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New SED Challenge

Discussion in 'Electronic Design' started by Tim Williams, Apr 2, 2007.

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  1. Tim Williams

    Tim Williams Guest

    Find the ratio of diameters (or radii, or whatever) of the large spheres to
    the largest sphere which fits in the space between four adjecent large

    A couple of vertices (if not explicitly a dot, then an intersection of two
    lines) and a radius are marked on the diagram. Remember this geometry
    problem is three-dimensional.

    Diagram is posted to alt.binaries.schematics.electronic.

  2. kell

    kell Guest

    For those of us who have no access to a.b.s.e.:
    Are the large spheres of equal size and stacked in a tetrahedron (as
    it were)?
  3. Tim Williams

    Tim Williams Guest


    Considering the simple base of the problem, I should've specified in text as
    well :)

  4. jasen

    jasen Guest

    spoiler follows.

    So we have spheres centrered at a=(1,1,1) , b=(1,-1,-1),
    c=(-1,-1,1) and d=(-1,1,-1) with radius measure(a,b)/2 = sqrt(8)/2
    = sqrt(2)

    and want to find the radius of the sphere at o=(0,0,0) that meets them
    measure(o,a)/2 less the radius of the other sphere. (sqrt(3) - sqrt(2))

    but you want it expressed in term of the 4 spheres so

    so: (sqrt(3) - sqrt(2)) / sqrt(2)

    so the answer is sqrt(3/2) - 1

    that was easy.

  5. Tim Williams

    Tim Williams Guest

    Odd, I got sqrt(5/3) - 1 (which is "close enough" by 6.7%, but a parsec
    mathematically) with a different method. Another response (from the
    original use of this problem) determined it experimentally as 0.224 times
    smaller (or sqrt(2.9964 / 2) - 1), which is within 0.1% of your result.

    So why is it that Jim screws with everyone and makes boastful threads miles
    long, yet this plain intellectual challenge goes almost unnoticed? Probably
    says something about the intellectual persuits of the rest of the group...

  6. Jim Thompson

    Jim Thompson Guest

    It's geometry, not circuits ;-)

    Seriously, I'm up to my eyeballs in a chip design, so I only have a
    few minutes here and there to comment.

    ...Jim Thompson
  7. I would have tried it 'experimentally' (more like a Spice sim) with
    Solidworks, but I recently found that program does not seem to
    understand how to 'mate' spheres on tangents. It's not entirely a
    theoretical question since machinists use tooling balls (almost
    perfect metal spheres of known diameter) to form datums for measuring
    tapers, so the designer has to specify the position of the tooling
    ball, but I couldn't get the program to properly mate a sphere in an
    assembly even in a simple round taper. Fortunately you can do it on a
    2-D section, but then it's just a circle meeting a line at a tangent
    and the math is easy for the program.

    Best regards,
    Spehro Pefhany
  8. Rich Grise

    Rich Grise Guest

    I've been watching the thread, and this sounds kind of interesting,
    given what the other posters have said, but I haven't seen the diagram;
    it probably expired off my server. Could you re-post, or put it up on
    your website somewhere?

  9. kell

    kell Guest

    Consider a tetrahedron with edge length of two, four spheres of unit
    radius centered at the vertices.
    My application of high school geometry, for better or ill, gives the
    distance from the center of the tet to each vertex as the square root
    of eleven sixths. Subtract one from that quantity, and you have your
    answer. I get the central sphere has a radius .3540064...
  10. Tim Williams

    Tim Williams Guest

    Sure enough of that answer to give a proof? ;-)

    So far I've heard...
    sqrt(5/3) - 1 (mine)
    sqrt(3/2) - 1
    sqrt(6)/4 - 1/2 (half the proceeding value)
    And now also sqrt(11/6) - 1.

  11. kell

    kell Guest

    I'll give it a shot. I've defined the tetrahedron having edge length
    of two. Looking at the base triangle and using the law of cosines to
    find the distance from the center of the base to one of the corners of
    that triangle, the distance equals the square root of four thirds.
    The tetrahedron measures sqrt(8/3) high from the center of the base to
    the top vertex (by the Pythagorean theorem).
    The center of the tetrahedron lies at a point on this imaginary line
    Now, the distance from the center of the tetrahedron's base to its
    geometric center, again by Pythagorean th., equals
    The distance from the center of the base to the vertex
    h = x + sqrt(x^2-(4/3))
    that is,
    h = sqrt(8/3) = x + sqrt(x^2-(4/3))
    I'll skip typing in the algebra; I get x = sqrt(11/6)
    result --> radius of central sphere = sqrt(11/6)-1
  12. kell

    kell Guest

    I made a mistake in the algebra. x = sqrt(3/2)
    r = sqrt(3/2)-1 = .22474487...
  13. And I got 3/sqrt(6)-1 by somewhat different logic, but where it's the
    same result.

