Meas. Sci. Technol. 10 (1999) 1015–1019. Printed in the UK
PII: S0957-0233(99)01233-3
The packing of circles on a
hemisphere
J Appelbaum and Y Weiss
Tel-Aviv University, Faculty of Engineering, Department of Electrical Engineering–Systems,
Tel-Aviv 699 78, Israel
Received 25 January 1999, in final form and accepted for publication 15 July 1999
Abstract. The problem of the closest packing of N equal nonoverlapping circles on a sphere
has been of interest in geometry, chemistry, biology, engineering and optimization. The
problem of packing N equal nonoverlapping circles on a hemisphere is a different problem of
the same type and is also of interest in various fields such as detecting signals in a
multisource neighbourhood and measuring solar radiation. The paper introduces a method
for locating N equal nonoverlapping circles on a hemisphere arranged along parallel rings.
The method provides information on the elevation and orientation angles of the circles and
circle size; and calculates the surface coverage by the circles. The circles may represent the
view angles of detectors arranged on a hemisphere. The solution is given for 2–40 detectors.
Keywords: packing of circles, detectors, signal detecting, mapping
1. Introduction
The problem of the closest packing of N equal
nonoverlapping circles (spherical caps) on a sphere so that
the size of the circles is maximized has been of interest in
geometry, chemistry, biology, engineering and optimization.
This problem is sometimes known as Tammes’ problem
[1] after the Dutch botanist and originated from a study
of the pattern of orifices in spherical pollen grains. The
closest packing of circles on a spherical surface was also
examined with the use of techniques that had been developed
to determine the stereochemical arrangement of atoms around
a central atom [2]. Packing problems were involved also in
coding theory, in which the distance between code words is
maximal [3]. The packing of circles can be presented by
graphs of packing [4, 5]. The references listed are only a few
of a greater list.
A different problem, of the same type, is that of
packing N equal nonoverlapping circles on a hemisphere
since circles are not allowed to cross the boundary of the
hemisphere. In another case, the sphere is manufactured from
two hemispheres. This happens to golf balls and geodynamic
satellites, for instance. The dimples on the golf ball [6] and
circular mirrors on the LAGEOS satellite [7, 8] are arranged
along parallel small circles of a sphere, similarly to the
arrangements introduced in this paper. LAGEOS (Laser
Geodynamics Satellite) is one of the first artificial satellites
developed exclusively for geodynamic measurements using
laser-ranging techniques. It carries an array of 426 prisms
called cube-corner retroreflectors, giving it the dimpled
appearence of a golf ball. Detectors arranged on a hemisphere
may measure directional radiation emanating from the sky,
which permits the calculation of the direct beam, diffuse and
global solar irradiances on inclined surfaces.
0957-0233/99/111015+05$30.00
© 1999 IOP Publishing Ltd
This paper introduces a method for locating N equal
nonoverlapping circles (spherical caps) on a hemisphere so
that the size of the circles is maximized when the circles are
packed in disjoint parallel rings. The method is based on a
numerical solution and provides information on the elevation
and azimuth angles of each circle and the percentage surface
coverage of the circles on the hemisphere. The solution is
given in this paper for 2–40 circles. The circles may represent
the viewing angles of detectors arranged on a hemisphere.
2. The disposition of detectors
2.1. The definition of the problem
We are given N equal detectors (i.e. having the same viewing
angles) to be arranged on a hemisphere of radius R. Each
detector is characterized by the angles αi , γi and η, where
αi is the elevation angle with respect to the basal plane
of the hemisphere, γi is the detector’s angular distance (or
azimuth) from a reference direction in the basal plane of the
hemisphere, e.g. S (South-positively in a clockwise direction)
and η is the viewing angle (see figure 1(a)).
The problem is this. Find angles
(αi , γi , η)
i = 1, 2, . . . , N
(1)
to obtain an optimal disposition of the detectors on the
hemisphere. An optimal disposition is defined as a set of
angles αi , γi and η, i = 1, 2, . . . , N that yields maximal
surface coverage without an overlapping of the coverage
regions of the different detectors. The requirement of
nonoverlapping is added because in measuring directional
solar radiation (or any other directional signal) it might be
difficult to compensate for the detectors’ overlapping regions.
