PROBLEMS
19.1. What is (in general) the shape of the silhouette of a sphere observed by a perspective camera?
Solution The silhouette of a sphere observed by a perspective camera is the intersection of the corresponding viewing cone with the image plane. By symmetry, this cone is circular and grazes the sphere along a circular occluding contour. The silhouette is therefore the intersection of a circular cone with a plane, i.e., a conic section. For most viewing situations this conic section is an ellipse. It is a circle when the image plane is perpendicular to the axis of the cone. It may also be a parabola or even a hyperbola branch for extreme viewing angles.
19.2. What is (in general) the shape of the silhouette of a sphere observed by an ortho-graphic camera?
Solution Under orthographic projection, the viewing cone degenerates into a viewing cylinder. By symmetry this cylinder is circular. Since the image plane is perpendicular to the projection direction under orthographic projection, the silhou-ette of the sphere is the intersection of a circular cylinder with a plane perpendicular to its axis, i.e., a circle.
19.3. Prove that the curvature κ of a planar curve in a point P is the inverse of the radius of curvature r at this point.
Hint: Use the fact that tan u≈ u for small angles.
Solution As P0 approaches P the direction of the line P P0 approaches that of the tangent T , and δs is, to first order, equal to the distance between P and P0, it follows that P M≈ P P0/ tan δθ≈ δs/δθ. Passing to the limit, we obtain that the curvature is the inverse of the radius of curvature.
19.4. Given a fixed coordinate system, let us identify points of E3 with their coordi-nate vectors and consider a parametric curve x : I ⊂ R → R3 not necessarily parameterized by arc length. Show that its curvature is given by
κ =|x0× x00|
|x0|3 , (19.1)
where x0 and x00 denote, respectively, the first and second derivatives of x with respect to the parameter t defining it.
Hint: Reparameterize x by its arc length and reflect the change of parameters in the differentiation.
Solution We can write
x0= d dtx=ds
dt d dsx=ds
dtt,
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82 Chapter 19 Smooth Surfaces and Their Outlines
and
x00= d
dtx0=d2s dt2t+ (ds
dt)2 d
dst= d2s
dt2t+ κ(ds dt)2n.
It follows that
x0× x00= κ(ds dt)3b, and since t and b have unit norm, we have indeed
κ = |x0× x00|
|x0|3 .
19.5. Prove that, unless the normal curvature is constant over all possible directions, the principal directions are orthogonal to each other.
Solution According to Ex. 19.6 below, the second fundamental form is symmet-ric. If follows that the tangent plane admits an orthonormal basis formed by the eigenvectors of the associated linear map dN , and that the corresponding eigen-values are real (this is a general property of symmetric operators). Unless they are equal, the orthonormal basis is essentially unique (except for swapping the two eigenvectors or changing their orientation), and the two eigenvalues are the maxi-mum and minimaxi-mum values of the second fundamental form (this is another general property of quadratic forms, see chapter 3 for a proof that the maximum value of a quadratic form is the maximum eigenvalue of the corresponding linear map). It follows that the principal curvatures are the two eigenvalues, and the principal di-rections are the corresponding eigenvectors, that are uniquely defined (in the sense used above) and orthogonal to each other unless the eigenvalues are equal, in which case the normal curvature is constant.
19.6. Prove that the second fundamental form is bilinear and symmetric.
Solution The bilinearity of the second fundamental form follows immediately from the fact that the differential of the Gauss map is linear. We remain quite informal in our proof of its symmetry. Given two directions u and v in the tangent plane of a surface S at some point P0, we pick a parameterization P : U×V ⊂R2→ W ⊂ S of S in some neighborhood W of P0such that P (0, 0) = P0, and the tangents to the two surface curves α and β respectively defined by P (u, 0) for u ∈ I and P (0, v) for v ∈ J are respectively u and v. We assume that this parameterization is differentiable as many times as desired and abstain from justifying its existence.
We omit the parameters from now on and assume that all functions are evaluated in (0, 0). We use subscripts to denote partial derivatives, e.g., Puv denotes the second partial derivative of P with respect to u and v. The partial derivatives Pu
and Pv lie in the tangent plane at any point in W . Differentiating N· Pu= with respect to v yields
Nv· Pu+ N· Puv= 0.
Likewise, we have
Nu· Pv+ N· Pvu= 0.
But since the cross derivatives are equal, we have Nu· Pv= Nv· Pu,
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or equivalently
v· dNu = u · dNv,
which shows that the second fundamental form is indeed symmetric.
19.7. Let us denote by α the angle between the plane Π and the tangent to a curve Γ and by β the angle between the normal to Π and the binormal to Γ, and by κ the curvature at some point on Γ. Prove that if κa denotes the apparent curvature of the image of Γ at the corresponding point, then
κa= κ cos β cos3α.
(Note: This result can be found in Koenderink, 1990, p. 191.)
Hint: Write the coordinates of the vectors t, n, and b in a coordinate system whose z-axis is orthogonal to the image plane, and use Eq. (19.6) to compute κa. Solution Let us consider a particular point P0 on the curve and pick the coor-dinate system (i, j, k) so that (i, j) is a basis for the image plane with i along the projection of the tangent t to the curve in P0. Given this coordinate system, let us now identify curve points with their coordinate vectors, and parameterize Γ by its arc length s in the neighborhood of P0. Let us denote by x : I⊂R→R3 this parametric curve and by y : I⊂R→R3its orthographic projection. We omit the parameter s from now on and write
( y= (x· i)i + (x · j)j, y0= (t· i)i + (t · j)j, y00= κ[(n· i)i + (n · j)j].
