Polar Zenithal Map Projections

POLAR
ZENITHAL
MAP
Part 1
PROJECTIONS
Five basic
projections
by Keith Selkirk, School of Education, University of Nottingham
Map projections used to be studied as part of the geography syllabus, but have disappeared from it in recent
years. They provide some excellent practical work in trigonometry and calculus, and deserve to be more widely
studied for this alone. In addition, at a time when world-wide travel is becoming increasingly common, it is unfortunate that few people are aware of the effects of map projections upon the resulting maps, particularly in the
polar regions. In the first of these articles we shall study five basic types of projection and in the second we shall
look at some of the mathematics which can be developed from them.
The zenithal and cylindrical projections are limiting cases of
Polar Zenithal Projections
the conical projection. This is illustrated in Figure 1. The
A football cannot be wrapped in a piece of paper without either
stretching or creasing the paper. This is what we mean when
semi-vertical angle of a cone is the angle between its axis and a
we say that the surface of a sphere is not developable into a
plane. As a result, any plane map of a sphere must contain distortion; the way in which this distortion takes place is determined by the map projection. There are three fundamental
types of projection, the zenithal, the conical, and the cylindrical.
In the first of these, the map is regarded as part of a plane, in
the second as part of a cone and in the third as part of a cylinder.
In the latter two cases the cone and the cylinder are unrolled
straight line on its surface through its vertex. Figure l(b) and
(c) shows cones of semi-vertical angles 600 and 300 respectively
resting on a sphere. If the semi-vertical angle is increased to 900,
the cone becomes a disc as in Figure l(a) and this is the zenithal
case. If the semi-vertical angle decreases, the vertex moves
and laid flat, these two surfaces, unlike the sphere's surface
towards the top of the diagram and in the limit, when it tends
to infinity, the cone tends to a cylinder as in Figure 1(d).
Because they start with a plane, the zenithal projections are
easier to visualise, and we shall restrict ourselves to them, even
though they are less common in practice than the other types.
being developable into planes.
The point where the plane touches the sphere is called the
Fig. 1 The conical projection and its limiting cases.
300
600o
(a) Zenithal
2
(b) Conical
(c) Conical
(semi-vertical angle 600)
(semi-vertical angle 300)
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(d) Cylindrical
March
1982
zenith, and this point may be anywhere on the sphere. There
is no fundamental difference in the projection wherever the
zenith is, but because of the way in which we define position
on the surface of the earth by means of latitude and longitude,
the case where the zenith is at a pole is the easiest to study from
show that NSQ= 1/2 (900 - 0), whence
NQ= 2R tan /2 (900 -0).
The important advantage of the stereographic projection is
that it is shape-preserving or orthomorphic. Although the scale
is not constant, the shape at any point is correct because the
distortion is the same in all directions. We shall show why this
is the case in the second article.
a mathematical point of view. These are the polar zenithal
projections we shall examine below.
The Gnomonic Projection
The Orthographic Projection
In the gnomonic projection the map may be considered as a
plane touching the earth at the north pole. A light at the centre
of the earth then casts a shadow of the earth's surface on the
If the light source is now removed a long way off so that a
parallel beam of light is produced, then Figure 4 is the result.
In this figure K is the foot of the perpendicular from P onto
the equatorial plane. In this diagram
plane map. Thus the word projection may be used literally, and
this is illustrated in the lower half of Figure 2. This shows a
cross-section through the earth and the map, the centre of the
earth being labelled O and the north pole N. The point P is a
NQ=R cos 0.
typical point in the northern hemisphere with latitude 0
The orthographic projection gives a view of the earth as seen
from outer space. The projection would be a suitable one to
(measured in degrees). The point P maps or projects onto the
point Q of the map. The upper part of the figure shows the
map itself with the pole N and the circle of latitude for P.
If R is the radius of the earth, then
use when mapping the visible surface of the moon for earthbound astronomers.
The Equidistant Projection
NQ=R cot 0
The above three projections are called perspective projections
since they are views of the northern hemisphere when seen
from particular points, namely the centre of the earth, the
south pole and an infinitely distant point in a southerly direction respectively. (In fact we are in these cases viewing the map
On the map, therefore, the radius of the circle of latitude 0 will
be R cot 0. (In practice this will be reduced by the scale of the
map, but we can ignore this for these articles, and imagine the
map drawn at full scale.)
The gnomonic projection has some grave disadvantages as
from behind.) The last two projections do not have this
we shall see, but it has one outstanding advantage. The shortest
distance between any two points on the surface of the earth is
an arc of a circle on the surface whose centre is at the centre
property and are therefore known as conventional projections.
The equidistant projection is designed in such a way that
distances measured from the zenith are correct. This means
of the earth. This is called a great circle to distinguish it from
smaller circles which may also be drawn on the earth's surface
that the concentric circles forming the circles of latitude on the
such as circles of latitude. Since the projection is from the
map have their radii equally spaced. This makes it a very
centre of the earth which is the centre of all great circles, any
natural map to use, since this is the most obvious way to draw
the basic grid of the map. In Figure 5 it means that
given great circle will have a shadow which is a plane, and
which must intersect the plane of the map in a straight line.
Thus great circles project into straight lines on the map, and
any straight line on the map is the projection of a great circle.
