PRACTICAL GEOMETRY.Geometric figuresFig. 1-9.
PRACTICAL GEOMETRY.
Fig. 1-9.
Geometric figuresFig. 10-14.
Fig. 10-14.
To divide a right angle into three equal parts.
FromBas a centre with any radius describe the arcA C. FromAwith the radiusA Bcut the arcA CinD, and with the same radius fromCcut it inE. Then through the intersectionsD, andEdraw the linesB D,B E, and they will trisect, or divide the angle into three equal parts.Fig. 5.
To find the centre of a circle.
Draw any chordA B, and bisect it by the perpendicularC D. DivideC Dinto two equal parts, and the point of bisectionOwill be the centre required.Fig. 6.
To describe an equilateral triangle.
From the pointsA,B, as centres, and withA Bas radius, describe arcs intersecting each other inC. DrawC A,C B, and the figureA B Cwill be the triangle required.Fig. 7.
To describe a square.
From the pointB, drawB Cperpendicular, and equal toA B. OnA, andC, with the radiusA B, describe arcs cutting each other inD. Draw the linesD A,D C, and the figureA B C Dwill be the square required.Fig. 8.
To inscribe a square in a circle.
Draw the diametersA B,C Dperpendicular to each other. Then draw the linesA D,A C,B D,B C; andA B C Dwill be the square required.Fig. 9.
To inscribe an octagon in a circle.
Bisect any two arcsA C,B Cof the squareA B C DinG, andE. Through the pointsG, andE, and the centreOdraw lines, which produce toF, andH. JoinA F,F D,D H, &c. and they will form the octagon required.Fig. 9.
On a line to describe all the several polygons, from the hexagon to the dodecagon.
BisectA Bby the perpendicularC D. FromAas a centre, and withA Bas a radius, describe the arcB E, which divide into six equal parts; and fromEas a centre describe the arcs 5F, 4G, 3H, &c. Then from the intersectionEas a centre, and withE Aas a radius, describe the circleA I D B, which will containA Bsix times. FromFin like manner as a centre, and withF Aas radius, describe the circleA K L B, which will containA Bseven times; and so on for the other polygons.Fig. 10.
To inscribe in a circle an equilateral triangle.
From any pointDin the circumference as a centre, and with the radiusD Oof the given circle, describe an arcA O Bcutting the circumferenceinA, andB. ThroughD, andOdrawD C. Then, joinA B,A C,B C; and the figureA B Cwill be the triangle required.Fig. 11.
To inscribe a hexagon in a circle.
Bisect the arcsA C,B CinE, andF, and joinA D,D B,B F, &c., which will form the hexagon. Or carry the radius six times round the circumference, and the hexagon will be obtained.Fig. 11.
To inscribe a dodecagon in a circle.
Bisect the arcA Dof the hexagon inG, andA Gbeing carried twelve times round the circumference, will form the dodecagon.Fig. 11.
To inscribe a pentagon, hexagon, or decagon, in a circle.
Draw the diameterA B, and make the radiusD Cperpendicular toA B. BisectD BinE. FromEas a centre, and withE Cas radius, describe an arc cuttingA DinF. JoinC F, which will be the side of the pentagon,C Dthat of the hexagon, andD Fthat of the decagon.Fig. 12.
To find the angles at the centre, and circumference of a regular polygon.
Divide 360 by the number of the sides of the given polygon, and the quotient will be the angle at the centre; and this angle being subtracted from 180, the difference will be the angle, at the circumference, required.
Table, showing the angles at the centre, and circumference.
To inscribe any regular polygon in a circle.
From the centreCdraw the radiiC A,C B, making an angle equal to that at the centre of the proposed polygon, as contained in the preceding table. Then the distanceA Bwill be one side of the polygon, which, being carried round the circumference the proper number of times, will complete the polygon required.Fig. 13.
Geometric figuresFig. 15-20.
Fig. 15-20.
To circumscribe a circle about a triangle.
Bisect any two of the given sides,A B,B Cby the perpendicularsE F,D F. From the intersectionFas a centre, and with the distance of any of the angles, as a radius, describe the circle required.Fig. 14.
To circumscribe a circle about a square.
Draw the two diagonalsA C,B Dintersecting each other inO. FromOas a centre, and withO A, orO B, as a radius, describe the required circle.Fig. 15.
To circumscribe a square about a circle.
