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19. By the 11th it appears, that the conical Superficies of a Right Cone, is equal to a Circle, whose Radius is a mean Proportional between its Side and the Radius of its Bafe.

Or, it is equal to a Triangle, whose Base is equal to the Circumference of the Base of the Cone, and its Altitude, to the Side of the Cone.

20. By the 12th it is manifest, that the conical Superficies, of the Fruitum of any Right Cone, is equal to a Circle, whose Radius is a mean Proportional between the Side of the Fruftum, and the Radius of the Bafe, added to the Radius of the oppofite Circle.

And every Circle is equal to a Triangle, whose Base is equal to its Circumference, and its Altitude to the Radius. Art. 8th.

Or, the conical Superficies is equal to a Rectangle, whose Sides are equal to the Side of the Frustum, and half the Sum of the Circumference of the Base, added to half the Circumference of the opposite Circle.

21. The Area of the Surface. of a Sphere is equal to a Circle, whose Radius is equal to the Diameter of the Sphere.

15.8.

And the Area of any spherical Segment is equal to a Rectangle, one Side of which is equal to the Circumference of the Sphere, the other to the height of the Segment.

For, it is equal to a cylindrical Surface of those Dimensions-16.8. The fame holds true of any portion of the Surface of a Sphere, intercepted between parallel Planes.

The spherical Surface, intercepted between parallel Planes, is therefore equal to a Rectangle, whose Sides are the Circumference of a large Circle of the Sphere, and the perpendicular distance between the parallel Circles.

These Rules, respecting curved Surfaces, being clearly understood, will be found extremely useful to Artificers, in measuring circular Halls, or Rotundas of any kind; and Domes, entire; or where a part is deficient by means of a Lanthorn, or otherwise, at the Top; the Surface between any two parallel Circles being determined by the last.

I

ΜΕΝ

A

MENSURATION

OF SOLIDS.

S, in Menfuration of Superficies, the whole Business is to find a Rectangle equal to the Figure proposed; and to determine how many Squares of a certain Dimension, the Figure is equal to; so, Menfuration of Solids confists in determining how many cubical Feet, &c. are contained in the proposed Solid.

In order to which, it is necessary to know the affinity and proportion between one Solid and another, as contained in the 7th and 8th Books of these Elements; and particularly Parallelopipeds, to which all other Solids must necessarily be reduced, in Mensuration; a right angled Parallelopiped being of the same importance, in respect of Solids, as the Rectangle amongst Plane Figures, Alfo, as a Square is the Criterion of superficial measure, so a Cube, being the most perfect Parallelopiped, is the standard, by which the Quantities of Solids are compared and estimated.

For, the Definition of a Cube, see Def. 10.7. 1. Let AGE be a Cube, whose Sides, AB, &c. are each equal to 3 feet; also, let acd be a Cube, whose Side is one foot, or inch, &c. Now, suppose the Cube acd to be the Unit of measure; it is required to know, how oft the Unit, acd, is contained in AGE.

b

a

B

a

h

A

It is evident, seeing, the Measure of AB, BG, &c. are each equal to three times ab, bc, &c. that the Face ABCD, contains the Square bd, 9 times; every other Face, BCFG, or CDEF, the fame; therefore, each small Cube abc, def, efg, &c. made by Sections of Planes through ag, gl, &c. as in the Figure, are equal to acd. Wherefore, fince each Surface contains 9 Squares, as BF; and if Ba, Cg, &c. be one foot in thickness, confequently, the Solid a Gl, contains the small Cube, acd, 9 times. But, AB is equal to three times ab; wherefore, the whole Cube AGE contains the small one 27 times.

For, if it be supposed to be cut, by parallel Planes, through agland bik, parallel to the Top and Base, the Parts a Gl, ail, and bDk are equal, seeing that their Surfaces are equal; and

G

k

E F

B

:

and being also of equal thickness, Cg, gi, iD; consequently, BgF, ail, and b D kare equal (15.7.) But, BgFG is equal to 9 times acb; therefore, AGE is equal to three times 9,=27.

C

E

2. It is manifest, that if the Solid ACE
(being a Parallelopiped) was longer, equal
D 5 times ab, having equal thickness, it would
contain the small Cube, acb, 45 times,
i.e. five times 9; and thus it will increase, as
often as the measure ab is added in length.
Suppose half a b be added; it will contain
half 9 Cubes, i. e. 4; seeing that the
Surface of the End, or Section through

CE, contains 9 Squares, and the Solid CE has but half a
Cube, i. e. half ab in thickness. Therefore, the whole Solid
ACE contains the Cube acb 49 times and a half.

Hence, the measure of a right-angled Parallelopiped is to multiply the Side AD by AB, which gives the Area of one Surface (the Top, or Bafe) and, that Product being multiplied by its height or thickness, AF, i. e. by the number of times it contains ab (as, in this Cafe, three times) the Product of that multiplication is the Solid Contents of the Parallelopiped AFH.

