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Maximum distance between things connected to a point (3D trig)

  1. Mar 3, 2005 #1

    ShawnD

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    I'm not exactly sure if this is a trig type thing. It just really seems like it.
    Suppose you attach a rope to the corner of a cube then suspend the cube. What will the angle be between the edges of the cube and the rope?

    If I hold a rubik's cube in my hand in a 2D perspective as just a square, the angle between the edges is 90 degrees, and the angle between either edge and where the rope would be should be the same. I would be able to say 2x + 90 = 360, x = 135 degrees. I can say that for all 3 ways of looking at that single corner of the cube (each corner has 3 adjacent planes). I can then draw 3 angles out from those 3 planes.
    What I want to know is how I can average those out or tie them together or something of that nature.

    Why was this moved to physics? It has essentially nothing to do with physics.
     
    Last edited: Mar 3, 2005
  2. jcsd
  3. Mar 4, 2005 #2
    Look, what you want to know, is not clear to me, but if you want to know the angle between any edge and the rope (which may not be the situation since you seem to have a solution for it), then I can give a solution.

    Let the point of suspension be O and let the three edges be OA, OB and OC. Also, let the required angle be a.(Of course, the angle that OA makes with the rope = angle that OB makes with the rope and so on) Also, let the cube be a unit cube. Don't let anything else.

    Produce the line along the rope downward to intersect the equilateral triangle ABC in G.
    From simple trigonometry, GA = GB = GC = sin(a)
    Also, since OA, OB, OC are mutually perpendicular, AB = BC = CA = square root of 2. (Pythagoras theorem)

    Again, from simple trigonometry in triangle ABC, AB/sin(60 degrees) = 2 * GA = sin(a)
    So, sin(a) = sqrt(2/3) And so a = arcsin (sqrt(2/3)), not 135.
     
  4. Mar 4, 2005 #3

    ShawnD

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    Thank you very much. You've been a tremendous help.
     
  5. Mar 4, 2005 #4

    BobG

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    Use the Pythagorean identity for cosines. For however many axes you have (2, 3, etc), if you measure the angle from a vector to each of the axes, the sum of the square of those cosines equals 1.

    In other words, if you have 2 axes (2-D) and the angle between the vector and the x-axis is α, and the angle between the vector and the y-axis is β, then:

    [tex]cos^2 \alpha + cos^2 \beta = 1 [/tex]

    With 3 axis and adding the angle γ for the angle between the vector and the z-axis, the Pythagorean identity becomes:

    [tex]cos^2 \alpha + cos^2 \beta + cos^2 \gamma = 1 [/tex]

    Since all three angles will be the same for the diagonal of a cube, your angle is:

    [tex]cos^{-1} (\sqrt{\frac{1}{3}}) [/tex]
     
  6. Mar 5, 2005 #5
    Oh, yes, that's a better way of doing it. Nice, bobG...
     
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