AnotherDilbert
SOC-14 1K
Except CT, MT, TNE, T4, and T5...
What X-Boat travels in your universe?
- Thread starter Swiftbrook
- Start date
Except CT, MT, TNE, T4, and T5...
Proneutron
SOC-13
(CT) Book 2 Drive potential table pg. 22
Straybow
SOC-14 1K
Don't forget GT only requires 5% hull per Jn for jump fuel, making J6 far more economical, and easily allowing M2+.
One could claim that the official X-boat network was political, rather than economical. But then you'd get an economic network emerging that would carry data at Cr1 minimum +Log2(GB) per pc making money by bucket-loads.
Private banking data would be the bread and butter, exchanging money transfers and confirmations back and forth. Perhaps it would even be a bank cartel network.
One could claim that the official X-boat network was political, rather than economical. But then you'd get an economic network emerging that would carry data at Cr1 minimum +Log2(GB) per pc making money by bucket-loads.
Private banking data would be the bread and butter, exchanging money transfers and confirmations back and forth. Perhaps it would even be a bank cartel network.
Proneutron
SOC-13
See US Const. written in 18th century and Article 1, Section 8 therein about creating post office and post roads. It has to do with knitting together a large polity and not directly economics. Although it no doubt facilitated economics...
The 3I gov running X-boat N/W is similar.
mike wightman
SOC-14 10K
It was political - the xboat system was laid on top of the existing trade routes with worlds often using as much political leverage as they could get away with to be included in the network.
AnotherDilbert
SOC-14 1K
See LBB5'79, p23; E.g. a jump drive-1 in a 100 Dt hull is 2 Dt.
Proneutron
SOC-13
S0? I'm talking about M-DRIVES
AnotherDilbert
SOC-14 1K
OK: See LBB5'79, p23; E.g. a [o]jump[/o] maneuver drive-1 in a 100 Dt hull is 2 Dt.
There's a formula for that too. I'd have to look it up, but 2% for 1G seems about right. 5% for 2G? 8% for 3G . . .
Proneutron
SOC-13
I'm talking LBB2
Minimum for a 100Td hull is 2G at 1Td.
There's no good way to extrapolate 1G in 100Td from LBB2 even if you wanted to.
The formula used to generate the m-drive sizes for the Drive Performance Table breaks down below 200 "G-tons"; at 100 "G-tons", the drive would be 0Td and MCr2.
I can almost accetpt that if using LBB2:'77 rules. In those, it seemed that the maneuver drive and power plant were mostly an integrated unit, while the jump drive was its own thing. It would just take house-ruling that this "maneuver drive" is a MCr2 modification to a half-A power plant (Pn-1 in 100Td) that concentrates its exhaust into a jet providing 100 G-tons of thrust. Doesn't make sense in the '81 rules, or in either version of HG. And it doesn't extend below 100 G-tons thrust, either.
Last edited:
Khan Trav
SOC-12
I love Traveller
IMTU We have used original style xboats and it seems unlikely anyone would actually use them. Having a ship that jumps into a system with no way to maneuver is really more trouble than it is worth. I still use it because that is the way I grew up knowing Traveller. I have used xboats to infiltrate players covertly into scout bases where the IISS is conducting investigations into base commanders etc..
Fun to be had by all...
IMTU We have used original style xboats and it seems unlikely anyone would actually use them. Having a ship that jumps into a system with no way to maneuver is really more trouble than it is worth. I still use it because that is the way I grew up knowing Traveller. I have used xboats to infiltrate players covertly into scout bases where the IISS is conducting investigations into base commanders etc..
Fun to be had by all...
mike wightman
SOC-14 10K
Yes there is.
A - 200u
B - 400u
C - 600u etc.
so you need a fractional drive - 1/2 of an A drive, or 1/4 of a B drive... etc
If a formula gives results of 0 for less than an A drive then all I do is change the formula to get the progression I want.
You can do an analysis of the drives that must be in the smallcraft which throws up some interesting results - some appear to use the A drive, some are A/2 drives, while the launch is A/10 and the fighter C/10.
Last edited:
Nope. The formula used for LBB2 maneuver drives looks like this (formulae for the rest of the drives as well are in the linked post).
Maneuver Drive:
M = Hull Tons, G = Acceleration in Gs
Conditions to ensure hull sizes are inside LBB2 constraints:
M must be between 100 and 5000 inclusive.
Note: values of M*G less than 100 yield negative drive tonnage!
