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in the present place, as it will be amply illustrated a little further on in treating of a general method of finding the contents of solids. See page 215.

The solid termed a Cylinder, or more properly a circular cylinder, derives its name from its resemblance to the shape of a roller ; and may be conceived to be formed by the revolution of a rectangle about one of its sides which remains fixed. The ends of a cylinder, like those of a prism, are equal and parallel figures,-in this case, circles. The lateral or convex surface of a cylinder is also circular and uniform throughont, so that if it were cut across at any part by a plane parallel to the base, the section would be a circle equal to the base.

The straight line joining the centres of the two ends, is called the Axis of the cylinder. If this axis be perpendicular to the base or end, the solid is called a right, or upright cylinder ; if otherwise, an oblique cylinder.

It is to be observed, that the circular cylinder is not the only kind of cylinder that occurs in the arts, &c., or presents itself in mathematical inquiries. There is the elliptic cylinder, which takes its name from the circumstance of its ends or bases being ellipses, and the uniformly elliptic character of all the sections parallel to those ends.

When the ends and all the intermediate parallel sections of a solid are equal and similar plane figures, the body is usually designated a cylindrical body, unless the lateral surfaces consist wholly of parallelograms, when, as previously defined, it receives the name of a prism.

To find the solidity of a cylinder, right or oblique.-RULE.-Multiply the area of either end by the perpendicular height of the figure.

Example (1.) The circumference of a circular cylinder is 44 inches; its height is 33.64 inches. Required its content.

Square of circum. x .07958 = Area of circular end

.07958 x 44 x 44 = 154.067, Area of base 154.067 X 33.64 5182.81, Solidity in cubic inches. Answer. Example (2.) What is the solid content of an oblique elliptic cylinder, 26 feet in perpendicular height, the transverse and conjugate diameters of the base being respectively 64 and 48 feet ?

Area of base 64 x 48 x .7854 = 2412.75

2412.75 x 26 62731-5 cubic feet. A circular cylinder may evidently be regarded as a prism, the base of which is a polygon having an infinite number of sides, that is, a circle (page 182). The volume of such a cylinder can be found, therefore, in the same manner as that of a prism, by multiplying the area of one end by the altitude of the figure. Similar reasoning applies, of course, to the elliptic cylinder. A cylinder of either kind may also be supposed to be built up of an infinite number of indefinitely thin plates of equal dimensions, laid one upon the other, and the perpendicular height must be the same whether these plates are piled obliquely

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Frustums of cylinders. A circular or elliptic cylinder cut by a plane parallel to its base, that is, straight across, is thus divided merely into smaller cylinders, which are evidently similar to the whole and to one another.

If a cylinder be cut by a plane passing parallel to its axis, or the direction of its length, the solid so formed may be termed “a frustum of a cylinder.” To find the content of such a body, all that is necessary is to compute the area of the base or end by the help of the Table of Circular Segments in the Appendix, and to multiply that area by the length of the frustum. The reason of the process is obvious.

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Example In the cylindrical frustum A LDC B K, let BP, tho height of the circular segment forming one of the equal ends, = 3 inches ; KC the chord, or edge of the seg. ment, 15 inches, and L K, the length, 40 inches. Required the solid content of the frustum.

1st. Having given B P and KC to find the diameter of the entire circle. As B F bisects KC (Euclid III., 3) K P B P x x P F (Euclid III., 35). Hence, PF=KP • BP= 7.59 ; 3= 18.75 the diameter ; then, 18.75 + 3 = 21.75; and the height divided by the diameter - 3 : 21.75 = .138.

Opposite to .138 in Table of Cir. Seg. is .0654-19 .065449 x 21.75 x 21.75 309.61, Area of segment B KO

309.61 X 40 = 1238.41 cubic inches. Answer. In the case of a frustum of an elliptic cylinder we should on a like principle, determine the area of the end by the Rule for segments of ellipses, page 196, and multiply this area into the length of the frustum.

A Cone, or rather a circular cone, is a round solid having a circle as its base, and a convex lateral surface tapering regularly to a point or vertex like the pyramid. A cone may be conceived to be described or traced out by a right-angled triangle when made to revolve around one of its perpendicular sides which remains fixed; the cono corresponds to a polygonal pyramid in the same

E a cylinder corresponds to a polygonal prism. Every section of a cone by a plane moving parallel to its base as, E f F, is a circle.

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A cone is said to be right or upright when the perpendicular line drawn from the vertex falls on the centre of the base, as Ä D. In other cases, the cone is termed oblique. There is also the elliptic cone, which differs from the circular in having the base and all sections parallel to it, ellipses. Generally, the name, conical body is applied to every solid that tapers to a point, and of which all the sections parallel to the base are figures similar to that base, except where the side-faces assume the form of triangles, and the solid consequently becomes a pyramid.

To find the solid content of a cone, right or oblique.—RULE. — Multiply the area of the base by one-third of the perpendicular height.