Draw the two diametersA B,C Dperpendicular to each other, through the pointsA,C,B,D, draw the tangentsE F,E G,G H,F H, andE G H Fwill be the square required.Fig. 16.
To reduce a map, or plan, from one scale to another.
Divide the given figureA Cby cross lines, forming as many squares as may be thought necessary. Draw a lineE F, on which set off as many parts from the scaleM, asA Bcontains parts of the scaleN. DrawE H, andF Gperpendicular toE F, and each equal to the proportional parts contained inA D, orB C. JoinH G, and divide the figureE Ginto the same number of squares as the originalA C. Describe in every square what is contained in the corresponding square of the given figure; andE F G Hwill be the reduced plan required. The same operation will serve either to reduce, or enlarge any map, plan, drawing, or painting.Fig. 17.
Mensuration is of three kinds, viz., lineal, superficial, and solid.
Lineal measurehas reference to length only.
Superficial measure(or the surface) includes length, and breadth.
Solid measure(or the content) comprehends length, breadth, and thickness.
MENSURATION OF PLANES.
The areaof any plane figure is the superficial measure contained within its extremes, or bounds. This area is estimated by the number of small squares that may be contained in it, the side of these measuring squares being an inch, a foot, or any other fixed quantity, and hence the area is said to be so many square inches, square feet, &c.Vide Table, Square measure.Page 275.
To find the area of a parallelogram, whether a square, rectangle, &c.
Multiply the length by the breadth, or perpendicular height, for the area required.
Example.—Required the area of a rectangle, whose length is 9 feet, and breadth 4 feet.
9 × 4 = 36 feet. The required area, or surface.
To find the area of a triangle, its base, and perpendicular height being given.
Multiply the base by the perpendicular height, and half the product will be the area.
Example.—Required the number of square yards contained in a triangle, whose base is 20 yards, and perpendicular height 14 yards.
20 × 142= 140 square yards. Area required.
To find the area of a triangle, whose three sides are given.
From half the sum of the three sides, subtract each side severally; multiply the half sum, and the three remainders together, and the square root of the product will be the area required.
Example.—Required the area of a triangle, whose sides are 50, 40, and 30 feet.
50 + 40 + 302= 60, half the sum of the three sides.60 - 30 = 30 First difference.60 - 40 = 20 Second difference.60 - 50 = 10 Third difference.30 × 20 × 10 × 60 = 360000.Square root of 360000 = 600. Area required.
Two sides of a right-angled triangle being given, to find the third side.
1. When the two sides forming the right angle are given, to find the hypothenuse, or side opposite the right angle.
Take the square root of the sum of the two sides squared for the side required.
Example.—Required the length of the interior slope of a rampart, whose perpendicular height is 17 feet, and the base of the slope 20 feet.
17 × 17 = 28920 × 20 =400
The square root of 689 = 26·24. The length required.
2. When the hypothenuse, and one of the perpendicular sides are given.
From the square of the hypothenuse, subtract the square of thegiven side, and the square root of the remainder will be the side required.
Example.—The hypothenuse being 5 yards, and the base 4 yards, required the other side.
5 × 5 = 254 × 4 =16
The square root of 9 = 3 yards. The side required.
To find the area of a trapezium, A B C D.
Draw the diagonalA C, upon which let fall from its opposite anglesB, andD, the perpendicularsB F,D E. Find by measurement the diagonalA C, and the perpendicularsB F,D E, then multiply the sum of the perpendiculars by the diagonal, and half the product will be the area of the trapezium.Fig. 18.
Example.—Required the area of the trapezium, whose diagonalA Cis 100 feet, and perpendicularsB F30 feet, andD E40 feet.
(30 + 40) × 1002= 3500 square feet. Area required.
Or, divide the trapezium into two triangles by a diagonal, then find the areas of these triangles, and add them together.
To find the area of a trapezoid, A B C D.
Multiply the sum of the parallel sidesA B,D Cby the perpendicular distanceE C, and half the product will be the area.Fig. 19.
Example.—Required the area of the trapezoidA B C D, of which the parallel sidesA B,D Care 120 feet, and 90 feet, and the perpendicular distanceE C40 feet.
(120 + 90) × 402= 4200 square feet. Area required.
To find the area of an irregular figure, or polygon.
Draw diagonals dividing the figure into trapeziums, and triangles; then, having found the area of each, add them together, and the sum will be the area required.
