Jumping at velocity?

[far-trader]

A normal jump has an exactly predictable time of emergence when the jump is computed.

But that would mean that the jump duration was established before the jump was initiated, at a time when the ship can still abort the jump and recalculate a new one. ""The calculated jump will take 180 hours. Abort and recalculate for jump in 20 minutes' time. How long? 164 hours? What are the odd of getting a shorter duration on another jump? Hum... We'll take this one then. Ready for jump!!"

See how that doesn't work?

Ref: "Well, where is your next jump headed?"

Player: "Back to Regina."

Ref: <rolls dice> "OK, you run the generate program and work out a jump of 180 hours duration."

Player: "That sucks, let's redo it and reroll... "

Ref: "No, that's not how it works. The roll represents the time it takes any ship jumping from this system to Regina under the current alignment. There may be some minor variation for different sizes of ships, different points of departure or arrival, and other factors but in general at this time all jumps from here to Regina take a definitive 180 hours. Running the generate plot again won't change that."

See how that works

...or alternatively (if you'd like to make player skills count for something):

Ref: "Sorry that's the best your Navigator can manage, he just doesn't have the skill required to make a better plot. Now if you had hired that old spice runner and his big furry sidekick back at the divey port bar, the one bragging about making the Regina run in less than 160 hours... "

...or maybe the gear should factor:

Ref: "...but he probably had his ship finely tuned and maintained as well. When was the last time you had the injector ports cleaned on this old trader?"

Or both. Still in any case, multiple rolls for the same trip are referee fiat nixed. It's not how it works, imo at least as that results in things not evidenced in the rules (like all jump times precalcuated to arrive at the shortest possible time, always).

And this can be done far ahead of time as evidenced by the use of jump cassettes. Only misjumps will have an unpredictable time factor.

Jump cassettes would be calculating the jump a REALLY long time in advance. Now you'll have System Control calculating thousands of jumps and selling the short ones to the highest bidder.

I think the above addresses this as well. When I say "far" I mean hours (plenty enough time to get to 100d), when you say "REALLY long" I read an exaggeration of my statement and implication of days or weeks or longer

For additional clarification jump cassettes in MTU don't cost as much as they do in CT (it just doesn't work at that cost, everyone would buy Generate programs). They are computed by SPA on demand and require the jumping ship's vital statistics* including as you do a specified point and time of jump entry among other specifics. Miss the window or mess up the rest and you may misjump.

* usually done via computer link, and rather than a self erasing cassette it's a one time use program download

I don't expect any conversions to happen, I'm just here to enjoy the debate and share points of view

 

[mike wightman]

Dan you pretty much nailed it - I was going to reply to Hans much along these lines too.

Note from an actually playing the game perspective it makes the pc skill relevant (good thing), and slightly improved or tuned equipment a good reward (see, I told you it was worth helping out that Droyne colony, the mods they've made to our jump drive are really paying off).

 

[nats]

Nats,
there is no such thing as "zero velocity"... unless you reference something else. If you are in system A at V=0 vs it's star, and exit jump in system B, at v=0 to its star, you have to account for the difference in V of the stars themselves....

Some systems may be 100G-hours vector difference or more... but usually only a few 10's of KPS... 10kps is 1000G/sec... roughly 15 min at 1G. So typical jumps will be correcting for an hour or so...

Yes I know - thats exactly how the Frontier PC game did it - you had a reference point that was usually the object with the largest gravity field relative to your ship. So on entering the system your reference pioint would usually be the star, once you were closing on the planet, your reference point would change to that planet.

But I am not interested in playing Traveller to that degree of realism. For me having a zero vector compared to the star is fine. And when you jump to a new system your ship would again have zero movement in relation to the new star. Thats good enough for me. Although I am aware its not realistic I play Traveller unrealistically (with Star Wars type asteroids etc) for the enjoyment, I dont play it to be realistic. Get enough of that in the real world.

 

[rancke]

Ref: "No, that's not how it works. The roll represents the time it takes any ship jumping from this system to Regina under the current alignment. There may be some minor variation for different sizes of ships, different points of departure or arrival, and other factors but in general at this time all jumps from here to Regina take a definitive 180 hours. Running the generate plot again won't change that."

