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- Thread starter spank
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There’s an article onMongoose Traveller has a nifty table for small craft drives, it is also in the SRD.

I’ve seen something similar on Freelance Traveller.

Was this ever published for CT?

No need for a table when an extremely obvious and simple formula works just as well (values inThere are threads here where people have theorycrafted that a plain old A plant/mdrive gets wedged into a small craft and gets the small craft results.

Maneuver-A = code:

Maneuver-B = code:

Maneuver-C = code:

Power Plant-A = 2 EP.

Power Plant-B = 4 EP.

Power Plant-C = 6 EP.

Drive-D and up "do not work" in hulls between 1-100 tons (as per LBB2.81), so you can't put a Power Plant-D into a small craft (for example) and get 8 EP out of it.

After that, everything is relatively straightforward, as far as the drives themselves are concerned.

Fuel consumption (per CT Beltstrike) is 1 ton of fuel per 2000 tons of craft per 7 days for "basic power" ... while EP (used for computer, weapons, agility, maneuver drive/agility, etc.) is 0.35 tons of fuel per 1 EP per 7 days. So a small craft can "get by" with a fuel tank of 1 ton of fuel reserve while not generating EPs (above "basic power" output) for an extremely long duration, just coasting/drifting, if necessary. However, longer journeys (more than 12-24 hours) will require Small Craft Cabin life support for crew/passengers. So you can "kinda sorta" throw away the 10Pn fuel requirement for LBB2.81 standard drives when installing them into small craft (since they don't jump and don't "need" the extra margin against mishap in foreign star systems that starships do).

Note: this and the next post are a reprise of something I posted here a while back. It was just easier to pull from my notes than to link it...

The basic problem is that interpolations based on the Size A Drives doesn't work properly below 200 rating points (Gs x tons or Pn x tons). Results below 200 rating points become increasingly implausible; at 100 rating points (1G in a 100 ton hull) the maneuver drive has zero tons displacement...

The underlying formulae for maneuver drives are:

__Maneuver Drive:__

M = Hull Tons, G = Acceleration in Gs

M must be between 100 and 5000 inclusive.

Note: values of M*G less than 100 yield negative drive tonnage! (Values below 200 get weird when you try to figure out Gs per ton of maneuver drive. Just don't go there, ok?)

If M=100, G must be divisible by two; if an odd G is desired, calculate drive size and cost using the next higher G divisible by 2. Drive capability and power requirements are based on the intended drive rating.

If M < 1000 but not evenly divisible by 200, round M up to the next multiple of 200 before calculating drive tonnage.

If M > 1000 but not evenly divisible by 1000, round M up to the next multiple of 1000 before calculating drive tonnage.

This rounding up does not affect fuel consumption calculations.

Maneuver Drive (tons) = (0.01*M*G)-1

Maneuver Drive (percent of M) = ((0.01*M*G)-1/M)*100

Cost = MCr0.02*M*G

Cost per ton = MCr2*(MD tons+1)

If M=2000 and G=1, subtract 2Td and MCr4.

If M*G>4000, consult table as exceptions may be present

------

For power plants:

__Power Plant:__

M = Hull tons, Pn = Power Plant Number

M must be between 100 and 5000 inclusive.

If M=100, Pn must be divisible by two; if an odd Pn is desired, calculate drive size and cost using the next higher Pn divisible by 2. Powerplant fuel consumption and output is based on the intended Pn.

If M < 1000 but not evenly divisible by 200, round M up to the next multiple of 200 before calculating drive tonnage.

If M > 1000 but not evenly divisible by 1000, round M up to the next multiple of 1000 before calculating drive tonnage.

This rounding up does not affect fuel consumption calculations.

Power Plant (tons) = (0.015*M*Pn)+1

Power Plant (percent of M) = ( (0.015*M*Pn+1)/M )*100

Cost = MCr0.04*M*Pn

Cost per ton = MCr2.67*(PP tons-1)

If M=2000 and Pn=1, subtract 3Td and MCr8.

If M*G>4000, consult table as exceptions may be present.

----

It's similar for Jump Drives, but that's off the topic.

The basic problem is that interpolations based on the Size A Drives doesn't work properly below 200 rating points (Gs x tons or Pn x tons). Results below 200 rating points become increasingly implausible; at 100 rating points (1G in a 100 ton hull) the maneuver drive has zero tons displacement...

The underlying formulae for maneuver drives are:

M = Hull Tons, G = Acceleration in Gs

M must be between 100 and 5000 inclusive.

