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warp drive active!
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Gor Yo
Taxes Shmuckses
#41 - 2014-08-14 09:36:29 UTC
Gully Alex Foyle wrote:
OP, Fozzie's formulas are fine.

x = e^kt is distance, in meters; t is time in seconds

v = k*e^kt is speed, in m/s; this formula is the derivative of the above formula, aka the definition of speed. it's correct.


Now, say Vmax is your maximum warp speed. Vmax = k*A where A is 150 billion meters (1 AU).

So if you want to know acceleration time, its Vmax = k*A = k*e^kt.

==> A = e^kt

==> ln(A) = kt

==> t = ln(A)/k = 25.7 / k


For example, a cruiser (3 AU/s) would take 8.6 seconds to reach max warp speed, an inty (8 AU/s) would take 3.2 seconds.

Sounds about right to me, though I CBA to test it.


Fozzie was slightly fuzzier on deceleration, I'll think about it and post back if I nail the formula.


where did you get this from???? it is stated in the update info that k is equal to max warp speed, no additional multipliers.
besides that, v=k*e^kt is NOT correct for the simple fact that v(0)=k, max warp speed already. there has to be an offset so that v(0)=0. that in turn will change the x(t) formula as well.
Gor Yo
Taxes Shmuckses
#42 - 2014-08-14 10:13:40 UTC
Ok, that was my initial reaction. Looking further into it, you may be right about it. Even disregarding boundary conditions, those formulas are wrong dimensions-wise. e is dimensionless, while x is meters, so x=e^anypower is wrong. the power of e should be dimensionless as well, but kt has distance dimensions, etc.

Anyway, in this light, Vmax as kA may be correct as it converts AU to meters. Then taking into account the corrected formula we do get warp acceleration time as
t_warp=ln(A+1)/k
and it sounds about right for my rigged badger with 6 point something warp speed.
So that part you got right. The distance though to reach max speed would be different.
11 seconds for deceleration though(same as interceptor, since k for deceleration is min(2,warp_speed/3)? I will check in game later today. It seems way to long. And it always looked to me like when the ship is out of warp, the speed gauge is at about 1/4. Could be different for different ships though, I will check in game later today!

Well, thank you Gully for staying on topic and actually contributing, unlike 20+ other responses in this thread! Blink
Gully Alex Foyle
The Scope
Gallente Federation
#43 - 2014-08-14 10:27:05 UTC
Gor Yo wrote:
Ok, that was my initial reaction. Looking further into it, you may be right about it. Even disregarding boundary conditions, those formulas are wrong dimensions-wise. e is dimensionless, while x is meters, so x=e^anypower is wrong. the power of e should be dimensionless as well, but kt has distance dimensions, etc.

Anyway, in this light, Vmax as kA may be correct as it converts AU to meters. Then taking into account the corrected formula we do get warp acceleration time as
t_warp=ln(A+1)/k
and it sounds about right for my rigged badger with 6 point something warp speed.
So that part you got right. The distance though to reach max speed would be different.
11 seconds for deceleration though(same as interceptor, since k for deceleration is min(2,warp_speed/3)? I will check in game later today. It seems way to long. And it always looked to me like when the ship is out of warp, the speed gauge is at about 1/4. Could be different for different ships though, I will check in game later today!

Well, thank you Gully for staying on topic and actually contributing, unlike 20+ other responses in this thread! Blink
Yw!

Tbh, I was about to post a useless but witty reply too, but then I noticed the server issues/extended downtime and decided it would be more fun to try to work out the maths. Big smile

The formulas I've posted give exactly the same figures of Fozzie's spreadsheet, so they should be quite ok.

Yes, it takes inties 11 seconds to decelerate. The deceleration constant is capped at 2 no matter the max warp speed - even if, say, you put 3 T2 warp speed rigs on an inty. It's pretty clear in the second graph here.

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Gor Yo
Taxes Shmuckses
#44 - 2014-08-14 10:47:58 UTC
Gully Alex Foyle wrote:
Yw!

Tbh, I was about to post a useless but witty reply too, but then I noticed the server issues/extended downtime and decided it would be more fun to try to work out the maths. Big smile

The formulas I've posted give exactly the same figures of Fozzie's spreadsheet, so they should be quite ok.

Yes, it takes inties 11 seconds to decelerate. The deceleration constant is capped at 2 no matter the max warp speed - even if, say, you put 3 T2 warp speed rigs on an inty. It's pretty clear in the second graph here.


Yeah, I calculated the proper acceleration distance and it is not much different from 1 AU, since the speeds and distances we are talking about here are so huge. The trick seems to be to convert meters to AU and back at certain points in calculations, and compare the results with actual numbers in game.

Well, it seems I made the right decision to go for agility as a 3rd rig, since deceleration takes a lot of time (half or more for shorter trips) and increasing warp speed above 6 AU/s doesnt affect deceleration times.
On the other hand, for ships with max sub-warp speed below 200m/s, increasing it to at least 200 might actually cut down the deceleration time a bit. It is a negative exponent and the tail comes down very slowly.
Gully Alex Foyle
The Scope
Gallente Federation
#45 - 2014-08-14 11:04:43 UTC
I just noticed j=k/3 up to k=6, so that means deceleration distance is exactly 3 AU up to 6AU/s max speed.

I previously thought that j (decel constant) was fixed for each hull type. I edited my previous post accordingly.


This also means that fitting 3x warp speed rigs to a cruiser actually does make it warp almost as fast as a frigate - interesting.

Make space glamorous! Is EVE dying or not? Ask the EVE-O Death-o-meter!
Gor Yo
Taxes Shmuckses
#46 - 2014-08-14 13:25:59 UTC
Gully Alex Foyle wrote:
I just noticed j=k/3 up to k=6, so that means deceleration distance is exactly 3 AU up to 6AU/s max speed.

I previously thought that j (decel constant) was fixed for each hull type. I edited my previous post accordingly.


This also means that fitting 3x warp speed rigs to a cruiser actually does make it warp almost as fast as a frigate - interesting.


Yep, deceleration coefficient is the minimum of 2 or k/3. And yeah, that little chart is really misleading when they indicate warp speeds by hull class. Heck, even t1 industrials have versions with 3 and 4.5 warp speeds.

So looks like acceleration distance is always ~1 AU, and deceleration distance is ~3 AU (or k/2 for warp speeds above 6). The "correct" formulas give only a very little adjustment - like A vs A+1 in one case, and 1 vs 1-v_exitwarp/kA.

Sorry, I had to be OCD about formulas. Any physicist would understand the need for honoring the boundary conditions :).
Gully Alex Foyle
The Scope
Gallente Federation
#47 - 2014-08-14 17:38:13 UTC  |  Edited by: Gully Alex Foyle
If you don't have enough distance to accelerate to maximum warp speed (you need exactly 4 AU up to 6 AU/s max warp speed; a bit more after that), the highest reachable warp speed should be:

Vmax = Dkj/(k+j)

You can use this new Vmax (instead of k*A) in the previous formulas, to calculate warp time over short distances.


Note: yes, I should've subtracted the exit warp speed in these and some of the previous formulas, but I simplified a bit since 100 m/s or so is much smaller than 1 or more AU/s :)

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