Triplanetary test game


The standard introductory game in the 2018 rules is a challenge to get from Mars to Venus (or vice versa) in the smallest number of turns. There’s no particular reason why this needs multiple players.

So, @denisbloodnok and anyone else who wants to try it: if you pick some colours and a planet to start at, you can try to beat the records (which so far is ten turns for me and 11 turns for @Lordof1) or just generally try to get the hang of the manoeuvre system.


I will be red, because red ones go faster, and hence logically start from Mars, taking off in direction C.


OK, that leaves you here at zero speed:

(There are some people who can see the National Express logo and not think “invasion from Mars”. I am not one of those people.)


Thrust C, saving overthrust for when I inevitably get myself into a comedic situation later.



I should have started from Venus, making it easier to surreptitiously spell a funny word rather than attempt the mission. Given the lack of vowels in this direction, I’ll have a D, please.


DEADBEEF simplifies to EEE…


I’ll have a C then a D (not an overthrust, just playing 2 moves in sequence).


After turn 4:

After turn 5:


“To land, a ship must be in orbit or stationary adjacent to a planet”, but it is not clear to me how I can be stationary adjacent to a planet unless I just took off from it, because I was moving when I entered the adjacent hex and the planet’s gravity can’t cancel that movement unless I am taking off. So I’d better think about how one gets into orbit.

I could enter Venus from the A-ward hex… gah. Find a Star Fleet Battles map (ETA: this is my own thought process, not an instruction to the reader). Of course, the hex grid is differently oriented (columns not rows). Put Venus in 4114. Naively I suppose one gets into orbit by arriving travelling parallel to the planet’s surface and counterthrusting. I could arrive in 4014 with velocity C, altered to CB by gravity, thrust E next turn, in orbit… but I could also arrive with velocity CD altered to CC by gravity, thrust F.

I might even be able to finish on an overload (I think “It is not possible to land as part of an overload manoeuvre” applies to the orbit->landed transition not to entering orbit; I’d be grateful if you’d confirm that) eg come in from 3812 to 4014 with CCD, CCC after gravity, thrust FF. Conveniently I’m pointed at 3812 already.

However, this journey isn’t a convenient size with the granularity of acceleration; that’s ten hexes away and if I accelerate to speed 5 to hit it in 2 turns I’m then going too fast when I arrive. I might as well save fuel and stress on the ship.

No thrust for 2 turns, then A, then F. Then a sitrep, if it’s not “we just crashed into Venus”, please.

(I think the optimal journey is overthrust CD, C, D (getting almost to where I am a turn early), C, F (making up the “almost” by thrusting in my coasting turns), then A, F to land as above. But I just eyeballed that so might be off a hex somewhere).


I have hex numbering available, though it clogs up the map somewhat…

Overloading to get into orbit is fine - the restriction is that, when you spend a thrust to land, that can’t be as part of an overload.

Turn 6

Turn 7

Turn 8

Turn 9


Thrust F into orbit, and I’m done?


Yes - and that’s turn 10, so you land on turn 11.

The only effective orbital entry patterns I’ve found so far are that cross-hex one coming in at speed 2, and moving along the row at speed 1.


I think there may have been an error here in that in the previous thread you took off on turn 0 with turn 1 being the first turn of thrust, but here my takeoff turn was turn 1. Hence I land on turn 10 in the previous thread’s terminology (and if I’m right, the optimal journey is 9 turns).


It is entirely likely that you are correct.