Download The Perimeter and Surface Area of Regular Polygons on the Sphere

This paper presents an overview of spherical geometry. We begin by considering the
trigonometry of the sphere in the first section and then move to a study of regular spherical polygons in
the second part. Ultimately, spherical geometry is the study of large-scale geometry on the earth.
Airplane flight planners know that the shortest distance from JFK International airport in New York and
Heathrow International airport in London is via the great circle route, which passes near the tip of
Greenland. Military engineers design Intercontinental Ballistic Missiles (ICBM) to account for the
curvature of the earth in their trajectories, and high school students who doubt their geometry teacher’s
lecture on triangles with angles sums greater than 180° never blink an eye when learning about latitude
and longitude in History class.
Spherical geometry is a form of non-Euclidean geometry, specifically it is a subset of elliptical
geometry. Euclid, the father of geometry, based his geometric system on five postulates, the fifth of
which states that through any point not on a line there exists a unique line parallel to that line. NonEuclidean geometry replaces this fifth postulate with a contradiction to it and proceeds to develop a
system based upon the original four postulates and the contradiction to the fifth. Hyperbolic geometry
replaces the fifth postulate with the statement: through any point not on a line there exist infinitely many
lines which are parallel to that line. Elliptical geometry replaces the fifth postulate with the statement:
through any point not on a line there do not exist any lines parallel to that line. Both hyperbolic and
elliptical geometry assume that space is curved, but they differ in that hyperbolic like Euclidean
geometry assumes that space is infinite while elliptical assumes that space is finite though immense.
(Gans 194)
In spherical geometry, lines have only finite length and are called great circles, being the
intersection of a plane with the sphere which passes through the origin. The length of all great circles is
the same being equal to the circumference of the sphere which is 2R . Every point has a point directly
opposite of it on the sphere called an antipodal point. Every point can be considered a pole. Each pole
defines a polar, which is the great circle on the sphere lying in the plane perpendicular to the axis of the
1
sphere through that pole. Two non-opposite points have one and only one shortest path between them
called a geodesic arc. A geodesic arc is a sub arc of a great circle. Two opposite points have an infinite
number of geodesic arcs connecting them. Since the length of a great circle is 2R and a point and its
opposite point divide the great circle into equal parts we conclude that the maximum distance between
any two points is R .
The common notion of angle measure involves a flat surface. In a spherical figure, the angle at
each vertex is curved, though the more closely you examine it the flatter it becomes. For the sake of
being definite, we say that the measure of the angle at a vertex of a spherical figure is equal to the
measure of the angle where the tangents to the arcs at that point intersect. (see figure below)
C
A
B
We now proceed to derive some trigonometric formulas for right spherical triangles. This form
of derivation follows that found in Palmer’s spherical trigonometry book found on p.187-195. Some
additional steps have been added and explained to aid in understanding. We consider the spherical
triangle with sides a, b, c and the angle opposite c is the right angle on a sphere of radius R . (see figure
below)
B

R
a
c
O
a
c

b
C
b
A
2
Let AOB  c, AOC  b, COB  a . Now form triangle BDE where D is on OA , E is on
OC , BD  OA , and BE  OC . (see figure below)
B

R
a
c
O
a
c
b

D
E 
A
C
b
Since ED  OA  ED is parallel to the tangent to AC at point A and since BD  OA  BD
is parallel to the tangent to BA at point A . Therefore BDE   . Note that we could have chosen to
form this triangle using A and the triangle perpendicular to OB . From the given triangle, the following
side lengths can be derived.

