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Switching the maneuver drive for the jump drive

Werner

SOC-13
This is an idea for an alternate traveller universe. What if we switched the maneuver drive for the jump drive? It always seemed to me that the maneuver drive used too little fuel and the jump drive was a fuel hog. This seems more realistic, and since an FTL drive is a bit of magic, we would just need the maneuver drive gulping fuel to get the ship to a spot where the FTL drive can be used, and then the ship would gulp more fuel to get the ship safely down to a planetary surface. How would the campaign setting be affected?
 
At first thought, given proper changes, the major difference would be much less advantage for non-jump spacecraft. The whole rationale for fighters, SBDs and Battle riders and the like would be reduced: instead of gaining use of the 15% to 60+% dedicated to Jump-drive and -fuel, you'd gain only a few %.
 
At first thought, given proper changes, the major difference would be much less advantage for non-jump spacecraft. The whole rationale for fighters, SBDs and Battle riders and the like would be reduced: instead of gaining use of the 15% to 60+% dedicated to Jump-drive and -fuel, you'd gain only a few %.

I guess it would be more like Star Trek, but what if there was no artificial gravity? That would put the advantage back with the fighters, as there is no way a capital ship will accelerate at 6g, but a fighter could, as pilots could wear g-suits as the sit in the acceleration couches of their fighter cockpits. Capital ships would be stuck accelerating at 1g. This would also change the floor plans of most of the ships.
 
I was thinking, in addition to getting rid of artificial gravity, we can also swap the jump distance of 100 planetary diameters for a jump velocity of 1% of the speed of light relative to the nearest and most massive body in a star system, which would be the star itself. It takes 3.5 days to achieve 1% of the speed of light at 1g and 3.5 days to slow down again, and the jump drive works by amplifying a picnic individual's teleport ability. Instead of just teleworking himself, with the ship drivers picnic amplifier, he can teleport the entire ship and its contents according to the ship's drive rating that number of parsecs instantaneously, velocity is preserved so the ship has to slow down at 1g for 3.5 days.
 
At first thought, given proper changes, the major difference would be much less advantage for non-jump spacecraft. The whole rationale for fighters, SBDs and Battle riders and the like would be reduced: instead of gaining use of the 15% to 60+% dedicated to Jump-drive and -fuel, you'd gain only a few %.
Another idea is to have different velocity requirement for different jumps. 1% of the speed of light is needed to do a jump-1, 2% of the speed of light is needed to do a jump-2, requiring a full week instead of 3.5 days to reach that velocity. Using a jump drive wears on the prion who's talent is being amplified by the drive. Fighters can have jump drives too, the pilot needs to be a psion, and if the fighter can accelerate at 6g, it takes 3.5 days to reach 6% of the speed of light and then he can make a jump. Only small ships can make these high g maneuvers, nothing over 800 tons, if passengers are carried, they often go in low berths to avoid feeling this acceleration at all.
 
The weird thing about the J drive is that there is no way to use all that LHyd as as fusion fuel that quickly so what the heck is it there for? M-Drive aren't chemical reaction drives so they use the Power plant (fusion power) to power their grav field gnerators. And even in P-P fusion plants you don't need but a small fraction of the LHyd that is currently slated for the J-drive
 
The jump drive uses some of the liquid hydrogen as coolant while the rest is used to power the jump drive. See MWM articles about the jump drive.

As to the maneuver drive, it was originally intended as a reaction drive, a fusion rocket to be precise - see High Guard 1st edition 1979, and the fuel burn rate in LBB2 77 edition..

As to switching them, yup, been there, done that. I've also used the jump drive/maneuver drive split to represent a long duration cruise engine and a rapid thrust launch/combat engine (I made the jump drive and jump fuel the combat drive, while the maneuver drive is the long duration cruise drive)
 
The weird thing about the J drive is that there is no way to use all that LHyd as as fusion fuel that quickly so what the heck is it there for?

WARNING TECHNOBABBLE FOLLOWS:

IMTU most of the LHyd is gravitically compressed into a metallic hydrogen matrix inside the jump drives which is both the source of the deep subatomic particles responsible for opening the jump field as well as the substrate within which they are manipulated to perform the desired jump (a jump field being analogous to an electromagnetic field and the jump matrix analogous to an inductor - don't push the analogy too hard.) As the jump progresses over a week, the matrix is consumed.
 
Here is what the 1977 edition of the LBBs says about fuel consumption, based on Book 2, pages 5-6.

1. Fuel. Starship fuel costs CR 500 per ton (refined) or CB 100 per ton (unrefined), at most starports. Fuel consumption is based on formulae related to the size of the starship power plant and the jump drive.