1015
J Appelbaum and Y Weiss
Thus it is obvious that the viewing angle of the detectors for
the optimal disposition is the maximal angle that satisfies the
condition of nonoverlapping.
2.2. A different problem of the same type
We are given N unequal detectors to be arranged on a
hemisphere of radius R. Each detector is characterized by
the angles αi , γi and ηi , where αi is the elevation angle with
respect to the base plane of the hemisphere, γi is the detector’s
azimuth with respect to a reference direction and ηi is the
detector’s viewing angle.
The problem is this. Find angles
(αi , γi , ηi )
i = 1, 2, . . . , N
to obtain a disposition satisfying the following conditions.
(i) There is no overlapping of the regions of coverage of the
different detectors.
(ii) The minimal viewing angle of the detectors is the largest
minimal viewing angle of all possible arrangements
satisfying condition (i), i.e. max{min(ηi )}.
(iii) It is required that the detectors have a maximal surface
coverage.
Figure 1. The detector: (a) characteristic angles and (b) the
surface area of a spherical sector.
Figure 2. The relationship between elevation and viewing angles.
The above definition determines uniquely the viewing angle
η but still leaves some degree of freedom for the positions
of the detectors, i.e. the angles (αi , γi ), i = 1, 2, . . . , N. To
solve this, another requirement is added, which says that,
under the above definition, the detectors are disposed such
that they are as far apart from each other as possible. This
ensures a more uniform coverage of the surface (for example
the sky in the case of solar-radiation measurements). Note,
however, that this requirement is redundant if, in the optimal
solution, the coverage regions of the detectors are tangential
in a way that the detectors cannot move.
The percentage of surface coverage (PSC) will now be
determined with the help of figure 1(b). From figure 1(b) we
have
h = R(1 − cos η).
(2)
The surface area of a spherical sector (cap) is
S = 2πRh = 2πR 2 (1 − cos η)
(3)
and the percentage of surface coverage of N equal detectors
is therefore
PN
2πR 2 (1 − cos η)
= N(1 − cos η). (4)
P SCeq = i=1
2πR 2
1016
Here, conditions (i) and (ii) ensure that the minimal ηi of a
detector is equal to η of the optimal solution for the problem
in section 2.1. Condition (iii) removes the degree of freedom
in the detectors’ positions because each detector is in contact
with at least one neighbour. Because the viewing angles of
the detectors are maximized (to obtain the maximal surface
coverage), each detector is positioned as far as possible from
its neighbours. The percentage of surface coverage of N
unequal detectors is
PN
N
X
2π R 2 (1 − cos ηi )
=
N
−
cos ηi . (5)
P SC = i=1
2π R 2
i=1
The solution to problem 2.2 provides the position of the
detectors (αi , γi ) and the minimal viewing angle (min(ηi )).
Note that the same positions of the detectors as those in
problem 2.2 is also the solution to problem 2.1, but with the
same minimal viewing angle ηi for all detectors. It should
be noted, in this case, that some detectors need not be in
contact with their neighbours; therefore, the PSC for N equal
detectors may be lower than that for N unequal detectors.
2.3. Relaxed assumptions
For practical engineering reasons we may want to place the
detectors on parallel circles on the surface of the hemisphere.
This requires relaxed assumptions for the solution of the
problem in section 2.2, bearing in mind a possible deviation
from the optimal solution. The assumptions are as follows.
(i) The detectors are arranged on the hemisphere in
concentric circles. All detectors in the same circle have
equal viewing angles. Detectors in different circles may
have different viewing angles. The number of detectors
in each circle is at least two (excluding the case that one
detector may be positioned at α = 90◦ ). The detectors
are equally distributed on the circumference of each
circle.
Circles on a hemisphere
(ii) The elevation angle between two successive circular
planes is equal to the sum of the viewing angles of two
detectors, one in each circle. This ensures that there is no
overlapping of detectors in adjacent circles (see figure 2).
