Notes by R. Bigoni
(from Note Didattiche)
Given a spherical surface Σ with center O and radius OP with measure R, two parallel planes φ and ψ such that their distance from the center O is smaller than R,
intersect Σ in two circumferences.
The region of Σ bounded by these circumferences is called spherical zone Ζ
. The circles enclosed by these circumferences are its bases.
To calculate the surface of a spherical zone we can proceed in the following way:
let's draw from O a straight line OQ perpendicular to OP;
we assume the straight line OQ as the axis of the abscissas and the line OP as the axis of the ordinates;
the points A and B are respectively the intersections in the first quadrant between γ and the planes φ and ψ; T is any point of γ;
the radiuses OA, OB and OT form, respectively, the angles α, β and θ with the positive direction of the x axis;
the coordinates of A, B and T are respectively
an infinitesimal segment dγ of γ has length
the area S of the zone Z can be conceived as the sum of the lateral surfaces of cylinders of infinitesimal height dγ and base radius with θ from α to β;
if we represent by δ the distance between the planes φ e ψ
we get
With reference to fig. 1
, the solid Ω bounded by the spherical zone Z and its bases is called spherical segment
.
With reference to fig. 3
and with a method similar to that used for the calculation of S, the volume V of Ω can be calculated
as the sum of the volumes of the cylinders with infinitesimal height dy and base radius
with y from yA to yB;
From the equation (4) we get
Moreover, if we apply the Pythagorean theorem
and subtract side by side
If we plug this expression for yA into the equation (12)
If in fig. 1
the plane most distant from the center is tangent to Σ in P, the resulting figure is said spherical cap
.
In fig. 4
, H is the center of the base of the cap, P is its vertex, the segment HK is the radius of the base with measure r,
the segment HP is the height of the cap with measure δ and OK is the radius of Σ with measure R.
The equation (5) allows the calculation of the outer surface of the cap
Th equation (14) with xA=r and xB=0 gives the volume of the cap
If we apply the Pythagorean theorem, we get
and, plugging this expression for r2 into the equation (15), we obtain
Given a spherical surface Σ with center O and radius R and the point E outside Σ at distance h from it, the portion of Σ observable from E is a cap whose area can be calculated as follows:
we represent the intersections of Σ and Γ with a plane passing through the line OE;
the section of Σ is a circumference; the section of Γ is an isosceles triangle with vertices E, A and B;
the height of the triangle lies on the line of the diameter EF, intersects the circumference in P and intersects the base in H;
the triangle EAO is similar to the triangle AHO
in the cap we have
and finally, from the equation (5),
The surface of Σ is S=4πR2, hence the fraction of Σ visible from E is
last updated: March 16, 2018