The great conversion of MegaTraveller starships back to High Guard

You do realise that a 13.5 cubic metre deckplan "ton" means you will need to allocate more volume to the fuel since 1000kg of liquid hydrogen requires 14 cubic metres...
Not in MT it doesn't...

Real LHyd is more than 14 m³/tonne.

All editions approximate.
 
I tried to reconstruct the System Defense Boat (SDB) from Supp 9 Fighting Ships. The problem is both its small size (200 D-tons) and high armor (HG 13). The way MT treats armor, every 3 MT factors equal 1 HG factor, so HG 13 = MT 40+3x13 = MT 79, which is a LOT of armor.
Note that MT has a limit of max armour = TL × 5, though how that is applied is hotly contested...
 
And I would be wrong again, LHyd is approximated to 0.07 tonne/m³ => 14.3 m³/tonne or 0.95 tonne/Dt.
0.07085 = 14.114, so pretty close. Much closer than most deck plans, especially streamlined ships where any tanks in wings of fins would be a lot less than 3 metres tall.
 
Except it isn't.
1000kg of liquid hydrogen fuel occupies 14 cubic metres.

So for deck plans 1.08 x jump fuel "displacement tonnage"
 
0.07085 = 14.114, so pretty close.
Sure, but MT stipulates 0.07. Good enough...


Much closer than most deck plans, especially streamlined ships where any tanks in wings of fins would be a lot less than 3 metres tall.
Most people probably can't be bothered to do all that geometry, and I agree. I generally try to get the hull volume about right, and deck area roughly right, and assume the rest is fuel.
 
So for deck plans 1.08 x jump fuel "displacement tonnage"
Ship fuel requirements are given in volume (except perhaps LBB2'79), so 1 Dt fuel takes 1 Dt volume aboard ship.

MT fuel requirements are specifically given in kl = m³, i.e. volume, so it's quite easy to approximate deck plan volume.
 
You still need 14 cubic metres which doesn't fit in a 13.5 cubic metre displacement ton, hence the conversion factor,
 
Lol, only an Engineer could approximate 14.3 to 13.5.
And the MT authors...
MT RM, p57:
Tons Displacement: A widespread method of determining a space vessel's size is to give its volume in terms of the amount of liquid hydrogen it would displace (as if it were immersed in a vast sea of liquid hydrogen). Tons displacement is not to be confused with the craft's weight in metric tons (that is, its actual mass). A starship that displaces 100tons may actually weigh over 1000 metric tons. A displacement ton is actually a measure of volume rather than weight; one displacement ton equals 13.5 kiloliters of volume.
Presumably that is done to make 1 Dt = 2 squares, to make deck plans simpler.

It's well within spec. as far as LBB2 is concerned:
LBB2'81, p21:
Deck Plans:
...
One ton of ship displacement equals approximately 14 cubic meters. Therefore one ton equals about two squares of deck space.
...
Finally, a leeway of plus or minus 10% to 20% should be allowed. If the final deck plans come within 20% of the tonnage of the ship specifications, then they should be considered acceptable.
LBB2 says 11.5-17 m³ is good enough...


MT approximates that 13.5 m³ fuel takes 13.5 m³ (=1 Dt), with a mass of ~0.95 tonnes.
 
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And the MT authors...

Presumably that is done to make 1 Dt = 2 squares, to make deck plans simpler.


MT approximates that 13.5 m³ fuel takes 13.5 m³ (=1 Dt), with a mass of ~0.95 tonnes
yes, the MT authors decided to base their displacement ton on deckplan scale and then tell porkies about liquid hydrogen density.
Science fiction is meant to be based on science and saying 13.5 is the new 14 is doublewrongspeak.
 
yes, the MT authors decided to base their displacement ton on deckplan scale and then tell porkies about liquid hydrogen density.
Science fiction is meant to be based on science and saying 13.5 is the new 14 is doublewrongspeak.
Jump drives and artificial gravity are OK, but 6% off on density is unacceptable?

Are you absolutely sure all internal combustion engines are exactly 1 tonne/m³ with no deviation, or is that doubleplusungood too?

It's just a game...
 
LBB2 says 11.5-17 m³ is good enough...
Yes.
It's not engineering drawing, but a playing surface. In any case, if Traveller (any version) is mostly concerned about volume and not mass, then, as someone pointed out, it would occupy less volume to store it as water (1/9 hydrogen at 14 times the compression - so 1.55 times as efficient).

In my opinion, Not a lot of point arguing about whether two old versions of Traveller did it 'right' - but there is a point here (sort of): Marc might say he is bound by canon, but it would be more correct to say 'guided by canon', you can't be bound by contradictory rules. And that's what this conversation seems to have become.

