-NRLF (p'^r William Collins, Sons, & Co.'s Educational Works. NEW CODE PROGRESSIVE READERS: FlRSl Seco: First Seco: Thir Four FiFTI SiXTI SiXTI SiXTI IN MEMORIAM FLORIAN CAJORl Old Testament History. By Rgv. C. Ivexs, - Nbw Tjestament Histojiy. By Ilov. C- Iyexs, • KS, s. d. - 2 - 3 - 5 - 7 - 9 - 1 - 1 3 - 1 6 » 1 6 L, 2 '■-' 1 6 to - 1 6 b, 1 6 - 1 1. - 1 - 1 r - 1 iG - 1 G y - 1 1 - 2 r," 1 - 1 6 - 1 6 - 2 - 2 - 2 ;s, 2 - 1 G - 1 G - 1 G - 1 - 1 2£-J^>z^ /fyc-h^ /^///^ — A X A TREATISE ^y2^ ftjOJuCtrv a.^^^^/9, MENSUEATION, FOR THE USE OF SCHOOL& v^/ GLASGOW Aiq'D LONDON: WILLIAM COLLINS, SONS, & OOMPANT. PREFACE TO SECOKD EDITIOS. To this Edition there is an AppendL"^, printed in a separate form, for the use of Teachers, containing the leading pro- perties of the Conic Sections, and the Demonstrations of the Eules of Mensuration. These were in the First Edition interspersed through the work, partly interwoven with the text, and partly in the shape of notes. It is hoped that the present arrangement will better suit the convenience of both Teachers and Pupils. Several other alterations have been made, which, it is hoped, will be found to be improve- ments. Teachers should direct their Pupils to learn only such portions of the work as may be necessary for their intended occupations ; for most pupils, the first and second sections, and a few problems in the fourth and sixth, will be quitf j^iifficJent CONTENTS. Mensuration of Sv^EnviciES— {continued). Given the side of a square, to find the diameter of a circle equal in area to the square . . , . Given the side of a square, to find the circumference of a circle whose area is equal to the square whose side ia given ...... To find the area of a sector of a circle . . To find the area of a segment of a circle . To find the area of a zone of a circle To find the area of a circular ring . To find the area of a part of a ring, or of the segment of a sector ....... To find the area of a lune . . . , To measure long irregular figures . . , Exercises in Mensuration of Superficies . • 36 3G 37 89 .40 40 40 41 42 SECTION III. CoNio Sections— Of the Ellipsis Of the Parabola Of the Hyperbola . • • SECTION IV. Mensuration of Solids— Definitions . To find the solidity of a cube Of a parallelopipedon . Of a prism Of a cylinder , To find the content of a solid, formed by a plane parallel to the axis of a cylinder To find the solidity of a pyramid Of a cone . . , Of the frustum of a pyramid , Of the frustum of a cone . Of a wedge .... Of a prismoid .... Of a cylindroid Of a sphere .... Of the segment of a sphere . • Of the frustum of a sphere . • Of a circular spindle . Of the middle frustum of a circular spindle Of a spheroid .... Of the segment of a spheroid . Of the middle zone of a spheroid Of a parabolic conoid . Of the frustum of a parabolic conoid . Of a parabolic spindle Of the middle frustum of a parabolic spindle Of a hyperbolic conoid Of a frustum of a hyperbolic conoid Of a frustum of an elliptical spindle Of a circular ring passing 45 49 52 57 69 59 60 50 61 61 62 62 63 63 64 64 65 66 66 67 68 ^9 69 70 70 71 71 72 72 73 73 74 CONTENTS. SECTION V. 'J'nB Five REOtiLAR Bodies— Definitions .... To find the solid contents of the regular bodies To find their superficial contents • 75 76 r SECTION VI. Surfaces op Solids — To find the surface of a prism . . . . 80 Of a pyramid .... 80 Of a cone 81 Of a frustum of a pyramid 81 Of a frustum of a cone 82 Of a wedge . . . , 82 Of the frustum of a wedge 82 Of a globe 83 Of a segment or zone of a sphere 83 Of a cylinder . 84 Of a circular cylinder . 84 Of a parallelopipedon . 1 84 SECTION VII. Mensuration of Timber and of Artificsbs* Work— Description of the carpenter's rule ... 85 Use of the sliding rule 86 Timber measure 88 Carpenters' and joiners' work— Of flooring .... 96 Of partitioning 97 Of wainscotting . 99 Bricklayers' work — Of tiling or slating . • 101 Of walling 102 Of chimneys 103 Masons' work , 104 Plasterers' work 105 Plumbers' work . . . , 106 Painters' work 107 Glaziers' work .... 108 Pavers' work 109 Vaulted and arched roofs . \ . 109 SECTION VIII. Specific Gravity— To find the specific gravity of a body • . 115 A table of specific gravities .... 117 To find the tonnage of ships .... 122 Floating bodies 123 rl C0NTEN1:3. SECTION IX. Weights and dimensions of balls and shells Piling of balls and shells Determining distances by sounds 125 128 131 SECTION X . I Gauging— ^ Ofthe Ganging rule ..... 132 Verie's sliding rule . . . . , 134 A table of multipliers, divisors, and gaugo -points, for squares and circles . . • , ,136 The gauging or diagonal rod . . . .141 Ullaging . 149 SECTION XL LAND-SURVEYING— Form of a field-book , , To measure land with the chain only To survey a field by means of the theodoliti To survey a field with crooked hedges To survey any piece of land by two stations To survey a large estate To survey a town or city . To compute the content of any survey 153 158 160 160 161 162 164 165 Miscellaneous Problems 167 A Table of the Areas 09 thu Ssomrnts ov a Circle, whose diameter is 1 • • . . • 171 MENSUEATION. SECTION I. \ \ A \ \ PEACTICAL GEOMETET. DEFINITIONS. 1. Geometry teaches and demonstrates the properties of all kinds of magnitudes, or extension ; as solids, surfaces, lines, and angles. 2. Geometry is divided into two parts, theoretical and practical. Theoretical Geometry treats of the various properties of extension abstractedly ; and Practical Geometry applies these theoretical pro- perties to the various purposes of life. When length and breadth only are considered, the science which treats of them is called Plane Geometry ; but when length, breadth, and thickness are considered, the science which treats of them is called Solid Geometry. 3. A Solid is a figure, or a body, having three dimen- Bions, viz., length, breadth, and thickness ; as A. The boundaries of a solid are surfaces or superficies. 4. A Superjicies^ or surface, has length and breadth i ~ < only ; as B. ^ I 1 The boundaries of a superficies are lines. ^ 5. A Line is length without breadth, and is q ^ formed by the motion of a point ; as C B. The extremities of a line are points. 6. A Straight or Right Line is the shortest distance between two points, and lies evenly* between these two points. 7. A Point is that which has no parts or magnitude ; it is indi- visible ; it has no length, breadth, or thickness. If it had length, it would then be a line ; were it possessed of length and breadth, it would be a superficies ; and had it length, breadth, and thickness, it would be a solid. Hence a point is void of length, breadth, and thickness, and only marks the position of their origm or termination in every instance, or of the direction of a line. 8. A Plane rectilineal Angle is the inclination of two right lines, which meet in a point, but are S-= Qot in the same direction ; as S. 8 PRACTICAL GEOMETRY. 9. One angle is said to be less than another, when the lines which form that angle are nearer to each other than those which form the other, B < measuring at equal distances from the points in which the lines meet. Take Bw, Bm, Ex, and Ew, equal to one another ; then if w n be greater than X w, the angle A B C is greater than the E - angle FED. By conceiving the point A to move towards C, till m n becomes equal to x w, the angles at B and E would then be equal ; or by conceiving the point F to recede from D, till x n becomes equal to m w, then the angles at B and E would be equal. Hence it appears that the nearer the extremities of the lines forming an angle approach each other, while the point at which they meet remains fixed, the less the angle; and the farther the extreme points recede from each other, the vertical point remaining fixed, as before^ the greater the angle. 10. A Circle is a plane figure contained by one line L^ called the circumference, which is everywhere equally distant from a point within it, called its centre, as o: and an arc of a circle is any part of its circumference ; as A B. 11. The magnitude of an angle does not consist in the length of the lines which form it : the angle C B G is less than the angle ABE, B< though the lines C B, G B are longer than AB, EB. 12. When an angle is expressed by three letters, as ABE, the middle letter always stands at the angular point, and the other two anywhere along the sides ; thus the angle A B E is formed by A B and B E. The angle A B G by A B and G B. 13. In equal circles, angles have the same ratio to each other as the arcs on which they stand (33. VI.) Hence also, in the same, or equal circles, the angles vary as the arcs on which they stand ; and, therefore, the arcs may be assumed as proper measures of angles. Every angle then is measured by an arc of a circle, described about the angular point as a centre ; thus the angle A B E is measured by the arc A E ; the angle A B G by the arc A F. 14. The circumference of every circle is generally divided into 360 equal parts, called degrees ; and every degree into 60 equal parts, called minutes; and each minute into 60 equal parts, called seconds. The angles are measured by the number of degrees contained in the arcs which subtend them ; thus, if the arc A E contain 4(>^e^rees, or the ninth part of the circumference, the angle A B E is said to measure 40 degrees. 15. When a straight line H 0, standing on another A B, makes the angle H A equal to the angle HOB; each of these angles is called a right angle ; and the line H is said to be a pei'pendicular to A B. PRACTICAL GEOMETRY. Tlie measure of the angle H A is 90 decrees, or the fourth part of 360 degrees. Hence a right angle is 90 degrees. 16. An acute angle is less than a right angle ; as A G, or G H. 17. An obtuse angle is greater than a right angle ; as GOB. IS. k plane Triangle is the space enclosed by three straight lines, and has three angles ; as A. 19. A right angled Triangle is that which has one of its angles right ; as A B C. The side B C, opposite the right angle, is called the hypothenuse ; the side AC is called the perpendicular; and the Bide A B is called the base. 20. An oltuse angled Triangle has one of its angles obtuse; as the triangle B, which has the obtuse angle A. 21. An acute angled Triangle has all its three angles acute, as m figure A, annexed to definition 18. 22. An equilateral Triangle has its three sides equal, and also its three angles ; as C. 23. An isosceles Triangle is that which has two of its sides equal ; as D. 24. A scalene Triangle is that which has all its / b sides unequal ; as E. 25.\ A quadrilateral figure is a space included by four straight lines. , If its four angles be right, it is called a rectangular parallelo- gramr« 26. A Parallelogram is a plane figure bounded by four straight lines, the opposite ones being parallel; that is, if produced ever so far, would never meet. 27. A Square ia a four sided-figure, having all its sides equal, and aJ^ its angles right angles ; as H. 10 PRACTICAL GEOMETRY. 28. An Ohlong^ or rectangle, is a right angled paral- lelogram, whose length exceeds its breadth ; as I. _J 29. A Rhomhus is a parallelogram having all its sides equal, but its angles not right angles; as K. 30. A Rhomboid is a parallelogram having , its opposite sides equal, but its angles are not right angles, and its lengtti exceuds its breadtli ; asM. 31. A 7Vapezium is a figure included by four straight lines, no two of which are parallel to each other ; as N. A line connecting any two of its opposite angles, is called a diagonal. 32. A Trapezoid is a four-sided figure having two / F \ of its opposite sides parallel ; as F. ^ i 33. Multilateral Figures, or Polygons, are those which have more than four sides. They receive particular names from the number ol their sides. Thus, a Pentagon has five sides ; a Hexagon has six sides; a Heptagon^ seven; an Octagon^ eight; a Nonagon, nine; a Decagon, ten; an Undecagon, eleven; and a Dodecagon has twelve sides. If all the sides and angles of each figure be equal, it is called a regular polygon ; but if either or both be unequal, an irregular polygon. 34. The Diameter of a circle is a straight line passing through the centre, and terminated both ways by the circumference ; thus A B is the diameter of the circle. The diameter divides the circle into two equal parts, each of which is called a semicircle ; the diameter also divides the circum- ference into two equal parts, each containing 180 degrees. Any line drawn from the centre to the ^\ circumference is called the radius, as A 0, B, or OS. If S be drawn from the centre perpen- dicular to A B, it divides the semicircle into two equal parts, A S and BOS, each of which is called a quadrant, or one-fourth of the circle ; and the arcs A S and B S contain each 90 degrees, and they are said to be the measure of the angles A S and BOS. 35. A Sector of a circle is a part of the cncle comprehended under PRACTICAL GEOMETRY. 11 two Radiu not forming one line, and the part of the circumference between them. From this definition it appears that a sector may be either greater or less than a semicircle ; thus A B is a sector, and is less than a semicircle; and the remaining part of the circle is a sector also, but is greater than a semicircle. 36. A Chord of an arc is a straight line joining its extremities, and is less than the diameter ; T S is the chord of the arc T H S, or ofthe arc TABS. 37. A Segment of a circle is that part of the circle contained between the chord and the circumference, and maybe either greater or less than a semicircle ; thus T S H T and T A B S T are segments, the latter being greater than a semicircle and the former less. 38. Concentric circles are those having the same centre, and the space included between their circumference is called a ring ; as F E. PROBLEM I. To bisect a given straight line A B ; that is, to divide it into two equal parts, O From the centres A and B, with any radius, greater than half the given line A B, describe two arcs intersecting each other at and S, then the line joining S will bisect A B. PROBLEM XL Through a given point x to draw a straight line CD paralla to a given straight line AB. In A B take any point 5, and with the C %^ ? -D centre s and radius s x describe the arc x; with X as a centre and the same ra- dius s a;, describe the arc s y. Lay the ^ extent ox taken with the compasses from s to y ; through xy draw C D, which will be parallel to A B. PROBLEM III. To draw a straight line C D parallel to A B, and at a given distance F from it. In A B take any two points x^f; ^ r ^ and from the two points as centres V^' ^^\ /-""^ ^\ with the extent F taken with the compasses, describe two arcs s, r; then draw a line C D touching these A * a jb aies at r and 5, and it will be at * the given distance fTOTn A B, and parallel to it. ^ 12 PRACTICAL GEOMETRT. PROBLEM IV. To divide a straight line AB into any numher oj equal parts. Draw A K making any angle with A B ; and through B draw B T parallel to AK; take any part A E and repeat it as often as there are parts to be in AB, and from the point B on the line BT, take BI, IS, SV, and VT equal to the parts taken on the line AK; then join AT, EV, GS, HI, and K B, which will divide the line A B into the number of equal parts required, as ^ AC, CD, DF, FB. PROBLEM V From a given point P in a straight li?ie AB to erect a perpendicular, 1. When the given point is in^ or near the middle of the line, D On each side of the point P take equal por- tions, P a;, V f; and from the centres, x, /, ivith any radius greater than P a;, describe two arcs, cutting each other at D ; then the line joining D P will be perpendicular to A B. Or thus: From the centre P, with any radius P w, describe an arc n x y ; set off the distance, P n from n to a:, and from x to y ; then from the points x and y with the same or any other radius, describe two arcs intersecting each other at D ; then the line joining the points D and P will be perpendicular to A B. 2. When the point is at the end of the line. From any centre q out of the line, and with the distance ^ B as radius, describe a circle, cutting AB in p; draw pa 0; and the line joining the points 0, B, will be perpendicular to AB. —3 A- Or thus: Set one leg of the cornpasses on B, and with any extent fi p describe an arc p x ; set off the same extent from p to q ; ]oinpq ; from 7 as a centre, with the extent jt? q as radius, describe an arc r ; produce p q to r^ and the line joining r B will be perpendicular to / PUACTICAL GEOMETRT. lb PROBLEM VI. From a given point D to let fall a perpendicular upon a given line A B. 1. When fhe point is nearly opposite the middle ofjhe given line. From the centre D, with any radius, describe an arc x y^ cutting A B in a: and y / from x and y as centres, and with the same distance as radius, describe two arcs cutting each other at S ; then the line join- ing D and S will be perpendicular to A B. -'•r- When the point is nearly opposite the end of the given line^ and when vie given line cannot be conveniently produced. Draw any line D x, which bisect in o; from o as a centre with the radius ox describe an arc cutting A B in 2/ ; then the line joining D y will be perpendicular to A B. PROBLEM VIL To draw a perpendicular^ from any angle of a triangle ABC, to its opposite side. Bisect either of the sides containing the angle from which the perpendicular is to be drawn, as BC in the point r; then with the radius r C, and from the centre r, describe an arc cutting A B (or A B produced if necessary, as m the second figure), in the point P; the line joining CP will be pendicular to A B, or to A B produced. per- A PROBLEM VilL Upon a given right line k^ to describe an equilateral triangle. From the centres A and B, with the given line A B as radius, describe two arcs cutting each other at C ; then the lines drawn from the point C to the points A and B will form, with the given line A B, an equilateral triangle, as A B C* u PRACTICAL GEOMETTir. PROBLEM IX. To make a triangle whose sides shall be equal to three given right lines A B, AD, and B D, any two oj which are greater tJian the third. From the centre A with the extent A D, describe an arc, and from the centre B with the radius B D describe another arc cutting the former at D; tiien join DA, D B, and the sides of the triangle » A B D will be respectively equal to the three given / right lines. f — ^ PROBLEM X. B ^ Two sides A B and BC of a right angled triangle being given, to Jind the hypo- thenuse. Place B C at right angles to A B ; draw A C, ^ and it will be the hypothenuse required. ^ PROBLEM XL The hypothenuse AB, and one side AC, of a riyht angled triangle being given, to find the other side. Bisect A B in a; ; with the centre x, and a; A as radius, describe an arc; and with A as a centre, and A C as radius, describe another arc cutting the former at C ; then ioin A C and C B ; and A B C a ^ ivill be a right angled triangle, and BC the re- ^_ quired side. PROBLEM XIL To bisect a given angle; that is, to divide it into two equal part:' c Let A C B be the angle to be bisected. From C as a centre, with any radius C x, de- scribe the arc xy; from the points x and y as cen- tres, wi^h the same radius, describe two arcs cutting each other at ; join C, and it will ^ bisect the angle A C B. PROBLEM XIIL At a given point A in a given right line A B, to make an angle equal to ihe given angle C. From the centre C with any radius C y, de- scribe an arc xy ; and from the centre A, with the same radius, describe another arc, on which take the distance mn equal to xy; then a line drawn from A through m will make the angleC w A n equal to the angle x C y. "o""^ PRACTICAi:. GEOMETRY. 15 M XV. \^ I a scale ofdhqri 'rt'nfni'yxy \ PROBLEM XIV. To make an angle containing any proposed number of degrees. 1. Vt^nthe required angle is less than a quadrant^ as 40 degrees, TakeHn the compasses the extent of 60 de- grees fromithe line of chords, marked ciio. on the scale ; aiad with this chord of 60 degrees as radius, and trh» centre A, describe an arc ^yt\^ take from the limiof chords 40 degrees, which set off from n to m ; nsmi A draw a line through m; and the angle m A n>^i\\ contain 40 degrees. 2. When the required^angle is greater than a quadrant^ as 120 degrees. / From the centre o, with ihe chord of 60 de- n greesas radius, describe the semicircle warwB; set off the chord of 90 degrees from. B to n, and the remaining 30 degrees from n to x ; join ox ; and the angle Boa; wiU contain 120^^- degrees ; or subtract 120 froijar 180 degrees, and set off the remainder (6^^ degrees) taken from the line of chorda from y to x: then join a: o^ and Box will contain 120 degreea as before. / PROBLEM An angle being gUfen, tojind^ by a scale ofehqrds^ how many degrees it contains. From the verWSc A as centre, with the chord of 60 degrees a/ radius, describe an arc xy; take the extent :i^y with the compasses, and setting one foot at'the beginning of the line of chords, "^ the others leg will reach to the number of degrees which the angle con- tains :iut if the extent a:?/ should reach beyond the scale, find the num- ber oldegrees in xy^ which deducted from 180, will leave the degrees iuyme angle B o x. See figure to the second case of the last Problem. / PROBLEM XVI. Upon a given right line A B, to construct a square. With the distance A B as radius, and A as a centre, describe the arc E D B ; and with the distance A B as radius, and B as a centre, describe the arc AFC, cutting the former in x; make x E equal to a; B ; join E B ; make x C and x D each equal to A F, or Fa:; then join AD, DO, OB, and A D C B will be the required square. Or thus. Draw B C at right angles to A B, and equal to it; then from the centres A and C, with the radius A B and C B, describe two arcs cutting each other at D ; join D A and D C, which will complete the square. PRACTICAL GEOMETRY. PROBLEM XVII. To make a rectangular parallelogram of a given length and breadth. , A" Let A B be the length, and B C the breadth. Erect B C at right angles to A B ; through C a and A draw C D and A D, parallel to A B and B BC. PROBLEM XVIII. To find the centre of a given circle. Draw any two chords AC, C B ; from the points, A, C, B, as centres, with any radius greater than half the lines, describe four arcs cutting in ra; and wv, draw rx and y V, and produce them till they meet in 0, >vhich will be the centre. PROBLEM XIX. Vpon a given right line A B, to describe a rhombus having an angle equal to a given angle A. c B. y^ Make the angle CAB equal to the angle at A ; make AC equal to A B ; then from C and B as centres, with the radius A B describe two arcs crossing each other at D ; join D C and D B, which will complete the rhombus. PROBLEM XX. To find a mean proportional between two given right lines A B and B C. Place A B and B C in one straight line ; bisect A C in o; froiiu o as a centre, with A o or C as radius, describe a semicircle A S C ; erect the perpendicular B S, and it will be a mean proportional between AB and BC; that is, A B : B s : : B S : B c. PROBLEM XXL To divide a given right line A B into two such parts^ as shall be to each other as xo to of. X ?i f From the point A draw A S equal to x o, and ^ produce it till F S becomes equal to o f; Jom F B, and draw S T parallel to F B ; tfien will A T : T B : : ar : (?/• TRACTICAL GEOMETRT. 17 PROBLEM XXII. To find a third proportional to two given right lines A B, A S. Place AB and AS so as to make any angle at A : from the centre A, with the distance A S, describe the arc S D ; then draw D x parallel to BS, and kx will be the third proportional required ; that is, A B : A S : : A S • Aa: PROBLEM XXIII. To find a fourth proportional to three given right lines^ A B, AC, and A D. Place the right lines A B and AC so as to make any angle at A ; on AB set off AD ; join B C ; and draw D S parallel to it ; then A S will be the fourth proportional required, viz. AB: AC:: AD: AS. PROBLEM XXIV. In a given circle to inscribe a square. Draw any two diameters A C, D B at right angles to each other ; then join their ex- tremities, and the figure ABCD will be a square inscribed in the given circle. If a line be drawn from the centre o to the middle of AB, and produced to/; the line joining /B will be the side of ah octagon mscribed in the circle. PROBLEM XXV. To make a regular polygon on a given right line^ A B. Divide 360 degrees by the number of sides contained in the polygon ; deduct the quotient from 180 degrees, and the remainder will be the number of degrees in each angle of the polygon. At the points A and B make the tingles oABandoBA each equal to half the angle of the polygon ; then from o as a centre, and with o A or o B as radius, describe a circle, in which place A B continually.* Or thus: Take the given line A B from the scale of equal parts, and multiply the number of equal parts in it by the number in the third column of the following table, answering to the given number of sides; the product will give the number of equal parts in the radius A o, or o B, which taken from the scale of equal parts in the compasses, will give * See Appendix, Demonstration 1. 18 TRACTICAL GEOMETRY. the radius, with which describe a circle, and place in it the line A B continually, as shewn in the first method.* TABLE I. When the side of the polygon is 1. No. of Name of the Radius of the circumscrib- Angle OAB, or sides. Polygon. ing circle. OB A. 3 Trigon •5773503 30 4 Tetragon •7071068 45 5 Pentagon •8506508 64 6 Hexagon 1, Side = radius. 60 7 Heptagon 1-1523825 64^ 8 Octagon 1-3065630 67^ 9 Nonagon 1-4619022 70 10 Decagon 1-6186340 72 11 Undecagon 1-7747329 73i^ 12 Dodecagon 1-9318516 75 PROBLEM XXVI. In a given circle to inscribe any regular polygon ; or^ to divide the circumference of a given circle into any number of equal parts. Divide the diameter A B into as many equal parts as the figure has sides ; erect the perpendicular o x, from the centre o ; divide the radius oy into four equal parts, and set off three of these parts from y to x; draw a line from x to the second division z, of the diameter A B, and produce it to cut the circumference at C ; join A C, and it will be the side of the required polygon. f PROBLEM XXVII. To draw a straight line equal to any given arc, of a circle^ AB. Divide the chord AB into four equal parts ; and set off one of these parts from B to D ; then join D C, and it will be equal to the length of half the given arc nearly.J * Set Api«ucL.v, £>duiousoniuoii 2. t Ibid. 3 X Ibid. 4. PRACTICAL GEOMETKY. 19 Or thus: From the extremity of the arc AB, whose length is required to be found, draw kom^ passing through the centre ; divide o n into four equal parts, and set off three of these parts from n to m ; draw m B, and produce it to meet A C drawn at right angles to A m ; then will A C be nearly equal in length to the arc AB.* PROBLEM XXVIII. To make a square equal in area to a given circle. First divide the diameter AB into fourteen equal parts, and set off eleven of them from A to S ; from S erect the perpendicular S C, and join A C, the square of which will be very nearly equal to the area of the given circle, f PROBLEM XXIX. To construct a diagonal scale. Draw an indefinite straight line ; set off any distance A E according to the intended length of the scale ; repeat A E any number of times, EG, G B, &c. ; draw C D parallel to A B at any convenient distance ; then draw the perpendiculars A C, E F, G H, B D, &c. Divide A E and A C each into ten equal parts; through 1, 2, 3, &c., draw lines parallel to AB, and through xy^ &c., draw a:F, ^Z, &c., as in the annexed figure. 8 r y z E wv V 9 8 7 6 5 W \ M ' 1 I 1 \ '■ p M ^1 MM 1 i 2 -L-UJuU- -i 2 1 11 11 M n 1 1 C '987654324] F ] S J) The principal use of this scale is, to lay down anj line from a given measure ; or to measure any line and compare it with others. — What- ever number C F represents, F Z will be the tenth of it, and the sub-^ divisions in the vertical direction FE v/ill be each one-hundredth part. Thus, if C F be a unit, the small divisions in C F, viz. F 2, Kc, will be lOths, and the divisions in the altitude will be the 100th * See Apjaendix, DemonstratioL. 6. t Ibid. Q. 20 PRACTICAL GEOMETRY. parts of a unit. If C F be ten, the small divisions F Z, &c., will be units, and those in the vertical line, tenths ; if C F be a hundred, the others will be tens and units.* To take any number off the scale, as suppose 2^^^^, that is, 2-38: place one foot of the compasses at D, and extend the other to the division marked 3 ; then move the compasses upward, keeping one foot on the line D B, and the other on the line 3 5, till you arrive at the eighth interval, marked 88, and the extent on the compasses will be that required. This, however, may express 2 "38, 23-8, or 288, according to the magnitude of the assumed unit. Note. — If C F were divided into 12 equal parts, each division would be 1 inch, and each vertical division 1-1 0th of an inch, by making C F one foot. PROBLEM XXX. To reduce a rectilinear figure to a similar one upon either a smaller or a larger scale. Take any point P in the figure ABODE, and from this as- sumed point draw lines to all the angles of the figure ; upon one of wiiich P A take P a agreeably to the proposed scale; then draw a h parallel to A B, 5 c to B C, &c., then shall the figure ahcde be similar to the original one, and upon the required scale. Or, measure all the sides and diagonals of the figure by a scale, and lay down the same measures respectively from another scale, in the required proportion. When the figure is complex, the reduction to a different scale ia best accomplished by means of the Eidograph, an instrument invented by Professor Wallace, or by means of the improved Pentograph. PROBLEM XXXr. 2^0 divide a circle into any number of equal parts^ having their perimeters equal also. Divide the diameter AB into the re- quired number of equal parts, at the points C, D, E, &c. ; then on one side describe the semicircles 1, 2, 3, 4, &c., and on the other side of the diameter describe the semicircles 7, 8, 9, 10, &c. on the diame- ters B F, B E, B D, B C, &c. ; so shall the parts 111, 2 10, 3 9, 4 8, &c. be equal both in area and perimeter. — Leslie's Geometry. • ft3ee Appendix, Demonstration 7. MENSURATION OF SUPERFICIES. 21 MENSUEATION OF SUPERFICIES. SECTIOISr II. I. Long Measure, 12 Inches . . 1 Foot. 3 Feet ... 1 Yard. 6 Feet ... 1 Fathom.^ 16^ Feet Eng. \ (1 Pole or 6 J Yards )* ( Perch. 40 Perches . . 1 Furlong. 8 Furlongs . 1 Mile. 1 Yard. 1 Fathom, f 1 Pole or \ Perch. 1 Furlong. 1 Mile. The area of any plane figure is tlie space contained within its boundaries, and is estimated by the number of square miles, square yards, square feet, &c. which it contains. II. Square Measure, 144 Inches ... 1 Foot. 9 Feet . . 36 Feet . . 272i Feet Eng. 1 30i Yards i 1600 Perches , 64 Furlongs . In Ireland 21 feet make 1 pole or perch, and 7 yards therefore will make a pole or perch. There are other measures used, for which eee Arithmetical Tables. Land is generally measured by a Cliain^ of 4 poles, or 22 yards ; it consists of 100 links, each link being -22 of a yard. See Section XL Surveying. Duodecimals are calculations by feet, inches, and parts, which decrease by twelves : hence they take their name. Multiplication of feet, inches, and parts, is sometimes called Cross Multiplication, from the factors being multiplied crosswise. It is used m finding the contents of work done by .artificers, where the dimensions are taken in feet, inches, and parts. Rule. I. Write the multiplier under the multiplicand in such a manner, that feet shall be under feet, inches under inches, &c. II. Multiply each term of the multiplicand by the number of feet in the multiplier, proceeding from right to left ; carry 1 for every 12, in each product, and set down the remainder under the term multi- plied. III. Next multiply the terms of the multiplicand by the number under the denomination inches, in the multiplier ; carry 1 for every 12, as before, but set down each remainder one place farther to the right than if multiplying by a number under the denomination feet. ly. In like manner proceed with the number in the multipliei 22 MENSURATION OF SUPERFICIES. under the denomination parts or lines, remembering to set down each remainder one place farther to the right than if multiplying by a number under the denomination inches. And so on with numbers of inferior denominations. V. Add the partial products thus placed, and their sum will be tho whole product. IN CROSS MULTIPLICATION IT IS USUAL TO SAY, Feet multiplied by feet, give feet. Feet by inches, give inciies. Feet by parts, give parts. Inches by inches, give parts. Inches by parts, give thirds. Inches by thirds, give fourths. Parts by parts, give fourths. Parts by thirds, give fifths. Parts by fourths, give sixths, &c * 1. Multiply 7 feet 9 inches by 3 feet 6 inches. F. I. 7.9 3 . 6 23 3 10 . 6 27 1 . 6Ans. F. I. P. F. I. P. 2. Multiply 240 . 10 . 8 by 9 . 4 . 6 9.4.6 2168 . , . 80 . 3 , . 6 . . 8 10 . , . 5 . . 4 2258 4.0. Ans. 9. 10. 11. Multiply 8 Multiply 9 Multiply 7 Multiply 4 Multiply 7 Multiplv 10 Multiply 75 Multiply 57 Multiply 75 I. P. '" 6 . 11. 6. 1 11 • 7 6. 10. 6.2.3. 0.6.6 8. 9 . 9. 11 . 3. * In multiplication, the multiplier must always be a number of times ; to talk of multiplying feet by feet, &c., is absurd, for what notion can be formed of 7 feet taken 3 times? However, since the above easily suggests the correct meaning, and is a concise method of expressing the rule, it haa been thought proper to retain it. See Appendix, Demonsti'&tiou « MENSURATION OF SUPERFICIES. 23 12. Multiply 321 13. Multiply 4 U. Multiply 39 16. 6'=| 2'=i I. p. F. I. P. F. I. p.'""" 7 . 3 by 9 . 3.6. Ans. 2988 . 2 . 10.4.6. 7 . 8by 9 . 6 . 44 . . 10. 10 . 7byl8 . 8 . 4. 745 . 4 . 10.2.4. be solved by the method of aliquot parts, thus :— F. ' " F. ' " Multiply 368 . 7 . 5 by 137 . 8 . 4 137 . 8 . 4 2576 1104 368 184 . 3 . 8 . 6 61 . 6 . 2 . 10 10 . 2 . 10 . 5,8 68 . 6 11 . 5 3.9.8 . 11 . 5 Ans. 50756 . 7 . 10 . 9.8 6 D C J PROBLEM I. To find the area of a square. Rule. Multiply the length of the side by Itself, and the product will be the area.* 1. Let the side of the square ABCD be 6: what is its area? Ans. 6x6 = 36, the area. 2. What is the area of a square whose side is 15 chains? Ans. 22^. 3. What is the area of a square whose side is 7 feet 9 inches? Ans. 60^. 4. What is the area of a square whose side is 4769 links? Ans. 22743361. PROBLEM II. To find the area of a rectangle. Rule. Multiply the length of the rectangle by its breadth, and the product will be the area.* * '" ^ 1. Let the sides of the rectangle A B C D be 12 and 9, what is its ^ea ? Ans. 12 x 9 = 108, the area. 2. What is the superficial content if a plank, whose length is 5 feet 6 inches, and breadth 7 feet 8 inches? Ans. 42 feet 2 inches. 3. What is the area of a field whose boundaries form a rectangle, its length being 176 links and Dread th 154 links ? Ans. -27104 of an acre. * See Appendix Demonstration* n 24 MENSURATION OF b'UPERFICIES. 