Recall that the curvature κa of y is, according to Ex. 19.4,
κa=|y0× y00|
|y0|3
By construction, we have t = (cos α, 0, sin α). Thus y0 = cos αi. Now, if n = (a, b, c), we have b = t× n = (−b sin α, a sin α − c cos α, b cos α), and since the angle between the projection direction and b is β, we have b = cos β/ cos α. It follows that
y0× y00= κ[(t· i)(n · j) − (t · j)(n · i)]k = κb cos αk = κ cos βk.
Putting it all together we finally obtain
κa= | cos β|
| cos3α|,
but α can always be taken positive (just pick the appropriate orientation for i), and β can also be taken positive by choosing the orientation of Γ appropriately.
The result follows.
19.8. Let κu and κv denote the normal curvatures in conjugated directions u and v at a point P , and let K denote the Gaussian curvature; prove that
K sin2θ = κuκv, where θ is the angle between the u and v.
84 Chapter 19 Smooth Surfaces and Their Outlines
Hint: Relate the expressions obtained for the second fundamental form in the bases of the tangent plane respectively formed by the conjugated directions and the principal directions.
Solution Let us assume that u and v are unit vectors and write them in the basis of the tangent plane formed by the (unit) principal directions as u = u1e1+ u2e2 and v = v1e1+ v2e2. We have
II(u, u) = κu = κ1u21+ κ2u22, II(v, v) = κv = κ1v12+ κ2v22, II(u, v) = 0 = κ1u1v1+ κ2u2v2.
According to the third equation, we have κ2 =−κ1u1v1/u2v2. Substituting this value in the first equation yields
κu = κ1
u1
v2(u1v2− u2v1) = κ1
u1
v2sin θ
since e1 and e2 form an orthonormal basis of the tangent plane and u has unit norm. A similar line of reasoning shows that
κv = κ2
v2 u1
(u1v2− u2v1) = κ2
v2 u1
sin θ,
and we finally obtain κuκv = κ1κ2sin2θ = K sin2θ.
19.9. Show that the occluding is a smooth curve that does not intersect itself.
Hint: Use the Gauss map.
Solution Suppose that the occluding contour has a tangent discontinuity or a self intersection at some point P . In either case, two branches of the occluding contour meet in P with distinct tangents u1 and u2, and the viewing direction v must be conjugated with both directions. But u1 and u2 form a basis of the tangent plane, and by linearity v must be conjugated with all directions of the tangent plane. This cannot happen until the point is planar, in which case dN is zero, or v is an asymptotic direction at a parabolic point, in which case dN v = 0.
Neither situation occurs for generic observed from generic viewpoints.
19.10. Show that the apparent curvature of any surface curve with tangent t is κa= κt
cos2α, where α is the angle between the image plane and t.
Hint: Write the coordinates of the vectors t, n, and b in a coordinate system whose z axis is orthogonal to the image plane, and use Eq. (19.2) and Meusnier’s theorem.
Solution Let us denote by Γ the surface curve and by γ its projection. We assume of course that the point P that we are considering lies on the occluding contour (even though the curve under consideration may not be the occluding contour).
Since κt is a signed quantity, it will be necessary to give κa a meaningful sign to establish the desired result. Let us first show that, whatever that meaning may be, we have indeed
|κa| = |κt|
cos2α.
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We follow the notation of Ex. 19.7 and use the same coordinate system. Since P is on the occluding contour, the surface normal N in P is also the normal to γ, and we must have N =∓j. Let φ denote the angle between N and the principal normal n to Γ. We must therefore have b =| cos φ| = cosβ/ cos α (since we have chosen our coordinate system so cos α≥ 0 and cos β ≥ 0), and it follows, according to Meusnier’s theorem and Ex. 19.7, that
|κa| = κ cos β
cos3α=|κt cos φ| cos β
cos3α= |κt|
cos2α.
Let us now turn to giving a meaningful sign to κa and determining this sign. By convention, we take κapositive when the principal normal n0 to γ is equal to−N, and negative when n0= N .
It is easy to show that with our choice of coordinate system, we always have n0= j:
Briefly, let us reparameterize y by its arc length s0, noting that, because of the foreshortening induced by the projection, ds0= ds cos α. Using a line or reasoning similar to Ex. 19.7 but differentiating y with respect to s0, it is easy to show that the cross product of the tangent t0= i and (principal) normal n0 to γ verify
t0× (κ0n0) = κ cos β cos3αk,
where κ0 is the (nonnegative) curvature of γ. In particular, the vectors t0 = i, n0=∓j, and k must form a right-handed coordinate system, which implies n0= j.
Therefore we must take κa> 0 when N =−j, and κa< 0 when N = j. Suppose that N = −j, then cos φ = N · n = −j · n = −b must be negative, and by Meusnier’s theorem, κt must be positive. By the same token, when N = j, cos φ must be positive and κt must be negative. It follows that we can indeed take
κa= κt cos2α.
Note that when Γ is the occluding contour, the convention we have chosen for the sign of the apparent curvature yields the expected result: κa is positive when the contour point is convex (i.e., its principal normal is (locally) inside the region bounded by the image contour), and κais negative when the point is concave.