The problem to solve is this. Given an arrangement of
detectors, i.e. the number of circles c and the number of
detectors in each circle nc , find the elevation of the circle
αc with respect to the plane of the hemisphere basis, the
viewing angle of the detectors in each circle ηc and the
angular distance (azimuth) of each detector from a reference
direction, all satisfying conditions (i)–(iii) in section 2.2.
The solution is obtained circle by circle, starting with
the lowest one. For each circle the elevation and viewing
angles (αc , ηc ) are found for known values of the previous
circle (αc−1 , ηc−1 ) and the number of detectors in the circle
nc . The angles of the circle are adjusted in such a way that the
detectors’ coverage regions just touch (without overlapping).
The following is the algorithm for finding the angles.
According to assumption (ii), the elevation angle αc of
circle c is related to the elevation αc−1 and viewing ηc−1
angles of the previous circle c − 1 and to the viewing angle
ηc (see figure 2), i.e.
αc = αc−1 + ηc−1 + ηc .
(6)
To solve equation (6) one needs first to determine ηc . The
maximum viewing angle ηc is determined by the tangentiality
(for nonoverlapping) of the regions covered by the detectors.
The coverage of a detector is a spherical sector (cap). The
arc angle of a circular cross section parallel to the base plane
and passing through the sector varies with the elevation angle
of the detector. This is described in figure 2. The arc angle
is denoted by 2ζ and is a function of the detector’s viewing
and elevation angles and the elevation angle of the arc αζ , i.e.
2ζ = f (ηc , αc , αζ ).
The angle ζ as a function of αζ has a maximum
determined by the tangentiality of two coverage regions
of two adjacent detectors lying on the same circle. The
condition of tangentiality is given by
ζm = π/nc .
(7)
The dependence of the angle ζm as a function of ηc will now
be determined. For this purpose we resort to the two vectors
A and B in figure 2 with polar coordinates (R, αc , γc ) for
vector A and (R, αζ , γζ ) for vector B .
The angle between vectors A and B is ηc by
definition and the azimuths are γc and γζ , respectively.
The Cartesian coordinates of the two vectors are
(R cos αc cos γc , R cos αc sin γc , R sin αc ) for vector A and
(R cos αζ cos γζ , R cos αζ sin γζ , R sin γζ ) for vector B and
the scalar product of these vectors is
R 2 cos ηc = R 2 cos αc cos γc cos αζ cos γζ
+R 2 cos αc sin γc cos αζ sin γζ + R 2 sin αc sin αζ .
To simplify, we assume that the azimuth of vector A is zero,
i.e. γc = 0; thus γζ = ζ and we obtain
cos ζ =
cos ηc − sin αc sin αζ
.
cos αc cos αζ
(8)
The maximal arc angle is obtained from the derivative
d cosζ /dαζ = 0 resulting in
sin αζ m =
sin αc
cos ηc
(9)
where αζ m is the elevation angle of a cross sectional
circular plane having a maximum arc (maximum angle)
of the spherical sector of the detector’s coverage region.
On substituting equation (9) into equation (8), after
manipulation, we get
cos ζm =
(cos2 ηc − sin2 αc )1/2
.
cos αc
(10)
By substituting equation (6) into equation (10) and equating
it to equation (7), we obtain
π
[cos2 ηc − sin2 (αc−1 + ηc−1 + ηc )]1/2
= cos
. (11)
cos(αc−1 + ηc−1 + ηc )
nc
The angles ηc−1 and αc−1 are known from the solution of the
previous circle. The solution of ηc (for the present circle) is
obtained numerically from equation (11) and the elevation
angle αc is then computed from equation (6). For a single
detector on top of the hemisphere αc = π/2, the viewing
angle is calculated by using equation (6) without resorting to
equation (11), i.e.
η90 = π/2 − ηc−1 − αc−1 .
(12)
Thus, equations (6) and (11) solve the elevation αc and
viewing ηc angles for the present circle c for known angles
αc−1 and ηc−1 of a previous circle c − 1, for a given number
of circles nc of the present circle.
