Szybkość przelotu statków

Low planetary orbit to moon (Terra-Luna, ~400,000 Km)
2 1/2 hours @ 2G or 1 1/2 hours @ 5G

Transfer between close planetary neighbours (Terra-Mars, 45,000,000 Km)
26 hours @ 2G or 17 hours @ 5G

Transfer from planet to low solar orbit, or 1 AU above planet (Terra-Sol, 150,000,000 Km)*
2 days @ 2G or 30 hours @ 5G

Transfer from planet to gas giant orbit (Terra-Jupiter, 600,000,000 Km)
4 days @ 2G or 2.5 days @ 5G

Outer Reaches of System (Terra-Neptune, 1,000,000,000 Km)**
5 days @ 2G or 3 days @ 5G

  • Dangerous-but-workable warp jumps at this point. Certainly warships with skilled navigators would risk this.
    • ‘Normal’ warp jumps at this point, especially chartist merchants with no navigator.
      If you were to accelerate constantly until halfway, and decelerate the rest of the way the following formula would give you travel time:

Time=2*Square root of Distance

For a trip of one AU at 1g that works out to 68 hours. That should be equivalent to traveling from Primary Biosphere to Inner Cauldron. Travel further out in the system has longer distances, and orbital position can easily double that again. I assume that most RTs aren’t going to be doing max accleration the whole time and instead allow orbital mechanics and momentum do a lot of the work for them (as we do now). Long burns like that are both hard on the drives and very expensive in terms of fuel. In an emergency you can do a pretty much straight line, full burn, trip to the Earth from the Kuiper Belt on the outer edge of the system in 230 hours (in your Sword class Frigate, accelerating at 4.5g). Note that you will be doing about 10% of light speed at midpoint transition. Even a single such maneuver would probably exhaust your ships fuel reserves, if they even lasted that long. The exact distance for transition from the Warp is a little fuzzy, so I expect it is less than that.

To travel 50 AU then, from standstill and to a complete stop, you would need to accelerate at 1 G for 10 days. This would take you 25 AU, and leave you with a top speed of some 3050 VU / hr. Then you simply hit the brakes (or more likely, turn the ship around and let the engines thrust as normal giving you the same deceleration as you had as acceleration). For the next 10 days you will cover the same distance, and end up stationary on the edge of our solar system after 20 days of travel.

If you did not need to slow down, you could be there after only 14 days at “full burn”, but then you’d be moving at almost 4300 VU/h

And this is at 1 G. The level of acceleration you experience if you jump of something here on earth. With fancy-gravplating and inertial dampeners and other sci-fi tropes you could easily increase this, drastically reducing travel times. At 2G you’d just need 7 days before reaching your mid-point. At 3G, you’re down to 6.

FORMUŁA OBLICZANIA CZASU LOTU, zakładając, że chcemy gdzieś zaparkować.
Time taken (in days) = Square root ((Distance * 8,2) / G)
t = √(s*8,2)/G t = dni s = AU g = 10 m/s2


Similarly, if you EXIT the warp space you could also assume that you have the same speed as when you entered it meaning that you would only need to decelarate to get to the correct point. Of course if you accelerated from a 50AU solar made a jump to the edge of a 30AU solar system, then you’ll be missing target by 20AU!!. Stuff can get a bit complex this way. Anyway, the corresponding formula assuming NO halfway point (just take out the 2* and 1/2 from the above steps):

[days] = SQRT


If I were to houserule:
1. Every system is 40 AU (our solar system which is fairly average)
2. THE planet (of your interest) is always at a distance of 1 AU from its star*
3. Your relative speed (RS) on entry to the closest gravitational system (CGS) is the same as your RS to the CGS you left.
4. That means to LEAVE a system you only need to ACCELERATE and to ENTER a system you only need to DECELERATE
That means that there is no half way point and that means that the used formula is [days] = SQRT with 40AU this becomes (a bit rounded) [days] = SQRT, for e.g. a merchant class raider travelling at 5g this becomes 5.7 days to exit or enter a system.

