Refining the 100D Jump limit

[whartung]

It's always faster to actively maneuver. If you can wait for something to leave you behind, you can run and stop to that spot faster than they can just leave you.

 

[whulorigan]

That doesn't change that it might be better, and quicker, to stop and let things move away from you than you try to move away from them...

Stop relative to what? The central star? The galactic center? Your initial "zero" reference frame is that of the originating planet, where you and it both have zero relative velocity to one another.

In either case you need to accelerate in order to "stop" or let the originating world move away (relative to some other arbitrary reference frame). Might as well continue to accelerate until you reach your optimal jump distance.

Is there a trajectory or orbital-geometry (or other consideration) that I am missing?

 

[Grav_Moped]

Of course, you could get a "boost" to orbit by taking off in the same direction as the planet's rotation about its own axis (the closer to the equator the better . . . ) .

Given Traveller maneuver drive capabilities, it'd be relatively insignificant.

It might well make a difference if all you've got is an Air/Raft though...

 

[RogerD]

Given Traveller maneuver drive capabilities, it'd be relatively insignificant.

Mercury has an orbital speed close to 50km/s. Depending on the timing (i.e. leaving when the vector is going exactly the wrong direciton), that could get doubled to 100km/s. At 1g=10m/s^2, you can compensate that in 10000s or a little under 3 hours. I'd have to agree that's relatively insignificant!

 

[Grav_Moped]

Mercury has an orbital speed close to 50km/s. Depending on the timing (i.e. leaving when the vector is going exactly the wrong direciton), that could get doubled to 100km/s. At 1g=10m/s^2, you can compensate that in 10000s or a little under 3 hours. I'd have to agree that's relatively insignificant!

My remarks were in the context of:

Of course, you could get a "boost" to orbit by taking off in the same direction as the planet's rotation about its own axis (the closer to the equator the better . . . ) .

The tangential velocity of Earth's rotation at the equator is approximately 460m/sec (1000 MPH). An Air/Raft has 0.1g discretionary acceleration* and would thus take 46 seconds to cancel this vector after a vertical ascent, once it's above the atmosphere and no longer constrained by aerodynamic drag.

While it's a bit more complicated than that, this shows the scale of the velocities involved.


*Reverse-engineered by cross-referencing the Air/Raft's performance from LBB3'81 with Striker's vehicle design rules.