Note: values of M*G less than 100 yield negative drive tonnage! (Values below 200 get weird when you try to figure out Gs per ton of maneuver drive. Just don't go there, ok?)

If M=100, G must be divisible by two; if an odd G is desired, calculate drive size and cost using the next higher G divisible by 2. Drive capability and power requirements are based on the intended drive rating.

If M < 1000 but not evenly divisible by 200, round M up to the next multiple of 200 before calculating drive tonnage.

If M > 1000 but not evenly divisible by 1000, round M up to the next multiple of 1000 before calculating drive tonnage.

This rounding up does not affect fuel consumption calculations.

Maneuver Drive (tons) = (0.01*M*G)-1

Maneuver Drive (percent of M) = ((0.01*M*G)-1/M)*100

Cost = MCr0.02*M*G

Cost per ton = MCr2*(MD tons+1)

If M=2000 and G=1, subtract 2Td and MCr4.

If M*G>4000, consult table as exceptions may be present

------

For power plants:

M = Hull tons, Pn = Power Plant Number

M must be between 100 and 5000 inclusive.

If M=100, Pn must be divisible by two; if an odd Pn is desired, calculate drive size and cost using the next higher Pn divisible by 2. Powerplant fuel consumption and output is based on the intended Pn.

If M < 1000 but not evenly divisible by 200, round M up to the next multiple of 200 before calculating drive tonnage.

If M > 1000 but not evenly divisible by 1000, round M up to the next multiple of 1000 before calculating drive tonnage.

This rounding up does not affect fuel consumption calculations.

Power Plant (tons) = (0.015*M*Pn)+1

Power Plant (percent of M) = ( (0.015*M*Pn+1)/M )*100

Cost = MCr0.04*M*Pn

Cost per ton = MCr2.67*(PP tons-1)

If M=2000 and Pn=1, subtract 3Td and MCr8.

If M*G>4000, consult table as exceptions may be present.

----

It's similar for Jump Drives, but that's off the topic.

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It's not clear how this would work

Energy points aren't tracked in this system, since they hadn't yet been invented... You can add tonnage to the power plant component of the "combined drives," or add an actual power plant, to generate them if you're porting this to a system that does use them.

This is one thread I started on the subject: LBB 2 Drives in Small Craft? (and LBB2 PP Fuel Rules).

A summary of the resulting "small craft drive" formulae is in post #31 of that thread.

A summary of the resulting "small craft drive" formulae is in post #31 of that thread.

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You can extrapolate downward for Jump drives and Powerplants, ignoring the 100 ton jump limit, to produce theoretical sizes of drives. Maneuver drives are the only ones that are really dinked. Even bundling them with the powerplant doesn't help, at 50 tons you'd essential be getting the maneuver drive for free, because the extrapolated mass of a maneuver drive is 0 for a 50 ton craft, and below 50 it actually make the powerplant+manuever drive smaller than the powerplant by itself.

I think it would be interesting to have the small drives be "bigger", relatively, than the extrapolated versions, but cheaper. And ideally you wouldn't make anything with more potential than about "B" or "C" on the starship chart.

Declaring a minimum "merged-drives size" of 1Td gets you there, pretty much. Scale the cost down based on output, but stick with a 1Td minimum size regardless of output.I think it would be interesting to have the small drives be "bigger", relatively, than the extrapolated versions, but cheaper. And ideally you wouldn't make anything with more potential than about "B" or "C" on the starship chart.

You can't really use anything bigger than Size C in a small craft, unless you need a lot of extra energy points in a 99Td shuttle... (which, in fairness, you might).

But, it would be handy to use small craft drives in starship to either reduce cost, at the expense of tonnage, or to produce intermediate sized ship such as a 250 ton ship.Declaring a minimum "merged-drives size" of 1Td gets you there, pretty much. Scale the cost down based on output, but stick with a 1Td minimum size regardless of output.

You can't really use anything bigger than Size C in a small craft, unless you need a lot of extra energy points in a 99Td shuttle... (which, in fairness, you might).

Don't assume a linear relationship sub 100t.

We call that a "tug" ... vastly overpowered for its size, intended to "hook up to and do stuff with" MUCH larger craft.For example a 10 ton pinnace would lose 50% of it's mass to drive if using a Drive A, and P-Plant A. And they would be vastly oversized for the performance you'd get.

Compelling argument.I think it would be interesting to have the small drives be "bigger", relatively, than the extrapolated versions, but cheaper.

Especially since one of the points in favor of standard maneuver+power plant drives (especially at lower tech levels!) is that they are already CHEAPER than the custom drive equivalents.

