Ken Turkowski Media Technologies: Graphics Software Advanced Technology Group Apple Computer, Inc.. Draft Friday, May 18, 1990 Abstract: We derive the expressions for first and second de
Trang 1Ken Turkowski Media Technologies: Graphics Software Advanced Technology Group Apple Computer, Inc.
(Draft Friday, May 18, 1990)
Abstract: We derive the expressions for first and second
derivatives, normal, metric matrix and curvature matrix for
spheres, cones, cylinders, and tori
26 January 1990
Apple Technical Report No KT-23
Trang 2The Differential Geometry of Parametric Primitives
Ken Turkowski
26 January 1990
Differential Properties of Parametric Surfaces
A parametric surface is a function:
where
is a point in affine 3-space, and
is a point in affine 2-space
The Jacobian matrix is a matrix of partial derivatives that relate changes in u and v to changes in
x, y, and z:
The Hessian is a tensor of second partial derivatives:
2(x,y,z)
∂(u,v)∂(u,v) =
∂2x
∂u2
∂2y
∂u2
∂2z
∂u2
∂2x
∂u∂v
∂2y
∂u∂v
∂2z
∂u∂v
∂2 x
∂v∂u
∂2 y
∂v∂u
∂2 z
∂v∂u
∂2 x
∂v2
∂2 y
∂v2
∂2 z
∂v2
=
∂2x
∂u2
∂2x
∂u∂v
∂2x
∂v∂u
∂2x
∂v2
J= ∂(x,y,z)
∂( )u,v =
∂x
∂u
∂ y
∂u
∂z
∂u
∂x
∂v
∂ y
∂v
∂z
∂v
=
∂x
∂u
∂x
∂v
u=[u v]
x=[x y z]
x=F u( )
Trang 3and establishes a metric of differential length:
so that the arc length of a curve segment, is given by:
The differential surface area enclosed by the differential parallelogram is approximately:
so that the area of a region of the surface corresponding to a region R in the u-v plane is:
The second fundamental matrix measures normal curvature, and is given by:
The normal curvature is defined to be positive a curve u on the surface turns toward the positive direction of the surface normal by:
The deviation (in the normal direction) from the tangent plane of the surface, given a differential displacement of is:
˙˙
x•n=uD˙ u˙t
u
κn= uD˙ u˙
t
˙
uG u˙t
D=n•H=
n• ∂2x
∂u2 n• ∂2x
∂u∂v
n• ∂
2x
∂v∂u n•∂
2x
∂v2
S= ( )G
R
δS≈( )G 12δuδv
δu,δv
s= ds
dt
t0
t1
t0
t1
t0
t1
∫ dt= t(uG˙ u˙t)
0
t1
∫
1 2 dt
u=u t( ), t0<t<t1
dx
( )2
=( )du G du( )t
Trang 4If the parametrization of the surface is transformed by the equations:
then the chain rule yields:
or
where
is the new Jacobian matrix of the surface with respect to the new parameters and , and
is the Jacobian matrix of the reparametrization
The new Hessian is given by
where
The new fundamental matrix is given by:
and the new curvature matrix is given by:
′
′
∂(u,v)
∂ ′u2
∂( )u,v
∂ ′u∂ ′v
∂(u,v)
∂ ′v∂ ′u
∂( )u,v
∂ ′v2
′
H =PHPT+QJ
P= ∂(u,v)
∂ ′(u,v′)=
∂u
∂ ′u
∂v
∂ ′u
∂u
∂ ′v
∂v
∂ ′v
′ v
′ u
′
J =∂(x,y,z)
∂ ′(u,v′)
′
∂(x,y,z)
∂ ′(u,v′) =
∂(u,v)
∂ ′(u,v′)
∂(x,y,z)
∂( )u,v
′
u = ′u u( ,v) and v′= ′v u( ,v)
Trang 5Change of Coordinates
For simplicity, we have defined several primitives with unit size, located at the origin Related
to the reparametrization is the change of coordinates , with associated Jacobian:
When the change of coordinates is represented by the affine transformation:
the Jacobian is simply the submatrix:
Regardless, the Jacobian and Hessian transform as follows:
The normal is transformed as:
The denominator arises from the desire to have a unit normal.