And to make it really easy here a table:

G (DAYS, both for exit and entry)
g (Days)
1 (13)
1,5 (10,5)
2 (9)
2,5 (8)
3 (7,5)
4 (6,5)
5 (5,5)
6 (5)


It’s unlikely that your players ever will take the chance, but ORKS just might do this. So here a suggestion of the percentile mishap (= BOOM)


- Warp exit/entry at edge: change of mishap 0% (concerning the gravity well)
- Warp exit/entry close to the star: chance of mishap 50% (again: concerning the gravity well). Initially I was thinking 100% but as habital planets are typically 1 AU away in a typically 40 AU solar system would mean that jumping close to the habital planet is about 100% chance on mishap. That with the fact that mishap = warpdrive explodes, in other words: end of story for the ship!! AND the reference earlier mentioned of the ork fleet jumping in on a star fleet in orbit, would have meant the instant decimation of the entire fleet. Hence: 50% chance on blow-up.

The formula to calculate this is very simple:

Chance on mishap = (7*DFE/SR)^2 %

here is:

SR = Solar system Radius (e.g. 30 AU, solar system of Terra)

DFE = distance from edge of the solar system where you jump in. E.g. with an with an SR=30, 0 is the edge and 30 is inside the sun. Entry at halfway = 15 AU:

Mishap = (7*15/30)^2 = 12%.

If you like to keep things simple and assume EVERY solar system of size 50AU then the formula becomes really simple:

Chance on Mishap = DFE^2 / 50 %

To resolve simply throw a 1d100 if you roll UNDER the set score: BOOM

Some chances (assuming SR = 50)

Jumping in at:
Distance from edge – %

0 (jump at the edge) – 0%
6 – 1%
10 – 2%
20 – 8%
30 – 18%
40 – 32%
49 (jumping close to a habital planet which are typically at 1 to 3 AU away from its sun) – 48%
50 – 50%


L4 and 5, which are stable and quite a bit larger then 1,2 and 3. If you would jump there then you would be at 1AU distance from the planet. L4 and 5 are actually pretty big, about 1AU length and half that in width so its not THAT hard. What makes it hard in general to find the right spot at all is the fact of the time distortion in the warp. In order to jump right next to a planet or in a lagrange point you need to know the time (as that decides the position of the planet and LP in regard of its star).

So basically you want to jump at a distance of 1AU from the star without much error, lets say max 10% so within 0.9 to 1.1 AU AND you need to know the exit time. Both these rolls are part of the 5 navigation steps.

First you need to know the time, now the LP are pretty big so there’s always a 30% chance that you jump into one. And I would rule that if you jump in from farther then from the edge of the solar system as beyond that its really completely random at what exact time you jump in. Should you jump first to the edge and then from the edge to the LP then I would rule a very hard Navigation Warp test (-30) for the correct estimate, however since there is a 30% chance that you get it right anyway I would simply rule a Navigation Warp Test (+0). Every degree of fail increases the difficulty of the ‘leaving the warp’ with one step, more then 3 DoF means your NOT jumping in the LP.

‘Charting course’ is negligible (again assuming jumping from edge) so ordinary warp
‘steering vessel’ will again influence the time and space exit which are vital so again very hard test (-30) navigation warp and again 3 DOF means NOT jumping in the LP and every DoF increases the difficulty with one step on leaving the warp.

When that’s done you need to roll ‘leaving the warp test’ which is usually at -20, this time modified with the DoF of the other rolls.

Should you actually succeed this test then this would half the chance of BOOM to 24% lowering it further with 1% for every degree of success. Fail would mean you arrive at roughly 1AU from the sun: 48%. Should you fail this warp with 6+ DoF then you jump straight into the sun.

Szybkość przelotu statków

Legacy of the Void LukaszWojcik