4. What 18 the superficial content of a floor, whose length is 40 feet 6 inches, and breadth 28 feet 9 inches ? Ans, 1164 feet, 4 inches, f> parts. PROBLEM III. To find the area of a rhombus. Rule. Multiply the length by the per- pendicular breadth, and the product will be the area.* ^ 1. What is the area of a rhombus, whose side is 16 feet, and perpendicular breadth 10 feet? Ans, 16 X 10 = 160 feet, the area. 2. What is the content of a field in the form of a rhombus, whose length is 7*6 chains, and perpendicular height 5*7 chains? Ans. 43'32 chain*!, - 3. What is the area of a rhombus whose side is 7 feet 6 inches, and perpendicular height 3 feet 4 inches ? Ans. 25 feet. 4. What is the area of a rhombus whose length is 3 yards, and perpendicular height 2 feet 3 inches ? Aiis. 20 feet 3 inches. PROBLEM IV. To find the area of a triangle* Rule. Multiply the base by the perpendicular height, and divide the product by two for the area.f 1. The base of a triangle is 76*5 feet, and perpendicular 92*2 feet; what is its area ? Ans. 76-6 x 92-2 -^ 2=3526-65 square feet, the area. 2. The base of a triangle is 72*7 yards, and the perpendicular height of 36*5 yards ; what is its area ? Ans. 1326*775 yards. 3. The base of a triangular field is 1276 links; and perpendiculai 976 links; how many acres in it? Ans. 6 acres 36*3008 perches. 4. The base of a triangle measures 15 feet 6 inches, and the per- pendicular 12 feet 7 inches ; what is its area ? Ans. 97 feet 6J inches. * See Appendix, Demonstration 0. t Ibid. 10. l^ENSURATION OF SUPERFICIES. 25 PROBLEM V. Having the three sides of any triangle given ^ to find its area. Rule I. From half the sum of the three Bides subtract each side separately, then multi- ply the half sum and the three remainders together, and the square root of the last pro- duct will be the area of the triangle.* Rule II. Divide the difference between the squares of two sides of the triangle by the third side ; to half this third side add half the quotient, and deduct the square of this sum from the square of the greater side, the re- mainder will be the square of the perpendicular, the square root of which, multiplied by half the base, will give the area of the triangle.f 1. Given the side AB = 9-2, BC = 7*5, and AC = 5*5 ; required the area of the triangle? 9-2 7-5 6-5 Sum 22-2 JSumIl-1— 9'2 = l-9) : then V (lMxl-9x 3-6x5-6) = )M— 7-5 = 3-6V V 425'1744 = 20-619 the area by ll-l— 5-5 = 5-6 ) Rule I. Again, 9-2^— 7-52 = 84-64— 56-25 = 28'39 ; then 28•39-^5•5 = B- 161818, quotient. Now (5-161818^2) + (5•5-^-2) = 2*580909 + 2-75 = 5-3309 = half quot. plus half third side: then 84-64 — 28-41849481 = 56-22150519, and V 56*22150519 = 7*498 = perpendicular; thep 7-498 X 2-75 = 20-619 the area as before. 2. What is the area of a triangle whose sides are 50, 40, and 30 ? Ans. 600. 3. The sides of a triangular field are 4900, 5025, and 2569 links; how many acres does it contain ? Arcs. 61 acres, 1 rood, 39*68 perches. 4. What is the area of an isosceles triangle, whose base is 20, and each of its equal sides 15? Ans. 111*803. 5. How many acres are there in a triangle, w^hose three sides are 380, 420, and 765 yards? Ans. 9 acres 38 poles. 6. How many square yards in a triangle, whose three sides are 13, 14, and 15 feet ? Ans. 9^ square yards. * See Appendix, Demonstration 11. t Ibid, 12. 26 MENSURATION OF SUPERFICIES. 7. How many acres, &c., in a triangle, whose three sides are 49 60-25, and 25*69 chains? Ans, 61 acres, 1 rood, 39*68 perches. PROBLEM VI. To find the area of an equilateral triangle. Rule. Square the side, and from this square deduct its fourth part ; then multiply the remainder by the fourth part of the square of the side, and the square root of the product will give the area.* Or multiply ab'^ by V ^ fo^ the area.f 4 1. Each side of a triangular field, ABC, measures 4 perches, what is its area? 42 = 16, then 16 -^ 4 = 4, and 16 — 4 = 12: then 12 x Y = 12x4 = 48, and V48 = 6-928, the area. 2. How many acres in a field of a triangular form, each of whose Bides measures 70 perches? Ans. 13 acres, 1 rood, 1 perch. 3. The perimeter of an equilateral triangle is 27 yards, what is its area? Ans. 35-074:. jjjote. — "When the triangle is isosceles, the perpendicular is equal to th« Bquare root of the difference between the squares of either of the equal sides, ftud half the base. PROBLEM yn. Given the area and altitude of a triangle^ to find the base. Rule. Divide the area by the altitude or per- pendicular, and double the quotient will give the base.J 1. Given the area of a triangle = 12 yards, and altitude = 4 ; what is its base? Ans. 12 H- 4 = 3 ; then 3x2 = 6 yards, the base, AB. 2. A surveyor having lost his field book, and requiring the base of a triangular field, wiiose content he knew from recollection was 14 acres, and altitude 7 yards, how much is the base ? Ans. 19360 yards. PROBLEM VIIL Given the area of a triangle, and its base, to find its altitude. Rule. Divide the area by the given base, and double the quo- ^tient will give the perpendicular. '" The reason of this rule is manifest, from the last. 1. Given the area of a triangle = 12, and its base = 6 ; what ia •Us perpendicular height ? ' Ans. 12-i-6 = 2 ; then 2 x 2 = 4 the altitude. • Soe Appendix, Demonstration 13. f Ibid. 14, I Ibid. 15 MENSURATION OF SUPERFICIES. 27 PROBLEM IX. Given any two sides of a right angled triangle^ to find the third side, and thence its area. Rule. I. To the square of tlie perpendicular add the square of the base, and the square root of the sum will give the hypothenuse. II. The square root of the difference of the squares of the hypo- thenuse and either side, will give the other. III. Or multiply the sum of the hypothenuse, and either side, by their difference ; and the square root of the product will give the other.* 1. Given the base ACS, the perpendicular C B 4 ; required the hypothenuse 'A B'?" 32 + 423^25; then V 25 = 5, the hypothenuse A B. 2. Given A B 5, A C 3 ; required C B ? 53 — 33 = 16 ; then V 16 = 4 the side B C ; or,, (5 + 3) X (5 — 3J = 8 X 2 = 16; then Vl6 = 4, as before. 3. Given A B 5, B C 4 ; required AC? 53_ 43 = 9 ; then V 9 = 3 the side AC; or (5 + 4) x (5 — 4) = 9x1 = 9; then V 9 = 3, as before. And 3x4-^2 = 6 the area of the triangle. 4. The wall of a building on the brink of a river is 120 feet, and the breadth of the river is 70 yards ; what is the length of the chord in feet that will reach from the top of the building across the river ? Ans. 241-86 feet. 5. A ladder 60 feet long, will reach to a window 40 feet from the dags on one side of a street, and by turning the ladder over to the other side of the street, it will reach a window 50 feet from the flags; required the breadth of the street ? Ans. 77*8875 feet. 6. The roof of a house, the side walls of which are the same height, forms a right angle at the top, the length of one rafter being 10 feet, and its opposite one 14 feet; what is the breadth of the house? Ans. 17*204. PROBLEM X. Given the base and perpendicidar of a right angled triangle^ to find the perpendicidar let fall on the hypothenuse from the right angle ; and also the segments into vAich the hypothenuse is divided by this perpendicidar. Rule. Find the hypothenuBe by Problem IX. Then divide the square of the greater side by the hypothenuse, and the quotient will give the greater segment, which deducted from the entire will give the less. Having found the segments, multiply them together, and the square root of the product will give the perpendicular.f * tsce Appendix, Demonstration 16. f IWd. IT. W MENSURATION OF SUPERFICIES. 1. Given AC 3 yards, and CB 4 yards; required the segments B D, DA, and the perpendicular D C. 33 + 42 = 25 ; then V 25 = 5 = A B. 42^5 = 16-T-5 = 3-2 =BD; then 5 — 3*2 = 1*8 = AD. Again, 32 x I'S = 576; then ^5*76 = 2*4 = D C. 2. The roof of a house whose side walls are each 30 feet high, forms a right angle at the top ; now if one of the rafters be 10 feet long, and its opposite yoke-fellow 12, required the breadth of the building, the length of the j)rop set upright to support the ridge of the root, and the part of the floor at which it must be placed? Ans. Breadth of the building 15*6204 feet, greater segment 9*2186 feet, lesser segment 6'4018 feet, and length of the prop 37*68 feet. PROBLEM XI. To find the area of a trapezium. Rule. Divide the trapezium into two triangles, by joining two of its opposite angles ; find the area of each triangle, and the sum of both areas will give the area of the trapezium. Or, Draw two perpendiculars from the opposite angles to the diagonal ; then multiply the sum of these perpendiculars by the diagonal, and half the product will give the area.* 1. In the trapezium ABCD, the diagonal AC is 100 yards, the perpendicular D E 35, and B F 30 ; what is its area? DE = 35 D BF=30 A<= 65 100 2)6500 3250 the area. 2. What is the area of a field, whose south side is 2740 links, easi side 3575 links, north side 3755 links, west side 4105 links, and the diagonal from south-west to north-east 4835 links ? Ans. 123 acres 11*8633 perches. 3. In the trapezium ABCD, the side AD is 15, DC 13, CB 14, and AB 12 ; also the diagonal A C 16 ; what is its area? A71S. 172-5247. 4. In the trapezium ABCD, there are given A B 220 yards, D C 265 yards, and AC 378 yards; also AF 100 yards, and E C 70 yards; what is its area ? Ans. 85342*2885 yards=17 acres, 2 roods, 21 perches. 5. In the trapezium ABCD, there are given A B 220 yards, D C 265 yards, BF 195*959 yards, DE 255*5875 yards; also FE 208 yards; required the area of the trapezium? Ans. 85342*2885 yards. 6. Suppose in the trapezium ABCD, on account of obstacles, I can only measure AB, DC, BF, DE, and FD, which are respectively 22 yards, 26 yards, 19 yards, 25 yards, and 32 yards ; required the area ? Ans. 840*55 square yards. * See Appendix, Demonstration 18. MENSURATION OF SUPERFICIEa PROBLEM XII. To find the area of a trapezium inscribed in a circle^ or of any one whose opposite angles are together equal to two right angles. Rule. Add the four sides together, and take half the sum, from this half sum deduct each side separately ; and the square root of the product of the four remainders will give the area of the trapezium.* 1. What is the area of a four-sided field, whose opposite angles are together equal to two right angles, the length of the four sides being as follows, viz. AB 12-5, AD 17, DC 17-6, and BC 8 yards? 12-5 17 A 17-6 8 2)55 27-5 27-5 27-6 27-5 12-5 17 17-6 15 X 10-5 X 10 X 19-5=30712-50; then 30712-50= 175-26, the area in yards. 2. There is a trapezium whose opposite angles are together equal to two right angles; the sides are as follows, viz. A B 25, AD 34, D C 35, and B C 16 ; required its area? Ans, 70099. PROBLEM XIII. To find the area of a trapezoid, Rqle. Multiply half the sum of the two parallel sides by the perpendicular distance between them, and the product will give the 1. Let AB C D be a trapezoid, the side AB = 40, DO=25, CP=18; re- \/ quired the area ? 40 G 25 A F PB 65-j-2 = 82-6x 18 = 585, area. 2. What is the area of a trapezoid, whose parallel sides are 750 and 1225 links, and the perpendicular height 1540 links? Ans. 15 acres 33-2 perches. 3. What is the area of a trapezoid whose parallel sides are 4 feet f> inches, and 8 feet 3 inches ; and the perpendicular height 5 feet 8 inches? Ans. 36 feet 1^ inches. 4. What is the area of a trapezoid whose parallel sides are 1476 and 2073 yards, and perpendicular height 976 yards? Ans. 220 acres, 3 roods, 25 perches, 7 yards Irish. * Boe Appendix, Demonstration 19l t I^id. 20. 30 MENSURATION OF SUPERFICIES. PROBLEM XIV. To find the area of an irregular polygon. Rule. Divide the figure into triangles and trapeziums, and find the area of each separately, by Problem IV. or XI. Add these areas together, and the sum will be the area of the polygon.* 1. What is the area of the irregular polygon ABCDEFGA, the following lines being given ? A0= 9 GB = 29 Cw =11 GC=28-4 Fa;=U-5 'w/\ Cw =id FD=36 E 2 = 7-4 G F A0= 9 Cn =11 2)20 sum 10 half 29diag.-GB 290=areaof ABCGA, C2/=13 E2;= 7.4 2)20-4 sum. 10^ 35 FD 357 area of FCDEF. Far = 14-5 i GC=14-2 205-9 areaofGFC. 290 =areaofABCGA 357 =areaofF CDE F 205-9 = area of GF C Ans. 852-9 = area ofABCDEFGA. 2. In a five-sided field G C D E F G there is G C = 28 perches, F x — 14 perches, C ?/ = 1 3 perches, 2 E = 7 perches, and F D = 35 perches ; required its area? Ans. 3 acres, 1 rood, 26 perches. * In finding the area of an irregular figure, draw a line through the extrema angles of the figure, on which let fall perpendiculars from all the other anglesr of the polygon, which will divide it into triangles and trapeasoids; then find tho area of these by Problems IV. and XIIL MENSURATION OF SUPERFICIES. 81 3. In the annexed figure, there are given in perches, AX = 15 XR ■ RT : . TD AP PS : SD : GX FR : ET BP : CS = Required the area ? 14 6 17 14 12 6 10 12 20 14 Ans, 4 acres, 3 roods, 19J perches. PROBLEM XY. To find the area of a regular polygon. Rule I. Add all the sides together, and multiply half the sum by tlie perpendicular drawn from the centre of the polygon t<\) the middle of one of the sides, and the product will give the area. This perpen- dicular is the radius of the inscribed circle. Rule II. Multiply the square of the side of the polygon by the number standing opposite to its name in the following table, under ihe word area, and the product will give the area of the polygon. Rule III. Multiply the side of the polygon by the number standing opposite to its name in the column of the following table, headed '^ Radius of Inscribed Circle," and the product will be the perpendicu- lar from the centre of the polygon to the middle of one of its sides ; tiien multiply half the sum of the sides by this perpendicular, and the product will give the area.* TABLE II. When the side of the polygon is 1. I [No. of Sitles. 3 Radius of inscrib- ed Circle. Area of Tolygon. 0-2886751 0-4330127 - f tan. 30° = ^V3 4 0-6000000 1-0000000 = 1 tan. 45°= 1 x 1 5 0-6881910 1-7204774 = f tan. 54°=|V(l+iV5) 6 0-8660254 2-5980762 = f tan. 60° = f V3 7 1-0382617 3-6339124 = 1 Um. 64°f 8 1-2071068 4-8284271 = f tan. 67°| = 2x(l+ V2) 9 1-3737387 6-1818242 = 1 tan. 70° 10 1-5388418 7-6942088 = V tan. 72°=:|V(3 + 2V^'i) 11 1-7028437 9-3656404 = V tan. 73VV 12 1-8660254 11-1961524 = V.tan. 75° = 3x(2+ V3) ♦ Eee Appeodix, Demonstration 21. C 32 MENSURATION OF SUPEKFICIES. Note. — ^Tlie radius of the circumscribed cii'cle, when the side of the polygon is 1, may be seen in Table I. The expressions in the fourth column may be seen in Trigonometry, to ivhich the pupil is referred for a full investigation of them. The tangents of the angle O a G iu the heptagon, nonagon, and undecagon, are extremely difficult to be found without a table of tangents. 1 . The side of a pentagon is 20 yards, and the perpendicular from the centre to the middle of one of the sides is 13*76382 ; required the area ? By Rule I. 20x5x 13'76382-^2 = 1376-3824-2 = 688-191. Ans, By Rule II. 20 x 20 x 1-720477 = 688*19, the area as before. 2. The side of a hexagon is 14, and the perpendicular from the centre 12*1243556 ; required the area? Ans. 509*2229352. 3. The side of an octagon is 5*7, required its area? Ans. 156*875596479. 4. The side of a heptagon is 19*38 yards, what is its area ? Ans. 1364*84. 5. The side of an octagon is 10 feet, what is its area ? Ans. 482*84271. f>. The side of a nonagon is 50 inches, what is its area ? Ans. 15454*5605. 7. The side of an undecagon is 20, what is its area ? Ans. 3746-25616. 8. The side of a dodecagon is 40 yards, what is its area ? Ans. 17913-84384. PROBLEM XYI. Given the diameter of a circle^ to find the circumference; or^ t'^^ circumference to find the diameter, and thence the area. Rule.* I. Say as 7 : 22 : : the given diameter : cir- cumference. Or, as 113 : 355 : : the diameter : the circum- ^ference. Or, as 1 : 3*1416 : : the diameter : the circum- ference. II. Say as 22 : 7 : : the given circumference : the diameter. Or, as 355 : 113 :: the circumference : the diameter. Or, as 3*1416 : 1 : : the circumference : the diameter. 1. The diameter of a circle. is 15, what is its circumference? 7 : 22 : : 15 : 22 X 15-r-7 = 330-T-7 = 47*142857. Or, 113 : 355 : : 15 : 355 X 15-j-113 = 5325-^113 = 47'124. Or, 1 : 3*1416 :: 15 : -3*1416x15 = 47*124. 2. The circumference of a circle is 80, what is its diameter ? 22 : 7 : : 80 : 7 X 80^22=25*45. 355 : 113 :: 80 : 113 X 80-^355 = 25*4647. 8*1416 ; 1 : : 80 : 80^3*1416 = 25-4647. ♦ See Appendix, Deraonstraticn 7lSi. MENSURATION OF SUPERFICIES. 33 8. What is the circumference of a circle whose diameter is 10 ? Ans. 31-4285. 4. What is the diameter of a circle whose circumference is 50 ? s Ans. 15-909. 5. The diameter of the earth is 7958 miles, what is its circum- terence? ^W5. 25000*8528 miles. 6. The circumference of the earth being 25000-8528 miles, what is its diameter V Ans, 7958 miles. PROBLEM XVII. To find the length of an arc of a circle. Rule I. Multiply tlie radius of the circle by the number of de- grees in the given arc, and that product by -01745329, and the last product will be the length of the arc* Rule II. From eight times the chord of half the arc, subtract the thord of the whole arc, one- third of the remainder will give the length o( the arc, nearly.f 1. If the arc AB contain 30 degrees, the radius being 9 feet, what is the length of the arc? 30x9 = 270, and 270 x -01745329 = 4-7124. 2. If the chord A D of half the arc A D B be 20 feet, and the chord A B of the whole arc 38 ; what is the length of the arc ? 20x8— 38= 122; then 122^3 = 401 feet. ^?2«. 3. The chord of an arc is 6 feet, and the chord of half the arc is 3J ; required the length of the whole arc ? Ans. 7^. 4. The chord of the whole arc is 40, and the versed sinej or height of the segment 15 ; what is the length of the arc? Ans. 53^. 5. The chord AB of the whole arc is 48*74, and the chord AD of half the arc 30-25 ; required the length of the arc? 6. AB = 30, DP=8; required the length of the arc? Ans, 35 J. PROBLEM XVIII. To find the area of a circle. Rule I. Multiply half the circumference by half the diameter, for the area.§ Rule II. Multiply the square of the diameter by -7854, for the area. II Rule III. Multiply the square of the circumference by -07958.1 Rum IV. As 14 to 11, so is the square of the diameter to the area. Rule V. As 88 to 7, so is the square of the circumference to the area. * See Appendix, Demons tJ-ation 23. f Ibid. 24. t By "versed sine," in works on mensuration, is not meant the trigono »netrical versed sine of the whole arc, hut of half the arc. g See Appendix, Demonstration 22, Jl Ibid. '28. % Ibid. 25. 84 MENSURATION OF SUPERFICIES. 1. To find the area of a circle whose diameter is 100 and circunr fereiice3U-J6. By Rule I. By Rule II. By Rule III. 314-16 -7854 98696*5 sq. cir. 100 100'^= 10000 -07958 4)31416 Area 7854 7854- Area. Area 7854 By Rule IV. By Rule V. 100^ = 10000 98696-5 sq. cir. 11 7 2)110000 8)090875-5 7)55000 11)86359-4 Area 7857 7850-85 2. What is the area of a circle whose diameter is 7 ? Ans. 38 J nearly. 3. How many square yards are in a circle whose diameter is l^yard? Ans. 1-069. 4. The surveying wheel turns twice in the length of 16| feet ; in going round a circular bowling-green it turns exactly 200 times ( how many acres, roods, and perches in it ? Ans. 4 acres, 3 roods, 35*8 perches. 6. The circumference of a fish-pond is 56 chains, what is its area! A71S. 249-56288. 6. What is the area of a quadrant, the radius being 100 ? Ans. 7854. ?. Required the length of a chord fastened to a stake at one end, and to a cow's horns at the other, so as to allow her to feed on an acre of grass and no more? . Aiis. Sd^ yards. 8. The circumference of a circle is 91, what is its area? Ans. 659-00198. 9. The diameter of a circle is 15 perches, what is its area? Ans. 176-715. 10. What is the area of the semicircle of which 20 is the radius ? Ans. 628-32. PROBLEM XIX. Given the diameter of a circle., to find the side of a square equal m area to the circle. Rule. Multiply the diameter by -8862269, and the product will be the side of a square equal in area to the circle.* 1. If the diameter of a circle be 100, what is the side of a square equal in area to the circle? Ans, 88-62269. 2. The diameter of a circular fish- pond is 200 feet, what is the side of a Rquare fish-pond equal in area to the circular one ? Ans. 177-24538. • See Appendix, Demonstration 26. MENSURATION OF SUPERFICIES. 36 PROBLEM XX. Given the circumference of a circle^ to find the side of a square ^ equal in area to the circle. Rule. Multiply the circumference by -2820948, and the product will be the side of the square.* 1. The circumference of a circle is 100, what is the side of a square equal in area to the circle? Ans. 28*20948. 2. The circumference of a round fish-pond is 200 yards, what is the side of a square fish-pond equal in area to the round one? Ans. 56-41896. PROBLEM XXL Given the diameter., to find the side of the inscribed square. Rule. Multiply the diameter by -7071068^^ t-^^ ^'^ and the product will give the side of the in- scribed square, f 1. The diameter of a circle is 100, what is the side of the inscribed square ? Ans. 70-71068. 2. The diameter of a circle is 200, what is the side of the inscribed square ? Ans. 141-42136. PROBLEM XXIL Given the area of a circle^ to find the side of the inscribed squa/ c Rule. Multiply the area by -6366197, and extract the square i'oot of the product, which will give the side of the inscribed square. J 1. The area of a circle is 100, what is the side of the inscribed square? Ans. 7-97884. 2. The area of a circle is 200, what is the side of the inscribed square ? 200 X -6366197= 127-3239400; then V127-3239400 = 11-2837. Ans. PROBLEM XXIIL Given the side of a square., to find the diameter of the circumscribea circle. Rule. Multiply the side of the square by 1-4142136, and the pro > duct will give the diameter of the circumscribed circle.§ 1. If the side of a square be 10, what is the diameter of the cir- cumscribed circle ? Ans. 14-142136. 2. If the side of a square be 20, find the diameter of the circum- scribed circle ? ^n5. 28-284272. PROBLEM XXIV. Given the side of a square., to find the circumference of the circumscrined circle. Rule/ Multiply the side of the square by 4*4428934, and the product will be the circumscribed circle ? 1| * See Appendix, Demonstration 27. t Ibid. 28. X Ibid. 29. § Ibid. 30. il Ibid. 31. 86 MENSURATION OF SUPERFICIES. 1. If the side of a square be 100, what is the circumference of tho circumscribed circle ? Ans. 444*28934. 2. If the side of the square be 30, what is the circumference of the circumscribed circle? Ans. 133'286802. PROBLEM XXV. Given the side of a square^ to find the diameter of a circle equal in area to the square. Rule. Multiply the side of the square by 1-1283791, and the product will be the diameter of a circle equal in area to the square whose side is given.* 1. If the side of a square be 100, what is the diameter of the circle whose area is equal to the square whose side is 100? Ans. 112-83791. 2. What is the diameter of a circle equal in area to a square whose side is 200 ? Ans. 225-67582. PROBLEM XXYI. Given the side of a square^ to find the circumference of a circle whose area is equal to the square whose side is given. Rule. Multiply the side of the square by 3-5449076, and the product will give the circumference of a circle equal in area to the given square.f 1. What is the circumference of a circle, whose area may be equal to a square whose side is 100? Ans. 354*49076. 2. Find the circumference of a circle equal in area to a square whose side is 300? Ans. 1063-47228. PROBLEM XXVII. To find the area of a sector of a circle. ^ Rule I. Multiply half the length of the arc by the radius of the circle, and the product is the area of the sector. J Rule II. As 360 is to the degrees in the arc of the sector, so is the area of the whole circle to the area of the sector.§ 1. Let A C B be a sector less than a semi- circle whose radius A is 20 feet, and chord A B 30 feet, what is the area ? Eirst, V (A 02 - A D2) = V 400 - 225) = 13-228 = OD; then C-0 D = 20- 13-228 = 6-772 = CD. Again, V (A Ds + C Da) = V 225 + 45-859984) = 16-4578 = AC, the chord of half the arc. Hence, by Problem XVII. the arc A B is 33-8874 ; then qo.OQ'T'J. ■ — 2 ^ ^^ = 338-874, the area required. * See Appendix, Demonstration a ; f Ibid. 33. t Ibid. 34. f Ibid. 35 MENSURATION OF SUPERFICIES. 2. Let aEFBOA be a sector greater than a semicircle, whose radius A is 20, the chord E B, 38, and chord B F of half E F B, ^23 ; required the area ? 23 = chord BF 8 87 184 38 = chord BE 3)146 48-666 &c. = arc BFE 20 973J area. 3. What is the area of a sector whose arc contains 18 degrees, the diameter being 3 feet ? •7854 18 : : 7*0686 : the area of the sector ; 1 : : 7-0686 : -35343. Ans. Then 360 ; Or, 20 : 4. What is the area of a sector whose arc contains 147 degrees 29 minutes, and radius 25 ? Ans. 804-3986. 5. What is the area of a sector whose arc contains 18 degrees, the radius being 3 feet? Ans, 1-41372/ PROBLEM XXVIII. Tofina the area of the segment of a circle. Rule I. Find the area of the sector having the same arc with tlie segment, by the last problem; find also the area of the triangle, formed by the chord of the segment and the two radii of the sector. Tlien add these two areas together, when the segment is greater than a semicircle, but find their difference when it is less than a semicircle, the result will evidently be the answer. 1. What is the area of the segment A C B D A, its chord A B being 24, and radius A E or E C 20? V(AE2 — AD2) = V (400 — 144) = 16 = DE; EC— ED = 20 — 16 = 4 = CD; V (AD2 + DC2) = V (144 4- 16) = 12-64911 = AC; then^^^ x 8) - 24 ^ = AC; 25-7309 = arc A CB, And 12-8654 = half arc 20 = radius 3 257-308 = area of sector E B CA. 192 =areaofAABE 192 = areaof AABB 65-308 = area of segment A B C A . 58 MENSURATION OF SUPERFICIES. 2. Let AG F B A be a segment greater than a semicircle, there are given the chord AB 205, FD 17-17, AF 20, FG 11-5, and AE 11*64 ; required the area of the segment? (FGx8)— AF (11-5x8) — 20 , , , . , ^ = ^^ ^ = 24 the length of the art AGF (Problem XYII.) ; then 24 x 11-64 = 279-36, area of sector AEBFGA (Problem XXVII.) Again, FD - EF = 17-17 — ,1^.1 K -o vT. .u ABxED 20-5x5-53 ^^ ^oo^ Il-64 = 5-o3 = LD; then ^ = o = 56 6825 the area of the triangle ABE, which being added to the area of the sector before found will give the area of the segment, viz., 279-36 + 56-6825 = 336-0425 the area of the segment A G F B A. Rule II. To two-thirds of the product of the chord and versed sine of the segment, add the cube of the versed sine divided by twice the chord, and the sum will give the area of the segment, nearly. When the segment is greater than a semicircle, find the area of the remaining segment, and deduct it from the area of the whole circle, the remainder will give the area of the segment.* 3. What is the area of the segment AC B, less than a semicircle, Us chord being 18*9, and height or versed sine DC 2-4? AB X DC = 18-9 x 2-4 = 45*36, and | AB x DC = | x 45-36 = 30-24; then --?^^^ = -36571 ; hence 30-24 + •36571 = ^ X lO V 30-60571 the area. Note. — If two cliords of a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the seg- ments of the other. This is the 35th Proposition of Book III. of Euchd. 4. Required the area of the segment AG FB whose height FD is 20, and chord AB 20? ^?=^= 10 = AD, and AD2= 100; but AD2=FD x DC '•^^ - FD-20-^- The area of the segment ACB is, by the last case, 69 7916; and the area of the whole circle, by Problem XYIIL, is 490*875; then 490-875— 69-7916 = 421-0834 = area of the segment AGFB. 5. W'hat is the area of the segment AGFB, greater than a semi- circle, whose chord AB is 12, and versed sine 18? Ans. 297*81034. Rule III. Divide the height of the segment by the diameter of the civcle, to three places of decimals. Find the quotient in the column Height of the table at the end of the practical part of this treatise, and take out the corresponding Area Seg., which multiply by the square of the diameter, and the product will be the area of the seg- ment required.^ Note. I. — If the quotient of the height by the diameter be greater than '5 sub- tract it from 1, and find the Area Seg. corresponding to the remainder, whicli subtract from •7864 for the correct Area Seg. * Se<9 Appendix, Demonstration 36. t Ibid. 37. MENSURATION OF SUPERFICIES. 89 Kote 11.— If the quotient of the height by the diameter does not terminato In three figures, find the Area Seg. corresponding to the first three decimal figures of the quotient, subtract it from tlie next greater Area Seg., multiply the remainder \^ the fractional part of the quotient, and add the product to tho area segment first, taken out of the table. When great accuracy is not required, the fractional part may be omitted. 6. Let the diameter be 20, and the versed sine 2, required the area of the segment ? ^2^= -1, to which answers -040875 Square of diameter, 400 « 16-35 area. 7. What is the area of a segment, whose diameter is 52, and versed sine 2 ? ^2^=-038y\ which is the tabular versed sine. Then to '038 an- swers '009763, and the difference between this area and the next is •000385, which multiplied by j\ gives -000177, which added to •009763 gives -009940, which is the area corresponding to the versed sine •038xV Then 52^ x 009940 = 26-87776 is the area required. PROBLEM XXIX. Tojind the area of a zone^ or the space included by two parallel chords and the arcs contained between them. Rule. Join the extremities of the parallel chords towards the same parts, and these connecting lines will cut off two equal segments, the areas of which, added to the area of the trapezoid then formed, wiU give the area of the zone. 1. Suppose the greater chord A B = 30, the less CD 20, and tlie perpendicular distance D a: = 25, required the area of the zone ABDC? i (AB-CD) = a:B=J (30-20) = 5: then V (x D2 + X B'O = D B = V (25^ + 52) = 25-49. A B— B a: = A a; = 30—5 = 25, and ( A a: X B a:) -^ D a; = F a; = (25 X 5) ~ 25 = 5. Da: + Fa;=DF = 25+5 = 30;i V(CD2 + iyY-2) =^CF=G2; = J V(20' + 30'-^) = 18-027, the radius of the circle; (DBxAa:)-r-2Da:=: G 2/* = (25-49 X 25) ~- {2 x 25) = 12-745 ; G z - G y = z y =z 18-027 - 12-745 = 5-282, the height of the segment A 2; C. 36 -05)5 '28(- 146, the tabular area seument answering to which is •071033, then '071033 x (36-05)^ = 92*315 = the area of the seg- ment AzC. J (A B + C D) X D a; = H30 + 20) X 25 = 625 the area of tho trapezoid ABDC: then 625 + 92.315 x 2 = 809-63 = the area of the zone. 2. Let the chord AB be 48, the chord CD 30, the chord AC 15*81 14 ; what is the area of the zone ABDC? Ans. Ti> diameter C F = 50, height of the segment A 2; 0=3 * Se© Appendix, Demonstration 33. 40 MENSURATION OF SUPERFICIES. 1-2829, area by the table of segments = 13-595. Area of the zone ABDC = 534-19? 3. Let AB = 20, C D = 15, and their distance = 17j: required the area? Ans. 395*4369. 4. Let AB = 96, CD = 60, and their distance = 26; required the area? Ans. 213Q'7627. PROBLEM XXX. Tojind the area of a circular ring^ or of the space included between two concentric circles. Rule. Multiply the sum of the two diameters by their difference, and the product arising by '7854 for the area of the ring.* 1. The diameter AB is 30, and CD 20; what is the area of the ring XX? 30 20 60 sum 10 difference 600 •7854 392-7000 area of the ring X X. 2. "What is the area of the circular ring, when the diameters are 40 and 30? Ans. 549-78. 3. What is the area of the circular ring, when the diameters are 60 and 45? Ans. 373-065. PROBLEM XXXI. To find the area of apart of a ring^ or of the segment of a sector Rule. Multiply half the sum of the bounding arcs by their dis* tance asunder, and the product will give the area.f 1. Let AB be 50, and ah 30, and the distance a A 10 ; what is the area cf the space a & B A ? . 50 + 30 ,^ ,^^ Ans. ^ — X 10 = 400 2. Let A B = 60, a & = 40, and a A = 2 ; re- quired the area of the space a 5 B A ? Ans. 100. 3. Let AB = 25, a5 = 15, and aA = 6; re- quired the area of the segment of the sector ? Ans. 120. PROBLEM XXXII. To find the area of a lune, or the space included between the intersecting arcs of two eccentric circles. Rule. Find the areas of both segments which form the lune, and deduct the less from the greater ; the remainder will evidently be the area required. • See Appendix, Demonstration 39* i Ibid. 40. MENSURATION OF SUPERFICIES. 41 1. Let the chord AB = 40, EC = 12, and E D = 4 ; what is the area of the lune ADBCA? By note, page 38, (A E^ ^ E C) + E C = diameter of the circle of which A C B is an arc ; and (AE'^ -4- E D) + E D = the diameter of the circle of which A D B is an arc ; hence (20" -^ 12) + 12 = 45-3; and (202-^4) + 4=104; are the two diameters. 12 +- 45-3 = -264. 4 -^ 104 = '038. The Area Seg. answering to -264 is -165780, and (45-3)2 x •165780 = 340-1954802 = area of the segment AE B C A. The Area Seg. answering to '038 is -009763, and (104)2 x -009763 = 105-596608 = area of the segment A E B D A ; then 340-1954802 — 105-596608 = 234-5988722 = the area of the lune. 2. Let the chord A B be 40, and the heights of the segments E C and E D 15 and 2 ; required the area of the lune? Ans. 388*5 PROBLEM XXXIIL TO MEASURE LONG IRREGULAR FIGURES. When irregular figures^ not reducible to any known figure^ present themselves^ their contents are best found by the method oj equi-distant ordinates. Rule. Take the breadths in several places, at equal distances, and divide the sum of the first and last of them by 2 for the arithmetical mean between those two. Add together this mean and all the other breadths, omitting the first and last, and divide their sum by the number of parts so added, the. quotient will give the mean breadth of the whole, which being multiplied by the given length will give the area of the figure, very nearly. It is not necessary sometimes to take the breadths at equal dis- tances, but to compute each trapezoid separately, and the sum of all the separate areas thus found will give the area of the entire nearly. Or, add all the breadths together and divide by the number of them for a mean breadth, which being multiplied by the length, as before, will give the area nearly. 1. Let the ordinate AD be 9-2, bf 7, eg 9, dh 10, BC S'^ and the length AB 30 ; required the area? 9-2 AD 8-8 BO 2)18 9 mean breadth of first and last. 7 bf 9 eg 10 dh A b 4)35 sum 8-75 mean 30 262-50 area of the whole figure 8-75 mean breadth of all. 30 42 EXERCISES IN 2. The length of an irregular figure is 39 yards, and its breiidths, in five equi-distant places, are 4-8, 5*2, 4*1, 7*3, and 7'2; what is its area? Ans. 220*35 square yards. 3. The length of an irregular figure is 50 yards, and its breadths, at seven equi-dlstant places, are 5*5, 6*2, 7*3, 6, 7*5, 7, and 8'8 ; what is its area? Ans. 342*9150 square yards. 4. The length of an irregular figure being 37*6, and the breadths, at nine equi-distant places. 0, 4*4, 6*5, 7*6, 5*4, 8, 5*2, 6 5, 6-1; what is the area? Ans. 218*315. EXEECISES. 