EB  R sin a

BD  R sin c

OE  R cosa

OD  R cosc
Also the four expressions of DE will be used in the following derivations. They are listed as follows.
1.
cot   
DE
DE

BE R sin a
 DE  R sin a cot  
2.
cos  
DE
DE

BD R sin c
 DE  R sin ccos 
3
3.
sin b 
DE
DE

OE R cosa
 DE  R cosasin b
4.
tan b 
DE
DE

OD R cosc
 DE  R cosc tan b
Now we can derive out the formulas for any part of the triangle as long as we are given two other
sides or angles. First, set equation 1 = equation 2:
R sin a cot    R sin c cos 
R sin c cos 
sin a 
R cot  
 cos  sin   

sin a  sin c

cos  
 1
sin a  sin csin  
Here we note that if we had used the other triangle mentioned during the construction of this one that we
could obtain the following similar result:
sin b  sin csin  
Next set equation 1 = equation 3:
R sin a cot    R cosasin b
R sin a cot  
sin b 
R cosa
sin a
sin b 
cot  
cosa
sin b  tan a cot  
A similar result is obtained by using the other triangle:
sin a  tan bcot 
4
Now set equation 1 = equation 4:
R sin a cot    R cosc tan b
R sin a cot  
cosc 
R tan b
sin a cot  
cosc 
tan b
tan b cot   cot  
cosc 
tan b
cosc  cot   cot  
Now set equation 2 = equation 3:
R sin c cos   R cosasin b
R cosasin b
cos  
R sin c
cosasin csin  
cos  
sin c
cos   cosasin  
A similar result is obtained by using the other triangle:
cos   sin  cosb
Now set equation 2 = equation 4:
R sin c cos   R cosc tan b
R cosc tan b
cos  
R sin c
cos   cot c tan b
A similar result is obtained by using the other triangle:
cos   cot ctan a
5
Now set equation 3 = equation 4:
R cosasin b  R cosc tan b
R cosasin b
cosc 
R tan b
 sin b cosb 

cosc  cosa

sin b 
 1
cosc  cosa cosb
Consider the relation between a , the angle measure, and a , the arc length. The ratio of the
angle to 2 equals the ratio of the arc length to the circumference.
a
a

2 2R
a
a 
R
We are now at the place where we can derive the Law of Sines and the Law of Cosines for
spherical triangles. From the first two equations previously derived for right triangles, substitute
a
in
R
a
b
sin  
sin  
b
 R  and sin     R  .
for a and
in for b to obtain sin   
R
c
c
sin  
sin  
R
R
Any oblique triangle can be subdivided into two right triangles. Merely construct an arc of a
great circle that passes through one vertex and is perpendicular to the opposite side. We assume this is
true without proof. Call this arc m . (see figures below)
B
a


C
B
b

c
D
a

1
b
m
c1
2
C

A
D c2
A
6
Using the formulas for right spherical triangles, we now have the following:
m
m
sin  
sin  
 R  and sin   
R
sin   
b
a
sin  
sin  
R
R
Dividing the one by the other produces the following relation.
m
a
sin   sin  
sin  
R
R
    
sin  
b
m
sin   sin  
R
R
a
sin  
sin  
R
  
sin  
b
sin  
R
sin  
sin  

a
b
sin   sin  
R
R
Instead of taking the projection of C onto AB , we could have taken the projection of A onto
BC . This derivation would produce the following. (see figure below)
B

a1
a2 C

m
c
b
A
Using the formulas for right spherical triangles, we now have the following:
m
m
sin  
sin  
R
R
sin   
and sin   
c
b
sin  
sin  
R
R
7
Dividing the one by the other produces the following relation.
m
c
sin   sin  
sin  
 R R

sin  
b
m
sin   sin  
R
R
c
sin  
sin  
R

sin  
b
sin  
R
sin  
sin  

c
b
sin   sin  
R
R
Combining these results together gives the Law of Sines for hyperbolic triangles.
sin  
sin  
sin  


a
b
c
sin   sin   sin  
R
R
R
The spherical triangle diagram introduced when we started is reproduced again to aid in the
derivation of the Law of Cosines for spherical triangles.
B

R
a
c
a
c
O
b

D
E 
A
C
b
Once again we derive two formulas for BE and equate them.

