Alien archeology and other mysteries: Thread 10 - The Canonn

[Maligno]

So to my mind the only sure way is to use mathematics to calculate a set of co-ordinates from the data set, and look to the nearest system to those co-ordinates. Once there is a mathematical method, it's just a case of simple arithmetical number-crunching. The difficult bit is working out the method. I'm not even sure it's possible.

It's possible, CMDR Wace has done it (https://forums.frontier.co.uk/showt...0-The-Canonn?p=5671013&viewfull=1#post5671013). You may want to dig deeper into this megathread to see further details from Wace.

In case you didn't know, here's a nice tool he put together that crunces all the numbers: http://jubjubnest.net:8001/

 

[Commander William Diffin]

It's possible, CMDR Wace has done it (https://forums.frontier.co.uk/showt...0-The-Canonn?p=5671013&viewfull=1#post5671013). You may want to dig deeper into this megathread to see further details from Wace.

In case you didn't know, here's a nice tool he put together that crunces all the numbers: http://jubjubnest.net:8001/

+1 Nice. Thanks. It looks like CMDR Wace has written a program that creates formulae for all of the points on the spheres surrounding Merope, HIP 14909 and Col 70 Sector FY-N c21-3 of identical radii to the given distances, then works out the points where all three circles intersect. I'd do it myself if I knew how.

This still isn't any kind of a game though.

 

[Thatchinho]

+1 Nice answer but not what I'm looking for. I've tried that and can't be bothered to sit around waiting for EDDB to work out the distances to Merope, or HIP 14909, or Col 70 Sector FY-N c21-3, and then guess which page of results the systems at the correct distance are on, then have to wait for EDDB to work out the distances for the systems on that page again, then find out it's the wrong page, then guess another page, etc., all whilst I should have been playing the game. And then what if a system isn't even on EDDB? It's not a foolproof method.

Actually flying to the system and checking the distances of surrounding systems is probably the best idea for avoiding the mathematics. But then you can't fly to Col 70 Sector FY-N c21-3 as it's Permit Only.

So to my mind the only sure way is to use mathematics to calculate a set of co-ordinates from the data set, and look to the nearest system to those co-ordinates. Once there is a mathematical method, it's just a case of simple arithmetical number-crunching. The difficult bit is working out the method. I'm not even sure it's possible.

So now it's an obsession. I want a purely mathematical method to calculate a set of co-ordinates from the UL data, and see if the co-ordinates don't match up with the results of those who could be bothered to use EDDB or whatever. So. I repeat. Are there any mathematicians who can tell me to how to solve these equations for x, y and z please?

1) x² + y² + z² + 157.1875x + 299.25y + 681.0625z = -127228.64
2) x² + y² + z² + 168.75x + 518.5625y + 717.3125z = -201766.3873
3) x² + y² + z² - 1374.125x + 725.0625y + 1394.125z = -426132.555

By subtracting 1) from 2), 1) from 3), and 2) from 3) I can eliminate the second degrees:

4) 11.5625x + 219.3125y + 36.25z = -74537.7473
5) -1531.3125x + 425.8125y + 713.0625z = -298903.915
6) -1542.875x + 206.5y + 676.8125z = -224366.1677

And I can eliminate one of the variables by multiplying an equation like so:

Multiply 5) by (36.25 / 713.0625):
5B) -77.84742308x + 21.64705496y + 36.25z = -15195.3958

Subtract 5B) from 4) to eliminate z:
7) 89.40992308x + 197.665445y = -59342.3515

But any other equation in terms of just x and y that I make using the other equation just yields a proportionally identical equation:

Multiply 6) by (36.25 / 676.8125):
6B) -82.63620833x + 11.06011635y + 36.25z = -12017.02625

Subtract 6B) from 4) to eliminate z:
8) 94.19870833x + 208.2523837y = -62520.72105

7) and 8) are practically identical equations and it isn't possible to deduce a value for x or y from them.

I need help.

Not a mathematician myself, but looks to me like you're probably most of the way there already.

There should be 2 solutions to the equations, and you've now got 2 equations for x (or y if you prefer!).

For the next steps I'd suggest now treating them separately and substituting them back into the equations above. Work through completely with x1 and see if you can find the first solution and then work through completely with x2 and find the second solution.

Disclaimer: I haven't actually checked this is correct, and haven't got time to do all the workings right at the moment, so there's a risk I could be sending you down the wrong path. It's the path I'd be taking myself though.