A power plant, to provide power for one trip (internal power, maneuver drive power, and other necessities) requires fuel in accordance with the formula: 10Pn. Pn is the power plant size rating, determined from the maximum drive potential table by cross-referencing power plant letter and hull size. The formula indicates amount of fuel in tons, and all such fuel is consumed in the process of a normal trip. A fully fuelled power plant will enable a starship an effectively unlimited number of accelerations (at least 288) if necessary to use the maneuver drive during the trip (as when miniatures combat is used to resolve a ship to ship encounter).

A jump drive requires fuel to make one jump (regardless of jump number) based
on the formula: 0.1MJn, where M equals the mass displacement of the starship and Jn equals the jump number of the drive. Drive jump number is determined from the maximum drive potential table by correlating drive letter and hull size. Thus a vessel with a type A jump drive and a type 200 hull has a jump number of 1. It requires 20 tons of fuel (0.1 x 200 x 1 = 20) for its jump drive. Jump fuel requirements are based on jump number rather than the size of the jump actually taken.

Fuel is also used by the maneuver drives of non-starships. When used in such
vessels displacing under 100 tons (ship's boats, shuttles, pinnaces, etc) 10 kilograms (1/100th of a ton) of fuel is sufficient for 1G of acceleration for 10 minutes.

No mention of fuel used by the maneuver drives, but the power plant fuel usage includes maneuvering, which would tend to make me think that some form of reactionless drive is being used for starships.

The following is from the 1981 edition of the LBBs, Book 2, pages 14-15,

D. Fuel: Total fuel tankage for a ship must be indicated in the design plans. There is no cost, but the capacity does influence how often the ship must refuel. At a minimum, ship fuel tankage must equal O.lMJn+lOPn, where M is the tonnage of the ship, Jn is the ship's jump number, and Pn is the ship's power plant rating. Power plant fuel under the formula (10Pn) allows routine operations and maneuver for four weeks. Jump fuel under the formula (O.1MJn) allows one jump of the stated level. Ships performing jumps less than their maximum capacity consume fuel at a lower level based on the jump number used.

Note that power plant fuel covers routine ship operations and also maneuver for four weeks. This again points to some form of reactionless drive.

The following comment on small craft fuel may be found on page 17 of the 1981 edition of the rules, Book 2.

The fuel tankage listed for each craft is sufficient for four weeks of operations

There is no correlation between fuel use and acceleration by a reaction drive. It is a bit fuzzy on the whole issue. No power plant is specified for small craft either.

What you do have is extremely inefficient use of fuel by a power plant for Jump, with the power plant fuel accounting for ship usage and also ship maneuver.

Now, the following are the energy yields of fission and fusion reactions, worked up by me from the energy yields for fission and fusion reactions given in The Effects of Nuclear Weapons, by Glasstone and Dolan.

The energy from the fusion of 235 units mass of Deuterium is equal to 582.8 MeV, or 2.914 times that of U-235 or Pu-239. The fission of 2 ounces or 57 grams of U-235 or Pu-239 will supply 132.4 kilowatts of power for every hour for a year. The fusion of an equivalent mass of Deuterium will supply 385.8 kilowatts of power for every hour of a year. To supply a constant power of 10 Megawatts for an entire year of 8760 hours would require the fission output of 151 ounces of U-235 or Pu-239 or 51.85 ounces of Deuterium.

When you consider how much energy is derived from the fusion of slightly over 3 pounds of deuterium, you get an understanding of how horribly inefficient the power plant for a Jump Drive is. The way that I am handling it in My Traveller/Cepheus Engine Universe is the Power Plant number must match the Maneuver Drive number, while the Jump Drive can be anything. The fuel for the Jump Drive stays the same, while the Maneuver Drive uses 10 dTons of fuel per Maneuver Drive number per Jump. If you think that this is too low, you could plug in a multiplier of the hull size divided by 100 to factor ship size into the fuel consumption. This would result, for example, of a 600 dTon ship with a Maneuver Drive number of 1 requiring 60 tons of maneuvering fuel. An auxiliary A-size power plant supplies ship's power, using the deuterium contained in a cubic meter of heavy water, which volume is included in the A-size power plant volume. I assume that the cubic meter of heavy water is good for at least a year of power for the ship, and very likely a lot more.
 
Your own quote of LBB2 77 edition shows that the maneuver drive uses fuel:
"A fully fueled power plant will enable a starship an effectively unlimited number of accelerations (at least 288) if necessary to use the maneuver drive during the trip (as when miniatures combat is used to resolve a ship to ship encounter)."
Add the quote from the non-starships section:
All non-starships consume fuel at the rate of 10 kg of fuel (1/100th of a ton) for each g of acceleration
and then of course there is the quote for LBB5 79 edition:
Fusion drives as weapons: any ship may use its maneuver drive as a weapon when at short range, provided the drive is operational, and fuel is available.

