A prismoid is a solid somewhat like the frustum of a pyramid, but its bases are disproportional. 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. If the diameters of the greater base of a cylindroid be 30 and 20 inches, the diameter of the less base 12, and length 60 inches; what is the solidity. 30×20=600 12x12=144 √144×600=293.9 1037.9 1037.9X 7854×20 (=60-3)= 16303-33, Ans. If the diameters of the greater base of a prismoid be 30 and 20 inches, the less base 20 by 10 inches, and length 30 inches: What is its solidity? 30×20=600 √600×200=346·4 1146 4 1146·4×10 (30-÷3)=11464 solidity in inches. Note. To find the solidity of a Wedge, add the length of the edge to twice the length of the base, and multiply the sum by the product of the height of the wedge and the breadth of the base, and one sixth of this product will be the solidity. Let the base of a wedge be 27 by 8, the edge 36, and the height 2×27+36×8×42 42; then 6 =5040, Ans. ART. 45. To measure a Solid Ring. RULE. Measure the internal diameter of the ring, and its girth, or circumference: then multiply the girth by 31831, and the product will be the diameter of the wire, which add to the internal diameter; multiply this sum by 3·1416, and the product will be the length of a cylinder equal to the ring of the same base. Then the area of a section of the ring multiplied by the length of the said cylinder will give the solidity of the ring. If an iron ring be 12 inches in girth, and its internal diameter be 20 inches; what is its solidity? 31831×12=3·8=ring's diameter. 20+3.8X3-1416-74 77 the length of a cylinder equal to the ring: And 3 8X3 8X 7854×74-77-847-97-solidity. ART. 46. To measure the Solidity of any irregular Body, whose dimensions cannot be taken. Take any regular vessel, either square or round, and put the irregular body into it: pour so much water into the vessel as will exactly cover the body, and measure the dry part from the top of the vessel to the water, then take out the body, and measure again from the top of the vessel to the water, and subtract the first measure from the second, and the difference is the fall of the water: then, if the vessel be square, multiply the side by itself, and that product by the fall of the water, and you will have the content of the body; but if it be a long square, multiply the length by the breadth, and that product by the fall of the water; or, lastly, if it be a round vessel, multiply the square of the diameter by 7854, and that product by the fall of the water, and you will have the content. EXAMP. 1. A body being put into a vessel 18 inches square, on taking out the body, the water sunk 9 inches; required the content of the body? 18 inch. 1.5 foot. 9 inch.75 foot. 1.5 X 1.5 X 75 = 1.6875 foot, content. EXAMP. 2. A body put into a cistern 4 feet by 3, on tak ing it out, the water fell 6 inches; required the content of the body? 4X3 X5 6 feet, content. EXAMP. 3. A body being put into a round tub, whose diameter was 1.5 foot, on taking out the body, the water fell 1.5 foot; what was the content of the body? 1.5X1.5X 7854×1·5= 2.65 feet, content. Of the five Regular Bodies. There are five solids contained under equal regular sides, which by way of distinction, are called the five regular bodies. These are the Tetraedron, the Hexaedron or Cube, the Octaedron, the Dodecaedron, and the Eicosiedron. The measuring of the cube was shewn at Art. 28. I shall now show how to measure the other four by the following Table, which is the shortest method. A Table of the solid and superficial content of each of the five bodies, the sides being unity, or 1. All like solid bodies being in proportion to one another as the cubes of their like sides, the solid content of any of these bodies may be found by multiplying the cubes of their sides by the numbers in the second column under Solidity; and their superficies, by multiplying the squares of their sides into the numbers in the third column under Superficies. OF THE TETRAEDRON. This solid is contained under four equal and equilateral triangles. that is, it is a triangular pyramid of four equal faces, the side of whose base is equal to the slant height of the pyramid, from the angles to the vertex. ART. 47. The side of the Tetraedron being 3, to find the solid and superficial content. Cube=3x3x3=27, and 27x11785=3·18195-solidity. OF THE OCTAEDRON. This solid is contained under eight equal and equilateral triangles, which may be conceived to consist of two quadrangular pyramids of equal bases joined together, the sides of whose bases are equal to the given sides of the triangles, under which it is contained. ART. 48. The side of an