I'm only six jumps or so away from the system, so I'll pop back by tonight and see how things look.
Exploration - When worlds colide
- Thread starter Ben Ryder
- Start date
I am not too far away - I'll pop back to verify in about an hour.
A couple of thoughts - assuming perfectly circular orbits the gap should be 134.297514 days. Over many, many readings it should also average out to 134.297514 days per collision. Eccentricity could affect any given collision time this up or down slightly, and axial tilt could make it slightly longer (but make a future collision seem slightly earlier, because of the relative lateness). I also think Oct 15 was closer to 15:28, and I'm not sure you're calcs take into account BST - the June and Oct ones are in fact GMT +1, but the last one was just GMT. Adding the additional hour and using the 15:28 time, gives a delta of 134.316 for latest collision. As that's above the calculated gap, I think we might see a slightly shorter gap until next collision. If we added 134.297514 to last timing, it comes out July 11th 06:11 (this is a BST time again). I think a better guess though is 05:44-05:49 (taking the second and first collision times and adding multiples of 134.297514). The more accurate times we record, the better we can estimate the next one.
Shame I missed the last one, and I'd say a safe bet I'll miss next one too
The time I came up with for the next collision is the In Game time. Not GMT or BST.
All times I used are from the In Game Clock, so no time zone or DST/BST conversion should be required.
The In game clock doesn't change for BST.
I used the following web site for date and duration Calculations
https://www.timeanddate.com/date/timeduration.html
https://www.timeanddate.com/date/timeadd.html
Which now that I think of it, the date calculator might have been inserting or removing an hour due to daylight savings.
But it doesn't have an option for time zone or daylight savings, and I just tested it by adding 180 days to an arbitrary time, the time of day remained the same so it's not adding or subtracting an hour. Which should be accurate for the in game clock.
The duration calculator provides absolute time, in a number of different units, not an actual date. No conversion required there either.
I used 15:23 for the time of the last collision from a cockpit screenshot I took showing my in game clock. I concede that from my perspective the moons may not have been touching at that point but they were very close and I could not see a gap from where I was.
I've tried to specify the time in IGT, or GMT where appropriate.
I would be happy to see someone else run the calculations to check if I'm right. I'm no mathematician so I would not be surprised to have made an error at some point.
We only have two intervals to work with so far, so I agree it's probably going to vary much more than 20 minutes due to the eccentricities and inclination.
We can only predict to within a few hours but an accurate window will help people know when to be there. It ends up being an all day thing anyway.
Where did you get 134.297514 days? Previous calculations based on orbital data provided by the game said 132.9 days.
Can you give a basic summary of how you calculated that number?
All times I used are from the In Game Clock, so no time zone or DST/BST conversion should be required.
The In game clock doesn't change for BST.
I used the following web site for date and duration Calculations
https://www.timeanddate.com/date/timeduration.html
https://www.timeanddate.com/date/timeadd.html
Which now that I think of it, the date calculator might have been inserting or removing an hour due to daylight savings.
But it doesn't have an option for time zone or daylight savings, and I just tested it by adding 180 days to an arbitrary time, the time of day remained the same so it's not adding or subtracting an hour. Which should be accurate for the in game clock.
The duration calculator provides absolute time, in a number of different units, not an actual date. No conversion required there either.
I used 15:23 for the time of the last collision from a cockpit screenshot I took showing my in game clock. I concede that from my perspective the moons may not have been touching at that point but they were very close and I could not see a gap from where I was.
I've tried to specify the time in IGT, or GMT where appropriate.
I would be happy to see someone else run the calculations to check if I'm right. I'm no mathematician so I would not be surprised to have made an error at some point.
We only have two intervals to work with so far, so I agree it's probably going to vary much more than 20 minutes due to the eccentricities and inclination.
We can only predict to within a few hours but an accurate window will help people know when to be there. It ends up being an all day thing anyway.
Where did you get 134.297514 days? Previous calculations based on orbital data provided by the game said 132.9 days.
Can you give a basic summary of how you calculated that number?
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Makes sense with IGT. I wonder if the forum is messing with timezone or something for me - both the oct one and the recent one have screenshots matching the forum time, but that shouldn't be right I think? Anyway. (I hate timezones, everything should be UTC)
So we know from journal data that the orbital periods are 1.397656883 and 1.412355505. This is a difference of 0.014698622. So for every orbit of the slower one, the faster one has done 0.014698622/1.412355505 "extra". Using that, we can do 1.412355505/0.014698622 to say that in 96.08761318 orbits of the faster body, it will have done an extra orbit, which we can work out to be (1.397656883 * 96.08761318) = 134.2975139 days.
Given f as the faster orbit, and s the slower one, f * (s / ( s - f ))
Happy to be corrected on it - but it makes sense for simpler cases, e.g. it gives 3.33333 days for next collision of a body with an orbit of 2 days and another of 5 days.