Note also that the designers of TNE are on record for stating that the switch back to a reaction drive HEPlaR engine was partly to match the intent of the original CT rules - ie maneuver drives are reaction drives not reactionless drives.

Personally the way I can reconcile the performance of Traveller ships with their fuel use is to postulate that the magic of null-grav technology can reduce inertial mass as well as gravitic mass.
 
It throws a rather large spanner into the game mechanics.

While I tend to ignore the more wilder technological options in the Mongoose editions, arguably these may more likely cater to your tastes.
 
Doesn't the Cepheus Engine starship book also have rules on using reaction drives as well for manuever drives?

And there is HOSTILE where the hyperdrive uses no fuel (and acts like a warp drive in travel mechanics) while the manuever drive [being a reaction engine] is where the majority of fuel usage goes (the rest being coolant for the power plant for a year or something).
 
As to the maneuver drive, it was originally intended as a reaction drive, a fusion rocket to be precise - see High Guard 1st edition 1979, and the fuel burn rate in LBB2 77 edition..

The '77 book has no indication of the M drive being reaction. And HG data would not work as insufficient LHyd for the Delta V obtained.

As for Using that much L-Hyd for J-drive coolant, wouldn't be possible to use that much in the short time period of transitioning to J-space. The basic problem is the J-drive fuel volume wasn't made that large for any logical drive reasons but to limit available tonnage in star ships. Hence the elephant in the room problem since day 1.
 
Doesn't the Cepheus Engine starship book also have rules on using reaction drives as well for manuever drives?

Nope. The design rules have the PP powering the M-drive and other ship systems for X amount of time independent of M drive usage. So not a reaction drive for standard M-drives.

As a side note I see that the C engine has continued using the illogical descriptions for Hull configurations. Written by a person who didn't think it through. :oo:
 
Nope. The design rules have the PP powering the M-drive and other ship systems for X amount of time independent of M drive usage. So not a reaction drive for standard M-drives.

As a side note I see that the C engine has continued using the illogical descriptions for Hull configurations. Written by a person who didn't think it through. :oo:
Er the Spacecraft Design Guide for CE, on page 36, literally mentions what fuel usage for reaction drives would be, along with referring to a table on page 29 of the same book.
 
Er the Spacecraft Design Guide for CE, on page 36, literally mentions what fuel usage for reaction drives would be, along with referring to a table on page 29 of the same book.

I'm looking at the SRD and nothing about reaction drive as the standard M drive in ship design section. I did a search for "reaction" in the book 2 srd and nothing for "reaction" drives. So, M-drives in the normal design are not reaction drives. No doubt there may be a low tech design option.
 
sometime we ought to design some 1950s style retro-sci fi starships instead of the Star Wars kind. In the Last Jedi, I think they took the World War II analogy a little too far with those gravity bombers. Most bombs in space would be delivered by missiles and wouldn't rely on the artificial gravity of the ship they were trying to attack to pull dropped bombs toward itself. Here's a dirty trick, the First Order simply reverses its artificial gravity and the drop bombs fall back up and hit the bombers.
 
Er the Spacecraft Design Guide for CE, on page 36, literally mentions what fuel usage for reaction drives would be, along with referring to a table on page 29 of the same book.

If the author of the CE Spacecraft Design Guide includes reaction drives for the maneuver drive, that does not change the System Reference Document, which I am using, and which has no mention of reaction drives, and this does include small craft. Small craft have a power plant and maneuver drive, with fuel usage confined to that used by the power plant.

As I tend to base a lot of my ideas on the works of H. Beam Piper, who does not have his starships using reaction drives, I am assuming something equivalent to the "Dean" drive as my maneuver drive.

As a further note on the Classic rule set, the Azhanti High Lightning supplement came out in 1980. The cruisers have 2G acceleration with "unlimited maneuver" (quote from page 5, General Specifications). Now, the power plant number is 5, and with the rules for power plant fuel consumption being 10 x PN (Power plant Number), the fuel required for the power plant would be only 50 tons, for "unlimited maneuver". This does not strike me as anywhere near correct, based on the size of the ship and the energy required to accelerate it at 2G. The fuel capacity is 32,000 tons, of which 30,000 is devoted to Jump fuel, leaving 2,000 tons of fuel for other purposes. Some of that will be used to fuel the small craft carried: 60 Rampart fighters, four 400 ton fuel shuttles, and five 40 ton launches/gunboats. If, however, you use the fuel consumption given for non-starships of 10 kilograms of fuel for 1G acceleration every 10 minutes for a ship not exceeding 100 tons, you get the following. With a size differential of 600 and 2G acceleration, the fuel consumption would increase to 1200 X 10 kilograms, giving 12,000 kilograms of fuel every 10 minutes, or 20 kilograms per second (approximately 44 pounds per second). I seem to remember that in the book Thrust Into Space the specific impulse for a fusion reaction drive being around 4 million (I do not have the book handy right now, so trusting my memory). That would equate to a thrust of 156,000,000 pounds. The sounds impressive, but the Saturn 5 liftoff thrust was 7.5 million pounds, so that is about 21 times greater, but with a lot smaller fuel consumption.