  14. Guest

    I've used a prototype version of MacSpice 3f5 2.9p34 to solve this
    problem. This version 2.9p34 has a built-in optimizer that uses a
    simplex method. Using default settings to minimize and n-parameter
    cost function, it starts by constructing a regular n-dimensional unit
    simplex with its centroid at the origin. I can extract these and solve
    the problem fairly easily, using this command file:

    *Gap between four spheres calculation
    echo "Get a regular simplex with four vertices on on a unit sphere:"
    optimize 'let cost = 1' npar 3 report -1 maxeval 3
    foreach v 0 1 2 3
    print line vertices[$v]
    echo "Rescale it so each point is separated by 2 units from the
    let delta = (vertices[2]-vertices[3])
    let separation = sqrt(length(delta)*mean(delta*delta))
    let rescaled = 2*vertices/separation
    foreach v 0 1 2 3
    print line rescaled[$v]
    echo "The distance from the origin to any of the rescaled vertices
    let d = sqrt(length(rescaled[0])*mean(rescaled[0]*rescaled[0]))
    print d
    echo "Hence the largest sphere to fit into the void has radius:"
    set numdgt = 15
    print (d-1.0)
    echo "Which is the same as:"
    print sqrt(3/2)-1.0

    Which produces the following results:

    *** MacSpice - 3f5 v2.9 PATCHLEVEL: 34
    *** Carbon Version by CDHW
    *** Date Created: Apr 7 2007

    Get MacSpice updates, information, userguide, tutorials from:


    Some useful Spice 3 commands:

    source <filename> : loads the named file as the new circuit.
    run : executes the specified analyses.
    edit : edits source with helper application.
    display : displays the list of output-vectors.
    plot <plotargs> : draws the results.
    set : displays all internal variables.
    rusage all : prints the resource usage.
    help : on-line help information (obsolete).
    quit : quit spice.

    MacSpice 1 -> source Primary:Users:cdhw:Documents:MacSpice:4-

    Get a regular simplex with four vertices on on a unit sphere:

    vertices[0] = ( -8.16497e-01 -4.71405e-01 -3.33333e-01 )
    vertices[1] = ( 8.164966e-01 -4.71405e-01 -3.33333e-01 )
    vertices[2] = ( 0.000000e+00 9.428090e-01 -3.33333e-01 )
    vertices[3] = ( 0.000000e+00 0.000000e+00 1.000000e+00 )

    Rescale it so each point is separated by 2 units from the others:

    rescaled[0] = ( -1.00000e+00 -5.77350e-01 -4.08248e-01 )
    rescaled[1] = ( 1.000000e+00 -5.77350e-01 -4.08248e-01 )
    rescaled[2] = ( 0.000000e+00 1.154701e+00 -4.08248e-01 )
    rescaled[3] = ( 0.000000e+00 0.000000e+00 1.224745e+00 )

    The distance from the origin to any of the rescaled vertices is:

    d = 1.224745e+00

    Hence the largest sphere to fit into the void has radius:

    (d-1.0) = 2.247448713915894e-01

    Which is the same as:

    sqrt(3/2)-1.0 = 2.247448713915889e-01
    MacSpice 2 ->

    Happy Easter

  15. jasen

    jasen Guest

    after a slow start this is shaping up to prove as controversial as
    the other challenges...

  16. kell

    kell Guest

    You found an elegant approach using analytical geometry.
    I used euclidean geometry, some elementary trig & algebra.
    Another guy used computer simulation of some sort.
    All in good fun, with no flames.
  17. Rich Grise

    Rich Grise Guest

  18. Tim Williams

    Tim Williams Guest

    The top and left spheres do not appear to be tangent, though it could just
    be the crummy way drawn (there isn't a good way to draw the surfaces while
    seeing everything in this problem).

  19. Rich Grise

    Rich Grise Guest


    **** me seven ways from Sunday - the "crummy" way it's drawn is that I
    didn't even get the fundamental equilateral tetrahedron right!

    Well, like they used to say, "Back to the ol' drawing board!"

    Thanks for not trashing me for such a horrendous blunder.

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