Example. The diameter of the base of a cone is 12 inches, and its perpendicular height 54 inches. Required its solidity.

Diam. sqd.

144 x 7854 X 18 = 2035.76 cubic inches. Answer. No example can be needed to illustrate the mode of calculating the content of an elliptic cone.

The reason of the rule for the solidity of the cone may be inferred from that given with respect to the pyramid, since the pyramid evidently merges into a cone when the triangular side-faces of the former become indefinitely narrowed by increasing the number of sides of its polygonal base.

To find the solidity of a frustum of a cone. -RULE.-Add together the areas of the two ends and the square-root of the product of these areas. Multiply the sum by one-third of the height of the frustum.

It will be readily perceived that this rule is founded on the same principle as the rule for a frustum of a pyramid given on page 207.

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Example. Let the dimensions of the conical frustum E BCF (last fig.) be,– BC, diameter of greater end, 28 inches; EF, diameter of smaller end, 20; and the altitude f D, 15 inches. What is the solid content ?

Area of greater end 28x 7854 = 615.754
Area of less end 202 x 7854

= 314.160 Mean proportional between 615.752 & 314.159

= 439.824 Or, 28 x 20 x .7854

Sum 1369.738 4 of 15 (altitude)

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6848.690 Cub. in. Answer. A shorter method is to add together the squares of the two diameters and the product of the diameters, and then to multiply this sum by •7854, and by one-third of the height, as before. For, the first part of the preceding operation is plainly equivalent to (28? + 20° + 28 x 20) * :7854. A little more accuracy is also gained by this process.

But a still more simple method is, to add together the diameters of the greater and less ends, square the sum, subtract from the result the product of the diameters, and multiply the remainder by :7854 and a third of the altitude. It may be easily shown in any instance that the square of the sum of two numbers diminished by their product is equal to the sum of the squares of these numbers increased by their product. Thus,

(28 + 20)2 = 48= 2304; 2304 — 28 X 20 = 1744

1744 x 7854 = 1369.738, as above. A Sphere or Globe is a round body, all the points in the surface of which are equally distant from a certain point within the solid called the centre of the sphere. Any straight line from this point to the surface is called a radius of the sphere ; and a line passing through the centre and meeting the surface in two opposite points, is a diameter.

Every section of a sphere, made by a plane, is a circle. A section through the centre is termed a great circle of the sphere; any other, a small circle. A sphere may be conceived to be described or generated, as it is said, by the revolution of a semi-circle about its diameter, which latter remains fixed while the semi-circumference revolves. This fixed line is called the axis of the sphere, and the extremities of it are termed, the poles.

It is proved in geometry,* that the surface of a sphere is equal to the circumference of one of its great circles multiplied by its diameter.

Now, the sphere may be conceived as composed of an infinite number of pyramids, the. bases of which make up the surface of the sphere, and the vertices of which all meet at the centre, the radius of the sphere being thus the altitude of each pyramid. Accordingly, the solidity of each pyramid being the product of its base by a third of its height, the solidity of the sphere must be the product of its surface by a third of the radius, or by a sixth of the diameter; and the surface being 3.1416 times the square of the diameter, it follows that the solidity of a sphere is equal to one-sixth of the diameter multiplied by 3-1416 times the square of the diameter.

It is also proved in geometry, (See Lardner's Euclid, -Solid Geom. Prop. XVI.) that a sphere is two-thirds of its circumscribing cylinder, that is, a cylinder having the same diameter and height as the sphere. But the diameter and height of a sphere are equal ; and since in finding the solidity of a cylinder we multiply the square of the diameter by •7854, and this again by the height, which in the case of a cylinder of equal diameter and altitude amounts to cubing the diameter and multiplying the result by ·7854, so in the case of a sphere, we should cube the diameter and multiply the result by ·5236, which is two-thirds of 7854. Hence, we derive the following rules :

To find the solid content of a sphere.—Rule I.—Multiply the diameter by the circumference and the product by one-sixth of the diameter.

* See Lardner's Euclid, Solid Geom., Prop. X.

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ROLE II.-Multiply the cube of the diameter by .5236.
Example. What is the solidity of a globe, the diameter of which is 30 inches ?
By Rule I.,-Diam x Circum. 30 x 30 x 3.1416 = 2827.44

And 2827.44 x 7th of 30 14137.2 cubic inches. Answer. By Rule II.,-Diam.” 30° =27000; 27000 x 5236 14137.2, as before. The simple and beautiful relation which exists between the solidities of the cone, the sphere, and the cylinder, may be familiarly illustrated in the following

Suppose ABC D to represent a cylindrical vessel, made of tin-plate, and E G F H, C F D, to represent respectively a globe and a cone of wood having the same diameter and height as the corresponding internal dimensions of the cylinder. Now, if the cylinder be filled with water and the cone fully immersed, one-third of the liquid will flow over; and if the cone be taken out, the vessel again filled with water, and the globe immersed, two-thirds of the contents will be displaced. Accordingly, it can thus be experimentally established, that the solidities of a cone, a sphere, and a cylinder having equal diameters and heights are to one another as the numbers 1, 2, and 3."