The problem is a fully fueled ship is carrying 32,000 tons of Liquid Hydrogen in addition to the ship's empty weight. That means that you have, figuring the fuel at 2205 pounds per metric ton, the fuel alone weights 70,560,000 million pounds. That is close to one-half the given thrust, not counting the mass of the ship, which is going to be considerably greater than 70 million plus pounds, 32 thousand metric tons. The fuel burn is not even going to give you 1G of acceleration. Quick estimate of mass of the cruiser: 840,000 cubic meters of displaced volume half-submerged in Liquid Ammonia at 681.9 kilograms/cubic meter (water is 1,000 kilograms/cubic meter), so roughly 420,000 cubic meters of hull displacing that volume of ammonia at 681.9/1000, which would give a mass of about 286,000 metric tons. Converted to pounds, which is what I am using for the thrust, that would be about 630 million pounds. Add in the mass of the Liquid Hydrogen carried, and you get over 700 million pounds with only 156 million pounds of thrust, so less than a quarter of a Gee.

Based on this, to give the Azhanti High Lightning class of cruisers 2G acceleration, you need to burn more like 100 or so metric tons of fuel every ten minutes. That is going to burn through that 2,000 tons of Liquid Hydrogen fairly quickly, so unlimited maneuver is out if you are using reaction drives.
 
The '77 book has no indication of the M drive being reaction. And HG data would not work as insufficient LHyd for the Delta V obtained.

As for Using that much L-Hyd for J-drive coolant, wouldn't be possible to use that much in the short time period of transitioning to J-space. The basic problem is the J-drive fuel volume wasn't made that large for any logical drive reasons but to limit available tonnage in star ships. Hence the elephant in the room problem since day 1.

Not the first time I've said this, but the Jump fuel burn rate feels an awful lot like the fuel burn rate of conventional rockets launched into orbit. Getting to orbit is (under Traveller physics) easy due to magic maneuver drives and gravity technobabble, but getting into Jump Space is HARD, hence the large fuel burn.

I'm still partial to the original "Jump Drive burns the fuel" with big Jump Drives (LBB2, leaning on the '77 rules) concept. I read the '81 rules requirement for the power plant as providing for surge/stabilization power rather than providing all the power used by the Jump Drive.*
Spoiler:
*I get there from HG's Jump Capacitor rules and the power required to Jump: Jump Drives include 18EP worth of capacitors per Jn per 100Td, but only require 2EP per Jn per 100Td to initiate Jump. This leaves a deficit of 16EP per Jn per 100Td, which can only come from the Jump Drive itself. If the Jump Drive put out 12.5% more energy, you wouldn't need the Power Plant at all (and we're back to LBB2 '77 again...) This isn't done, which means that an in-universe reason exists for not doing it.

My explanation is that the Jump Drive's integral power plant, in addition to being very high output and very inefficient, also has a very unstable power output. If you attempted to use it to charge the Jump Capacitors on its own, it would very likely either undercharge them (failed Jump) or overcharge them (misjump if you're lucky, explody if you're not). So the Jump Drive's power plant is designed to provide 89% of the energy needed for Jump (and might produce 100% -- but not more), and the power plant is sized to make up any shortfall.
Thus, the Jump Drive incorporates a very powerful but horribly inefficient power plant that can only be used to power the Jump Drive itself (maybe it generates alternating current at some insane frequency set by the Zuchai Crystals?). In theory, you could build a power plant with the same fuel burn and power output characteristics to power ship systems, but it'd be so inefficient as to be pointless.

(Out of universe, that requirement was probably to address the issues that led to High Guard's Energy Point rules without having to completely rewrite LBB2's ship design rules.)
 
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My explanation is that the Jump Drive's integral power plant, in addition to being very high output and very inefficient, also has a very unstable power output. If you attempted to use it to charge the Jump Capacitors on its own, it would very likely either undercharge them (failed Jump) or overcharge them (misjump if you're lucky, explody if you're not). So the Jump Drive's power plant is designed to provide 89% of the energy needed for Jump (and might produce 100% -- but not more), and the power plant is sized to make up any shortfall.

Yes, this is a good explanation
 
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