See how that works

Yes, I do. And I can see how that could work out fine if that was the way it worked.

I also see that it's not the way it works in the OTU, since the jump duration is determined when the jump is initiated, not when the jump solution is generated, as the previously quoted line from Marc Miller's essay plainly states. (And also as the MT rules about the problems of fleets jumping together plainly and indisputably shows.)

I can also tell you that I personally, as a referee, really like the ability to have the PC ship and the ship their rivals are using jump more or less at the same time and yet have one arrive in the destination system six hour early and the other arrive six hours late. It provides me with such wonderful opportunities to complicate the lives of the PCs.


Hans

 

[rancke]

For me having a zero vector compared to the star is fine. And when you jump to a new system your ship would again have zero movement in relation to the new star.

But that's not what the rules have you doing. The rules have you achieving a vector that is zero in relation to the departure world, not the departure star. And the odds against that vector also being neutral vis-a-vis the destination world are literally astronomical.


Hans

 

[BytePro]

From MWM Jumpspace article:

"The duration of a jump is fixed at the instant that jump begins..."

Says to me the duration is fixed once you jump, not before. Once you jump, "the instant that jump begins", you cannot change it, you are committed to the jump and it will take that long to complete.

Or it says that if you know the moment jump is to begin you can pre-calculate the jump duration.

Don't have the article, so the rest of the context is missing - though I suspect it is no more definitive. I like CT's loosely defined jump aspects anyway.

Can see 'jump tapes' (how quaint) facilitating calculating jump duration based on an initial position and time of jump. Since it takes time for the jump plot normally, the option to re-roll means a loss of pre-jump time for the opportunity to obtain another result.

Thus, re-running plots may be fine when leisurely cruising and hoping to shave some time at the destination - not so easy a choice when being pursued out of the system.

 

[nats]

But that's not what the rules have you doing. The rules have you achieving a vector that is zero in relation to the departure world, not the departure star. And the odds against that vector also being neutral vis-a-vis the destination world are literally astronomical.


Hans

Do the rules go into that much detail? If so I cant remember seeing it. Either way makes no difference to me. As far as I am concerned zero vector is for all intents and purposes not moving in relation to everything else in that star system. In a simple role playing game like Traveller I dont want to get bogged down by astronomical relationships and orbital velocities. Thats far too complex for me.

 

[BytePro]

Actually they go into that little detail

There are no provisions for 'zero in relation to a star'... that would require astronomical relationships and orbital velocities - or the presumption that objects don't move in a system.

The rules require making up some distance (min to max orbital distance) and handwaving away orbital movement.

Simplicity and limiting yourself to the rules would just mean summing the 100D of the departure and destination worlds and using the midpoint formula/table for that distance with a given acceleration.

 

[dean]

Do the rules go into that much detail? If so I cant remember seeing it. Either way makes no difference to me. As far as I am concerned zero vector is for all intents and purposes not moving in relation to everything else in that star system. In a simple role playing game like Traveller I dont want to get bogged down by astronomical relationships and orbital velocities. Thats far too complex for me.

There cannot be a state in which "zero vector" exists as you describe it (not moving in relation to "everything else" in the system). In trying to "simplify" you are actually making it much more complex than the rules make it.

Just handwave the whole problem and say the jump program handles all the needed details.

 

[nats]

Actually they go into that little detail

There are no provisions for 'zero in relation to a star'... that would require astronomical relationships and orbital velocities - or the presumption that objects don't move in a system.

The rules require making up some distance (min to max orbital distance) and handwaving away orbital movement.

Simplicity and limiting yourself to the rules would just mean summing the 100D of the departure and destination worlds and using the midpoint formula/table for that distance with a given acceleration.

Where does it say this in the rules, because I cant see it? The only thing I can see is on page 10 of Book 2 it says you decelerate to arrive at the destination at rest. The diagram above that says destination world but that is not actually described in the text and a 'destination' could just as easily be an asteroid belt or a jump point or anything else, not just a destination 'world'. Nowhere can I see it mention anything about arranging to achieve a zero vector relative to your destination world or your origin world.

