MATHEMATICS AND ORIGAMI

(Dana P.) #1
Mathematics and Origami

From that drawing we get the angle of the hyperboloid ́s asymptotic cone as

= 2 arctg = 84 , 7369 º
b

a
α

Note that the angle is smaller than that of the cones in Fig. 6 (100,57º). This has being
designed purposely to elude that the vertices of both cones will get in contact before their sur-
faces do on the hyperboloid bases.


To finish, let ́s set forth the question of what the tangent plane on a point of the hyper-
boloid will be.
For that we shall do some simple changes in equation (2):

2 2 2

(^1) 





 = −





 −





a
y
b
z
a
x
(difference of squares)





 −





= +





 −





 +
a
y
a
y
b
z
a
x
b
z
a
x
1 1
u
b
z
a
x
a
y
a
y
b
z
a
x






  • =

    − 1
    1
    ; v
    b
    z
    a
    x
    a
    y
    a
    y
    b
    z
    a
    x







  • − 1
    1
    (4)
    0
    1 1



  • − z−u=
    b
    y
    a
    u
    x
    a
    0
    1 1
    − − z−v=
    b
    y
    a
    v
    x
    a
    (5) (6)
    1 0
    1
    − + z− =
    b
    u
    y
    a
    x
    a
    u
    1 0
    1 1



    • z− =
      b
      y
      a
      x
      a
      v
      Finally we get equations (5) and (6) that mean:





  • The four of them are equations of planes.

  • Either pair (5) or (6) represent the intersection lines of each of those two planes.

  • Equations (5) are parametric in u, and (6) in v, what means that for each value of u
    we get a straight line (5) and for values v the right line we obtain is (6).

  • Do not mix up lines u, v with the plückerian co-ordinates of Point 13.2. If we de-
    velop equations (5) and (6) (what we shall not do), we would arrive to the asymp-
    totic or hiperboloidal co-ordinates that define parametrically the hyperboloids.

  • Getting a new generatrix for each value of u recalls what was said earlier when each
    rotation of an angle of 17,142857º yielded, also, another generatrix.

  • Generatrices v are the symmetric of u with respect to a plane containing the hyper-
    boloid axis and the mid-point of a generatrix u. This way, Fig. 11 has been drawn
    from Fig. 4.

  • Note that to get Fig. 11 as a paper construction, it suffices to begin with a Fig. 1
    changed in such a way that the mountain / valley crease would be symmetric with

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