Obviously ignoring eccentricity, inclination etc - but imagine it over thousands of collisions, and with constant orbital periods you can see how the average will be 134.2975139 days. Right?
FWIW, I pretty much agree with your calcs (but also not a mathematician!). Next one will be interesting additional data.
So we know from journal data that the orbital periods are 1.397656883 and 1.412355505. This is a difference of 0.014698622. So for every orbit of the slower one, the faster one has done 0.014698622/1.412355505 "extra". Using that, we can do 1.412355505/0.014698622 to say that in 96.08761318 orbits of the faster body, it will have done an extra orbit, which we can work out to be (1.397656883 * 96.08761318) = 134.2975139 days.
Given f as the faster orbit, and s the slower one, f * (s / ( s - f ))
Happy to be corrected on it - but it makes sense for simpler cases, e.g. it gives 3.33333 days for next collision of a body with an orbit of 2 days and another of 5 days.
Obviously ignoring eccentricity, inclination etc - but imagine it over thousands of collisions, and with constant orbital periods you can see how the average will be 134.2975139 days. Right?
FWIW, I pretty much agree with your calcs (but also not a mathematician!). Next one will be interesting additional data.
Forum post time stamps in the top left corner are converted to local time. Your post above shows Today, 10:40 AM. Mine's Today, 10:49 AM
I'm pretty sure they also convert for DST/BST
In game screenshots or YouTube streams showing the in game clock are the best reference.
I got 22:02 for the last collision from a live stream showing the actual contact point at the time of contact.
Actually I think it was 22:02 and a few seconds, but I'm ok with limiting accuracy to minutes.
So just to confirm, 134.2975139 days is 134 days, 7 hours, 8 minutes.
Which makes the next collision more likely to be on July 11, 2018 at ~5:10 am.(in game time) Give or take a couple of hours.
Just slightly later than my previously posted 134.291 days but I agree your number will be more accurate on average.
I was just basing my numbers on empirical data and we don't have enough data points to be that accurate.
I'm pretty sure they also convert for DST/BST
In game screenshots or YouTube streams showing the in game clock are the best reference.
I got 22:02 for the last collision from a live stream showing the actual contact point at the time of contact.
Actually I think it was 22:02 and a few seconds, but I'm ok with limiting accuracy to minutes.
So just to confirm, 134.2975139 days is 134 days, 7 hours, 8 minutes.
Which makes the next collision more likely to be on July 11, 2018 at ~5:10 am.(in game time) Give or take a couple of hours.
Just slightly later than my previously posted 134.291 days but I agree your number will be more accurate on average.
I was just basing my numbers on empirical data and we don't have enough data points to be that accurate.
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If I view a post from October now though - is it showing me the time as it was in Oct, or relative to my current local time. I think the latter, which I think is what's messing with my head!
Quite! Could do with a few more that show the clock though...
Yep. (and 25.2 secs...)
To have worked it out differently and be, what, ~9 minutes apart is impressive - especially with the lack of data points, as you note
I wonder if it's possible to work out the maximum divergency on time based on the tilt/eccentricity. Hmm.
An eccentric orbit will always be slower for one half and faster for the other half, averaging out over one full orbit.
Since this game does not consider other bodies in it's orbit calculation, there is no gravitational interaction between the two moons.
So the maximum variation would have to be limited to the combined maximum variation of the two moons over a single orbit, no? (or half orbit?)
Worst case is the inside moon at it's slowest point in it's orbit and the outside moon at it's fastest, or vis-versa.
Inclination may have an effect on the perceived linear approach speed and the overlap of the two spheres, but unless I'm missing something, it shouldn't have an affect on the timing.
Since this game does not consider other bodies in it's orbit calculation, there is no gravitational interaction between the two moons.
So the maximum variation would have to be limited to the combined maximum variation of the two moons over a single orbit, no? (or half orbit?)
Worst case is the inside moon at it's slowest point in it's orbit and the outside moon at it's fastest, or vis-versa.
Inclination may have an effect on the perceived linear approach speed and the overlap of the two spheres, but unless I'm missing something, it shouldn't have an affect on the timing.
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It's at times like this I wish I was still playing KSP... I think you're right, and furthermore we don't know at which points our two moons have been at during the known contacts.
You mean that I would get to do this kind of math if I were playing KSP? I keep meaning to pick it up when it's on sale but I hardly ever play anything except Elite:
You might profit from seeing some of Scott Manley's KSP videos or streams.
I already watch all of his videos. Fly Safe.
KSP is very much "what you make of it" - you can approach it as a deadly serious exercise in (semi-realistic) orbital mechanics and spaceflight or you can install a dozen mods and have an autopilot fly around a replica of the Enterprise. Or any point in between.
I ended up making rocket-powered quadrigas. Quadrigae? Whatever.
I tried to calculate their orbits and this is what I come up with.
