Jump vectors

[McPerth]

While discussing about the X-boats, the issue of the jump vectors appeared, and I think it deserves a thred by itself.

In other games (e.g. Starfire) the points from where you leave a system are fixed ones, leading to fixed other points. In Traveller, they have always been jump vectors that have to be calculated by the Navigator (or an equivalent computer program).

Are those vectors more or less fixed?

I understand they are not, but they must be calculated for each jump as for where, in which entry vector and when, at a specific time span (that I asume quite short).

See that if we asume them to be more or less fixed:

• You don't need a Navigator in your ship, as they will be widely known (at least in "civilized" space.
• You can tell where another ship went by knowing it, even without the restrictions I've read TNE sets to this knowledge (AFAIK othr versions don't mention the issue).
• You could jump only from more or less set places, not from anywhere in the system just looking for an available vector to your destination (as told in many Traveller references)

And sure, there would be other implications I cannot think about right now...

 

[whartung]

No, they're not fixed.

The stars and systems move through space, and jump is directed to absolute coordinates that simply happen to be near where you want to go. Jump is "Line of sight" from the source to the destination, in that masses can be "in the way" of the jump (ala jump masking and shadowing). For example, can't be on the Rimward side of a star, and jump directly coreward, your ship will end up at the 100D of the star a week later. So, in theory navigators need to worry about those nagging details. But give the "vast emptiness" of space, this is likely not an issue.

Also, the jump direction is unrelated to the normal space vector of the ship.

Jump is like a scratched cue ball on a pool table. You can place the ball anywhere you want.

By the same token, consider the cue ball is moving, but maintains it's vector after you move it.

So, you can place the ball anyplace on the field but once it lands, it keeps going in the same direction as it was when you picked the ball up.

Your ship can be rocketing spinward, and then jump perpendicularly rimward. When you exit jump the ship is still traveling spinward.

All that said, it is quite likely that trade traffic from one system to another may well exit the planet all using a similar vector, and they can use that vector to minimize arrival time, and since all pilots want to do that, and all pilots are lazy, they'll all use similar vectors.

So, story wise, it's far to assume that if you see a ship flying off 20 degrees anti-spinward, you may well assume that it's heading to planet X, because that's the common vector for ships heading to that planet. But note, that vector will vary based on time of year and destination orbit etc.

But, that's simply presumption on best practice. If you have random ship trying to just evade and jump, there's no way to tell where they went, and I certainly wouldn't necessarily apply these assumptions to military traffic.

 

[mike wightman]

You may be confusing real space velocity vector with jump line.

I can have zero velocity and jump in any direction I want up to the range of my jump drive and the rules for 100D, I can have a real space vector of 100km/s towards rimward and yet jump - up to the limit of my jump drive - in any direction I want, including returning to where I started the jump, a few thousand km behind my jump initiation point, doesn't matter.

Real space vector has no bearing on jump.

It does have a bearing on which way I am pointing at my destination and how much I will have to use my maneuver drive when I get there.

 

[aramis]

There are several factors to consider.

1: Solar motion
Sol is cruising 790,000. km/h through the galaxy
Apparent speeds can range from 0 to 10x that... up to 2190km/s

219 km/s, a decent "medium" speed (yes, pun intended), is 6.08 G-hours (6:05:00), or a quarter day. Just to give a clear idea. 10x that is 4 days...

2: World motions
A few minutes accounts for that.

Intervening position objects. Sometimes, you need to change angle to hit the target, because stuff's in the way on your end. Figure several days on average.

Intervening stuff at the far end? hey will have to meet you.

 

[AnotherDilbert]

Combining what Mike and Aramis says we can get a good clue to the destination. We cannot jump through through big objects, so some destinations are ruled out. If we assume that the observed ship is accelerating for a minimum time intercept with a target planet, some destinations are more likely than others.

In recent editions there are Jump Filters that allows you to detect where an observed ship is jumping.

Edit: Sorry, whartung already said that better.


In CT we can buy jump tapes, instead of having a Navigator.

 

[aramis]

Combining what Mike and Aramis says we can get a good clue to the destination. We cannot jump through through big objects, so some destinations are ruled out. If we assume that the observed ship is accelerating for a minimum time intercept with a target planet, some destinations are more likely than others.

In recent editions there are Jump Filters that allows you to detect where an observed ship is jumping.

Edit: Sorry, whartung already said that better.


In CT we can buy jump tapes, instead of having a Navigator.

Note that if you come in with a 6-G•Hour vector, you're going to need 6 hours in a merchant to stop relative to the star, a few minutes to match the world's vector, and are likely to be shot down if you aren't obviously going to be close to orbital on intercept predicted location... Ideally, you want to be zero-target-relative before jumping, because Gravitic Maneuver is off in J-Space.

There's also the observable "direction of jump" - it's explicit in TNE's Regency Sourcebook that the observable entry to jump gives at least a general direction, and often can be reverse engineered to determine where they went.

Couple observation of entry (J-Space Vector) with observation of N-Space vector, and with potential obscurations, and you can probably pin down where they're bound to about ±2 AU...