1. Find the area of a square whose side is 35*25 chains. Ans. 124 acres, 1 rood, 1 perch. 2. Find the area of a rectangular board, whose length is 12J feet, and breadth, 9 inches. Ans. 9| feet. 3. The sides of three^uares being 4, 5, and 6 feet, what is the length of the side ofjj(™fe which is equal to all three? '^^ \ Ans. 8*7749 feet. 4. Required the area of a rtomboid whose length is 4^1, chains, and breadth, 4*28 chains? ^A7is. 4 acres, 1 rood, 39']^erclies. 5. There is a triangle whose base is 12*6 chains, and altitude 6*4 chains, wliat is its area? Ans. 40-32. 6. Find the area of a triangle whose sides are 30, 40, and 50 yards. Ans. 600 square yards. 7. There is a triangular corn-field whose sides are 150, 200, and 250 yards, determine the number of acres contained in the field, and the expense of reaping the corn at 9s. 6d. per acre. Alls. Content of the field, 3 acres 15 perches ; expense of reaping, £1, 9s. 5d. 8. What must the base of a triangle be to contain 36 square feet, whose vertex is to be 9 feet from the base? Ans. 8 feet. 9. What must be the altitude of a triangle equal in area to the last, whose base is 12 feet? Ans. 6 feet. 10. The height of a precipice standing close by the side of a river is 103 feet, and a line of 320 feet will reach from the top of it to the opposite bank; required the breadth of the river? Ans. 302 97 feet. 11. A ladder 12|- feet in length stands upright against a wall, how far must the bottom of it be pulled out from the wall so as to lower the top 6 inches? Ans. 3 J feet. 12. A person wish - :*ng to measure the distance from a point A, at one side of a canal, to an object 0, at the other, and having no mstrument but a book, placed a MENSURATION OF SUPERFICIES. 43 comer of it on the point A, and directed an edge of it, as in tha figure, in a straight line with the object 0, and drew the straight lines A B, A C ; he then placed the book so that a comer of it rested on the point B, at the distance of eight times its length from the point A, and directed an edge of it, as before, to the object 0, and drew the straight line BC which met AC at the distance of three times the length of the book from A ; how many times the length of the book is the object tVom the points A and 6 ? Ans. 21 i and 22*78 times. 13. What is the area of a trapezium whose diagonal is 70*5 feet, and the two perpendiculars 26*5 and 30*2 feet? Ans, 1998*675 square feet. 14. "What is the area of a trapezium whose diagonal is 108 feet 6 inches, and the perpendiculars 56 feet 3 inches, and 60 feet 9 inches? Ans. 6347 feet 36 inches. 15. What is the area of a trapezoid whose two parallel sides are 75 and 122 links, and the perpendicular distance 154 links? Ans. 15169 square links. 16. A field in the form of a trapezoid, whose parallel sides are 6340 and 4380 yards, and the perpendicular distance between thena 121 yards, lets for £207, 14s. per annum; what is that per acre? Ans. £1, lis. 17. Two opposite angles of a four-sided field are together equal to t^70 right angles,, and the sides are 24, 26, 28, and 30 yards ; what is its area? Ans. 723*99 square yards, nearly. 18. Kequired the area of a figure similar to that annexed to the first question under Problem XIV., whose dimensions are double of those there given ? Ans. 3411-6. 19. What is the side of an equilateral triangle equal in area to & Bquare, whose side is 10 feet? Ans. 15*196 feet, nearly. 20. Required the area of a regular nonagon, one of whose sides is 8 feet, and the perpendicular from the centre = 10*99 feet? Ans. 395*64 square feet. 21. Required the area of a regular decagon, one of whose sides is 20*5 yards? Ans. 3233*491125 square yards. 22. A wheel of a car turns round 4400 times in a distance of 10 jniles; what is its diameter? Ans. 3*819708 feet. 23. If the diameter of a circle be 9 feet, what is the length of the circumference? Ans. 28f feet, nearly. 24. Required the length of an arc of 60°, the radius of the circle being 14 feet? Ans. 14*660772 feet. 25. The chord of an arc is 30 feet and the height is 8 feet, what is the length of the arc? Ans. 35^ feet, nearly. 26. The diameter of a circle is 200, what is the area of the quad- rant? Ans. 7854. 27. The diameters of two concentric circles are 15 and 10, what is the area of the ring formed by those circles ? Ans. 98*175. 28. The circumference of a circle is 628*32 yards, what is the radius of a concentric circle of half the area ? Ans. 70*71. 29. What is the side of a square equal in area to the circle whose diameter is 3? Ans 2*6586807 44 EXERCISES IN MENSURATION OF SUPERFICIES. 30. The two parallel chords of a zone are 16 and 12, and theif perpendicular distance is 2, what is the ai'ea of the zone? Ans. 28-376. 31. The length of a chord is 15, and the heights of two segments of circles on the same side of it are 7 and 4 ; wliat is the area of the lune formed by those segments? Ans. 38, nearly. 32. The base and perpendicular of a right-angled triangle are each 1, what is the area of a circle having the hypothenuse for its diameter? Ans. 1*5708. 33. If the area of a circle be .^00, what is the area of the inscribed SQuaie? dns. 63-66. CONIC SECTIONS. 45 OOmO SECTIONS. SECTION III. (or THE ELLIPSIS*^ PROBLEM I. llie transverse and conjugate diameters of an ellipsis being given ^ to find the area. Rule. Multiply the transverse and conjugate diameters together, and the product arising by '7854:, and the result will be the area.f 1. Let the transverse axis be 35, and the conjugate axis 25; required the area? . 35 x 25 x •7854 = 687'225. Ans. 2. The longer diameter of an ellipse is 70, and the shorter 50 ; what is the area? ^W5. 2748-.9. 3. What is the area of an ellipse whose longer axis is 80, and shorter axis is 60? Ans. 3709-92. 4. What is the area of an ellipse, whose diameters are 50 and 45? Ans, 1767-15. PROBLEM IL To find the area of an elliptical ring. Rule. Find the area of each ellipse separately, and their difference will be the area of the ring. Or, From the product of the two diameters of the greater ellipse deduct the product of the two diameters of the less, and multiply the remainder by -7854: for the area of the ring. J Q 1. The transverse diameter A B is 70, and the conjugate CD 50; and the transverse diameter E F of another ellipse having the same centre 0, is 35, and the conjugate GH is 25; required the area of the elliptical space between their circumferences ? ^"^or definitions of the ellipsis (or, as it is frequently written, ellipse) and k. lEier Conic Sections, see Appendix, Properties of the Conic Sections. t JSeti Appendix, Demonstration 41. J Ibid. 42. 4C CONIC SECTIONS. 70 X 50 X -7854 = 27489 ; and 35 x 25 x -7854 = G87-226 ; tbcn 2748'9 — 687*225=: 2061-675 = area of the elliptical ring. 70x50 = 3500 35x25= 875 2625 X -7854 = 2061 '675 = area 2. The transverse and conjugate diameters of an elh'pse are 60 and 40, and of another 30 and 10 ; required the area of the space between their cu'cumferences? Ans. 1649 •34. 3. A gentleman has an elliptical flower garden, whose greater diameter is 30, and less 24 feet ; and has ordered a gravel walk to be made round it of 5 feet 6 inches in width ; required the area of the walk? Ans. 371-4942. PROBLEM III. Given the Tieiglit of an elliptical segment., whose base is parallel to either of the axes of the ellipse] and the two axes of the ellipse^ to find the area. Rule. Divide the height of the segment by that diameter of which it is a part, to three places of decimals, find the quotient in the column Height of the table referred to in page 38, and take out the correspondent Area Seg. Multiply the Area Seg. thus found and both the axes of the ellipsis together, and the result will give the area required.* 1. Required the area of an elliptical seg- ment R A Q, whose height A P is 20 ; the transverse axis AB being 70, and the tonjugate axis CD 50 ? 20 -7- 70 = •285f = the tabular versed sine, the corresponding segment answering .^0 which is -185166; then •]85166x70x 50=648-081, the area. 2. What is the area of an elliptical segment cut off by a chord parallel to the shorter axis, the height of the segment being 10, and the two diameters 35 and 25 ? Ans. 162-0202. 3. What is the area of an elliptical segment cut off by a chord parallel to the longer axis, the height of the segment being 10, and the two diameters 40 and 30? Ans. 275-0064. 4. What is the area of an elliptical segment cut off by a chord parallel to the shorter diameter, the height being 10, and the two diameters 70 and 50 ? ^?is. 240-884. PROBLEM IV. To find the circumference of an ellipse., by having the two diameters given. Rule. Multiply the sum of the two diameters by 1*5708, and the product will give the circumference nearly; that is, putting t for the transverse, c for the conjugate, andjo for 3*1416 j the circumference will be {t+c}x^p.f * Soe Apftendis, Demonstration 43. ♦ TbM. 44. OF THE ELLIPSIS. 47 1. Let the transverse axis be 24, and the conjugate 18; required the ^B^ circvwierence- (24 + 18) X 1-5708 = 42 xl'5708 = 65'9736 is the circumference Dearly. 2. Required the circumference of an ellipse whose transverse axis fs 30, and conjugate 20? Ans. 78*54. 3. Required the circumference of an ellipse whose diameters are 60 and 40? Ans. 157-08. 4. What is the circumference of an ellipse whose diameters are 6 and 4? Ans, 15-708. 5. What is the circumference of an ellipse whose diameters are 3 and 2? Ans. 7*854. PROBLEM V. To find the length of any arc of an ellipse. Rule. Find the length of the circular arc xy^ intercepted by C, B, and whose radius is half the sum of C, OB: and it will be equal to the elliptical arc B C nearly.* Note. — ^The nearer the axes of the ellipse apin-oach towards equality, the uiore exact the result of the operation by this R\ile ; and the less the elliptical arc, the nearer its exact length will approach the arc x y. 1. Let the axis AD be 24, CK 18, and T 3 ; required the length of the arc B C ? Here we have TD = 9, and AT=15; then from the property of the ellipsis, we have A02 : C2 : : ATxT D : T Ba= 9^x9x15 9x9x15 , ^ ^j //rkT2. -i2^rT2-=— 16-'^"^^^=^^^^ + T B2)= V (9 + ^^^f^) =9-21616, the radius of the circle of which G B is an arc ; but C is the radius of the circle of which C V is an arc; therefore the radius of the circle of which a; V is an arc, is \ C + ^-O B = 9-10808. But by Trigono- me^r?/,tHB-^OB = 3-4-9-21616 = -325515, is the sine of the angle COB, or arc xy^ to the radius 1, answering to 18*9968 degrees. Therefore, by Problem XYII. Rule I., the length of the arc xy is •01745 X 18-9968 x 9-10808 = 3-0192, which is also equal to the length of the elliptical arc CB, nearly. 2. Given A D 30, C K 20, and T 5 ; required the length of the arc B C ? Ans. 5-03917786255. 3. Given AD 40, CD 30, and T 5 ; required the length of the arc B C? Ans. 5-033880786. * See Appendix, Demonstration 45. t It may be done without Trigonometry, by first finding the lencrth of tho aic B by Rule XL Prob. XVII. Sec. 2, tbeu OG:Oy::GB:YZ. D 48 CONIC SECTIONS. PROBLEM VI. Given the diameter and abscissas^ tojind the ordinate. Rule. Say, as the transverse is to the conjugate, so is the square root of the rectangle of the two abscissas, to the ordinate.* 1. In the ellipse A C D K, the transverse diameter A D is 100, the conjugate diameter C K 80, and the abscissa DT 10; required the length of the ordinate T B? 100 : 80 : : V (90 x lO) : T B = 24. (See the last figure.) 2. Let the transverse axis be 35, the conjugate 25, and the abscissa 7 ; required the ordinate. Ans. 10. 3. Given the two diameters 70 and 60, and the abscissa 10 ; re- quired the ordinate? Ans. 209956. PROBLEM VII. Given the transverse axis^ conjugate and ordinate^ to find the abscissas. Rule. As the conjugate is to the transverse diameter, so is the equare root of the difference of the squares of the ordinate and semi- conjugate, to the distance between the ordinate and centre. Then this distance being added to, and subtracted from, the semi- diameter, will give the two abscissas, f 1. Let the diameters be 35 and 25, and the ordinate 10 ; required the abscissas? By the Bale f + |f V ([f ]' - 10«) = — ^^ = 28 ami 7, the two abscissas. 2. Let the diameters be 120 and 40, and the ordinate 16; re- quired the abscissas ? i4?i5. 96 and 24. PROBLEM VIII. Given the conjugate axis^ ordinate^ and abscissas^ to find the transverse axis. Rule. Find the square root of the difference of the squares of the semi- conjugate axis and the ordinate, which add to, or subtract from, the semi-conjugate, according as the less abscissa or greater is given. Then say, as the square of the ordinate is to the rectangle of the conjugate, and the abscissa, so is the sum or difference found above to the transverse required. J 1. Let the ordinate be 10, and the less abscissa 7; what is the diameter, allowing the conjugate to be 25? V( [yT"~^^^ ) = 7-5 ; then 7*5 + 12-5 = 20 ; then 10^ : 25 X 7 : : 20 : 35, the transverse required. 2. Let the ordinate be 10, the greater abscissa 28, and the con- jugate 25 ; required the transverse diameter ? Ans. 35. * S©8 Appendix, DomoaHtiation -Id. \ Ibid. 47. \ J bid. 48- OF THE PARABOLA. 40 PROBLEM IX. Given the transverse axis, ordinate, and abscissa, to find the conjugate. 't Rule. The square root of the product of the two afflpissas is to the ordinate, as the transverse axis is to the conjugate. mB ': 1. Let the transverse axis be 35, the ordinate 10, anl^p ^scissaa . 28 and 7; required the conjugate ? //oo rrx 1A n^ 35x10 35x10 ^. ,, *! V (28 X 7) : 10 : : 35 : ^.^g^y. = ~~II~ =^^' ^^® conjugate. 2. Let the transverse diameter be 120, the ordinate 16, and the abscissas 24 and 96; required the conjugate? * Ans. 40* OF THE PAEABOLii. PROBLEM X. Given the base and height of a parabola, to find its area. Note. — Any double ordinate AB, to the axis of a parabola, may be callecJ. ita base, and the abscissa O D, to that ordinate, its height. q Rule. Multiply the base by the height, and § of the product will be the area.f 1 . Required the area of a parabola, whose height qI is 6 and base 12 ? / 6 X 12 X 1 = 48 the area. A d b 2. What is the area of a parabola, whose base is 24, and height 4? Ans. 64. 3. What is the area of a parabola, whose base is 12, and height 2 ? Ans. 16. PROBLEM XI. To find the area of the zone of a parabola, or the space between two parallel double ordinates. Rule I. When the two double ordinates, their distance, and the altitude of the whole parabola are given ; find the area of the whole parabola, and find also the area of the upper segment, their diflference will be the area of the zone. II. When the two double ordinates and their distance are given; to the sum of the squares of the two double ordinates, add their pro- duct, divide the sum by the sum of the two double ordinates, multip'y the quotient by | of the altitude of the zone, and the product will be the area of the zone. J 1. Given A B 20, S T 12, and D j; 8 ; what is the area of the zone ASTB, the altitude D being 12-5? « See Appendix, Demonstration 49. f Ibid. 50. X Ibid. 51. % BO CONIC SECTIONS. (20 X 12-6) x| = 166| = area of the parabola ABO, and (12-5—8) X 12 = 64, and 54x1 = 36; hence 166|— 36 = 130|, the area. III. When the altitude of the whole parabola is not given. 2. Suppose the double ordinate A B = 10, the double ordinate ST =6, and their distance Da;=4: what is the area of the zone ASTB? ^^'^1^0 + 6^"^^ ^ 12i ; then 12i x 4 x | = 32|, the area as before. 3. Let the double ordinate AB = 30, CP = 25, and their distance D G= 6 ; required the area of the zone A B P C? Ans. 165j\ PROBLEM XIL 7'o Jind the length of the curve^ or arc of a ftarahola^ cut off by a double ordinate to the axis. Rule. L Divide the double ordinate by the parameter, and call the quotient q. IL Add 1 to the square of the quotient ^, and call the square root ot the sum s. in. To the product of q and 5, add the hyperbolic logarithm of their sum, then the last sum multiplied by half the parameter, will give the length of the whole curve on both sides of the axis. Putting c for the curve, q for the quotient of the double ordinate divided by the parameter, s for V (1 + 9'^) ^^^ ^ for half the para- meter; then c^ay. {75 + hyp. log. of (<7 + s.)}* Note.— The common logarithm of anv number multiplied by 2*302586093 gives the hyperbolical logarithm of the same number. 1. What is the length of the curve of a parabola, cut off by a double ordinate to the axis, whose length is 12, the abscissa being 2 ? y^ y x—2 and y — Q\ then a = — = 3^« = 9, and g = - = f = |^ f Iso 5 = V (1 + ^^) = V (1 + t) = V (¥) = k V (13) = 1-2018504=5. Then | + 1-2018504 = 1-868517, whose common logarithm is -271497, which being multiplied by 2-302585093, produces -6251449 for its hyperbolic logarithm; and also |x 1-2018504= -8012836; the sum of these two is 1 4263785, there- fore 9x1-4263785 = 12-8374065, is the length of the curve re- quired. Rule II. Put y equal to the ordinate, and q equal the quotient arising from the division of the double ordinate by the parameter, or from the division of doubio the abscissa by the ordinate ; then the length of the double curve will be expressed by the infinite series. "^^^^V 2.3 2.4.5^2.4.6.7'^^ ) * See Appendix* Demons 52. OF THE TARABOLA. 51 Note. — This series 'will converge no longer than till 3=1. For when q is greater than 1, the series will diverge. Let the last example be resumed, in wliich the abscissa is 2, and the ordinate 6. Hence, 2 x 2-7-6 = f =5'; then employing f instead of q in the last series, we get 12x (l+QI"— 114 +3xh^?^) = 12'837 the length of the curve as before. Rule III. To the square of the ordinate, add ^ of the square of the abscissa, and the square root of the sum will be the length of the single curve, the double of which will be the length of the double tourve, nearly.* Note. — The two first rules are not recommended in practice. The practical application of this is much simpler, and is therefore to be employed in prefer- ence to either. Eetaining the same example, in which a: =2, and 2/ = 6, we shall geti;=V(2/^ + #^')= V(36+V) = 6-1291, and C = 12-8582, nearly. 2. Required the length of the parabolic curve, whose abscissa is 3, and ordinate 8 ? Ans. 17*4:35. PROBLEM XIIL Grtven any two abscissas and the ordinate to one ofthem^ to find the corresponding ordinate to the second abscissa. Rule. Say, as the abscissa, whose ordinate is given, is to the square of the given ordinate, so is the other given abscissa to the square of its corresponding ordinate.f 1. If the abscissa a;0 = lO, and the ordinate x S = 8, what is the ordinate AD, whose abscissa D is 20 ? a; : X S2 : : D : A D2, viz. 10 : 64 : : 20 : 128, the square root of which is 11-313, &c.,=AD. 2. If 6 be the ordinate corresponding to the abscissa 9, required the ordinate corresponding to the abscissa 16? Ans, 8. PROBLEM XIV. Given two ordinates^ and the abscissa corresponding to one of them, tojind the abscissa corresponding to the other. Rule. Say, as the square of the ordinate whose abscissa is given, is to the given abscissa, so is the square of the other ordinate to its corresponding abscissa.J 1. Given Sa;=6, a; = 9, and AD- 8; required the abscissa CD? 36:9: : 64: 16=0D. 2. GivenSa;=8, a;0=10, and AD-^9; required OD? Ans. 12-656. * See Appendix, Demonstration 63. f Ibid. 54- t Ibid. 64, 52 CONIC SECTIONS. PROBLEM XV. Jiven two ordinates perpendicular to the axis and their distance^ tojina the corresponding abscissas. Rule. Say, as the difference of the squares of the ordinates is to their distance, so is the square of either of them to the corresponding abscissa.* 1. Given Sa:=6, A D = 8, and a;D = 7; required the abscissas? (64 — 36) : 7 : : 64 28 : 7: :64: 16 = 0D, and 28 :7: :36: 9=0 x. 2. Given Sa; = 3, AD = 4, and a;D = 2; required the abscissas? Ans, 4f and 2^. OF THE HYPERBOLA. PROBLEM XVI. Given the transverse and conjugate diameters^ and any alscissa^ to find the corresponding ordinate. Rule. As the transverse is to the conjugate, so is the mean pro^ portional between the abscissas to the ordinate.^ 1. If the transverse be 24, the conjugate 21, and the less abscissa ADS; required the ordinate ? Note. — The less abscissa added to ttie transverse gives the greater. 24:2i::vr32x8):y^^^ip^ = i4the ._ ordinate. 2. If the transverse axis of an hyperbola be 120, the less abecissa 40, the conjugate 72 ; required the ordinate? Ans. 48. 3. The transverse axis being 60, the conjugate 36, and the less abscissa 20 ; what is the ordinate ? Ans. 24. PROBLEM XVII. Given the transverse.^ conjugate^ and ordinate^ to find the abscissa/ Rule. To the square of half the conjugate, add the square of the ordinate, and extract the square root of the sum. Then say, As the conjugate is to the transverse, so is that square root to half the sum of the abscissas. * See Appendix, Demonstration 56. T Ibid 56. OF THE HYPERBOLA. 53 Then to tliis half sum, add half the transverse, for the greater ahscissa ; and from the half sum take half the transverse for tlie less abscissa.* 1. If the transverse be 24, and the conjugate 21 ; required the abscissas to the ordinate 14 V 10-5=. J conjugate 14 = ordinate 10-5 14 110-25 196 196 306*25, the square root of which is 17*5 ; then 21 : 24 : : 17*5 : 20 = half sum, 20 H- 12 = 32 the greater abscissa, and 20 — 12=8 the less abscissa. 2. The transverse is 120, the ordinate 48, and the conjugate 72; required the abscissas ? Ans. 40 and 160. PROBLEM XVIII. (riven the conjugate^ ordinate^ and abscissas, tojind the transverse. Rule. To or from the square root of the sum of the squares of the ordinate and semi-conjugate, add or subtract the semi-conjugate, according as the less or greater abscissa is used; then, as the square of the ordinate is to the product of the abscissa and conjugate, so is the sum or difference, above found, to the transverse. f 1. Let the conjugate be 21, the less abscissa 8, and its ordinate 24 » required the transverse ? 21x8x V(142 + ^) + 104 IP ^ 1 X V(3* + 42) + 3) = 3 X (5 + 8) = 24 the transverse. 2. The conjugate axis is 72, the less abscissa 40, the ordinate 48 i -,equired the transverse? Ans. 120. 3. The conjugate is 36, the less abscissa 20, and its ordinate 24 ; required the transverse? Ans. 60. PROBLEM XIX. Given the ahscissa, ordinate, and transverse diameter, to find the conjugate. Rule. As the mean proportional between the abscissas is to the ordinate, so is the transverse to its conjugate % 1. What is the conjugate to the transverse 24, the less abcissa being 8, and its ordinate 14 ? 24 X 14 ;^,^3^3^^ = 21 the conjugate. 2. The transverse diameter is 60, the ordinate 24, and the lesa abscissa 20 ; what is the conjugate ? Ans. 36. * See Appendix, Demonstration 57. f Ibid. 58. | ibid. 59. ^4 CONIC SECTIONS. PROBLEM XX. divert any two abscissas^ X, x, and their ordinates^ F, y^ to find the transverse to which they belong. Rule. Multiply each abscissa by the square of the ordinate belong- ing to the other; multiply also the square of each abscissa by tlie square of the other's ordinate ; then divide the diffepence of the latter products by the difference of the former ; and the quotient will be the transverse diameter to which the ordinates belong.* 1. If two abscissas be 1 and 8, and their corresponding ordinates 4r| and 14, required the transverse to which they belong? ^ S'^x 41x41—1^x14^ ^ 35x35-14x1 4 5x5 — 2x2 ® Ixl42-8x4|x4| ~14xl4-35x4|~2x2-5x| = — = — =24, the transverse. PROBLEM XXL To find the area of a space AN OB, bounded on one side by thi curve of a hyperbola, by means of equi- distant ordinates. Let AN be divided into any given number of equal parts, AG, CE, EG, &c., and let perpendicular ordinates AB, CD, EF, &c., b« erected, and let these ordinates be terminated by any hyperbolic curve BDF, &c. ; and let A=AB + N0, B = CD+GH4-LM,&c.,and C = EF+IK, &c. ; then the com- mon distance A C, of the ordinates, being multiplied by the sum arising ^ ^ from the addition of A, 4 B, and 2 C, A C' E G~ and one-third of the product taken will be the area, very nearly.. ^, ,. A + 4B + 2C -p, ,, ... T^ * n , That IS, ^ X D = the area, puttmg D = A C.f 1. Given the lengths of 9 equi-distant ordinates, viz., 14, 15, IC, 17, 18, 20, 22, 23, 25 feet, and the common distance 2 feet ; required the area? Ans. 300| feet. 2. Given the lengths of 3 equi-distant ordinates, viz., AB = 5, CD = 7, and EF=8, also the length of the base AE 10; what is the area of the figure A B F E ? Ans. 68^ feet. 3. If the length of the asymptote of a hyperbola be 1, and there be 11 equi-distant ordinates between it and the curve, the common distance of the ordinates will then be ^, and from the nature of the curve their lengths will be ^, ^, 4-f, ^, ii, ih \h ^, H, rh U; Mhat is the area of the'curved figure? Ans. -69315021. This formula will answer for finding the area of all curves by using the sections peq)endicular to the axis. The greater the num- ber of ordinates employed, the more accurate the result ; but in real practice three or five are in most cases sufficient. * See Appendix, Demonstraticai 6Q» f Ibid. 6L OF THE HYPERBOLA. 65 PROBLEM XXII. To find tjit length of any arc of an hyperbola beginning at the vertex. Rule. I. To 19 times the square of the transverse, add 21 times the iquare of the conjugate ; also to 9 times the square of the transverse add, as before, 21 times the square of the conjugate, and multiply each of these sums by the abscissa. II. To each of these two products, thus found, add 15 times the product of the transverse and the square of the conjugate. III. Then, as the less of these results is to the greater, so is the ordinate to the length of the curve, nearly.* 1. In the hyperbola BAG, the transverse diameter is 80, the conjugate 60, the ordinate BD 10, and the abscissa AD 2 ; required the length of the arc B A C ? (Fig. p. 52.) Here 2 (19x802 + 21 x 602) = 2 (121600 + 75600) = 394400. And 2 (9x802 + 21x602) = 2 (57600 + 75600) = 266400. Whence 15 x 80 x 60« +394400 = 4320000 + 314400 = 4714400. And 1 5 X 80 X 602 + 266400 = 4320000 + 266400 = 4586400. Then- 4586400 : 4714400 : : 10 : ^^^^ = 10-279 = AB. 458641^ Hence A FC = 10-279 x 2 = 20*558. 2. In the hyperbola BAG, the transverse diameter is 80, the conjugate 60, the ordinate BD 10, and the abscissa AD 2-1637; required the length of the arc A B ? Ans. 10-3005. PROBLEM XXIII. Given the transverse axis of a hyperbola^ the conjugate^ and the abscissa, to find the area. Rule. I. To the product of the transverse and abscissa, add f of the square of the abscissa, and multiply the square root of the sum by 21. II. Add 4 times the square root of the product of the transverse and abscissa, to the preceding product, and divide the sum by 75. III. Divide 4 times the product of the conjugate and abscissa by the transverse ; this quotient, multiplied by the tormer quotient, will give the area of the hyperbola, nearly.f 1. In the hyperbola BAG (see figure, page 52), the transverse axis is 30, the conjugate 18, and the abscissa A D is 10 ; what is the area? * See Appendix, Demonstration 62. t Ibid. 63. 56 CONIC SECTIONS. Hpre 21 V(30 X 10 + ^ X 10^) = 21 V (300 + 71-42857) = 21 V (371-42857) = 21 x 19272 = 404-712 ; And^ V (30 X 1 0) + 404-712 ^ 4 x 17-3205 + 404-712 _^ 75 ~ 75 ^ C9-282 + 404-712 473994 _ ^^^^ ~-T5 = ~75- = ^■^^^^• Whence -i^^^^^^x 6-3199 = 24 x 6-3199= 151-6776, the area re- quired. 2. What is the area of an h}^erbola whose abscissa is 25, the transverse and conjugate being 60 and 30? Ans. 805 0909. 3. The transverse axis is 100, the conjugate 60, and abscissa 60 j rti L 30 inches ; what is its solidity? 202-^15 = 261, then 26| + 15 = 41|, the dia- meter of the circle. Again, -~ — -= 5|, the central distance. Now 15-^4:l| = •36, the area segment corres- ponding to V hich is '254550, which multiplied by the squaie of 41|, produces 441-92708 the area of the generating segment ABC, the half of which is 220*96354. Lastly, (20'-^ 3)— (5|x 220-96354) = 1377-7 1268, and this multi- plied by 12-5664 produces 17312-88862 cubic inches, the solidity required. 2. The axis of a circular spindle is 48, and the middle diameter 36; required the solidity of the spindle? Ans. 29916-6714. PROBLEM XVIL 7^0 find the solidity of the middle frustum of a circular spindle. Rule. I. Find the distance of the centre of the middle frustum, from the centre of the circle. IL Find the area of a segment of a circle, the chord of which is equal to the length of the frustum, and height half the difference l)etween its greatest and least diameters ; to which add the rectangle of the length of the frustum and half its least diameter ; the result will be the generating surface. in. From the square of the radius subtract the square of the central distance, the square root of the remainder will give half the length of the spindle. IV. From the square of half the length of the spindle take one- third of the square of half the length of the middle frustum, and multiply the remainder by the said half length. V. Multiply the central distance by the generating surface, and subtract this product from the preceding ; the remainder, multiplied by 6-2832, will give the solidity.* 1. Required the solidity of the middle frustum of a circular spindle, the length D E being 40, the greatest diameter Q F 32, and the least diameter P S 24 ? First, 202-^4=100, and 100 + 4=104, the diameter of the circle. Again, 52—16 = 36, the central distance. Also, ^ (32— 24) = 4, and 44- 104 = -0381^2 ^^^ ^'^^^ segment corresponding to which ia •009940, which, multiplied by the square of 104, produces 107*51104, the area of PLQ ; and 40 x 12 = 480 the area ot the rectangle P D E L. Hence 107*51104 + 480=587*51104 the area of the generating surface P D L E. * See Appendix, Demonstration 78. MENSURATION OF S0LID3. 69 Next y (522 _ 362) = ^ (1408) = 8 V (22) = B half the length of the spindle ; And (U08 — ^)x 20 = 25493^. Then 36 x 587-51104 = 21150-39744, and (25493J — 21150-39744) X 6-2832 ==27287-5347, the required so- lidity. 2. What is the solidity of the middle frustum P S R L of a circular spindle, whose middle diameter F Q is 36, the diameter P S of the end 16, and its length D E 40? A72S. 29257-2904. PROBLEM XYIII. To find the solidity of a spheroid. Rule. Multiply the square of the revolving axis by the fixed axis, and this product again by -5236 for the solidity.* 1. What is the solidity of a prolate spheroid whose longer axis A B is 55 inches, and shorter axis CD 33? Here 33^ x 55 x -5236 = 31361-022 cubic uiches, the answer. 2. What is the solidity of an oblate spheroid, whose longer axis is 100 feet, and shorter axis 6? Ans. 31416 cubic feet. 3. What is the solidity of a prolate spheroid, whose axes are 40 and 50? ^ Ans. USSS. 4. What is the solidity of an oblate spheroid, whose axes are 20 and 10? Ans. 2094-4. PROBLEM XIX. To find the solidity of the segment of a spheroid^ the base of the segment being parallel to the revolving axis of the sj)heroid. CASE I. Rule. From three times the fixed axis, deduct twice the height of the segment, multiply the remainder by the square of the height, and that product by -5236. Then say, as the square of the fixed axis is to the square of the revolving axis, so is the product found above to the solidity of the spheroidal segment. f 1. What is the content of the segment of a prolate spheroid, the height C being 5, the fixed axis 60, and the revolving axis 30? — See last Jgure. 50x3-5x2 = 150-10 = 140; then 140x52=3500, and 3500 X -5236= 1832-6; then 25 : 9 :: 1832-6 : 659-736, the answer. CASE n. When the base is elliptical.) or perpendicular to the revolving axis. Rule. From three times the revolving axis, take double the * See Appendix, Demonstration 79. * 'bid. 80. 70 MENSURATION OF SULIDS. height; multiply that difference by the square of the height, and the product again by •5236. Then as the revolving axis is to the fixed axis, so is the last pro- duct to the content.* 2. What is the content of the segment of a sphe- roid, whose fixed axis is 50, revolving axis 30, and height 6? 30x3 — 2x6 = 90 — 12 = 78; Then 78x62=2808; and 2808 x •5236 = 1470-2688: Then 30 : 50 : : 1470-2688 : 2450-448, the answer. 3. In a prolate spheroid, the transverse or fixed axis is 100, the conjugate or revolving axis is 60, and the height of the segment, whose base is parallel to the revolving axis, is 10: required the solidity? Ans. 6277-888 4. If the axes of a prolate spheroid be 10 and 6, required the con- tent of the segment, whose height is 1, its base being parallel to tlie revolving axis? Ans, 6-2"7888. PROBLEM XX. To find the solidity of the middle zone of a spheroid^ the diameter of the ends being perpendicular to the fixed axis ^ the middle dia^ meter ^ and that of either end being given^ together with the Ungth of the zone. Rule. To twice the square of the middle diameter, add the square of the diameter of the end; multiply the sum by the length of tha zone, and the product again by -2618 for the solidity. f 1. What is the solidity of the middle zone of an oblate spheroid, the middle diameter being 100, the diameter of the end 80, and the length 36 ? 1002 X 2 + 802 ^ 26400 ; then 26400 x 36 = 950400, and 950400 x ^2618 = 248814-72, the answer. 2. What is the solidity of the middle frustum of a spheroid, the greater dia- meter being 30, the diameter of the end 18, and the length 40? Ans, 22242-528. PROBLEM XXI. To find the solidity of a parabolic conoid. Rule. Multiply the square of the diameter of its base by ;3927, and that product by the height; the last product will be the solidity.^ 1. What is the solidity of the parabolic conoid, whose height Is 10 feet, and the diameter of its base 10 feet? 102 X -3927 = 39-27; then 39-27x 10 = 392-7, the solidity required. * See Appendix, Demonstratiou 81. t Ibid. 82. X Ibid. 83. MENSURATION OF SOLIDS. 71 2. WhAt is the solidity of a parabolic conoid, ^hose height is 30, and the diameter of its base 40? Ans. 18849-6. 3. What is the content of the parabolic conoid, whose altitude is 40, and the diameter of its base 12 ? Ans. 2261-952. ^ 4. Required the solidity of a parabolic conoid, whose height is 30, and the diameter of its base 8 ? Ans. 753 -984. PROBLEM XXII. To find the solidity of the frustum of a parabolic conoid. Rule. Multiply the sum of the squares of the diameters of the two ends by tlie height, and that product by '3927 ; the last product will be the solidity.* 1 . The greater diameter of the frustum is 10, T and the less diameter 5 ; what is the solidity, the length being 12? 103=100 63= 25 "125. Then 125x12 = 1500, and ^.1 1500 X •3927 = 589-05, the solidity. 2. The greater diameter of the frustum of a parabolic conoid is 20, the less 10, and the height 12 ; what is the solidity? Ans. 2357*4. 3. The greater diameter of the frustum of a parabolic conoid is 30, the less 10, and the height 50; required the solidity? Ans. 19635. 4. The greater diameter of the frustum of a parabolic conoid is 15, the less 12, and the height 8 ; required the solidity ? Ans. 1159-8408. PROBLEM XXIII. To find the solidity of a parabolic spindle. Rule. Multiply the square of the middle diameter by -7854, and that product by the length; then ^ of this product will be \he solidity, t 1. The middle diameter CD, of a parabolic ^^-^"^ ^"-^^ spindle is 10 feet, and the length A B is 40 ; re- A«^— — 4— — ^B quired its solidity ? D 103x •7854x40 = 3141-6 feet. Then ^^x 3141*6 = 1675-52 feet, the answer. 