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PROPOSITION XVII. THEOREM.

Every segment of a sphere is measured by the half sum of its bases multiplied by its altitude, plus the solidity of a sphere whose diameter is this same altitude.

B

M

Let BE, DF, be the radii of the two bases of the segment, EF its altitude, the segment being described by the revolution of the circular space BMDFE about the axis FE. The solid described by the segment BMD is equal to .ВD.EF D (Prop. XVI.); and the truncated cone described by the trapezoid BDFE is equal toπ.EF.(BE2+DF2+ BE.DF) (Prop. VI.); hence the segment of the sphere, which is the sum of those two solids, must be equal to 7.EF.(2BE2+2DF2+2BE.DF+BD) But, drawing BO parallel to EF, we shall have DO=DF—BE, hence DO2=DF_2DF.BE+BE2 (Book IV. Prop. IX.); and consequently BD2=BO2 + DO2=EF2 + DF2—2DF.BE+BE2. Put this value in place of BD2 in the expression for the value of the segment, omitting the parts which destroy each other; we shall obtain for the solidity of the segment,

¦πEF.(3BE2+3DF2+ EF2),

one ¡Ã.EF.(3BE2+3DF2), or EF.(TM.BE2+7.D

2

an expression which may be decomposed into two parts; the being the half sum of the bases multiplied by the altitude; while the other.EF represents the sphere of which EF is the diameter (Prop. XIV. Sch.): hence every segment of a sphere, &c.

A

E

F

Cor. If either of the bases is nothing, the segment in question becomes a spherical segment with a single base; hence any spherical segment, with a single base, is equivalent to half the cylinder having the same base and the same altitude, plus the sphere of which this altitude is the diameter.

General Scholium.

Let R be the radius of a cylinder's base, H its altitude: the solidity of the cylinder will be "R2x H, or "RH.

Let R be the radius of a cone's base, H its altitude: the solidity of the cone will be R2x }H, or ¦¬RoH.

Let A and B be the radii of the bases of a truncated cone,

II its altitude: the solidity of the truncated cone will be π.H. (A2+B2+AB).

Let R be the radius of a sphere; its solidity will be R. Let R be the radius of a spherical sector, H the altitude of the zone, which forms its base: the solidity of the sector will be RoH.

Let P and Q be the two bases of a spherical segment, H its P+Q altitude: the solidity of the segment will be .H+. H3.

2

If the spherical segment has but one base, the other being nothing, its solidity will be PH+д¬Н3.

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Definitions.

1. A spherical triangle is a portion of the surface of a sphere, bounded by three arcs of great circles.

These arcs are named the sides of the triangle, and are always supposed to be each less than a semi-circumference. The angles, which their planes form with each other, are the angles of the triangle.

2. A spherical triangle takes the name of right-angled, isosceles, equilateral, in the same cases as a rectilineal triangle. 3. A spherical polygon is a portion of the surface of a sphere terminated by several arcs of great circles.

4. A lune is that portion of the surface of a sphere, which is included between two great semi-circles meeting in a common dia.neter.

5. A spherical wedge or ungula is that portion of the solid sphere, which is included between the same great semi-circles, and has the lune for its base.

6. A spherical pyramid is a portion of the solid sphere, included between the planes of a solid angle whose vertex is the centre. The base of the pyramid is the spherical polygon intercepted by the same planes.

7. The pole of a circle of a sphere is a point in the surface equally distant from all the points in the circumference of this circle. It will be shown (Prop. V.) hat every circle, great or small. has always two poles.

PROPOSITION I. THEOREM.

In every spherical triangle, any side is less than the snm of the

other two.

Let O be the centre of the sphere, and ACB the triangle; draw the radii OA, OB, OC. Imagine the planes AOB, AOC, COB, to be drawn; these planes will form a solid angle at the centre O; and the angles AOB, AOC, COB, will be measured by AB, AC, BC, the sides of the spherical triangle. But each of the three plane angles forming a solid angle is less than the sum of the other two (Book VI. Prop. XIX.); hence any side of the triangle ABC is less than the sum of the other two.

A

Let ANB be the arc of a great circle which joins the points A and B; then will it be the shortest path between them.

1st. If two points N and B, be taken on the arc of a great circle, at unequal distances from the point A, the shortest distance from B to A will be greater than the shortest distance from N to A.

PROPOSITION II. THEOREM.

