That's fairly accurate yes, there should be more information we can use to narrow it down to an exact location but until that clever person comes along and shows us how to do it that may be the only option we have.
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Alien archeology and other mysteries: Thread 9 - The Canonn
- Thread starter Ian Phillips
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That's fairly accurate yes, there should be more information we can use to narrow it down to an exact location but until that clever person comes along and shows us how to do it that may be the only option we have.
But... it sets over the larger circle. A known, fixed point. Or will be doing very soon*.
*Well, I say that, but its starting to look like another 4 hours. I *should* already be in bed, I'm due at work in 9 hours!
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not quite, the moons pass at an angle over the ruins, not overhead.
Ok, so I did some back-of-the-napkin calculations based on the idea that the sites share moon rise/set angles.
First overlay a protractor on the site center, read about 150 degrees between the rise and set points.
This creates an isosceles triangle with 15 degrees at each other point, and the base (long side) running along the moon's orbital path.
Now I try to measure the triangle's sides:
Starting with the radius of the planet for the first site (1,122km) I determine a planetary circumference of 7,049km. Divided by 360 degrees, I get 19.5km per degree. Since the site is -31 degrees off the equator, I multiply 19.5 by 31 and get 606km.
So now to verify this, I would expect to locate the first ruins site at around 606km off the 0 degree mark. In-game, I am not sure how to measure this (see "Remaining Problems 2" below).
Completing the triangle, solving for the base: Approximately 4,523km.
So how to advance with proving this is a valid analysis corresponds with the 2nd ruins site:
Find the circumference of the 2nd known ruins site (presently I do not know it's radius). Create a proportion vs. the base of the 1st ruins site triangle, to find the base of the 2nd ruins site triangle:
7,049km (1st ruins site planet circumference) / 4,523km (1st ruins triangle base) : 2nd ruins site circumference / 2nd ruins site triangle base
Then calculate the height of the 2nd triangle, as with the first.
If the calculated 2nd site triangle height is near the second ruins site distance from the moon's orbital path, it's looking good. The concept is holding true. Now take it to other candidate planets.
Remaining problems:
1. Measuring a reliable distance offset from the moon's orbit in-ship/in-game. Maybe a wing of two players do this. Have one wing member land at the moon's orbit path, then a 2nd CMDR heads out, reading the distance to the other CMDR to know his range?
2. Flying a steady "small circle" based on an inclined moon orbit, which is not as easy as flying the equator. The heading would need to change steadily as the route circles the planet.
3. Perhaps most importantly... All my calculations above are based off the equator. Well, at the 1st ruins planet, the moon has a -6.8 degree inclination. So how many km above/below the equator is this at the ruin's -128.97 longitude? This variance will likely impact the triangle height calculation a bit, and from that, everything after. Solving this will take a good deal of legwork.
But ultimately, if this triangle holds true for all ruins sites, it seems likely that the search areas could be narrowed greatly by calculating the height of the triangle stretching between the base (running along the moon orbit) and the opposite angle vertex (centered on the ruins site).
As ever, if I've messed something up, please correct it. Thanks.
First overlay a protractor on the site center, read about 150 degrees between the rise and set points.
This creates an isosceles triangle with 15 degrees at each other point, and the base (long side) running along the moon's orbital path.
Now I try to measure the triangle's sides:
Starting with the radius of the planet for the first site (1,122km) I determine a planetary circumference of 7,049km. Divided by 360 degrees, I get 19.5km per degree. Since the site is -31 degrees off the equator, I multiply 19.5 by 31 and get 606km.
So now to verify this, I would expect to locate the first ruins site at around 606km off the 0 degree mark. In-game, I am not sure how to measure this (see "Remaining Problems 2" below).
Completing the triangle, solving for the base: Approximately 4,523km.
So how to advance with proving this is a valid analysis corresponds with the 2nd ruins site:
Find the circumference of the 2nd known ruins site (presently I do not know it's radius). Create a proportion vs. the base of the 1st ruins site triangle, to find the base of the 2nd ruins site triangle:
7,049km (1st ruins site planet circumference) / 4,523km (1st ruins triangle base) : 2nd ruins site circumference / 2nd ruins site triangle base
Then calculate the height of the 2nd triangle, as with the first.
If the calculated 2nd site triangle height is near the second ruins site distance from the moon's orbital path, it's looking good. The concept is holding true. Now take it to other candidate planets.
Remaining problems:
1. Measuring a reliable distance offset from the moon's orbit in-ship/in-game. Maybe a wing of two players do this. Have one wing member land at the moon's orbit path, then a 2nd CMDR heads out, reading the distance to the other CMDR to know his range?
2. Flying a steady "small circle" based on an inclined moon orbit, which is not as easy as flying the equator. The heading would need to change steadily as the route circles the planet.
3. Perhaps most importantly... All my calculations above are based off the equator. Well, at the 1st ruins planet, the moon has a -6.8 degree inclination. So how many km above/below the equator is this at the ruin's -128.97 longitude? This variance will likely impact the triangle height calculation a bit, and from that, everything after. Solving this will take a good deal of legwork.