The rules are very vague on the matter and I think people here are assuming a lot that just is not in the rules at all. It certainly does not say that you are to arrive at the jump point at rest in respect to any world whatsoever and as far as I am concerned being at rest relative to the main star would be far more sensible. But the rules state nothing about it per se.

Which is why I was asking originally. If anyone knows anything else about it that is covered perhaps elsewhere in a JTAS or something I would be interested to hear about it but certainly my rules dont seem to say anything about it in detail, I dont know what game rules you lot are reading from! Dont you just love the way people make things up in this forum!!

 

[far-trader]

Where does it say this in the rules, because I cant see it? The only thing I can see is on page 10 of Book 2 it says you decelerate to arrive at the destination at rest.

It is an over-simplification. Nothing is at rest (zero vector) except relative to something else.

Our galaxy has a vector through the universe. Our sun has a vector through our galaxy. Every planet, moon, rock, and speck of dust in the solar system has a unique vector.

If you leave planet A (starting at rest, relative to the surface, a zero vector) and thrust out for planet B you have to account for the vector planet A adds and adjust for the difference planet B introduces. Otherwise you'll never get there.

In the over-simplified Traveller system you thrust from Planet A at X-gees for Y-hours, flip, and thrust at X-gees for another Y-hours and miraculously arrive at Planet B with zero vector.

In reality you would be off from it by some degree, with a zero vector relative to Planet A.

...more or less. I've also over-simplified. The problem becomes more complicated when you jump into a star system where the star and the planet you hope to arrive at both have (possibly huge) different vectors than the ones you just left.

It certainly does not say that you are to arrive at the jump point at rest in respect to any world whatsoever...

Actually that is exactly what the rules describe. A zero vector relative to the world you left. Not the star of the system. Not the world you're jumping to. Not even another world in the same system if that was your destination.

 

[rancke]

Where does it say this in the rules, because I cant see it? The only thing I can see is on page 10 of Book 2 it says you decelerate to arrive at the destination at rest.

The travel formula. I don't know if it's in Book 2, but it is in The Traveller Book. It has the ship accelerate for half the time, then decelerate at the same rate for the other half of the time. That places you at zero velocity to only one thing, namely your original vector, which would be the vector of the origin world.


Hans

 

[aramis]

The travel formula. I don't know if it's in Book 2, but it is in The Traveller Book. It has the ship accelerate for half the time, then decelerate at the same rate for the other half of the time. That places you at zero velocity to only one thing, namely your original vector, which would be the vector of the origin world.


Hans

That's the beauty of CT & MT....

as a kid (or an astrophysics-ignorant adult), one isn't confronted with calculations for zeroing out the vector from the system, nor any references to same...

But as one learns a little astrophysics (maybe, just maybe, to level 0), one suddenly realizes that leaves an interesting amount of speed in some cases, and a minor one in many... and one can immediately implement that if they choose.

The rules state you retain your vector... so you retain your vector.

 

[BytePro]

Sorry, nats, not sure what you are trying to say or what you are reading into what you quoted - but I do know I certainly didn't make up any rules in what you quoted.

Directly from CT reprint of Book 2 pg 10 - 'The three travel formulae assume constant acceleration to midpoint, turnaround, and constant deceleration to arrive at the destination at rest, as shown in the diagram above.' (emphasis mine)

The 'diagram above' is titled 'A Typical Interplanetary Journey' and depicts an 'Origin World' and a 'Destination World'. The table is part of the description as it is explicitly stated in the sentence. ('World' could be left out of those terms, but the table is about 'Interplanetary' trips. <shrug>)

Mathematically, since the final velocity equals the initial velocity, the trip began 'at rest' as well. This is explicitly backed up in Book 2.

From pg 4 of Book 2 (Classic Reprints) - 'All of the formulae ... assume that the ship is undertaking a journey from rest ... to rest again.'