The shortest path from one point to another, on the surface of a sphere, is the arc of the great circle which joins the two given points.

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B

For, about A as a pole describe a circumference CNP. Now, the line of shortest distance from B to A must cross this circumference at some point as P. But the shortest distance from P to A whether it be the arc of a great circle or any other line, is equal to the shortest distance from N to A; for, by passing the arc of a great circle through P and A, and revolving it about the diameter passing through A, the point P may be made to coincide with N, when the shortest distance from P to A will coincide with the shortest distance from N to A: hence, the shortest distance from B to A, will be greater than the shortest distance from N to A, by the shortest distance from B to P.

If the point B be taken without the arc AN, still making AB greater than AN, it may be proved in a manner entirely similar to the above, that the shortest distance from B to A will be greater than the shortest distance from N to A.

If now, there be a shorter path between the points B and A, than the arc BDA of a great circle, let M be a point of the short

est distance possible, then through M draw MA. MB. arcs of great circles, and take BD equal to BM. By the last theorem, BDA <BM+MA; take BD-BM from each, and there will remain ADAM. Now, since BM=BD, the shortest path from B to M is equal to the shortest path from B to D: hence if we suppose two paths from B to A, one passing through M and the other through D, they will have an equal part in each; viz. the part from B to M equal to the part from B to D.

But by hypothesis, the path through M is the shortest path from B to A: hence the shortest path from M to A must be less than the shortest path from D to A, whereas it is greater since the arc MA is greater than DA: hence, no point of the shortest distance between B and A can lie out of the arc of the great circle BDA.

PROPOSITION III. THEOREM.

The sum of the three sides of a spherical triangle is less than the circumference of a great circle.

Let ABC be any spherical triangle; produce the sides AB, AC, till they meet again in D. The arcs ABD, ACD, will be semicircumferences, since two great circles always bisect each other (Book VIII. Prop. VII. A Cor. 2.). But in the triangle BCD, we have the side BC<BD+CD (Prop I.); add AB+ AC to both; we shall have AB+AC+BC<ABD+ACD, that is to say, less than a circumference.

B

D

E

E

C

PROPOSITION IV. THEOREM

The sum of all the sides of any spherical polygon is less than the circumference of a great circle.

Take the pentagon ABCDE, for example. Produce the sides AB, DC, till they meet in F; then since BC is less than BF+CF, the perimeter of the pentagon ABCDE will be less G than that of the quadrilateral AEDF. Again, produce the sides AE, FD, till they meet in G; we shall have ED<EG+DG; hence the pe rimeter of the quadrilateral AEDF is less than that of the triangle AFG; which last is itself less than the circumference of a great circle; hence, for a still stronger reason, the perimeter of the polygon ABCDE is less than this same circumference.

A

D

C

F

B

E

Scholium. This proposition is fundamentally the same as (Book VI. Prop. XX.); for, O being the centre of the sphere, a solid angle may be conceived as formed at O by the plane angles AOB, BOC, COD,&c., and the sum of these angles must be less than four right angles; which is exactly the proposition here proved. The demonstration here given is different from that of Book VI. Prop. XX.; both, however, suppose that the polygon ABCDE is convex, or that no side produced will cut the figure.

A

PROPOSITION V. THEOREM.

Let ED be perpendicular to the great circle AMB; then will E and D be its poles; as also the poles of the parallel small circles HPI, FNG.

For, DC being per-. pendicular to the plane A AMB, is perpendicular to all the straight lines CA, CM, CB, &c. drawn through its foot in this plane; hence all the arcs DA, DM, DB, &c. are quarters of the circumference. So likewise are

The poles of a great circle of a sphere, are the extremities of that diameter of the sphere which is perpendicular to the circle; and these extremities are also the poles of all small circles parallel to it.

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D

N

M M

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B

all the arcs EA, EM, EB, &c.; hence the points D and E are each equally distant from all the points of the circumference AMB; hence, they are the poles of that circumference (Def. 7.).

Again, the radius DC, perpendicular to the plane AMB, is perpendicular to its parallel FNG; hence, it passes through O the centre of the circle FNG (Book VIII. Prop. VII. Cor. 4.) ; hence, if the oblique lines DF, DN, DG, be drawn, these oblique lines will diverge equally from the perpendicular DO. and will themselves be equal. But, the chords being equal.

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