Sum 89X15X1309-174,7515 answer.

General Rule for the foregoing figure.

Find a diameter half way between the middle and end; multiply it it by two, square the product, and add the square to the square of a diameter taken in the middle; multiply the sum by the length, and the product again by,1309, the last product will be the solidity nearly.

PROBLEM 40.

To find the solidity of the middle frustum of an elliptic spindle.

RULE.

To the square of twice a diameter, taken half way between the middle and the end, add the squares of the middle and end diameters; multiply the sum by the length, and again by,1309, the last product will be the solidity required.

EXAMPLE.

What is the solidity of the middle frustum, whose length is 10, diameter at the end 4, diameter at the middle 6, and diameter taken half way between the middle and end, equal to 5?

5X2=10X10=100

6 X 6 36

4X 4= 16

Sum 152X10X,1309-198,968 ans.

PROBLEM 41.

To find the solidity of a parabolic conoid.

This solid may be generated by the revolution of a semiparabola about its abscissa or altitude.

RULE.

As a parabolic conoid is half of its circumscribing cylinder, of the

same base and altitude, multiply the area of the base by half the height and the product will be the solidity.

EXAMPLE.

If the diameter of the base of a parabolic conoid be 40 inches, and its height 42, what is the solidity?

40X40X,7854X2126389,44 ans.

PROBLEM 42.

Of the lower frustum of a Parabolic Conoid.

This solid is made by a plane, passing through the conoid, parallel to its base.

RULE.

Multiply the sum of the squares of the diameters of the bases by ,7854, and that product by half the height, the last product will be the solidity.

EXAMPLE.

If the diameter of a frustum of a parabolic conoid be 40 and 30 inches, and its height 20 inches, required its solidity? 40X40+30X30X,7854 X 10-19635 ans.

PROBLEM 43.

Of a Parabolic Spindle.

This solid is formed by an obtuse parabola, turned about its greatest ordinate; to find the solid content thereof this is the

RULE.

Multiply the area of its middle or greatest diameter by of its perperpendicular length, and it will give its solidity.

EXAMPLE.

If the diameter at the middle of a parabolic spindle be 20 inches, and its length 60, required its solidity?

20X20X,7854 X32=10053,12 ans.

N. B. Every parabolic spindle is of its least circumscribing cylinder.

PROBLEM 44.

Of the middle frustum of a Parabolic Spindle.

This is a cask like solid, wanting two equal ends of said spindle. To find the solid content thereof this is the

RULE.

To the sum and half sum of the squares of the two diameters add three tenths of the difference of their squares, which multiply by a third of the length, and the product will be the solidity.

EXAMPLE.

If the middle and end diameters of the middle frustum of a parabolic spindle be 40 and 30 inches, and its length 60, what is its solidity?

PROBLEM 45.

Of a Cylinderoid.

Then

A cylinderoid is a solid somewhat like the frustum of a cone, one base being elliptical and the other circular. To find the solid content thereof this is the

RULE.

The same as for the frustum of a cone or pyramid : Or, to the areas of both bases add a mean area, that is the square root of the product of the two bases, then multiply that sum by a third of the height or length, and it will give the solidity.

EXAMPLE.

If the diameters of the greater base of a cylinderoid be 30 and 20 inches, the diameter of the less base 12, and length 60 inches ; what is the solidity?

Measure the internal diameter of the ring, and its circumference; then multiply the circumference by ,31831, and the product will be the diameter of the ring, which add to the internal diameter; multiply this sum by 3,1416, and the product will be the der equal to the ring of the same base. Then the of the ring, multiplied by the length of the said cylinder, will give the solidity of the ring.

EXAMPLE.

length of a cylinarea of a section

If an iron ring be 12 inches in circumference, or girth, and its internal diameter be 20 inches, what is its solidity?

20+3,8×3,1416-74,77 the

,31831 X12=4,8= ring's diameter. length of the cylinder, equal to the ring: And 3,8×3,8×,7854×74,77 847,97

solidity required.

PROBLEM 47.

To find the solidity of any irregular body, whose dimensions cannot be taken; such as the bones in a horse's head, a thorn, &c.

RULE.

Take any vessel of a regular form, either square or round, and therein put the irregular solid; then pour in as much water as will cover the solid: this being done, take out the solid, and observe how far the water sinks or falls on the side of the vessel, and compute the solidity by the foregoing problems.

Or thus :-Take any sort of vessel, fill it with water to the brim, then immerse the irregular body; receive the water that runs over, and pour it into some vessel of a regular form, and then proceed to find the content as above.

OF THE FIVE REGULAR OR PLATONIC BODIES.

There are five solids contained under equal regular sides, which, by way of distinction, are called the five regular badies. These bodies are the Tetraedron, the Hexaedron orCube, the Octaedron, the Dodecaedron, and the Eicosiedron.

PROBLEM 48. Of the Tetraedron

A tetraedron, or solid, is contained under four equal eqilateral triangles or it is a pyramid on a triangular base, to find its solidity observe the Rule in Problem 30. Or the solid and superficial content of this and the following bodies may be found from the following

The table exhibits the solid and superficial content of any of the above bodies, the side being unity. In using which observe the following

RULES.

1. Multiply the proper tabular number by the cube of the given; the product will be the solidity.

2. Multiply the proper tabular number by the square of the given side, the product will be the superficial content.

EXAMPLE.

The side of a Tetraedron being given equal 3, to find the solid and superficial content?

PROBLEM 49. Of the Oetaedron.

An octaedron is a solid, contained under eight equal and equilateral triangles, which may be conceived to consist of two pyramids on the same square base, and of the same height; and consequently its solidity or area may be found by Prob. 30, or the tabular numbers.

EXAMPLE.

The side of an octaedron being 3, to find the solid and superficial content?

Cube of 3=27X,47144-12,7278 solidity.
Square of 3=9X3,4641=31,176 area.

PROBLEM 50. Of the Dodecaed on.

A dodecaedron is a solid contained under 12 equilateral pentagons, having pentagonal bases, whose common vertex is in the middle point or centre. Its solid and superficial content may be found by Prob. 30, or by the tabular numbers.

EXAMPLE.

The side of a dodecaedron being 3, to find the solid and superficial content?

PROBLEM 51.

Of an Eicosiedron.

An eicosiedron is a solid contained under twenty equal and equilateral triangles, and may be conceived to consist of twenty equal triangular pyramids, whose vertices all meet in the centre; its solidity may be found by prob. 30, or by the tables.

EXAMPLE.

The side of an eicosiedron being 3, to find the solid and superficial"

content.

Cube of 3=27X2,18169-58,90563 solidity, Square of 39x8,66025-77,94225=uperficial content.