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Siigari Kitawa
Gallente The Aduro Protocol
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Posted - 2009.11.10 12:40:00 -
[1]
I was just moving about with my corpmate and he jumped through a gate and moved to investigate a wreck on the gate. When I jumped in, I was 422 meters away from him and the wreck.
We laughed.
That made me wonder about how that never ever happens (cloakers gettin' away!). So, I tried my hand at some complex math problems but the numbers were pretty astronomical so here I am to ask you all:
What is the chance of landing next to somebody within 2km of him if the radius of the gate drop is 15,000 meters?
:O
I'd love an answer!
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Kazuo Ishiguro
House of Marbles
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Posted - 2009.11.10 12:48:00 -
[2]
Edited by: Kazuo Ishiguro on 10/11/2009 12:51:51 Assuming you can spawn anywhere within the 15km sphere with equal probability, a good approximation would be (volume of 2km sphere) / (volume of 15km sphere) = 2^3 / 15^3 = 1 in 422. However, if there are multiple people in the 15km sphere and you want the chance of avoiding all of them, it gets a bit more complicated.
Note that if you land 13-15km away from the gate, the chances of a decloak become smaller, as you only have a hemisphere to land in, not a full 2km sphere, so the odds are actually slightly worse than 1 in 422. --- 34.4:1 mineral compression ISRC Racing, Season 7 - schedule |
Cpt Branko
The Scope
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Posted - 2009.11.10 12:54:00 -
[3]
Approximately 1:2250, using very rough math.
Sig removed, inappropriate link. If you would like further details please mail [email protected] ~Saint |
Kazuo Ishiguro
House of Marbles
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Posted - 2009.11.10 12:55:00 -
[4]
Originally by: Cpt Branko Approximately 1:2250, using very rough math.
Please explain your working, if it's really as simple as you claim. --- 34.4:1 mineral compression ISRC Racing, Season 7 - schedule |
Nikla Uthaan
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Posted - 2009.11.10 12:57:00 -
[5]
Edited by: Nikla Uthaan on 10/11/2009 12:58:57 It was also my impression that, while you do physically appear "somewhere" on jump-in, you aren't actually "on the grid", although you do see your start position. Until you move or the cloak times out, you aren't actually there.
Some also documented some effects this had on align times of freighters and such, as similarly since you aren't "anywhere", while there's direction to how you appear, you aren't actually facing in any direction.
Whether my explanation is correct or not, I've definately had people come in at bubbled gates within even just 1000m of me and not decloak me immediately after jump-in.
EDIT: To get slightly more on topic though, what about different gates? Surely the regional gates have a drastically reduced chance of coming that close? Since your 15km from the gate isn't a 15km true radius, but a 15km radius on top of some other spherical radius that is the border of the object. |
Siigari Kitawa
Gallente The Aduro Protocol
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Posted - 2009.11.10 13:01:00 -
[6]
Originally by: Nikla Uthaan Edited by: Nikla Uthaan on 10/11/2009 12:58:57 It was also my impression that, while you do physically appear "somewhere" on jump-in, you aren't actually "on the grid", although you do see your start position. Until you move or the cloak times out, you aren't actually there.
Some also documented some effects this had on align times of freighters and such, as similarly since you aren't "anywhere", while there's direction to how you appear, you aren't actually facing in any direction.
Whether my explanation is correct or not, I've definately had people come in at bubbled gates within even just 1000m of me and not decloak me immediately after jump-in.
EDIT: To get slightly more on topic though, what about different gates? Surely the regional gates have a drastically reduced chance of coming that close? Since your 15km from the gate isn't a 15km true radius, but a 15km radius on top of some other spherical radius that is the border of the object.
I can answer this one :)
When you jump in you are under a grace period and session change timer. For sixty seconds you cannot be targeted, decloaked or smartbomb'd until the time expires or you move.
Once you move all bets are off and you are "on the grid." |
Lady Spank
Amarr Sekret Kool Klubb
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Posted - 2009.11.10 13:06:00 -
[7]
Statistics of the 'landing sphere'
radius = 15km diameter = 30km Circular circumference = 94.248km Spherical surface area 2827.433km
to land 2km from someone, you have to land within a 'circle' (on the surface of the sphere so its not a cirle really) of 1km radius. Thats a circular circumference of ...
to be continued... |
Kazuo Ishiguro
House of Marbles
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Posted - 2009.11.10 13:06:00 -
[8]
Originally by: Cpt Branko Edited by: Cpt Branko on 10/11/2009 12:59:03 Edited by: Cpt Branko on 10/11/2009 12:57:00 Approximately 1:2250, using very rough math, to land in 2km range of a given ship assuming you always spawn 15km from gate.