BE   OB   OE 

BE   BD   DE 
2
2
2
2
2
2
 2  OB  OE  cosa
* from the flat triangle OEB
 2  BD  DE  cos 
* from flat triangle DEB
8
OB   OE   2  OB  OE  cosa  BD   DE   2  BD  DE  cos 
 2  OB  OE  cosa  OB   OE   BD   DE   2  BD  DE  cos 
2  OB  OE  cosa  OB   OE   BD   DE   2  BD  DE  cos 
2  OB  OE  cosa  OB   BD   OE   DE   2  BD  DE  cos 
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
Note that OB and BD are the hypotenuse and leg respectively of the right triangle OBD . The same can
be said regarding OE and DE in right triangle OED .
   
2  OB  OE  cosa   2OD   2  BD  DE  cos 
2OD   2  BD  DE  cos 
cosa  
2  OB  OE  cosa   OD  OD  2  BD  DE  cos 
2
2
2
2
2  OB  OE
OD OD BD DE
cosa  



 cos 
OB OE OB OE
cosa   cosc  cosb   sin c sin b  cos 
a
c b
c b
cos   cos  cos   sin   sin   cos 
R
R R
R R
Similar derivations prove the following.
b
c a
c a
cos   cos  cos   sin   sin   cos 
R
R R
R R
c
b a
b a
cos   cos  cos   sin   sin   cos 
R
R R
R R
9
This section of the paper conducts an investigation into regular spherical polygons. A regular
spherical polygon is a closed concave figure consisting of sides of equal length and having vertex angles
with the same measure. We will assume that any regular polygon is centered at a pole of the sphere.
When we define a regular spherical polygon we need to provide the number of sides, n , and the flat
length of each side, s , and the radius, R , of the sphere on which the polygon is constructed. A brief
explanation is needed for the variable s . Imagine a flat square inside a sphere having vertices on the
surface of the sphere and parallel to the plane that defines the polar of the sphere (see figure below). The
side length of the flat Euclidean square that defines these vertices is s . However, when the square is
actually constructed on the sphere it will have curved edges.
s
Note that as the vertices of the square approach the polar the sides will begin to form a circle if
viewed from above the pole looking down at the trace of the figure. We also see that the biggest square
which can be circumscribed by a “slice” of the sphere is located in the plane that defines the polar. For
any side length except that corresponding to the side length of a square with vertices on the polar there
exists two possible options for the square (see figure below). These figures are congruent in that they
have the same perimeter and surface area and so we will handle only those on top of the sphere.
10
s
s
Now we seek to derive some formulas for these figures with the ultimate goal of determining the
perimeter and the area of any regular polygon. First, we note that the flat length of a side corresponds to
a unique arc of a great circle since it is a chord of the circle. We create an Isosceles triangle with the
chord of length s at its base. Then we split this triangle into two right triangles and solve for  , the
central angle of the great circle.
s
s
2
s
R
R
R


s
 
   2
sin   
R
2
s
 
sin   
 2  2R

 s 
 sin 1 

2
 2R 
 s 
  2 sin 1  
 2R 
11

2
Now the length of the arc corresponding to the chord of length s , is the ratio of angle B to 2
times the circumference of the sphere. Let Ls  denote the arc length corresponding to a chord of length
s . Then Ls  


 s 
1  s 
2R  R   2 sin 1 
  R  2 R sin 
 . Since each edge arc is of the same
2
 2R  
 2R 

size, the perimeter of the polygon with n edges of length s , is given by P( s, n)  n  L( s) .
The problem that now arises is that on a sphere of a given radius, polygons have a maximum side
length. We have already noted that the square of maximum perimeter, and obviously maximum area as
well, has its vertices on the polar. This holds true for all polygons. Consider a polygon with n sides,
when the polygon’s vertices lie on the polar they evenly subdivide the circumference of the sphere, as
such, the central angle that marks them off has a measure of
2
. Using this fact we can construct an
n
Isosceles triangle as before, but this time we will solve for the maximum side length, denoted s max . We
can use the formula that we developed before by letting  
2
.
n
 s max 

