Path curvatures on a convex roof
Except where reference is made to the work of others, the work described in this
dissertation is my own or was done in collaboration with my advisory committee.
This dissertation does not include proprietary or classified information.
Robert Ford
Certificate of Approval:
Andras Bezdek
Professor
Mathematics and Statistics
Krystyna Kuperberg, Chair
Professor
Mathematics and Statistics
Gary Gruenhage
Professor
Mathematics and Statistics
Piotr Minc
Professor
Mathematics and Statistics
George T. Flowers
Interim Dean, Graduate School
Path curvatures on a convex roof
Robert Ford
A Dissertation
Submitted to
the Graduate Faculty of
Auburn University
in Partial Fulfillment of the
Requirements for the
Degree of
Doctor of Philosophy
Auburn, Alabama
December 17, 2007
Path curvatures on a convex roof
Robert Ford
Permission is granted to Auburn University to make copies of this dissertation at its
discretion, upon the request of individuals or institutions and at
their expense. The author reserves all publication rights.
Signature of Author
Date of Graduation
iii
Vita
Robert Ford was born December 20th, 1968 to parents Jo and Ralph Ford, both
mathematicians at Auburn University. He is the youngest of four siblings and his two
brothers still live in Auburn. He has spent time in Pennsylvania, Ohio, and England,
but has always returned to his hometown. In 1995, he graduated with a BA in History
at Auburn University. He added a Masters in Mathematics four years later. He has
never married and has no children.
iv
Dissertation Abstract
Path curvatures on a convex roof
Robert Ford
Doctor of Philosophy, December 17, 2007
(Masters of Mathematics, Auburn University)
(Bachelor of Arts, History, Auburn University)
40 Typed Pages
Directed by Krystyna Kuperberg
Given a set of rectangles, R1 through Rk, where Ri and Ri+1 share a common
edge and these common edges are congruent and parallel to each other. The resulting
?roof? is part of the surface of a convex body. We?ll consider paths from one corner
of this ?roof? to the opposite corner. Extending the common edges to lines we?ll call
ridges and rectangles to planar strips, we can allow such paths to go ?off the roof?.
Path curvature is computed for a polygonal path by simple adding up the curvatures
of each intermediate vertex. More general paths use a sequence of polygonal approx-
imations to compute curvature. We?ll discern when the path of least curvature goes
off the roof or when it is the shortest path.
v
Acknowledgments
I would like to acknowledge Krystyna Kuperberg for her tireless support. Her
patience in guiding my research made this material possible.
vi
Style manual or journal used Journal of Approximation Theory (together with the
style known as ?aums?). Bibliograpy follows van Leunen?s A Handbook for Scholars.
Computer software used The document preparation package TEX (specifically
LATEX) together with the departmental style-file auphd.sty.
vii
Table of Contents
List of Figures ix
1 Introduction 1
2 Convex Paths 4
3 Path Curvature on a convex roof. 11
3.1 The Simplest Roofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3.2 More Complex Roofs . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
Bibliography 31
viii
List of Figures
1.1 Milnor?s Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2.1 Convex Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2.2 Rotating Part of a Path . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.3 Partial Rotation of the New and the Old . . . . . . . . . . . . . . . . 6
2.4 The Iterative Process . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3.1 Simple Roof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
3.2 A Simple Folding of the Longer Path . . . . . . . . . . . . . . . . . . 14
3.3 The Longer Path has Greater Curvature . . . . . . . . . . . . . . . . 15
3.4 In the Wider Case the Longer Path also has Greater Curvature . . . 16
3.5 The Optimal Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.6 Three Local Extrema . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.7 Before Folding with More Ridges . . . . . . . . . . . . . . . . . . . . 21
3.8 After Folding with More Ridges . . . . . . . . . . . . . . . . . . . . . 22
3.9 A Partial Folding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.10 Optimal Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3.11 The Closer Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
3.12 After Partial Folding . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
ix
Chapter 1
Introduction
The total curvature of a C2 path C parameterized by arc length s in Rn, r(s), is
defined as integraltextC |rprimeprime(s)|ds. W. Fenchel proved in [4] for R3 and K. Borsuk in [3] for any
Rn that the total curvature of a closed curve is bounded from below by 2pi, with the
equality holding only for convex simple closed curves in R2. The total curvature of a
polygonal path P = [z0,z1,...,zn] is defined as
t(P) =
n?1summationdisplay
i=1
(pi ? negationslash zi?1zizi+1).
