Fleet pursuit mechanic broken?

[Vince]

I am not quite sure if it is a bug or by design, but from my experience the Aliens can magically pull some acceleration out of their behinds (pun intended). Engaging fleets should be straightforward and obey Newtonian physics. If ship A has higher acceleration value then ship B it should be impossible for the ship B to escape even if it has a higher ∆v. Especially if both are already close to each other.

My fleet consisting of ships with a cruise acceleration of 50 miliG and combat acceleration of 1 G can never catch an Alien fleet with 5 miliG cruise acceleration and 300 miliG combat acceleration.

So on average a 5 times slower fleet can evade whenever it wishes to do so. That doesn't seem right. Am I missing something here?

[CA53-Scourge]

引用自 Arcimbaldo：

引用自 Rashka：

Screenshot? Whenever I've seen the bidding screen it always let me bid up to the remaining delta-v of my fleet.

There's not much to show in a screenshot. The fleet has about 1000 kps, but each time it says only 5 kps is available for the pursuit maneuveur.

Although I eventually managed to destroy the ship anyway by using a trick i saw mentioned in another thread, which is just send two separate fleets to intercept simultaneously - it seems like it can only evade one at a time (don't know if this is an exploit or working as designed?)

I'm having this issue as well, although the amount varies dependant on from what I can tell is the total DV that should be available, but it is always WAY off, by ~10^2

[steeldog]

broken... unbroken... what is the dfifference? the main question is how to calculate it?
for example i want to catch fleet with 2.1g and 500dV
it's obciously that 4g with 100500dV will be enought. but would be egrate to have some predictable behaviour instead of just checking wasting 20 mins for test because of impossibility to fast time scrolling and enormouse time of loading

[corisai]

引用自 steeldog：

broken... unbroken... what is the dfifference? the main question is how to calculate it?
for example i want to catch fleet with 2.1g and 500dV
it's obciously that 4g with 100500dV will be enought. but would be egrate to have some predictable behaviour instead of just checking wasting 20 mins for test because of impossibility to fast time scrolling and enormouse time of loading

Go for 4G combat acceleration in that case. ~20 dV will be enough to catch most of (if not all) alien ships.

[steeldog]

引用自 corisai：

Go for 4G combat acceleration in that case. ~20 dV will be enough to catch most of (if not all) alien ships.

actually 20 wasn't enough, but 27 - was. but still. is there some formula?

i think something like:
bid^2/2g1 * (g1 - g2) == critDist.

[gimmethegepgun]

引用自 steeldog：

引用自 corisai：

Go for 4G combat acceleration in that case. ~20 dV will be enough to catch most of (if not all) alien ships.

actually 20 wasn't enough, but 27 - was. but still. is there some formula?

i think something like:
bid^2/2g1 * (g1 - g2) == critDist.

Note that everything I am about to type is based on my understanding of physics, not pulled from the game.

I don't remember the rules offhand, but I think the game tells you the starting and ending conditions for pursuit.
With these conditions, you use the equation

d = v(i) * t + 1/2 * (a(1) - a(2)) * t^2

d is the difference between your starting distance and the distance you need to reach for it to complete the pursuit, in meters (these are based on the maximum weapon range of any ship in the fight). t is time, in seconds. v(i) is the initial relative velocity, in m/s, but I think that's 0? Or maybe it's 1000 m/s, which is the starting relative velocity between 2 fleets that engage each other? a(1) is your combat acceleration, a(2) is theirs, both measured in m/s^2 (1 g = 9.81 m/s^2). Solve for t.

Then, do

DV = a * t

Where t is the solution of the previous problem, in seconds, and a is your combat acceleration, in m/s^2.
This should be the maximum DV expenditure of a successful pursuit.

[steeldog]

引用自 gimmethegepgun：

引用自 steeldog：

actually 20 wasn't enough, but 27 - was. but still. is there some formula?

i think something like:
bid^2/2g1 * (g1 - g2) == critDist.

Note that everything I am about to type is based on my understanding of physics, not pulled from the game.

I don't remember the rules offhand, but I think the game tells you the starting and ending conditions for pursuit.
With these conditions, you use the equation

d = v(i) * t + 1/2 * (a(1) - a(2)) * t^2

d is the difference between your starting distance and the distance you need to reach for it to complete the pursuit, in meters (these are based on the maximum weapon range of any ship in the fight). t is time, in seconds. v(i) is the initial relative velocity, in m/s, but I think that's 0? Or maybe it's 1000 m/s, which is the starting relative velocity between 2 fleets that engage each other? a(1) is your combat acceleration, a(2) is theirs, both measured in m/s^2 (1 g = 9.81 m/s^2). Solve for t.