Technically, 'at rest' means zero velocity with respect to some body. In this case, the bodies are implicitly the origin and destination worlds, though one could choose to interpret things otherwise (a bit of a stretch, but so be it). If you do, then you are stating that the worlds are also at rest to that something else, or that the journey begins and ends with different velocities with respect the worlds or other objects (probably not so desirable, eh).

Unless the worlds (or other objects) are at rest with respect to each other (or to each other and the star if one adds that complexity), then that's a lot of slamming into objects, or watching destinations fly away.

As I stated, 'The rules require making up some distance (min to max orbital distance) and handwaving away orbital movement.'

Using this simplification directly from the rules, for jump one can make that 'some distance' the sum of the 100D limit from origin and the 100D limit from destination and be done with it. Jump happens when the origin 100D is reached - the velocity is irrelevant (to the rules) and only addressed as 'at rest'.

And that last is the key. Traveller interplanetary rules cover Acceleration (G's), Distance, and Time - Velocity's only mention is 'at rest'...

JTAS 24 may offer more insight into the whole 'Jump at zero velocity' thing (that others have referred to). Doing so adds complexity - i.e. 2 calculations vs. one - not to mention compounds a handwave that is already made regarding ignoring orbital velocities...

 

[shadowcat20]

I could not resist giving this ball a swift kick.....

How does it work for the scout service blind jumping into a new system?

They may have some info from long range telescopes but little knowledge of the system thay are jumping into.

No pre plotted vectors or accurate distances of planets vs sun(s). No clue if there may be a stray rock belt at the end of their jump.

I figured stopping before a jump was a safe way to enter a new system. Military units and ships one step ahead of that missile salvo may be a little less cautious.

 

[dean]

I could not resist giving this ball a swift kick.....

How does it work for the scout service blind jumping into a new system?

They may have some info from long range telescopes but little knowledge of the system thay are jumping into.

No pre plotted vectors or accurate distances of planets vs sun(s). No clue if there may be a stray rock belt at the end of their jump.

I figured stopping before a jump was a safe way to enter a new system. Military units and ships one step ahead of that missile salvo may be a little less cautious.

Why do you think the scouts' survival rolls are so marginal?

 

[rancke]

Why do you think the scouts' survival rolls are so marginal?

They are a lot worse than the odds of jumping blind into a system and winding up in the direct path of any object without time to get out of its way. THOSE odds are WAY low.


Hans

 

[dean]

They are a lot worse than the odds of jumping blind into a system and winding up in the direct path of any object without time to get out of its way. THOSE odds are WAY low.


Hans

Hence the smilie to indicate an amusing rather than completely serious statement.

 

[Grognard]

I allow both standing jumps (0-velocity) and running jumps.
However, Running jumps are more difficult, and can cause a higher chance of misjump.

Also, a running jump has to be calculated for the differences in jump point distances (travel time at xG) arrival at the destination world. Full accel from a size 2 world to jump point is a lot less distance than to a size 9 world. Oft times this means coasting, (if going to a smaller world) or longer accel (if going to a larger world). Jumping outside of the 100d factors in as well. When jumping from a large planet to a small planet, sometimes the ship will need to vector around and return to orbit after having passed it by at high velocity.

Please also consider a planet well inside a Star Gravity well as an example of a origination world that would require weeks of acceleration just to reach the Jump Point.
That same amount of decel will be needed at the exit point....and could at this point, depending then on how long it took to get back to the destination world (possibly passing it by why decelerating!) could come close to burning an entire load of fuel and Life Support reserves. (Menorb in Regina/Spinward Marches comes to mind)

Standing Jumps are easier to calculate travel times for, but running jumps are more cinematic.

 

[Grognard]

I also argue that Velocity is necessary when trying to calculate how long it takes to catch another ship, or reach a space station or whatever....

ignoring velocity would be foolish.

Suppose during in system transit the M drive is disabled....how fast is the ship moving? It can no longer continue to decelerate to arrive at the destination world, and will so pass it by completely.......what vector, velocity and how long does the rescue team have to prepare? How long will it take to match vectors and return?

giving this info to the players to devise a rescue is critical.