What I am trying to find out is what factor is used to calculate how fast a ship will move across the surface of a planet?
I know using the Landing Time, and Take Off Time from the ship's turn report I can work out how many TUs it would take to go into orbit from xy position 1, and to then land again at xy position 2, but what I am interested in is performing the manuever without takingoff and landing.
If I am at xy position 1, and I issue the order for the ship to move to xy position 2, then the ship moves sector by sector, similar to a ground party. For a ground party however there is a clear 'Ground' movement speed on the turn report, but this doesn't appear to be the case for a ship.
How do I work it out?
Many thanks,
Phil
Phil
For the most part Mica/David seem to use 'real-life' forumlas, so it's probably mass, thrust and gravity that matter in this case.
In both cases the Gravity rating of the planet involved is 1g.
Example 1:
| QUOTE |
| Starting Location: Landed on xxxxxx at {7,12} in Mountains >TU 112: Move to Starbase {xxxxx} {Dock - No} Move to 6,13 - Ice Move to 5,13 - Ice Move to 4,13 - Ice Move to 3,13 - Ice Move to 2,13 - Ice Move to 1,13 - Craters Move to 35,13 - Ice Total TU cost for this action is 45 | Manoeuvre Speed: 1.6 g Orbit Time: 6 tus | | Landing Time: 31 tus Takeoff Time: 31 tus | | Surface Area: 42 Embarking Size: 3024 mus | |
Example 2:
| QUOTE |
| Starting Location: Landed on xxxxxx at {35,13} in Ice >TU 227: Move to XY {7} {12} Move to 1,12 - Ice Move to 2,12 - Mountains Move to 3,12 - Ice Move to 4,12 - Dust Move to 5,12 - Ice Move to 6,12 - Ice Move to 7,12 - Mountains Total TU cost for this action is 15 | Manoeuvre Speed: 5.2 g Orbit Time: 2 tus | | Landing Time: 10 tus Takeoff Time: 10 tus | | Surface Area: 42 Embarking Size: 3024 mus | |
"Orbital time" is used to move up and down the columns, while 1.5"Orbital time to change column:
Start 7, 12. Move to 6.13 1.5*orbital time.
Start 6.13, move to 1.13 then 35.13 ( Which is assumed to be next sector down) 6*orbital speed.
Total 7.5 *Orbital time
Orbital time 6
7.5 * 6 = 45.
Appears to work for your first example.
Since it is a 1G planet no idea if gravity effects it.
| QUOTE |
| >TU 182: Land at XY {2} {5} Landing at 2,5 - Craters. >TU 121: Move to XY {3} {5} Move to 3,5 - Crust Total TU cost for this action is 2 >TU 69: Move to XY {2} {2} Move to 2,4 - Broken Move to 2,3 - Broken Move to 2,2 - Mountains Total TU cost for this action is 7 | Manoeuvre Speed: 4.6 g Orbit Time: 2 tus | | Landing Time: 11 tus Takeoff Time: 11 tus | | Surface Area: 77 Embarking Size: 7650 mus | |
On a planet with a 0.1g rating.
| QUOTE (Auld Nick on Jan 3 2006 @ 07:37 PM) |
| "Orbital time" is used to move up and down the columns, while 1.5"Orbital time to change column: |
So modifing Nick's forumla slightly to be 1.5 x "Orbital Time" on diagonal movement and 1 x "Orbital Time" on column or row movement seems to work nicely, regardless of the planet's gravity.
Many thanks all. So the gravity rating of the planet affects whether you can land / takeoff, and the orbit time specifies the time across the planet's surface.
Thanks again,
Phil
(IC: Sees With Knowledge - Mohache)
For the following example
| Manoeuvre Speed: 0.5 g Orbit Time: 20 tus |
| Landing Time: 98 tus Takeoff Time: 98 tus |
And orders as follows:
Starting Location:
Docked at zzzz (xxxx) - hmmmmm(yyy)
>TU 246: Move to Starbase {xxxx} {Dock - No}
Move to 14,11 - Forest
Move to 14,10 - Forest
Move to 14,9 - Forest
Move to 14,8 - Forest
Move to 14,7 - Forest
Move to 14,6 - Forest
Move to 14,5 - Tundra
Move to 14,4 - Tundra
Move to 14,3 - Ice
Total TU cost for this action is 132
Here we have 9 rows that have traversed.
Now 9 * Orbital Time = 9 * 20 = 180 Tus.
The planet has a gravity rating of 0.8g
180 * 0.8 = 144 which exceeds the actual TU usage.
Note also the ship efficiency level at the time was 144% at the end of the turn.
| QUOTE (Archangel @ Jan 5 2006, 11:54 AM) |
| | Manoeuvre Speed: 0.5 g Orbit Time: 20 tus Starting Location: Docked at zzzz (xxxx) - hmmmmm(yyy) >TU 246: Move to Starbase {xxxx} {Dock - No} Move to 14,11 - Forest ... Move to 14,3 - Ice Total TU cost for this action is 132 Note also the ship efficiency level at the time was 144% at the end of the turn. |
Assuming the first movement is from a starbase at 14,12 to the forest in 14,11 then yes the forumla is broken and we are missing something.
However if the starbase is at 13,12 or 15,12 then the first movement is diagonal.
This takes it to 1square * 1.5 * 20 + 8squares * 20 = 190tus
190tus @ 100% efficiency = x @ 144%
x = 190/144 * 100 = 131.944
Round up to 132tus for 144%
Can you just confirm the x,y location of the starting (starbase) square?
| QUOTE (ptb @ Jan 5 2006, 01:49 PM) |
| [Assuming the first movement is from a starbase at 14,12 to the forest in 14,11 then yes the forumla is broken and we are missing something. However if the starbase is at 13,12 or 15,12 then the first movement is diagonal. This takes it to 1square * 1.5 * 20 + 8squares * 20 = 190tus 190tus @ 100% efficiency = x @ 144% x = 190/144 * 100 = 131.944 Round up to 132tus for 144% Can you just confirm the x,y location of the starting (starbase) square? |
Very well done indeed PTB, that solution had not occurred to me.
The starbase was in fact at 15,12 as you guessed.
So it appears the original algorithm suggested still works and more importantly is independant of the planetary gravity as well.
Regards
Archangel