The first and second fundamental matrices are then calculated as:
Not very pretty But certain types of transformations can be applied easily For a uniform scale with arbitrary translations,
r 0 0
0 r 0
0 0 r
=r I
′
D = ′H• ′n =( )HC •(nC−1 t)
nC−1 t
C−1nt
( )21
− 1nt
nC−1 t
C−1nt
( )12
nC−1 t
C−1nt
( )12
nC−1 t
C−1nt
( )12
′
G = ′J J′t=JCCtJt
′
− 1 t
nC−1 tC−1nt
( )21
′
J =JC, H′=HC
xx yx zx
xy yy zy
xz yz zz
xx yx zx
xy yy zy
xz yz zz
xo yo zo
C=∂ ′x
∂x =
∂ ′x
∂x
∂ ′y
∂x
∂ ′z
∂x
∂ ′x
∂y
∂ ′y
∂y
∂ ′z
∂y
∂ ′x
∂z
∂ ′y
∂z
∂ ′z
∂z
′
x = ′x x( )
Trang 6so that
For rotations (and arbitrary translations), the Jacobian matrix C=R is orthogonal, so the inverse
is equal to the transpose, yielding:
Combining the two, we have the results for a transformation that includes translations, rotations and uniform scale:
or in terms of the composite matrix :
′
J =JC, H′=HC, n′= nC
C
( )13 , G′=( )C 23G, D′=( )C 13D
C=r R
′
J =rJR, H′=rHR, n′=nR, G′=r2G, D′=rD
′
J =JR, H′=HR, n′=nR, G′=G, D′=D
′
J =rJ, H′=rH, n′=n, G′=r2G, D′=r D
Trang 7Given the spherical coordinates:
we have the Jacobian matrix:
the Hessian tensor:
the first fundamental form:
the normal:
and the second fundamental form:
2+y2
0 −r
r
y r
z r
2+y2
0
0 r2
∂2
x,y,z
∂ θ( ,φ)∂ θ( ,φ) =
x2+y2
xz
x2+y2 0
x2+y2
xz
x2+y2
0
∂(x,y,z)
∂ θ( ,φ) =
xz
x2+y2
yz
x2+y2 − x2+y2
x y z
[ ] =[rsinφ cosθ rsinφsinθ rcosφ]
Trang 8Unit Sphere
Angle Parametrization
Given the unit spherical coordinates with , we parametrize the sphere:
This yields the Jacobian matrix:
the Hessian tensor:
the first fundamental form:
the normal:
and the second fundamental form:
Angle Parametrization
With the reparametrization , we have the Jacobian:
Applying the chain rule, we have:
0 π
θ =2πu,ϕ = πv
Dθφ = − x
2+y2
0 −1
n=[x y z]
Gθφ = x
2+ y2 0
0 1
Hθφ =
x2+y2
xz
x2+y2 0
x2+y2
xz
x2+y2
0
Jθφ=
xz
x2+y2
yz
x2+y2 − x2+y2
x y z
[ ] =[sinφcosθ sinφsinθ cosφ]
0≤ θ <2π, 0≤ ϕ < π
Trang 9Changing coordinates to yield a sphere of arbitrary radius, we find that the expressions for the
Jacobian, the Hessian, and the metric matrix remain the same, because x, y, and z scale linearly with r The curvature matrix changes to:
Duv= −
4π2(x2+y2)
0 −π2r
Duv= −4π
2
x2+y2
0 −π2
Guv= 4π
2
x2+y2
0 π2
Huv=
4π2[−x −y 0] 2π − yz
x2 +y2
xz
x2+y2 0
2π − yz
x2+y2
xz
x2+ y2
0
Trang 10Angle Parametrization
Given the unit conical parametrization:
we have the Jacobian matrix:
the Hessian tensor:
the first fundamental form:
the normal:
and the second fundamental form:
Unit Parametrization
For the parametrization:
we have:
2πy 2πx 0
x y z
[ ] =[rvcos2πu rvsin2πu vh]
Dθz= −
z
2 0
0 0
nθz= x
z 2
y
z 2 − 1
2
Gθz=
x2+y2 0
2+y2+z2
z2
=
z2 0
0 2
Hθz=
z
x
z 0
−y z
x
z 0
Jθz= −
y x 0 x
z
y
z 1
x y z
Trang 11Duv= −
4π2 rz
1+h2
0
1+h2
h2x rz
h2y
rz −1
Guv=
4π2
x2+y2
2
x2+y2+z2
z2
Trang 12Angle Parametrization
Given the cylindrical parametrization:
we have the Jacobian matrix:
the Hessian tensor:
the first fundamental form:
the normal:
and the second fundamental form:
Unit Parametrization
With the parametrization:
we have the Jacobian matrix:
the Hessian tensor:
Huv= −4π
2x −4π2y 0
0 0 0
Juv= −2πy 2πx 0
x y z
[ ] =[rcos2πu rsin2πu hv]
Dθφ = −1 0
0 0
n=[x y 0]
0 1
Hθφ = [−x −y 0] [0 0 0]
0 0 0
Jθφ= −y x 0
0 0 1
x y z
Trang 13and the second fundamental form:
Duv= −4π
2r 0
0 0
r
y
r 0
Trang 14Angle Parametrization
Given the torus parametrization:
we have the Jacobian matrix:
the Hessian tensor:
the first fundamental form:
the normal:
and the second fundamental form:
Dθφ = −x
2+y2
r 1− R
x2+y2
2−x2−y2+z2−r2
1− R
x2+y2
1− R
x2+y2 r
z r
Gθφ = x
2+ y2
0
0 r2
Hθφ =
x2+y2 − xz
x2+y2 0
yz
x2+y2 − xz
x2+y2
0
x2+y2
x2+y2
−z
Jθφ=
x2+y2 − yz
x2+y2 x2+y2−R
x y z
[ ] =[(R+rcosφ)cosθ (R+rcosφ)sinθ rsinφ]