 2R 
2
s 
 2 sin 1  max 
n
 2R 
  2 sin 1 

s 
 sin 1  max 
n
 2R 
  s
sin    max
 n  2R
 
s max  2 R sin  
n
Notice that when we let s  s max we expect to obtain the circumference of the sphere for any given
radius and for any given number of sides as the perimeter of that polygon. The following calculation
shows that this is indeed the case.
12
Ps max , n   n  Ls max 

 
 2 R sin   
 n 
 n  2 R sin 1 


2R




   
 n  2 R sin 1  sin   
  n 
 n  2R 

n
 2R
We now turn our attention to finding the area of a regular spherical polygon. As we did when
considering the perimeter of a spherical polygon, we will split the spherical polygon into triangles,
determine the area of the triangles and then multiply the result by n , the number of sides. The area of a
spherical polygon, denoted A is given by the following formula: A    n  2 R 2 , where  is the
angle sum of the polygon. To find the angle at each vertex of the triangle we will need the tangents to
the vertices. To find the tangents at each point we will need to know the planes containing the great
circles that these points lie on. To simplify matters in our derivation we will consider the spherical
triangle to be lying on it’s side on the plane z  0 .
z
A
OM  d
FP  r
R
P
OP  t1
PA  t 2
FA  a
N
O
B
AM  h
ON  d 
y
R
M
F
x
13
s
Since A, B, F are all points on the surface of the sphere, then OA  R, OB  R, OF  R . Also
since F and B are vertices of the spherical polygon then FB  s , the side length. Let M be the
midpoint of FB , then AM  FB since AF  AB . Consider triangle OMF . (see figure below)
O
R
F
s
d  R  
2
d
2
2
M
s
2
Let P  ProjOA F , then P is the center of the flat polygon. Consider triangle FPB . (see figure
below) Since  is a central angle of the flat polygon then we know that  
2
.
n
P
P

r
r
F
s
F
B
s
 
   2 
sin   
r
2
  2

n
sin  
 2




  s
 2r


  s
sin   
 n  2r
r
14
s
 
2 sin  
n
s
2
M

2
Now that we know r , consider triangle FPO . (see figure below)
t1
O
P
t1  R 2  r 
2
r
R
F
This makes PA  R  t1  t 2 . Consider triangle FPA . (see figure below)
A
a  r 2  t 2 
a
t2
F
r
P
2
Now consider triangle FMA . (see figure below)
A
a
F
s
h  a  
2
h
s
2
2
2
M
Now A lies in the plane x  0 , and dist O, A  R as well as dist M , A  h . Consider the circle
of radius R centered at O , the equation is z 2  y 2  R 2 . Also there is a circle of radius h centered at
M , the equation of which is z 2   y  d   h 2 . (see figure below) Arranging both equations so they are
2
equal to zero and setting them equal produces the following.
15
z
A
h
R
y
M
O
d
z 2  y 2  R 2  z 2   y  d   h2
2
y 2  R 2   y  d   h2
2
y 2  R 2  y 2  2dy  d 2  h 2
 R 2  2dy  d 2  h 2
R 2  2dy  d 2  h 2
R 2  d 2  h 2  2dy
R 2  d 2  h2
y
2d
d 
R 2  d 2  h2
2d
Let N  Pr ojOM A . Now consider triangle ONA . (see figure below)
A
zh  R2  d 
2
R
zh
O
d
N
We now know the coordinates of the vertices A, F , B . (see table below)
Point
A
F
B
x
0
s
2
s