One way you might compute these angles would be to replace the segments of
the path with vectors, all heading in the same direction along the path. The angle
between consecutive vectors would give you the curvature at each point. Recall the
angle between two vectors, u and v, is the arccos( u?v|u||v|). For more general paths,
approximate polygonal paths can be used to approximate the curvature. As you
develop a sequence of polygonal paths which better approximate the curve, you get
a sequence of curvatures that approach the curvature of the path. This is essentially
the definition for path curvature.
Polygonal paths form the building blocks of all path curvature. Just as polygonal
paths may approximate a general path, polyhedral surfaces approximate a general
surface. We know the path of least curvature on a convex body can not be greater
1
than 2pi. Currently the total curvature of the shortest path between two points on
the surface of a convex body is not known to have an upper bound as discussed in
[1], [6], and [7]. It has recently been proved in [2] that such an upper bound, should
it exist, must be greater than 2pi.
Figure 1.1: Milnor?s Proof
J. Milnor proved in [5] if a segment of a polygonal path is replaced by two
segments then the curvature of the path will either increase or remain the same. His
argument is a very simple one. There are three places where the curvature can change:
At the endpoints of the replaced segment and at the point common to the two new
segments. The new point of curvature will obviously add to the curvature of the path
(assuming an actual change has occurred). The old points of curvature can be altered
at most by the measure of the angle between the corresponding new segment and the
2
replaced segment. Since these angles will add up to the curvature of the new point,
the net decrease in the curvature at these points can not be more than the increase
in curvature at the new point. For paths of least curvature on these polyhedral roofs,
we can focus our attention on polygonal paths comprised of the minimum number of
segments.
3
Chapter 2
Convex Paths
Figure 2.1: Convex Path
We?ll call a planar polygonal path convex if by adding a line segment between
the initial and final point of the path we get a convex polygon. The interior angle at
a point of curvature is supplemental to that curvature. The interior angles of an n
sided polygon add up (n ? 2)pi. If we add the total curvature to the corresponding
interior angles we get (n?2)pi. So the two interior angles of the polygon formed by
the new segment will have a sum equal to the curvature of the original path.
In addition to proving curvature is increased or preserved when replacing one
segment with two in a polygonal path, Milnor also observed, but did not prove, that
4
if the segments involved in the change in the curvature were not co-planar then the
curvature would be increased.
To see this, we?ll analyze the curvature a two segmented path as one segment is
rotated about a line. The rotated segment can be seen as the slant height of a cone.
Let ? be the angle between the slant height and height (the line of rotation for the
segment). The other segment will be attached to the vertex of the cone. Orienting
this segment and the line of revolution into the xy-plane so the vertex is the origin
and the attached segment has non-positive x-coordinates, we?ll let ? be the angle
between the attached segment and negative x-axis. Finally, let ? be the angle of
rotation of the slant height from the xy-plane.
b
a
f
Figure 2.2: Rotating Part of a Path
5
Given this orientation we can compute unit vectors parallel with the two segments
as < cos?,?sin?,0 > and < ?cos?sin?,?cos?,sin?sin? >. So the curvature is
arccos[?cos?cos? sin? + sin? cos?]. If we fix ? between 0 and pi and ? between
?pi2 and pi2, then this becomes a decreasing function of ? from ? = 0 to ? = pi.
Figure 2.3: Partial Rotation of the New and the Old
In Milnor?s picture (figure 1.1), if the segment immediately preceding the re-
placed segment is not coplanar with the two new segments (whose plane automatically
contains the replaced segment) then it can be rotated about the line containing the
replaced segment. If it?s rotated away from the first of the two new segments then the
new path?s curvature will decrease while the old path?s curvature will remain the same.
If we signify the old path with O and the new path with N and let Oprime and Nprime represent
their rotated versions then we observe that cur(N) > cur(Nprime) ? cur(Oprime) = cur(O).
