876

THEORY AND PRACTICE OP ENGINEERING.

Book II.

Fig. 1150.

Fig.1151.

Fig. 1152.

To find the Area of an Oval —Multiply the length of thegreater diameter CD by the lesser EF; their product willserve for the third term, whose first is 14 and second 11 ;the quotient will be the area sought: or, multiply the twodiameters together, and their product by 785, which dividedby 1000 gives the area: or, multiply the two diameters toge-ther, and extract the square root, which will be the diameterof a circle whose area is equal to that of the ellipsis.

To find the Area of a Sphere or Globe -Multiply its cir-

cumference by the diameter; the product will be the superficialcontent; or, the surface of a sphere is equal to the diameter squared,multiplied by 3*141593, and the surface of a spherical segment isequal to the circumference of the sphere, multiplied by the heightof the segment: the sphere is a solid bounded by one curvesurface, which is everywhere equally distant from the centre ; itmay be conceived to be generated by the rotation of a semicircleabout its diameter, which remains fixed : the axis of the spherebeing a right line about which the semicircle revolves, and thecentre the same as that of the revolving circle.

Archimedes gives another rule for finding the superficies ofa globe, which is that of multiplying the superficies of their greatcircles by 4 ; thus in the globe A the area of its greatest circleis 176 ; this multiplied by 4 gives 704 for the superficial content.

The diameter squared, its product multiplied by 314, and thendivided by 100, the quotient will he the superficies of a sphere.

There are some peculiar properties belonging to spheres ; fortwo will cut one another, or touch one another, or one of themwill fall wholly without the other, according as the distancebetween their centres is , first, less than the sum, and greater thanthe difference of their radii; secondly, equal to the sum, or tothe difference of their radii; or thirdly, greater than the sum, orless than the difference of their radii; for the sections of twospheres made by a plane passing through their centre, and throughany point which is supposed to be common to the two surfaces,will be circles having the same radii and centres with the spheresrespectively.. If two spheres cut one another, they do so in acircle, the plane of which is perpendicular to the line joining theircentres, and its centre is that line. If three spheres be describedwhose centres do not lie in the same straight line, the surfaces ofthe three cannot have more than two points in common, whichpoints lie in a straight line perpendicular to the plane of thecentres, and at equal distances on either side of that plane. Ifthe centres of three spheres lie in the same straight line, theircircles of intersection cannot meet one another, because theirplanes are perpendicular to this straight line, and thereforeparallel: and accordingly, the surfaces can have no point in com-mon, unless each of them passes through the same point of thestraight line in which the centres lie, for then each of them willtouch the other two in that point. When a point is equally dis-tant from the three angles of a triangle, it must lie in a perpen-dicular to the plane of the triangle, which passes through thecentre of the circumscribing circle.

To find the superficial Content of Segments of Spheres .—Multiplythe circumference of the entire sphere by the axis A of F G forthe area. p ig

To measure the superficies of regular zones, as of BKD E, wemust multiply the circumference by the height of the segment.

The superficial content of the irregular zone BCED may* be also found by ascertaining,first, those portions which are of a regular breadth, and then those which are not.

Suppose it is required to ascertain the superficial content of the zone BCDE, thediameter of which is 15 feet, and the greatest circumference 47 feet 1 inch 8 parts. If wefollow the rules already laid down, we shall find the area of the great segment C E F, whichis formed from the segment B D G, and the irregular zone BCED, by multiplying thecircumference of the globe B C E D F G hv the length of the part of the diameter FIcomprised within the great segment C E F, which will give 565 superficial feet for itscontent. Then, after subtracting the superficial content already found, of the smallersegment B D G from that of the great segment C E F. there will remain 329 superficial feetfor the area of the irregular zone B C E D. which surrounds the globe.