To find the area of a figure, having a part bounded by a curve.
Draw a right line joining the extremities of the curve, then find the area of the trapezium. On the right line let fall as many perpendiculars as the several windings of the curve may require. Find their lengths, and divide their sum by the number of perpendiculars, and the quotient will be the mean breadth; which being multiplied by the length of the right line, will give the area of the curved part. Thisarea being added to that of the trapezium will give the area of the required figure.
To measure long irregular figures.
Measure the breadth at both ends, and at several placesat equal distances. Add together all these intermediate breadths, and half the two extremes, which sum multiply by the length, and divide by the number of parts for the area. If the perpendiculars, or breadths,be not at equal distances, compute all the parts separately, as so many trapezoids, and add them all together for the whole area.
Example.—The breadths of an irregular figure at five equi-distant places being 8, 2, 7, 9, 4, and the whole length 40, required the area.
8 + 4 = 1212 ÷ 2 = 66 + 2 + 7 + 9 = 2424 × 404= 240. Area required.
To find the number of square acres in any of the preceding figures.[56]
Divide the superficial content in feet by 43560, and the quotient will be the number required.
To bring square chains to acres.
Of square chains strike off two decimal places to the right, and the rest of the figures will be acres.
To bring square links to acres.
Of square links cut off five of the figures on the right hand, for decimals, and the rest will be acres; then multiply these decimals by 4, for roods, cutting off five figures as before; and the decimals of these again by 40, for perches, when five figures are again to be struck off.
To find the area of a regular polygon.
Multiply theperimeter(or sum of the sides) of the polygon by theperpendicular drawn from its centre on one of its sides, and take half the product for the area.
Or, multiply the area of one of the triangles by the number of sides of the polygon, and the product will be the area of it.
Example.—Required, the area of a regular hexagon, whose side is 40 feet, and the perpendicular 34·64 feet.
40 × 6 = 240 the perimeter.240 × 34·642= 4156·8 square feet. Area required.
To find the diameter, and circumference of any circle, the one from the other.
Use either of the following proportions:
or, instead of dividing the diameter by 3·1416, multiply it by ·3183, for the circumference.
Example 1.—Required, the circumference of a circle, whose diameter is 20 feet.
As 7 : 22 :: 20 : 62·857 feet. Circumference required.
Example 2.—Required, the diameter of a circle, whose circumference is 36 inches.
As 22 : 7 :: 36 : 11·45 inches. Diameter required.
To find the diameter of a circle, the area being given.
Divide the area by ·7854, and the square root of the quotient will be the diameter required.
Example.—Required, the diameter of a circle, whose area is 176·715 square feet.
176·715 ÷ ·7854 = 225.Square root of 225 = 15 feet. Diameter required.
To find the area of a circle.
1. Multiply half the circumference by half the diameter, or multiply the whole circumference by the whole diameter, and take ¼ of the product.
2. Or, square the diameter, and multiply that square by ·7854 for the area.
3. Or, square the circumference, and multiply that square by 0·7958.
Example 1.—Required the area of a circle, whose circumference is 55·548 inches, and its diameter 18 inches.
55·5482= 27·774 half circumference.182= 9 half diameter.27·774 × 9 = 249·966, square inches. Area required.
Example 2.—Required the area of a circle whose diameter is 12 feet.
12 × 12 = 144, square of the diameter.·7854 × 144 = 113·0976 square feet. Area required.
Example 3.—Required the area of a circle, whose circumference is 22 feet.
22 × 22 = 484.484 × ·07958 = 38·51672 square feet. Area required.
To find the area of a circular ring,
or space included between the circumferences of two circles, the one within the other.
1. Subtract the square of the less diameter from the square of the greater, and multiply their difference by ·7854.
2. Or, find the area of each circle separately, and subtract one from the other, for the area required.
3. Or, multiply the sum of the diameters by the difference of the same, and that product by ·7854 for the area.
Example.—Required the area of a ring, the diameters of whose bounding circles are 10, and 20.
By Rule 3.
20 + 10 = 30, sum of diameters.20 - 10 = 10, difference of diameters.30 × 10 × ·7854 = 235·62. The area.
To find the length of any arc of a circle.
1. As 360° is to the number of degrees in the arc, so is the circumference to the length of the arc.
2. Or, multiply the degrees in the given arc by the radius of the circle, and the product by ·01745 for the length of the arc.