[IMGUR]tXMEZ[/IMGUR]
Be aware that axis are not to scale and vertical axis is not to scale to other two. Inclination difference is under one degree so they are very close to each other most of the time. Major problem is that I don't know what is value of Longitude of Ascending Node. Another unknown is mean anomaly which define where planet is on given time. Without it I can just plot whole orbit.
When I assume that LAN is zero these two orbits come out. It seems like maximum possible separation is around 800km so the moons will always collide.
[IMGUR]tXMEZ[/IMGUR]
Be aware that axis are not to scale and vertical axis is not to scale to other two. Inclination difference is under one degree so they are very close to each other most of the time. Major problem is that I don't know what is value of Longitude of Ascending Node. Another unknown is mean anomaly which define where planet is on given time. Without it I can just plot whole orbit.
When I assume that LAN is zero these two orbits come out. It seems like maximum possible separation is around 800km so the moons will always collide.
I might have found another two moons which collide at various points. I have them Bookmarked to look at in the future:
https://imgur.com/NaSLFon
Currently on my way to Beagle Point, so at the moment I am going to keep the system private.
https://imgur.com/NaSLFon
Currently on my way to Beagle Point, so at the moment I am going to keep the system private.
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I suggested this on the Distant Worlds 2 scouting post, but I'm missing information on where the best basecamp would be that would be close to the collision zone but not in it. (beside it would be nice)
Would anyone be able to help me out determine the next 2 conjunctions to a higher accuracy?
Hi i've been doing these calculations as well and I'm wondering if the discrepancy has something to do with the inclination and eccentricity that delayed the contact.
Is there anyway we can get these values? what program do you use to model this?
Would anyone be able to help me out determine the next 2 conjunctions to a higher accuracy?
Hi i've been doing these calculations as well and I'm wondering if the discrepancy has something to do with the inclination and eccentricity that delayed the contact.
Is there anyway we can get these values? what program do you use to model this?
We've observed three collisions so far and have two time intervals between them.
MattG's post at the top of page 28 has the formula and calculation down to (I believe) a very accurate 134.2975139 days between collisions.
The next collision should be on July 11, 2018 at ~5:10 am.(in game time)
The following collision should be at (or very close to) November 22, 2018 at ~12:18 Game-Time
There's a bit of unpredictability to the exact contact time due to the orbital inclination and eccentricities, but peak collision times should be less variable.
I don't think it will vary by much more than a half hour. The last two collisions were predicted to within 20 minutes.
The collision lasts at least 4 hours between contact and separation.
That's an interesting question. The moons are both tidally locked so rotation won't affect where you park.
When I was there I chose to park on the outer moon so that the B moon would pass in front of the gas giant. I thought the screenshots would look better with the giant in the background.
Then I ended up observing the collision from the air in order to get shots for video. My Video is here https://youtu.be/d8HJFXRxV8w
At 9:29 in that video the collision is well underway and the B moon is moving sideways to my position. Coordinates 8.99 by 9.94. It should be a generally safe area to be close to the collision but not impacted by it.
On page 22 of this thread, CMDR ANCALAGON found geysers on the C moon at ~ 7.55 by -161.05. Geysers are always a fun spot to have meetups.
Warnings for pilots.
Gravity can shift on you if you get closer to the other moon, flipping your instrument orientation up side down.
Your relative motion frame of reference can get locked to the wrong moon so that even if your ship is not moving, it can actually be moving toward the ground at around 115m/s.
Several ships were lost in the last couple of collisions because of this. If this basecamp is at the end of the expedition, I would jot back to Colonia and sell my data beforehand.
IIRC...
SRVs on the surface will clip through the other moon. I don't think you can jump your SRV between moons. Several have tried but I haven't read about any success.
Ships landed (down and locked) on the surface should clip through the other moon.
Ships in the air and/or occupied ships can impact with the other moon but some have also clipped through, especially if caught in the intersection point.
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I used octave (free Matlab alternative) to calculate positions of both moon and just plot them. Someone posted orbit parameters in this thread. But LAN is missing so it is not accurate.
Moon B is almost but not quite tidally locked - its rotational period is 15.17 seconds longer than its orbital period. This results in a synodic day that is about 7960.3 orbits, or about 30.46 earth years.
Moon C is almost but not quite tidally locked - its rotational period is 15.33 seconds longer than its orbital period. This results in a synodic day that is about 7962 orbits, or about 30.79 earth years.
This results in the average collision point moving about 4.3455 and 4.2994 degrees eastward for b and c respectively each time the moons meet.
The orbit line of B should always be outside the collision zone, but the orbit line of C may fall inside the collision zone.
Moon C is almost but not quite tidally locked - its rotational period is 15.33 seconds longer than its orbital period. This results in a synodic day that is about 7962 orbits, or about 30.79 earth years.
This results in the average collision point moving about 4.3455 and 4.2994 degrees eastward for b and c respectively each time the moons meet.
The orbit line of B should always be outside the collision zone, but the orbit line of C may fall inside the collision zone.