Problem 13.

To find the Length of an Arch of an Hyperbola, the Bafe and Height being given.

Many different Hyperbolas may be cut from the fame Cone; but all of them will be less (in Area) than a Parabola, and greater than a Triangle, whence we may deduce this

Rule.

First, find the Length of the Curve as if it were a Parabola; and afterwards find the Length of the Hypothenuse as a Triangle of the same Base and Altitude; then add the two Sums together, and half this last Sum will give the Length of the Hyperbolic Arch, near enough in Practice.

Example.

Suppose the Base A B of the following Hyperbola be 20 Inches, and the Height VC 2.163; what is the Length of the Arch V A, and also the whole Arch AV B?

The Length of the Arch V A as a 10.307 by Problem 8th

Parabola

The Length of the Hypothenuse

VA as a Triangle

10.230

the Sum = 20.537

the Half = 10.268 = the Hyper

bolic Arch V A, which doubled, gives the Length of the whole Arch A V B required.

Problem 14.

To find the Area of an Hyperbola.

In the foregoing Problem we observed, that every Hyperbola is greater than Half, and less than two-Thirds of its circumfcribing Parallelogram: It will therefore be accurate enough to take all Hyperbolas at a Mean; that is, fuppofing every Hyperbola is of its circumfcribing Parallelogram, whence this

Rule.

☐ Multiply the Base by its Height, and that Product multiply again by 5, and divide this last Product by 8, the Quotient will be (nearly) the Area required. *

Example.

Let ABC be an Hyperbola, whose Base AC is 22 Inches, and the Height BD 14 Inches, what is its Area?

Operation. AC 22 × BD 14 = 308 × 5 = 1540÷ 8192.5 Inches, the Area.

* To determine the Length of an Arch, as well as Area of an Hyperbola, to Mathematical Precision, other Lines, besides the above, must be taken into the Calculation, which will be difficult for the Learner to understand, till he has made some Progress in Algebra and the Conic Sections.

Def. A Spheroid is a Solid, approaching to the Figure of a Sphere, having one of its Diameters longer than the other, and is formed by the Revolution of a Semi-Ellipfis about its Axis. If the Spheroid be formed by the Rotation of a Semi-Ellipfis round its tranverse Axis, it is called an oblong Spheroid; if generated by the Rotation of a SemiEllipfis round its conjugate Axis, it is called an oblate Spheroid.

Rule.

Find the Area of the Circle in the Middle, and multiply it by the Length; then multiply that Product by 2, and divide by 3, the Quotient will give the Solidity. *

3

* Every Sphere and every Spheroid is equal to of a Cylinder of the fame Diameter and Length.

See another Method of solving this Problem at Page 150 in Stereometry. Example.

Suppose ABCD be an oblong Spheroid, whose Diameter in the Middle CD is 33 Inches, and whose Length A B is 55 Inches, what is its Solid Content?

A

C

D

Operation.

CD 33×33 = 1089 ×.7854 = 855.3006, the Area in the Middle, which × 55 = 47041.5330, which multiplied by 2, and divided by 3, gives 31361.0220 inches, the Solidity required.

Another Way of finding the Solidity of a Spheroid.

Multiply the Square of the revolving Axe, by the fixed Axe, and this Product again by .5236, * and it will give the Solidity required.

of 7854, the Area of

a Circle whose Diameter is

Problem 2.

To find the Solidity of a Segment of a Spheroid. Def. A Segment of a Spheroid is a Part cut off by a Plane perpendicular to one of its Axes or Diameters. If the Segment be cut off perpendicular to the Tranfverse Axe, obferve this

Rule.

Divide the Square of the Conjugate Axe by the Square of the Transverse Axe; multiply the Quotient by the Difference between three Times the Transverse Axe, and twice the Height of the Segment; multiply the Product by the Square of the said Height, and this Product multiplied again by .5236, will give the Solidity required.

Example.

Suppose in the following Spheroid, the Transverse Axe AB be 50 Inches, and the Conjugate CD 30, and the Height of the Segment A G5 Inches, what is the Solidity of the Segment E AF cut off perpendicular to the Transverse Axe?

Operation. of Conjugate 900÷2500 of Tranfverse is.36; this x by 140 (the Difference between three Times the Transverse and twice the Height of the Segment) is = 50.40, which x by 25 of the Height is = 1260; this by .5236 gives 659.7360 for the Solidity required.

Note. If the Segment be cut off perpendicular to the Conjugate Axis, then divide the Transverse Axe by the Conjugate Axe; multiply the Quotient by the Difference between three Times the Conjugate Axe, and twice the Height of the Segment; multiply the Product by the Square of the faid Height, and this laft Product multiplied again by .5236, will give the Solidity of the Segment fought.