Example.—Rule 2.—Required the length of an arc of 30°, the radius being 9 feet.
30 × 9 × ·01745 = 4·7115. Length of arc.
To find the area of the sector of a circle.
Multiply the radius by the arc, and half the product will be the area.
Example.—Required the area of the sector, whose radius is 30 inches, and the length of the arc 36·6 inches.
36·6 × 302= 549 square inches. Area required.
To find the area of the segment of a circle.
Find the area of the sector, by the preceding rule. Then find the area of the triangle formed by the chord of the segments, and the radii of the sector. Then, if the segment be less than a semicircle, subtract the area of the triangle from it; or, if the segment be greater than a semicircle, add the area of the triangle to it; for the area of the segment.
Example.—Required the area of a segment less than a semicircle, the radius being 20 inches, the chord 22·42 inches, the length of the arc 24·43 inches, and the perpendicular 16·56 inches.
24·43 × 202= 244·3 square inches. Area of the sector.22·42 × 16·562= 185·6376 square inches. Area of the triangle.244·3 - 185·6376 = 58·6624 square inches. Area required.
To find the area of a semicircle.
1. Multiply ¼ of the circumference by the radius, and the product will be the area.
2. Or, multiply the square of the diameter by ·7854, and half the product will be the area.
Example.—Rule 2.—Required the area of a semicircle, the diameter being 50 inches.
50 × 50 × ·78542= 981·75 square inches. Area required.
To find the area of an ellipsis, or oval.
Multiply the longest diameter, or axis, by the shortest, then multiply the product by ·7854 for the area.
Example.—Required the area of the ellipse, whose diameters are 25 inches, and 18 inches.
25 × 18 × ·7854 = 353·43 square inches. Area required.
To find the area of a parabola, or its segment.
Multiply the base by the perpendicular height, and take two-thirds of the product for the area.
Example.—Required the area of a parabola, whose base is 20 feet, and height 12 feet.
20 × 12 = 240⅔ of 240 = 160 square feet. Area required.
MENSURATION OF SOLIDS.
A solidis a body containing length, breadth, and thickness.
Solids are measuredby cubes, whose sides are each an inch, a foot, a yard, &c., and the solidity, capacity, or content of any figure is computed by the number of such cubes as are contained in it.—Vide Cubic measure,page 276.
A cubeis a solid contained by six equal square sides.
A pyramidis a solid whose sides are all triangles meeting together in a point, the base being any plane figure whatever. It is called a triangular pyramid when its base is a triangle; a square pyramid when its base is a square, &c.
The segment of a pyramid, cone, or any other solidis a part ofD E F Gcut off from the top by a planeD E F, parallel to the baseA B C.—VideFig. 21, Plate 2,Heights, Distances, andPractical Geometry.
A frustrum, or trunk, is a partA B C D E F, that remains at the bottom after the segment is cut off.
A coneis a round pyramid, of which the base is a circle.
The axis of a solidis a line from the vertex (or point) to the centre of the base, or through the centres of the two ends. When the axis is perpendicular to the base, it is a right prism, pyramid, or cone; otherwise it is oblique.
A sphereis a solid contained under one convex surface, and is described by the revolution of a semicircle about its diameter, which remains fixed.
The centre of the sphereis such a point within the solid as is everywhere equally distant from the convex surface, or circumference of it.
The diameter(or axis)of a sphereis a straight line, which passes through the centre, and is terminated by the convex surface.
A segment of a sphereis a part cut off by a plane, the section of which is always a circle, called thebase of the segment.
A sector of a sphereis that which is composed of a segment (less than an hemisphere) and of a cone.
A prismis a solid, the sides of which are parallelograms, having its ends equal, and similar plane figures.
Prisms are namedaccording to the number of angles in the base.
A cylinderis a solid, the two ends of which are circular; and it is described, or formed, by the revolution of a right-angled parallelogram about one of its sides, which remains fixed.
To find the superficies of a prism, or cylinder.
Multiply the perimeter of one end of the prism by the length, or height of the solid, and the product will be the surface of all itssides. To which add also the area of the two ends of the prism, when required.
Or, compute the areas of all the sides, and ends separately, and add them all together.
Example.—Required the surface of a cube, whose sides are each 5 inches.
5 + 5 + 5 + 5 = 20, perimeter of one end.20 × 5 = 100, surface of sides.5 × 5 = 25, area of one end.100 + 25 + 25 = 150 square inches. Surface of cube.