The solution for a given arrangement (c, nc ) starts with
the initial conditions αc−1 = ηc−1 = 0, thus obtaining the
angles for the first circle, i.e. αc=1 and ηc=1 . These values
serve now to calculate the angles for the second circle. Note
that, if the viewing angle of the second circle is smaller
than that of the first circle, then the viewing angle of the
first circle is decreased iteratively until the viewing angles
of the two circles are equal. This is done in order to
maximize the smallest angle in the arrangement to satisfy
condition (ii) in section 2.2. After the viewing angle of
the second circle has been determined, the viewing angle
of the third circle is calculated. Again, if the viewing angle
is smaller, adjustments to the previous circles are made, etc.
The procedure described finds the best (αc , ηc ) for a given
arrangement (c, nc ).
The angular distance (azimuth) with respect to a
reference direction of each detector has not yet been
determined. The azimuth angle has not been taken into
account directly in the determination of the best arrangement
described above because assumption (ii) in section 2.2
precludes overlapping of the viewing angles of detectors in
the different circles. This degree of freedom (azimuth angle)
may impair the optimality of the solution to some degree.
The azimuths of the detectors are determined as follows. For
each circle it is sufficient to assign an azimuth to one detector
only. The other detectors in the same circle are then evenly
distributed. In adjacent circles, the azimuths of the detectors
1017
J Appelbaum and Y Weiss
Table 1. Dispositions of detectors for N = 2–40.
Circle 1
Circle 2
N
c
n1
α1
η1
γ1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
1
1
1
1
1 + top
1 + top
1 + top
2
2
2
2
2
2
2 + top
2 + top
2 + top
2 + top
2 + top
2 + top
3
3
3
3
3
3
3
3
3 + top
3 + top
3 + top
3 + top
3 + top
3 + top
3 + top
3 + top
3 + top
4
4
4
2
3
4
5
5
6
7
7
7
8
8
9
9
9
9
10
10
11
11
11
11
12
12
12
12
13
13
13
13
13
14
14
14
15
15
15
15
15
16
45.0
40.9
35.3
30.5
30.0
26.6
23.5
22.5
21.6
20.9
20.2
18.9
18.9
18.0
17.6
17.2
16.4
15.7
15.4
15.0
14.6
14.5
14.4
14.0
13.6
13.5
13.4
12.8
12.8
12.8
12.5
12.2
12.1
11.7
11.6
11.4
11.3
11.1
11.0
45.0
40.9
35.3
30.5
30.0
26.6
23.5
22.5
21.6
20.9
20.2
18.9
18.9
18.0
17.6
17.2
16.4
15.7
15.4
15.0
14.6
14.5
14.4
14.0
13.6
13.5
13.4
12.8
12.8
12.8
12.5
12.2
12.1
11.7
11.6
11.4
11.3
11.1
11.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
n2
1
1
1
2
3
3
4
4
5
5
6
6
7
7
8
8
8
8
9
9
10
10
10
10
11
11
11
11
12
12
12
13
13
13
13
α2
90.0
90.0
90.0
67.5
64.8
64.1
60.7
59.1
56.7
54.0
52.8
52.2
49.3
48.3
46.2
45.0
43.8
43.7
43.3
42.0
40.8
40.5
40.2
38.4
38.4
38.4
37.9
36.6
36.3
35.6
34.8
34.2
33.9
33.4
33.2
η2
30.0
36.8
43.0
22.5
21.6
22.3
20.3
21.3
18.9
18.0
17.6
17.8
16.5
16.9
15.4
15.0
14.6
14.7
14.5
14.0
13.6
13.5
13.4
12.8
12.8
12.8
12.9
12.2
12.1
12.2
11.6
11.4
11.3
11.2
11.2
Circle 3
γ2
0.0
0.0
0.0
12.9
8.6
7.5
22.5
5.0
4.0
4.0
10.0
6.0
2.6
2.3
2.1
2.1
2.1
7.5
5.0
5.0
3.0
1.4
1.4
1.4
1.3
1.3
1.2
1.2
2.1
3.0
3.0
0.9
0.9
0.9
0.9
