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I'm confused. It's based on arch but not really? Is it arch based or not? Does it use any arch package manager? The post raises a number of new questions
The answers to that seem pretty obvious to me: yes, it is based on arch. No, it does not come with a package manager. Presumably, they use Arch packaging tools and package definitions behind the scenes in some way, but the end result is a premade immutable system image.
I really hope not, that feels like crypto all over again, with inconsistent payouts and varying electricity prices... And on top of that probably awful service since people tend to have the weirdest internet connections.
Though if you remove the part where it's used to stream games to other players, that sounds too niche to be viable, but could be cool. If going in that direction, I'd imagine it more likely to be gaming servers for businesses, like VR gaming spots, where they have multiple gaming computers hooked up to headsets.
I will point out that it generally takes more effort to make one sandwich, than it takes to make one sandwich out of a hundred. Getting fresh bread and (fresh?) spiced mayo is extra work that you only need to do once per multiple sandwiches, but it doesn't really get easier when only making one.
All I'm saying is, enjoy things made by specialized professionals, economies of scale mean that it's more efficient for one person to make sandwiches for their surroundings anyways!
I'm not a soulslike fan myself, but I don't think hollow knight is very soulslike - the combat is very snappy, avoiding locking you into animations or making you consider your momentum, and I have the impression soulslikes also tend to be way more environmentally lethal, so to speak.
It might have some of that visual/lore/exploration vibe though.
So, the issue is, as far as I know the calculations are dead simple - you "enter" and "exit" the planet's influence at the same distance from the planet, which means your potential gravitational energy didn't change, so from the orbital mechanics point of view, from the planet's frame of reference, your velocity should stay the same.
As you "fall" in the orbit around the planet, you're converting potential gravitational energy to kinetic energy, but as you "climb" you convert it back into potential gravitational energy, ending with the same amount of each kind of energy. The only change is that the velocity is redirected.
With that in mind, it's why, from my knowledge, the equations are really simple, with the only complications being trigonometry (to resolve the angles) and pythagoras (squaring, adding and getting the square root make the result unintuitive).
Going back to your graph, if I were to do the math, according to my theory:
If you want an intuitive example of what I'm referring to, consider a planet approaching you as you are stationary relative to the sun. If we assume ideal, presumably impossible, entry and exit angles of 0° and 180°, leaving the planet's gravity field moving in the exact opposite direction than what you entered, you'll note you'll be gaining speed on exit either way, despite not moving towards the planet on the approach and "catching up".
The graph doesn't really show anything other than illustrate your thoughts - but there's absolutely nothing backing that as being true :/
Either way, it does feel like we're going around in circles, and I don't want to be taking up your time unnecessarily. If you have something to disprove my math (maybe my understanding of orbital dynamics is wrong, and it's not that simple), that'd be a starting point to try to figure out what's wrong; if you're interested, I could try to make diagrams, though I feel like they might kind of look the same, just with different numbers based on calculations.
I guess one last thing I can offer is a video somebody replied to me with elsewhere in the thread, explaining this idea: https://youtube.com/shorts/kD8PFhj_a8s
They said cheaper than vanilla extract, the synthetic stuff is probably still cheaper. Though there is a possibility, since the materials and production will have a cost...
Well, relative motions are more intuitive to me - they make sense, and I can use calculations for them.
In the first example, you presented 101 speed - this means only 1 speed relative to the planet, and that's all that's getting redirected (in the planet's frame of reference your enter and exit velocity should be the same, since that's how orbits work). The number is just too small, but your velocity would be planet velocity + 1 on a different vector, which will be less than 101 total.
If we estimate the angle on the picture is about 60 degrees from the velocity vector, and the speed to be 100+v1, the speed from the planet's frame of reference is v1 - so, the exit velocity will have components of (100+v1cos(60°)) and (v1sin(60°)), so the final speed relative to the sun should be
sqrt((100+x*cos(60°))^2+(x*sin(60°))^2)
Wolfram alpha suggests this simplifies to sqrt(x(x+100)+10000), and comparing the equation by appending <x+100 gives the solution of x>0
This means, if my math is correct, with an entry angle of 0° and exit angle of 60°, you always lose speed.
I could try replacing the angle with a variable and setting a constraint of x>0 and see if the free version of wolfram alpha would spit out something, but just replacing the 60 with y is spitting out some convincing solutions, since in those x is never greater than 0.
PS: Thanks for taking the time to explain, the fact that you went out of your way to draw the diagrams is not lost on me. I unfortunately still don't see it, but I really do appreciate the effort!
I'm sorry, but this comment thread genuinely makes me feel like I'm going insane. You seem to have explained exactly the same thing as me, with the same example, and none of it includes the "fall for longer before you catch up" bit.
As for the orbit not curving, yeah, I think you're right - the obvious case is if you're sitting stationary on the planet's orbit, but the curious case is if you're approaching from the sun, with the planet's velocity plus velocity away from the sun. If I'm not mistaken, in that case you'd end up with the same velocity (minus what you might have lost to the sun's gravity), but on the other side of the planet's gravity well, which means you still gained energy.
Ayy, I'm not crazy, that sounds like exactly what I described... The only question is, is the explanation of "you spend longer falling" is bs, or if it makes sense if you conceptualize it differently?
Right, but as I explained, it's the how that doesn't make sense to me - the explanation that you "fall for longer" doesn't make sense, since 1. with how orbits work, it takes the same energy and time to "fall" as it does to ascend, and 2. at these scales you can use the planet as an inertial frame of reference, so the angle of approach doesn't matter for how "long" you "fall", it'll be the same regardless of whether you're moving towards or away from the planet.
I'm curious - was it also a checkbox that immediately applied when toggled, instead of not actually applying until you press save?