But very rough calculation - assumed spawn position as (X deg,Y deg) pair which are randomly picked from 0 to 360 (which almost certainly isn't true). You need to be witin 7.5 degrees or so in both coordinates (actually, the chance is somewhat less, probably closer to 1:2500-1:3000).
Heh, I think the trouble is that there are several possible right answers here, depending on the method CCP uses to assign a pseudo-random position.
Your method appears not to account for radial distance from the gate, and your two co-ordinates could be chosen from different ranges - a standard definition for spherical polars is (r, theta, phi) with theta in the range [0¦,360¦) and phi in [0¦,180¦) |
Maia Blaze
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Posted - 2009.11.10 13:08:00 -
[9]
I've been flying a blockade runner for around a year, and have only had it happen to me once. It can happen, but it seems to be very rare.
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Siigari Kitawa
Gallente The Aduro Protocol
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Posted - 2009.11.10 13:11:00 -
[10]
Originally by: Lady Spank to land 2km from someone, you have to land within a 'circle' (on the surface of the sphere so its not a cirle really) of 1km radius. Thats a circular circumference of ...
to be continued...
Wait
o = target, x = me
0km....2km....4km -------o-------X
??
Shouldn't that double because you have a 2km radius reaching in every direction along a 2km surface area on the sphere?
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Kazuo Ishiguro
House of Marbles
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Posted - 2009.11.10 13:19:00 -
[11]
Edited by: Kazuo Ishiguro on 10/11/2009 13:22:53 Heh, didn't notice initially that you specified the jump point as always being 15km from the gate. If a point is chosen uniformly from the whole surface area, then one can approximate via (area of 2km radius circle) / (area of 15km radius sphere) and that gives 1 in 225.
If it's done via randomly choosing standard spherical co-ordinates for theta and phi as I set out above, you need to be within a maximum of approximately arc-sin(1/15) = 7.65¦ for each ordinate, so an upper bound would be (7.65 / 180) * (7.65 / 360) = 1 in 1109. --- 34.4:1 mineral compression ISRC Racing, Season 7 - schedule |
Siigari Kitawa
Gallente The Aduro Protocol
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Posted - 2009.11.10 13:23:00 -
[12]
Originally by: Kazuo Ishiguro Heh, didn't notice initially that you specified the jump point as always being 15km from the gate. If a point is chosen uniformly from the whole surface area, then one can approximate via (area of 2km radius circle) / (area of 15km radius sphere) and that gives 1 in 225.
If it's done via randomly choosing standard spherical co-ordinates for theta and phi as I set out above, you need to be within a maximum of 7.65¦ for each ordinate, so an upper bound would be (7.65 / 180) * (7.65 / 360) = 1 in 1109.
Wow! Can you show your work? I am trying to learn.
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Cpt Branko
The Scope
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Posted - 2009.11.10 13:40:00 -
[13]
Edited by: Cpt Branko on 10/11/2009 13:42:08
Originally by: Kazuo Ishiguro Edited by: Kazuo Ishiguro on 10/11/2009 13:21:47 Edited by: Kazuo Ishiguro on 10/11/2009 13:09:09
Originally by: Cpt Branko
Approximately 1:2250, using very rough math, to land in 2km range of a given ship assuming you always spawn 15km from gate.
But very rough calculation - assumed spawn position as (X deg,Y deg) pair which are randomly picked from 0 to 360 (which almost certainly isn't true). You need to be witin 7.5 degrees or so in both coordinates (actually, the chance is somewhat less, probably closer to 1:2500-1:3000).
Heh, I think the trouble is that there are several possible right answers here, depending on the method CCP uses to assign a pseudo-random position.
Your two co-ordinates should be chosen from different ranges - a standard definition for spherical polars is (r, theta, phi) with theta in the range [0¦,360¦) and phi in [0¦,180¦).
With your system there are two different ways of specifying any particular point, so it's actually twice as likely as you think.
You're correct, yes, I just did it with 2d and then expanded to 3d in a blunt straighforward manner. Back of the envelope solutions and error margin, heh.
Obviously there is a slight error since being 7.5 degrees off on both coordinates will not produce <2km distance.
Sig removed, inappropriate link. If you would like further details please mail [email protected] ~Saint |
Kazuo Ishiguro
House of Marbles
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Posted - 2009.11.10 13:57:00 -
[14]
Originally by: Cpt Branko There is a slight error since being 7.5 degrees off on both coordinates will not produce <2km distance and we didn't correct for that, so the real value would approximately be 1:1400 or so.
I think it's a reasonably good approximation, since it really does seem that ships do appear 15km of the center of normal, non-regional stargates.