2
y
d
z
zh
d
0
d
0
We will now find the equation of the plane that passes through O, A, F , call it the front plane,
and the equation of the plane that passes through O, A, B , call it the back plane.
16
Since the origin is in our plane we can use the coordinates of A and F as vector components.
First we find the normal vector to the plane by crossing the vector A with the vector F , call it N f .
N f  A F

s
 s 
 d 0   z h d , z h    00 , 0d   d  
2
 2 


 s   s 
  z h d , z h  , d  
 2   2 

Now the equation of the tangent plane is N f 0x  x0   N f 1 y  y 0   N f 2z  z 0   0 , where
x0 , y0 , z 0  is some point in the plane. Since the origin is in the plane use that as the point and you the
following equation is produced for the front plane.
s
s
 z h d x  z h   y  d   z  0
2
2
A similar derivation of the back plane, produces the following.
 s
 s
 z h d x  z h    y  d    z  0
 2
 2
The equation of the whole sphere is x 2  y 2  z 2  R 2 . The gradient of this equation evaluated
at any point on the surface will give us the normal vector to the sphere at that point.
f x, y, z   2x  2 y  2z
A summary of the four vectors needed is presented below.
v1
Normal to sphere at F
iˆ
s
v2
Normal to front plane
 z h d 
v3
Normal to sphere at A
0
v4
Normal to back plane
 z h d 
vector
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ĵ
2d 
s
zh  
2
2d 
 s
zh   
 2
k̂
0
s
d  
2
2z h 
 s
d   
 2
We now want to find the tangent at F along the great circle connecting F to A . We do so by
taking the cross product of the normal vector to the sphere at point F with the normal vector to the front
plane. (see diagram below)
z
A
B
O
y
F
x
The blue vector is the normal vector to the sphere, the red vector is the normal vector to the plane
and the green vector is the resulting cross product of the first two. It is tangent to the great circle through
F and A at F since it is in the plane and the normal to the sphere was evaluated at point F . Using the
normal vector to the plane z  0 and the normal vector to the sphere at point F , we obtain the tangent
vector to the great circle passing through F and B at F . We use the vector angle formula to find the
angle between these two tangent vectors which is equal to the spherical angle at F . This angle is also
equal to the angle at B , since AF  AB .
We repeat a similar process to find the angle at A . Cross the normal vector to the sphere at A ,
v3 , with the normal vector to the front plane, v2 . This gives us a tangent vector. Similarly crossing the
normal vector to the back plane, v4 , with v3 gives the other tangent vector. These cross products must
be taken with the right hand rule in mind to provide the tangent in the correct direction. A summary is
presented below.
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vector
u1
u2
u3
formula
v2  v1
0,0,1 v1
v3  v 2
u4
v 4  v3
Once these angles are found we can use the formula for the area of a polygon where n  3 since
we are working with a triangle and if  is the angle sum given by the above proceedings we have
A    n  2 R 2 . However, this is the area of one piece of the regular polygon and so we must
multiply this area by n to obtain the area of the whole polygon. Each of these variable derivations can
be written as function for the TI-89. Then the functions can be combined together to produce a single
spherical polygon area program that accepts a side length, s , the number of sides, n , and the radius of
the sphere, R , and returns the area of the spherical polygon. To keep this simple each function should
accept s, n, R and call the other functions to compute the various pieces in each formula.
The maximum area of a spherical polygon is half the surface area of the sphere. Since the surface
area of a sphere is 4R 2 , the maximum area of a spherical polygon is 2R 2 . This should correspond to
the polygon with maximum side length. Using the TI-89 functions mentioned previously and the
 2
 
  5 2 one
maximum side length for a square on a sphere of radius 5, which is 2(5) sin    10

4
 2 
obtains a surface area of 50 for both the half sphere and the area of the spherical polygon.
In closing, I would like to thank two people who assisted me in writing this paper. First, I thank
my brother, Matt Black, who spent several hours helping me visualize the surface area problem and
suggested tipping the triangle onto the plane to derive the formulas. Second, I thank Dr. Guthrie who
helped me wrestle with an attempt to integrate the spherical triangle region. Though we were
unsuccessful in finding the surface area this way his help on the use of the vectors was much
appreciated.
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