To find the surface of a pyramid, or cone.
Multiply the perimeter of the base by the slant height, or length of the side, and half the product will be the surface of the sides; to which add the area of the base when required.
Example.—Required the upright surface of a triangular pyramid, the slant height being 20 feet, and each side of the base 3 feet.
3 + 3 + 3 = 9, perimeter of base.9 × 202= 90 feet. Surface required.
To find the surface of the frustrum of a pyramid, or cone.
Add together the perimeters of the two ends, multiply their sum by the slant height, and take half the product.
Example.—How many square feet are in the surface of the frustrum of a square pyramid, whose slant height is 10 feet, each side of the base 3 feet, and each side of the less end 2 feet.
3 + 3 + 3 + 3 = 12, perimeter of base.2 + 2 + 2 + 2 = 8, perimeter of less end.(12 + 8) × 102= 100 feet. Surface required.
To find the solid content of a prism or cylinder.
Find the area of the base, or end, and multiply it by the length of the prism, or cylinder. Fora cube, multiply its side twice by itself; and for a parallelopipedon, multiply the length, breadth, and depth together for the content.
Example.—Required the solid content of a cube, whose side is 24 inches.
24 × 24 × 24 = 13824 square inches. Content required.
To find the content of the solid part of a hollow cylinder.
From the content of the whole cylinder considered as a solid, subtract the content of the hollow part, also considered as a solid, and the difference will be the solidity required.
Example.—Required the content of the solid part of the hollow cylinder whose exterior diameter is 12 inches, the interior diameter 8 inches, and height 20 inches.
12 × 12 × ·7854 = 113·0976, area of base of cylinder.113·0976 × 20 = 2261·952, solidity of whole cylinder.8 × 8 × ·7854 = 50·2656, area of base of hollow cylinder.50·2656 × 20 = 1005·312, content of hollow part.2261·952 - 1005·312 = 1256·64 cubic inches. Solidity required.
To find the solidity of the frustrum of a cylinder.
Multiply the area of the base by half the greatest, and the least lengths, and the product will be the solidity.
Example.—Required the solidity of a frustrum, whose diameter is 24 inches, the greatest length 36 inches, and the least length 20 inches.
24 × 24 = 576. Square of the diameter.576 × ·7854 = 452·3904. Area of the base.452·3904 × (36 + 20)2= 12666·9312 Cubic inches. Solidity required.
To find the content of a pyramid, or cone.
Find the area of the base, and multiply that area by the perpendicular height, and take ⅓ of the product.
Example.—Required the solidity of a square pyramid, each side of its base being 30, and its perpendicular height 25.
30 × 30 = 900, area of base.900 × 253= 7500, solidity required.
To find the solidity of the frustrum of a cone, or pyramid.
Add into one sum the areas of the two ends, and the mean proportional between them: take ⅓ of that sum for the mean area, which multiply by the perpendicular height, or length of the frustrum.
Note.—To find a mean proportional.
As one of the sides of the base is to the homologous, or corresponding side of the other end, so is the area of the base to the mean proportional required.
Example.—Required the number of solid feet in a piece of timber,whose bases are squares, each side of the greater end being 15 inches, and each side of the less end 6 inches; also the length of the perpendicular altitude 24 feet.
15 × 15 = 225, area of the base.6 × 6 = 36, area of the top.As 15: 6 :: 225: 90, mean proportional.24 feet = 288 inches.(225 + 36 + 90) × 2883= 33696 cubic inches = 19½ cubic feet.
To find the surface of a sphere, or any segment.
Multiply the circumference of the sphere by its diameter, which will give the whole surface.
Or, square the diameter, and multiply by 3·1416.Or, square the circumference, and multiply by ·3183;or divide by 3·1416.
Note.—For the surface of the segment, or frustrum, multiply the whole circumference of the sphere by the height of the part required.
Example.—Required the superficies of a globe whose diameter is 24 inches.
24 × 24 × 3·1416 = 1809·5616 square inches.
To find the solidity of a sphere, or globe.
1. Multiply the surface by the diameter, and take ⅙ of the product.
Or, multiply the square of the diameter by the circumference, and take ⅙ of the product.
2. Cube the diameter, and multiply by ·5236.
3. Cube the circumference, and multiply by ·01688.
Example.—Required the content of a sphere, whose axis is 12.