Well, what we really need is P(√(Δθ¦+ΔΦ¦)) < 2 arcsin(1/15), but that's rather tricky to work out. I suppose you're correcting via (area of square, side 2) / (area of circle, radius 2). --- 34.4:1 mineral compression ISRC Racing, Season 7 - schedule |
Lady Spank
Amarr Sekret Kool Klubb
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Posted - 2009.11.10 13:58:00 -
[15]
Edited by: Lady Spank on 10/11/2009 14:02:40
Originally by: Siigari Kitawa
Originally by: Lady Spank to land 2km from someone, you have to land within a 'circle' (on the surface of the sphere so its not a cirle really) of 1km radius. Thats a circular circumference of ...
to be continued...
Wait
o = target, x = me
0km....2km....4km -------o-------X
??
Shouldn't that double because you have a 2km radius reaching in every direction along a 2km surface area on the sphere?
Where I was going was mapping the surface of the sphere with 1km radius circular areas. I say 1km because once mapped out, we need to calculate the odds of both landing in the same area.
1km radius = 2km circumference. within a 4km circumference circle there is the chance for you both to be anywhere from 2 - 4km away from each other. Therefore we need the surface area of a circle of 1km radius.
using your example
0km....2km....4km ---0---~-------X
/o\
so a 1km radius circle has a surface area of 3.142km
so sherical surface area divided into 'zones' of 1km radius circles.
2827.43 / 3.142 = 899.882
so erm... 900 'zones' to land in.
chance of one ship landing in a specific zone, 1 in 900, or 0.1111111% chance of two ships landing in a specific zone, (CRIES) um 0.1111111^2? so 0.0123457% chance.
I can't do maths and I'm meant to be working.
EDIT:
That or its 899.992^2 = 1 in 809,986 chance of 2 people landing within 2km.
The quality of my replies is directly related to the QQuality of the opÆs comments |
zzCoins
UK1 Zero
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Posted - 2009.11.10 13:59:00 -
[16]
Edited by: zzCoins on 10/11/2009 14:01:04 area circle pi r2 decloak radius is 2km, so a ship decloaks 4pi of area
area sphere is 4 pi r2 jumping through a gate puts you 15km from the gate, so if the gate is a point, then you appear on a sphere with area 4 pi 15x15 decloak probability 1 in 225 (gate is a point)
But gates are not points, theh have a size, region gates are big, constallation gates are medium pirate gates are extra small. I have not measured the radius of a gate, but I think some of them are 15 km which gives a sphere of radius 30km - area of 4 pi 30x30 decloak probability 1 in 900 (gate has 15km radius)
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Kazuo Ishiguro
House of Marbles
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Posted - 2009.11.10 14:09:00 -
[17]
You want to work out the area on a sphere of radius 15km enclosed by a circle of radius 2km. This is not easy to find, but it will be so close to the area of the circle itself that you might as well just approximate. It's not as if we need pinpoint accuracy for the purposes of this problem. --- 34.4:1 mineral compression ISRC Racing, Season 7 - schedule |
Soi Mala
Whacky Waving Inflatable Flailing Arm Tubemen
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Posted - 2009.11.10 14:09:00 -
[18]
I spawned 1.5km from a falcon once, and i must have jumped roughly 1 gazillion times in the 2.5 years i've been playing.
1 in a gazillion.
ps. He got away.
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Jamyl TashMurkon
Amarr
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Posted - 2009.11.10 15:01:00 -
[19]
But your all calculations are wrong because you didn't take the gate model size itself into thought.
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Jowen Datloran
Caldari Science and Trade Institute
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Posted - 2009.11.10 15:14:00 -
[20]
Originally by: Jamyl TashMurkon But your all calculations are wrong because you didn't take the gate model size itself into thought.
That doesn't mean the calculations are wrong, far from it.
In fact, they answer the given question correctly according to the premises provided. ---------------- Mr. Science & Trade Institute
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Lady Spank
Amarr Sekret Kool Klubb
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Posted - 2009.11.10 15:46:00 -
[21]
Wait a minute... I got this one right?
The quality of my replies is directly related to the QQuality of the opÆs comments |
Cpt Branko
The Scope
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Posted - 2009.11.10 16:03:00 -
[22]
Edited by: Cpt Branko on 10/11/2009 16:06:01
Originally by: Kazuo Ishiguro
Well, what we really need is P(√(Δθ¦+ΔΦ¦)) < 2 arcsin(1/15), but that's rather tricky to work out. I suppose you're correcting via (area of square, side 2) / (area of circle, radius 2).
Yes, I just multiplied by 1.27.
Originally by: Lady Spank
chance of one ship landing in a specific zone, 1 in 900, or 0.1111111% chance of two ships landing in a specific zone, (CRIES) um 0.1111111^2? so 0.0123457% chance.
Well, obviously one ship has landed somewhere, and it chance of being in its own position are 1, so calculating whether two ships land at a predeterminate position and multiplying the probabilities is wrong here.