6
This same argument works with the segment immediately after the replaced segment
in the path.
Milnor?s first observation,that adding vertices never decreases the curvature of
the path, is easily extendable to replacing a single segment with multiple segments.
Likewise the relation is reversible (replacing multiple segments with one segment will
not increase the curvature). Milnor?s second observation, that the segments involved
in the change in the path must be coplanar to preserve curvature, requires a simple
argument to see it extends to multiple segments.
Theorem 2.1 Replacing a single segment with multiple segments only preserves cur-
vature in the path if the new segments along the segments immediately before and after
are coplanar.
Proof. Imagine we have a path and we replace one segment with multiple segments
(more than two). We?ll call the shorter path O and the longer N. For simplicity,
further suppose N has the minimum number of added vertices, b0 to bk, which is
to say the path does not have curvature zero at any new vertex. Last we?ll assume
cur(N) = cur(O). Working backwards, starting with N and ending in O, an iteration
will consist of replacing two segments with one segment. While it?s possible that N
has no segment immediately following or preceding the change in the path, it is not
possible for both segments to be missing. We will label the segment which for certain
exists a0,a1. We?ll then label the added vertices that follow or precede depending
b0,b1,...,bk. If there exist the other segment immediately following or preceding it
7
will be labeled a2,a3, otherwise a2 will exist as the initial or final point of the path.
So the first iteration will replace a0,a1,b0,b1,b2 with a0,a1,b1,b2, the second iteration
will replace a0,a1,b1,b2,b3 with a0,a1,b2,b3, and so forth.
?
?
?
?
?
?
?
?
?
a
a
a
a
b
b
b
b
b
0
1
2
3
1
2
?
b
0
?
b
i
i +1
i +2
i +3
Figure 2.4: The Iterative Process
For curvature to be preserved, it must be preserved at each iteration. This means
the vertices of each iteration must be co-planar.
Case 2.1: There are three or more iterations.
Assume there are three or more iterations, but the first iteration is collinear. If
a path is collinear then each point of curvature is either zero of pi. Since we assumed
the curvature at b0 and b1 are both positive, they each must be pi. If the curvature is
to be preserved then the new curvature at a1 and b1 must also add up to 2pi. So in the
second iteration, a1, b1, and b2 initially add up to more than 2pi since the curvature
8
at b2 can not be zero. This leads to the second iteration failing to preserve curvature.
So the first iteration is not collinear.
For the induction hypothesis, let the i+1st iteration be non-collinear, but copla-
nar with all of the vertices from previous iterations, and is not the second to last
iteration.
The i + 2nd iteration is not the last iteration. The i + 1st and i + 2nd itera-
tions share four vertices in common a0,a1,bi+1, and bi+2. Suppose these vertices are
collinear. In the i+1st iteration, a1, bi, and bi+1 must then initially have positive cur-
vature. When bi is removed, curvature at a1 and bi+1 must be increased for curvature
to be preserved. Since the remaining vertices are collinear, the new curvatures at a1
and bi+1 must add up to 2pi. This will prevent the i + 2nd iteration from preserving
curvature. Since the four common vertices to the two consecutive iterations can not
be collinear, they must uniquely determine a plane. This means the i+2nd iteration
is not collinear and is coplanar with all of the vertices of the previous iterations.
For the last iteration then we can consider three subcases. If a3 does not exist
then we?re done. If a3 does exist but is not collinear with the four common vertices
then the second to last iteration argument can be employed to show the four common
vertices cannot be collinear. Finally, if a3 exists and is collinear with the four common
vertices then we?re done.
Case 2.2: There are two iterations.
9
This is an easier case. If the first iteration is collinear then it will be coplanar
with the last iteration. If the first iteration is not collinear then we use the same
argument as above for the last iteration.
10
Chapter 3
Path Curvature on a convex roof.
