egg changed the topic of #principia to: Logs: https://esper.irclog.whitequark.org/principia | <scott_manley> anyone that doubts the wisdom of retrograde bop needs to get the hell out | https://xkcd.com/323/ | <egg> calculating the influence of lamont on Pluto is a bit silly…
<paculino> Does the spamming of ChanServ mode sets indicate a problem?
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<umbralraptop> Uh, was I causing join-spam issues?
<queqiao-> ⟨lpg⟩ no, rp-1 does
<queqiao-> ⟨zephyr⟩ Isn't it tracking station once mission control twice
<queqiao-> ⟨zephyr⟩ * twice?
<paculino> Sorry raptop, I thought it might have meant something since there was a gap in messages from the bridge around the same time.
<queqiao-> ⟨lpg⟩ ⟪zephyr⟫ Isn't it tracking station once mission […] ⮪ err. yes, that
<queqiao-> ⟨Champ0220⟩ ⟪lpg⟫ no, rp-1 does ⮪ Uh I’m not playing rp-1 I’m playing mods and one of them is principia
<queqiao-> ⟨lpg⟩ answer's still no
<queqiao-> ⟨zephyr⟩ ⟪Champ0220⟫ Uh I’m not playing rp-1 I’m playing […] ⮪ It's also a _stock_ ksp thing
<queqiao-> ⟨zephyr⟩ There's a reason visual moonshots are so popular there
<queqiao-> ⟨nvv⟩ https://forum.kerbalspaceprogram.com/index.php?/topic/99228-win-flyby-finder-for-real-solar-system-mod-v085-rss120-for-ksp122/ would this still _somewhat_ work with principia?
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<queqiao-> ⟨ZombieZilla⟩ I think KSPTOT works
<queqiao-> ⟨TheSquirrelPatrol⟩ Is there any way to get a really long flight plan? I've set steps and tolerance both to max, but can't get it to show the saturn encounter in 5 years that I'm trying to get from a TWP plot
<queqiao-> ⟨test_account⟩ One reason could be that you are out of RAM? How much do you have?
<queqiao-> ⟨TheSquirrelPatrol⟩ 32gb, and I'm only at 18gb used
<queqiao-> ⟨Stonesmile⟩ Is the plan length set to the long time too?
<queqiao-> ⟨TheSquirrelPatrol⟩ That's not actually max, ive been playing around
<queqiao-> ⟨TheSquirrelPatrol⟩ it may be that I hadn't actually put in the manuever yet
<queqiao-> ⟨TheSquirrelPatrol⟩ Yeah, once i'm not just orbiting earth, it seems much happier
<queqiao-> ⟨test_account⟩ Max steps 1024 seems really low and tolerance 10km probably too high
<queqiao-> ⟨test_account⟩ What I do is gradually increase plan length and max steps per segment
<queqiao-> ⟨Mayo⟩ ⟪TheSquirrelPatrol⟫ unknown.png ⮪ 1024 is really low segment
<queqiao-> ⟨Mayo⟩ and the tollerance is fuckhuge
<queqiao-> ⟨Mayo⟩ i can get to like 16 thousand without my pc dying
<queqiao-> ⟨Butcher⟩ Smart maths people (hi egg).
<queqiao-> I have a vector, v; I apply two rotations by forming two unit quaternions, q1, q2 and multiplying them to give q3, then using q3 to rotate v to give v'.
<queqiao-> This all seems fine, however from this I would expect that v.v' would be the same as 2 * acos(q3.w), but it's a little off. What have I messed up?
<queqiao-> ⟨§κyℓαr ♀⟩ you missed the cat chewing on your maths /j
<queqiao-> ⟨Stonesmile⟩ ⟪Butcher⟫ Smart maths people (hi egg). I have a […] ⮪ Is w the same in unity as your source?
<queqiao-> ⟨Butcher⟩ Yes, w is the real component of the quat.
<queqiao-> ⟨Stonesmile⟩ It does seem like it should be the same... is it a small difference, or completely wrong?
<queqiao-> ⟨Butcher⟩ About a degree off, so non-trivial.
<queqiao-> ⟨egg⟩ ⟪Butcher⟫ Smart maths people (hi egg). I have a […] ⮪ that v.v' would be the same as 2 * acos(q3.w)
<queqiao-> The inner product is the product of the norms times the cosine of the angle; if the cosine of some angle is the arccosine of something, something is weird. you want cos (2 arc cos w), which is 2w²-1.
<queqiao-> ⟨Butcher⟩ The discrepancy varies with the amount of rotation applied by q2.
<queqiao-> ⟨egg⟩ * 2w²−1.
<queqiao-> ⟨Butcher⟩ egg Sorry, yes you are correct, i was actually comparing, acos(v.v') with 2 * acos(q3.w).
<queqiao-> ⟨Butcher⟩ v and v' both being unit vectors.
<queqiao-> ⟨egg⟩ can you check that q3 is actually normalized? Silliness with quaternions drifting away from the unit sphere has bitten us in Principia
<queqiao-> ⟨test_account⟩ I know nothing about quaternions, however, could it be that this phrase from wikipedia contains the answer: "Multiplication of quaternions is noncommutative." 🤔
<queqiao-> ⟨egg⟩ No.
