1. Introduction
Figure 1.
Suppose a circle is drawn on a sphere as shown in Figure 1,
and a light placed at the North Pole. Then the projection of g
onto the plane parallel to the XZ plane and tangent to the sphere
(this will be called the projection plane) is a circle. The proof
presented below is based on a construction involving a cone with an elliptical cross section
with vertex at the North Pole. The intersection of this cone with the sphere is precisely the
circle drawn on the sphere. One can see this by first recalling the definition of an ellipse as a
conic section, i.e. the intersection of a plane with a circular cone. Replacing the circular cone
with an elliptical one, it is easy to see that the intersection will be an ellipse whose eccentricity
is a continuous function of the angle that the plane makes with the axis of the cone. In fact, there exists an angle such
that the intersection is a circle.
It will be shown that the plane of projection cuts the elliptical cone at precisely the supplement of that
special angle, resulting in a circular projection.
2. Preliminary Results
A few definitions are in order. A circular cone is a surface
generated by a straight line which moves so that it always intersects
a given circle c and passes through a fixed point V
not in the plane of the circle. Each position of the generating
line is defined as an element of the cone, and the fixed
point is the vertex of the cone. The vertex divides the
cone into two nappes, each of which is generated by one
of the half-lines. If the line joining the vertex of the cone
to the center of the circle c is perpendicular to the plane
of c, the cone is a right circular cone; otherwise
it is an oblique circular cone.[1]
An elliptical cone will be referred to as a surface generated
by a straight line which moves so that it always intersects a
given ellipse e and passes through a fixed point V;
in addition, the axis of the cone (the line joining the
center of e and V ) is perpendicular to the plane
of e. It is clear that an elliptical cone possesses symmetries
about both planes passing through its axis and the major/minor
axis of the ellipse e.
The intersection of an oblique circular cone with any plane is a conic.[2] When a plane cuts all the elements of one nappe of an oblique cone and does not pass through the vertex, the resulting intersection is an ellipse. In fact, an oblique circular cone can also be considered an elliptical cone. The reason is as follows.
Figure 2.
Figure 2 shows an oblique circular cone Q which was generated
by the line through V moving so that it intersects the circle
b. The center of the coordinate system
coincides with the center of b, and
b lies in the XZ plane. Q is
symmetric about the YZ plane.
Points, T, C, V, and S lie in the YZ plane. Point S was chosen
such that SV = VT, and C is the midpoint of ST. The intersection
of the cone Q with any plane passing through ST and perpendicular
to the YZ plane is an ellipse. This makes Q an elliptical cone
with axis VC, according to the definition stated earlier.
These results will now be applied to the situation in Figure 1.
The rays of light from the origin through the circle g
drawn on the sphere are the elements of an oblique (in general)
circular cone. As shown in the above paragraph, this cone of light
can also be considered elliptical.
The coordinate system in Figure 1 was chosen such that the circle g is symmetric about the XY plane. The X axis, which was not shown in Figure 1, points away from the viewer. Figure 3 shows the intersection of the sphere with the XY plane, as well as some auxiliary lines. The chord AB is a diameter of the circle g. Since C is the midpoint of the arc AB, OC bisects the angle AOB. (This is a consequence of the well known fact that the size of the angle subtended by a chord of a circle does not depend on the position of the vertex of the angle, as long as the vertex stays on the circumference of the circle. My own proof of this fact follows later in this document.) Therefore, OC is the axis of the cone. Thus the plane of the circle g intersects the elliptical cone at a special angle, q with respect to OC, such that the intersection is a circle. Since the cone is symmetric about OC, the intersection will be a circle regardless of where the cut is made, as long as this angle is kept constant. Moreover, because of the symmetry of the cone about the plane through OC and perpendicular to the XY plane, q will produce the same cross section as 180° - q. It will now be shown that if the plane of the circle g is at the angle q with respect to OC, then the projection plane is at the angle 180° - q with respect to OC, causing the projection to be a circle.
3. Proof
Figure 3.
CP is the line tangent to the circle at C and parallel to the
chord AB, so that CP and AB are both at the same angle with respect
to OC. SP is the line tangent to the circle at the point opposite
to O. Thus OS is a diameter, so that OCS is a right angle.
Since CP and SP are both tangent to the circle, angle PCS = angle PSC.
But q = angle PCS + 90°, and the
measure of the angle that SP makes with OC, f,
is 90° - angle PSC. Thus q = 180°
- f.
It was shown that the circle g drawn on the sphere subtends an elliptical cone of light which is intersected by the plane of g at the angle required to make the intersection a circle. The cone is intersected by the projection plane at the same angle with respect to its axis. The result is a circular projection.
Figure 4.
In Figure 4 the lines l, m, and n are tangent to the cirlce at points B, C, and A, respectively. Angle b is independent of the position of the point B on the circumference of the circle because it is a function of h + f, which in turn is a function of P+R. The latter quantity is just 180°-Q, which does not change as B slides along the circle. More specifically,
b = 180°-(h +
f).
But
h = (180°- P)/2
and f = (180°- R)/2,
so that
h +
f= 180°-(P+R)/2.
Now P+R=180°-Q, so
b=(P+R)/2=(180°-Q)/2.