 

[whartung]

There's also the observable "direction of jump" - it's explicit in TNE's Regency Sourcebook that the observable entry to jump gives at least a general direction, and often can be reverse engineered to determine where they went.

Couple observation of entry (J-Space Vector) with observation of N-Space vector, and with potential obscurations, and you can probably pin down where they're bound to about ±2 AU...

What does T5 say about "observable direction of jump", because JTAS was pretty explicit about this.

 

[AnotherDilbert]

Note that if you come in with a 6-G•Hour vector, you're going to need 6 hours in a merchant to stop relative to the star, a few minutes to match the world's vector, and are likely to be shot down if you aren't obviously going to be close to orbital on intercept predicted location... Ideally, you want to be zero-target-relative before jumping, because Gravitic Maneuver is off in J-Space.

I was assuming we could accelerate out from the origin, jump so that our velocity vector pointed towards the target, and decelerate to intercept. Hence our velocity when we jump would give a big clue as to where we are jumping.

There's also the observable "direction of jump" - it's explicit in TNE's Regency Sourcebook that the observable entry to jump gives at least a general direction, and often can be reverse engineered to determine where they went.

Where?

Couple observation of entry (J-Space Vector) with observation of N-Space vector, and with potential obscurations, and you can probably pin down where they're bound to about ±2 AU...

Normally we would only have to compare the jumping ships vector and possible jump effects with a few nearby standard targets to get the most likely fits.

 

[Whipsnade]

McPerth,

As Mike suggested, you're confusing real space velocity vector with jump line. You're not the first and you won't be the last. That's because the terminology is confusing.

There is a real space vector you carry through jump. It doesn't effect your jump line at all.

There is a jump line which is a line of sight between your jump's start point and destination. It doesn't effect your real space vector at all.

Read whartung's post. That is perhaps the best and most succinct explanation of what we've been talking about in the X-boats thread. It's something I wish I was smart enough to write.

Read whartung's post. It will allow you to understand the situation.

 

[aramis]

There's also the observable "direction of jump" - it's explicit in TNE's Regency Sourcebook that the observable entry to jump gives at least a general direction, and often can be reverse engineered to determine where they went.

Where?

Page 79 "Penetration"

 

[kilemall]

McPerth,

As Mike suggested, you're confusing real space velocity vector with jump line. You're not the first and you won't be the last. That's because the terminology is confusing.

There is a real space vector you carry through jump. It doesn't effect your jump line at all.

There is a jump line which is a line of sight between your jump's start point and destination. It doesn't effect your real space vector at all.

Read whartung's post. That is perhaps the best and most succinct explanation of what we've been talking about in the X-boats thread. It's something I wish I was smart enough to write.

Read whartung's post. It will allow you to understand the situation.

Then there are freaks like me that go WAY the hell off the reservation.

 

[AnotherDilbert]

Page 79 "Penetration"

Thanks, I hadn't seen that.

 

[McPerth]

McPerth,

As Mike suggested, you're confusing real space velocity vector with jump line. You're not the first and you won't be the last. That's because the terminology is confusing.

There is a real space vector you carry through jump. It doesn't effect your jump line at all.

If I read you right here, I'm not confusing them.

I know your jump entry vector needs not to be pointing your target, but I always understood that it must be quite precise, both in ector (angle and speed), place, and time.

After all, calculating it the navigator job (and matching it is pilot's)..

In CT we can buy jump tapes, instead of having a Navigator.

And, IIRC they only serve you for a jump, and I guess for a limited span of time, due to the needed jump varying with time.

 

[McPerth]

Let's see if I can explain how I've always understood jump vectors, so that you can tell me what (if anything) I'm doing right and what wrong...

Entry in jump can be achieved from nearly everywhere (the farther from gravity Wells, the safer) if you match the correct vector. See that this vector has nothing to do with the one pinting to your target, but is the correct one to enter the jump simension you want.

A vector is defined by an angle and a speed, due to gravity influence, I've also assumed that this vector is relative to the highest gravity well where you're looking for it (so, if a ship wannts to go from Earth orbit (so Earth is the highest gravity well) to Barnard, and another wants to also jump to Barnard but from Júpiter orbit, their vectors (relative to Sun) would be different, as they has to match the vector relative to Earth and Júpiter respectivelly, who also have different vector relative to sun. This vector (again, relative to the highest gravity well) is the one you will have at your jump exit.

Also, as planets (that use to be the highest gravity wells) vectors, relative to their stars, are constantly varying, the needed vectors (relative to the star) are also constantly varying, though as variation is small (e.g. Earth vector changes about a radian in angle every two months, and does not change in speed; Jupiter's vector will vary even less, as its angular speed is slower), a given vector is valid for (at least) some hours or days...

Add to this the possibility of intervining objects, and you'll have the Navigator's job to be done.

The span of arrival time will depend on the exact vector you take, the closer to ideal one, the shorter the jump will be. This is what allows for coordinated jumps, that allow you to reach the destination in a smaller span of time, as the vectors used are more close to each other...