Anyway, the real question which could use testing would be how exactly are ships positioned after jump-in. It's very certain it is not 15km from the edge of the stargate model. For small gates it does seem it's either 15km from the center or 15km from a central line of the stargate model (I'm not surem but from my recollection of gatecamping I would say the former is probably more true).
Sig removed, inappropriate link. If you would like further details please mail [email protected] ~Saint |
Jarne
Increasing Success by Lowering Expectations
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Posted - 2009.11.10 16:28:00 -
[23]
I'll go with 1:225.
The surface of a 15km sphere is about 2827 square kilometers (surface of a sphere with radius r: 4*pi*r^2).
The area of a circle with 2km radius (decloak range) is about 12.5 square kilometers (area of a circle with radius r is 2*pi*r^2).
The ratio of the two is 1:225.
I think, however, that CCP might use a server-tick-based random number generator to generate the spawn position, because it appears to me when I land on top of someone it's often that we both jumped through the gate at the same time... pure speculation on my part, of course. - Success=Achievements/Expectations
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Lubomir Penev
Dark Nexxus
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Posted - 2009.11.10 16:41:00 -
[24]
Originally by: Kazuo Ishiguro
Assuming you can spawn anywhere within the 15km sphere with equal probability, a good approximation would be (volume of 2km sphere) / (volume of 15km sphere) = 2^3 / 15^3 = 1 in 422. However, if there are multiple people in the 15km sphere and you want the chance of avoiding all of them, it gets a bit more complicated.
Note that if you land 13-15km away from the gate, the chances of a decloak become smaller, as the de-cloaker only has a hemisphere to land in, not a full 2km sphere, so the odds are actually slightly worse than 1 in 422.
The gate is not a point. And the spawning surface is not a sphere. The spawning surface looks to be the surface where you are about 12km from EDGE of gate model. Hence this surface is pretty small on regular gates, bigger on constellation gates, huge on region gate (you even have a pretty good chance or spawning out of disruptor range on a region gate, and they cannot be bubbled by a single dictor either), and you must also account for racial variation, e.g. Minmatar gates are huge.
So short answer : depend on the gate, but small. -- 081014 : emoragequit, char transfered to a friend, 090317 : back to original owner blog |
Siigari Kitawa
Gallente The Aduro Protocol
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Posted - 2009.11.10 21:54:00 -
[25]
Good morning o/
Wow, looks kind of divided on the answer :|
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Dominatus Crispus
Gallente Nation of Muppets
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Posted - 2009.11.10 21:59:00 -
[26]
Originally by: Siigari Kitawa I'd love an answer!
42
Corp/Allaince Setup & Tweaking / Faction Standings Boost [Details] |
Lucas Avidius
Einherjar Rising Cry Havoc.
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Posted - 2009.11.10 22:37:00 -
[27]
I just skimmed, so I dont know if this has been mentioned or not, but if you want the real chance of this happening, you need to take into account that the 15 KM begins at the edge of the gates 0m sphere. For instance, amarr gates are the smallest, but still have at least a few KM of radius, while gallente gates in 0.0 (for instance the Orvolle gate in PF-346) are so huge that the 15km-from-the-gate sphere would be several times that of a standard amarr gate's sphere.
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Kazuo Ishiguro
House of Marbles
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Posted - 2009.11.10 22:50:00 -
[28]
Originally by: Lubomir Penev The gate is not a point. And the spawning surface is not a sphere. The spawning surface looks to be the surface where you are about 12km from EDGE of gate model.
...
So short answer : depend on the gate, but small.
Each model in the physics engine has a centre point as well as edges iirc, and distances can be measured from either - a mix-up of this kind was what led to the 'no hits at 0km' turret bug.
If, as I suspect, placement is done by choosing 2 random angular co-ords, it seems likely that the centre of the gate is being used as the origin, so even if the resulting jump-in surface is not convex it will at least be star-like. It's only at worst a moderately distorted sphere, so I expect the numbers quoted for that method are accurate to within an order of magnitude. --- 34.4:1 mineral compression ISRC Racing, Season 7 - schedule |
Mel'h N'gih
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Posted - 2009.11.10 23:17:00 -
[29]
The odds are higher than you think and a lot more variable since the distance between two objects/ships in eve isn't center to center but edge to edge :
ie: a freighter jumps into a gate where a cov ops is sitting cloaked at the gate drop range (jumped through, let his cloak expire, activate his cloak) vs a pod jumps into our same cloaker... |
Illectroculus Defined
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Posted - 2009.11.10 23:39:00 -
[30]
Of course this can also apply to other questions like
'If an RR battleship gang jumps through a gate, what's the odds that a gang member will be in RR range of another pilot'
Which leads to
"How many RR BS's does your fleet have to contain before the chances of there being an 'isolated' pilot drop below 50% on the other side of a gate"
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