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Circumscribing a sphere with a tetrahedron

  1. Apr 8, 2007 #1

    DaveC426913

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    I wish to drill four evenly-spaced holes in a ball. How do I form my construction lines so that my marks are accurate?

    Obviously, if I could circumscribe a tetrahedron inside (or outside) the ball its vertices (or face-centres) would mark my holes. But I can't do that. I need to scribe my marks on the ball.

    I suppose my first start would be to know the angle between two vertices (perhaps organic chemists might know the bonding angles of a carbon atom?)

    Is that the dihedral angle (which Wiki sez for a tetrahedron is 70.5degres)?

    Does that mean any two holes would be 70.5 degrees apart? Cuz I think I can do that.
     
    Last edited: Apr 8, 2007
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  3. Apr 8, 2007 #2

    Hurkyl

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    Note that you can validate any set of marks: once you've placed four dots on your sphere, you can then use a length of string to measure the distance between each pair of dots to ensure they are the same distance apart.


    As for the actual measurement... hrm...

    Each of the four spherical triangles you can draw from your dots have to have the same area -- once quarter of the surface area.

    IIRC, the angular excess of a spherical triangle is proportional to the area of the triangle. Taking the triangle with three right angles (e.g. one point at the north pole, and two points on the equator, one-quarter circumference apart) as a reference, we see that there's a 90 degree excess for one eighth of the sphere. So, your triangles have to have a 180 degree excess.

    So, the angle of each of the spherical triangles must be 120 degrees.

    I don't know how to compute the length of the side of one of these triangles -- I can never remember spherical trigonometry. :frown: But if I follow the law of cosines from wikipedia correctly, I think that the length of one side of a triangle (i.e. the angular distance between points) has to be 109.47 degrees.



    Let me try saying that again...


    You may not be able to inscribe the tetrahedron, but you can scribe the next best thing on the surface of your ball. You can mark the four vertices on the ball, and then draw the great circles joining the points. This will produce four spherical triangles. If I've done my mensuration correctly, each of those triangles will have:

    (1) Area 1/4 of the surface
    (2) Side arclength of 109.47 degrees
    (3) Angles of 120 degrees


    edit: I redid my computations.
     
    Last edited: Apr 8, 2007
  4. Apr 8, 2007 #3

    StatusX

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    In R^4, the coordinates (1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1) form the four vertices of a tetrahedron, which lies in the hyperplane x+y+z+w=1. By symmetry, the center of the tetrahedron is the intersection of this hyperplane with the line x=y=z=w, or (1/4,1/4,1/4,1/4). So if we subtract this off so that the tetrahedron is centered at the origin, we get the vertices at (3/4,-1/4,-1/4,-1/4), etc, and the norm of these vectors is sqrt(9+1+1+1)/4=sqrt(3)/2. Thus the cosine of the angles between two vertices is just 4/3 times the dot product of two of these vectors, ie, 4/3(-3/16-3/16+1/16+1/16)=4/3(-1/4)=-1/3. So the angle you want is cos^(-1)(1/3), which is about 109.5 degrees.
     
  5. Apr 8, 2007 #4

    DaveC426913

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    Yes. Um. Without doing all the math there, I realized the, moment I closed my laptop that the angle of the vertices would be the complement of this dihedral angle. i.e if the dihedral angle (angle between two planes) is 70.5, then the angle between two vertices would be 180 - 70.5 = 109.5 - just as Status X says.

    Sometimes I wish I had the formal math to back up my spatial and deductive skills...


    P.S. Interesting that Status X makes use of hyperobjects in his explanation, since what I'm using this to construct is in fact a hyperobject.
     
    Last edited: Apr 8, 2007
  6. Apr 8, 2007 #5

    DaveC426913

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    Hurykl, as I say, I don't have the formal math skills but I can do a sanity check on your answer.

    If three of the holes were 120 degrees apart, they could be equidistant on a great circle of the sphere (i.e. in a plane). Thus, the distance between them must be LESS than that.
     
  7. Apr 8, 2007 #6

    Hurkyl

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    Spherical triangles still have sides and angles. The measures of the angles are 120 degrees. The lengths of the sides are 109.47 degrees.
     
  8. Apr 8, 2007 #7

    DaveC426913

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    Oh, I thought you were saying that two points would be 120 degrees apart. You're saying that the edges would radiate from the vertices at 120 degrees. Which, of course, they'd have to in a symmetrical object.
     
  9. Apr 9, 2007 #8

    DaveC426913

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    Well, I have a bit of egg on my face. Considering the spheres are 3/8" in diameter, a well-placed guess or two does the trick as good as any microscopic construction lines I might be able to make.
     
  10. Apr 9, 2007 #9

    StatusX

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    It's not that hard to do it pretty accurately if you can draw great circles on the sphere (say, rotate it with something while holding a pencil against it) and measure distances along the great circles (say, with a piece of string).
     
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