I hope I clarified my views and assumptions (be them right or wrong) on it, instead of confusing you and giving you bad headaches.

 

[mike wightman]

If I read you right here, I'm not confusing them.

You confuse them in your very next sentence... well, sort of

I know your jump entry vector needs not to be pointing your target, but I always understood that it must be quite precise, both in ector (angle and speed), place, and time.

After all, calculating it the navigator job (and matching it is pilot's)..

Real space velocity vector and jump line are different things.
While you can use the expression jump vector in place of jump line you are confusing the real space vector.
Note that standard operating practice rules as written is to use your m-drive to halfway to the jump distance, then decrease velocity for second half of the journey so you have a velocity vector of zero when you jump.
You can if you choose jump with a vector but that vector will be transferred to the destination.
Note you can plot a jump so you arrive one week later at your departure point, or a few thousand km behind your departure point, or a few thousand km anywhere in a circle (sphere?) around your departure point. You will still be pointing in the same direction and still have the same vector you started the jump with.

Unless you have a reaction drive and use it in jump to alter your real space vector

And, IIRC they only serve you for a jump, and I guess for a limited span of time, due to the needed jump varying with time.

We are not told exactly why they are time limited, but it is a good guess that the jump cassette uses the relative motion of the departure system and the destination system and if you wait too long the calculations are no longer valid.

But note you can use a jump cassette with any real space velocity vector, and since the cassette program is pre-written and can not know what your chosen vector upon departure is the implication is that real space vector doesn't matter to the jump parameter calculation.

 

[whartung]

I think the phrase "jump line" is much more appropriate than "jump vector".

The jump line is literally taking a piece of string from the source to the destination. The navigators job is to ensure that there's nothing along that piece of string that will cause the ship to emerge from jump.

This is primarily an issue at the source and destination system, avoid the local stars, planets and giants while at the same time getting your ship in to the destination system while avoiding its stars, planets, and giants.

While there can certainly be a black body out in deep space, the basic assumption is that they simply are not common enough to matter. If there is some dark, undetected body floating around, then, yea, it will pull the ship out of jump. If you only have enough fuel for a single jump, then you're in for a bad day.

The ideal jump leaves the 100D limit of the source world and arrives at the 100D limit of the destination world. This offers the shortest practical trip. Sometimes, intervening stars and giants simply get in the way. The simplistic case is that your source position is rimward of the systems star, and the destination is coreward, and the 3D relative position of the destination system isn't enough to allow the ship to jump without the local star being within "line of site", or "line of jump". So, the ship needs to fly around to one side of the star or the other until the line of jump is clear.

Jump masking and shadowing complicate the process enough without adding in the third dimension, but it certainly exists for those willing to work out the details. Local stars are far more "in the way" than the destination stars, but if you jump above of below the ecliptic at the destination, you may be able to arrive closer to your destination by target above or below it, vs arriving on the other side of the star.

But again, for the bulk of operations, it's really pedantic, needless detail. G:ISW, or maybe it's G:FT, has a procedure to basically add on to the travel time in an abstract sense, "sometimes" to offer the flavor of the effect without having to go through all of the detail.

But, in the end, there is no "jump vector", there's the "jump line", and the navigators job is to route the ship to a proper jump point so as to best arrive at the destination system. Beyond simply getting to the proper point in space, which can be as simple as 100D from "anything", to complicating it routing around the stars and what not, maneuver has no effect on jump. It simply gets you do a good place to start.

 

[Whipsnade]

I think the phrase "jump line" is much more appropriate than "jump vector".

Please take whartung's suggestion, McPerth. Re-read what he and Mike have posted too.

Forget what you "know" because it's a hopeless jumble. Forget it all and begin again by reading the explanations whartung and Mike have kindly shared with you.

But again, for the bulk of operations, it's really pedantic, needless detail. G:ISW, or maybe it's G:FT, has a procedure to basically add on to the travel time in an abstract sense, "sometimes" to offer the flavor of the effect without having to go through all of the detail.

Very well put.

 

[Enoki]

Although we don't know how jumping would really work, the way I've tried to explain it is that the ship opens a "well" ahead of it whose base (the point at which you emerge) is a gravitic singularity. That is, the ship opens a sort of gravity well in Nth dimensional space, "falls" into it, and emerges at the singularity point.

The J6 limit is because you can only "stretch" the singularity so far and that's the limit.

The reason you need to be away from other gravity fields is these distort the singularity well resulting in it not forming the singularity point where you thought it would be.

The navigator's job is to pick the point correctly. He has to account for the movement of the system being jumped to, the position of the world at the time of the jump, and avoid the well being near any gravity sources.

The engineer's job is to properly charge the system for the intended jump so the well's depth is the proper amount. The higher the jump number / distance to be jumped, the stronger the jump field has to be to create a deeper well.

The pilot's job is to properly align the ship so it's jump field will generate the well in the right direction, as well as to compensate for any movement / momentum the ship might have that will effect the point where the well forms. I make it easier and safer if the ship stops to jump rather than does it with momentum, although you can.