Given a set of rectangles, R1 through Rk, where Ri and Ri+1 share a common
edge and these common edges are congruent and parallel to each other. Further
suppose the resulting ?roof? is part of the surface of a convex body. We?ll consider
paths from one corner of this roof to the opposite corner. Extending the common
edges to lines we?ll call ridges and rectangles to planar strips, we can allow such paths
to go off the roof. For a polygonal path, curvature is computed by simple adding
up the curvatures of each intermediate vertex. More general paths use a sequence of
polygonal approximations to compute curvature. We?ll attempt to discern when the
path of least curvature goes off the roof or when it is the shortest path.
3.1 The Simplest Roofs
We?ll begin with k = 2. The roof is simple enough, it?s comprised of two rect-
angles with dimension a ? c and b ? c and has a dihedral angle of ?. We?ll even
consider extreme scenarios like the roof being impossibly thin (c = 0) or impossibly
steep (? = 0). In the case where ? = 0 the two half planes would coincide, but we?ll
consider each side of the roof distinct and any path plotted has to go through the
ridge line.
11
Figure 3.1: Simple Roof
Let S be the shortest path and P be the path of least curvature. If we flatten
the roof, S would become a single line segment connecting the opposite corners of a
rectangle.
Theorem 3.1 S is the unique path of least curvature in four basic cases: When
? = pi, when a = b and 0 < ? < pi, when b < a, c = 0 and ? ? arccos(ba), and when
a = b, ? = 0, c > 0.
Proof.
Case 1
? = pi
12
If the roof is actually flat then S has curvature of zero. In truth, this is really
the k = 1 case.
Case 2
a = b and 0 < ? < pi
We?ll call a two segmented path isosceles if the two segments are of equal length.
If we compare two isosceles paths with the same initial point and final point, then the
one with the greater total length has the greater curvature. All that?s left to show
is that an isosceles path has less curvature than an non isosceles path of the same
length.
Let x and y be the lengths of the two segments of our path. Let z be the distance
from the initial point to the final point. By the law of cosines z2 = x2+y2?2xycos?
(where pi ?? is the curvature).
z2 + 2xy = (x + y)2 + 2xycos(pi ??)
z2 ?(x + y)2
2xy + 1 = cos(pi ??)
arccos(z
2 ?(x + y)2
2xy + 1) = pi ??
pi?? is minimized when xy is maximized. Given x+y is fixed, xy is maximized
when x = y.
13
If a = b then the shortest path is also isosceles. A longer path will be non
isosceles so its curvature is greater than an isosceles path with the same length which
in turn has greater curvature than the shortest path.
Case 3
b < a, c = 0 and ? ? arccos(ba)
Figure 3.2: A Simple Folding of the Longer Path
We can visualize this case by orienting the roof in the following way. Let the
ridge line be the z axis. Let the short side of length b lie on the negative, y-axis. if
? is equal to the arccos(ba) then the initial point on the long side will have the same
y-coordinate as the final point on the short side, so the line passing through those two
points is parallel to the x-axis. If we plot a longer two segmented path, one whose
ridge point has a positive or negative z-coordinate, then we can revolve that path
14
about the line passing through the initial and final point until it also lies in the xy
plane. Its curvature is clearly greater.
a b
Figure 3.3: The Longer Path has Greater Curvature
If ? > arccos(ba) then the initial point has a greater y-coordinate than ?b, but
if we rotate the roof about the z axis until the initial and final points have matching
y-coordinates then the proof follows as before.
Case 4
a = b, ? = 0, c > 0.
If we orient the path in the xy-plane so that the ridge point is the origin and
the initial and final points have the same negative y-coordinate then we?ll observe
the x-coordinates of the initial and final points have the same absolute value, but
are on opposite sides of the y-axis. So the center of the circle passing through the
15
a b
Figure 3.4: In the Wider Case the Longer Path also has Greater Curvature
ridge, initial, and final points has an x-coordinate of zero. This means the ridge line
is tangent to the circle and so all other paths have greater curvature.
Observe when c = 0, a = b, and ? = 0, all two segmented paths have the same
curvature in this case. So S is a path of least curvature, but not the only path of
least curvature.
