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Irregular tetrahedron (coordinate of vertices)

  1. Aug 1, 2012 #1
    This is my first post and I wish to get help in finding an analytically way to get the coordinate of an irregular tetrahedron.

    let ABCD be the 4 vertices of the tetrahedron in 3D, all vertices have different (x,y,z).

    the coordinate of vertex D is known (Xd,Yd,Zd), and the 3 angle between faces at vertex D are also known angle adb , adc, cdb.

    Coordinate of the other 3 vertices A, B, C are known on X, Y but not on Z. (ie Za,Zb,Zc are unknown).

    As I can expect this system will give me 2 solutions +&-.
    I've started by developing system of equation for the sides using cosines and as a function of Za,Zb,Zc, where (Zb-Za)^2= [Lda^2+Ldb^2-2Lda*Ldb*cos(adb)]-[(xb-xa)^2+(Yb-Ya)^2]

    Any suggestions on how to approach this problem would be appreciated.
     
  2. jcsd
  3. Aug 2, 2012 #2

    haruspex

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    Lda = √((Xd-Xa)2+ etc.) etc.
    So you can get three equations involving Za, Zb, Zc. For each pair of unknowns, Z1, Z2, you get an equation of the form
    (quadratic in Z1, Z2) = √(quadratic in Z1)√(quadratic in Z2)
    Squaring those gives you three quartics. If lucky, there may be some useful cancellation.
    There may be a better way, but I doubt it.
     
  4. Aug 2, 2012 #3
    The vertices should all be on the surface of a sphere. I don't know if that makes the problem easier, just a thought.
     
  5. Aug 2, 2012 #4

    haruspex

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    Why? We're told it is irregular (or, at, least, not necessarily regular).
     
  6. Aug 2, 2012 #5
    Thank you for the contribution,
    coolul007: unfortunately its not the case in irregular tetrahedron.

    haruspex: It looks like I must go through these equations and see what i get, they look very ugly though.
    do you think adding another equation using the volume of the tetrahedron, which is 1/6 of the volume of parallelepiped, with α, β, and γ are the internal angles between the edges, the volume is equations are given by ( http://en.wikipedia.org/wiki/Parallelepiped )
    V=1/6 [Lad.Lbd.Lcd * √(1+2cos(α) cos(β) cos(γ)-cos(α)^2-cos(β)^2-cos(γ)^2)]

    it is also equivalent to the absolute value of the determinant of a three dimensional matrix built using a, b and c:
    V=1/6 [a.(b*c)]

    thanks for the help
     
  7. Aug 2, 2012 #6

    haruspex

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    My experience of 3D trig is that quartics are par for the course. Working with volumes is likely to make matters worse, introducing 6th powers.
     
  8. Aug 3, 2012 #7
    I hope I am not riding a dead horse here, but it seems that the bottm triangle is on a sphere and all that is left is the vertex to intersect with a sphere. I haven't tried finding it yet but it seems there has to be one.
     
  9. Aug 3, 2012 #8

    haruspex

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    Sorry, you're right. Just as every triangle defines a circle through its vertices, so there is a unique sphere through the vertices of a tetrahedron, regular or not.
    However, I don't see that it helps.
     
  10. Aug 3, 2012 #9
    Yes, coolul007 is correct, that would have helped alot if the center of the sphere can be defined.
     
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