But ultimately, if this triangle holds true for all ruins sites, it seems likely that the search areas could be narrowed greatly by calculating the height of the triangle stretching between the base (running along the moon orbit) and the opposite angle vertex (centered on the ruins site).
As ever, if I've messed something up, please correct it. Thanks.
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Given that the design allowed "perfect alignment" at the alpha site, and that simply choosing the correct location and orientation would allow similarly "perfect" alignment at the beta site, and given that this perfect alignment does not appear to exist, I would theorise that it's not going to work out mathematically.
Next question then - does Barnard's Loop land neatly over the large mound at both sites? Or the star (through the same wall gap at both)?
o7
Try this from the last page.
Its a wacky angle, but the moon and BLoop are racing diagonally towards the roughly same spot of horizon.
And I don't think the variance in angles between the two moon orbits is going to be that big a deal, I think it might still narrow the search into a managable sized area.
And 2 CMDR's winged up, one super-cruising the moon orbit line, the other close to the ground travelling the identifid lat line (-29/-30 ish degrees?) may be able to do it pretty quick.
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Oooh. Now, see, I have a vague recollection that Barnard's Loop sets neatly over the large mound at the alpha site - am I remembering correctly?
I have this image of BLoop at site one - its labelled rising, but my memory keeps saying, no, it was setting.
Either way this is BLoop on the horizon at SITE ONE, in line with the triangular platform holding obelisk cluster G:
UPDATE:
I collect images like these at both sites when I can, in the hope that any revealed alignments might help locate further sites, and/or back engineer a way to get the 4 named systems out of studying the ruins (not saying the latter is possible, just hoping).
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Right. I'm going to sleep on it all. I think I'm too tired
o7
I know the feeling, going to try and squeeze another hour in, see if the moon wins that race
He was within 2 degrees of each at the new site... i think he is on on to it....
Synuefe ZL-J d10-119 Is going to set or rise where the new site moon will rise...
I believe this site and his coords will net us another site
I like the moon rising/setting idea, but doesn't it vary a lot like it does here on earth : https://www.timeanddate.com/moon/canada/london
My video a few pages back shows zl-j behind me as i face that platform with the moon in the sky as well behind me... bernards loop rises and sets in nearly the same points ad the first site... so this and his coords should get us a site in that system
That's a great idea because then you can keep a consistent distance.
Probably, but if it only varies such that it fits within the areas marked by the OUTER circles around groups A and F, I think its enough to narrow the search down significantly.
Along LAT line -29/-30, under the moons orbital track.
I have to sleep soon, otherwise would look to wing up with someone and go look for site 3. I'm hoping some erstwhile CMDR's will manage it before I wake
Site 4 may be harder, as there are no suitable moons in the last system, iirc, but I have a feeling that site may be "special" some how. Not super-duper magic kingdom special, but slightly different from the two we have so far, in some way (not necessarily 'physical').
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In order for the moon to rise and set at the circles the ruins have to be offset from the orbit but located at a point that's along a parallel path.
I'm trying to figure out the ideal distance using existing sites. Unless all the bodies have the same radius I suspect it will be an angle from the center rather than kilometers.
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-29.1 in Lat (site 2) to -32 (preddicted), is that makes 2 degrees ? ...
Ok, we can "approximate" ... but until when ? 15degrees ? Oh wait, that would match the -64 ... Mmmhh maybe you're right.
EDIT : I will wait what the 3rd site has to say before validating or invalidating anything ...
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That is fantastic so we can go out and wings and to have one person fly the path of the moon and the other person offset.
My next task will be to create a formula in Google sheet that can be used for the radius and it will give you the correct offset distance
Alright, so my last post claimed that the sound strings that the obelisks emit are time-aligned. It since then has turned out that while yes they mostly are time-aligned, it seems to have no importance. After even more inspection, I *finally* noticed that there are only 7 different tokens, which are demonstrated below.
At first I thought that L3 also had a variant, which I named L4 and I was already jumping with joy, because base8 numbers fit exactly into 3 binary bits (note the number 3), which would have made a huge amount of sense... alas, it was not to be and I can confirm that indeed there are only the 7 tokens + the [END] block which always stays the same.
So with those tokens we can divide the messages like shown below:
The above string would be: SL1VL2L1SBSBL3L2VL1L2BL3VBHL2L3HL1VHBL1L2L3S[END]
It is interesting to note that my rule about there never being 2 tokens of the same type in a row still holds true! So, CMDR Farran speculated that perhaps it is indeed a language. We know that the Guardians used sign language to communicate, making two of the same sign in a row makes no sense, right? so naturally you can't have 2 tokens in a row... in addition, 7 gestures of some sort can be arranged into strings of gestures/tokens - aka. words! So we think that the strings transmitted by the obelisks are indeed words (or even sentences). So, now we definitely need some linguistics people working on this!