12 × 12 × 12 × ·5236 = 904·7808. Content required.
To find the solidity of an hemisphere.
Find the solidity of the sphere, and half the content will be that of the hemisphere.
Note 1.—Any sphere, or globetwicethe diameter of another containsfour timesthe superficies, or area of the other, andeight timesthe solid content. Hence the superficies of spheres are as the squares, and the solidity as the cubes of their diameters.
Note 2.—The cube of the diameter of a sphere in inches, multiplied by ·00188, will give the number ofimperial gallons it will contain.
To find the solid content of a spherical segment.
1. From three times the diameter of the sphere, take double the height of the segment; then multiply the remainder by the square of the height, and this product by ·5236.
2. Or, to three times the square of the radius of the segment’s base add the square of its height; then multiply the sum by the height, and the product by ·5236.
Example.—Required the content of a spherical segment 2 feet in height, cut from a sphere of 8 feet diameter.
(3 × 8) - (2 × 2) = 2020 × 22× ·5236 = 41·888 cubic feet. Content required.
To find the diameter of a sphere, its solidity being given.
Divide the solidity by ·5236, and take the cube root of the quotient.
Example.—The solidity of a sphere being 113·0976 solid inches, what will be its diameter?
113·0976·5236= 216, the cube root of which is 6 inches, the diameter required.
To find the weight of an iron shot, its diameter being given.
Take ⅛ of the cube of the diameter, and ⅛ of that eighth, and the sum of these two quotients will be the weight in pounds.
Or, as 64 is to 9 lb. so is the diameter cubed to its weight.
Example.—Required the weight of an iron shot whose diameter is 3·5 inches?
3·5 cubed = 42·875, cube of diameter.42·8758= 5·3595·3598= ·6695·359 + ·669 = 6·028 pounds. Weight required.
To find the weight of a leaden ball, its diameter being given.
Take ⅓ of the cube of the diameter, and from it subtract ⅓ of this third, and the remainder will be the weight, nearly.
Or, take 3/14 of the cube of the diameter.
Example.—What is the weight of a leaden ball whose diameter is 3·3 inches?
3·3 cubed = 35·937, cube of diameter.35·9373= 11·97911·9793= 3·99311·979 - 3·993 = 7·986 pounds. Weight required.
To find the diameter of an iron shot, its weight being given.
Multiply the cube root of the shot’s weight by 1·923 for the diameter.
To find the diameter of a leaden ball, its weight being given.
To 4 times the weight add half the weight, and 3/100 of half the weight; and the cube root of this sum will be the diameter in inches, nearly.
Example.—What is the diameter of a leaden ball, whose weight is 8 pounds?
8 x 4 = 3282= 43100of 4 = ·12.32 + 4 + ·12 = 36·12, of which the cube root is 3·3 inches, nearly.Diameter required.
To find the weight of an iron shell, its interior and exterior diameter being given.
Take964of the difference of the cubes of the external and internal diameters, for the weight of the shell in pounds.
Example.—What is the weight of a shell whose exterior diameter is 12·85 inches, and interior diameter 8·75 inches?
12·85 cubed = 2121·8241,8·75 cubed = 669·9218.2121·8241 - 669·9218 = 1451·9022.964of 1451·9022 = 204·1737 pounds. Weight required.
To find the quantity of powder a shell will contain.
Divide the cube of the interior diameter in inches by 57·3, and the quotient will be the weight in pounds.
Or, multiply the cube of the diameter by 11, and divide by 21 for the quantity in half ounces.
Example.—How much powder will fill a shell, whose internal diameter is 7 inches?
7 cubed = 343.34357·3= 6 pounds nearly. Powder required.
To find the side of a cubical box to contain a given quantity of powder.[57]
Multiply the weight in pounds by 30, and the cube root of the product will be the side of the box in inches.
Example.—Required the side of a cubical box to hold 50 pounds of powder?
50 × 30 = 1500, the cube root of which is 11·44, which will be the side of the box in inches.
To find the quantity of powder to fill the chamber of a mortar, or howitzer.
Multiply the content of the chamber in inches by 55, and divide the product by 1728, and the quotient will be the quantity of powder in pounds.
Note.—The chamber of a mortar, or howitzer, is formed of a hollow frustrum of a right cone, and of a hollow hemisphere.