n3
1
1
1
1
1
1
2
3
3
3
4
4
4
5
5
5
6
6
7
7
7
8
8
8
9
9
α3
η3
90.0
90.0
90.0
90.0
90.0
90.0
75.0
73.1
73.1
72.7
70.0
68.9
68.9
67.0
64.1
64.1
64.0
63.6
61.0
60.6
60.2
58.2
57.5
56.5
55.8
55.6
18.0
19.6
20.0
24.2
24.8
28.4
15.0
14.7
14.7
14.9
14.0
14.5
14.9
13.4
12.9
12.9
12.8
12.9
12.2
12.2
12.4
11.8
11.9
11.3
11.2
11.2
Circle 4
γ3
0.0
0.0
0.0
0.0
0.0
0.0
22.5
7.5
7.5
20.0
5.0
9.0
9.0
18.0
18.0
3.3
2.7
2.7
2.3
2.1
2.1
7.5
1.7
1.7
1.5
1.5
n4
1
1
1
1
1
1
1
1
1
2
2
2
α4
90.0
90.0
90.0
90.0
90.0
90.0
90.0
90.0
90.0
79.1
78.4
78.3
η4
13.0
13.0
13.2
13.4
17.0
17.2
17.4
20.0
20.6
11.3
11.4
11.5
γ4
PSC
η
PSCeq
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
22.5
10.0
10.0
58.578
73.244
73.545
69.185
80.385
83.434
84.923
68.508
70.224
75.074
74.050
75.847
75.480
73.415
76.008
79.475
78.300
80.492
80.254
71.556
71.172
74.228
76.455
74.261
74.427
77.000
76.228
72.339
74.824
77.194
78.814
76.639
78.045
79.172
78.084
78.828
73.664
73.905
75.310
45.0
40.9
35.3
30.5
30.0
26.6
23.5
22.5
21.6
20.9
20.2
18.9
18.9
18.0
17.6
17.2
16.4
15.7
15.4
15.0
14.6
14.5
14.4
14.0
13.6
13.5
13.4
12.8
12.8
12.8
12.5
12.2
12.1
11.7
11.6
11.4
11.3
11.1
11.0
58.578
73.244
73.545
69.185
80.385
74.092
66.352
68.508
70.224
72.375
73.808
70.089
75.480
73.415
74.845
76.026
73.235
70.886
71.809
71.556
71.040
73.260
75.400
74.261
72.901
74.601
76.228
72.067
74.552
77.037
75.853
74.528
75.537
72.720
73.529
72.997
73.664
72.959
73.491
are determined by maximizing the minimal distance between
detectors. The azimuth of the first detector in each circle is
denoted by γc . Figure 3 shows the azimuthal arrangement of
the detectors in two circles for N = 21, n1 = 12 and n2 = 9,
where the azimuths are in degrees with respect to the South
(S) direction (positively in the clockwise direction).
2.4. The arrangement of N detectors on the hemisphere
To determine the disposition of the detectors on the
hemisphere, one has first to find all possible arrangements
for a given N detectors. A possible arrangement is defined
by c circles having nc detectors in the circle. For N = 6 for
example, there are eight possible arrangements:
(i) one circle with n1 = 6;
(ii) one circle with n1 = 5 and n2 = 1 (α2 = 90◦ );
(iii) two circles with n1 = 4 and n2 = 2;
(iv) two circles with n1 = 3 and n2 = 3;
(v) two circles with n1 = 2 and n2 = 4;
(vi) two circles with n1 = 3, n2 = 2 and n3 = 1
(α3 = 90◦ );
(vii) two circles with n1 = 2, n2 = 3 and n3 = 1
(α3 = 90◦ ); and
(viii) three circles with n1 = 2, n2 = 2 and n3 = 2.
For the above possible arrangements the elevation αc ,
the viewing ηc and the azimuth γc angles are determined
according to section 2.3. For each arrangement the smallest
1018
Figure 3. A top-plan view of N = 21 (n1 = 12, n2 = 9)
detectors.
viewing angle is recorded, i.e. ηr = min(ηi ). Finally, the
preferred arrangement (among the eight possibilities) for a
given N is determined by the largest minimal viewing angle.
Circles on a hemisphere
N = 17, for example, the viewing angle is η = 17.2◦ and they
cover 76.026% of the surface of the hemisphere (equation 4).
