-------- Original Message --------
 On January 25, 2018 12:24 AM, Tim <xxxxxx@little-possums.net> wrote:

>
>
>>The most difficult part though, is that the difference of initial
>> velocity of the Hamiltonian with the expected final velocity (that
>> of Hot chi in 780366 seconds) is 22730.9 meters/second, requiring an
>> additional 38m37s worth of thrust, and somewhat screwing up all
>> previous calculations. Obviously this can be ignored for all but the
>> most technical of games.
>>
> You can still do this in a "turnover" trajectory with the acceleration
> in the second half not being precisely opposite the first.  The nice
> part is that the obvious midpoint correction is also the best linear
> approximation to the true requirement for this type of trajectory, so
> the error at the end would be very minor.

Traveller 5 changed the way maneuver drives worked that complicate it somewhat, at least for this case. The drive must be within 1000 diameters of a "gravitational source" to be effective, otherwise the drive drops to 1/100th rated acceleration. The bubbles around Home where a maneuver drive is effective looks like this:

https://i.imgur.com/baNW7hi.png

So you don't get constant acceleration from start to destination in transitions from inner to outer planets.

> You can still do this in a "turnover" trajectory with the acceleration
> in the second half not being precisely opposite the first.

You can, it's a little more difficult when you're asking for programmed thrust (which we do in our system ; e.g., 10251 seconds in the direction of {-0.995037,0.0995016,0}, then 10251 seconds in the direction of {0.995037,-0.0995016,0}, the task is to find a vector which will lead to a particular position at a particular time with a particular velocity.

>