Note: values of M*G less than 100 yield negative drive tonnage!
Conditions to force drive sizes to resolve to existing letter drives, and hull sizes to match LBB2 listed sizes:
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.
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 (percent of M) = ((0.01*M*G)-1/M)*100
Cost = MCr0.02*M*G or
Cost per ton = MCr2*(MD tons+1)
Exceptions:
If M=2000 and G=1, subtract 2Td and MCr4.
If M*G>4000, consult table as exceptions may be present.
These exceptions are:
1. an exception at 1G for 2000Td
2. the increased effectiveness of Drives W-Z -- they gave them a bump to keep from needing to extend the table beyond Z (or maybe they just wanted TL15 to be better?)
If M*G>4000, consult table as exceptions may be present.
These exceptions are:
1. an exception at 1G for 2000Td
2. the increased effectiveness of Drives W-Z -- they gave them a bump to keep from needing to extend the table beyond Z (or maybe they just wanted TL15 to be better?)
Use the formula to do 1.5Gs in 100Td (or 3G in 50Td): 0.5Td. That's not implausible.
1.25Gs in 100Td? 0.25Td. That's implausible.
Now try at 1.0G in 100Td: 0Td. How does that even work?
But 1G in 50Td is negative 0.5Td! That's simply inexplicable.
Last edited:
Jump drives being 5+(steps*5)Td and MCr(10*Steps) and so a half-A is 7.5 Td and MCr5.
mike wightman
SOC-14 10K
You do know that just because you 'discovered' that formula it doesn't make it right?
You are placing artificial constraints on your formula and so get the results you describe - change the formula and you will suddenly find that fractional drives are allowed.
Anything less than an A drive is a fractional drive - an A drive is 200 thrust units so an A/2 delivers 100 thrust units, an A/10 delivers 20.
The question is what tonnage and cost you wish to allocate to an A/2 drive and an A/10 drive.
The best fit I can get for a 10t 6g fighter is a C/10 drive for example.
You are placing artificial constraints on your formula and so get the results you describe - change the formula and you will suddenly find that fractional drives are allowed.
Anything less than an A drive is a fractional drive - an A drive is 200 thrust units so an A/2 delivers 100 thrust units, an A/10 delivers 20.
The question is what tonnage and cost you wish to allocate to an A/2 drive and an A/10 drive.
The best fit I can get for a 10t 6g fighter is a C/10 drive for example.
But that's kind of the point here, isn't it?
You simply can't extrapolate the characteristics of maneuver drives below Size A from the formula that created the Drive Table, because the results rapidly become nonsensical.
Anything smaller than that is just making things up from scratch. That doesn't mean you can't do it, it just means you should be consistent about how you go about making things up.
And not everyone will agree on how that should go.
The OTU answer is to use LBB5 design rules.
I wouldn't object to using a Size A Maneuver Drive and Power Plant in a small craft, but the rules-as-written fuel use rates would make it impractical. (4G in a 50Td craft, but it would be Pn-4 thus would need 40Td fuel... That might be why LBB2:'77 had small craft with limited delta-v "burns"!)
Last edited:
Condottiere
SOC-14 5K
Current Mongosian formulations follow minimum default jump drive tonnage of ten, with five tonnes overhead included, derived from Tee Five.
Interestingly enough, there was never a minimum for manoeuvre drives.
Power plants remain at one tonne minimum, and fuel tanks also at one tonne minimum.
Interestingly enough, there was never a minimum for manoeuvre drives.
Power plants remain at one tonne minimum, and fuel tanks also at one tonne minimum.
mike wightman
SOC-14 10K
Yes you can and no they don't, not if you refine the formula for the fractional drives
Hence the analysis of the smallcraft in LBB2 81 to get some sense of what may be the fractional drive formula.
Consensus can rarely be reached with such a fractured fan base and inconsistent and often contradictory rules, which have often been revised in the past without the benefit of the forty years of analysis we have since given them.
Yup, but there is fun to be had in dissecting the LBB info and making stuff up - that's what the game originally encouraged
This is another example of unintended consequence of a revision.
77 smallcraft have limited endurance thanks to the fuel use rate, then comes 81 with the flat 4 weeks of operation but at only a fraction of the ship power plant fuel requirement to break things. The CT LBB2 power plant fuel formula for ships has never made sense.
Proneutron
SOC-13
Glad this was fixed in MgT HG SRD. You can design using a straight formula for any hull size. PROGRESS!
Merry Christmas to you and yours!
Similar threads
