Spherical Cap

Consider a sphere of radius rho and a similar sphere of unit radius. Each angle will be the same. Each linear measure of the former will be rho times that of the latter. Each aerial ... rho squared.... Each volumetric ... rho cubed.... For ease of writing the formulae, we will do so for the latter, only. The areas and the volumes are obtained by integration. Where the required integration is less than obvious, we provide a hint.

Sphere

For the forthcoming spot-checks, we will need the area of the sphere = 4 pi. Hence, the area of the hemisphere = 2 pi. And the volume enclosed by the sphere = 4 pi / 3. Hence, the volume enclosed by the hemisphere = 2 pi / 3.

Right Circular Cone

From the center of the sphere, construct a right-circular cone, of half-angle alpha, with its base being the plane of the small-circle intersection of the cone with the sphere.

flat bottom

The altitude of the cone terminates at the center of its base. The length of the altitude is cos(alpha). The length of the radius of the base of the cone is sin(alpha). The circumference of the base is 2 pi sin(alpha). The area of the base is pi (sin(alpha))^2. The area of the (surface of the) cone is pi sin(alpha). At alpha = pi / 2, this area becomes pi, as it should be. The volume enclosed by the cone is pi (sin(alpha))^2 cos(alpha) / 3.

For a hyperbolic surface, radius would be sinh(alpha). Hence, the circumference would be 2 pi sinh(alpha).

caped bottom

Here -- for the cone -- it is easy; because, the circumference of the cap lies on the same plane as that of the flat-bottom.

The area of the spherical cap is the integral, from zero to alpha, of the circumference, with respect to alpha. It is 2 pi (1 - cos(alpha)). At alpha = pi / 2, this area becomes 2 pi, as it should be. Likewise, at alpha = pi, ... 4 pi.... The volume bounded between the cone and the spherical cap is pi (2 - 3 cos(alpha) + (cos(alpha))^3) / 3, as shown elsewhere. Then the volume of the spherical cap is the difference of the volumes; that is, the volume of the cone and the spherical cap less the volume of the cone.

Relative Area of Cap

To obtain the relative area of the cap,
we divide it -- 2 pi (1 - cos(alpha)) -- by the area of the sphere -- 4 pi --, to obtain

relative-area = (2 pi (1 - cos(alpha))) / (4 pi) =
= (1 - cos(alpha)) / 2 =
= hav(alpha) =
= (sin(alpha / 2))^2

where we have utilized the definition of the haversine.
This formula answers the question,
"What is the relative size of the circum-polar region?", where the angle alpha is the latitude.

Area of Cap in a Bowl

Here we will include the radius of the sphere, explicitly.

Let r be the radius of the base of the cone.
Let h be the height of the cap (above the base of the cone).
Let b be the distance from a point on the circumference of the base of the cone to the center of the cap.

Then:

b^2 = r^2 + h^2
r = rho sin(alpha)
rho - h = rho cos(alpha).

Square each of the last two equations, and add, to obtain

r^2 + (rho - h)^2 = rho^2 ((sin(alpha))^2 + (cos(alpha))^2).

Hence, employing the identity theorem for the sine function,

r^2 + rho^2 - 2 h rho + h^2 = rho^2.

Substitute into the first equation, to obtain

b^2 = 2 h rho.

The area A of the cap had been given by

A = 2 pi (1 - cos(alpha)) rho^2.

Substitute the value of cos(alpha), to obtain

A = 2 pi (h / rho) rho^2 = 2 pi h rho = pi b^2,

which is independent of either rho or alpha.

Take a point P. Draw a circle of radius b and center at P. Draw a vertical line through P.
Consider the family of circles, each of which passes through P and has its center on the aforementioned line.
Each of these circles may be revolved about that center-line, generating a family of spheres.
The original circle also will generate a sphere, which we call "the bowl" to distinguish it.
The set of arcs inside the original circle generates a family of spherical caps.
Then, each of these caps has the same area A = pi b^2, which is one-fourth of the area of the bowl.
At one extreme, we have a plane circle (of radius b) on the horizontal plane through P.
At the other extreme, we have a sphere (of radius b/2).
In between, we have a hemi-sphere (of radius b/sqrt(2)).
Since there is nothing special about the center-line, we may generalize further and say that
any spherical cap which passes through the point P, and is bounded by the bowl, has the same area.