Also observe that case 3 is not optimal. There are smaller ? that allow for
agreement, but the proof of this is more involved and will held to the end. We can
however establish a visual criteria for agreement. If the ridge point is not co-linear
with the initial and final points, then these points are co-circular. If this circle does not
intersect the positive y-axis then S has the minimum curvature. It is also interesting
to notice when c = 0 and S is not the path of least curvature that there are two paths
16
of least curvature. If a path whose ridge point has a positive z-coordinate is a path
of least curvature then by symmetry there is a path whose ridge point has a negative,
but same absolute valued, z-coordinate with the same curvature.
Figure 3.5: The Optimal Case
Once we?ve finalized case 3, in all other cases S will not be the path of least
curvature. If the ridge line is oriented to be parallel to the z-axis, we give the initial
point a z-coordinate of 0 or more specifically make it the origin, and give the ridge
of the roof a positive orientation then the z-coordinate of S?s ridge point is aca+b.
Computing the path of least curvature?s ridge point will be more involved.
If a > b, c > 0 and ? = 0 then the z-coordinate of P?s ridge point is less then
the z-coordinate of S?s ridge point.
17
Figure 3.6: Three Local Extrema
The ridge point which is collinear with the initial point and final point produces
the largest possible curvature, pi. As z goes to infinity or negative infinity, the curva-
ture also tends to pi. This suggests at least one local minimum to either side of the
collinear ridge point.
If a ridge point is not collinear with the initial and final point, then it is co
circular with them. If the center of the circle should have the same z-coordinate as
the ridge point, then the ridge line would be tangent to the circle. So the ridge points
immediately to the left or right would produce larger curvatures, making such a ridge
point a local minimum. If the z-coordinates are different, then such a ridge point is
neither a local minimum nor maximum. The equation of the circle, (y?h)2+(z?k)2 =
18
(a?h)2, reduces to a quadratic when we plug in (0,0) and (a?b,c) and eliminate h
as a variable:
(a?b)k2 ?2(ac)k + (ac2 ?ba2 + ab2) = 0.
k = ac?
radicalBig
(ab((a?b)2 + c2))
a?b .
Observe both local minimums are equidistant to the collinear ridge point or local
maximum. The two circles tangent to the ridge line are not congruent. The left circle
has a smaller radius and produces a smaller curvature in the path. So the left local
minimum is an absolute minimum and the right local minimum is not.
In the specific case where a = b, c = 0, and ? = 0, S is a path of least curvature,
but all paths comprised of two segments have the equally bad curvature of pi.
For a > b, c > 0, and 0 < ? < pi, we must look at the curvature in three
dimensions. Let the initial point again be the origin and the ridge line be parallel to
the z-axis and lie in yz-plane with a positive orientation. Then the coordinates of
the final point will be (?b(sin(?)),a ? b(cos(?)),c). Given a ridge point of (0,a,z),
we can compute the vectors and then the curvature.
Let u =< 0,a,z >, and v =< ?b(sin(?)),?b(cos(?)),c ? z >. Let ? equal the
curvature.
cos(?) = (u?v)/(|u||v|) ?? ? = arccos((u?v)/(|u||v|)). If we differentiate ? with
respect to z we get ?prime = (?1/
radicalBig
1?(cos(?))2)(cos(?)prime)
19
?1/
radicalBig
1?(cos(?))2 is never zero and is undefined when ? = 0 or pi, which can
only happen when ? = pi or 0. cos(?)prime, once simplified, has a denominator based on
the magnitudes of u and v and cannot be zero. So the only critical points we get
come from the numerator which simplifies to:
(2abcos(?)?a2 ?b2)z3 + (3a2c?3abccos(?))z2
+(abcos(?)(a2 + b2 + c2)?3a2c2 ?2a2b2)z + (a2c3 + a2b2c?a3bccos(?))
If we plug in S?s ridge point, we do not get zero, so disagreement exists in this
case. We?re also interested in when P tries to leave the roof. This solvable cubic
produces one local extremum when ? ? pi/2. We see this by taking the derivative
of the cubic, we get a quadratic with no real roots. If we plug in x = 0 and x = c
into the cubic we?ll observe the local extreme is a minimum and occurs on the roof
when ? ? pi/2. By analyzing this cubic we can see c ?
radicalBig
b(a?b) also guarantees P
stays on the roof. Finally, shifting back to c = 0, we can optimize our original Case 3
for agreement. Notice the coefficient of z is zero in this case when ? = arccos( 2aba2+b2).
