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:
- z− =
- 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