 

[viper1209]

+1 Nice answer but not what I'm looking for. I've tried that and can't be bothered to sit around waiting for EDDB to work out the distances to Merope, or HIP 14909, or Col 70 Sector FY-N c21-3, and then guess which page of results the systems at the correct distance are on, then have to wait for EDDB to work out the distances for the systems on that page again, then find out it's the wrong page, then guess another page, etc., all whilst I should have been playing the game. And then what if a system isn't even on EDDB? It's not a foolproof method.

Actually flying to the system and checking the distances of surrounding systems is probably the best idea for avoiding the mathematics. But then you can't fly to Col 70 Sector FY-N c21-3 as it's Permit Only.

So to my mind the only sure way is to use mathematics to calculate a set of co-ordinates from the data set, and look to the nearest system to those co-ordinates. Once there is a mathematical method, it's just a case of simple arithmetical number-crunching. The difficult bit is working out the method. I'm not even sure it's possible.

So now it's an obsession. I want a purely mathematical method to calculate a set of co-ordinates from the UL data, and see if the co-ordinates don't match up with the results of those who could be bothered to use EDDB or whatever. So. I repeat. Are there any mathematicians who can tell me to how to solve these equations for x, y and z please?

1) x² + y² + z² + 157.1875x + 299.25y + 681.0625z = -127228.64
2) x² + y² + z² + 168.75x + 518.5625y + 717.3125z = -201766.3873
3) x² + y² + z² - 1374.125x + 725.0625y + 1394.125z = -426132.555

By subtracting 1) from 2), 1) from 3), and 2) from 3) I can eliminate the second degrees:

4) 11.5625x + 219.3125y + 36.25z = -74537.7473
5) -1531.3125x + 425.8125y + 713.0625z = -298903.915
6) -1542.875x + 206.5y + 676.8125z = -224366.1677

And I can eliminate one of the variables by multiplying an equation like so:

Multiply 5) by (36.25 / 713.0625):
5B) -77.84742308x + 21.64705496y + 36.25z = -15195.3958

Subtract 5B) from 4) to eliminate z:
7) 89.40992308x + 197.665445y = -59342.3515

But any other equation in terms of just x and y that I make using the other equation just yields a proportionally identical equation:

Multiply 6) by (36.25 / 676.8125):
6B) -82.63620833x + 11.06011635y + 36.25z = -12017.02625

Subtract 6B) from 4) to eliminate z:
8) 94.19870833x + 208.2523837y = -62520.72105

7) and 8) are practically identical equations and it isn't possible to deduce a value for x or y from them.

I need help.

I myselfe ever prefered to solve calculations like this by substitution. So transforming one equation to one of the variables and substitute this variable in the other two equations by this.
This way you will get two equations with two variable. if you now again transform one of them to one of the variables and substitute again you get the last variable.
Now you can calculate all other values.

To make things short I solved your equations using Wolfram Alpha and get this results:
1) x=-55.3206 - 3.92052 i, y=-275.193 + 1.77337 i, z=-373.651 - 9.47836 i
2) x=-55.3206 + 3.92052 i, y=-275.193 - 1.77337 i, z=-373.651 + 9.47836 i

As for the variables power 2 you get a complex result but as you are calculating "real" physics you probably can just forget about the imaginary part leading to this:
x=-55.3206, y=-275.193, z=-373.651

 

[sircalist]

Can't find HIP 18368 2H

CMDR sircalist here,
I'm quite new to the exploration, so I may say something stupid, but I can't find the planet HIP 18368 2H, where a Thargoid surface site should be. The system has only one planet, HIP 18368 1, with several satellites. This confuses me.
Am I missing something?

 

[Jmanis]

I myselfe ever prefered to solve calculations like this by substitution. So transforming one equation to one of the variables and substitute this variable in the other two equations by this.
This way you will get two equations with two variable. if you now again transform one of them to one of the variables and substitute again you get the last variable.
Now you can calculate all other values.

To make things short I solved your equations using Wolfram Alpha and get this results:
1) x=-55.3206 - 3.92052 i, y=-275.193 + 1.77337 i, z=-373.651 - 9.47836 i
2) x=-55.3206 + 3.92052 i, y=-275.193 - 1.77337 i, z=-373.651 + 9.47836 i

As for the variables power 2 you get a complex result but as you are calculating "real" physics you probably can just forget about the imaginary part leading to this:
x=-55.3206, y=-275.193, z=-373.651

Not doubting the correctness of your maths here, but I would like to highlight that the recorded location of that particular signal being used in the example is x=-57.125, y=-274.128, z=-373.687

Point being there is a margin of error large enough to mean you'll still have multiple candidate systems and need to do some trial and error for several cases.

 

[Thatchinho]

I myselfe ever prefered to solve calculations like this by substitution. So transforming one equation to one of the variables and substitute this variable in the other two equations by this.
This way you will get two equations with two variable. if you now again transform one of them to one of the variables and substitute again you get the last variable.
Now you can calculate all other values.

