General Is there a way to figure out the jump shadow of a sun?

[AnotherDilbert]

Barring the "assuming you don't die" part, this is true.

After CT, misjump rules are much more lenient, and generally involve skills, so, as Condottiere pointed out, with great skills you can fairly safely jump from at least 10 D.

 

[Condottiere]

1. Depending on how complicated you want to make your campaign.

2. Or, there's a supplement that comprehensively maps out, say, a subsector.

3. Plot point potential.

4. At this point, the early jump advantage realizes it's potential.

5. Ten to thirty percent distance shortening.

6. True, you hit the wall and bounce off a jump shadow, but it's still at seventy diameters.

7. Balanced off by transitioning after seven diameters, if you have ticked all the boxes.

8. Unless it's a well worn jump lane, the further out the hundred diameters are, the less likely you'll hit traffic.

 

[Nathan Brazil]

A question. Computationally, 100 x diameters is easy to compute for the not too math inclined (like me without my A-game these days). What if something else were used? Like 100 x Diameters x Terrestrial Density (Terra = 1.0)

Would not some other standard distance based something involving mass or gravity of the stellar/planetary body? IRL mass and gravity seem to be a better fit, because object's mass and density are better descriptors of the deformation of space and jump drive seems to want "wrinkle free" space.

Gravity was used to describe describing limitations of stutterwarp. Narratively, Long before that, Known Space Hyperdrive I and Hyperdrive II operated avoiding the "mass shadow" of stars.

Of course:
Using a distance relationship involving density will make gas giants shardows much smaller as they are less dense than a terrestrial body like Terra
Using a distance relationship involving mass will make Jump shadows other than the star itself of lesser consequence.

 

[AnotherDilbert]

Would not some other standard distance based something involving mass or gravity of the stellar/planetary body? IRL mass and gravity seem to be a better fit, because object's mass and density are better descriptors of the deformation of space and jump drive seems to want "wrinkle free" space.

Yes, it's explicitly a gravity effect that is approximated by 100 D range. Exactly how it works is "only dimly understood", or something like that.

JTAS#24, p30:

Entering jump is possible anywhere, but the perturbing effects of gravity make it impractical to begin a jump within a gravity field of more than certain specific limits based on size, density, and distance. The general rule of thumb is a distance of at least 100 diameters out from a world or star (including a safety margin), and ships generally move away from worlds and stars before beginning a jump. The perturbing effects of gravity preclude a ship from exiting jump space within the same distance. When ships are directed to exit jump space within a gravity field, they are precipitated out of jump space at the edge of the field instead.

Of course:
Using a distance relationship involving density will make gas giants shardows much smaller as they are less dense than a terrestrial body like Terra
Using a distance relationship involving mass will make Jump shadows other than the star itself of lesser consequence.

Actually mass is higher since volume ∝ radius³.

Example: Jupiter has ten times the radius of Earth, but 300 times the mass.
https://en.wikipedia.org/wiki/Jupiter

The Sun has ~100 times the Earths radius, but ~300 000 times the mass.

On the other hand the gravity field decreases by the inverse square of the distance, g ∝ ¹/_d²

 

[Putraack]

Wow, that was diverting.

I guess the better question would have been, "how can I easily determine the radius of a star, given the information in some system profiles?"

 

[whartung]

Of course, to play the game we have to ignore all this, it's much too complicated to care about at the table. It can be used as a plot device, perhaps?

I'm sure that's why FT used the mechanic that they did, to make the effect "real" without having to break out the geometry.

 

[Grav_Moped]

Barring the "assuming you don't die" part, this is true.

As a GM and player, I prefer the mis-jumps at less 100 and even 10 diameters rather than "Your ship is destroyed." For NPC's and such, mis-jumps are effectively death sentences "off-screen". Not many ships have fuel capacity to "double jump" even when they know where they are at after mis-jump.