(Me, I just want to be un-bound by the Virus* and Empress Wave† - but the best way non-contradictory to do that is to set a campaign in the era prior to Strephon's assassination, an event that didn't bind Loren Wiseman at all...)

*On a personal note, as much as I found the Virus annoying, and unnecessary for the eventual start of a new age of decaying splinter states, the thing about TNE that annoyed me the most was the 70ish year timeframe between the start of the Virus and the rise of the Space Vikings. Way too short. Most of the system names changed, lots of knowledge was forgotten, but that's less of a time than from today back to World War II, when my still living parents were alive (and have vague memories of the occupation - they were born a month before the Nazis invaded Norway) and my grandfathers' families fought in the resistance - I grew up with a bunch of those stories, some even verifiably true.

†And Empress Wave, I still think the 1 parsec per year thing was just an editing oversight, but the wave rolls on so it hits when it hits. At least to will be gone faster that way. Now there is an event where 70ish years could result in enough brain scramble and widespread damage where memories of the 'Before Times' where garbled.
 
Irrelevant:
HG Agility is calculated on power production - weapons power, hence on power for the m-drive. HG Agility is limited by m-drive. It's all about the m-drive.

MT Agility is calculated on excess power when the m-drive is already fully powered, it's not limited by m-drive. MT Agility has nothing to do with the m-drive, and that's ridiculous.
No, it's supposed to be a power to weight ratio, like all other actual physical systems. MT Agility is a power to weight ratio. HG doesn't "use" weight, but if you amazingly use the D-ton mass assumption I used you miraculous get the MT formula from the HG formula. Clearly a freak accident. /end sarcasm.

More seriously, Striker uses a power to weight ratio (which MT also uses), to determine the velocity of vehicles. Higher power to weight = more velocity, NOT just power. Also, I never mentioned the M-drive. Both HG and MT use excess power not powering something else to determine Agility.
 
Now that I think about it.

Is the wave spreading evenly in all directions and dimensions?

In which case, why isn't it diluted?

While I wouldn't mind having differing fuel densities, as long as players don't mind calculating out fuel mixture formulas, and possible increased chance of misjumps.
 
While we continue to discuss the density of LH, I have one more comment on the revised assumptions I'm using to get MT back closer to HG. I knew there was still one missing assumption that would eventually produce a failure like the SDB. I did not use this assumption because it is the worst part of the HG system. But I should still lay out the problem and let you all discuss it.

The problem, as simply as I can put it, is: 1 HG power plant unit = 1 MT power plant unit (EPs only) + "free" power for all the drives once the PP is "large enough".

In other words, there is no explicit energy cost for M/J drives in HG, only that the PP be large enough. The EPs provided by a HG PP are almost exactly (within rounding) the same as the MWs (x250) provided by MT PPs.

The final step to "exactly" replicate HG in MT is as follows:
1) MT PPs still need to be made as fuel efficient as HG, so multiply needed MT PP fuel by 0.15, and use the HG formula for jump fuel of %age ship volume = 10% x jump number.
2) MT maneuver and jump units are fine, except eliminate the MW cost for maneuver units. This will be covered by the size of the overall power plant.
3) We must compute a PP number for MT, same as HG in the same way. This is total volume of PPs as %age of hull divided by a TL factor of 4 for TL 7-8, 3 for TL 9-12, 2 for TL 13-14, and 1 for TL 15. This is identical to the Power Plant Table from HG pg. 23.
4) The computed PP number must exceed both the maneuver number and jump number. There are NO other requirements except fuel for powering the PPS and j-drive.
5) All available MWs of power from the PP are available for all other systems, including excess power to compute Agility.

It is now easy to see the difference/problem. In MT, you need to build more power plant to power the drives. In HG, drives are powered for free if the PP is large enough, and all EPs/MWs are available for other systems. I have a table that quantifies this precisely, but it is approximately correct that HG PPs implicitly produce almost 2x the energy as MT PPs of the same size.

I didn't add this calculation because I don't like it. All power is explicit in MT, and I don't want an exception for space drives. So my original proposed solution means there will be some limitations on Agility as ships become more massive. I suspect this diminishes as size increases, as is the case for the SDB. I have a MT version of one of the very large capital ships from Supp9, so I'll convert that next and see what we get.

Anyway, this is all hammering away at a dead, dead system, but for me it was a very useful exercise. I'm probably going to implement these changes in my game as I like the scaling effects. All but the problem discussed here, or course, and if anyone has a good explanation of why drive power should be "free" for large enough PP, I am very interested to hear it.

Thanks as always, fellow gearheads!
 
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