Then, do

DV = a * t

Where t is the solution of the previous problem, in seconds, and a is your combat acceleration, in m/s^2.
This should be the maximum DV expenditure of a successful pursuit.

haven't i wrote the same?
the problem is that we don't know starting distance. (at least i don't). guess starting velocity the same for both so it doesn't matter
and i'm not sure about metrics of game. dD of 27 - what does means? 27m/s? i'm not sure

[Pawleus]

27 kps so km/s

[steeldog]

引用自 Pawleus：

27 kps so km/s

logicaly, but it doesn't helps )

[gimmethegepgun]

引用自 steeldog：

haven't i wrote the same?

No? The equation you posited bears no resemblance to what I wrote.

引用自 steeldog：

引用自 Pawleus：

27 kps so km/s

logicaly, but it doesn't helps )

Why not?

[steeldog]

引用自 gimmethegepgun：

No? The equation you posited bears no resemblance to what I wrote.

ahahah. really? ahahaha... you are majestic ))))))))))))))

引用自 gimmethegepgun：

Why not?

haven't you try to solve it?
just try to and may be you can to understand ;)

[gimmethegepgun]

引用自 steeldog：

引用自 gimmethegepgun：

No? The equation you posited bears no resemblance to what I wrote.

ahahah. really? ahahaha... you are majestic ))))))))))))))

引用自 gimmethegepgun：

Why not?

haven't you try to solve it?
just try to and may be you can to understand ;)

Your posted equation can't be correct. The units don't match up. The bid is in m/s, so bid^2 is m^2/s^2, whereas the distance is in m, and the units of the g's cancel each other.

[steeldog]

引用自 gimmethegepgun：

引用自 steeldog：

ahahah. really? ahahaha... you are majestic ))))))))))))))


haven't you try to solve it?
just try to and may be you can to understand ;)

Your posted equation can't be correct. The units don't match up. The bid is in m/s, so bid^2 is m^2/s^2, whereas the distance is in m, and the units of the g's cancel each other.

you are right mr teacher, there is mistake, but it isn't "bears no resemblance", right? )))
the only one mistyping of ^2 ;)

but still. have you tries to solve it? if you did, you would have smth like f(a) = a^2/( a - c ) which is nonmonolithyc and is against of common sense, whicj says that the bigger difference in a the less effort you need.
it happends because your (and my as whell) easiest formula means acceleration untill reach distance. but actually it isn't necassary.

[gimmethegepgun]

引用自 steeldog：

引用自 gimmethegepgun：

Your posted equation can't be correct. The units don't match up. The bid is in m/s, so bid^2 is m^2/s^2, whereas the distance is in m, and the units of the g's cancel each other.

you are right mr teacher, there is mistake, but it isn't "bears no resemblance", right? )))
the only one mistyping of ^2 ;)

Completely changing the equation does indeed make it bear no resemblance.
Also, neither removing nor relocating the ^2 fixes the unit issue.

引用自 steeldog：

but still. have you tries to solve it? if you did, you would have smth like f(a) = a^2/( a - c ) which is nonmonolithyc and is against of common sense, whicj says that the bigger difference in a the less effort you need.

Which is the result you get from solving it, because a larger difference in acceleration means less time needed, and less time at that acceleration means less DV expenditure.

[silburnl]

Note that the latest beta (0.77 at time of writing) is changing the interception mechanics significantly. There will now be an automatically calculated 'envelopment' tactic that the intercepting fleet will execute to bring the enemy to battle (assuming they have the delta-V advantage, natch) and, crucially, the difference in cruise acceleration between the two fleets will determine how many taskforces the intercepting fleet will need to split into to pull this off successfully.

If you don't have enough hulls to devote to the taskforces you need for this then it won't work (there will be a 'hole' in the envelopment that the enemy fleet will be able to slip through); and if it does work, then the resulting interception will be between one of these taskforces and the enemy fleet.

I haven't looked at the numbers on this yet, but early-ish rockets should be a viable design approach for point defence interceptor fleets. Later midgame high delta-V engine techs tend to be pretty poor for cruise acceleration, so you will need to overbuild your fleets if you want to use them in the defence role.

[steeldog]

引用自 gimmethegepgun：

Also, neither removing nor relocating the ^2 fixes the unit issue.

Oh really?

引用自 gimmethegepgun：

d = v(i) * t + 1/2 * (a(1) - a(2)) * t^2

DV = a * t

v(i) - let's think is 0. so d = 1/2 * (a(1)-a(2))*t^2

t = DV/a => d = 1/2*(a(1)-a(2))* DV^2/a^2 => DV^2/2a^2 * (a(1)-a(2)) = d

no absolutelly different drom

引用自 steeldog：

bid^2/2g1 * (g1 - g2) == critDist.

as i said you are just majestic... XXXXX ;)

引用自 gimmethegepgun：

Which is the result you get from solving it, because a larger difference in acceleration means less time needed, and less time at that acceleration means less DV expenditure.

are you so stuppid? just try to look at how this function looks like. according to that function larger A doesn't means less DV.