2. The middle diameter C D, of a parabolic spindle is 12 feet, and the length AB is 30; required the sohdity? Am. 1809-5616. 3. The middle diameter of a parabolic spindle is 3 feet, and the length 9 feet; required its solidity? Ans, 3392928. * See Appendix, Demonstration 84. t IbicL 85* 72 MENSURATION OF SOLIDS. 4. The middle diameter of a parabolic spindle is 6 feet, and the length 10; required its solidity ? Ans. 150'7908. 5. The middle diameter of a parabolic spindle is 30 feet, and the length 60; required its solidity ? Ans. 18849-6. PROBLEM XXIY. Tojind the solidity of the middle frustum of a parabolic spindle. Rule. To double the square of the middle diameter, add the square of the diameter of the end ; and from the sum subtract y*^ of the square of the difference between these diameters ; the remainder multiplied by the length, and that product by '2618, will be the solidity.* 1. In a parabolic spindle, the middle dia- meter of the middle frustum is 16, the least diameter 12, and the length 20 ; required the solidity of the frustum? Here 2 x 16^ + 122- 3*5- x 42= 512 + 144— 6-4 = 649-6 ; hence 649*6 x 20 x '2618 = 3401-3056, the solidity. 2. The bung diameter of a cask is 30 inches, the head diam.eter 20 inches, and the length 40 inches ; required its content in ale gallons, allowing 282 cubic inches to be equal to one gallon? Ans. 80-211 gallons. 8. The bung diameter of a cask is 40 inches, the head diametei BO inches, and tlie length 60 ; how many Mnne gallons does it con- lain, 231 cubic inches being equal to one gallon ? Ans, 276-08 gallons. PROBLEM XXV. Tofnd the solidity of a hyperbolic conoid. Rule. To double the height of the solid add three times the trans- verse axis, multiply the sum by the square of the radius of the base, and that product by the E height, and this last product by -5236; the result divided by the sum of the height and transverse axis, will give the solidity.'!' 1. Required the solidity of an hyperbolic conoid, whose height Y m is 50, the diameter A B 103-923048, and the transverse axis VEIOO? H«re(2x 50 + 3 X 100) X (103-923048)^ = 400x 2700 = 1080000; and 1080000 X 50 X -5236 150 : 188496, the solidity. 2. What is the content of an hyperboloid, whose altitude is 10| the radius of its base 12, and the transverse 30 ? Ans. 2073-451151369. • See AiBDendix, Demonstration dfi. t Ibid. 87. MENSURA'nON OF SOLIDS. 73 PROBLEM XXVI. To find the solidity of the frustum of an hyperholoid, or hyperbolic conoid. Rule. To four times the square of the middle diameter, add the Bum of the squares of the greatest and least diameters; multiply the result by the altitude, and that product by -1309, for the solidity.* 1- Required the solidity of the frustum A C E H I) B of an hyperbolic conoid, whose f»;reatest diameter A B is 96, least diameter E II 54, middle diameter C D 76*4264352, and the altitude m n 25? Here 4 C D^ + A B^ + E H^ = (5841 x 4) + 9216 + 2916 = 35496, and 35496 x 25 x -1309= 116160-66, the answer. 2. What is the solidity of an hyperboloidal cask, its bung diameter being 32 inches, its head diameter 24, and the diameter in the middle between the bung and head f V^IO, and its length 40 inches? Ans. 24998-69994216 inches. PROBLEM XXVII. To find the solidity oj a frustum of an elliptical spindle^ or awj other solid formed hy the revolution of^a conic section about Rule. Add together the squares of the greatest and least diameters, and the square of double tlie diameter in the middle between the two ; multiply the sum by the length, and the last product by '1309 tor the solidity.f 1. What is the content of the middle frustum C D I H of any spindle, the length OP being 40, the greatest, or middle iliameter E F 32, the least, or diameter at either end C D 24, and the diameter GK 30-157568? Here 322 +(2x30-157568)^+ 242 = 5237-89 sum; Then 5237-89x40 = 209515-6, and 209515-6 X -1309 = 27425-7, the answer. 2. What is the content of the segment of any spindle, the length being 20, the greatest diameter 10, the least diameter at either end 5, jaud the diameter in the middle between these 8? Ans. 997-458. ^ * See Appendix, Demonstration 88. t Ibid, m 74 MENSURATION OF SOLIDS. PROBLEM XXVIII. To find the solidity of a circular ring. Rule. To the thickness of the ring add the inner diameter; multiply the sum by the square of the thickness, and the product by 2-4674, for the solidity.* 1. The thickness of a cylindrical ring is 2 inches, and the diameter CDS inches ; required its solidity ? (2 + 5)x4=:28; then 28x2-4674 = 69-0872 cubic inches, the answer. 2. Required the solidity of an iron ring whose axis forms the circumference of a circle ; the diameter of a section of the ring 2 inches, and the inner diameter, from side to side, 18 inches ? Ans. 197-3925 cubic inches. 3. The thickness of a cylindrical ring is 7 inches, and the innev diameter 20 inches; required its solidity ? Ans. 3264-3702. 4. What is the solidity of a circular ring, whose thickness is 2 inches, and its diameter 12 inches ? Ans. 138*1744 cubic inches. * Bkso Appendix. Domonetration 90. THE FIVE REGULAR BODIES. 75 THE FIYE EEGULAE BODIES. SECTION V. DEFINITIONS. A regular body is a solid contained under a certain number of similar and equal plane figures. Only five regular bodies can possibly be formed. Because it is proved in Solid Geometry that only three kinds of equilateral and equiangular plane figures joined together can make a solid angle. 1. The tetraedron^ or equilateral pyramid, is a solid having four triangular faces.* 2. The Tiexaedron^ or cube, six square faces. is a solid having 3. The Qctaedron is a regular solid having eight tiiangular faces. * If figures similar to those annexed to the definitions, be drawn on paste- board, and cut out by cutting through the bounding liues, and if the othex lines be cut half through, and then the parts be turned up and glued together, the bodies defined will be formed. 76 THE FIVE REGULAR BODIES. 4r. The dodecaedron has twelve pentagonal faces. 6. The icosaedron has twenty triangular faces. PROBLEM I. To find the solidity of a tetra&dron. Rule I. Multiply -^ of the cube of the lineal side by the sqaare root of 2, and the product will be the solidity. Rule II. Multiply the cube of the length of a side of the body by tlie tabular solidity, and the product will be the solidity of the body.* This rule is general for all the regular bodies. 1. If the side of each face of a tetraedron be 1 ; C required its solidity? Here ^x l»x V2 = tVx V2=*11785113, the solidity. 2. The side of a tetraedron is 12; what is its / fL solidity? ^W5. 203*6467. /C^ ^^ \ PROBLEM II. To find the solidity of a hexaedron^ or a cube. Rule. Cube the side for its solidity.f 1. If the linear side of a hexaedron be 3, what is its content? Ans. 3x3x3 = 27. PROBLEM III. To find the solidity of an octaedron. Rule. Multiply the cube of the side by the square root of J?, and ^ of the product will be the content. J 1. What is the solidity of an octaedron, when the linear side iol? ♦ See Appendix, Demonstration 91. t Ibid. 61 t Ibid, 92. THE FIVE REGULAR BODIES. It^ 1' X V2 X J = ^2 = -4714045. 2. What is the solidity of the octaedron, whose linear side is 2 ? Ans, 3 "7712. PROBLEM IV. To find the solidity of a dodecaedron. Rule. To 21 times the square root of 5 add 47, and divide the gum by 40; multiply the root of the quotient by 5 times the cube of the lineal side, and the product will be the solidity.* 1. If the lineal side of the dodecaedron be 1, what is its solidity? Here A=l, consequently 5 A'V Tri" = 7-66311896, the content. 2. The side of a regular dodecaedron is 12 inches ; how many cubic inches does it con- tain? Ans, 13241-8694592. PROBLEM y. To find the solidity of an icosaedron. Rule. To 7 add three times the square root of 5, take half the sum, multiply the square root of this half sum by f of the cube oi the lineal side, and the product will be the solidity.+ 1. What is the solidity of an icosaedron, whose lineal side is 1? Let the side be denoted by A. Then A= 1, and consequently I AV^^ = iV^^t|V5 = 2-18169499, the content. 2, What is the solidity of an icosaedron, whose lineal side is 12 fuet? Ans, 3769-9689 feet. iVo/e. — The following table may be collected from the examples given in the forep:oing rules, each of which has been demonstrated under its particular head. It has also been demonstrated that the cube of the lineal side of any regular solid multiplied by the tabular number corresponding to the figure, will give its content. It is particularly recommended to the pupil to employ the general rule given in Problem I. whenever the content of any of the five rogular bodies is required. * See Appendix Demonstration 93. t Ibid. 91 THE FIVE REGULAR BODIES. TABLE IIL Shewing the solidity of the Jive regular bodies^ the length in each being 1. No of Bides. Names. Solidity. 4 6 8 20 12 Tetraedron Hexaedron Octaedron Icosaedron Dodecaedron •1178511 10000000 •4714045 2-1816950 7-6631189 PROBLEM VI. To find the surface * of a tetraedron. Rule I. Multiply the square of the linear side by the square root of 3, and the product will be the whole surface.f Rule II. Multiply the square of the length of a side of the body, by the tabular area corresponding to the figure, and the product will be the surface of the body. This is a general rule for finding the surfaces of the regular bodies. 1. If the side of a tetraedron be 1, what is its surface? Here, Px V3= V3 = 1*7320508 = the whole surface. 2. The side of a tetraedron is 12 ; what is its surface ? Ans, 249-415316^. PROBLEM VII. To find the surface of a hexaedron^ or cube. Rule. Square the side and multiply it by 6, and the product will be the surface.^ 1. If the side be 1, what is the surface of a hexaedron? 12x6 = 6 the whole surface. 2. If the side be 4, what is the surface of a hexaedron? Am. 9G. PROBLEM VIII. To find the surface of an octaedron. Rule. Multiply the square of the side by the square root of 3, and double the product will be the surface. § ♦ Though the next section treats exclusively of the surfaces of solids, and would therefore seum to be the proper place for this problem and the following ones in this section, yet it has been thouglit more convenient to place together the rules both for finding the solidities and surfaces of those curious bodies. t Sco Appendix, Demonstration 95. J Ibid. 96. § Ibid. Vt THE FIVE REGULAR BODIES. /9 1. If the side of an octaedron be 1, what is its surface? 2x 12V3 = 2 V3 = 3-4641016 = the wliole surface. 2. If the side of an octaedron be 12, what is its superficies ? Ans. 498-8300304. 3. If the side of an octaedron be 4, what is its surface ? Ans, 55-4256256. PROBLEM IX. 7 find the superficies of a dodecaedron. KULE. To 1 add \ of the root of 5 ; multiply tlie root of the sum by 15 times the square of the lineal side, and tlie product will be the surface.* 1. If the lineal side be 1, what is the surface of a regular dode- caedron V Herel2xl5V(l + IV5) = 15V(l + iV5) = 20-645728807,thesur- ace. 2. What is the surface of a dodecaedron, whose lineal side is 2? Ans. 82-58292. PROBLEM X. To find the superficies of an icosaedrou. Rule. Multiply five times the square of the lineal side by the equare root of 3, and the product will be thu surface.f 1. The side of an icosaedron is 1, what is its surface? SxPx V3 = 5V3 = 8-66025403. 2. What is the surface of an icosaedron whose side is 2? Ans. 34-641. 3. What is the surface of an icosaedron whose side is 3 ? Ans. 77-9423. j^ote. — In finding the superficial content of the regular bodies, it is particu- larly recommended to employ the general rule given in Problem VI. in prac- tice, in preference to any other. The particular rules given for each solid are introduced merely to find the tabular numbers by which the pupil is to work. From the examples given in the preceding rules, in which the lineal side of each regular solid is 1, the following tabular numbers may be collected. TABLE IV. Shewing the surfaces oj the five regular todies., when the linear side is 1. Number of sides. Names. Surface. 4 6 8 12 20 Tetraedron Hexaedron Octaedron Dodecaedron ... Icosaedron 1-7320508 6-0000000 3-4641016 20-6457288 8-6602540 " tJee Ai'pQudix, Demonstration 98. t ibid. 9d. 80 SLTIFACKS OF SOLIDS. SURFACES OF SOLIDS. SECTION VL H I PROBLEM I. To find the surface of a prism. Rule. Multiply the perimeter of the end of the solid by its length, ko the product add the area of the two ends, and the sum will be the surface.* 1. If the side H I of the pentagon be 25 feet, and /'Ax. height I D 10, what is its surface? A|< ^/ \^ '^ 25 X 5 = 125, the perimeter ; Then 125 x 10=1250 = the upright surface; 252 X 1-720477= 1075-298125 = tlie area of one end; And 1075-298125 x2 = 2150-596250=the area of both ends ; Then 2150-596250 + 1250 = 3400-59625 = the entire surface. 2. If the side of a cubical piece of timber be 3 feet 6 inches, what is the upright surface and whole super- ^ ficial content? ^ ^ A (49 feet upright surface. * (73 feet 6 in. whole superficial content. 3. If a stone in the form of a parallelopipedon be 12 feet 9 inches lon^, 2 feet 3 inches d«ep, and 4 feet 8 inches broad, what is the upright surface and whole superficial content ? A (176 feet 4 in. 6 sec. upright surface. * (197 feet 4 in. 6 sec. whole sup. content. PROBLEM n. To find the surface of a pyramid. Rule. Multiply the slant height by half the circumference of the base, and the product will be the surface of the sides, to which add the area of the base for the whole surface.f Note. — The slant height of a pyramid is the perpendicular distance from the vertex to the middle of one of the sides, and the perpendicular height is a Btraight line drawn from the vertex to the middle of the base, * See Appendix, Demonstration 100. t itld. lOL SURFACES OF SOLIDS. 81 1. The slant height of* a triangular pyramid is 10 feet, and each side of the base is 1 ; what is its surface ? Half circumference = f Slant height =10 Upright surface = 15 Area of the base = -433013 The entire surface = 15-433013 B 2. The perpendicular height of a heptagonal pyramid is 13*5 feet, and each side of the base 15 inches ; required its surface? Ans. 65-0128 feet. PROBLEM III. To find the surface of a cone. Rule. Multiply the slant height by half the circumference of the base, and the product, with the area of the base, will be the whole surface.* 1 . What is the surface of a cone whose side is 20, and the circumference of its base 9 ? Here 20x|=90=the convex surface. 92 X -07958 = 6-44598 = the area of the base. Then 90 + 6-44598 = 96-44598 = the whole sur- face, 2. The perpendicular height of a cone is 10-5 feet, and the circumference of its base is 9 feet; what is its superficies? Ans, 54-1336 feet. PROBLEM IV. To find the superficies of the frustum of a rights regular pyramid. Rule. Add the perimeters of the two ends together, and multiply half the sum by the slant height, the product will be the upright surface ; to which add the areas of both ends, and the sum will be the whole surface. f 1. What is the superficies of the frustum of a square pyra- mid, each side of the greater base A B being 10 inches, and each side of the less base C D 4 inches, and slant height 20 inches? Here 10 x 4 = 40 the perimeter of the greater base. And 4 X 4 = 16 the perimeter of the less end. Sum 56, the half of which is 28. Then 28 x 20 = 560 = the upright surface. 10 X 10= 100 = the area of the greater base. 4 X 4= 16 = the area of the less end. Hence 560+ 100+ 16 = 676 = the whole surface. 2. What is the superficies of the frustum of an octagonal pyramid, each side of the greater base being 9 inches, each side of the less base 5 inches, and the height 10'5 feet? Ans. 52-59 feet. * See Appendix, Demonstratiou 10i2. f Ibid. 103. 82 SURFACES OF SOLIDS. PROBLEM V. To find the superficies of the frustum of a cone. Rule. Add the perimeters of both ends together, and multiply half the sum by the slant height, to which add the areas of both ends, for the whole superficies.* ^ ^^rr^^ a 1. If the diameters of the two ends CD and AB are 7 and 3, and the slant height D B 9, what is the whole surface of the frustum A B C D ? ^^x 3-1416x9 = 141-372, the convex surface, d 7x7x -7854 = 38-4846, the area of the base CD 3x3x-7854= 7-0686, the area of tlie end A B. Then 141-372 + 45-6532= 186-9252 = the whole surface of the frustum. 2. What is the superficies of the frustum of a cone, whose greater diameter is 18 inches, and less diameter 9 inches, and the slant height 171-0592 inches? Ans, 7672-981. PROBLEM VL To find the superficies of a wedge. Rule. Find the area of the back, which is a right*angled paral- lelogram; find the areas of both ends, which are triangles; and also of both sides, which are trapezoids ; all these areas added together will evidently be the whole surface.f ^ ^ 1. The back of a wedge is 10 inches long, dnd 2 inches broad, each of its faces is 10 inches from the edge to the back ; required its whole surface? 10x2 = 20= the area of the back. 10 x 10 X 2 = 200 the areas of both faces. V(A£2-Ex2)z=V(100--l) = 9-949 = Aic;then J 9-949 X 2= 19-898 = areas of both ends. * Hence 200 + 20 + 19-898 = 239-898 = the whole surface of the wedge. 2. The back of a wedge is 20 inches long, and 2 inches broad ; each of its faces is 10 inches from the back to the edge; what is its whole surface? Ans. 459*898. PROBLEM VIL To find the area of the frustum of a wedge. Rule. Find the areas of the back and top sections ; of the two faces ; and of the two ends ; the sums of all the separate results will evidently be the whole surface. * See Appendix, Demonstration 104. t Ibid* 106. SURFACES OF SOLIDS. 83 1. The length and breadth of the back are 10 and 2 inches, the length and breadth of the upper section are 10 and 1 inches, the length of the edge from the back to the upper section is 10 inches; required the whole surface? 10x2 = 20 = the area of the back. 10 X 1 = 10 = the area of the upper section. 10 X 10 X 2 = 200= the areas of both faces. ?^=i = -5, andV(100--25) = 9-98 = B2^. Then(2 + l)x9-98 = 29-94 = areasofbothends. Hence 20 + 10+ 200+ 2994 = 259-94 inches, the answer. 2. The length and breadth of the back are 10 and 4, the length and breadth of the upper section are 5 and 2, and the length of each of the faces is 20; required the whole superficies? Ans, 470* 78» PROBLEM VIII. To find the surface of a glohe or sphere. Rule. Multiply the diameter of the sphere by its circumference, and the product will be its convex surface.* 1. What is the surface of a globe, whose diameter is 24 inches ? 24x3-1416 = 75-3984, the circumference: 75-3984x24=1809-5616 inches, the answer. 2. What is the surface of the earth, its diameter being 7957|, and circumference 25000 miles? Ans. 198943750 square miles. PROBLEM IX. To find the convex surface of any segment, or zone of a sphere. Rule. Multiply the circumference of the whole sphere by the neight of the segment, or zone, and the product will be the convex surface.f 1. If the diameter of the earth be 7970 miles, the height of the frigid zone will be 252-361283 miles, what is its surface? Here 7970 x 3-141 6 = the circumference ; then 7970 X 3-1416 x 252-361283 = 6318761-107182216 miles. 2. If the diameter of the earth be 7970 miles, the height of the temperate zone will be 2143*6235535 miles ; what is its surface? Ans. 53673229-812734532 miles. 3. If the diameter of the earth be 7970 miles, the height of the torrid zone will be 3178 030327 miles; what is its surface? A?is. 79573277-600166504 miles. Note. — By adding the surfaces of both frigid zones and both temperate zonea^ to the surface of the torrid zone, the sum 199557259 -44, is the surface of the earth in square miles. 4. The diameter of a sphere is 3, the height of the segment 1 ; what is its convex surface ? Ans. 9 4248. 5. The circumference of a sphere is 33, the height of the segment la 4 ; what is its convex surface? Ans. 132. * Bee Appendix, Demonstration 107, t Ibid. 108. 84 SURFACES OF SOLIDS. PROBLEM X. To find the surface of a cylinder. Rule. Multiply the circumference by the length, and the product will be the convex surface ; to which add the area of the two ends, and the sum will be the surface of the entire solid.* 1. What is the entire surface of a cylinder, whose length is 10 feet, and its diameter 5 feet ? 3-1416 5 15-7080, then 15-708 x 10=157'08 the convex surface. 5 X 5 X -7854: = the area of the base ; then 2 X 5 X 5 X -7854 = 50 x -7854 = 39 -2700 = the area of both bases; then 157*08 + 39-27= 196-35, the answer. 2. Required the superficial content of a cylinder, whose diameter is 21-5 inches, and height 16 feet? Ans. 95-1 feet. 3. What is the surface of a cylinder whose diameter is 20-75 inches, and its length 65 inches? Ans. 29*595 feet. PROBLEM XL 2^0 find the superficies of a circidar cylinder. Rule. Add the inner diameter to the thickness of the ring, multi- ply the sum by the thickness, and that product by 9*8696 for the superficies, t 1. The thickness AC of a cylindrical ring is 2 inches, the inner diameter CD 5 inches ; required its superficial content ? Here (2 + 5) x 2=14 ; then 14 x 9-8696 = 138-1744 square inches. PROBLEM XIL To find the surface of a parallelopipedon. Rule. Find the area of the sides and ends, and their sum will be the surface. 1. What is the surface of a parallelopipedon, whose length is 10 feet, breadth 4, and depth 2 ? Ans. 136 teet. 10 X 4 = 40=the area of one face. 10x4=40 = the area of its opposite face. 10x2 = 20 = the area of one face. 10 X 2 =20= the area of its opposite face. 4x2= 8 = the area of one end. 4x2= 8 = the area of its opposite end. 136=the surface of the whole solid. 2. The length of a parallelopipedon is 5, breadth 4, and depth 3 , v/hat is its S'arface ? Ans. d^. * &QQ Appendix, Demonstration 109. % I^id. 110. MENSURATION OP TIMBER AND OF ARTIFICERS' WORK, 85 MENSUEATION OF TIMBER AND OF AETIFIOERS' WORK. SECTIOlSr VII, DESCEIPTION OF THE CARPENTERS' RULE. ^ This instrument is sometimes called the sliding rule, and is usea in measuring timber and artificers' work. By it dimensions are taken, and contents computed. It consists of two eq^ual pieces of box-wood, each one foot long, connected by a folding joint.. One face of the rule is divided into inches and half-qnarters or eighths. On the same side or face are several plane scales divided by diagonal lines into twelfths; these are chiefly used in planning dimensions which are taken in feet and inches. The edge of the | rule is divided decimally ; that is, each foot is divided into 10 equal I parts, and each of these again into 10 equal parts. By means of this I last scale, dimensions are taken in feet, tenths, and hundredths ; and I then multiplied as common decimal numbers. In one of these equal pieces, there is a slider on which are marked the two letters B, C; on the same face are marked the letters A, D. The same numbers serve for both these two middle lines, the one being above the numbers, and the other below. Three of these lines, viz.. A, B, C, are called double lines, as they proceed from I to 10 twice over. These three lines are exactly alike l30th in division and numbers, and are numbered from the left hand towards the right, 1, 2, 3, 4, 5, 6, 7, 8, 9 to 1, which stands in the middle ; the numbers then go on, 2, 3, 4, 5, 6, 7, 8, 9 to 10, which stands at the right-hand end of the rule. These four lines are logarithmic ones ; the lower line D, is a single one, proceeding from 4 to 40, and is called the girt line, from its use m finding the content of timber. Upon it are also marked W G at 17*15, A G at 18*95, and I G at 18*8. These are the wine, ale, and imperial gauge points. On this face is a table of the value of a load, or 50 cubic feet, of timber, at all prices from 6 pence to 2 shillings per foot. To ascertain the values of the figures on the rule, which have no determinate value of their own, but depend upon the value set on the anit at the left hand of that part of the rule marked 1, 2, 3, &c. : if 86 MENSURATION OF TIMBER AND OF ARTIFICERS WORK. » the first unit be called 1, the 1 in the middle will be 10, the other figures that follow will be 20, 30, 40, &c., and the 10 at the right- hand end will be 100. If the left-hand unit be called 10. the 1 in the middle will be 100, and the following figures will be 200, 300, 400, 500, &c. : and the 10 at the right-hand end mil be 1000. L the 1 at the left-hand end be called 100, the middle 1 will be 1000, and the following figures will be 2000, 3000, 4000, &c., and the 10 at the right hand will be 10,000. From this it appears that the values of all the figures depend upon the value set on the first unit. The use of the double line A, B, is to find a fourth proportional, and also to find the areas of plane figures. The use of the several lines described here is best learned in practice. If the rule be unfolded, and the slider moved out of the grove, the back part of it will be seen divided like the edge of the rule, all measuring 3 feet in length. Some rules have other scales and tables delineated upon them; such as a table of board measure, one of timber measure, another foi Bhewing what length for any breadth will make a square foot. There is also a hue shewing whal length for any thickness will make a soli^ foot. — ^--- '" " ' '^"^'— ' .,^«,«.*.«*«!*,...*«K. THE USE OF THE SLIDING EULE. PROBLEM I. To multiply numbers together. Set 1 on B to the multiplier on A ; then against the multiplicanci on B, stands the product on A. 1. Multiply 12 and 18 together. Set 1 on B, to 12 on A ; then against 18 on B, stands the product 216 on A. 2. Multiply 36 by 22. Set 1 on B, to 36 on A ; then as 22 on B goes beyond the rule, look for 2-2 on B, and against it on A stands 79*2 ; but as the real multiplier was divided by 10, the product 79*2 must be multij)lied by 10, which is effected by taking away the decimal point, leaving the product 792. PROBLEM II. To divide one number by another. Set the divisor on A, to 1 on B ; then against the dividend on A, stands the quotient on B. 1. Divide 11 into 330. Set the divisor 11 on A, to 1 on B ; then against the dividend 330 on A, stands the quotient 30 on B 2. Divide 7680 by 24. THE USE OF THE SLIDING RULE. 87 Set 24 on A, to 1 on B ; then because 7680 goes beyond the rule on A, look for 768 (the tenth of 7680), on A, and against it stands 82 on B ; but as the tenth of the dividend was taken that the number should fall within the compass of the scale A, the quotient 32 must be multiplied by 10, which gives 320 for the answer. PROBLEM III. To square ariy number. Set 1 upon C, to 10 upon D ; then if you call the 10 upon D 1, the 1 on C will be 10 ; if you call the 10 on D, 10, then the 1 on C will be 100; if you call the 10 on D, 100, then the 1 on C will be 1000 ; this being understood, you will observe that against every number on D, stands its square on C. 1. What are the squares of 25, 30, 12, and 20? Proceeding according to the above direction, 625 stands against 25, 900 against 30, U4 against 12, 400 against 20. PROBLEM IV. To extract the square root of a number. Set 1 or 100, &c., on C, to 1 or 10, &c., on D ; then against every number found on C, stands its root on D. 1. What are the square roots of 529 and 1600? Proceeding according to the above directions, opposite 529 stands 23 J opposite 1600 stands 40, and so on. PROBLEM V. To find a mean proportional between two numbers^ as 9 aiid 25. Set the number 9 on C, to the same 9 on D ; then against 25 on V, stands 15 on D, the required mean proportional. The reason of this may be seen from the proportion, viz., 9: 15 :: 15:25. 1. What is the mean proportional between 29 and 430 ? Set one number 29 on C, to the same on D ; then against the other number 430 on C, stands 112 on D, which is the mean proportional, nearly. PROBLEM VL To find a third proportional to two numbers^ as 21 and 32. Set the first number 21, on B, to the second, 32, on A; then against the second, 32, on B, stands 48 8 on A, which is the required third proportional. PROBLEM VII. To find a fourth proportional to three given numbers. Set the first term on B, to the second on A ; then against the third term on B, stands the fourth on A. If either of the middle numbers fidl beyond the line, take one-tenth part of that number, and increase the fourth number found, ten times. 1. Find a fourth proportional to 12, 28, and 114. Set the first term, 12, on B, to the second term, 28, on A; then against the third term, 114, on B, stands 266 on A, which is the answer. HENSURATION OF TIMBER AND OF ARTIFICERS' WORK. TIMBEE MEASUEE. PROBLEM I. To find the superficial content of a hoard or plank. Rule. Multiply the length by the breadth, ar.d the product will be the area. Uote. — When the plank is broader at one end than at the other, add both ends together, and take half the sum for a mean breadth. BY THE carpenters' RULE. Set 12 on B, to the breadth in inches on A; then against the length in feet, on B, will be found the superficies on A, in feet. 1. If a board be 12 feet 6 inches long, and 2 feet 3 inches broad, how many feet are contained in it ? 12 . 6 12-5 2 . 3 2-25 25 . 625 3.1.6 250 28.1 . 6 Ans. ^^^ 28-125 Ans, BY THE carpenters' RULE. As 12 on B : 27 on A : : 12-5 on B : 28-125 on A. 2. What is the value of a board whose length is 8 feet 6 inches, and breadth 1 foot 3 inches, at 5c?. per foot? Ans. 45. 5c?. 3. What is the value of a board whose length is 1 2 feet 9 inches, and breadth 1 foot 3 inches, at 5o?. per foot? Ans. Qs. l\d, 4. What is the value of a plank whose breadth at one end is 2 feet, and at the other end 4 feet, at Qd. per foot, the length being 12 feet? Ans. 185. 5. How many square feet in a board, whose breadth at one end is 15 inches, and at the other 17 inches, the length being 6 feet? Ans. 8. 6. How many square feet in a plank, whose length is 20 feet, and mean breadth 3 feet 3 inches ? Ans, 65. PROBLEM II. To find the solid content of squared or four sided timber. Rule. Take half the sum of the breadth and depth in the middle, (that is, the quarter girt,) square this half sum, and multiply it by the length for the solid content.* * This rule, which is generally employed in practice, is far from bcina correct, when the breadth and depth differ materially from each other, and the Limber does not taper. TIMBER MEASURE. 89 BY THE carpenters' RULE. As 12 on D : length on C : : quarter girt on D : the solid content onC. 1. If a piece of squared timber be 3 feet 9 inches broad, 2 feet 7 inches deep, and 20 feet long ; how many solid feet are contained therein ? 3.9 2. 7 2)6 .4: 3 . 2 quarter girt. 3. 2 , 6 6.4 10 . . 4 square of the quarter girt. 20 length of the piece. 200 . 6 . 8 solid content. BY THE carpenters' RULE. As 12 on D : 20 on C : : 38 on D : 200^ on C. 2. A squared piece of timber is 15 inches broad, 15 inches deep, and 18 feet long; how many solid feet does it contain? Ans. 28^ feet, wliich is the accurate content, as the breadth and depth are equal. 3. What is the solid content of a piece of timber, whose breadth is 16 inches, depth 12 inches, and length 12 feet? Ans. 16 feet. Rule II. Multiply the breadth in the middle by the depth in the middle, and that product by the length, for the solidity.* 4. The length of a piece of timber is 18 feet 6 inches ; the breadths at the greater and less end 1 foot 6 inches, and 1 foot 3 inches, and the thickness at the greater and less end 1 foot 3 inches, and 1 foot ; what is the solid content ? 1-5 ^ 1-25 1-25 ■** 1 2)2-75 2)2-25 1*375 mean breadth. 1*125 mean depth. 1-125 mean depth. 1 375 mean breadth. 1-546875 18-5 length. 28-6171875 solid content. * This rule is correct when the timber does not taper; but when the timber takers considerably, and the breadth and depth are neany equal, the rule is very erroneous. The measurer, therefore, ought to consider the shapo of tha timber he is about to measure before he appUes either of the above rules. 90 MENSURATION OF TIMBER AND OF ARTIFICERS' WORK., BY THE SLIDING RULE. B A B A As 1 : 13i: : 16i : 223 the mean square. CD CD As 1 : 1 :: 223 : 14-9 quarter girt. C D D C As 184 : i2 : :14'9 : 28-6 the content. Note. — When the piece to be measured tapers regularly from one end to th» other, either take the mean breadth and depth in the middle, or take the dimensions at both ends, and half their sum for the mean dimension. This, however, though very easy in practice, is but a very imperfect approximation. When the piece to be measured does not taper regularly, but is thick in some i^arts and small in others, in this case take several dimensions ; add them all together, and divide their sum by the number of dimensions so taken and use the quotient as the mean dimension. Rule III. Multiply the sum of the breadths of the two ends by the sum of the depths, to which add the product of the breadth and depth of each end ; one-sixth of this sum, multiplied by the length, will give the exact solidity of any piece of squared timber tapering regularly.* 5. How many feet in a tree, whose ends are rectangles, the length and breadth of one being 14 and 12 inches, and the corresponding dimensions of the other 6 and 4 inches ; also the length 30^ feet ? 14 12 12x14=168 6 4 6x 4= 24 20x16 = 320 612 square inches =-^ square feet. Then i x y x 30| = 18^ feet, the solidity. 6. How many solid inches in a mahogany plank, the length and breadth of one end being 91| and 55 inches, the length and breadth of the other end 41 and 29^ inches, and the length of the plank 47i inches? Ans, 126340-59375 cubic inches. PROBLEM III. Given (he breadth of a rectangular plank in inches^ to find how much in length will make afoot^ or any other required quantity. Rule. Divide 144, or the area to be cut off, by the breadth in inches, and the quotient will be the length in inches. The Carpenters' Rule is furnished with a scale which answers the • This rule is correct, being that given for finding the solidity of the pria- moid — which see. Let B and 6 be the breadths of the two ends, D and d the depths, and L the length : | (B D + (B + 6) x (D + d) + 6 rf) x L =the true solidity, a^ iu the rule for the prismoid. 20 16 TDIBER MEASURE. 91 purpose of this rule. It is called a Table of Board Measure, and ia in the following form : — 5 Si 6 Inches. 12 6 4 3 2 2 1 1 Feet. 1 2 3 4 6 6 7 1 8 Breadth. If the breadth be 1 inch, the length standing against it is 12 feet ; if the breadth be 2 inches, the length standing against it is 6 feet ; ir the breadth be 5 inches, the length is 2 feet 5 inches, &c. When the breadth goes beyond the limits of the table on the rule, it must be shut, and then you are to look for the breadth in the line of board measure, which runs along the rule from the table of board measure, and over against it on the opposite side, in the scale of inches, will be found the length required. For example, if the breadth be 9 inches, you will find the length against it to be 16 inches; if the breadth be 11 inches, the length will be found to be a little above 13 inches. 