<queqiao-> ⟨Butcher⟩ Yes, I have verified that q3 is normalized.
<queqiao-> ⟨Butcher⟩ Actually, let me double check that.
<queqiao-> ⟨egg⟩ Hm. Can you give me an example of a q3, v, and v′ exhibiting the weirdness?
<queqiao-> ⟨Butcher⟩ v= 0,-1,0
<queqiao-> v'= -0.121979430317879,-0.701124727725983,-0.702527701854706
<queqiao-> q3= 0.374441915922073,-0.0730170878691503,-0.0960776455097884,0.919364368742346
<queqiao-> ⟨egg⟩ ⟪Butcher⟫ Smart maths people (hi egg). I have a […] ⮪ oh wait, the issue isn’t your calculations (aside from the already-discussed typo in that message), it’s your expectation
<queqiao-> ⟨egg⟩ The angle between the two vectors is measured in the plane that they span, but the angle of the rotation is measured about its axis. So this just tells us that v isn’t perpendicular to the axis of q3.
<queqiao-> ⟨egg⟩ (indeed if v were the axis of q3, the angle between v and v′ would be 0)
<queqiao-> ⟨Butcher⟩ Hrm. I thought it might be something like this.
<queqiao-> ⟨Butcher⟩ I'm trying to solve an inverse problem really, I thought this might help. Guess not.
<queqiao-> ⟨egg⟩ (whenever you are confused by quaternions, you can bet that you are really being confused by rotations, regardless of their representation; rotations are weird.)
<queqiao-> ⟨Butcher⟩ Essentially I'm trying to solve the problem of - given a target vector (pointing at say the moon) and a latitude, I would like an orbital plane that with inclination equal to my latitude that contains the target vector.
<queqiao-> ⟨rnlahaye⟩ This can be solved using spherical trigonometry.
<queqiao-> ⟨Butcher⟩ Possibly, but that's not really my speciality. :-p
<queqiao-> ⟨Stonesmile⟩ Wouldn't applying the rotation to a unit vector and comparing that to the original unit vector give you the wanted angle?
<queqiao-> ⟨Butcher⟩ I don't know how much rotation to apply.
<queqiao-> ⟨egg⟩ It isn’t really pretty, but in equatorial coordinates, if (x, y, z) are the coordinates of the normal to your plane, and (cos α cos δ, sin α cos δ, sin δ) are those of the (earth-centred) target vector, and i is your desired inclination, it looks like you get something like
<queqiao-> x = ± |sin α| √(cos 2δ − cos 2i) / ( √2 cos δ ) − cos i cos α tg δ
<queqiao-> y = ∓ |sin α| √(cos 2δ − cos 2i) / ( √2 tg α cos δ ) − cos i sin α tg δ
<queqiao-> z = cos i
<queqiao-> (solving for (x, y, z) being at an angle i from the polar axis (0, 0, 1), orthogonal to the target vector (cos α cos δ, sin α cos δ, sin δ), and normalized.)
<queqiao-> ⟨egg⟩ That should simplify to
<queqiao-> x = ± sin α √(cos 2δ − cos 2i) / ( √2 cos δ ) − cos i cos α tg δ
<queqiao-> y = ∓ cos α √(cos 2δ − cos 2i) / ( √2 cos δ ) − cos i sin α tg δ
<queqiao-> which should be a little less ugly.
<queqiao-> ⟨nvv⟩ When I calculate my LAN, what would the reference line be?
<queqiao-> ⟨Butcher⟩ Payment for egg.
<queqiao-> ⟨siimav⟩ The cat looks unamused by that bribe.
<umbralraptop> I feel like egg would prefer EUR or CHF rather than GBP
<umbralraptop> (dunno about how the cat feels)
<queqiao-> ⟨Butcher⟩ The cat prefers meat.
<queqiao-> ⟨sichelgaita⟩ That's a trap, these banknotes will be retired soon.
<queqiao-> ⟨QeDelphyn (Xšayāršā/-ām/-āyā)⟩ >fiat is worth anything
<queqiao-> ⟨QeDelphyn (Xšayāršā/-ām/-āyā)⟩ >fiat
<queqiao-> >worth anything
<queqiao-> ⟨nvv⟩ When I calculate my LAN, what would the […] ⮪ ?
<umbralraptop> I'd assume the first point of aries
<queqiao-> ⟨test_account⟩ But what does it mean - calculate my LAN? It is shown by various tools, no need to calculate 🤔
<queqiao-> ⟨nvv⟩ I know, but I'm asking in general. Where is it measured from?
<queqiao-> ⟨test_account⟩ It should be first point of aries, but this is quite meaningless in game, I don't think anything shows that direction in the interface at least
<queqiao-> ⟨test_account⟩ -at least
<queqiao-> ⟨Butcher⟩ In RSS it's the first point of Ares.
<umbralraptop> Unfortunately, Ares tends to move around a lot when compared with Aries
<queqiao-> ⟨test_account⟩ Aries isn't in Aries either...
<umbralraptop> That's fishy (pun intended)
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