The gnomonic projection is therefore of great use in navigation
since it makes it easier to plan the shortest route on the surface
of the earth. We shall return to this point in the second article.
NQ=arc NP
_ (900 - 0) 2tnR
3600
This property makes it suitable for use in cases where distances
from the zenith need to be compared, remembering that the
zenith can be placed anywhere we wish.
The Stereographic Projection
The Equal Area Projection
If the light (or centre of projection) is placed at the south pole,
S, Figure 2 changes to Figure 3 and the stereographic projection is obtained. To determine NQ in this case, we must first
In the equal area projection, equal areas on the earth correspond, as might be expected, to equal areas on the map.
Fig. 2 Polar zenithal
Fig. 3 Polar zenithal
gnomonic projection.
Fig. 4 Polar zenithal
stereographic projection.
orthographic projection.
N
0
N
ao
N
A
N
N
x
o
N
x ---- p
X
x
N
Fig. 6 Polar zenithal
equal area projection.
N 0Q
i
N
NcN
0
Q
Fig. 5 Polar zenithal
equidistant projection.
a/
p
0 il
'0
Xt p
0
0
0
K
0
0
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3
Figure 6 illustrates this as far as is possible on a plane diagram.
The crux of the matter is that for all points P, the area of the
spherical "cap" of the earth bounded by the circle of latitude
on which P lies must equal the area of the circle of radius NQ
on the map, where Q is the point onto which P projects. The
cap is indicated by the shaded area of Figure 6; clearly because
of the curvature of the earth, the radius XP of its bounding
circle is less than NQ.
The area of the cap is impossible to find without the use of
the integral calculus, or some method such as the Greeks used
which effectively assumes the methods of the integral calculus.
The result is, however, comparatively well known and rather
surprising. It states that the area of the cap is equal to the area
of the curved surface of the cylinder of radius R and height
XN. Thus the area of the cap is
27tR .XN
(The same result for the whole sphere shows that the surface
area of the sphere is 4itR2 and is illustrated in the crest of the
Mathematical Association.) Now
XN= R(1 - sin 0)
Hence
7tNQ2= 2r R2 (1 - sin 0)
and
NQ=R,2 (1--sin 0)
This type of projection is of great importance and should be
used whenever symbols are used to indicate the density of a
feature on the earth's surface. If this is not done, a very misleading impression of the relative densities in different parts of
the map may be given.
Getting the answer wrong
by Lesley R. Booth, SESM Project, Chelsea College
Teachers are quite used to the idea that children sometimes get
mathematics questions wrong. Sometimes the wrong answer is
due to a "careless" mistake which is easily corrected; often
however, the teacher feels that the error is symptomatic of some
and-pencil tests of understanding in different topic areas, such
as ratio, generalised arithmetic, measurement, fractions, graphs
and so on. Examples of the kinds of wrong answer which were
found are shown in Figure 1.
misunderstanding on the part of the child. If only we could
find out why the child had made that error, we might be able
to help the child restructure his or her concept of the problem, and so avoid the error. Unfortunately, teachers usually
don't have time to diagnose every child's mistakes in this way.
Nor, of course, do research workers, but they may have a
chance of studying "common errors" - particular wrong
programme, it was possible to suggest reasons why these particular errors may have occurred.
In essence it was suggested that many of these errors might
be due to children's use of intuitive "child-methods" which are
answers which are made by large numbers of children.
This is the task which the Strategies and Errors in Secondary
perfectly adequate for handling "easy" problems, but which
do not generalise to harder questions, where success really
Mathematics (SESM) Project1 has set itself. Funded by the
Social Science Research Council and based at Chelsea College,
this project aims to investigate particular mathematical errors
identified by the earlier Concepts in Secondary Mathematics
and Science (CSMS) Project2'34,"5. The errors chosen for study
were those which were made by a large proportion (in some
cases 50% or more) of the children tested by CSMS on paperPercentage
CSMS item "Error" giving
(abridged) answer answer
(13 year olds)
2+3
7
4
5
30
11
Add 4 onto 3n 3n4, 7n 45
By carefully examining these wrong answers, and the responses
which children had given when asked about their answers in
interviews conducted as part of the CSMS test development
requires the use of the "proper" mathematical methods taught
in the classroom6,7. In addition, of course, it was thought
highly likely that there would be specific kinds of misconception
associated with the different areas of mathematics - algebra8,
number""., fractions", and so on. Consequently, the SESM
Project set out to examine these ideas by interviewing individual children on relevant problems, in order to discover
precisely which kinds of misunderstanding contribute most to
the particular errors under study.
While the results of this (still continuing) examination seem
to support the notion that children often do have their own
methods which break down when the questions get harder, the
investigation into children's work on beginning algebra, or
"generalised arithmetic", has also revealed another source of
error. It seems that in this topic at least, some of the mistakes
which children make are due not to their misinterpretation of
the question or their ideas of what letters mean, nor to the
methods which they use to solve the problem, but rather to
their misconceptions concerning the way in which the answer
Percentage
CSMS "Error" giving
Perimeter s s 2u556, 2ul 6 46
item
answer
answer
(13 year olds)
6
c C and D are 48
LD I 1 same length
912
13
1. What can you write for 5e2, e 10, 10e, 42
the area of this e + 10
rectangle:
44
21 I
e
I Fig. 1
4
21
Fig. 2
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1982