1. If a board be 6 inches broad, what length of it will make a square foot? Ans. 2 feet. 2. If a board be 8 inches broad, whi^ length of it will make 4 square feet? Ans. 6 feet. 3. If a board be 16 inches broad, what length of it will make 7 square feet ? ■ Ans. 5^ feet. When the board is broader at one end than at the other, proceed according to the following Rule. To the square of the product of the length, and narrow end, add twice the continual product of tuese quantities, viz., the length, the difference between the breadths of the ends, and the area of the part required to be cut off; extract the square root of the sum ; .from the result deduct the product of the length and narrow end, and divide the remainder by the difference betw^een the breadths of the ends.* If it were required to cut off 60 square inches from the smaller end of a board, A D being 3 inches, C E 6 inches, and A B 20 inches. 1 Here kx= -(V{(ABxAD)2 + > ' B s 1 ^r-—--^.^ B "2 BG^ 4BCxABx60}-ABxAD) = J (V{(20x3)2 + 6x20x60}~20x3) = 14-64, the length required. ' PROBLEM IV. To find how much in length will make a solid foot., or any other required quantity., of squared timber ., of equal dimensions from end to end. Rule. Divide 1728, the solid inches in a foot or the solidity to be cut off, by the area of the end in inches, and the quotient will be the end in inches. • See Appendix, Demonstration 111. 92 MENSURATION OF TIMBER AND OF ARTIFICERS' WORK. 1. If a piece of timber be 10 inches square, how much in length will make a solid foot? 10 X 10=100 the area of the end ; then 1728^100 = 17*28. Ans, 2. If a piece of timber be 20 inches broad, and 10 inches deep, how much of it will make a solid foot? Ans. 8^ inches. 3. If a piece of timber be 9 inches broad, and 6 inches deep, how much of it will make 3 solid feet? Ans. 8 feet. On some Carpenters' Rules, there is a table to answer the purpose t'lf the last rule ; it is called a Table of Timber, and is in the following form : — 11 3 9 Inches. ' 144 36 16 9 5 4 2 2 1 Feet. 1 2 3 4 5 6 7 8 9 Side of square. PROBLEM V. To find the solidity of round or unsquared timber. Rule I. Gird the piece of timber to be measured round the middlr with a string, take one-fourth part of tlie girth, and square it, an multiply this square by the length for the solidity. BY THE SLIDING RULE. As the length on C : 12 or 10 on D : : quarter girt, in 12ths or lOths on D : content on C. Note — When the tree is ^eiy irregnlar, divide it into several lengths and find the solidity of each part separately; or add all the girths together, and divide the sum by the number of them. 1. Let the length of a piece of round timber be 9 feet 6 inches, and its mean quarter girt 42 inches ; what is its content ? 3.5 quarter girt. 3 . 6 quarter girt. 3.5 3.6 12-25 9-5 length. 10 1 6 9 116-375 content. 12 9 3 6 length. 110 6 3 1 .6 116 . 4 . 6 content. BY THE SLIDING RULE. As 9-5 on C : 30 on D : : 35 on D : 116| on C ; Or 9-5 : 12 :: 42 : iie^. Rule II. Multiply the area corresponding to the quarter girt in inches, by the length of the piece in feet, and the product will be tho solidity. Note. — It may sometimes happen that the quarter girt exceeds the limits of Ihe table : in this case, take half of it» aud four times the content thus found will give the required coutenc. TIMBER MEASURE. A TABLE FOR MEASURING TIMBER. 93 Quarter Girt. Area. Feet. 250 272 294 ■317 340 364 390 417 444 472 •501 531 562 594 626 659 694 730 766 803 840 ■878 918 959 Quarter . Girt. ^'^®*- Feft. 1-000 1-042 1-085 1-129 1-174 1-219 1-265 1-313 1-361 1-410 1-460 1-511 1-662 1-615 1-668 1-722 1-777 1-833 1-890 1-948 2-006 2-066 2-126 2-187 Quarter Girt. Inches. 18 18^ 19 19i 20 20i 21 21J 22 22i 23 23i 24 25 25J 26 26A 27 27J 28 28J 29 29^ Fett. 2-250 2-376 2 506 2-640 2-777 2-917 3-062 3-209 3-362 3.516 3-673 3-835 4-000 4-168 4-340 4-516 4-694 4-876 5-062 6-252 6-444 6-640 5-840 6-044 2. If a piece of round timber be 10 feet long, and the quarter girt I2i inches; required the solidity ? Ans. lO'^b. To find the solid content by this table, look for the quarter girt 12 J, in the column marked Quarter Girt, and in adjoining column, marked Area, will be found 1-085, which multiplied by the length, 10 feet, will give 10-85 feet for the solid content. 3. A piece of round timber is 20 feet long, and the quarter girt 14^; how many feet are contained therein? Ans. 28*2 feet. 4. How many solid feet are contained in a tree 40 feet long, its quarter girt being 9 inches? Ans. 22*48. 5. How many solid feet in a tree 32 feet long, its quarter girt being 8 inches. Ans. 14-208. 6. How many solid feet in a tree 8J feet long, its quarter girt being 7i inches. Ans. 3-316 feet. ^4 MENSURATION OF TIMBER AND OF ARTIFICERS' WORK. 7. Required the content of a tree, whose length is 40 feet, and qiiai'ter girt 27i inches? Ans. 210*08 feet. 8. What is the content of a tree, whose length is 30 feet 6 inches, and quarter girt 27^ inches? Ans. 160-186 feet. 9 Required the content of a piece of timber, whose length is 25 feet 9 inches, and quarter girt 12| inches? Ans. 29 071 feet. 10. What is the solid content of a piece of timber, whose length ia 12 feet, and quarter girt 13^ inches? Ans. 15*18 feet. 11. What is the solid content of a piece of timber, whose quarter girt is 14f inches, and length 38 feet? Ans. 57*418 feet. When the square of the quarter girt is multiplied by the length, the product gives a result nearly one-fourth less than the true quantity m the tree. This rule, however, is invariably practised by timber merciiants, and is not likely to be abolished. W^hen the tree is in the form of a cylinder, its content ought to be found by Prob. lY. Sec. IV., which gives the content greater than that found by the last rule, nearly in the proportion of 14 to 11. Notwithstanding that the true content is not found by means of the square of the quarter girt, yet some allowance ought to be made to the purchaser on account of the waste in squaring the wood so as to be fit for use. If the cylindrical tree be reckoned no more than what the inscribed square will amount to, the last rule, which is said to give too little, gives too mucli. When the tree is not perfectly circular, the quarter girt is always too great, and therefore the content, on that account, will be too great. Dr Button recommends the following rule, which will give the Dontent extremely near the truth : — Rule. Multiply the square of one-fifth of the girt, or circumfer- ence, by twice the length, and the product will be the content. BY THE SLIDING RULE. As double the length on C : 12 or 10 on D : : ^ of the girt, in I2ths or lOths on D : content on C. 12. Required the content of a tree, its length being 9 feet 6 inches^ and its mean girt 14 feet. 14-5 = 2-8 ft. 2-8 9. 2*8 ft i = 2 in. 6 2 . in. p. .9. 7= J- of the girt; then 2.9.7 2.9.7 7-84 19 . 19 • 6 : 6.7. 2 2.1. 2 . 3 1 . 7. 7.1 148*96 content. 7. 9 . 11 . 10. 1 19 C As 19 : 10 :: 28 Or 19 : 12 :; 33- 148 . 9 . 8 . 11 . 7 content. D D C 149 content by the Sliding Rule, 149 content without it. TIMBER MEASURE. 96 Dr Gregory recommends the following rules given by Mr Andrews : — Let L denote the length of the tree in feet and decimals, and G the mean girt in inches. Rule I. Making no allowance for bark. L Gr^ T r ^ 2oqt= cubic feet, customary; and -^—= cubic feet, true content. Rule II. Allowing J for bark. L G^ L r^ gTr^= cubic feet, customary; oufiO^^^^^^^ ^^^*' *"^® content. Rule III. Allowing ^ for bark. LQ.2 LG^ ^g|^= cubic feet, customary; 22oT= cubic feet, true content. Rule IV. Allowing y\ for bark. 2^^-=^ cubic feet, customary; ^y—= cubic feet, true content. What is the solid content of a tree, whose circumference, or girt, is 60 inches, and length 40 feet ? By Rule I. QOQ4 = o2J cubic feet, customary. 40x602 ^^„ ,. n . — -.orvy = 79 f cubic feet, customary. By Rule II. 40x60^ ^,7 OK 1. r . . -^^Q— = 47*85 cubic feet, customary. 40x602 ^, ;. r ^ , gRfio ~ cubic feet, true content. By Rule HI. • ^^.^ =50-61 cubic feet, customary 40x60* nt ^A v.- c ^ . • =64-54 cubic feet, true content By Rule IV. 40x602 ^„ _ , • r . ■ =52-47 cubic feet, customary- 40 X 60^ 2150 = 66-97 cubic feet, true content. Then rlie two ends are very unequal, calculate its content by the I given for finding the solidity of the frustum of a cone, and tict the usual allowance from the result. When it is required to find the accurate content of an irregular body, not reducible to any figure of which we have already treated, provide a cylindrical or prismatic vessel, capable of containing the a 96 MENSURATION OF ARTIFICERS' WORK. solid to be measured ; put the solid into the vessel, and pour in water to cover it, marking the height to which the water reaches. Then take out the solid, and observe how much the water has descended in consequence of its removal ; calculate the capacity of the part of the vessel thus left dry, and it will evidently be equal to the solidity of the body whose content is required. AETIFICERS' WOEK. Artificers compute their works by several different measures : — Glazing and masonry by the foot. Plastering, painting, paving, &c., by the yard of 9 square feet. Partitioning, roofing, tiling, flooring, &c., by the square of 100 square feet. Brick work is computed, either by the yard of 9 square feet, or by the perch, or square rood, containing 272^ square feet, or 30^ square yards ; 272^ and 30;^ being the squares of 16^ feet and 5 J yards re- spectively. CAEPENTEES' AND JOINEES' WOEK OF FLOORING. To measure joists, multiply the breadth, depth, and length toge- ther for the content.* 1. If a floor be 50 feet 4 inches long, and 22 feet 6 inches broad; how many squares of flooring are in that room? 50-333 60 . 4 22-5 22 . 6 251665 1107 . 4 100666 25 . 2 100666 100)1132-4925 100)11-32 . 6 11-3249 squares. 11-325 Ans. 11 squares 32^ feet. 2. If a floor be 51 feet 6 inches long, and 40 feet 9 inches broad; low many squares are contained in that floor? Ans. 20*986 squares. 3. If a floor be 36 feet 3 inches long, and 16 feet 6 inches broad, how many squares are contained in that floor ? Ans. 5 squares 98g feet. 4. If a floor be 86 feet 11 inches long, and 21 feet 2 inches broad ; how many squares are contained in it? Ans. 18-3972. 5. In a naked floor the girder is 1 foot 2 inches deep, 1 foot broad, and 22 feet long ; there are 9 bridgings, the scantling of each (viz., * Joists receive various names from their position ; such as girders, bindingf- joists, ti-imming-joists, common-joists, ceiling-joists, &c. When girders and joists of flooring are designed to bear considerable weight, they should be lot irto the wall at each end about two-thirds of the thickness of the waU. carpenters' and joiners* work 97 bieadth and depth) being 3 inclies, by 6 inches, and length 22 feet; 9 binding joists, the length of each being 10 feet, and scantlings 8 inches by 4 inches ; the ceiling joists are 25 in number, each 7 feet long, and their scantlings 4 inches by 3 inches ; what is the solidity of the whole? ^^^5. 85 feet. 6. What would the flooring of a house three storeys high come to, at £5 per square ; the house measures 30 feet long, and 20 broad ; there are seven fire-places,* two of which measure, each 6 feet by i feet ; two others, eacli 6 feet by 5 feet 6 inches ; two, each of 5 feet 6 inches by 4 feet ; and the seventh 5 feet by 4 ; the well-hole for the stairs is 10 feet by 8 ? Ans, £69, 2s. OF partitioning. Partitions are measured by squares of 100 feet, as flooring ; their dimensions are taken by measuring from wall to wall, and from floor to floor ; then multiply the length and height for the content in feet, which bring to squares by dividing 100, as in flooring. When doors and windows are not included by agreement, deductions must be made for their amount, f 1. A partition measures 173 feet 10 inches in length, and 10 feet 7 inches in height ; required the number of squares in it ? Ans. 18*3972 squares. 2. A partition between two rooms measures 80 feet in length, and 60 feet 6 inches in height ; how many squares in it? Ans. 40f squares. 3. If a partition measure 10 feet 6 inches in length, and 10 feet 9 inches in height ; how many squares in it? Ans. 1 square 12^ feet. 4. What is the number of squares in a partition, whose length is 50 feet 6 inches, and height 12 feet 9 inches ? Ans. 6 squares, 43 feet, 10| inches. In roofing, the length of the rafters is equal to the length of a string stretched from the ridge down the rafter till it meets the top of the wall. To find the content, multiply this length by the breadth and depth of the rafters, and the result will be the content of one rafter ; and that multiplied by the number of them will give the content of all the rafters 4 1. If a house within the walls be 42 feet 6 inches long, and 20 feet 3 inches broad ; how many squares of roofing in that house ? * Fire-places, Yhen the height of the dome is equal to the radius of its base, (the curved sides being circular or elliptical quadrants), or to half the mean proportional between the two axes of its elliptical base, the above i-ule will answer pretty vrcll ; but with any other dimensions it ought not to be used. * This rule is correct only when the dome is circular, and its height oqual to the radius of the base.— See Appendix. Demonstration 112. VAULTED AND ARCHED ROOFS. 113 arch of 2 feet radiuvS, springing over a rectangular room of 20 feet long and 16 feet wide? 23 X -7854 = 3-1416 = area of the quadrant CDAF. 2 x2-^2 = 2 = areaofthe triangle CDF; then 3-1416— 2 = 1*1416 = area of the segment D AF. Now, 2x2 = 4= area of the rectangle CDEF; then 4— 3-1416= -8584 = area of the section DEFAD. V(22 + 22)= V8 =2-8284271. 2 X 16 + 2 X 20 = 72 = the compass within the walls, i (2-8284271 — 2) = -4142136 = ES and 2-8284271 : -4142136 : : 2 : -2928932 = E ?/; hence 72 = (-2928932 x 8) = 69-6568544 = the circumference of the middle of the solid part of the saloon ; therefore 69-6568544 X •8584 = 59-79344381696 = the content of the solid part of the saloon. 20 X 16 = 320 the area of the roof or floor, and 320 x 2 = 640 = the »?olidity of the upper part of the room; then 640— 59*79344 = i80-20656 feet, the solidity of the saloon. 2. If the height D E of the saloon be 3*2 feet, the chord J) F=4'5 feet, and its versed sine = 9 inches ; what is the solid content of the lolid part, the mean compass being 60 feet? Ans. 138-26489 feet. PROBLEM VI. To find the superficies of a saloon. Rule. Find its breadth by applying a string close to it across the Burface ; find also its length by measuring along the middle of it, quite round the room ; then multiply these two dimensions together for the superficial content. 1. The girt across the face of the saloon is 5 feet, and its mean compass 100 feet ; what is its superficial content ? 100 x 5 = 500, the answer. 2. The girt across the face of the saloon is 12 feet, and its mean x>mpa8s 98 ; required its siuface? Ans, 1176 feet. 114 SPECIFIC GRAYITT. SPECIFIC GRAVITY. SECTION YIII. 1. The specific gi-avity of a body is the relation which the weight of a given magnitude of that body has to the weight of au equal magnitude of a body of another kind. In this sense a body is said to be specifically heavier than another, when under the same bulk it weighs more than that other. On the contrary, a body is said to be specifically lighter than another, when under the same bulk it weighs less than that other. Thus, if there be two equal spheres, each one foot or one inch in diameter, the one of lead and the other of wood, then since the leaden sphere is found to be heavier than the wooden one, it is said to be specifically, or in Bpecie, heavier, and the wooden sphere specifically lighter. 2. If two bodies be equal in bulk, their specific gravities are to each other as their weights, or as their densities. 3. If two bodies be of the same specific gravity or density, thei\ nbsolute weights will be as their magnitudes or bulks. 4. If two bodies be of the same weight, the specific gravities will be reciprocally as their bulks. 6. The specific gravities of all bodies are in a ratio compounded of the direct ratio of their weights, and the reciprocal ratio of their magnitude. Hence, again, the specific gravities are as the densities. 6. The absolute weights or gravities of bodies are in the compound ratio of their specific gravities and magnitudes or bulks. 7. The magnitudes of bodies are directly as their weights, and reciprocally as their specific gravities. 8. A body specifically heavier than a fluid, loses as much of its weight, when immersed in it, as is equal to the weight of a quantity of the fluid of the same bulk or magnitude ; if the body be of equal density with the fluid, it loses all its weight, and requires no force but the fluid to sustain it. If it be heavier, its weight in the fluid will be only the difference between its own weight and the weight of the same bulk of the fluid ; and therefore it will require a force equal to this difl\3reuce to sustain it. But if the body immersed be liglitei- than the fluid, it will require a force equal to the difterence between its own weight and that of the same bulk of the fluid, to keep it from rising in the fluid. 9. In comparing the weights of bodies, it is necessary to consider SPECIFIC GRAVITY. 115 soiTie one as the standard with which all other bodies may be com- pared. Rain water is generally taken as the standard, it being found to be nearly alike in all places. A cubic foot of rain water is found, by repeated experiments, to weigh 62J pounds avoirdupois, or 1000 ounces, and a cubic foot containing 1728 cubic inches, it follows that a cubic inch weighs •03616898148 of a pound. Therefore if the specific gravity of any body be multiplied by •03616898U8, the product will be the weight of a cubic inch of that body in pounds avoirdupois ; and if ttds weight be multiplied by 175, and the product be divided by 144, the quotient will be the weight of a cubic inch in pounds troy, 144 pounds avoirdupois being exactly equal to 175 pounds troy. 10. Since the specific gravities of bodies are as their absolute gravities under the same bulk, the specific gravity of a fluid will be to the specific gravity of any body immersed in it, as the part of the weight lost by the solid is to the whole weight. Hence the specific gravities of different fluids are as the weights lost by the same solid immersed in them. PROBLEM I. To find the specific gravity of a lady. Case I. When the tody is heavier than water. Weigh the body first in water, and afterwards in the open air; the difference will give the weight lost in water ; then say, as the weight lost in water is to the absolute weight of the body, so is the specific gravity of water to the specific gravity of the body. Case II. When the body is lighter than water. Fix another body to it, so heavy as that both may sink in water together as a compound mass. Weigh the compound mass and the heavier body separately, both in the water and open air, and find how much each loses in water, by taking its weight in water from its weight in the open air. Then say, as the difference of these remainders is to the weight of the lighter body in air, so is the specific gravity of water to the specific gravity of the lighter body. Case III. For a fluid of any kitid. Weigh a body of known specific gravity both in the fluid and open air, and find the loss of weight, by subtracting the weight in water from the weight out of it. Then say, as the whole, or absolute weight, is to the loss of weight, so is the specific gravity of the solid to the specific gravity of the fluid. The usual way of finding the specific gravities of bodies is the following, viz. : — On one arm of a balance suspend a globe of lead by a fine thread, and to the otl>er arm of the balance fasten an equal weight sufficient to balance it in the open air ; immerse the globe into the fluid, and observe what weight balances it then, by which the lost weight is ascertained, which is proportional to the specific gravity. 116 SPECIFIC GRAVITT. Immerse the globe successively in all the fluids whose proportional Bpecific gravity you require, observing the weight lost in each ; then these weights lost in each will be the proportions of the fluids sought. Examples. — Case I. 1. A piece of platina weighed 83*1886 pounds out of water, and in water only 79 '5717 pounds; what is its specific gravity, that of water being 1000? 83-] 886— 79-5717 = 3-61G9 pounds, which is the weight lost m water ; then 3-6169 : 83*1886 : : 1000 : 23000 the specific gravity, or the weight of a cubic foot of metal in ounces. 2. A piece of stone weighed 10 pounds in the open air, but in water only 6 J pounds ; what is its specific gravity ? Arts, 3077. Examples. — Case II. 3. If a piece of elm weigh 15 pounds in the open air, and that a piece of copper, which weighs 18 pounds in open air, and 16 pounds in water, is affixed to it, and that the compound weighs 6 pounds in water ; required the specific gravity of the elm ? Copper. Compound. 18 in air. 33 16 in water. 6 ~2 loss. 27 2 As 25 : 16 : : lOOO : 600, the specific gravity of the elm. 4. A piece of cork weighs 20 pounds in open air, and a piece of granite being affixed to it, which weighs 120 pounds in air, and only 80 pounds in water, the compound mass weighs 16| pounds in water ; required the specific gravity of the cork V Ans 240. Examples. — Case III. 5. A piece of cast iron weighed 259*1 ounces in a fluid, and 298*1 ounces out of it ; required the specific gravity of the fluid, aflowing the specific gravity of the cast-iron to be 7645 ? 298-1—259*1=39 loss of weight in the iron; then 298*1 : 39 : : 7645 : 1000, the specific gravity of the fluid ; shewing the fluid to be water.* 6. A piece of lignum vitje weighed 42| ounces in a fluid, and 166| ounces out of it ; what is the specific gravity of the fluid, that of the lignum vitae being 1333 ? Ans. 991 is the specific gravity of the fluid, which shews it to be liquid turpentine or Burgundy wine. * In this manner may the species of a fluid or solid be ascertained, by meana of its specific gravity, and the annexed table. This table has been taken from Gregoiy's work for practical men. SPECIFIC GRAVITY. 117 TABLE OF SPECIFIC GRAVITIES. Platina Do. hammered Cast zinc Cast iron Cast tin Bar iron Hard steel Cast brass Cast copper Pure cast silve Cast lead Mercury Pure cast gold Amber Brick . Sulphur Cast nichel Cast cobalt Paving stones Common stone Flint and spar Green glass White ghm Pebble Slate . Pearl . Alabaster Marble Chalk . Limestone Wax . Tallow Camphor Bees wax Honey Bone of an ox Ivory . Air at the earth's surface Liquid turpentine *^ 'Vw^^ Olive oil Burgundy wine ^- r-^^i^ Distilled water . Sea water Milk . Spec. grav. 19500 20S3Q 7190 7-207 7-291 7-788 7-816 8595 .^. 10474 11^352 13668 19258 ^ 1078 2000 2033 7-807 imi 2416 2520 2594 2642 2892 2664 2672 2684 2730 2742 2784 3179 897 945 989 965 1456 1659 1822 14 >- 991- 915 991- 1-000 1028 1030 wt. cub. in. oz. 11-285 11-777 4-161 4-165 4-219 4-507 4-523 4-858 5-085 6-061 6-569 7-872 11-145 wt. cub. ftw 125-00 127 06 4513 4520 151-00 157-50 162-12 166-50 lbs 167-00 171-38 174-00 193-68 IIS Beer .. Cork . Poplar . Larch . Elm and West India fir Mahogany- Cedar . Pitch pine Pear tree Walnut Elder tree Beech . Cherry tree Maple and Riga fir Ash and Dantzic oak Apple tree Alder . Oak, Canadian Box, French Log'W'ood Oak, English Oak, 60 years old Ebony . Lignum vitae PROBLEM IL 2'he specific gravity of a hody^ and its weight, heing given, to Jind it's solidity. HuLE. Say, as the tabular specific gravity of the body is to its weight, in ounces avoirdupois, so is 1 cubic foot to the content. 1. What is the solidity of a block of marble that weighs 10 tons, its specific gravity being 2742? First, 10 tons =200 hundreds =22400 pounds =358400 ounces; then 2742 : 358400 : : 1 1 SPECIFIC GRAVITY. Spec. grav. wt. ctib. ft lbs. 1-034 240 , 15-00 383 . 2-394 544 . 34-00 556 . 34-75 660 . 35-00 696 . 37-25 660 . 41-25 661 . 41-31 671 . 41-94 695 . 43-44 696 . 43-50 715 . 44-68 750 . 46-87 760 . 47-50 793 . 49-56 800 . 60-00 872 . 64 50 912 . 67-00 913 . 67-06 970 . 61-87 H70 . 73-12 1-331 . 83-18 1-333 . 83-31 2742)358400(130xV7T- 2742 8420 8226 ^2742^'^^^ 2. How many cubic inches in an irregular block of marble which weighs 112 pounds, allowing its specific gravity to be 2520? Ans. 12281^^ cubic inches. 3. How many cubic inches of gunpowder are there in 1 pound weight, its specific gravity being 1745? Ans. 15 J, nearly. SPECIFIC GRAVITY. 119 4. How many cul)ic feet are there in a ton weight of dry oak, \i% Bpecific gravity being 925 ? Ans. 38f ||-. PROBLEM III. The linear dimensions^ or magnitude of a hody^ heing given^ and also its specific gravity^ tojind its weight. Rule. One cubic foot is to the solidity of the body, as the tabular gpocific gravity of the body is to the weight in avoirdupois ounces. 1. What is the weight of a piece of dry oak, in the form of a parallelopipedon, whose length is 56 inches, breadth 18 inches, and depth 12 ? 56 X 18 X 12= 12096 cubic inches, the solid content. Then 1728 : 12096 :: 932 ; 6524 ounces = 407| pounds, the weight required. 2. What is the weight of a block of dry oak, which measures 10 feet long, 3 feet broad, and 2J feet deep , its specific gravity being 925 ? Ans. 4335^^ pounds. 3. What is the weight of a block of marble, whose length is 53 feet, and its breadth and thickness, each 12 feet ? Ans. 694xVV tons. PROBLEM ly. Tojind the quantities of two ingredients in a given compound. Rule. Take the difference of every pair of the three specific gravities, viz., of the compound and each ingredient; and multiply the difference of every two by the third. Then as the greater product is to the whole weight of the com- pound, so is each of the other products to the weights of the two ingredients.* 1. A composition of 112 pounds being made of tin and copper, whose specific gravity is found to be 8784 ; what is the quantity of each ingredient, the specific gravity of tin being 7320, and of copper 9000? 9000 9000 8784 7320 8784 7320 1680 216 1464 diff. 8784 7320 9000 14757120 1581120 13176000 Then ij.7K7i9n • 110 •• J 13176000 : 100 pounds copper. i4/i>/izu . iiz .. -J 1581120 : 12 pounds tin. 2. Hiero, king of Sicily, furnished a goldsmith with a quantity of gold, to make a crown. When it came home, he suspected that the goldsmith had used a greater quantity of silver than was neces- sary in the composition ; and applied to the famous mathematician, Archimedes, a Syracusian, to discover the fraud, without defacing the crown. * For the reason of this rule, see Alligation Total in the Second Book of /Lrithmetic, published by the Commissioners. 120 SPECIFIC GRAVITY. To ascertain the quantity of gold and silver in the crown, ha procured a mass of gold and another of silver, each exactly of the same weight with the crown ; justly considering that if the crown were of pure gold, it would be of equal bulk, and therefore displace an equal quantity of water with the golden mass ; and if of silver, it would be of equal bulk and displace an equal quantity of water with the silver mass ; but if of a mixture of the two, it would displace an intermediate quantity of water. Now suppose that each of the three weighed 100 ounces ; and that on immersing them severally in water, there were displaced 5 ounces of water by the golden mass, 9 ounces by the silver mass, and 6 ounces by the crown ; what quantity of gold and silver did the crown contain ? . j 75 ounces of gold. Ans. j 25 ounces of silver. JVo/e.— Questions relating to specific gravities may be wrought by the rulets of Alligation in Arithmetic, as well as by any Algebraic process that might be employed. PROBLEM V. To find Jiow many inches a floating body will sink in a fluid. Rule. Find, by Problem III. the weight of the floating body from its solidity and specific gravity, and that will be the weight of the fluid which it will displace. Then say, as the specific gravity of the fluid is to 1728 cubio inches, so is the weight of the body, in ounces, to the cubic inches immersed. The depth will be found from the given dimensions. 1. Suppose a piece of dry oak, in the form of a parallelopipedon, whose length is 56 inches, breadth 18, and depth 12, is to be floated upon common smooth water, on its broadest side ; how many inches will it sink, its specific gravity being 932? By Problem III., the weight of the piece of oak is 6524 ounces, which by the preliminary part of this section, is the weight of water displaced. Then 1000 : 1728 : : 6524 : 11273-472 cubicinches of oak immersed. Therefore, 11273-472-t-(56x 18) = 11-184 inches, the depth it will sink. To find how far it will sink, allowing it to float on its narrower side, 11273'472-f-(56x 12) = 16-776 inches. 2. How many inches will a cubic foot of dry oak sink in common water, allowing the specific gravity of the oak to be 970? Ans. 11-64. PROBLEM VI. To find what weight may he attached to a fioating body^ so that it may he just covered with a given fluid. Rule. Multiply the cubic feet in the body by the difference between its specific gravity and that of the fluid, and the product will be the weight in ounces avoirdupois, just sufficient to immerse it in the fluid. . 1. What weight must be attached to a piece of dry oak, 56 inches SPECIFIC GRAVITY. 121 lon^, 18 inches broad, and 12 inches deep, to keep it from rising above the surface of a fresh-water lake ; the specific gravity of the water being 1000, and that of the oak 932 ? Here 56 x 18 x 12= 12096 cubic inches. Then 120964-1728 = 7 feet. Then (1000—932) x 7 = 68 x 7 = 476 ounces = 29 pounds 12 ounces. 2. What weight fixed to a piece of dry oak, 9 inches long, 6 inches broad, and 3 inches deep, will keep it from rising above the surface of common water, the specific gravity of water being 1000, and that of the oak 970 ? Ans. 2^^ ounces. 3. A sailor had half an anker of brandy, the specific gravity of the liquor was 927, the cask was oak, and contained 216 cubic inches, and its specific gravity was 932 ; to secure his prize from the custom- house officers, he fixed just as much lead to the cask as would keep it under water, and then threw it into the sea; what weight of lead was necessary for his purpose ? Ans. The cask of brandy contained 1371 cubic inches, the weight of sea- water of an equal bulk was 81 7*20486 ounces, the cask weighed 116-5 ounces, the brandy 619*609375, both together weighed 736*19375 ounces. The difference between the specific gravity of lead and sea-water is to this remainder, as the specific gravity of lead to its weight in ounces, which will be found to be 89 '09495 ounces, or 6 pounds 9 ounces. PROBLEM VII. I'd find the solidity of a oody^ lighter than a fluids which will he sufficient to prevent a body much heavier than the fiuid^ from sinking. Rule. Find the solidity of the body to be floated, from its weight and specific ^avity, by Problem II. Find also the weight of an equaj bulk of the fluid by Problem J II. Then say, as the difference between the specific gravity of the fluid, and that of the body lighter than the fluid, is to the difference between the weight of the body to be floated and the weight of an equal bulk of the fluid, so is 1728 to the soli- dity of the lighter body in cubic inches. 1. How many solid feet of yellow fir, whose specific gravity is 657, will be sufficient to keep a brass cannon, weighing 56 cwt., afloat at sea, the specific gravity of brass being 8396, and of sea- water 1030? First, 56 cwt. = 100352 ounces, weight of the body to be floated. Then, 8396 : 100352 : : 1728 : 20653*675 cubic inches in the cannon. And, 1728 : 20653-675 : : 1030 : 12310*9289, the weight of sea-water equal in bulk to that of the cannon. Hence, 1030—657 : 100352 — 12310-9289 ;; 1728 : 407868*5545 cubic inches = 236*036 feet, the answer. 2. The specific gravity of lead is 11325, of cork 240, and of sea- water 1030; now it is required to know how many cubic inclies of cork will be sufficient to keep 49| pounds of lead afloat at sea? Ans, 1570-84 cubic inches. 122 t<;nnage of ships. TO FIND THE TONNAGE OF SHIPS. 1st.— VESSELS AGROUND. By the Parliamentary Rule, PROBLEM VIII. For a ship or vessel, the length is to be measured on a straight line along the rabbet of the keel, from a perpendicular, let fall from the back of the main post, at the height of the wing-transom, to a perpendicular at the height of the upper deck (but the middle deck of three-decked ships), from the fore-part of the stern ; then from the length between these perpendiculars subtract three-fifths of the extreme breadth for the rake of the stern, and 2J inches for every foot of the height of the wing-transom above the lower part of the rabbet of the keel, for the rake abaft ; and the remainder will be the length of the keel for tonnage. The main breadth is to be taken from the outside of the outside plank, in the broadest part of the ship, either above or below the wales, deducting therefrom all that it exceeds the thickness of the plank of the bottom, which shall be accounted the main breadljh ; so that the moulding breadth, or the breadth of the frame, will then be less than the main breadth, so found, by double the thickness of the plank of the bottom. Then multiply the length of the keel for tonnage, by the main breadth, so taken, and the product by half the breadth, then divide the whole by 94, and the quotient will give the tonnage. In cutters and brigs, where the rake of the stern-post exceeds 2J inches to every foot in height, the actual rake is generally subtracted instead of the 2J inches to every foot, as before mentioned. 1. Let us suppose the length from the fore-part of the stern, at the height of the upper deck, to the after-part of the stern-post, at the height of the wing-transom, to be 155 feet 8 inches, the breadth from out to outside 40 feet 6 inches, and the height of the wing- transom 21 feet 10 inches, what is the tonnage? ft. 40-6 breadth. ^ deduct 3 40*3 3 5)120-9 24-lf = 24-15 21*10 height of wing-transom. 2|- multiply. 12)54tV 4-55-t-24-]5 = 28-70 155-66-28-70 = 126-96 = length. 126-96 X 40-25 + 20-125 ,^^, ,, ^7 = 1094, the answer. 94 FLOATING BODIES. 123 2. Suppose the length of the keel to be 50*5 feet, breadth of the midship -beam 20 feet ; required the tonnage? Ans. 107*4. 3. If the length of the keel be 100 feet, and the breadth of the beam 30 feet ; what is the tonnage? Ans, 478. 2d.