Octaedron being 3, to find the solid and superficial content. Cube 3x3x3=27, and 27x-4714-12 7278=solidity. OF THE DODECAEDRON. This solid is contained under 12 equilateral pentagons, and may be conceived to consist of twelve pentagonal pyramids, of equal bases and altitude, whose vertices meet in the centre of the dodecaedron. ART. 49. The side of a Dodecaedron being 3, to find the solid and superficial content. Cube=3x3x3=27, and 27×7·663119=206.904. OF THE EICOSIedron. This solid is 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. ART. 50. The side of an Eicosiedron being 3, to find the solid and superficial content. Cube 3x3x3=27, and 27×2.18169-58 90563=solidity. As the figures of some of these bodies would give but a confused idea of them, I have omitted them; but the following figures, cut out in pasteboard, and the lines cut half through, will fold up inte the several bodies. Tetraedron." Hexaedron. Octaedron. Dodecaedron. Eicosiedron. OF CASK GAUGING. Among the many different canons drawn from Stereometry, for Gauging casks, the following is as exact as any. Take the dimensions of the cask in inches, viz. the diameter at the bung and head, and length of the cask; subtract the head diameter from the bung diameter, and note the difference. If the staves of the cask be much curved or bulging between the bung and the head, multiply the difference by 7; if not quite so curve, by 65; if they bulge yet less, by 6; and if they are almost or quite straight, by ·55, and add the product to the head diameter; the sum will be a mean diameter, by which the cask is reduced to a cylinder. Square the mean diameter, thus found, then multiply it by the length; divide the product by 359 for ale or beer gallons, and by 294 for wine gallons. Note 1. The length is most conveniently taken by callipers, allowing, for the thickness of both heads, 1 inch, 11 inch, or 2 inches, according to the size of the cask; but if you have no callipers, do thus; measure the length of the stave, then take the depth of the chimes, which with the thickness of the head, being subtracted from the length of the stave, leaves the length within. : Note 2. You must take the head diameter, close to its outside, and, for small casks, add three tenths of an inch for casks of 30, 40, or 50 gallons, 4 tenths, and for larger casks, 5 or 6 tenths, and the sum will be very nearly the head diameter within. In taking the bung diameter, observe, by moving the rod backward and forward, whether the stave, opposite the bung, be thicker or thinner than the rest, and if it be, make allowance accordingly. By the Sliding Rule. On D is 18.94, the gauge point for ale or beer gallons, marked AG, and 17·14, the gauge point for wine gallons, marked WG: set the gauge point to the length of the cask on C, and against the mean diameter, on D, you will have the answer in ale or wine gallons according to which gauge point you make use of. By the Scale. Take the extent from the guage point to the mean diameter, set one foot of the dividers in the length, and turning them twice over, they will point out the content. ART. 51. Required the content in ale and wine gallons, of a cask, whose bung diameter is 35 inches, head diameter 27 inches, and length 45 inches? Bung diameter 35 Square of the diameter=1062-76 Length= 45 ART. 52. A round mash tub is 42 inches diameter at the top, within, and 36 inches at the bottom, and the perpendicular height 48 inches; required the content in beer and wine gallons? This being the lower frustum of a cone, to the product of the di ameters add of the square of their difference; multiply this sum by the length, and it will give the solidity in such parts as the dimensions are taken in. If they be taken in inches, divide by 359 for beer, and 294 for wine gallons. 42×36+ 42 -36×42-36 3 X48 359-203 ale gallons. ART. 53. Let the difference of diameters of this tub be 6 inches, the height 48 inches, and the content 2033 gallons, to find the diameters? Multiply the content, if beer measure, by 359; if wine measure, by 294, and divide the product by the length from the quotient subtract of the square of the difference of the diameters; to this remainder add the square of the difference of the diameters, and extract the square root of the sum; from the square root subtract the difference of the diameters, and it will give the least diameter to great exactness, to which add the difference of the diameters, and the sum is the greatest diameter. The content of any vessel in gallons, &c. may be thus found: measure the inside of the vessel, according to the rule of the figure, and find the content in cubick inches; then, |