Notice also that 2aba2+b2 = ba a2+a2a2+b2 > ba if b < a. So arccos( 2aba2+b2) < arccos(ba), making
the following theorem an improvement.
Theorem 3.2 S and P will agree if c = 0 and ? ? arccos( 2aba2 + b2).
20
3.2 More Complex Roofs
If we add to the number of ridges, we can still partially determine when P and
S will agree.
Theorem 3.3 Let c = 0 and suppose we have k ridges where k > 1. Further suppose
that the shortest path has the following properties: 1. It is a convex path. 2. When
we orient the path in the xy-plane so that the y-coordinates of the initial and final
point are the same and less than the y-coordinates of the interior points of the path,
and no two points of this path share the same x-coordinate. Under these conditions
S and P will agree.
Figure 3.7: Before Folding with More Ridges
21
Figure 3.8: After Folding with More Ridges
Proof.
Let S be the shortest path and L be an arbitrary longer path.
Subcase 3.1: Let the initial and final segment of L be a planar.
We?ll do two things to the path L. First we?ll replace the intermediate segments
with one segment, and then fold L about the line passing through the initial and
final point, a line which is parallel to the x-axis, into the xy-plane above the line of
revolution. We?ll call the new path Lprime. If the original L was planar then cur(L) ?
cur(Lprime) and the folding allows us to easily see Lprime has greater curvature than S (the
two interior angles whose sum is the curvature of the path are larger for the polygon
formed by Lprime). If the original path was not planar then cur(L) > cur(Lprime) and the
folding allows us to see Lprime has curvature no less than S.
22
Subcase 3.2: Let the initial and final segments of L be non planar.
L can once again be polygonally approximated with a planar path, but this
time it will take three steps. First we can replace the intermediate segments with
one segment (this may or may not produce a change in L). This step will reduce
the curvature or maintain it depending. Second we?ll rotate the path about the line
passing through the initial and final point until the final ridge point lies in the xy-
plane with a y-coordinate greater than the initial or final point (again this may or
may not produce a change in the perspective of the path). This step produces no
change in curvature. At this point, the initial segment is not in the xy-plane. We?ll
rotate the initial and intermediate segments about the line passing through the initial
point and the last ridge point until the first ridge point of L lies in the xy-plane in
the half plane not containing the final segment of the path. This last step is certain
to be non-trivial (it will produce a change in the path). We?ll call this new path Lprime. If
we consider the intermediate segment of Lprime to be the slant height of a cone centered
on the axis of revolution in the third step then we see that third step decreased the
curvature at the final ridge point. So cur(Lprime) < cur(L).
The first ridge point of Lprime must fall in the interior of the half plane formed by
the line passing through the initial point and last ridge point which does not contain
the last segment of the path. Since the axis of revolution has a positive slope, the
x-coordinates of the effected points in step three will decrease. Since the x-coordinate
of the first ridge point had been unaffected during the first two steps, the first ridge
point of Lprime must be to the left of the vertical passing though the first ridge point of
23
Figure 3.9: A Partial Folding
S. Finally the length of the initial segment of Lprime is not less in length than the initial
segment of S, there fore the first ridge point of Lprime may not lie within that distance
of the initial point of the path. So the angle the initial segment of Lprime makes with the
segment between the initial and final point is greater than the corresponding angle
for S. The other angle needed to compute the curvature for Lprime is no less than its
corresponding angle for S. So cur(S) < cur(Lprime).
In the one ridge case we were able to expand our folding argument to include thin
roofs with some overhang. We used both folding and a circle to see this improvement.
With multiple ridges, folding and partial folding along with the circle does not always
work to see S is also P, but we can get a partial result using this argument. Imagine we
orient the shortest path as above, but that some points share the same x-coordinate.
