Page 86 - Geometric Modeling and Algebraic Geometry

84 R. Krasauskas and S. Zube
where x 4 plays a role of x 0 . Let λ(C) be an infinite line of the 2-plane spanned by
the conic C. Different signatures of the 2-plane σ = (++), (+0), (+−) correspond
to different number of intersection points #(λ(C) ∩ Ω)=0, 1, 2, respectively. The
proof now follows from the following simple observations:
(i) the projection π preserves affine type of conics;
(ii) the tangent line to the conic C in any point must be positive, since tangents to
π(C) are positive.
Indeed, according to (ii) all cases can be directly chosen from Fig. 2-4 with one
[1]
exception: H cannot be included in the list, since λ(C) is tangent to Ω in the point
+0
[2]
that already belongs to C. This means that π(C) should be a parabola P , that is
+0
impossible according to (i).
5.4 Rational parametrizations

Any rational curve C ⊂ R with non-negative tangents defines a canal surface
4
1
Env(C) with a rational spine curve s(t)=(C 1 (t),C 2 (t),C 3 (t)) and a rational ra-
dius function r(t)= C 4 (t). It is known that there is a rational parametrization of
such canal surface in the form

M(t, u)= s(t)+ r(t)N(t, u), (5.4)
where N(t, u) defines a rational Gaussian map to the unit sphere S. Let c(t) and
n(t, u) be a homogeneous form of C(t) and N(t, u), i.e. C(t)=(c 1 /c 0 ,...,c 4 /c 0 )
and N =(n 1 /n 0 ,n 2 /n 0 ,n 3 /n 0 ). In Laguerre geometry it is natural to consider
a slightly different variant of a Gaussian map with the image at infinity ˜n =
(0,n 1 ,n 2 ,n 3 , −n 0 ). Note that ˜n(t, u) ∈ Ω.
Lemma 3. An isotropic hypersurface Γ(C) ⊂ R can be parametrized by 2-
4
1
parameter set of isotropic lines connecting c(t) with ˜n(t, u) for all t, u ∈ RP .
1
Proof. Any of such lines can be parametrized v 0 c(t)+ v 1 ˜n(t, u) with homogeneous
coordinates (v 0 : v 1 ). The intersection with the hyperplane x 4 =0 gives the condi-
tion v 0 c 4 (t) − v 1 n 0 (t, u)=0. Hence choosing v 0 = n 0 (t, u) and v 1 = c 4 (t) we get
the parametrization of the intersection Γ(C) ∩{x 4 =0}:

n 0 (t, u)c(t)+ c 4 (t)˜n(t, u)=(n 0 c 0 ,n 0 c 1 + c 4 n 1 ,n 0 c 2 + c 4 n 2 ,n 0 c 3 + c 4 n 3 , 0).
Switching to the cartesian coordinates in R we get exactly the parametrization of
3
the canal surface Env(C) (5.4).
Using this lemma one can easily find a rational parametrization of any PE trans-
form of the canal surface. It is enough to transform the curve c(t) and the Gaussian
map ˜n(t, u) separately, and then intersect the resulting isotropic hypersurface with
the hyperplane x 4 =0. Hence it remains to find the Gaussian map in all canonical
cases of interest. Here we remind some definitions and results from [3].