But, all Parallelopipeds, having equal Bases and Altitudes, are equal (17.7.) Therefore, whether it be right or acute angled, having obtained the Area of any one Face, in square measure (Art. 12.) that Product being multiplied by the perpendicular Distance, between that Face and its oppofite, gives the folid Contents of the Parallelopiped. e. g.

7-9

3. Let ABCD be an acute angled Parallelopiped whose Base AD is 7 Feet, 9 Inches, on one Side, AE; the other, ED, 5 Feet, 4 Inches, its Area in square meafure, is 41 Feet, 4 Inches.

5-4

38-9

2-7

41-4
3-8

124-0 27-6-8 151-6-8

The height of the Parallelopiped, BF, being 3 Feet, 8 Inches, it is obvious, that the whole Solid contains 3 times 41-4, and; i. e. as often as the measure of a Foot is contained in the height, BF, so often the Solid contains cubical Feet, as its Base contains square Feet. But, the Perpendicular BF, is 3 Feet and 8 Inches; consequently, the Solid will contain 41-4, 3 times and two thirds, (8 Inches, being two thirds of a Foot.) Wherefore, 41 Feet, 4 Inches multiplied 3 times, and two thirds, equal 151 Feet, 9 Inches, and 8 Parts, is the

true Area, or folid Contents of the Parallelopd. ABCD.

A

Note. What is here meant by an Inch, in solid measure, is the twelfth part of a Cube Foot; viz. a Foot square, and one Inch in thickness; or, that Quantity in any other Figure. And the Part, in Solid measure, is 12 cubical Inches; or 12 Inches in length, and one Inch in breadth and thickness: a cubical Inch is, consequently, a Second of a cubical Foot.

It is obvious, that the Side BG, being inclined to the Base AD, is longer, than the Perpendicular BF, and consequently cannot give its true Area, for (joining GF) BFG is a right angled Triangle, of which BG is the Hypothenuse, therefore it is longer than B F (12. 1.)

Hence it is manifest, that all Parallelopipeds, or Prisms whatever, (for Parallelopipeds are Prisms) having equal height, have that Proportion to each other, which is between their Bases. And, having equal Bases, they are consequently, as their Altitudes. The Rule, therefore, for measuring any Prism whatever, is to multiply the superficial Area of its Base, by its perpendicular height. And consequently, the fame Rule is applicable for Cylinders-5.8.

Every Pyramid is equal to the third part of a Prism having equal
Bases and equal Altitudes.

Th. 4. 8.

Also, every Cone is equal to the third part of a Cylinder - 5.8. And confequently, Pyramids and Cones, having equal Bases and Altitudes, are equal.

Hence the Contents of Pyramids and Cones are obtained; viz. by multiplying the Areas of their Bases by one third of their height,

Or, if multiplied by the whole height, it gives the Contents of a Prism or Cylinder, of equal height; consequently, a Pyramid, or Cone, is one third part of such a Prism, or Cylinder.

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5. To find the folid Contents of a Sphere.

Having obtained its Diameter; find the Area of a Circle of

that Diameter (Art. 8.) which being multiplied, by the Diameter,

gives the Contents of a Cylinder, whose height and Diameter

are equal.

The Sphere is equal to two thirds of such a Cylinder-Th. 9. 8.

Or, having obtained the Area of a large Circle, of the Diameter
of the Sphere; multiply the Product by two thirds of the Diameter.

Otherwife. Every Sphere is equal to a Cone, whose Base is equal

to the Surface of the Sphere, and its Altitude to the Radius -17.8.

Therefore, having obtained the Area of its Surface, multiply

that Area by one fixth part of the Diameter.

See an Example, in each, in the Margin.

Let the Diameter of a Sphere be 15; the Circumference of a

Circle, of that Diameter, is 47,14, nearly; by the Ratio of 7 to 22.

Then, by Art. 8. the Semicircumference (23, 57) being mul-

23,57 tiplied by the Radius (7,5, half 15) gives the Area

7,5
of a great Circle of that Sphere, equal 176,775;

which multiplied by 10 (two thirds of the Diameter)

11785 gives 1767,75, or three fourths of the Integer.
16499

176,775

1767,75

Multiplying by 10, is only giving one place more of
Integers; feeing that, a Cypher, being added, makes
no difference in the Decimal; 750 thousandth parts,
being equal to 75 hundreth parts; i. e. equal to
3 fourths.

Secondly. 15 being the Diameter of the Sphere, find the
Area of its Surface (Art. 21.) which is equal to a Circle whose
Radius is equal to the Diameter of the Sphere.

The Circumference of every Circle has the fame Ratio to its

Diameter (Cor. 1. 14. 6. El.) consequently, the Circumference

of a Circle, whose Area is equal to the Surface of the Sphere,

is double the Circumference of the Sphere; half of which, is

equal to the whole of the other, equal 47,14.

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