A Spheroid is such a solid as would be described by the rotation of an ellipsis about one of its axes or diameters. If this be the transverse or major axis, the solid is called a prolate (oblong) spheroid ; if it be the conjugate or minor axis, the solid is called an oblate spheroid.

A prolate spberoid is of the form of an egg which has its ends equal: an oblate spheroid resembles an orange. Vessels in the form of an oblate spheroid are rarely, if ever met with in the practice of gauging.

To find the solid content of a prolate spheroid.—RULE.—Multiply the square of the conjugate axis by the transverse axis, and the product by ·5236.

Example. Required the solidity of a prolato spheroid having its axes respectively, = 55 and 33 inches. 332

1089; 1089 x 55 x 5236 = 31361 cubic inches. Answer. The solidity of an oblate spheroid is determined by multiplying the square of the transverse axis by the conjugate axis, and the product by ·5236 as before.

The truth of these rules will appear in the following account of a general method for the mensuration of solids and the frustums of solids, irrespective of their geometrical outline.

It is evident that the principle of the system of equi-distant ordinates, as explained on page 198, may be applied to the mensuration of solids as well as of surfaces, by using the areas of sections parallel to the axis of the solid instead of drawing simple lines or ordinates as in the case of superficies.

Suppose the solid A B D C to be divided into any number of equi-distant parallel slices or sections, by the planes a c, b d, &c. Then, if each of the prismoidal frustums Ac, ad, &c., be conceived to be completed into a a parallelopiped by the addition of a triangular prism at top and bottom, it will at once be seen that the volume of the solid A d

B

K may be expressed approximately, by taking one-third of the total N volume of the scveral rectangular prisms corresponding to the parallelograms included in the formula given on

page 199, and similarly, as regards each pair of tho remaining portions of the solid. But the cubical content of a prism is found by multiplying the area of the base by the height or length of the figure; and if A C, a c, &c., be considered the bases of prisms, then the volumes of such prisms will be denoted by the product of A C and N h, a cand N h, &c., since the widths, N h, h k, &c. of the several frustums are the same.

Ritchie's Principles of Geometry.

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If the given solid have such a shape that its ends and sections present the form of circles, ellipses, or other curvilinear or rectilinear figures, it will only be necessary to compute the area of each surface according to the proper rule, regarding the intervening portions as so many circular or elliptic cylinders, &c., and then to multiply by the common height or distance as in the case of a prismoidal or cylindrical solid. From these considerations we have the following most useful

GENERAL RULE, FOR DETERMINING THE SOLID CONTENT OF ANY BODY, REGULAR OR IRREGULAR, OF WHICH TWO OPPOSITE ENDS ARE PARALLEL PLANE FIGURES.- Let the perpendicular distance between the opposite parallel ends of the solid be divided into an even number of equal parts, through which let planes parallel to the ends be conceived to be drawn, thus resolving the solid into an even number of slices, and let the area of each end, and of the section made by each of these planes, be ascertained from the proper external or internal dimensions as the case may be.

Then add together four times the sum of the sectional areas at the odd points of division,—the 1st, 3rd, 5th, &c.; twice the sum of the areas at the even points of division,—the 2nd, 4th, 6th, &c., and the areas of the two parallel ends or boundaries of the figure. The result multiplied by one-third of the common distance between every pair of consecutive sections, that is, by one-third of the uniform thickness of the slices, will express the solid content of the body.

This rule, as will appear from some of the subjoined examples, is applicable to all complete solids with rounded or tapering outlines, such as the sphere or spheroid, as well as to what may be considered as the frustums, segments, or zones of regular solids; for as regards a complete solid, it is evident that we may so divide its length or height as to tako sections within as short a distance as we please of either end, and therefore may treat the ends themselves as having areas equal to 0. The rule gives a result absolutely exact when used to determine the content of any solid such as would be generated by the revolution of one of the conic sections-a triangle, a circle, an ellipsis, &c., -or a right line with every species of pyramid. In all other instances, it furnishes a very close approximation to the truth, if the various sections be taken with judgment, according as tho figure deviates more or less from known geometrical forms.

Example (1.) In the solid represented in the diagram above, let the areas in square inches of the two ends A. C, B D, and of the five equi-distant sections a c, bd, &c., as obtained by measuring the length and breadth of each prismoidal frustum, and regarding the section as a rectangle, be as follows :

Intermediate sections. AC=10; 1st (ac)=12; 2nd (6d)= 15; 3rd=17; 4th=16; 5th=13; B D=11. And let the length NO= 26 inches. Required the content of the entire solid. Areas of ends. 10 12

15
11
17

16
Sum 21

13

Sum 31
Sum 42

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62
168
62

21 Common distance

251 == 4

4

End.

End.

Areas of odd sections,

Areas of even sections.

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