AnotherDilbert
SOC-14 1K
A 100dt hull gets you M2 as a minimum (in most versions.)
Except CT, MT, TNE, T4, and T5...
A 100dt hull gets you M2 as a minimum (in most versions.)
Except CT, MT, TNE, T4, and T5...
One could claim that the official X-boat network was political, rather than economical.
(CT) Book 2 Drive potential table pg. 22
See LBB5'79, p23; E.g. a jump drive-1 in a 100 Dt hull is 2 Dt.
S0? I'm talking about M-DRIVES
S0? I'm talking about M-DRIVES
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 . . .
Minimum for a 100Td hull is 2G at 1Td.I'm talking LBB2
Yes there is.There's no good way to extrapolate 1G in 100Td from LBB2 even if you wanted to.
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.
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 and A/2 drive while the launch is A/10 and the fighter C/10.
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.
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 - 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.
Yes you can and no they don't, not if you refine the formula for the fractional drivesYou 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.
Hence the analysis of the smallcraft in LBB2 81 to get some sense of what may be the fractional drive formula.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.
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.And not everyone will agree on how that should go.
Yup, but there is fun to be had in dissecting the LBB info and making stuff up - that's what the game originally encouragedThe OTU answer is to use LBB5 design rules.
This is another example of unintended consequence of a revision.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"!)
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:
Spoiler:M must be between 100 and 5000 inclusive.
Note: values of M*G less than 100 yield negative drive tonnage!