— VESSELS AFLOAT. Drop a plumb-line over the stern of the ship, and measure the distance between such line and the after-part of the stern-post, at the load water-mark : in a paraM direction with the water, to a perpendicular point immediately over the load water- mark, at the fore-part of the main-stern, subtracting from such measurement the above distance, the remainder will be the ship's extreme length ; from which is to be deducted three inches for every foot of the load draught of water for the rake abaft, and also three-fifths of the ship's breadth for the rake forward, the remainder shall be esteemed tlie just length of the keel to find the tonnage ; and the breadth shall be taken from outside to outside of the plank, in the broadest part of the ship, either above or below the main-wales, exclusive of all manner of sheathing or doubling that may be wrought upon the sides of the ship ; then multiply the length of the keel, taken as before directed, by the breadth, as before taught, and that product by half the said breadth, and dividing the product by 94, the quotient IS the tonnage. 3d.— STEAM VESSELS. The length shall be taken on a straight line, along the rabbet ol the keel, from the back of the main- stern-post to a perpendicular line from the fore-part of the main-stem under the bow-sprit ; from which deducting the length of the engine-room, and subtracting Ihree-fifths of the breadth, the remainder shall be esteemed the just length of the keel to find the tonnage ; and the breadth shall be taken from the outside of the outside plank in the broadest place of the ship or vessel, be it either above or below the main-wales, exclusively of all manner of doubling planks that may be wrought upon the sides of the ship or vessel ; then multiply the length and breadth so fjund together, and that product by half the same breadth, and dividing by 94, the quotient will be the tonnage, according to which all such vessels ^hall be measured. Note. — Under certain penalties nothing but the fuel can be stowed in the engine-room. Some divide the last product by 100, to find the tonnage of king's ehips, and by 95, to find that of merchants' ships. FLOATING BODIES. 1. The buoyancy of casks, or the load which they will carry with- out sinking, may be estimated by reckoning 10 pounds avoirdupois to the ale gallon, or 8^ potmds to the wine gallon 124 PLOATINQ BODIES. 2. The buoyancy of pontoons may be estimated at about half a lumdred weight, or 56 pounds for each cubic foot. Iherefore a pontoon which contained 96 cubic feet, would sustain 48 hundred- weight before it would sink. N.B. — This is an approximation, In which the difference between -^1 and J, viz , 1 of the whole weight, is allowed for that of the pontoou itself. 3. The principles of buoyancy are very ingeniously applied in the self-acting flood-gate, which, m the case of common sluices to a mill-dam, prevents inundation when a sudden flood occurs. By means of the same principle, it is that a hollow ball attached to a metallic lever of about a foot long, is made to rise with the liquid in a water-cask, and thus to close the cock and stop the supply from the pipe, just before the time when the water would otherwise run over the top of the vessel. The property of buoyancy has also been successfully employed in raising ships which had sunk under water, and in pulling up old piles in a river wlien the tide ebbs and floiws. A large barge is brought over a pile as the water begins to rise; a strong chain which has been previously fixed to the pile by a ring, &c., is made to gird the barge, and is then firmly fastened ; then, as the tide rises, the barge rises also, and by means of its buoyant force draws up the pile with it. In a case which actually occurred, a barge of 50 feet long, 12 feet wide, 6 deep, and drawing 2 feet water, was employed. Then 50 X 12 X (6-2) X 4^ 50x12x16 ^ ^^2 x 7f = 1344 + 27f = 1371f cwt. = 66 J tons, nearly, which is the measure of the force with which the barge acted in pulling up the pfle> WEIGHT AND DIMENSIONS OF BALLS AND SHELLS. 126 WEIGHT AND DIMENSIONS OF BALLS AND SHELLS. SECTION IX. The foregoing problems furnish rules for finding the weight and dimensions of balls and shells. But they may be found much easier by means of the experimental weight of a ball of a given size, and from the well-known geometrical property, that similar solids are aa the cubes of their diameters. PROBLEM I. H j' To find the weight of an iron hall from its diameter. ^ i* Rule. Nine times the cube of the diameter being divided by 64, will express the required weight in pounds.* 1. The diameter of an iron shot is 5 inches ; required its weight? 6 X 5 X 5 = 125 = cube of the ball's diameter. Then 125x 9-^64: = 17f} pounds, the answer. 2. The diameter of an iron shot being 3 inches; required ita weight? Ans. 3-8 pounds. 3. The diameter of an iron shot is 5*54 inches; what is its weight? Ans, 24 pounds. PROBLEM II. ^d ' To find the weight of a leaden ball^ by having its diameter given, ^ Rule. Multiply the cube of its diameter by 2, and divide the pro- duct by 9, and the quotient will give the weight in pounds.f 1. AVhat is the weight of a leaden ball of 5 inches diameter? 5x5 x 5= 125 cube of ball's diameter. Then, 125 x 24-9 = 250-^9 = 27J pounds, answer. 2. What is the weight of a leaden ball, whose diameter is 6*6 inches? Ans. 63*888 pounds. 3. What is the weight of a leaden ball, whose diameter is 3*5 inches? Ans. 9*53 pounds. 4. What is the weight of a leaden ball, whose diameter is 6 inches? Ans. 48 pounds. * See Appeudix, DemonstJ^tioii 113. f Ibid. 111. '126 WEIGHT AND DIMENSIONS OF BALLS AND SHELLS. PROBLEM III. Having the weight of an iron ball^ to determine its diameter. Rule. Multiply the weight by 7J, then take the cube root of the product for the diameter.* 1. What is the diameter of an iron ball, whose weight is 42 pounds ? 42x7|-=298|. Then, ^298 = 6*685 inches, the answer. 2. Required the diameter of an iron ball, whose weight is 24 pounds? ^725. 5 '54 inches. 3. What is the diameter of an iron ball, whose weight is 3-8 pounds ? Ans. 3 inches. PROBLEM IV. Having the weight of a leaden ball, to determine its diameter. Rule. Multiply the weight by 9, and divide the product by 2 ; and the cube root of the quotient will express the diameter.f 1. What is the diameter of a leaden ball, whose weight is 64 pounds? 64x9 = 576. Then, 576-^2 = 288. Hence, >^288 = 6*6 inches, the answer. 2. Required the diameter of a leaden ball, whose weight is 27{ pounds? Ans. 6 inches. 3. What is the diameter of a leaden ball, whose weight is 63-888 pounds? Ans. 6*6 inches. PROBLEM V. ^ Having given the external and internal diameter of an iron shell, to find its weight. Rule. Find the difference between the cubes of the two diameters, iind multiply it by 9 ; divide the product by 64, and the quotient will express the weight in pounds. J 1. What is the weight of an 18-inch iron bomb-shell, whose mean thickness is Ij inches? 18— 2j= 15| = mternal diameter. Then, 18^ = 5832 the cube of external diameter. (15-5)^ = 3723 -875 the cube of internal diameter. And, 6832-3723-875 = 2108-125 = difference of cubes. Hence, 2108-125 x 9^-64 = 296*45 pounds, the answer. 2. What is the weight of a 9 -inch iron bomb-shell whose mean thickness is 1^ inch? Ans. 72-14 pounds. 3 AVhat is the weight of an iron bomb- shell, whose external diameter is 9-8 inches, and internal diameter 7 inches? Ans. Si I pounds. * This niTe is obvious from Problem I., being the converse thereof, t Tliis rule is manifest from Problem III,, haing its converse, i bee Appendix, Demonstration 1 S WEIGHT AND DIMENSIONS OF BALLS AND SHELLS. 127 PROBLEM VI. To find 7iow much powder will fill a shell of given dimensions. Rule. Divide the cube of the internal diameter in inches, by 57*3, and the quotient will express the answer.* 1. What quantity of powder will fill a shell, whose internal diameter is 10 inches? First, 10 X 10xlO=1000=cube of diameter. 57'3)1000(17-45 pounds, answer. 673 4270 4011 2590 2292 115, &c. f^ Kote.— In some recent works, the cube of the diameter is divided by 59*32, I y for the weight of powder in pounds. L J 2. How many pounds of gunpowder are required to fill a hollow ( I shell, whose internal diameter is 13 inches? j >^ Ans. 37 pounds, according to the note. ^ jf^ ^. Required the number of pounds of powder that will fill a shell, whose internal diameter is 7 inches ? Ans. 6 pounds, by the rule in the text. PROBLEM VII. To find how much powder will fill a rectangular box oj^ given dimensions. Rule. Multiply the length, breadth, and depth together in inches, and the last result by '0322, and the last product will give the weight in pounds.f 1. How many pounds of powder will fill a rectangular box, whose length is 16 inches, breadth 12 inches, and depth 6 inches? 16x12x6=1 152 = content of the box. Then, 1152 x •0322 = 37*0944, the answer. 2. How many pounds of powder will fill a rectangular box, whose length is 10 inches, breadth 5 inches, and depth 2 inches? Ans. 3-22 pounds. 3. How many pounds of powder will fill a rectangular box, whose length is 5 inches, breadth 2 inches, and depth 10 inches? Ans. 3-22 pounds. PROBLEM YIII. Having the length and diameter of a cylinder^ to determine how many pounds of gunpowder will fill it. Rule. Multiply the square of the diameter by the length, and divide the product by 40, for the weight in pounds. J * bee Appendix, Demonstration 116. f Ibid. 117. X Ibid. 11 1 128 PILING OF BALLS AND SHELLS. 1. The diameter of a hollow cylinder is 10 inches, and the length 14 inches ; how many pounds will it hold ? 10 X 10 = 100 = square of diameter. Then, 100x14=1400. Hence, 1400-^40 = 35 pounds, the answer. 2. The diameter of a hollow cylinder is 5 inches, and its length 40 inches ; how much powder will it hold ? Ans. 25 pounds. 3. The diameter of a hollow cylinder is 5 inches, and the length 12 inches ; how many pounds will it hold? Ans. 7*5 pounds. PROBLEM IX. To find wJiat portion of a cylinder will be occupied by a given quantity of powder , the diameter of the cylinder being given. Rule. Multiply the given weight of powder by 40, and divide the product by the square of the diameter of the cylinder, and the quotient will be the pounds required.* 1. The diameter of a hollow cylinder is 10 inches; how much of it will hold 50 pounds of powder ? 50x40 = 2000. Then, 2000-7-100=20 inches, the answer. 2. How much of a cylinder of 14 inches diameter will hold 10 pounds of powder? Ans. 2*05 inches. 3. How much of a cylinder, 12 inches in diameter, will hold 144 pounds of powder ? Ans. 40 inches. PILING OF BALLS AND SHELLS. Iron-shot and shells are usually piled in horizontal courses, eithel in a pyramidical or in a wedge-like form ; the base being either an equilateral triangle, a square, or a rectangle. Those piles whose bases are triangles or squares, terminate in ono ball at the top ; but piles whose bases are rectangles terminate in a single row of balls. In triangular and square piles, the number of horizontal rows or courses, is always equal to the number of balls in one side of the bottom row. And in rectangular piles the number of rows is equal to the num- ber of balls in the breadth of the bottom. Also the number in the top row or edge, is one more than the dif- ference between the length and breadth of the bottom row. PROBLEM I. ^ • ^. .^^ To find the number of balls in a mtstoB^ttar pile. Rule. Multiply the number in one side of the bottom row, by that number increased by 1, and the result by the same number increased by 2 ; then the one-sixth of the last product will give the number of balls required.f ♦ See ADDendijL Demonstration 119. t Ibid. 120. PILING OF BALLS AND SHELLS. 129 1. Required the number of shot in a complete triangular pile, one f J whose sides contains 22 balls? 22 = the number in one side of base. 23 = the numbers 1. 66 44 "We 24= the number +2. 2024 1012 6)12144 2024 = the number of shot in the pile. 2. Required the number of sliot in a complete triangular pile, ono Bide of whose base contains 15 balls ? Ans. 680 balls. 3. Required the number of balls in a triangular pite, each side of the base containing 30 balls ? Ans, 4960. PROBLEM II. Tojind the number of halls in a square pile. Rule. Multiply continually together the number in one side of the bottom course, that number increased by 1, and double the same number increased by 1 ; then one- sixth of the last product will be the answer.* M ^(^^^^ ♦^/^gt^^vM 1. How many balls are in a square pile of 30 rowsr" " *^ 30 = number in one side. ^ 31 = number in one side + 1. 930 61 = twice the number in one side + 1. 6)56730 9455 answer. 2. Required the number of shot in a complete square pile, one side of whose base contains 19? Ans. 2470. 3. How many shot in a finished square pile, when a side of the base contains 21 shot? Ans, 3311. PROBLEM III. Tojind the numter of shot in a finished rectangular pile. Rule. Adi 1^ to three times the number of shot contained in the length of the base, subtract the number of shot in the breadth of the base, multiply the remainder by the said number increased by 1, and this result again by the number in the breadth ; then one-sixth of the last result will give the number of shot in the rectangular pile.f • See Appendix, Demonstration 121. f Ibid. 122. tC^c^ 130 PILING OF BALLS AND SHELLS. 1. Required the number of shot in a finished rectangular pile, the length of the base containing 59, and its breadth containing 20 balls? 69 = the number of shot in the length. 3 177; then 177 + 1 = 178, and 178-20 = 158. 158x21 = 3818, and 3318x20 = 66360. Hence 66860-5-6 = 11060, the answer. 2. How many balls are in a rectangular complete pile, the length of the bottom course being 46, and its breadth 16 V Ans. 4960. PROBLEM IV. To determine the number of balls contained m a pile which is not jinished^ ike highest course being complete^ and the number oj balls in each side thereof being given. Rule. Find the number of shot whicli would be contained in the pile if it were complete. Find also the number in that complete pile, each side of whose base contains one shot fewer than the correspond- ing side of the uppermost course of the unfinished pile, and the difference between these results will evidently give the number of balls in the unfinished pile. 1. How many shot are there in an unfinished triangular pile, a Bide of whose base contains 23, and a side of the uppermost course 7 shot? 23 = number of balls in the base. 24 = number of balls in the base + 1. "652 25 6)13800 2300= number of the pile when complete. 6 7 42 8 6)336 66 number of balls in the imaginary pile. Therefore, 2300 — 56 = 2244, the answer. 2. How many balls in an incomplete square pile, the side of the base being 24, and of the top 8? Ans. 4760. 3. How many balls are there in the incomplete rectangular pile of 12 courses, the length and breadth of the base being 40 and 20? Ans. 6146. DETERMINING DISTANCES BY SOUND. 131 DETERMINING DISTANCES BY SOUND. The velocity of sound, or the space through which it is propa- gated in a given time, has been very differently estimated by phi- losophers who have written on this subject. We shall, however, take it to be 1142 feet in a second. From repeated experiments it has been ascertained that sound moves uniformly, or to speak more philosophically, that the pulses of air which excite it move uniformly. The velocity of sound is the same with that of the aerial waves, and does not vary much whether it go with the wind or against it. By the wind, no doubt, a certain quantity of air is carried from one place to another, and the sound is somewhat accelerated while its waves move through that part of the air, if their direction be the same as that of the wind. But as the velocity of sound is vastly swifter than the wind, the accelera- tion it will thereby receive is but inconsiderable, being at most but ^ of the whole velocity. The chief effect perceptible from the wind is, that it increases and iliminishes the space tnrough which sound is propagated. The utmost distance at which sound has been heard is about 200 miles. It is said that the unassisted human voice has been heard from Old to New Gibraltar, a distance of about 12 miles. Dr Derham, placing cannon at different distances, and causing them to be fired off*, observed the intervals between the flash and report, by means of which he found the velocity of sound to be as above stated. 1. Having observed the flash of a cannon, I noticed by my watch that 5 seconds elapsed previous to my hearing the report ; determine iny distance from the gun. 1U2 5 5710 feet, the answer. 2. Bemg at sea I saw the flash of a cannon, and counted 8 seconds between the flash and the report ; required the distance? dns. 1^ mile. 132 GAUGINa GAUGING. SECTIOIsr X. Gauging is the art of measuring the capacities of vessels, such as casks, vats, &c. The business of gauging is generally performed by means of two instruments, namely, the gaugmg or sliding rule, and the gauging or diagonal rod. 1. OF THE GAUGING RULE.— LEADBETTER'S. By this instrument is computed the contents of casks, &c., after the dimensions have been taken. It is a square rule, having various logarithmic lines on its four faces, and jjhree sliding pieces capable of being moved through grooves in whicTthey fit, in three of these faces. On the first face are delineated three lines, namely, two marked A.B, on which multiplication and division are performed; and the third marked M D, signifies malt depth, and serves to gauge malt. The middle one B is on the slider, and is a kind of double line, being marked at both edges of the slider, for applying it to both the lines A and M D. These three lines are all of the same radius, or distance from 1 to 10, each containing twice the length of the radius. A and B are numbered and placed exactly alike, each commencing at 1, which may be either 1, or 10, 100, &c., or -1, or -01, or -001, &c. "Whatever the 1 at the beginning is estimated at, the middle division 10 will be 10 times as much, and the last division 100 times as much. But 1 on the line MD is opposite 2220, or more exactly 2218*2 on the other lines, which number 2218*2 denotes the cubic inch in an imperial malt bushel ; and its divisions numbered retrograde to those of A and B. On these two lines are also several other marks and letters ; thus on the line A are M B, or sometimes only B, for malt bushel, at the number 2218*2, and A for ale, at 282, the cubic inches in an old ale gallon ; and on the line B is W, for wine, at 231, the cubic inches in an old wine gallon. These marks are now usually omitted upon the rule, since the late new act of parliament for uniformity of weights and measures, and G for gallon is put at 277*274 the inches in an imperial gallon,* whether of ale, wine, or spirits. * Until 5 George IV., in which a uniform system of weights and measures was established under the denomination of imperial weights and measures, there were, amongst other sources of inconvenience, different measures, though GAUGING. 133 On many sliding rules are also found s 2, for square inscribed, at ♦707, the side of a square inscribed in a circle whose diameter is 1 ; ge^ for square equal, ^t '886, the side of a square which is equal to the same circle ; and c, for circumference, at 3 '14 16, the circumference of the same circle. On the second face, or that opposite the first, are a slider and four lines marked D, C, D, E, at one end, and root square, root cube at the other end ; the lines C and D containing, respectively, the squares and cubes of the opposite numbers on the lines D, D ; the radius of D being double to that of A, B, C, and triple to that of E ; therefore whatever the first 1 on D denotes, the first on C is its square, and the first on E its cube ; that is, if D begin with 1, C and E will begin with 1 ; but if D begin with 10, C will begin with 100, and E with 1000 ; and so on. On the line are marked o c at -0796, for the area of the circle whose circumference is 1 ; and o d, at '7854, for the area of the circle whose diameter is 1. On the line D are marked G S, for gallon square, at 16*65, and G R for gallon round at 18-789 ; also MS for malt square at 47*097, and M R for malt round at 53*144. These are the respective gauge-points for gallons and bushels. The first 16*65 is the side of a square, which at an inch depth holds a gallon; the second 18*789, the diameter of a circle, which at an mch depth holds a gallon ; the third 47*097 the side of a square, which at an inch depth holds a bushel; the fourth 53*144, the diameter of a circle, which at an inch depth holds a bushel. On the third face are three lines : one on a slider marked N ; and two on the stock, marked S S and S L, for segment standing and segment lying, which serve ullaging, standing and lying casks. And on the fourth side, or opposite face, are a scale of inches, and three other scales, marked spheroid, ©r 1st variety, 2d variety, 3d variety ; the scale for the 4th or conic variety, being on the inside of the slider in the third face. The use of these lines is, to find the mean diameter of casks. On the inside of the two first sliders, besides all those already described, are two other lines, being con- tinued from one slider to the other. The one of these is a scale of inches, from 2J to 36, and the other is a scale of ale gallons, between the corresponding number 435 and 3*61, which form a table, to shew, in ale gallons, the contents of all cylinders whose diameters are from 12j to 36 inches, their common altitude being 1 inch. of the same name, for ale and wine. A gallon of ale contained 282 cubic inches, and a gallon of wine 231; a bushel of malt contained 2150 •42 cubic inches. To reduce old measure into new, say, as the number of cubic inches in the imperial standard is to the number of cubic inches in the old standard, so is the number of gallons or bushels, &c., old measure, to the number of gallons, &c., imperial measure. When great accuracy is not required, old wine gallons may be reduced to imperial gallons by dividing by 1'2; and old ale gallons may be reduced to imperial gallons by multiplying by 60, and dividing the product by 69 ; and old, or Winchester bushels, maybe reduced to imperial bushels by multiplying by 31, and dividing the product by 32. x 134 GAUGINa. verie's sliding rule. Tills rule is in the form of a parallelopipedon, and is generany made of box. 1. The line marked A, on the face of this rule, is called Gunter's line, and is numbered 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. At 2218-192 is fixed a brass pin, marked IM, B, signifying the cubic inches in an imperial bushel; at 277*274 is fixed another brass pin, marked IM, G, denoting the number of cubic inches in an imperial gallon. 2. The hue marked B is on the slide, and is divided exactly like that marked A. There is another slide B, on the opposite side, which is used along with this. The slide on the first face is called the second radius^ and that on the opposite face, the first radius. The two brass ends, when placed together, make a double radius, numbered from the left hand towards the right. At 277*274, on the second radius, is a fixed brass pin, marked IM, G, denoting the cubic inches in an imperial gallon; at 314 is fixed another brass pin, marked C, signifying the circumference of a circle whose diameter is i; "These lines are used and read exactly as the lines A and B on tlie Carpenter's Rule, which have been already described. 3. Tlie back of one slide or radius, marked B, has the dimensions for imperial gallons, and bushels, green starch, dry starch, hard soap hot, hard soap cold, green soft soap, white soft soap, flint glass, &c., &c., as in Table, page 136. The back of the other slide or radius, marked B, contains the gauge- points corresponding to these divisors, where S denotes squares, and C circles. 4. The line M D on the rule, denoting malt depth, is a line of numbers commencing at 2218*192, and is numbered from the left to the right hand 2, 10, 9, 8, 7, 6, 6," 4, 3. This rule is used in malt- gauging. 5. The two slides B, just described, are always used together, either with the line A, M D, or the line D, which is on the opposite face of the rule to that already described. The line D is numbered from the left hand towards the right, 1, 2, 3, 31, to 32, which is at the right-hand end ; it is then continued from the left-hand end of the other edge of the rule, 32, 4, 5, 6, 7, 8, 9, 10. A.t 16*651 is a brass pin G S, signifying a gauge square^ being the square gauge*' point for imperial gallons. ' At 18*789 is fixed a brass pin, marked G R, denoting gauge rounds or circular gauge-point for imperial gallons. At 47*097, M S signifies malt square^ the square gauge- point for malt bushels. At 53*144, ME, denotes malt rounds the round or circular gauge-point for malt bushels. The line D on this rule is of the same nature as the line marked D on the Carpenter's Rule, which has been already described. The line A and the two glides B, are used together, for performing multiplication, division, simple proportion, &c. ; and the line D, and the same slides B, are used together for extracting the square and cube roots. 6. The other two slides belonging to this rule are marked C, and are divided in the same manner, and used together, like the slides B. The back of the first slide or radius, marked C, is divided, next tlic edge, into inches, and numbered from the left hand towards the QAUGINO. 185 right 1, 2, 3, 4, 5, &c., and these inches are again subdivided into ten equal parts. The second line is marked spheroid, and is numbered from the left hand towards the right 1, 2, 3, 4, 5, 6, 7, 8. The third line is marked second variety, and is numbered 1, 2, 3, 4, 5, 6. These Mnes are used, vv^ith the scale of inches, for finding a mean diameter. The back of the second slide or radius, marked C, has several factors for reducing goods of one denomination to others of equiva- lent values. Thus I X. to VI. 6. | signifies that to reduce strong beer at 85. per barrel, to small beer at Is. M. you are to multiply by 6. I VI. to X. 17. I signifies that to reduce small beer at Is. 4^. per barrel to strong beer at 8s. per barrel, you are to multiply by -17. I C 4 ^ to X. 27. I signifies that 27 is the multiplier for reducing cider at 4s. per barrel to another at 8s., &c. 7. The two slides C, just described, are always used together, with the lines on the rule marked Seg. St., or SS, segments stand- ing ; and Seg. L ?/ or S L, segments lying ; for uUaging casks. The former of these lines is numbered 1, 2, 3, 4, 5, 6, 7, 8, which stands at the right-hand end ; it then goes on from the left hand on the other edge 8, 9, 10, &c., to 100. The latter is numbered in the same manner 1, 2, 3, 4, which stands at the right-hand end ; it then goes on from the left hand on the other edge, 4, 6, 6, 7, &c., to 100. PROBLEM I. To find the several multipliers^ divisors^ and gauge-pointSy belonging to the several measures now used, MULTIPWERS FOR SQUARES. As 277*274 solid inches are contained in one imperial gallon, and 2218*192 solid inches in an imperial bushel ; then it is obvious that if 1 be divided by 277*274, and 2218*192, respectively, the quotients will be the multipliers for imperial gallons and bushels respectively. Hence the method of finding the following multipliers is obvious : — 277*274)1 -OOOOOC-OOSGOeS multiplier for imperial gallons. 2218*192)l*00000(-0004508 multipher for imperial bushels. Now it is manifest that if the solid inches contained in any vessel be multiplied by the first of these multipliers, the product will be the imperial gallons that vessel will contain ; and if multiplied by the other, the product will be the imperial bushels. MULTIPLIERS AND DIVISORS FOR CIRCLES. It has been shewn that when the diameter of a circle is 1, the area of that circle is -785398, &c., or -7854, nearly ; then by dividing the solid capacity of any figure by '7854, the quotient will be the proper iivisor for the square of the diameter of a circular figure. Then to reduce the area at one inch deep into gallons, divide '7854, or •785398, &c., by 277*274, and 2218*192, and the quotients will give the multipliers for imperial gallons and bushels respectively ; and •7S54 divided into 277*274 and 2218*192, will give the divisors for the imperial gallons and bushels. 277'274)'785398(. 002832 multiplier for imperial gallons. 2218'192)'785398(.00354 multiplier for imperial bushels. •785398)277 •274(3530362 divisor for imperial gallons. •785398)2218*192(2824-2897 divisor for imperial bushels. 136 GAUGING. The gauge-points are found bv extracting the square root of tho divisors. GAUGE- POINTS FOR SQUARES. V 277-274 = 16-651 imperial gallons. V 2218-192 = 47-097 imperial bushels. GAUGE-POINTS FOR CIRCLES. V 353-0362 = 18-789 imperial gallons. V 2824-2897=53-144 imperial bushels. In this manner the numbers in the following Table were calculated G PQ ^ 3 ^ H « < Ci5 -?^ s c I «- 00 -rHr-'O5'-'* c3 e8 (fq T^ o b- »b ih »b »b »b »b i* CO CO M 1 tH ^ ^ ^ ^ ->* •^H^0tiC5O1000'>*G0r-^.-(-rH S^ C^ CO^Oa CO CO 1 "o ^ C^ CO (MCOCO(MC<«'-il:^»0 E- -C3 •S si||aaaaaaaag.a 1 1 1 GAUGING. 137 Kote. — ^It very often happens in the practice of prauc^ng-, that when the on.3 given number is set to the gauge-point on the sUding rule, the other given number will fall ofif the rule ; hence in many cases it will be necessary to find a second, or new gauge-point. The second gauge-points are the square roots of ten times the divisors in the above table. Thus, for squares, the new gauge- point for imperial gallons is 52-65, for bushels 14.8*93 ; and for circles, the new gauge-point for gallons is 59*42, for malt bushels 168 05. PROBLEM II. Tojind the area^ in imperial gallons^ of any rectilineal plane figure. Rule. By the rules given in Mensuration of Superficies, find the area of the figure in inches, which bein^ divided by 277*274, or mul- tiplied by '0036065, w^ill give the area in gallons.* 1. Suppose a back or cooler in the form of a parallelogram to be 100 inches in length, and 40 in breadth ; required the area in impe- rial gallons ? 100x40 = 4000 the area in inches, which divided by 277*274, the quotient 14 '426 = the number of imperial gallons; or if we multiply 4000 by "0036065, the product 14.426 is the number of imperial gal- lons as before. BY THE SLIDING RULE. On A. On B, On A. On B. As 277*274 : 40 : : 100 : 14-4, nearly. 2. If the side of a square be 40 inches, what is the area in imperial gallons? ^ ^??5. 5*77 gallons. 3. If the side of a rhombus be 40 inches, and its perpendicular Dreadth 37 inches ; required its area in wine gallons? Ans. 5*41. 4. What is the area of a square cooler, in imperial gallons, the side being 144 inches? Ans. 74*785. 5. Allowing the side of a hexagon to be 64 inches, and the per- pendicular from the centre to the middle of one of the sides 55*42 inches ; required its area in imperial gallons and malt bushels ? A no (38*38 imperial gallons. ^^^' \ 4*8 malt bushels. PROBLEM III. The diameter of a circular vessel being given in inches^ to find its area in imperial gallons^ Sfc. Rule. Multiply the square of the diameter by '002832 ; or divide the square of the diameter by 353*036, the product or quotient will give the area in imperial gallons. When it is required to find the area in any other denomination than imperial gallons, use the proper multiplier or divisor for the required denomination, as given in the Table, page 136. * The areas of plane figures, in gauging, are expressed in gallons, or bushels. For there will be as many solid inches in any vessel of one inch deep, as there are superficial inches in its base. What is called in gauging a surface or area, is in reality a solid of one inch deep, which multiplied by the height will givo the whole content in gallons or bushels. 138 GAUGING 1. The diameter of a circular vessel is 32 '6 inches ; required the area in imperial £:allons ? (32-6y=1062'76. Then, 1062-76 X •002832 = 3-01 gallons. Or, 1062-76-r-353-036 = 3-01. BY THE SLIDING RULE. As 18 '78 is the circular gauge-point for imperial gallons, say, On D. On B. On D. On B. As 18-78: l::32-6 :3 2. If the diameter of a circular vessel be 10 inches, what is the area in imperial gallons ? Ans, -283. 3. Suppose the diameter of a circular vessel is 30 inches, what is its area in imperial gallons ? Ans. 2-548. 4. What is the area in imperial gallons of a round vessel, whose diameter is 24 inches? Ans, 1*631. PROBLEM IV. Given the transverse and conjugate diameter of an elliptical vessel^ to find its area m imperial measure. Rule. Multiply the product of the two diameters by '002832 ; or divide the product of the two diameters by 353 '036; the product or quotient will give the imperial gallons required. When any other denomination is required, the proper multiplier or divisor in the table is to be employed. 1. Suppose the longer diameter of an elliptical vessel is 10, and Vat shorter diameter 6, required the area in ale and wine gallons ? Here, 10x6 = 60. Then, 60 x -003832 = -17 of a gallon. 2. The transverse or longer diameter of an elliptical vessel is 20, and the conjugate or shorter diameter 10 inches ; what is the area in imperial measure ? Ans, '666 of a gallon. On A. On B. On A. On B. As 353 : 20 : : 10 : -566 of a gallon. 8. Suppose the transverse diameter of an elliptical vessel is 70 inches, conjugate 50 inches; required its area in imperial gallons and malt bushels? ^^_ (9*914 gallons. ^^^' "11-24 malt bushels. Note. — As vessels are seldom or never made truly elliptical, being generally ovals, the area found by the above rule is not correct, except the vessel be a truly mathematical ellipsis ; when the vessel is of an oval form, the area is best found by the method of equidistant ordinates. Let A B C D be the oval vessel whose area is required, and let AB and CD be the transverse and conjugate diameters, at right angles to each other, the former behig 102-8 inches. Divide this transverse (102-8) by some even number which will leave a small remainder, the quotient will be the distance of the ordinates ; which distance may be laid off on both sides of the conjugate diameter GAUGING. 139 a number of times equal to half the even number by which the transverse was divided ; then with chalk and a parellel ruler, draw the ordinates through the points 1, 2, 3, 4, &c. Then, by Problem XXL, Sec. III., the area may be found, which being multiplied or divided by the proper tabular number?, will give the area in gallons, &c. Or, 1st, Add together the first and last ordinates. 2d, Add together the even ordinates, that is, the 2, 4, 6, 8, 10, &c., and multiply the sum by 4. 3d, Add together the odd ordinates, except the first and last ; that IS, add the ordinates, 3, 5, 7, 9, &c., and multiply the sum by 2. 4th, Multiply the sum of the extreme ordinates by their distance from the curve. 5th, Add the three first found sums together, and multiply the Bum by the common distance of the ordinates, and to the product add the fourth found sum, and divide the total by 3, and the quotient resulting by 277*274, or 2218*192, for the area in imperial gallons, or malt bushels, respectively. First, 102*8-h10 = 10 the distance of the ordinates asunder, and the remainder 2 8 is double the distance of the extreme ordinatea from the curve; that is, 1*4 = A 1, or B 11. Now let us suppose the lengths of the ordinates to be 20, 40*2, 57, 66*6, 73, 75, 73, 66*6, 67, 40*2, 20, respectively beginning at 1, and proceeding to 11. 1st, {.l: = 20 20 40 inches, sum of the firat and last 1-4 fl: 2A, -! 