24
Agreement between S and P is still provable using a folding argument if the following
conditions are added. If we replace the intermediate segments with one segment then
last ridge point has curvature less than or equal to pi/2. We?ll call this condition 1. If
we extend the initial and final segments into lines then they intersect at a point whose
y-coordinate is greater than the y-coordinate the initial or final point. We?ll call this
condition 2. If we label the distance between the initial point and intersection point
a and the distance between the final point and intersection point b then the included
angle at point of intersection is greater than or equal to arccos( 2aba2+b2). We?ll call this
condition 3.
Theorem 3.4 Let c = 0 and suppose we have k ridges where k > 1. Further suppose
that the shortest path has the following properties:
1. It is a convex path.
2. If the path is oriented in the xy-plane so that the y-coordinates of the initial
and final point are the same and less than the y-coordinates of the interior of the path
then conditions 1, 2, and 3 are met.
Under these conditions S and P will agree.
Proof.
Suppose L is an arbitrary longer path and S is the shortest path. In each case
we can replace the intermediate segments of both paths with single segments, thus
creating Lprime and Sprime. The curvature of S is equal to the curvature of Sprime.
Case 4.1 Suppose Lprime has the same initial and final segments as Sprime.
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Figure 3.10: Optimal Case
In this case, cur(L) > cur(Lprime) = cur(Sprime) = cur(S), since L would be non-
coplanar.
Case 4.2 Suppose the initial and final segments of Lprime are coplanar but do not lie
in the xy-plane.
If we construct a pseudo ridge line where the lines containing the initial and
final segments of Sprime meet then the lines containing the initial and final segments of
Lprime also intersect at the pseudo ridge, but not on the xy-plane. Now a simple folding
argument works as it did with the one ridge case (here the pseudo ridge is the one
ridge).
Case 4.3 Suppose the initial and final segments of L are not coplanar.
For purposes of closeness, we?ll look at the plane containing a ridge point, the
initial point, and final point and compute it?s dihedral angle with the xy-plane. The
smaller the dihedral angle, the ?closer? the ridge point will be. The planar path for
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a ridge point will be the intersection of the plane containing that ridge point, the
initial and final points with the roof, but with intermediate segments replaced with
one as usual (this last step does not change the curvature of the planar path). If we
compute the planar paths for the initial and final ridge points, the one for the closer
ridge point will be a shorter path than the original path. That is it?s other ridge
point will be closer to the xy-plane. So the corresponding segment, either initial or
final, is shorter in the planar approximation than in the original longer path. So let
Lprimeprime be the ?closer? planar approximation. cur(Lprimeprime) ? cur(Sprime), with equality occurring
when Lprimeprime is S. We?ll fold Lprimeprime into the xy-plane and fold Lprime by the same amount. The
common segment to both will lie in the xy-plane, but the rest of Lprime will not. If the
final segment was the closer segment then a second partial folding, similar to case 3.2,
will allow us to see cur(Lprime) > cur(Lprimeprime). If the initial segment is closer then a more
difficult argument is necessary.
We?ll again let Lprimeprime be the planar approximation of Lprime, but this time using the
initial segment and final point to form the plane. We?ll then do a partial folding first.
We?ll fold the intermediate segment and final segment about the line passing through
the first ridge point and final point until they are in the same plane as Lprimeprime, but away
from the first segment. If we look at the projected image of this partial folding in
the xy-plane, then we?ll see the shadow of the final ridge point falls underneath the
shadow of the line containing the final segment of Lprimeprime. This partially folded path, we?ll
call Lprimeprimeprime, has curvature less than Lprime but more than Lprimeprime. Since we know the curvature
of Lprimeprime is at least as great as S, we?re done.
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Unlike in the one ridge case, congruency of the rectangles does not generally
yield agreement between S and P, even if we add congruency of the dihedral angles
as an added condition. In the case of a two ridge symmetric roof, we can see the
problem when we divide the path into two halves. If c > 0 then neither the first nor
the second half of the shortest path represents the path of least curvature between
the midpoint and initial or final point. Because of the symmetry in fact, the paths of
least curvature between the midpoint and the initial or final point add no curvature
at the midpoint if joined. So it is easy to see a path with less curvature than S. It is
also possible to construct counter examples with more than two ridges.
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Figure 3.11: The Closer Plane
29
Figure 3.12: After Partial Folding
30
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