To make things short I solved your equations using Wolfram Alpha and get this results:
1) x=-55.3206 - 3.92052 i, y=-275.193 + 1.77337 i, z=-373.651 - 9.47836 i
2) x=-55.3206 + 3.92052 i, y=-275.193 - 1.77337 i, z=-373.651 + 9.47836 i

As for the variables power 2 you get a complex result but as you are calculating "real" physics you probably can just forget about the imaginary part leading to this:
x=-55.3206, y=-275.193, z=-373.651

Not disputing the maths for that solution either, but pretty sure there should be a second solution which has gone missing somewhere.

Why: (Spoilered for the sake of those not interested in this kind of stuff!)

Spoiler

Geometrically in 3D, co-ordinates & distance defines a sphere (Except special case: d=0 defines a point). Then, (assuming the spheres do actually intersect):

A. The intersection of 2 spheres defines a circle.

B. Intersection of 3 spheres = intersection of a sphere and a circle, which defines two points.

Special cases obviously exist which define a single point (for example, when the sum of the radii of 2 spheres exactly equals the distance between their centres.) but in general there should be 2 points.

 

[viper1209]

Not disputing the maths for that solution either, but pretty sure there should be a second solution which has gone missing somewhere.

Why: (Spoilered for the sake of those not interested in this kind of stuff!)

Spoiler

Geometrically in 3D, co-ordinates & distance defines a sphere (Except special case: d=0 defines a point). Then, (assuming the spheres do actually intersect):

A. The intersection of 2 spheres defines a circle.

B. Intersection of 3 spheres = intersection of a sphere and a circle, which defines two points.

Special cases obviously exist which define a single point (for example, when the sum of the radii of 2 spheres exactly equals the distance between their centres.) but in general there should be 2 points.

Your are right, there should be a second solution. Totally forgot about it.

Math spoiler again^^

Spoiler

I can imagine where it went missing. I used those equations:

1) x² + y² + z² + 157.1875x + 299.25y + 681.0625z = -127228.64
2) x² + y² + z² + 168.75x + 518.5625y + 717.3125z = -201766.3873
3) x² + y² + z² - 1374.125x + 725.0625y + 1394.125z = -426132.555

but he already did some math since the original ones:

√ ( (x - (-78.59375))² + (y - (-149.625))² + (z - (-340.53125))² ) = 131.52
√ ( (x - (-84.375))² + (y - (-259.28125))² + (z - (-358.65625))² ) = 34.84
√ ( (x - 687.0625)² + (y - (-362.53125))² + (z - (-697.0625))² ) = 814.40

(x + 78.59375)² + (y + 149.625)² + (z + 340.53125)² = 131.52²
(x + 84.375)² + (y + 259.28125)² + (z + 358.65625)² = 34.84²
(x - 687.0625)² + (y + 362.53125)² + (z + 697.0625)² = 814.40²

...

When resolving the square root he should have gotten to this:

+-((x + 78.59375)² + (y + 149.625)² + (z + 340.53125)²) = (+-131.52)²
+-((x + 84.375)² + (y + 259.28125)² + (z + 358.65625)²) = (+-34.84)²
+-((x - 687.0625)² + (y + 362.53125)² + (z + 697.0625)²) = (+-814.40)²


So the +- gives you two sets of equations. Technically it gives you 6 single equations with 3 variables, this is where things get tricky as you have more equations then variables which can lead to problems when resolving.

 

[Squid Gumbo]

CMDR sircalist here,
I'm quite new to the exploration, so I may say something stupid, but I can't find the planet HIP 18368 2H, where a Thargoid surface site should be. The system has only one planet, HIP 18368 1, with several satellites. This confuses me.
Am I missing something?

I'll give it a look when I get home! I would cross-check multiple sources, and if need be move to a different system.

 

[G_Bison]

CMDR sircalist here,
I'm quite new to the exploration, so I may say something stupid, but I can't find the planet HIP 18368 2H, where a Thargoid surface site should be. The system has only one planet, HIP 18368 1, with several satellites. This confuses me.
Am I missing something?

That is the location to US #30 2H Would be 2nd planet, H moon designation. Coords Lat -21.5287 // Long -27.1165

 

[Factabulous]

CMDR sircalist here,
I'm quite new to the exploration, so I may say something stupid, but I can't find the planet HIP 18368 2H, where a Thargoid surface site should be. The system has only one planet, HIP 18368 1, with several satellites. This confuses me.
Am I missing something?

