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I have been able to easily create a mesh of a regular tetrahedron thanks to this answer:

enter image description here

However, as you can see, it looks like it's sitting on one of its edges, what I was looking for was for it to be orientated like an egyptian pyramid, i.e. sitting on one of its faces:

enter image description here

But as @YiFan cleverly pointed out, a pyramid is a pentahedron ! So my picture showing an egyptian pyramid is in fact wrong.

Back to my tetrahedron, I can correct somehow the orientation manually but it's imprecise, I rotate on X axis by ~35 degrees and on Z axis by 45 degrees.

Question:

How can I rotate a regular tetrahedron so it has the same orientation as an egyptian pyramid, so it looks like sitting on the floor ?

Eric Cartman
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  • What do you mean by "the same orientation" ? Do you mean "the same aspect" ? – Jean Marie Feb 15 '19 at 23:31
  • I have added a better picture that shows the initial orientation when using the formula in the link I've posted. – Eric Cartman Feb 15 '19 at 23:35
  • @Aybe by orientation do you mean you'd like it to be rotated so that one of the faces is parallel with the horizontal plane (XZ plane) according to your program? – Andrew Feb 15 '19 at 23:46
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    Basically, you can see that by looking at the orientation widget at top-right, the pyramid looks like it's sitting on one of its edges. What I would like is to rotate it so it looks like it's sitting on one of its face instead. Hope it's clearer now :) – Eric Cartman Feb 15 '19 at 23:54

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I don't know what software that is nor how to use it, but your request is impossible since the shape of the Egyptian pyramids are not regular tetrahedra. They have $5$ vertices! That is called a pentahedron, and a pentahedron one of whose faces lie on the $xy$-plane can be given by the coordinates for vertices $(-1,-1,0),(-1,1,0),(1,-1,0),(1,1,0)$ and $(0,0,1)$.

YiFan Tey
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  • I see ... definitely makes sense, the shape I've shown is not correct :) – Eric Cartman Feb 15 '19 at 23:57
  • Okay now thanks to your answer I just figured out that it's much simpler to orientate such shape properly, by -90 degrees on the X axis and we're set! – Eric Cartman Feb 16 '19 at 00:19
  • @YiFan I don't agree. Lateral views give you four vertices at a time in general, not five. See the figure in my "Answer" – Jean Marie Feb 16 '19 at 00:19
  • @YiFan forgot to mention, while indeed your answer shows that I've used the wrong example, a pyramid, would you still have an idea about rotating a tetrahedron instead ? – Eric Cartman Feb 16 '19 at 00:55
  • @Aybe well, the vertices of such a tetrahedron must be $(1,0,0)$, $(-1/2,\sqrt3/2,0)$, $(-1/2,-\sqrt3/2,0)$ and one more on the $z$ axis that is equidistant from the three previous points (because on the $xy$-plane, they must form equal angles viewing from the origin; exactly $120^\circ$). Can you find this point? – YiFan Tey Feb 16 '19 at 00:57
  • I don't know how to, trying to find the solution online ... – Eric Cartman Feb 16 '19 at 01:08
  • @Aybe try to do it yourself! First step: find the distance $r$ between two vertices. Then the set of points $r$ away from each of the three vertices we already know form three spheres that intersect in exactly two points. Any one of these can be taken as your vertex. – YiFan Tey Feb 16 '19 at 01:10
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    Okay, give me a few minutes please! – Eric Cartman Feb 16 '19 at 01:10
  • It's about the second part that I'm left in the dust, 'the set of points r away from each of the three vertices', what does that mean? – Eric Cartman Feb 16 '19 at 01:24
  • I have been able to visualize it on how it looks by putting 3 spheres at points then inflating them until they intersect, but I just don't understand how to compute this intersection. – Eric Cartman Feb 16 '19 at 01:32
  • @Aybe you could find the equations $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ of the sphere, so you have three simultaneous equations for three variables to solve. – YiFan Tey Feb 16 '19 at 02:29
  • If I know what x,y,z,a,b,c are I can always give it a try :D – Eric Cartman Feb 16 '19 at 05:41
  • @YiFan btw have you seen my answer ? it does solve the problem though I didn't do the math by myself :D – Eric Cartman Feb 16 '19 at 05:42
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This is not an answer. It is just a demand of clarification with a scientific wording.

Here is an imaginary view of a tetrahedron (resting on one of its faces) and of an egyptian pyramid as seen from an observer at the ground level.

enter image description here

The question of how to rotate "in the best way" these two solids around their vertical axis in order that "they look the same" needs to be precised in the following way.

We have to transform this "best way" into an objective criteria. I propose for example to minimize the gap between the resp. apparent angles BAC and B'A'C', on one side and angles CAD and C'A'D' on the other. Is it a good criteria ? Are there other parameters ?

Jean Marie
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    My interpretation of the OP's question was that they wanted to find the exact angles to rotate the tetrahedron so that it exactly overlaps with the pyramid. It's a reasonable question, if not for the fact that they are different shapes. This is a good answer, but I don't think it's what the OP was looking for. – YiFan Tey Feb 16 '19 at 00:23
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    @YiFan Indeed, exact overlap is illusory... A rather good approximation is maybe possible but I doubt that the illusion can last a long time... – Jean Marie Feb 16 '19 at 00:27
  • Honestly I don't understand everything in your request by lack of knowledge ... let me edit my question further a bit, doing some better screen caps. – Eric Cartman Feb 16 '19 at 00:31
  • I have seen that you have said now in your text "same orientation" : thus, if the summital angles BAC $\approx$ B'A'C' and CAD $\approx$ C'A'D' , ($\approx$ meaning "approximately equal") as seen by an observer from the ground level, would you be satisfied ? – Jean Marie Feb 16 '19 at 00:35
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    I have edited my question with a video which definitely explains the problem better, hopefully. – Eric Cartman Feb 16 '19 at 00:50
  • But could you answer my recent question ? This is important not only on the mathematical point of view, but also for software that might have to do this process in a blind way. – Jean Marie Feb 16 '19 at 00:52
  • Sorry I forgot, basically, yes ! – Eric Cartman Feb 16 '19 at 00:57
  • I will do some calculations but not now (I must go to bed : 2 o'clock in the morning, Central European Time) – Jean Marie Feb 16 '19 at 01:11
  • A last point : has your egyptian pyramid a height such that its faces are equilateral, or are you free to give it the height you want ? – Jean Marie Feb 16 '19 at 01:14
  • Ideally I'd like it to be equilateral :) Okay take your time, there's no hurry ! 2am here too :) – Eric Cartman Feb 16 '19 at 01:25
  • Taking your problem again, I think that the best similitude is achieved plainly when the two triangles in both views (I refer to the drawings in my text) are symmetrical with respect to a vertical line). The illusion would be perfect with a "higher" pyramid (i.e., with equilateral triangles remplaced by elongated isoceles triangles for its faces). – Jean Marie Feb 19 '19 at 09:42
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Currently I've found a way on how to solve my problem though I didn't figure out the math involved by myself, rather, a simple idea that ended up working:

A quaternion that rotates from/to and then transform all the points with it.

For the tetrahedron it is:

Quaternion.FromToRotation(Vector3.one, Vector3.up)

enter image description here

While it does work I would have loved to achieve it by myself!

I am posting this solely as a reference and not accepting my own answer, while indeed it solves the problem it doesn't really explain how :)

Thanks for anyone who took the time to help me out!

Eric Cartman
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