Nazalassa

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4-vertex triangles are called tetrahedrons.

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If it's Hyperedit, I know it exists for KSP 0.22 (and where to find it). However, I think the problem is because of KSP using simple-precision floats, which are not that precise when you go that far.

They're playing KSP Is @cantab around?

He's playing Number Wars @Zozaf Kerman?

Haha N -18

Ah, I must've misunderstood. But if this is a cheat that works in KSP 1.11, I'm going to try it in 1.11 -- maybe the SSTV is here.

It may be a good setting for a fantasy/SF story

We're all What's the mod anyways?

N -18 I like -∞ better

I got bored in class yesterday, and let my imagination tell me stories about physics, rockets, wizards and magic, exploration, discoveries... And then I somehow imaginated a tetrahedral planet. A big tetrahedron in space, orbiting a star. My math brain immediately wanted to know how gravity (and that sort of things) would work on such an hypothetical planet. So I took a sheet of paper, and started doing math, searching for triangles in the tetrahedron, applying the Pythagorean theorem... And I got these results. But first, let's define some things... Let's call center [of the tetrahedron] the tetrahedron's center of mass (center of gravity too, which is equidistant to all of the tetrahedron's vertices, and also equidistant to the centers of all of the tetrahedron's faces, and also equidistant to the centers of all of the tetrahedron's edges, you see what I mean). By "edge" I mean the middle of the edge, by "face" I mean the center of the face, and by "adjacent vertex" I mean vertex that is linked to the current vertex by an edge, vertex that belongs to the current edge, or vertex that belongs to the current face (depending wether I'm talking of a vertex, an edge, or a face). If one unit = the side of the tetrahedron The distance between a vertex the center the center the center an edge a face a face and an adjacent vertex a vertex an edge a face an adjacent vertex an adjacent vertex one of its edges is 1 √(6) / 4 √(2) / 4 √(6) / 12 √(3) / 2 √(3) / 3 √(3) / 6 If one unit = the distance between the center of the tetrahedron and one of its faces The distance between a vertex the center the center the center an edge a face a face and an adjacent vertex a vertex an edge a face an adjacent vertex an adjacent vertex one of its edges is 2 * √(6) 3 √(3) 1 3 * √(2) 2 * √(2) √(2) If one unit = the distance between the center of the tetrahedron and one of its vertices The distance between a vertex the center the center the center an edge a face a face and an adjacent vertex a vertex an edge a face an adjacent vertex an adjacent vertex one of its edges is 2 * √(6) / 3 1 √(3) / 3 1 / 3 √(2) 2 * √(2) / 3 √(2) / 3 (Note that there may be errors, and that you can go from one table to another with a simple multiplication.) So, as gravity is inversely proportionnal to the distance to the center of mass, that would mean (if we look at the second table) that the gravity at the center of a face is nine times stronger than the gravity at a vertex? Cool... Now you can "walk" into space, and also to a place were gravity is waaay weaker. However, as you progress towards the edges (and vertices) of a face, the apparent "slope" will increase and increase, right? Because the angle between the plane orthogonal to the direction of gravity and the actual surface increases. I have calculated some slopes: Sloppy slopes, yey The slope at a vertex an edge a face is (roughly) arccos(1 / 3) ≈ 70.53° arccos(√(3) / 3) ≈ 54.74° arccos(1) = 0° So a vertex looks like a mountain now. That's impractical for rocket launches. Fortunately, the gravity there is a ninth of the gravity at the center of a face.

Meow. @Maria Sirona how are you?

I suggest you give it a test! If you do, don't forget to record the whole thing so we can make a video after -- and decode ther thing too :)

Not anyone have KSP 0.17.

Wait, we're on Danny's page and nobody said it?

Procedural music!

No, in 80 minutes

I meant, I have to install Steam on my computer. I already have an account but I lost the Steam application itself a while ago... I don't use Steam now.

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3571

Just to see what the forums think. Nothing more. Happy Xplosions!

EDIT: idea may take a bit more time due to the need of a Steam client.

3566 = 2 * 1783

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