I recall there was some good discussion of this in a thread which I can't find at all.
Someone had pointed out that because
A) stars move in relation to each other and
B) planet certainly move as well,
therefore C), a ship with no vector relative to planet Y jumping into the vicinity of planet Z will arrive with a vector equal to the relative speed of those two planets in relation to one another.
On a day to day basis, this is not really something that PCs and Referees really need to be worrying their little heads about. But from time to time, it might be relevant. SO:
QUICK & DIRTY STELLAR RELATIVE SPEEDS:
(Gentlemen, please check my numbers and assumptions)
Thought about this a bit, and did some googling. Near as I can tell, stars move anywhere from a few KPS to 500+kps; planets in our system move from a few kps to somewhere near 50.
Now, taking this to Book 2 combat speeds, where each turn is 1000 seconds and a vector produced over a turn's worth of 1G acceleration is 10,000kilometers long (a ship traveling on such a vector is then moving 10kps.)
The really fast stars seem to be in the very core of the galaxy; slow ones are in globular clusters. So on average, they ought to be between, on the low side. Right?
for system Y, (6D6)*10 speed in kps. (60-360kps.)
Roll D6; make Y's speed a negative number if 1-3
for system Z, (6D6)*10 speed in kps. (60-360kps.)
Roll D6; make Z's speed a negative number if 1-3
Add the speeds of systems Y and Z; the resulting number is the relative speed of the two systems. It will vary from This should be recorded, as it will not change over the course of the game.
For a given jump, do the same process with 1D6, and add that result to the stellar relative speed to take planetary movement into account.
So say that system Y has a speed of 300 (moving away from Z at 300kps)
and system Z has a speed of -250 (moving towards from Y at 250kps)
they'll have a relative speed of 50kps.
Figure planetary speeds for y and z at 30 and 50 kps; add that for a total of 130kps.
A free trader jumping from one planet to the other from a relative standstill will arrive moving at 130kps, or in book 2 game terms, will have a vector 130,000km per turn. It will take that free trader about 13 turns to decelerate and match speeds with the planet in order to achieve orbit.
On the other hand, the free trader might have decided instead to match speeds with her destination at launch, and spent 13 turns accelerating prior to jump. A scout could do this in half the time, either way.
This would have interesting tactical repercussions, if a fleet's navigator chose to come out of jump moving towards the target at half a light-second per turn. No?
It also means that ships coming in and out of system can be expected to be moving at absolutely hellacious speed about half the time.
Someone had pointed out that because
A) stars move in relation to each other and
B) planet certainly move as well,
therefore C), a ship with no vector relative to planet Y jumping into the vicinity of planet Z will arrive with a vector equal to the relative speed of those two planets in relation to one another.
On a day to day basis, this is not really something that PCs and Referees really need to be worrying their little heads about. But from time to time, it might be relevant. SO:
QUICK & DIRTY STELLAR RELATIVE SPEEDS:
(Gentlemen, please check my numbers and assumptions)
Thought about this a bit, and did some googling. Near as I can tell, stars move anywhere from a few KPS to 500+kps; planets in our system move from a few kps to somewhere near 50.
Now, taking this to Book 2 combat speeds, where each turn is 1000 seconds and a vector produced over a turn's worth of 1G acceleration is 10,000kilometers long (a ship traveling on such a vector is then moving 10kps.)
The really fast stars seem to be in the very core of the galaxy; slow ones are in globular clusters. So on average, they ought to be between, on the low side. Right?
for system Y, (6D6)*10 speed in kps. (60-360kps.)
Roll D6; make Y's speed a negative number if 1-3
for system Z, (6D6)*10 speed in kps. (60-360kps.)
Roll D6; make Z's speed a negative number if 1-3
Add the speeds of systems Y and Z; the resulting number is the relative speed of the two systems. It will vary from This should be recorded, as it will not change over the course of the game.
For a given jump, do the same process with 1D6, and add that result to the stellar relative speed to take planetary movement into account.
So say that system Y has a speed of 300 (moving away from Z at 300kps)
and system Z has a speed of -250 (moving towards from Y at 250kps)
they'll have a relative speed of 50kps.
Figure planetary speeds for y and z at 30 and 50 kps; add that for a total of 130kps.
A free trader jumping from one planet to the other from a relative standstill will arrive moving at 130kps, or in book 2 game terms, will have a vector 130,000km per turn. It will take that free trader about 13 turns to decelerate and match speeds with the planet in order to achieve orbit.
On the other hand, the free trader might have decided instead to match speeds with her destination at launch, and spent 13 turns accelerating prior to jump. A scout could do this in half the time, either way.
This would have interesting tactical repercussions, if a fleet's navigator chose to come out of jump moving towards the target at half a light-second per turn. No?
It also means that ships coming in and out of system can be expected to be moving at absolutely hellacious speed about half the time.