6 = dd, 56 2 = 40-2 66*6 750 8 = 66*6 10=40*2 288*6x4 = 1154'4 3 = 57 5 = 73 7 = 73 9 = 57 "260 >< 2=620 HO GAUGING. Then, 40+ 11 54 4 + 520 = 171 4 -4 sum of first three sums. 10 17144 56 "3)17200 5733-3; then, 5733-3-+277-274 = 20-64 gallons. 5733-3-T-2218-192 = 2-58 malt bushels. When the vessel is not circular, or elliptical, it is best to measure the equidistant ordinates, which though ever so unequal, will, by proceeding as above, serve to find the area of the base. Whenever the vessel is an irregular curved figure, the area should be invariably found by the method of equidistant ordinates, as the true result cannot be found by any other method. 4. What is the area, in imperial measure, of an ellipse, whose . transverse axis is 24, and conjugate 18? Ans, 1-2234 gallons. PROBLEM V. To find the content ofaprism^ in imperial gallons. Rule, Find the area of the base, by Problem II. in Gauging, which being multiplied by the depth within, will give the content in gallons. Or, find the solid content by mensuration, and divide that content by 277-274 for imperial gallons. A vessel, whose base is a right-angled parallelogram, is 49-3 inches in length, the breadth 36*5 inches, and the depth 42*6 inches ; required its content in imperial gallons ? Here, 49-3 x 36-5 x 42-6 = 76656-57. Then, 76656-57-t-277-274 = 276'465 gallons. And 76656-57-r-2218-192 = 34-558 malt bushels. BY THE SLIDING RULE. OnB. OnD. OnB. 49-3: 49-3:: 36-5: 42-42. OnD. OnB. OnD. 16-65> . .o-e . . 42-42 • |27-6 gallons. 46-371 . *^ *> . . *^ ^^ . -(34.5 malt bushels. 2. Each side of the square base of a vessel is 20 inches, and its depth 10 inches, what is the content in old ale gallons ? Ans. 14-28 gallons. 3. The side of a vessel in the form of a rhombus is 20 inches, breadth 15 inches, and depth 10 inches ; required the content in old ale gallons? Ans. 10*638 gallons. 4. What is the content, in old wine gallons, of a vessel in the form of a rhomboid, whose longest side is 20 inches, breadth from side to Bide 8 inches, and depth 10 inches? Ans. 6-88 wine gallons. GAUGING. - 14| PROBLEM VI. To find the content of any vessel, whose ends are squares or rectangles, of any dimensions. Rule. Multiply the sum of the lengths of the two ends, by the 8um of their breadths, to which add the areas of the two ends ; this sum, multiplied by one-sixth of the depth, will give the solidity in cubic inches ; then divide by 277-274, or 2218'192 for the content in imperial gallons, or malt bushels. 1. Suppose the top and bottom of a vessel are parallelograms, the length of the top is 40 inches, and its breadth 30 inches ; the length of the bottom is 30 inches, and its breadth 20 ; and the depth 60 inches ; required the contents in imperial gallons ? 40 + 30 = 70 sum of the lengths. 30 + 20 = 50 sum of the breadths. 3500 product. 40 X 30 = 1200 area of the greater base. 80 X 20= 600 area of the lesser base. 5300 10 one-sixth of the depth. 53000 solidity in cubic inches. Then, 53000-277-274= 191 '146. BY THE SLIDING RULE. Find a mean proportional (V(40 x 30) = 34*64,) between the length and breadth at the top, and a mean proportional ( V(^0 x 20) = 24-49,) between the length and breadth at the bottom; the sum of these is 59*13, twice a mean proportional between the length and breadth in the middle. Then, On D. On B. On D. On B. :-64 : (34-6 :: ^24-4 (59-1 16*65 : V : i •< 24-49 : }- sum 191*146 imperial gallons. •13 : ) 2. Suppose the top and bottom of a vessel are parallelograms, the length of the top is 100 inches, and its breadth 70 inches ; the length of the bottom 80, and its breadth 56, and the depth 42 inches ; what is its content in imperial gallons? Ans, 862*59 imperial gallons. THE GAUGING OR DIAGONAL ROD. The diagonal rod is a square rule, having four faces, and is gene- rally 4 feet long. It folds together by joints. This instrument is employed both for gauging and measuring casks, and computing their contents; and that from one dimension only, namely, the diagonal of the cask, or the length from the middle of the bung-hole to the meeting of the cask with the stave opposite the bung ; being 142 GAUGING. the longest line that can be drawn from the middle of the bung-hole to any part within the cask. On one face of the rule is a scale of inches for measuring this diagonal; to which are placed the areas, in ale gallons, of circles to the corresponding diameters, in like manner as the lines on the under sides of the three slides in the Sliding Rule. On the opposite face, there are two scales of ale and wine gallons, expressing the contents of casks having the correspondent diagonals. All the other lines on the instrument are similar to those on the Sliding Rule, and are used in the same manner. Example. The diagonal, or distance between the middle of 7he bung-hole to the most distant part of the cask, as found by the dia- gonal rod, is 34*4 inches ; what is the content in gallons? To 34*4 inches correspond, on the rod, 90| ale gallons, or 111 wme gallons, 92 J imperial gallons, the content required ? Note. — ^The contents shewn by the rod answer to the most commou form of casks, and fall in between the 2d and 3d varieties following. OF CASKS AS DIVIDED INTO VARIETIES. Casks are usually divided into four varieties, which are easily distinguished by the curvature of their sides. 1. The middle frustum of a spheroid belongs to the first variety. 2. The middle frustum of a parabolic spindle belongs to the second rariety. 3. The two equal frustums of a paraboloid belong to the third Variety. 4. And the two equal frustums of a cone belong to the fourth variety. If the content of any of these be found in inches by their proper rules, and this divided by 277*274, or 2218*2, the quotient will be the content in imperial gallons, or bushels, respectively. PROBLEM VII. To find the content of a vesselin the form of the frustum of a cone. Rule. To three times the product of the two diameters add the square of their difference; multiply the sum by one-third of the depth, and divide the product by 353-0362 for imperial gallons, and by 2824-289 for malt bushels. 1. What is the content of a cone's frustum, whose greater dia- meter is 20 inches, least diameter 15 inches, and depth Ai inches ? 20x15x3 =900 20— 15 = 5 and 62= 25 "925x7=6475. Then, 353'0362)6475(18'34 imperial gallons. 294-12)6475(22-01 wine gallons. 2. The greater diameter of a conical frustum is 38 inches, the less diameter 20*2, and depth 21 inches ; what is the content in old ale gallons? Ans. 51*07 gallons. GAUGINa. 143 PROBLEM VIII. To find the content of the frustum of a square pyramid, RuLF To three times the product of the top and bottom sides, add the square of their difference, multiply their sum by one-third of the depth, and divide the product by 282 and 231, for old ale and wine gallons, respectively ; and by 277*274, for imperial gallons. 1. Suppose the greater base is 20 inches, the less base 15 inches, and depth 21 inches ; required the content in old wine measure ? 20x15x3 = 900 20—15 = 5 Then, 5x5= 25 925 X 7-=-231 = 27-8 gallons. Note. — ^The content of the frustum of a pyramid is found just like that of a cone, with the exception of the tabular divisor, or multiplier, the cone require ing the circular factor, and the pyramid the square one. PROBLEM IX. To find the content of a globe. Rule. Multiply the diameter of the globe by its cir^jumference, and the resulting product by one-sixth of the diameter; then the last product multiplied or divided by the circular factor, will give the content in gallons. 1. Let the diameter be 34 inches, what is its content? 34 X 34 X 34 X •5236 = 20579-5744. Then, 20579-5744-t-282 = 72-9772 old ale gallons. And, 20579-5744-f-231 = 89-08 old wine gallons. Rule. II. Or cube the diameter of the globe, which multiply by •001888 (I of -002832) for the content in imperial gallons. 343 = 39304; then 39304 x •001888 = 74-2 imperial gallons. 2. What is the coirtent of a globe in old ale and wine measme, the diameter being 20 inches ? . J 14-848 old ale gallons. '^^^' I 18-128 old wine gallons. 3. Required the content of a globular vessel, whose diameter is 100 inches? An's. 1888^ imperial gallons. PROBLEM X. To find the content of the segment of a sphere^ as the rising crowr of a copper stilly ^c. Rule. Measure the diameter, or chord of the segment, and the altitude just in the middle. Multiply the square of half the diameter by 3 ; to the product add the square of the altitude ; multiply this 8um by the altitude, and the product again by •001856, or -002266, for old ale or wine measure, respectively, and by -001888 forimperiaj gallons. 144 GAUGING. 1. The diameter of the crown of a copper still is 27*6, its depth 9'2 ; required its content? Here 27-6-7-2= 13-8. Then 13-8 x 13-8 x 3 = 571*32 9-2x9-2= 84-64 655-96 sum. 9-2 depth. 6034-832 X -001888 = 13-39 imperial gallons. PROBLEM XL To gauge a copper having either a concave or convex bottom; or what is called a falling bottom, or rising crown. Rule. If the side of the vessel be straight, with a falling bottom, find the content of the segment Q y D, by Prob. X. ; find also the content of the upper part ABDC, by Prob. VIL ; the sum of both ^^ will give the content of the '3opper. When the copper has a ris- ing crown, find the content of ABCD, by Prob. VIL; from which deduct the content of the segment C a; D, and the remainder will be the content of the vessel A B D a; C. PROBLEM XII. • To gauge a vessel whose side is curved from top to bottom. Take the diameters at equal distances of 2, 3, 4, or 5 inches, according as the case may require ; if the side of the vessel be con- siderably curved, the number of diameters that will be required will be considerable ; the less the curvature of the side, the less the num- ber of diameters that will be required. To gauge the vessel, or cop- per, ABDC, fasten a piece of pack-thread at A and B, as AFB; then with some con- venient instrument find the distance aC of the deepest part of the copper, which let us suppose to be 47 inches. By means of the same in- strument measure the distance oF from the top of the crnwn to F the middle of A B ; which let us suppose to be 42 inches, this de- ducted from a C, 47, wiU leave 5 ( = o G) the height of the crown. To find the diameter CD, of the bottom of the crown. Measure the top diameter AB, which suppose to be 99 inches: then hold a thread, so that a plummet attached to the end thereof may hang just over C, and measure Aa=B F, each of which let U3 GAUGING. Hff admit to be 17*5 inches; add these together, and deduct their sura (85) from 99, and the remainder (64) will evidently be equal to CD, the diameter at the bottom of the crown. Measure the diameter mon^ which touches the top of the crown, which suppose is 65 inches. Now, as this copper is not considerably curved, the diameters may be taken in the middle of every 6 inches of the depth, whicfi suppose to be as in the second column of the following table; to each diameter find the area in imperial gallons, by Prob. III., which write in the third column ; find also the content of every 6 inches, corresponding to these diameters, which write in the fourth column of the table ; lastly, find the content of the crown by Prob. X., and subtract it from the content of A B D G C, the remainder will give the capacity of the copper. Or thus, CD being 64 inches, the area answering to it is 11'6022 ; this multiplied by half the altitude of the crown, viz., by 2 -5, gives .^9*0055 gallons, the content of the crown. The content of the part mnDC is 58'9222 gallons, from which the content of the crown being deducted, the remainder (29*9167 gallons) is the quantity of Vquor which covers the crown. Parts of the depth. Diameters. Areas. Content of every 6 inches. 6 95-3 25-7257 154-3542 6 90-1 22-9948 137-9688 6 85- 20-4653 122-7918 6 80- 18-1284 108-7704 6 75-2 16-0183 96-1098 6 70-5 14-0786 84-4716 6 (jQ- 12-3387 74-0322 The sum 778-4988 To cover crown 29-9167 The whole content 808'4155 PKOBLEM yill. To find the content of any close cash. Whatever be the torm of the cask, the following dimensions must be taken ; that is, The bung diameter, The head diameter. The length of the cask. On account of the difficulty in ascertaining the figure of the cask. It is not, in many cases, easy to find the exact contents of casks. > within. 146 GAUGING. In taking the dimensions of a cask, it is essential that the bung* hole be in the middle of the cask, and also that the bung-stave, and the stave opposite to it, are both regular and even within. It is likewise essential that the heads of casks are equal and truly circular ; and if so, the distance between the inside of the chimb to the outside of the opposite stave will be the head diameter within the cask, nearly. From the variety in the forms of casks, no general rule could be given to answer every form ; two casks may have equal head dia- meters, equal bung diameters, and equal lengths, and yet their con- tents may be very unequal. PROBLEM XIV. To find the content of a cask of the first variety. Rule. To the square of the head diameter add double the square of the bung diameter, and multiply the sum by the length of the cask. Then multiply the last product by -OOOOf , or divide by 1059 '1, the product or quotient will be the content in imperial gallons. 1. What is the content of a spheroidal whose length is 40 inches, bung diameter 32 inches, and head diameter 24 inches ? 24x24= 576 32x32=1024 2 2624x40=104960 •00091 944640 34987 11662 99*1289 imperial gallons. BY THE GAUGING RULE. Set 40 on C, to the G R 18-79 on D, against 24 on D, stands 64-99 on C, 32 on D, stands 116-2 on C, + 116 2 3)297-39 99 13 gallons. GAUGTNa. m 2. What is the content of a spheroidal cask, whose length is 20 inches, bung diameter 16 inches, and head diameter 12 inches? 4 (12-36 old ale gallons. '^ ^* \U'86d old wine gallons. Tojind the content of a cask hy the mean diameter. Rule. Multiply the difference of the head and bung diameters by '68 for the first variety ; by -62 for the second variety ; by -55 for the third ; and by "5 for the fourth, when the difference between the head and bung diameters is less than 6 inches ; but when the difference between these exceeds 6 inches, multiply that difference by '7 for the first variety ; by '64 for the second ; by '57 for the third ; and by '52 for the fourth. Add this product to the head diameter, and the sum will be a mean diameter. Square this mean diameter, and multiply the square by the length of the cask; this product multiplied, or divided, by the proper multiplier or divisor^ will give the content. By resuming the last example but one, we have Bung diameter 32 29-6 mean diameter. Head diameter 24 29*6 876-16 square. 40 length. 6-6 24 859-5)35046-40 97-6 gallons. Mean diameter 29*6 In the same manner the content for the second variety will be 94-46 ale gallons; for the third variety 90*87 ale gallons; and for vhe fourth variety 83*34 gallons. PROBLEM XV. Tojind the content of a cask of the second variety. Rule. To the square of the head diameter add double the square- of the bung diameter, and from the sum deduct two- fifths of the square of the dif- ference of the diameters* multiply the remainder by the length, and the product again by -00091 for the content in imperial gallons. 1 . What is the content of a cask, whose length is 40 inches, bung diameter 32 inches, and head diameter 24 inches ? 32—24 = 8; then 8^ = 64, and f of 64 = 25*6 242 = 576, and 32^ = 1024, then 1024x2 = 2048 2048 + 576 = 2624, and 2624-256 = 2598*4 40 103936 X •00091 = 98*1617 gallons. 103936 148 GAUGINQ. PROBLEM XVI. To find the content of a cask of the third variety. Rule. To the square of the bung diameter add the square of the head diameter ; multiply the sum by the length, and the last product Vy '001416 for the answer in imperial gallons. Let us resume the last example : thus 322= 242= = 1024 = 576 1600x40= 64000 •001416 90*624 imperial gallons. PROBLEM XVII. To find the content oj a cask of the fourth variety. Rule. Add the square of the difference of the diameters to 3 timej' the square of their sum ; multiply the sum \ by the length, and the last product by ^^^ ^^■^-^ I •000236 for the content in gallons. Resuming still the last example, 32 + 24 = 56, and 562x3 = 3136x3 = 9408, and 82= 64, then 9408 + 64 = 9472 ; then 9472 x 40 = 378880, and 378880 x •000236 = 89*41668 imperial gallons. PROBLEM XVIII. To find the content of any cask by Dr Hutton''s general rule. Rule. Add into one sum, 39 times the square of the bung dia- meter, 25 times the square of the head diameter, and 26 times the product of the two diameters ; then multiply the sum by the length, and the product again by -000311 for the content in gallons. 1. What is the content of a cask, whose length is 40 inches, and the bung and head diameters 32 and 24 ? 322=1024 242=576 32x24 = 768 39 25 26 39936 14400 19968 14400 19968 74304 X 40 = 2972160 •000311- 93*4579 gallona. QAUGINCr. 149 ULLAGING. PROBLEM XIX. To ullage a lying cask. This is the finding what quantity of liquor is contained in a cask when partly empty. To ullage a lying cask, the wet and dry inches must be known, as also the content of the cask and bung diameter. Rule. Take the wet inches, and divide them by the bung dia- meter ; find tlie quotient in the column of versed sines, in the Table at the end of the practical part of this book, and take out its corresponding segment; multiply this segment by the whole content of the cask, and the product arising by IJ for the ullage required, nearly. 1. Find the ullage for 8 wet inches, the bung diameter being 32 inches, and the content 92 ale gallons ? 32)8('25, whose tabular segment is -ISSS^G. Then, •153546x92=14-126232. And, 14-126232 X 1^=17-65779 gallons. PROBLEM XX. To ullage a standing cask. Rule. Add together the square of the diameter at the surface of the liquor, the square of the diameter of the nearest end, and the square of double the diameter taken in the middle between the other two ; multiply the sum by the length between the surface and nearest end, and the product arising by -000472 for the gallons in the less part of the cask whether empty or filled. 1. What is the ullage for 10 wet inches, the three diameters being 24, 27, and 29 inches? 242= 576 43330 292= 841 -000472 (2x27)^=^2916 ^^ 4333 303310 10 173320 43330 20-45176 gallons. PROBLEM XXL To find the content of an ungula^ or hoof, of the frustum of a cone. Rule. For the less hoof, multiply the product of the less diameter and height, by the product of the greater diameter multiplied by a 150 GAUGING. mean proportional between both diameters, less the square of the less diameter, and this last divided by three times the circular factor multiplied by the difference of the diameters, gives the content of the less hoof. 1. CD = 30, AB = 40, C(Z=:20; required the ^ ZZ — --v^ b content of the less hoof ? 40x30 =1200, and V 1200 = 34-6 mean. 80 X 20 = 600, 1st product. 40 X 34-6 = 1384, 2d product. 30x30 = 900 **? ., 484 remainder. 484x600 = 290400 40->^30=10, then 359 x 3x 10 = 10770Vaiiii':r;i7rO7?<'i^"^ 290400-^10770 = 26-96 gallons. \^ ^J "^—^ Rule. For the greater hoof multiply the product of the greater diameter and the height of the frustum, by the square of the greater diameter made less by the product of the less diameter multiplied by a mean proportional between those diameters ; this remainder, divided by three times the circular divisor multiplied by the differ- ence of the diameters, gives the content of the greater hoof. Resuming the last example, vre have ^ 40x40 = 1000 20x40 = 800, 1st product. 40x30=1200, and V 1200 = 34-6 34-6 X 30 = 1038, 2d product. / *- ] 40-30=10. lA/JndLJ^J^ Then 1600-1038 = 562 -/-"7?^^ 800 859 X 3 X 10= 10770)449600, last product. 41*74 old ale gallons. PROBLEM XXII. To gauge a still. Fill the still with water, and draw it off in another vessel of some regular form, whose content is easily computed. This is by far the most accurate method that can be employed. Or gauge the shoulder by itself, and gauge the body by taking a greater number of diameters at near and equal distances throughout, first covering the bottom, if there be any cavity, with water, the uantity of which is known. LAND-SURVEYING. IBl LAND-SUEYEYING. /^.-sp.-.. .^'^ ,/ 71 SECTION XI. Land-surveying is that art which enables us to give a true plan or representation of any field or parcel of land, and to determine the superficial content thereof. In measuring land, the area or superficial content is always ex- pressed in acres, or in acres, roods, and perches ; each acre containing 4 roods, and each rood 40 perches. Land is measured with a chain, called Gunter's chain, of 4 poles or 22 yards in length, which consists of 100 equal links, each link being ^^V of a yard long, or ,% of a foot, or 7-92 inches. 10 square chains, o: 10 chains in length and 1 in breadth, make an acre ; or 4840 square yards, 160 square poles, or 100,000 square links make an acre. The length of lines measured with a chain are generally set down in links as integers ; every chain being 100 links in length. Therefore, after the content is found, it will be in square links, and as 100,000 square links make an acre, it will be necessary to cut of! five of the figures on the right-hand for decimals, and the rest will be acres. The decimals are reduced to r6ods by multiplying by 4, and cutting off" five figures as before for decimals, which decimal part is reduced to perches by multiplying by 4^, and cutting ofi*five figures from the product. As an example : — Suppose the length of a rectangular piece of ground to be 792 links, and its breadth 385 ; required the number of acres, roods, and perches it contains ? 792 3-04920 385 4 3960 -19680 6336 40 2376 . ■ 7-87200 304920 Ans. 3 acres, roods, 7 perches. The statute perch is 5 J yards, but the Irish plantation perch is 7 yards ; hence the length of a plantation link is 10*08 inches. PROBLEM I. To measure a line or distance on the ground, two persons are employed ; the foremost, for the sake of distinction, is called the leader, and the hindermost, the follower,.. Ten small arrows or rods, to stick in the ground at the end of each 152 LAND-SURVEYING. chain, are provided; also some station -staves, or long poles with coloured flags, to set up in the direction of the line to be measured, if there do not appear some marks naturally in that direction. The leader takes the 10 aiTOvs^s in one hand, and one end of the chain by the ring;, in the other ; the follower stands at the beginning of the line, holding the ring at the end of the chain in his hand, while the leader drags forward the chain by the other end of it, till it is stretched straight, and the leader directed by the follower, by moving his hand, to the right or left, till the follower see him exactly in a line with the mark or direction to be measured to ; then both of them holding the chain level and stretched, the leader sticks an arrow upright in the ground, as a mark for the follower to come to, and advances another chain forward, being directed in his position by the follower standing at the arrow, as before, as also by himself, now and at every succeeding chain's length, by moving himself from side to side, till the follower and back-mark be in a direct line. Having then stretched the chain, and stuck down an arrow, as before, the follower takes up the arrow, and thus they proceed till the 10 arrows are employed, or in the hands of the follower, and the leader, without an arrow, is arrived at the end of the eleventh chain-length. The follower then sends or brings the 10 arrows to the leader, who puts one of them down at the end of his chain, and advances with his chain, as before. And thus the arrows are changed from one to the other at every 10 chains' length, till the whole line is finished, if il exceed 10 chains; and the number of changes shews how many times 10 chains the line contains, to which the follower adds the arrows he holds in his hand, and the number of links of another chain over to the mark or end of the line. Thus, if the whole line measure 36 chains 45 links, or 3645 links, the arrows have been changed three times, the follower will have Arrows in his hand, the ^ leader 4, and it will be 45 links from the la^ arrow, to be taken up by the follower, to the end of the line. In works on Surveying, it is usual to describe the various instru- ments used in the art. The pupil, however, will best learn the use of these instruments when actually engaged in the practice. The chief instruments employed are the chain, the plane table, tlie theo- dolite, the cross, the circumferentor, the offset staff, the perambulator, used in measuring roads, and other great distances. Levels, with telescopic or other sights, are used to find the levels between two or more places, or how much one place is higher or lower than the other. Besides all these, various scales are used in protracting and measuring on paper ; such as plane scales, line of chords, protractor, compasses, reducing scales, parallel and perpendicular rulers, &c. THE FIELD-BOOZ. In surveying with the plane table, a field-book is not required, as everything is drawn on the table immediately when it is measured. But when the theodolite, or any other instrument, is us^, some sort of a field-book is used in order to register all that is done relative to the survey in hand. This book every one contrives and rules as he r LAND-SURVEYING. 153 thinks fit. It is, however, usuall}' divided into three columns. The middle column contains the different distances on tlie chain-line, angles, bearings, &c., and the columns on the right and left are for the offsets on the right and left, which are set against their cor- responding distances in the middle column ; as also for such remarks as may occur, and may be proper to note in drawing the plan ; sucn as houses, ponds, castles, churches, rivers, trees, &c. &c. But in smaller surveys, an excellent way of setting down the work is, to draw by the eye, on a piece of paper, a figure resembling that which i? to be measured ; and then write the dimensions, as they are found, against the corresponding parts of the figure. This method may be practised even in larger surveys, and is far superior to any other at present practised. A specimen of this plan will be seen further on. FORM OF THE FIELD-BOOK. Offsets and remarks on the left. stations, Bearings, and Distances. Offsfets and remarks on the right. D 1 104° 25' 00 Cross a hedge 24 67^ Brown's bam. a brook 30 120 u ' . /'^ 734 954 Tree. w.- "-1^ ./ .. .i' y 736 67 stile. 82 62"' 25' 00 .L House comer €1 40 67 Foot path 15 84 95 44 467 14 Spring. J \, 1 ' 976 ns '^ W .,-:,. ., 1 54*= ir 62 20 Pond. 124 Clayton's hedge 24 630 767 767 30 Stile. ' <. '^' 305 760 - I5i LAND-SURVEYING. In this form of a field-book D 1 is the first station, where the angle or bearing is 104° 25'. On the left, at 67 links in the distance or principal line, is an offset of 24 ; and at 120 an offset of 30 to a brook on the right ; at 67 Brown's barn is situated ; at 954 ia an offset of 20 to' a tree, and at 736 an offset to a stile. And so on for the other stations. A line is drawn under the work, at the end of every station, tO prevent confusion. PROBLEM II. To take angles and bearings. Let it be required to take the bearings of the two objects B, C, from the station A. In this problem it is required to measure the angle ^ at A, formed by two lines, passing from the station A, through two objects B and -G. 1. By measurement with the chain., ^c. Measure with the chain any distance along the two lines A B, AC, as A 6, Ac; then measure the distance be; and this being done^ transfer the three sides of the triangle Abe to paper, on which measm-e the angle c A &, as in Problem XV., Practical Geometry. 2. With the magnetic needle and compass. Turn the instrument, or compass, so that the north end of tho needle may point to the flower-de-luce. Then direct the sights to a mark at B, noting the degrees cut by the needle. Next direct the sights to another mark at C, noting the degrees cut by the needle as before. Then their sum or difference, as the case may be, will give the number of degrees in the angle GAB. 3. With the theodolite, ^c. Direct the fixed sights along the line A B, by turning the instru- iiient about till you see the mark B through these sights, and in that position screw the instrument fast. Then turn the moveable index about till, through its sights, you see the other mark G. Then the degrees cut by the index, on the graduated limb or ring of the instru- ment, shew the number of degrees in the angle GAB. 4. With the plane table. Having covered the table with paper, and fixed it on its standi plant it at the station A, and fix a fine pin, or a point of the compass, in a proper point of the paper, to represent the station A. Glose by the side of this pin, lay the fiducial edge of the index, and turn it about, still touching the pin, till one object B can be seen through the sights ; then by the fiducial edge of the index draw a line. By a similar process draw another line in the direction of the object C. And it is done. LAND-SURVEYING. 155 PROBLEM IlL To measure the offsets. Let Abed efg be a crooked hedge, river, or brojok, &c., and A Gr a base line. Begin at the point A, and measure towards G; and when yon come opposite any of the corners bcd^ &c., which is ascertained by means of the cross- staff, measure the offsets B &, C c, D J, &c., with the chain, and register the dimension, as in the annexed field-book. FIELD-BOOK. 91 57 98 70 84 62 785 = AG. 634 510 340 220 45 D A go North. , Offsets Left. Base line A G, or □ Station. Offsets Right. 2'o lay down the plan. Draw the line AG of an indefinite length ; then by a diagonal scale, set off A B equal to 45 links ; at B erect the perpendicular B b equal to 62 links taken from the same scale. Next set oft' A C equal to 220 links, or 2 chains 20 links, and at C erect the perpendicular C c, equal to 84 links ; in the same way set off A D equal to 340 links, or 3 chains 40 links, and at D erect the perpendicular D d equal to 70 links. Proceed in a similar manner with the remaining offsets, and straight lines joining the points A, &, c, d^ e, &c., will complete the figure. To find the content. Some authors direct to add up all the perpendiculars B&, Cc, &c.^ and divide their sura by the number of them, tlien multiply the quo- tient by the length AG. This method, however, should never be used, except when the offsets B &, Cc, &c., are equally distant from each other. When the offsets are not equally distant from each other, which indeed is generally the case, this method is erroneous ; therefore the following method ought to jje employed. 159 LAND-SURVEYINa. Find the content of the space A B & as a triangle, by Problem V., Section II. Find the contents of the figures BCcb^ C D c? c, &c., as trapezoids, by Problem XIII., Section II., the sum of all these separate results will be the content of the figure A G gfe dch A. The actual calcwlation is as follows : — CALCULATION. AB= 45 b6= 62 AC = 220 AB= 45 AD = 840 A = 220 AE = 510 AD = 340 AF=634 AE=510 AG = 785 AF = 634 90 BC=175 CI>=120 DE = 170 EF = 124 GF=151 270 2790 Bb=; 63 Cc= 84 Cc= 84 Dc^= 70 Dd= 70 Ee= 98 Ee= 98 ?/= 57 F/= 57 Gg= 91 Pr Sum 146 BC=175 S»m 154 CD=120 Sum 168 DE = 170 Sum 155 EF=124 Sum 148 FG=151 od. 25550 18480 28560 19220 22348 These respective products are evidently double the true contents of the respective figures A B &, B C c 6, C D df c, &c., that is, 2790 = double area of A B 6. 25550 = double area of B C c &. 18480 = double area ofCJ>de. 28560= double area of D E e d. 19220 = double area of E F/e. 22348 = double area of F G gf. 2)116948= double area of the whole in square linkb. 68474 = area in square links. y<(> •58474= area in acres =0a., 2r., 13 -5584?. ^r^ 2. Required the plan and ccaitent of part of a field, from the fol- lowing field-book : — A t> AC 45 62 C^ A 6/ 220 84 di Ae 340 70 ek A/ 510 88 fl Acr 634 AB785 57 am 91 Bn Ans, Oa., 2r., 12P. h- LAND-SURVEYINO. 157 PROBLEM IV. To measure ajield of a triangular form, — 1. By the chain. Set up marks at the three corners A, B, C, and c measure with the chain, the distance A D, D being the point at which a perpendicular demitted from C, would meet the line A B ; measure also the distance D B ; hence you have the measure of A B. Next measure the perpendicular D C ; then from the two dimensions A B and D C, the contei^t may A,/ ^ ^F be found by Problem IV., Section II. ^ Let AD = 794, AB=1321, DC = 826 links. 1321 X 826-7-2 = 545573 links. Then 545573-^-100000 = 5-45573 acres. 45573 X 4 = 1 -82292 roods. 82292x40 = 32-91680 perches. Hence the answer is 5a., 1r., 33p., nearly. 2. What is the area of a triangular field, whose base is 12*25 chains, and height 8-5 chains? Ans. 5A., Dr., 33p. 2. By taking one or more of the angles. Measure two sides A B, AC, and the angle A, included between them; then half the continual product of the two sides, and the natural sine of the contained angle will give the area.* Or, measure the two angles A and B, and the adjacent side A B, ) from which the figure may be planned, and the perpendicular C D found, which perpendicular being multiplied by half the base A B, will give the area. Or by measuring the three sides of the triangle, its area may be found by Problem V., Sectiou II. PROBLEM V. To survey a four -sided field. — 1. By the chain. Measure the diagonal A C, and, as be- fore directed, measure the perpendiculars D E and B F ; then the area of each of the triangles A B C, A D C n>ay be found, as in the last problem, and both are^s being added together, wjU give the ccHi- tent of the four-sided Sgure A B C D. 1. Let AC = 592, DE=210, BF=30e links. 592 X 210= 124320 double area of A D C. 592x 306 = 181152 double area of ABC. 2)305472 double area of A B C D. 1-52736 = area of A B C D. 4 2-10944 40 4-37760 Hence lA., 2r., 4r., th« answer. ' See Appendix, Demonstratioa 11. 158 LAND-SURVEYING. 2. By talcing one or more of the angles. Measure the diagonal AC, also the sides AD and AB. Next measure the angles DAG and BAG; then the area of each of the triangles ABC and ADC may be found by case 2, last problem. 2. Required the plan and content of a field by the following field- book : — FIELD- BOOK. 1360 = AB. 1190 600 n D go East. 625 342 Offsets Left. Station D? or base line. Offsets Right. \J" Ans. 6a., 2r., 12p. 3. How many acres are there in a four-sided field, whose diagonal is 4*75 chains, and the two perpendiculars falling on it, from its opposite angles, 2-25 and 3*6 chains, respectively? Ans. lA., IR., 22-3P. PROBLEM VL m. To survey afield of many sides by the chain only. Let ABCDEFG be the field whose content is required. Set up marks at the corners of the field, if there be none there naturally. Con- sider how the field may be best divided into trapeziums and triangles ; measure them separately, as in the two last problems ; and the sum of all the separate results will give the area of the whole field. In this way of measurmg with the chain, the field should be divided into trapeziums and triangles, by drawing diagonals from corner to corner, so that all the perpendiculars may be within the figure. The last figure is divided into two trapeziums A B C G, G D E F, and the triangle G C D. In the first trapezium measure the diagonal A C, and the two perpendiculars G m and B n. In the triangle G C D, measure the base G C, and the perpendicular D q. Finally, measure the diagonal F D, and the two perpendiculars G o and Ep. Having drawn & rough figure resembling the field, set all these measures against the corresponding parts of the figure. Or set them down thus : — LAND-SURVEYING. 