Right Regular Pyramid

From the center of the sphere, construct a right-regular pyramid, of half-angle alpha (measured to the ribs of the pyramid), with its base being the plane of the regular-polygon intersection of the pyramid with the sphere. The sides of this polygon are great-circle arcs. Let the natural number n be the amount of sides of this polygon or pyramid. We define beta as the half-angle measured to the faces of the pyramid. We will have to find a relationship between these two angles. One may analyze an irregular pyramid in a similar manner; however, then either the set of vertices or the set of altitudes of the base would have to be given explicitly.

flat bottom

The altitude of the pyramid terminates at the center of its base. The length of the altitude is cos(alpha). The length of the line from the center of the base to each of the vertices of the polygon, that is the radius (of the circumscribing circle) of the polygon is sin(alpha). The length of the altitude to each of the sides of the polygon is sin(alpha) cos(pi / n). The length of each semi-side of the polygon is sin(alpha) sin(pi / n). For future reference, this is the sine of the semi-side of the pyramid. The perimeter of the base is 2 n sin(alpha) sin(pi / n). Its limit, as n increases without bound, is 2 pi sin(alpha) -- the circumference, above, as it should be. The area of the base is the area of the polygon is one-half of its altitude times the perimeter, namely n (sin(alpha))^2 sin(2 pi / n) / 2). Its limit, as n increases without bound, is pi (sin(alpha))^2 -- the area, above, as it should be. The slant-height of the pyramid is the square-root of the sum of the squares of the altitude -- cos(alpha) -- of the pyramid and the altitude -- sin(alpha) cos(pi / n) -- of the base of the pyramid. Thus, the slant-height of the pyramid is sqrt(1 - (sin(alpha) sin(pi / n))^2). Looks good. Then the area of the pyramid is one-half of the product of this slant-height by the perimeter -- 2 n sin(alpha) sin(pi / n) -- of the base of the pyramid. Thus, the area of the (surface of the) pyramid is n sqrt(1 - (sin(alpha) sin(pi / n))^2) sin(alpha) sin(pi / n). Its limit, as n increases without bound, is pi sin(alpha) -- the area of the cone, above, as it should be.

caped bottom

Here -- for the pyramid -- it is difficult; because, the circumference of the cap now lies above the plane of the flat-bottom. We know that the area enclosed by any curve upon the surface of a sphere is 2 pi minus the cumulative external-angle along the curve.

Let us employ the Mollweide's Formulae. The spherical cap of the pyramid is a regular spherical-polygon. Drop an altitude from the center of this cap to one of the sides. By symmetry, it bisects the side. Also, by symmetry, its central angle is C = pi / n. The angle between the altitude (b = beta) is at right angles (by construction) A = pi / 2 to the base, whose half-length is sin(c). The hypotenuse a = alpha is related to b = beta by cos(C) = tan(b) ctan(a). Substituting, we have cos(pi / n) = tan(beta) ctan(alpha). Since either alpha xor beta is given, we may solve for the other. The half-length of the base sin(c) is given by sin(c) = sin(a) sin(C) or sin(b) = tan(c) ctan(C). Substitution yields sin(c) = sin(alpha) sin(pi / n) or tan(c) = sin(beta) tan(pi / n). From the later, we obtain sin(c) = 1 / sqrt(1 + (csc(beta) cot(pi / n)^2). Finally, half of the interior angle of the polygon is B. It is given by cos(B) = cos(b) sin(C) or cos(a) = ctan(B) ctan(C). Substitution yields cos(B) = cos(beta) sin(pi / n) or tan(B) = sec(alpha) ctan(pi / n). Thus, the perimeter of the cap is 2 n sin(c) = 2 n sin(alpha) sin(pi / n) = 1 / sqrt(1 + (csc(beta) cot(pi / n)^2). The area of the cap is 2 pi - n (pi - 2 B), where the half the interior angle B has been given already. Finally, the volume bounded between the pyramid and the spherical cap is the area of the cap times rho^3 / 3. The relationship between the area and the volume was shown.

Right Irregular Pyramid

For an irregular right-pyramid, the unit normal to each face would have to be given. For each pair of adjacent faces, solve the two half-triangles between them, on the base and on the spherical cap. The angle beta is given by cos(beta) = (unit normal to a face) . (unit vector from the apex of the pyramid, towards the center of its base). The pseudo-vector along an edge (rib) of the pyramid is the crp = cross-product of the unit-normal vectors of the adjacent faces. Then, the angle alpha is given by cos(alpha) = crp . (unit vector from the apex of the pyramid, towards the center of its base) / || crp ||.