LBB5.80 basically does this already in its small craft design sequence. No drive system or fuel quantity can be less than 1 ton minimum.Declaring a minimum "merged-drives size" of 1Td gets you there, pretty much. Scale the cost down based on output, but stick with a 1Td minimum size regardless of output.

Therefore, the smallest possible small craft design is:

- 1 ton maneuver drive
- 1 ton power plant
- 1 ton fuel capacity
- 1 ton model/1 computer
- 0.5 ton acceleration couch for 1 crew

Take out the (expensive) model/1 computer + acceleration couch and replace it with no computer + bridge (which includes 2 acceleration couches) and you're up to a 7 ton minimum size, but a cheaper craft ... and you wind up with a (remarkably cheap)

That's actually a problem. If they're better (overall) than the already existing starship drives, why aren't they being used instead of them on every ship? (Which is to say, if you're house-ruling stuff into existence, you need to consider the effects of doing so on the rest of the rules.)But, it would be handy to use small craft drives in starship to either reduce cost, at the expense of tonnage,

Easy enough. Go with the formulae I gave in the linked post upthread, and ignore the parts I put in there to force teh results to resolve to the "letter drives" from the Drive Performance Table from LBB2. Just drop the requirement to "round performance down" for hull sizes between the ones listd on the chart.or to produce intermediate sized ship such as a 250 ton ship.

As a nod to game consistency, drives that do not exactly correspond to those in LBB2 are custom, and spares for them are not stocked at starports of TLs lower than that necessary to build them. (They're not stocked at higher TLs either, but they can be locally produced.)

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Probably not, but whatever math you use should have results that don't diverge too far from the trend lines above 100Td. IMHO there's nothing special about 100Td except that it's the smallest size for which Jump is possible. Everything else should scale past that point with reasonable smoothnesDon't assume a linear relationship sub 100t.

It all depends on what you consider "linear".Probably not, but whatever math you use should have results that don't diverge too far from the trend lines above 100Td. IMHO there's nothing special about 100Td except that it's the smallest size for which Jump is possible. Everything else should scale past that point with reasonable smoothnes

The size of CT LBB2 drives are not what I would call linear. Their size is offset by a couple Tons from a straight percentage of the hull. As a result the smaller the hull the bigger the drive, as a fraction of the hull. If you project the standard size drives downward they continue to take up a larger percentage of the hull, but they are nowhere near as bad as using full sized starship drive.

Their size is linear plus-or-minus a constant. This yields the nonlinear percentages you noted.

M-drives are different, but only because they are paired with the large constant-size power fuel allocation (explicitly maneuver fuel in '77, relabeled but keeping the same formula on '81).

This scale efficiency was a core element of the LBB1-3 tech ladder: bigger is better (and for spaceships, the*only* way to get better efficiency), and only TL lets you build bigger. Ships of the same size had the same efficiency regardless of their TL (except for the Size W-Z drives.)

In HG and later, higher TL is always better regardless of size.

M-drives are different, but only because they are paired with the large constant-size power fuel allocation (explicitly maneuver fuel in '77, relabeled but keeping the same formula on '81).

This scale efficiency was a core element of the LBB1-3 tech ladder: bigger is better (and for spaceships, the

In HG and later, higher TL is always better regardless of size.

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Yep. As in LBB2'81, butA point, the smallcraft of CT's Traveller Book are straight Book5 2nd.

Other editions have handled them differently.

Not quite straight LBB5: The bridges were all 4 Dt, not 20% of hull.A point, the smallcraft of CT's Traveller Book are straight Book5 2nd.

Example:

LBB2'81:

Pinnace:Using a 40-ton hull, the pinnace is capable of 5-G acceleration, carries 2 tons of fuel, and has a crew of two. It may mount two lasers, and any remaining weapons must be missile racks or sandcasters. It has 22.4 tons of excess space, and costs MCr20.

Code:

```
KK-0105501-000000-00000-0 MCr 20,6 40 Dton
bearing Crew=1
batteries TL=9
Cargo=22 Fuel=2 EP=2 Agility=5
Single Occupancy 22,4 25,7
USP # Dton Cost
Hull, Streamlined Custom 0 40
Configuration Needle/Wedge 1 4,8
Manoeuvre D 5 1 5,6 2,8
Power Plant 5 1 6 18
Fuel, #J, #weeks J-0, 4 weeks 2
Bridge 1 4 0,1
Cargo 22,4
Nominal Cost MCr 25,70 Sum: 22,4 25,7
Class Cost MCr 5,40 Valid ≥0 ≥0
Ship Cost MCr 20,56
```

The bridge should have been 8 Dt...

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