Example.—Required the quantity of powder to fill the chamber of a 13-inch mortar in which the diameterA Bis 9·5 inches, the diameterC E6·5 inches, and the lengthD G21·5 inches.VideFig. 22. Plate 2.Heights and Distances, andPractical Geometry.
The content of the chamber must be found by finding the content of the hollow frustrum of the cone, and that of the hemisphere (vide preceding rules): which in this example will be 999·9741875.
Then999·9741875 × 551728= 31 pounds, nearly.
To find the quantity of powder to fill a rectangular box.
Divide the content (viz., length × breadth × depth) of the box in inches by 30 for the pounds of powder.
Example.—How much powder will fill a box, the length being 15 inches, the breadth 12, and the depth 10 inches.
15 × 12 × 1030=180030= 60 pounds. Number required.
To find the quantity of powder to fill a cylinder.
Multiply the square of the diameter by the length, then divide by 38·2 for the pounds of powder.
Example.—How much powder will the cylinder contain, whose diameter is 10 inches, and length 20 inches?
10 × 10 × 2038·2= 52⅓ pounds, nearly.
To find the size of a shell, to contain a given weight of powder.
Multiply the pounds of powder by 57·3, and the cube root of the product will be the diameter in inches.
Example.—Required the diameter of a shell to contain 6 lb. of powder?
6 × 57·3 = 343·8, the cube root of which is 7, the diameter required, in inches.
To find what length of a cylinder (or bore of a gun) will be filled by a given weight of powder.
Multiply the weight in pounds by 38·2, and divide the product by the square of the diameter in inches, for the length.
Example.—What length of a cylinder 8 inches in diameter will be filled with 20 lb. of powder?
20 × 38·28 × 8= 111516inches.
To find the content, and weight of a piece of ordnance.
Divide the length of the gun into as many sections as may be found necessary. Find the content of each (by preceding rules) and from their sum subtract the content of a cylinder, whose length is equal to that of the bore, and its diameter equal to that of the calibre of the piece; multiply the difference (if it be a brass gun) by 5·0833, (if an iron gun) by 4·2968, and the product will be the weight in ounces.
To find the content of a cask.
Multiply half the sum of the areas of the two interior circles, viz. at the head, and bung, by the interior length, for the content.
Or, to the area of the head add twice the area at the bung, multiply that sum by the length, and take one-third of the product.
Example.—Required the content of a cask, its greatest interior diameter being 24 inches, its least interior diameter 20 inches, and the interior length 30 inches.
24 × 24 × ·7854 = 452·3904, area of large circle.20 × 20 × ·7854 = 314·1600, area of small circle.452·3904 + 314·16002= 383·2752, half sum.
Then 383·2752 × 30 = 11498·256, the content; which being divided by 277¼ (the number of cubic inches in a gallon) will give the number of gallons contained in the cask.
Thus11498·256277·25= 41·4725, &c. Number of gallons required.
Note.-The content of any vessel in cubic feet, multiplied by 6·232 (or if in inches by ·003607) will give the number ofimperial gallons it will contain.
OF THE CIRCLE, CYLINDER, SPHERE, ETC.
1. The circle contains a greater area than any other plane figure, bounded by an equal perimeter, or outline.
2. The areas of circles are to each other as the squares of their diameters; any circle twice the diameter of another contains four times the area of the other.
3. The diameter of a circle being 1, its circumference equals 3·1416.
4. The diameter of a circle is equal to ·31831 of its circumference.
5. The square of the diameter of a circle being 1, its area equals ·7854.
6. The square root of the area of a circle, multiplied by 1·12837, equals its diameter.
7. The diameter of a circle, multiplied by ·8862, or the circumference multiplied by ·2821, equals the side of a square of equal area.
8. The sum of the squares of half the chord, and versed sine, divided by the versed sine, the quotient equals the diameter of the corresponding circle.
9. The chord of the whole arc of a circle taken from eight times the chord of half the arc, one-third of the remainder equals the length of the arc.
10. Or, the number of degrees contained in the arc of a circle, multiplied by the diameter of the circle, and by ·008727, the product equals the length of the arc in equal terms of unity.
11. The length of the arc of the sector of a circle multiplied by its radius, half the product is the area.
12. The area of the segment of a circle equals the area of the sector, minus the area of a triangle whose vertex is the centre; and base equals the chord of the segment.