As a GM I see players doing this as a message to me the GM "You made this to hard", "We were stoopid greedy. Sorry." Whether mis-jump or destroyed, I have to start a campaign elsewhere or roll new characters and start a new campaign.

As a player, if I'm involved in deciding, I'm not going to die a "no chance of surviving" death by Navy or local pirates, so I would risk the mis-jump and pray on GM mercy. TPK is devastating to GMs and players alike,

This is version-dependent!

LBB2 '81 misjumps from within the 10-100D zone can be survivable. Just be sure to have J1 fuel left over afterwards!
Over half the time (58%), the jump is normal.
Most of the rest of the time (33%), you come out of Jump intact at a random location less than 36 parsecs* from the starting point (median is 11 parsecs).
Odds of the exit point being within J1 range of a world are almost 90% (I counted -- tediously -- once.)

There's only an 11% chance of an immediate "ship destroyed" result.

So if you start with a fuel surplus of J1 fuel range -- which for small ships in LBB2 isn't a huge marginal increase over the already required powerplant load (the Type S already has it) -- you can self-rescue.

------
* A 36-parsec misjump only results from rolling a 6, 6 times in a row...

 

[Spinward Flow]

I guess the better question would have been, "how can I easily determine the radius of a star, given the information in some system profiles?"

LBB6 p45 has a chart of stellar radii in which a value of 1.0 equals the radius of Sol.
So if you know the radius of Sol (in km), multiply by 2 (to get the diameter), multiply by 100 (to get the jump shadow) and multiply by the radii multiplier from the stellar radii chart on LBB6 p45.

 

[whartung]

A question. Computationally, 100 x diameters is easy to compute for the not too math inclined (like me without my A-game these days). What if something else were used? Like 100 x Diameters x Terrestrial Density (Terra = 1.0)

That's that whole "the 100D limit is a Tidal/Gravity force" thing.

Using the universal gravity equation, F = G (m1 m2) / r^2, for Earth we get an F of about .00027 at 100D. To get that value from the sun, it's about 500 diameters. Even Jupiter is inside of that envelope.

 

[whulorigan]

That's that whole "the 100D limit is a Tidal/Gravity force" thing.

Using the universal gravity equation, F = G (m1 m2) / r^2, for Earth we get an F of about .00027 at 100D. To get that value from the sun, it's about 500 diameters. Even Jupiter is inside of that envelope.

The numbers work better for the tidal-force model (Ft = GMm/r^3) -OR- (gt = GM/r^3). Under the tidal-force model stars and planets have minimum jump distance values that are much closer to one another.

Tidal force - Wikipedia

en.wikipedia.org

It also gives a nice rationale for mis-jumps from within 100D: If the gravitational force differs along/across the length/diameter of the ship above a certain tolerance, different parts of the ship have a variance in their "jump-vector" across the surface of the ship, leading to uncertain jump results. If the tidal force is large enough (e.g. for <10D), the ship can get torn apart as different parts of the ship attempt to enter jump space along different vectors.

 

[whulorigan]

The numbers work better for the tidal-force model (Ft = GMm/r^3) -OR- (gt = GM/r^3). Under the tidal-force model stars and planets have minimum jump distance values that are much closer to one another.

BTW, I made a typo above, and have exceeded the edit-timeframe.

The Tidal Acceleration should be: gt ≈ 2∆r GM/r^3

​

where "R" is the distance from the gravitating body and "∆r" is the dimension of the ship measured along the radial axis pointing toward the star/planet.​

Tidal Force would be: Ft ≈ 2∆r GMm/r^3.

 

[Putraack]

LBB6 p45 has a chart of stellar radii in which a value of 1.0 equals the radius of Sol.
So if you know the radius of Sol (in km), multiply by 2 (to get the diameter), multiply by 100 (to get the jump shadow) and multiply by the radii multiplier from the stellar radii chart on LBB6 p45.

I think that's closer to what I was looking for, thank you.