One may conclude from table 1 that the practical solution
of equal detectors does not, in general, significantly impair
the solution for PSC for unequal detectors. We compare
also the PSC of the disposition of detectors on a hemisphere
having 79.475% coverage for unequal detectors and 75.8%
coverage for equal detectors with the maximum PSC (optimal
solution) for N = 34 equal detectors on a sphere having
80.4% coverage [9] (corresponding to N = 17 detectors
on a hemisphere). The disposition and surface coverage
of 31 identical detectors, for example, on a hemisphere are
shown in figure 4.
3. Conclusions
Figure 4. The disposition and surface coverage of 31 identical
detectors on a hemisphere.
For practical reasons one may prefer to use equal
detectors (ones with equal viewing angles). The viewing
angle is then the smallest of all viewing angles satisfying
the condition of nonoverlapping, i.e. η = min(ηi ), i =
1, 2, . . . , N. Table 1 lists the dispositions of the detectors
on a hemisphere for N = 2–40 detectors. In table 1 N
denotes the number of detectors, c is the number of circles
on the hemisphere, n is the number of detectors in each
circle, α is the elevation angle with respect to the base plane,
η is the detector viewing angle; γ is the azimuth of the
first detector in the given circle with respect to the South
direction and PSC is the percentage of surface coverage,
all for N unequal detectors. The η and P SCeq in the last
two columns are the viewing angle of the detectors and the
percentage surface coverage, respectively, for a hemisphere
having equal detectors. For example, a hemisphere with
N = 17 detectors is arranged in two circles. In circle 1
there are n1 = 10 detectors with elevation angle α1 = 17.2◦
and viewing angle η1 = 17.2◦ ; in circle 2 the number
of detectors is n2 = 6 with elevation angle α2 = 52.2◦
and viewing angle 17.6◦ ; and one detector is on the top of
the hemisphere, having a viewing angle of 20◦ . The first
detector in circle 1 is positioned at azimuth γ1 = 0◦ and
the detectors are uniformly distributed 36◦ apart. The first
detector in circle 2 is positioned at azimuth γ2 = 6◦ and the
detectors are uniformly distributed 60◦ apart. The percentage
surface coverage is 79.475 (equation (5)). Table 1 lists also
the arrangements of equal detectors on a hemisphere. For
The problem of locating N equal detectors along small circles
in disjoint zones of a hemisphere such that no overlapping
of the viewing angles of the detectors occurs was solved
mathematically. This type of solution may be of interest
in various fields, such as in signal detecting and solar
energy. The solution is given for numbers of detectors
in the range 2–40, providing information on the elevation,
azimuth and viewing angles as well as the percentage surface
coverage. The criterion of nonoverlapping guarantees
directional detection and the largest possible viewing angles
provide optimal surface coverage. The disposition of the
detectors is in concentric circles on the hemisphere. Each
circle contains a different number of detectors.
References
Tammes P M L 1930 On the origin of number and arrangements of
the places of exit on the surface of pollen-grains Recueil
Travaux Botaniques Néderlandais 27 1–84
Clare B W and Kepert D L 1986 The closest packing of equal
circles on a sphere Proc. R. Soc. A 405 329–44
Hardin R H and Sloane N J A 1995 Codes (spherical) and design
(experimental) Proc. Symp. Applied Mathematics vol 50,
pp 179–206
Tarnai T and Gaspar Zs 1983 Improved packing of equal circles on
a sphere and rigidity of its graph Math. Proc. Camb. Phil.
Soc. 93 191–218
Danzer L 1986 Finite point-sets on S 2 with minimum distance as
large as possible Discrete Math. 60 3–66
Tarnai T 1996 Symmetry of golf balls Katachi U Symmetry
ed T Ogawa et al (Tokyo: Springer) pp 207–14
1976 Since May of this year the Earth has an artificial moon Laser
Elektro-Opt. No 4, pp 24–7
Cohen S C and Smith D E 1985 LAGEOS scientific results:
introduction J. Geophys. Res. 90 9217–20
Kottwitz D A 1991 The densest packing of equal circles on a
sphere Acta Crystallogr. A 47 158–65
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