13. The sum of the diameters of two concentric circles multiplied by their difference, and by ·7854, equals the area of the ring, or space contained between them.
14. The sum of the thickness, and internal diameter of a cylindric ring multiplied by the square of its thickness, and by 2·4674, equals its solidity.
15. The circumference of a cylinder multiplied by its length, or height, equals its convex surface.
16. The area of the end of a cylinder multiplied by its length, equals its solid content.
17. The area of the internal diameter of a cylinder multiplied by its depth, equals its cubical capacity.
18. The square of the diameter of a cylinder multiplied by its length, and divided by any other required length, the square root of the quotient equals the diameter of the other cylinder of equal solidity, or capacity.
19. The square of the diameter of a sphere multiplied by 3·1416 equals its convex surface.
20. The cube of the diameter of a sphere multiplied by ·5236, equals its solid content.
21. The height of any spherical segment, or zone, multiplied by the diameter of the sphere, of which it is a part, and by 3·1416, equals the area, or convex surface of the segment;
22. Or, the height of the segment multiplied by the circumference of the sphere of which it is a part, equals the area.
23. The solidity of any spherical segment is equal to three times the square of the radius of its base, plus the square of its height, and multiplied by its height, and by ·5236.
24. The solidity of a spherical zone equals the sum of the squares of the radii of its two ends, and one-third the square of its height, multiplied by the height, and by 1·5708.
25. The solidity of the middle zone of a sphere equals the sum of the square of either end, and two-thirds the square of the height, multiplied by the height, and by ·7854.
26. The capacity of a cylinder 1 foot in diameter, and 1 foot in length, equals 4·895 imperial gallons.
27. The capacity of a cylinder 1 inch in diameter, and 1 foot in length, equals ·034 of an imperial gallon.
28. The capacity of a cylinder 1 inch in diameter, and 1 inch in length, equals ·002832 of an imperial gallon.
29. The capacity of a sphere 1 foot in diameter, equals 3·263 imperial gallons.
30. The capacity of a sphere 1 inch in diameter, equals ·001888 of an imperial gallon.
31. Hence the capacity of any other cylinder in imperial gallons is obtained by multiplying the square of its diameter by its length; or the capacity of any other sphere by the cube of its diameter, and by the number of imperial gallons contained as above in the unity of its measurement.
OF THE SQUARE, RECTANGLE, CUBE, ETC.
1. The side of a square equals the square root of its area.
2. The area of a square equals the square of one of its sides.
3. The diagonal of a square equals the square root of twice the square of its side.
4. The side of a square is equal to the square root of half the square of its diagonal.
5. The side of a square, equal to the diagonal of a given square, contains double the area of the given square.
6. The area of a rectangle equals its length multiplied by its breadth.
7. The length of a rectangle equals the area divided by the breadth; or the breadth equals the area divided by the length.
8. The side, or end of a rectangle, equals the square root of the sum of the diagonal, and opposite side to that required, multiplied by their difference.
9. The diagonal in a rectangle equals the square root of the sum of the squares of the base, and perpendicular.
10. The solidity of a cube equals the area of one of its sides multiplied by the length of one of its edges.
11. The edge of a cube equals the cube root of its solidity.
12. The capacity of a 12-inch cube equals 6·232 gallons.
The tabular surface multiplied by the square of the linear edge, the product equals the surface required:
Or, the tabular solidity, multiplied by the cube of the linear edge, the product is the solidity required.
OF TRIANGLES, POLYGONS, ETC.
1. The complement of an angle is its defect from a right angle.
2. The supplement of an angle is its defect from two right angles.
3. The sine, tangent, and secant of an angle, are the cosine, cotangent and cosecant of the complement of that angle.
4. The hypothenuse of a right-angled triangle being made radii, its sides become the sines of the opposite angles, or the cosines of the adjacent angles.
5. The three angles of every triangle are equal to two right angles; hence the oblique angles of a right-angled triangle are each other’s complements.
6. The sum of the squares of the two given sides of a right-angled triangle is equal to the square of the hypothenuse.
7. The difference between the square of the hypothenuse, and given side of a right-angled triangle is equal to the square of the required side.
8. The area of a triangle equals half the product of the base multiplied by the perpendicular height;
9. Or, the area of a triangle equals half the product of the two sides, and the natural sine of the contained angle.
10. The side of any regular polygon multiplied by its apothem, or perpendicular, and by the number of its sides, half the product is the area.