Now, employ the plane Mollweide's Formulae to solve each half-triangle, upon the flat base. The angle B, at a vertex of the base, between a face and a line towards the center of the base, is sin(B) = sin(beta) csc(alpha). The angle C, at the center of the base, between an altitude to a face and a line to a vertex of the base is C = pi / 2 - B. The length of a half-side c of an edge of the base is ||c|| = rho sin(alpha) sqrt(1 - (sin(beta) csc(alpha))^2).

Now, employ the spherical Mollweide's Formulae to solve each half-triangle, upon the spherical cap. Each of the following are parts of a spherical triangle. The angle C, at the center of the cap, between a face and an edge is cos(C) = tan(beta) ctan(alpha). The angle B, at a vertex of the cap, between a face and a great-circle line towards the center of the cap, is sin(B) = sin(beta) csc(aalpha). The half-side c of an edge of the cap is cos(c) = cos(alpha) sec(beta). Now that you have each of the 2 n half-triangles solved, add up whatever you desire.

Radius of the Sphere

We present four methods for finding the radius of a sphere. The first two do not employ the embedding space. The first three are elegant -- "cool". The last one is crude -- "gross". Unfortunately, the first two methods require excessive precision in the measurements; thus, they are impractical. The third method is marginally practical. Observe that each method measures a local value for the radius of curvature and its reciprocal, the curvature.

In each of the first three methods, a certain ratio is computed, from the measurements. If this ratio is equal to one, we have a plane geometry and the radius of curvature of the space in infinite. If this ratio is strictly less than one, we have a spherical geometry. The radius is computed as shown. If, however, the radius is strictly greater than one, we have a hyperbolic geometry. The radius would be computed from the analogous formulae: Just replace the circular sine by the hyperbolic sine; i.e., sinh. Replace the circular cosine by minus the hyperbolic cosine, i.e., -cosh.

circumference

The circumference of a spherical cap was shown to be 2 pi rho sin(alpha). The radius of the spherical cap is just rho alpha. Then the ratio is circumference / (2 pi radius) = 2 pi rho sin(alpha) / (2 pi rho alpha) = sin(alpha) / alpha. Solve this transcendental equation for alpha. Since this equation is transcendental, its solution, of necessity, has to be numerical For instance, binary search, Raphson or Newton methods, or inversion of the Taylor's series. Then, substitute back into either of the original equations, for instance, the second one gives rho = radius / alpha.

area

The area of a spherical cap was shown to be 2 pi rho^2 (1 - cos(alpha)). The radius of the spherical cap is just rho alpha. Then the ratio is area / (pi radius^2) = 2 pi rho^2 (1 - cos(alpha)) / (pi (rho alpha)^2) = 2 (1 - cos(alpha)) / alpha^2. Solve this transcendental equation for alpha. Since this equation is transcendental, its solution, of necessity, has to be numerical For instance, binary search, Raphson or Newton methods, or inversion of the Taylor's series. Then, substitute back into either of the original equations, for instance, the second one gives rho = radius / alpha.

One could measure this area of the spherical cap area in two ways. The cap may be of any shape. First as 2 pi rho^2 times the cumulative exterior angle of the cap. Second directly. Then the radius of the sphere is rho = sqrt( (directly measured area) / (2 pi times the cumulative exterior angle of the cap) ).

spherometer

A spherometer consists of three legs -- in an equilateral triangle--, each terminated in a pin, and a micrometer in the center, pointing parallel to the legs, and in the same direction. The radius of the circumscribing circle of the triangle is rho sin(alpha). Then the length of the side -- the distance between a pair of legs -- is d = rho sin(alpha) sqrt(3). The height of the micrometer, above the plane of the triangle is h = rho (1 - cos(alpha)). Solve this set of simultaneous equations for rho, to yield rho = (d^2 / 3 + h^2) / (2 h). Then, K, the curvature, which is defined as the reciprocal of the radius of curvature, is K = 2 h / (d^2 / 3 + h^2). A value of zero for h -- hence, also for K -- indicates a flat space. A strictly negative value indicates a hyperbolic space.

vise

Clamp the sphere in a vise. Measure the distance between its jaws. Half that distance is the radius of the sphere. This is, by far, the easiest and most precise method. However, it is applicable to neither a flat space nor a hyperbolic space.