viable (except under subsidy ... and sometimes not even then!). However, there are some really fascinating possibilities that can be explored in this (for lack of a better term) "prototype realm" of drive performances, particularly when taking a holistic "end user" approach to the entire endeavor.

Add in the fact that it's "dirt simple" to calculate external towing capacity (J4@600 = J3@800 = J2@1200 = J1@2400) with standard drives (you just need a rules framework for doing so) and you're basically all set. With flexibility comes OPTIONS ... and with more options available to you, there are more paths to success and profitability that open up beyond the one single well worn rut carved out by the A1 Free Traders.

 

[kilemall]

Or you could go another way- that LBB2 PP doesn’t go by LBB5 EP formula. Maybe something like an extra 10-20%, and that partially explains the increased fuel use.

 

[Spinward Flow]

Or you could go another way- that LBB2 PP doesn’t go by LBB5 EP formula. Maybe something like an extra 10-20%, and that partially explains the increased fuel use.

Nah.
That just creates more problems of intellectual inconsistency than it solves.

Simple? Sure.
USEFUL? Not so much ...

 

[Grav_Moped]

Or you could go another way- that LBB2 PP doesn’t go by LBB5 EP formula. Maybe something like an extra 10-20%, and that partially explains the increased fuel use.

That's one way to look at it, and it could be useful. This comes back to my notion that ships that serve similar narrative functions but constructed under different rule sets are functionally equivalent from an RPG perspective (for example, the Gazelle is the LBB5 equivalent of LBB2's Type T Patrol Cruiser). The Patrol Cruiser should have decent agility and armor in a system where those matter, regardless of how a straight transposition of its stats into LBB5 values looks. The Gazelle's particle accelerator is probably an oversized (4x? 6x?) beam laser turret with double-fire capability in LBB2, but it doesn't have armor in that system. You get the idea -- it's hand-waving, with the idea that it's not the numbers that matter, but what purpose that ship is meant to have in the game system.


I still have significant reservations about LBB2 '81's power plant fuel rules, in both directions. On one hand, changing them would have significant effects on canon designs and game balance. On the other, they're a crude patch to the credulity-straining fuel consumption paradigm from LBB2 '77, that doesn't actually fix the problem from a design (or rather, fictional engineering) standpoint.

 

[Spinward Flow]

I still have significant reservations about LBB2 '81's power plant fuel rules, in both directions. On one hand, changing them would have significant effects on canon designs and game balance. On the other, they're a crude patch to the credulity-straining fuel consumption paradigm from LBB2 '77, that doesn't actually fix the problem from a design (or rather, fictional engineering) standpoint.

Which is fair ... it's a radically different paradigm from LBB5.80.

As I've mentioned elsewhere, the LBB2 standard drive paradigm is Code: 1 @ 200 ton increments (so Drive-A is code: 1 in a 200 ton hull), which then uses simple division (and dropping fractions) to determine codes in smaller hull sizes (thus, Drive-A is code: 2 in a 100 ton hull).

So a LBB2 standard Drive-M is:

1. Code: 1 in a 2400 ton hull
2. Code: 2 in a 1200 ton hull
3. Code: 3 in a 800 ton hull
4. Code: 4 in a 600 ton hull
5. Code: 5 in a 480 ton hull
6. Code: 6 in a 400 ton hull

However, if that's a Power Plant-M standard drive, it will have different rates of fuel consumption per 4 weeks:

1. Code: 1 is 10 tons of fuel per 4 weeks in a 2400 ton hull
2. Code: 2 is 20 tons of fuel per 4 weeks in a 1200 ton hull
3. Code: 3 is 30 tons of fuel per 4 weeks in a 800 ton hull
4. Code: 4 is 40 tons of fuel per 4 weeks in a 600 ton hull
5. Code: 5 is 50 tons of fuel per 4 weeks in a 480 ton hull
6. Code: 6 is 60 tons of fuel per 4 weeks in a 400 ton hull

The only difference then is ... how big is the hull you're installing that drive into?
Yes, you get "higher code" performance from the same drive in a smaller hull, but you also consume more fuel in a smaller hull ... so what gives?

For me, the answer is (relatively) simple.
Standard drives are designed to be installed as Code: 1 in their default hulls (in the above example, 2400 tons for a Drive-M), meaning those Code: 1 hull sizes are what the drives are optimized for. However, as the hull displacement goes down, power "density" remains the same (what LBB5.80 would call EP per ton of power plant drive) while the surface area to (safely) radiate any waste heat away from goes down with the hull size. So in order to provide enough cooling capacity for the power plant, fuel consumption has to increase in order to carry away waste heat by means of conduction/convection in addition to radiators radiating heat away from the (smaller) hull. This then helps explain why as the hull size goes down "around" a standard drive power plant, the rate of fuel consumption goes up ... even though the drive itself is not being altered ... just the context of the hull (size) that it is being installed into.



Extra bonus points for anyone who realizes that in letter codes missing I and O (to remove opportunities to confuse them with 1 and 0 in print) this means that Z=24 ... and 24 * 200 = 4800 (not 5000 like the LBB2 table erroneously claims).

4800 / 4 = 1200 tons ...

Meaning that the Kinunir that appeared in LBB A1 @ 1200 tons was designed to use TL=15 Drive-Z(s) and a model/7 computer ... the maximums available under LBB2 when the adventure was being written ... hence why it "builds" the way it does.

It was only when trying to recompute the design using LBB5.80 rules (that operate on a different paradigm) that the "hull bloat" retcon to 1250 tons became necessary in order to make everything fit (as seen in LBB S9, p19).