EDIT: I will post the strings I have tokenized -
VBL1HVBL1SHL2BL2VL2HBL1VL1BL1BHBHVL3SVSHSHBVL2L3L2VL2VL1[END]
SL1VL2L1SBSBL3L2VL1L2BL3VBHL2L3HL1VHBL1L2L3S[END]
SVSL3HVBSL3BL1BSL1L2BL2HBL3L1L3SHL2L3HL1HL3BL1L2L3SL1V[END]
VBL3BHVL2SBVBL1[END]
HL3HL3HVL1L3HBHL2LBL1HVBHBL1VHL2SLL2HVL1SHL2L1SL3VL3HVL2L1HLHSL1B[END]
VBVL2SHVSL3VHL1L2LL1L2BL3SHSL3HBHSV[END]
SVHL3L1L2L3B[END]
HSL2L1L3SL2SVHL2HL2L3VSHSL1L3L1VSVL1VL1VBVL1L3BL2HS[END]
L2L1BL2SL2VL3BVBSL3SL3L1L3L1VHVL3L2BSL3L1L2L3SL2VL2VBL3L1VL1H[END]
L1VL3BL2L3BSBVH[END]
L2L3L1VHL1HL2VBL3SVL3VSVL1L3SL2L3VHB[END]
L1SBL1L3SBL2BVL3H[END]
L1HVHBL3L1BL3SVL3VHSL3VL3VL3HL2HL2BL2L1[END]
VHL1L3L1BSVSL3VL2[END]
L2HL1L3VL3L2SL1L3L1B[END]
BHL3L1SVL1HL3HL2[END]
L1L2HBSVL3L1VSL3L2VHB[END]
VL3HL3BSBL3BL1SL3HL2[END]
HVHBL2SL2L3SHVHSHSL1BL1BL1L3L1HBVHL1HL3BSBSL1VHBL2[END]
VL3L2HL2HL3L1VSL3VHVB[END]
VL3BL2HSVL1[END]
L3L1SVBHL3SL1BL1HL1SVL1BL3BL2VSVL3VL2BSBH[END]
~CMDR Clanga
At first I thought that L3 also had a variant, which I named L4 and I was already jumping with joy, because base8 numbers fit exactly into 3 binary bits (note the number 3), which would have made a huge amount of sense... alas, it was not to be and I can confirm that indeed there are only the 7 tokens + the [END] block which always stays the same.
So with those tokens we can divide the messages like shown below:
The above string would be: SL1VL2L1SBSBL3L2VL1L2BL3VBHL2L3HL1VHBL1L2L3S[END]
It is interesting to note that my rule about there never being 2 tokens of the same type in a row still holds true! So, CMDR Farran speculated that perhaps it is indeed a language. We know that the Guardians used sign language to communicate, making two of the same sign in a row makes no sense, right? so naturally you can't have 2 tokens in a row... in addition, 7 gestures of some sort can be arranged into strings of gestures/tokens - aka. words! So we think that the strings transmitted by the obelisks are indeed words (or even sentences). So, now we definitely need some linguistics people working on this!
EDIT: I will post the strings I have tokenized -
VBL1HVBL1SHL2BL2VL2HBL1VL1BL1BHBHVL3SVSHSHBVL2L3L2VL2VL1[END]
SL1VL2L1SBSBL3L2VL1L2BL3VBHL2L3HL1VHBL1L2L3S[END]
SVSL3HVBSL3BL1BSL1L2BL2HBL3L1L3SHL2L3HL1HL3BL1L2L3SL1V[END]
VBL3BHVL2SBVBL1[END]
HL3HL3HVL1L3HBHL2LBL1HVBHBL1VHL2SLL2HVL1SHL2L1SL3VL3HVL2L1HLHSL1B[END]
VBVL2SHVSL3VHL1L2LL1L2BL3SHSL3HBHSV[END]
SVHL3L1L2L3B[END]
HSL2L1L3SL2SVHL2HL2L3VSHSL1L3L1VSVL1VL1VBVL1L3BL2HS[END]
L2L1BL2SL2VL3BVBSL3SL3L1L3L1VHVL3L2BSL3L1L2L3SL2VL2VBL3L1VL1H[END]
L1VL3BL2L3BSBVH[END]
L2L3L1VHL1HL2VBL3SVL3VSVL1L3SL2L3VHB[END]
L1SBL1L3SBL2BVL3H[END]
L1HVHBL3L1BL3SVL3VHSL3VL3VL3HL2HL2BL2L1[END]
VHL1L3L1BSVSL3VL2[END]
L2HL1L3VL3L2SL1L3L1B[END]
BHL3L1SVL1HL3HL2[END]
L1L2HBSVL3L1VSL3L2VHB[END]
VL3HL3BSBL3BL1SL3HL2[END]
HVHBL2SL2L3SHVHSHSL1BL1BL1L3L1HBVHL1HL3BSBSL1VHBL2[END]
VL3L2HL2HL3L1VSL3VHVB[END]
VL3BL2HSVL1[END]
L3L1SVBHL3SL1BL1HL1SVL1BL3BL2VSVL3VL2BSBH[END]
~CMDR Clanga
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