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On COMPARISON AND MEASUREMENT OF SPHERICAL SURFACES.

751. DEF. A Lune is a part of the surface of a sphere included between two semi-circumferences of great circles. 752. DEF. The Angle of a lune is

Α the angle included by the semi-circumferences which forms its boundary. Thus 2 CAB is the angle of the lune.

753. DEF. A Spherical Ungula, or Wedge, is a part of a sphere bounded by a lune and two great semicircles.

754. DEF. The Base of an ungula is the bounding lune.

755. Def. The Angle of an ungula is the dihedral of its bounding semicircles, and is equal to the angle of the bounding lune.

756. DEF. The Edge of an ungula is the edge of its angle.

757. Def. The Spherical Excess of a spherical triangle is the excess of the sum of its angles over two right angles.

758. DEF. Three planes which pass through the centre of the sphere, each perpendicular to the other two, divide the surface of the sphere into

eight tri-rectangular triangles. Thus B the three planes C A D B, CEDI' and A EBF divide the surface of the sphere into the eight tri-rectangular triangles C E B, D E B, C BF, DBF, etc.

As in Plane Geometry the whole angular magnitude about any point in a plane is divided by two straight lines perpendicular to each other into four right angles, and each right angle is measured by a quadrant, or fourth part of a circumference described about that point as a centre with any given radius ; so, if, through a point in space, three planes be made to pass perpendicular to one another, they will divide the whole angular magnitude about that point into eight solid right angles, each of which is measured by an eighth part of the surface of a sphere described about that point with any given radius.

And, as in Plane Geometry, each quadrant which measures a right angle is divided into 90 equal parts called degrees, so each of the eight tri-rectangular spherical triangles is divided into 90 equal parts called degrees of surface. Hence, the whole surface of the sphere is divided into 720 degrees of surface.

D

PROPOSITION XXIX. LEMMA. 759. The area of the surface generated by the revolution of a straight line about another line in the same plane with it as an axis, is equal to the product of the projection of the line on the aris by the circumference whose radius is perpendicular to the revolving line erected at its middle point and terminuteil by the axis.

YI
Let the straight line A B revolve about the axis Y Y

in the same plane; let EF be its projection on
the axis; and Co the perpendicular to A B at its
middle point C, and terminated in the axis.
We are to prove area A B= E F X 2 4 0 C.

The surface generated by A B is the lateral surface of the frustum of a cone of revolution.

Draw CHI, and A D II, to Y Y'.
Then
area A B=AB X 2 CH,

$ 662 (the lateral area of a frustum of a cone of revolution is equal to the slant

height multiplied by the circumference of a section equidistant from its bases).

The A A B D and C O H are similar;

. AD: A B ::CH:C0.
But CH:CO:: 2 CH: 2 0 C 0, $ 375
(circumferences of ® have the same ratio as their radii).

... A D : A B :: 2 m C H : 2 = 0 (0.
.. A D X 2 * CO)= A B X 27 C H.
.. area of A B = A D X 2 7 C 0.
Now A D= EF.

§ 135 ... area A B = EF X 2 C 0.

Q. E. D. 760. SCHOLIUM. If either extremity of A B be in the axis Y Y', A B generates the lateral surface of a cone of revolution; and if A B be parallel to the axis Y Y', it generates the lateral area of a cylinder of revolution. In either case the formula holds good.

ilar ;

$ 287

EXERCISES.

1. If, from the extremities of one side of a spherical triangle, two arcs of great circles be drawn to a point within the triangle, the sum of these arcs is less than the sum of the other two sides of the triangle.

2. On the same sphere, or on equal spheres, if two spherical triangles have two sides of the one equal respectively to two

5 of the one equate of the first greater sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first will be greater than the third side of the second.

3. To draw an arc perpendicular to a given spherical arc, from a given point without it.

4. At a given point in a given arc, to construct a spherical angle equal to a given spherical angle.

5. To inscribe a circle in a given spherical triangle.

6. Given a spherical triangle whose sides are 60°, 80°, and 100°; find the angles of its polar triangle.

7. The volume of a pyramid is 200 cubic feet; find the volume of a similar pyramid which is three times as high.

8. Find the centre of a sphere whose surface shall pass through three given points, and shall touch a given plane.

9. Find the centre of a sphere whose surface shall pass through three given points, and shall also touch the surface of a given sphere.

10. Find the centre of a sphere whose surface shall touch two given planes, and also pass through two given points which lie between the planes.

PROPOSITION XXX. THEOREM. 761. The area of the surface of a sphere is equal to the product of its diameter by the circumference of a great circle.

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Let A B C D E be the circumference of a great circle,

and A D the diameter, and 0 A the radius of a sphere.

We are to prove

surface of sphere = AD X 2 = 0 A.

Let the semicircle and any regular inscribed semi-polygon revolve together about the diameter A D.

The semi-circumference will generate the surface of the sphere,

and the semi-perimeter a surface equal to the sum of the surfaces generated by the sides A B, BC, C D, etc.

Draw from the centre 0, Is O H, 0 I and OK to the chords A B, BC, C D, etc.

These 15. bisect the chords and are equal; $ 185

... area A B = AP X 2. 0 H; § 759

area BC= PR X 2 a 0 I;

and area CD= RD X 2 0 K.

Adding, and observing that 0 H, 0 I and 0 K are equal,

ard

area A B CD= (A P + PR + RD) X 2 = 0 H.

.. area A BCD= A D X 2, 0 H.

Now, if the number of sides of the regular inscribed semipolygon be indefinitely increased, the surface generated by the semi-perimeter will approach the surface of the sphere as its limit, and 0 H will approach 0 A as its limit.

.. at the limit we have

surface of the sphere = AD X 2 * 0 A.

$ 199

Q. E. D.

762. COROLLARY 1. If R denote the radius of the sphere, then A D will equal 2 R, and 0 A will equal R. Hence the surface of a sphere equals 2 R X 2 * R = 4 a RP.

763. Cor. 2. Since the area of a great circle of a sphere is equal to a R2 ($ 381), and the area of the surface of a sphere is equal to 4* R?, the surface of a sphere is equal to four great circles.

764. CoR. 3. If we denote the surfaces of two spheres by S and S', and their radii by R and R', we have S : S::47 R2 : 4a R, or S :S :: Ro : R^2; that is, the surfaces of two spheres have the same ratio as the squares on their radii.

765. Cor. 4. Since S= 4a Ro= q (2 R)?, the surface of a sphere is equivalent to a circle whose radius is equal to the diameter of the sphere.

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