139 A n 41o> A C 550) 180 w B ^SS^SO^D F^ 288^20 130 + 180 550-^2 = 275, 275 X 310 = 440x230-^2 = 120 + 80 = 200, 5204-2 = 260, 260 X 200 = CALCULATION. 310, 85250=ABCG. 50600 = CGD. 52000=DEFG. ABCDEFG. 1-87850 = 4 3 51400 40 20-56000 lA., 3r., 20'56p., answer. Other methods will naturally present themselves to an ingenious practitioner who has read the preceding part of this work, or who lias been previously acquainted with the principles of Mathematics. Every surveyor ought to be well acquainted with Plane Geometry at least. This, with a knowledge of Trigonometry, would be sufficient for the purpose of most surveyors. The content of the last figure may be found by measuring the sides AB, BC, CD, DE, E F, FG, GA; and the diagonals A 0, CG, G D, D F, by which the figure is divided into triangles, the content of each of which may be found by Problem V., Section II. 2. Required the plan and content of a field of an irregular form from the following FIELD-BOOK. 268 900 = EG 550 D E, go S.W. 280 1100 = HE 790 350 n H, go East. 410 140 1180=CH 710 350 D C, go S.W. 280 200 900 = AC 430 300 a A, go S.E. 450 Oflfsets Left. 1- Stations, D» or Base Lines. Offsets Right. Ans. IOA.. IB., 24'64P. L 160 lAND- SURVEYING. PROBLEM VII. To survey a field with the theodolite^ ^c. 1. From one point or station. When all the angles can be seen from one point, as suppose C. Having placed the instrument at C, turn it about till, through the fixed sights, the mark B may be seen. Fixing the instrument in this posi- tion, turn the moveable index about, till the mark A is seen through the sights, and note the degrees on the instrument. In the same manner, turn the index successively to the angles E and D, taking care to note the degrees cut off at each ; by which you have all the angles, viz., B C A, B C E, B C D. Now, having obtained the angles, measure the lines C B, C A, C E, CD; entering the respective mea- sures against the corresponding part of a rough figure, drawn to resemble the figure. 2. By going round the field. Set up marks a!, B, C, D, &c. Place the instrument at the point A, and turn it about till the fixed index be in the direction AB, and then screw it fast : turn the moveable index in the direction A F, and the degrees cut off will be the angle A; next measure AB, and planting the instrument at B, mea- sure, as before, the angle B ; measure the line B C, and the angle C : and so proceed round the figure, always mea- suring the side as you go along, as also the angles. The 32d Proposition of the 1st Book of Euclid affords an easy method of proving the work : thus, add all the internal angles. A, B, C, &c., of the figure together, and their sum must be equal to twice as many right angles as the figure has sides, wanting four right angles. But when the figure has a re-enterant angle as F, measure the external angle, which is less than two right angles, and deduct it from four right angles, or 360 degrees, the remainder vsqll give the internal angle (if such it may be called), which ia greater than 180 degrees. When the field is surveyed from one station, as in the first case shewn above, the content of the figure is found as in'the second case of Prob. IV., since we have two sides and the angle included between them in each triangle of the figure. PROBLEM VIII. To survey a field with crooked hedges. Measure the lengths and positions of lines running as near tho sides of the field as you can ; and, in proceeding along these lines, measure the offsets to the different corners, as before taught, and LAND-SURVEYIxS'G. 101 join the ends of the offsets ; these connecting lines will represent the required fi^nire. When the plane table is used, the plan will be truly represented on the paper which covers it. But wlien the survey is made with the theodolite, or other instrument, the different measures are to be noted in the field-book, from which the sides and angles are laid down on a map, after returning from the field. In surveying the piece ABCDEFGHIKLM,set up marks at 5 E Fa;. Begin at the station 5, and measure the lines sE, E^?, px^ xs^ as also their posi- tions, or the angles E 5 a:, ^EF, Ejoar, and pxs; and in going along the four-sided figure s Ep x^ measure the offsets at «, &, d^ g^ k, Z, m^ as before taught. By means of the figure sEpx, and of the offsets, the ground is easily planned. When the principal lines are taken within the figure, as in tha Kbove case, the contents of the ex- terior portions sCBA, ODE, &c., must be added to the area of the quadralateral 5a:FE. But when the principal lines are taken outside the figure, the portions included oetween them and the boundaries of the field. are to be deducted from the content of the quadralaieral, and the remainder will give the true content of the field. When there are obstructions within the jvater, hills, &c., measure the lengths and positions of the four- sided figure atccl^ taking care to measure the offsets from the different corners as you go along. PROBLEM IX. To survey any puce of land by two stations. Choose two stations, from which all the corners of the ground can be seen, if possible ; measure the distance between the stations ; at each station take the angles formed by every object, from the station line, or distance. Then, the station line and these different angles being laid down from a regular scale, and the external points of in- tersection connected, the connecting lines will give the boundary. The two stations may be taken within the bounds, in one of the «;ides, or without the bounds of the ground to be surveyed. Let m and n be two stations, from which all the marks A,* B, C figure, such as wood, iG2 LAND-SURVEYING. &c., can be seen, plant the in- strument at m, and by it mea- sure the angles kmn^ ^mn^ C m 72, &c. Next measure m w, and planting the instrument at 71, measure the angles knm^ BwTTi, Cwm, &c. These ob- servations being planned, the lines joining tlie points of ex- ternal intersection will give a true map of the ground. The method of finding the content will be shewn further on. The principal objects on the ground may be delineated on the map, by measuring the angles at each station, which every object makes with the station line m n. When all the objects to be surveyed cannot be seen from two stations, then three or four may be used, or as many as may be found necessary ; taking care to measure the dis- tance from one station to another ; placing the instrument at every station, and observing the angles formed by all the visible objects with the respective station line ; then the intersection of the lines forming these respective angles will give the positions of all the re- markable objects thus observed. In this manner may very extensive surveys be taken; and the positions of hills, rivers, coasts, &c., ascertained. PROBLEM X. To survey a large estate. The following method of surveying a large estate was first given by Emerson, in his '' Surveying," page 47. It has been followed by Hutton and Keith. When the estate is very large, and contains a great number of fields, it cannot be accurately surveyed and planned by measuring each field separately, and then adding all the separate results together; nor by taking all the angles, and measuring the boundaries that enclose it. For in these cases the small errors will be so multiplied as to render it very much distorted. 1. Walk over the estate two or three times, in order to get a per- Tect idea of its figure. And to help your memory, make a rough draft of it on paper, inserting the names of the different fields within it, and noting down the principal objects. 2. Choose two or more elevated places in the estate for your stations, from which you can see all the principal parts of it ; and let these stations be as far distant from each otlier as possible, as the fewer stations you have to command the whole, the more exact the work will be. In selecting the stations, care should be taken that the lines which connect them may run along the boundaries of the estate, or some of the hedges, to which ofi^sets may be taken when necessary. 8. Take such angles, between the stations, as you think necossary. LAND-SURVEYING. 1 63 and measure the distance from station to station, always m a right line ; these thinj]js must be done till you get as many lines and angles as are sufficient for determining all the station points. In measuring any of these station distances, mark accurately where these lines meet with any hedges, ditclies, roads, lanes, paths, rivulets, &c., and where any remarkable object is placed, by measuring its distance from the station line ; and where a perpendicular from it cuts that line ; and always mind, in any of these observations, that you be in a right line, which you may easily know by taking a back-sight and fore-sight, along the station line. In going along any main station line, take offsets to the ends of ail hedges, and to any pond, house, mill, bridge, &c., omitting nothing that is remarkable. All these things must be noted down; for these are the data by which the places of such objects are to be determined on the plnn. Be careful to set up marks at the intersections of all hedges with the station line, that you may know where to measure from when you come to survey the particular fields that are crossed by this line. These fields must be measured as soon as you have completed your station-line, whilst they are fresh in your memory. In this manner ail the station lines must be measured, and the situations of all adjacent objects determined. It will be proper to lay down the work on paper every night, that you may see how you go on. 4. With respect to the internal parts of the estate, they must be determined by new station lines ; for, after the main stations are determined, and everything adjoining to them, then the estate must be subdivided into two or three parts by new station lines : taking the inner stations at proper places, where you can have tlie best view. Measure these station lines as you did the first, and all their intersec- tions with hedges, ditches, roads, &c., also take offsets to the bends of hedges, and to such objects as appear near these lines. Then proceed to survey the adjoining fields, by taking the angles which the sides make with the station line at the intersections, and measur- ing the distances to each corner from these intersections ; for every station line will be a basis to all future operations, the situation of every object being entirely dependent on them ; and therefore the^- should be taken of as great length as possible ; and it is best io< them to run along some of the hedges or boundaries of one or raorj fields, or to pass through some of their angles. All things being determined for these stations, you must take more inner stations, and continue to divide and subdivide, till at last you come to single fields ; repeating the same work for the inner stations as for the outer ones, till the whole is finished. The oftener you close vour work, and the fewer lines you make use of, the less you will be liable to error. 5. An estate may be so situated that the whole cannot be surveyed together, because one part of the estate may not be seen from another. In this case you may divide it into three or four parts, and survey these parts separately, as if they were lands, belonging to diff"erent persons, and at last join them together. 6. As it is necessary to protract or lay down the work as you proceed in it, you must have a scale of due length to do it by» Tc 164 LAND-SURVEYING. ^et such a scale, measure the wliole length of the estate in chams ; then consider how many inches long the map is to be : and from t hese you will know how many chains you must have in an inch ; then make your scale accordingly, or choose one already made. 7. The trees in every hedge-row may be placed in their proper pituation, which is soon done by the plane table ; but may be done by the eye without an instrument ; and being thus taken by guess in a rough draft, they wili be exact enough, being only to look at ; except it be such as are at any remarkable places, as at the ends of hedges, at stiles, gates, &c., and these must be measured or taken with the plane table, or some other instrument. But all this need not be^ done till the draft is finished. And observe, in all hedges, what side the gutter or ditch is on, and to whom the fence belongs. PROBLEM XT. To survey a town or city. To survey a town or city, it wdll be proper to hax^e an instrument for taking angles, such as a theodolite or plane table ; the latter is a very convenient instrument, because the minute parts may be drawn upon it on the spot. A chain of 50 feet long, divided into 60 links, will be more convenient than the common surveying chain, and an offset staff of 10 feet long will be very useful. Begin at the meeting of two or more of the principal streets, through which you can have the longest prospects, to get the longest station lines. There having fixed the instruments, draw lines of direction along these streets, using two men as marks, or poles set in wooden pedestals, or perhaps some remarkable places in the houses at the farther ends, as windows, doors, corners, &c. Measure these lines with the chain, taking offsets with the staff, at allcomers of streets, bendings, or windings, and to all remarkable objects, as churches, markets, halls, colleges, eminent buildings, &c. Then remove the instrument to another station, along one of these lines, and there repeat the same process as before. And so continue until the w-hole is finished. Thus, fix the instrament at A, and draw lines in the directions of all the streets meeting there ; then measure A C, noting the street at X. At the second station C, draw the directions of all the streets meeting there ; measure from C to D, noting the place of the street K, as you pass by it. At the third station D, take the direction of all the streets meeting there, and measure DS, noting the cross street at T. Proceed in like manner through all the principal streets; after which proceed to the smaller and intermediate Ftreets ; and last of all to the lanes, alleys, courts, yards, and every other place which i| may be thought proper to represent in the plan. LAND-SURVEYING. IfiS PROBLEM XII. To compute the content of any survfi.y, 1. In small and separate pieces, the method generally employed is, to compute their contents from the measm^es of the lines taken in purveying them, without drawing any correct map of them : rulet for this purpose have been given in the preceding part of the work. But in large pieces, and whole estates, consisting of a great number of fields, the usual method is, to make an unfinished but correct plan of tlie whole, and from this plan, the boundaries of which include the whole estate, comjjute the contents quite independent of the measure, of the lines and angles that were taken in surveying. Divide the plan of the survey into triangles and trapeziums, by drawing new lines through it : measure all the bases and perpendiculars of all these new figures, by means of the scale from which the plan was drawn, and from these dimensions compute the contents, whether triangles or trapeziums, by the proper rules for finding the areas of such figures. The chief difficulty in computing consists in finding the contents of land bounded by curved or very irregular lines, or in reducing ' such crooked sides or boundaries to straight lines, that shall enclose an equal area with those crooked sides, and so obtain the area of the curved figure by means of the right-lined one, which in general wilJ be a trapezium. The reduction of crooked sidee to straight ones is easily performed, thus : — Apply a hors€-hair or silk thread across th^ crooked sides in such a manner, that the small parts cut off from the crooked figure by it, may be equal to those taken in. A iittle practice will enable you to exclude exactly as much as you include ; then, with a pencil draw a line along the thread, or horse-hair. Do the same by the other ^ides of the figure, and you will thus have the figure reduced to a straight-sided figure equal to the curved one : the content of which, 1 eing computed as before directed, will be the content of the curved figure proposed. The best way of using the thread or horse-hair is, to string a small slender bow with it, either of whalebone or wire, which will keep it stretched. If it were required to find the contents of the following crooked- sided figure ; draw the four dotted straight lines A B, B C, CD, and D A, excluding as much from the survey as is _^ y" \ B , taken in by the straight lines; by which the crooked figure is re- duced to a right-lined one, both equal in area. Then draw the diagonal B D, which being mea- sured by a proper scale, and multiplied by half the sum. of the perpea- 266 LAND-SURVEYINO. diculars let fall from A and C upon BD (measured on the same scale), will odve the area required. Many other methods might have been given for computing the contents of a survey, but they are omitted, the above being, perhaps, tho most expeditioijs. }^ MISCELLANEOUS PROBLEMS. 167 MISCELLANEOUS PEOBLEMS. 1. The three sides of a triangle are 12, 20, and 28; what is the area ? ^ Ans. 60 V 3. 2. Find the difference between the area of a triangle whose sides are 3, 4, and 5 feet, and the area of an equilateral triangle having an equal perimeter. Ans. '928 of a square foot. 3. There is a segment of a sphere, the diameter of whose base is 24 inches, and its altitude 10 inches ; required its solidity ? Ans. 2785-552 inches. 4. There is a bushel in the form of a cylinder, whose depth is 8» inches, and breadth 18^ inches ; required to determine the breadth of another cylindrical vessel of the same capacity as the former, whose depth is only 7j inches? Ans. 19-107 inches. 5. A ladder, 40 feet long, may be so planted that it shall reach a window 33 feet from the ground on one side of the street ; and by only turning it over, without moving the foot out of its place, it will do the same by a window 21 feet high on the other side ; what ia the breadth of the street? j.'Lns. 56 feet 7| inches. 6. In turning a one-horse chaise within a ring of a certain diameter, it was observed that the outer wheel made two turns while the inner made but one ; the wheels were both 4 feet high ; and supposing them fixed at the statutable distance of 5 feet asunder on the axle-tree, what was the circumference of the track described by the outer wheel ? Ans. 63 feet, nearly. 7. A cable which is 3 feet long, and 9 inches in compass, weigiis; 22 pounds ; what will a fathom of that cable weigh which measures a foot about? Ans. 78f pounds. 8. How many solid cubes, a side of which equals 4 inches, may be cut out of a large cube, whose side is 8 inches? Ans. 8. 9. Determine the areas of an equilateral triangle, a square, a hexagon, the perimeter of each being 40 feet. Ans. 76-980035- 100- -115-47. 10. A person wants a cylindrical vessel 3 feet deep, that shall contain twice as much as another cylindrical vessel whose diameter is 3 J feet, and altitude 5 feet ; find the diameter of the required vessel. Ans. 6*39 feet. 11. Three persons having bought a conical sugar-loaf, wish to divide it into three equal parts by sections parallel to the base ; it is required to find the altitude of each person's share, the altitude of the loaf being 20 inches. Ans. Altitude of the upper part= 13-867 of the middle part=s 8*604, of the lower part = 2-528 inches. 1C8 MISCELL.ANEOUS PROBLEMS. 12. There is a frustum of a pyramid, whose bases are regular octagons; each side of the greater base is 21 inches, and each side of the less base 9 inches, and its perpendicular length 1 5 feet ; how many solid feet are contained in it? Aiis. 119-2 feet. 13. Requiring to find the height of a May-pole, I procured a staff 5 feet in length, and placing it in the sunshine, perpendicular to the horizon, I found its shadow to be 4*1 feet. Next I measured the shadow of the May-pole, whicli I found to be 65 feet ; from this data the heiglit of the pole is required. Ans. 79*26 feet. 14. Given two sides of an obtuse-angled triangle, whicli are 20 and 40 poles ; required the third side, that the triangle may contain just an acre of land? Ans. 58-876 or 23-099. 15. A circular fish-pond is to be made in a garden, that shall take np just half an acre; what must be the length of the cjiord that strikes the circle ? Ans. 27f yards. 16. A gentleman has a garden 100 feet long, and 80 feet broad •, now a gravel walk is to be made of an equal width all round it ; what must the breadth of the walk be, to take up just half the ground? Ans. 12-9846 feet. 1 7. A silver cup, in form of a frustum of a cone, whose top diameter is 3 inches, its bottom diameter 4, and its altitude 6 inches, being filled with liquor, a person drank out of it till he could see the middle of the bottom ; it is required to find how much he drank? Ans. -152127 ale gallons. 18. I have a right cone, which cost me £5, 135. 7(1.^ at 10^. u cubic foot, the diameter of its base being to its altitude as 5 to 8 ; and would have its convex surface divided in the same ratio, by a plane parallel to the base ; the upper part to be the greater ; required the slant height of each part ? A (3-9506486, the slant height of the upper part. ^ns. -11.0854612, the slant height of the under part. 19. How many acres of the earth's surface may be seen from the top of a steeple whose height is 400 feet, the earth being supposed to be a perfect sphere, whose circumference is 25000 miles? Ans. 12120981-338267112 acres. 20. Two boys meeting at a farm-house, had a tankard of milk set down to them ; the one being very thirsty drank till he could see the centre of the bottom of the tankard ; the other drank the rest. Now, if we suppose that the milk cost 4id, and the tankard measured .4 inches diameter at the top and bottom, and 6 inches in depth ; it is required to know what each boy had to pay, proportionable to th-^! quantity of milk he drank. . (14-1 802815 farthings, for the first, ^^^- \ 3-8197185 farthings, for the second. 21. If the linear side of a certain cube be increased one inch, the surface of the cube will be increased 246 square inches : determine the side of the cube. Ans, 20 inches. 22*. If from a piece of tin, in the form of a sector of a circle, whose radius is 30 inches, and the length of its arc 36 inches, be cut another sector whose radius is 20 inches ; and if then the remaining frustum be rolled up so' as to form the frustum of a cone; it is MISCELLANEOUS PROBLEMS. 109 required to find its content, supposing one- eighth of an inch to be allowed off its slant height for the bottom, and the same allowance of the circumference, of both top and bottom, for what the sides fold over each other, in order to their being soldered together. Ans. 685-3263 cubic inches. 23. Three men bought a grinding-stone of 40 inches diameter, which cost 20.9., of which sum the first man paid 9^., the second 6s. ^ and the third 55. ; how much of the stone must each man grind down, proportionably to the money he paid ? Ans. The first man must grind down 5*167603 inches of the radius ; the second 4*832397 inches, and the third 10 inches. 24. There is a frustum of a cone, whose solid content is 20 feet, and its length 12 feet ; the greater diameter is to the less as 5 to 2 ; what are the diameters ? .^,^ (2-02012 feet. ^"*- t -80804 feet. 25. A farmer borrowed of his neighbour part of a hay-rick, which measured 6 feet in length, breadth, and thickness ; at the next Lay- time he paid back two equal cubical pieces, each side of which was I feet. Has the debt been discharged? Ans. No ; 88 cubic feet are due. P6. There is a bow-' in form of the segment of an oblong spheroid, whose axes are to each other in the proportion of 3 to 4, the depth of the bovjl one-fourth of the whole transverse axis, and the diameter of its top 20 inches ; it is required to determine what number of glasses a company of 10 persons would have in the contents of it, when filled, using a conical glass, whose depth is 2 inches, and the diameter of its top an inch and a half. Ans. 114*0444976 glasses each. 27. If a cubical foot of br^ss were to be drawn into wire, of -^ of an inch in diameter; it is required to determine the length of the 8aid wire, allowing no loss in the metal. Ans. 55f miles. 28. How many shot are there in an unfinished oblong pile, the length and breadth of whose base being 48 and 30, and the length and breadth of the highest course being 24 and 6 ? Ans. 17356. 29. How many shot are there in an unfinished oblong pile of 12 courses ; length and breadth of the top contain 40 and 10 shot respectively ? Ans. 8606 shot. 30. Of what diameter must the bore of a cannon be cast for a ball of 24 pounds weight, so that the diameter of the bore may be -j^ of an inch more than that of the ball? Ans. 5*757098 inches. 31. What is the content of a tree, whose length is 17^ feet, and which girts in five different places as follows, viz., in the first place 9-43 feet, in the second 7 '92, in the third 6*15, in the fourth 4*74, and the fiflh 3-16? Ans. 42*5195. 32. What three numbers will express the proportions subsisting between the solidity of a sphere, that of the circumscribing cylinder, and circumscribing equilateral cone ? Ans. 4, 6, 9. ■" 33. Given the side of an equilateral triangle 10, it is required to find the radii of its circumscribing circle. Ans. 5-7786. 34. Given the perpendicular of a plane triangle 300, the sum of 170 MISCELLANEOUS PROBLEMS. the two sides 1150, and the difference of the segment of the base 495 ; required the base and the sides? Ans. 945, 375, and 780. 35. A side wall of a house is 30 feet high, and the opposite one 40, the roof forms a right angle at the top, the lengths of the rafters are 10 feet and 12 ; the end of the shorter is placed on the higher wall, and vice versa; required the length of the upright which supports the ridge of the roof, and the breadth of the house ? Ans. 41-803, length of upright, and 12 feet the breadth of thj house. AREAS OF THE SEGMENTS OF A CIRCLE. 171 A TABLE OF THE AREAS OF THE SEGMENTS OF A CIRCLE, IVhose diameter is 1, and supposed to he divided into 1000 equal parts. Height. Area Seg. Height. Area Seg. H eight. Area Seg. Height. Area Seg. •001 •000042 •044 •012142 087 •033307 •130 •059999 •002 •000119 •045 •012554 • 088 •033872 •131 •060672 •003 •000219 •046 •012971 089 •034441 •132 •061348 •004 •000337 •047 •013392 090 •035011 •133 •062026 •005 •000470 •048 •013818 091 •035585 •134 •062707 •006 •000618 •049 •014247 092 •036162 •135 •063389 •007 •000779 •050 •014681 093 •036741 •136 •064074 •008 •000951 •051 •015119 094 •037323 •137 •064760 •009 •001135 •052 •015561 095 •037909 •138 •065449 •010 •001329 •053 •016007 096 •038496 •139 •066140 •Oil •001533 •054 •016457 097 •039087 •140 •066833 •012 •001746 •055 •016911 098 •039680 •141 •067528 •013 •001968 •056 •017369 099 •040276 •142 •068225 •014 •002199 •057 •017831 100 •040875 •148 •068924 •015 •00243S •058 •018296 101 •041476 •144 •069625 •016 •002685 •059 •018766 102 •042080 •145 •070328 •017 •002940 •060 •019239 103 •042687 •146 •071033 •018 •003202 •061 •019716 104 •043296 •147 •071741 •019 •003471 •062 •020196 105 •043908 •148 •072450 •020 •003748 •063 •020681 •106 •044522 •149 •073161 •021 •004031 •064 •021168 107 •045139 •150 •073874 •022 •004322 •065 •021659 108 •045759 •151 •074589 •023 •004618 •066 022154 109 •046381 •152 •075306 •024 •004921 •067 •022652 110 •047005 •153 •076026 •025 •005230 •068 •023154 111 •047632 •154 •076747 •026 •005546 •069 •023659 •112 •048262 •155 •077469 •027 •005867 •070 •024168 •113 •048894 •156 •078194 •028 •006194 •071 •024680 114 •049528 •157 •078921 •029 •006527 •072 •025195 115 •050165 •158 •079649 •030 •006865 •073 •025714 116 •050804 •159 •080380 •031 •007209 •074 •026236 117 •051446 •160 •081112 •032 •007558 •075 •026761 •118 •052090 •161 •081846 •033 •007913 •076 •027289 119 •052736 •162 •082582 •034 •008273 •077 •027821 •120 •053385 •163 •083320 •035 •008438 •078 •028356 121 •054036 •164 •084059 •036 •009008 •079 •028894 •122 •054689 •165 •084801 •037 •009383 •080 •029435 123 •055345 •166 •085544 •038 •009763 •081 •029979 124 •056003 •167 •086289 •039 •010148 •082 •030526 125 •056663 •168 •087036 •040 •010537 •083 •031076 126 •057326 •169 •087785 •041 •010931 •084 •031629 •127 •057991 •170 •088535 •042 •011330 •085 •032180 128 •058658 •171 •089287 •043 •011734 •086 •032745 •129 •059327 •172 •090041 172 AREAS OF THE SEGMENTS OF A CIRCLE. Height. 2 \.rea Seg. Height. Area Seg. Height. Area Seg. H eight. Area Seg. •173 •090797 •224 •131438 •275 •175542 •326 •222277 I'M •091554 •225 •132272 •276 •176435 327 •223215 175 •092313 •226 •133108 •277 •177330 •328 •2^24154 176 •093074 •227 •133945 •278 •178225 •329 •225093 177 •093836 •228 •134784 •279 •179122 330 •226033 178 •094601 •229 •135624 •280 •1C'0019 331 •226974 179 •095366 •230 •136465 •281 •180918 332 •227915 180 •096134 •231 •137307 •282 •18:81'/ . 333 •228858 •181 •096903 •232 •138150 •283 •182718 334 •229801 182 •097674 •233 •138995 •284 •183619 335 •230745 183 •098447 •234 •139841 •285 '184521 336 •231689 184 •099221 •235 •140688 •286 •185425 337 •232634 185 •099997 •236 •141537 •287 •186329 338 •233580 186 •100774 •237 •142387 •288 •187C34 339 •23:^526 187 •101553 •238 •143238 •289 •188140 340 •235473 188 •102334 •239 •144091 •290 •189047 341 236421 189 •103116 •240 •144944 •291 •189955 342 •237369 190 •103900 •241 •145799 •292 •190864 343 •238318 191 •104685 •242 •146655 •293 •191775 344 •239268 192 •105472 •243 •147512 •294 •192684 345 •240218 193 •106261 •244 •148371 •295 •193596 346 •241169 194 •107051 •245 •149230 •296 •194509 347 •242121 195 •107842 •246 •150091 •297 •195422 348 •243074 •196 •108636 •247 •150953 •298 •196337 349 •244026 •197 •109430 •248 •151816 •299 •197252 350 •244980 •198 •110226 •249 •152680 •300 •198168 351 •245934 •199 •111024 •250 •153546 •301 •199085 352 •246889 •200 •111823 •251 •154412 •302 •200003 353 •247845 •201 •11262J •252 •155280 •303 •200922 354 •248801 •202 •113426 •253 •156149 •304 •201841 355 •249757 •203 •114230 •254 •157019 •305 •20-2761 356 •250715 •204 •115035 •255 •157890 •306 •203683 - 357 •251673 •205 •115842 •256 •158762 •307 •204605 358 •252631 •206 •116650 •257 •159636 •308 •205527 359 •253590 •207 •117460 •258 •160510 •309 "206451 1)60 •254550 •208 •118271 •259 •161386 •310 207376 361 •255510 •209 •119083 •260 •162^263 •311 •208301 362 •256471 •210 •119897 •261 •163140 •312 •209227 363 •257433 •211 •120712 •262 •164019> •3i3 ;210154 364 •258395 •212 •121529 •263 •164899 •314 /211082 365 •259357 •213 •12-2347 •264 •165780 •315 •212011 366 •260320 •2U •123167 265 •166663 •316 •212940 367 •261284 •215 •123988 •266 •167546 •317 •213871 368 •262248 •216 •124810 •267 •168430 •318 •214802 369 •263213 •217 •125634 •268 •169315 •319 •215733 370 •264178 •218 •126459 •2q^ •170202 •320 •216666 371 •265144 •219 •127285 •270 •171089 •321 •217599 372 •266111 •220 •128113 •271 •171978 •322 •218533 373 •267"7S •221 •128942 •272 •172867 •323 •219468 374 •268045 •222 •129773 •273 •173758 •324 •220404 375 •261013 •2-23 •130605 •274 •174649 •325 •221340 •376 •269982 AREAS OF THE SEGMENTS OF A CIRCLE. 173 Height. Area Seg. Height. Area Seg. Height. Area Seg. Height. Area Seg. •377 •270951 •408 •301220 •439 •331850 •470 •362717 •378 •271920 •409 •302203 •440 •332843 •471 •363715 •879 •272890 •410 •303187 •441 •333836 •472 •364713 •380 •273861 •411 •304171 •442 •334829 •473 •365712 •381 •274832 •412 •305155 •443 •335822 •474 •366710 •382 •275803 •413 •306140 •444 •336816 •475 •367709 •383 •276775 •414 •307125 •445 •337810 •476 •368708 •384 •277748 •415 •308110 •446 •338804 •477 •369707 •385 •278721 •416 •309095 •447 •339798 •478 •370706 •386 •279694 •417 •310081 •448 •340793 •479 •371705 •387 •280668 •418 •311068 •449 •341787 •480 •372764 •388 •281642 •419 •312054 •450 •342782 •481 •373703 •389 •282617 •420 •313041 •451 •343777 •482 •374702 •390 •283592 •421 •314029 •452 •344772 •483 •375702 •391 •284568 •422 •315016 •453 •345768 •484 •376702 •392 •285544 •423 •316004 •454 •346764 •485 •377701 •393 •286521 •424 •316992 •455 •347759 •486 •378701 •394 •287498 •425 •317981 •456 •348755 •487 •379700 •395 •288476 •426 •318970 •457 •349752 •488 •380700 •396 •289453 •427 •319959 •458 •350748 •489 •381699 •397 290432 •428 •320948 •459 •351745 •490 •382699 •398 •291411 •429 •321938 •460 •352742 •491 •383699 •399 •292390 •430 •322928 •461 •353739 •492 •384699 •400 •293369 •431 •323918 •462 •354736 •493 •385699 •401 •294349 •432 •324909 •463 •355732 •494 •386699 •402 •295330 •433 •325900 •464 •356730 •495 •387699 •403 •296311 •434 •326892 •465 •357727 •496 •388699 •404 •297292 •435 •327882 •466 •358725 •497 •389699 •405 •298273 •436 •328874 •467 •359723 •498 •390699 •406 •299255 •437 •329866 •468 •360721 •499 •391699 •407 •300238 •438 •330858 •469 •361719 •500 392699 PRINTED BY BALLANTYNE AND COMrANY EDINBURGH AND LONDON :* ) "William Collins, Sons, & Co'a Educational Works. COLLINS' SERIES OF SCHOOL ATLASES, Carefidh/ Constructed and Engraved from the best and latest AuthoritiiL and Beautifully Printed in Colours^ oi Superfine Cream Wove Faper, (^'n - MODERN GEOaRAPHY-Crown Series. «. |«i.^* The Primauy Atlas, consisting of 16 Maps, 9 inches by 7| inches, 4to, Stiff Wrapper, « - - - - - - - -Ol^ ^ The Junior, or Young Child's Atlas, consisting of 16 Maps, \f / j 4to, with 16 pp. of Questions on the INIaps, in Neat AVrappcr, - 1 m'^t The School Board Atlas, consisting of 24 Maps, Crown 4to, in 'ow Neat Wrapper, - -- 10 The Progressive Atlas, consisting of 32 Maps, 9 inches by 1\ inches, 4to, cloth lettered, -20 The National Atlas, consisting of 32 Maps, 4to, with a Copious \ Index, cloth lettered, - - - -\- -^-26 ^"^-i?- — — Th Th ^ Thi Th: Th; Th] /-n a ' th< (yinH'ip' n M3051SM I /grg: THE UNIVERSITY OF CALIFORNIA LIBRARY HISTORICAL AND CLASSIGASi GEOGRAPHY, Pocket Atlas op Historical Geography, 16 Maps, Imp. 16mo, The Crown Atlas of Historical Geography, 16 Maps, with Lettei^press, by Wm. F. Collier, LL.D., Imperial 16mo, cloth. Pocket Atlas of Classical Geography, 15 Maps, Imp. 16mo, cl.. The Crown Atlas op Classical Geography, 15 Maps, with Letterpress, by Leon. Schmitz, LL.D., Imperial 16mo, cloth^ v^ w. r^ SCRIPTURE GEOGRAPHY. The Atlas op Scripture Geography, 16 Maps, with Questio^g^ on each Map, Stiff Cover, - /^-l London, Edinburgh, and Herriot Hill Works, Glasgow.