You probably don't have an Advanced Discovery Scanner - the Basic scanner only 'sees' 500 ls, so you can't see the 2nd planet. Either go to a closer site (CG thread has some references), or buy an Advanced Discovery Scanner

Edit: Found my post in the CG thread:

Sounds like you only have a basic discovery scanner - you want an Advanced Discovery Scanner to see the whole system.

You could try US121 : HYADES SECTOR PC-V B2-3 Planet 2 A LAT -7.0555 LNG 144.1756 Edit: No - try US173 - ARIES DARK REGION PY-R B4-1 Planet A 4 A
LAT -44.6778 LNG -96.0452 - you should be able to see it with a basic scanner

Good luck!

 

[Jorki Rasalas]

You probably don't have an Advanced Discovery Scanner - the Basic scanner only 'sees' 500 ls, so you can't see the 2nd planet. Either go to a closer site (CG thread has some references), or buy an Advanced Discovery Scanner

Edit: Found my post in the CG thread:



Good luck!

@sircalist. There is a way to find planets visually without an Advanced Discovery Scanner ( we all had to do it in the early days). See https://m.youtube.com/watch?v=itkIhPRxAPQ

edit: I have a vague recollection of FD (MB or DB?) saying that some bodies are not discoverable by the scanner, but I can't find the reference. Does anyone else remember this?

 

[Jmanis]

edit: I have a vague recollection of FD (MB or DB?) saying that some bodies are not discoverable by the scanner, but I can't find the reference. Does anyone else remember this?

That was mentioned in the same breath as things like "Raxxla is in the game" and the likes iirc.

 

[Factabulous]

@sircalist. There is a way to find planets visually without an Advanced Discovery Scanner ( we all had to do it in the early days). See https://m.youtube.com/watch?v=itkIhPRxAPQ

edit: I have a vague recollection of FD (MB or DB?) saying that some bodies are not discoverable by the scanner, but I can't find the reference. Does anyone else remember this?

As if by magic: Raxxla found: https://forums.frontier.co.uk/showthread.php/370871-Bug-or-Feature-Planet-that-doesn-t-exist (yeah, ok, probaby not Raxxla)

(and while I'm sure Jorki is right and you can find planets using mk I eyeball I don't believe this was used to find any of the Shipment locations, so my 'buy an ADS' comment is more correcterer )

 

[Lori Jameson]

As if by magic: Raxxla found: https://forums.frontier.co.uk/showthread.php/370871-Bug-or-Feature-Planet-that-doesn-t-exist (yeah, ok, probaby not Raxxla)

(and while I'm sure Jorki is right and you can find planets using mk I eyeball I don't believe this was used to find any of the Shipment locations, so my 'buy an ADS' comment is more correcterer )

If it is, I'd be sorely disappointed... Also, I've yet to disprove Shintara as Raxxla. It is either correct or I have yet to find info that points fully elsewhere. It annoys me.

 

[Jorki Rasalas]

That was mentioned in the same breath as things like "Raxxla is in the game" and the likes iirc.

That's what I thought. I've been dipping into the Quest for Raxxla thread & looking out that visual planet finding link for Sircalist brought it to mind.

 

[Jorki Rasalas]

As if by magic: Raxxla found: https://forums.frontier.co.uk/showthread.php/370871-Bug-or-Feature-Planet-that-doesn-t-exist (yeah, ok, probaby not Raxxla)

(and while I'm sure Jorki is right and you can find planets using mk I eyeball I don't believe this was used to find any of the Shipment locations, so my 'buy an ADS' comment is more correcterer )

Oh I agree, an ADS takes the hard work out of the equation. But if Sircalist can'tyet afford one & is trying to find a US on a distant planet that he can't locate then it can be done by visiual search.

Then I thought maybe the RaXXlers should be using that same technique around the bubble....

 

[Jorki Rasalas]

If it is, I'd be sorely disappointed... Also, I've yet to disprove Shintara as Raxxla. It is either correct or I have yet to find info that points fully elsewhere. It annoys me.

Well all the lore on RaXxla seems to be retconned, but it was supposed to be a planet rather than a system, so an undetectable (by ADS) planet within Shinrarta Dezhra?

 

[Thatchinho]

If it is, I'd be sorely disappointed... Also, I've yet to disprove Shintara as Raxxla. It is either correct or I have yet to find info that points fully elsewhere. It annoys me.

I'm obviously missing a lot of prior discussion, but... why Shinrata? Doesn't make sense to me as a location for Raxxla.

 

[Jorki Rasalas]

I'm obviously missing a lot of prior discussion, but... why Shinrata? Doesn't make sense to me as a location for Raxxla.

This is the wrong thread, but a place that only Elites can access...well-stocked with equipment....Dark Wheel...