5S 1 B ^k "X 1 M i , . ,,1 i " i J^\" H_^. r } ? i f- '. t. B ,. *.'' i ' ' ^ /.//// <'/' Tn /m en ts A TREATISE ox BOX OF INSTRUMENTS AXD THE SLIDE-RULE. FOR THE USE OF GACGERS, ENGINEERS, SEAMEN, AND STUDENTS. BY THOMAS KENTISH. PHILADELPHIA : HEXRY CAREY -BATED, INDUSTRIAL PUBLISHER. No. 406 WALNUT STREET. 1872. PREFACE. IN the first edition of this Treatise, its utility, in the absence of other works upon the subject, was assigned as an apology for its publication. The in- struments, whose uses it explains, are often so little understood, that scarcely half of 'them are of any service to their possessor. The Sector, in particular, the most important in the box, is generally regarded as unintelligible. The Slide-rule is briefly noticed in some of the treatises on Mensuration ; but, as the pupil is presented merely with a few formal precepts how to use it, without knoAving why, he never under- stands its nature, never understands the method of determining the real value of any result, and, ac- cordingly, soon lays it by with dissatisfaction, and banishes it from his memory. The steady sale which the first edition has met with has convinced the Author that his labours were not in vain, and that he has extended among many thousands a knowledge of intrinsic value to all em- ployed in the delineation of mathematical figures. 3 4 PREFACE. No attempts, however, are perfect in the beginning ; and much was wanted in the first edition to render the work complete. This additional information has been supplied. Several problems have been prefixed, requiring only the compasses and ruler, which, together with those that follow, embrace all that are truly useful, and preclude the necessity of referring to other works on Practical Geometry. In books upon this subject, it is not usual to annex reasons for any of the operations, but it has been thought advisable to do so, in a few instances, with the more difficult problems ; with the rest it is not attempted, because, to have entered fully upon the subject, would have been to transcribe the whole of the Elements of Euclid, a work which is within the reach of every one, and which every one must study, who desires thoroughly to understand Geometry. The part relating to Trigonometry, though concise, will be found to comprehend every thing necessary to enable the student to obtain a clear conception of the subject, and when carried out in connection with the portion devoted to Navigation, will render its acquirement alike easy, pleasing, and useful. The chapter on Logarithms is written simply to show the mode of adapting them to instrumental computation ; a purpose to which every part of tho work is, as a matter of course, as much as possible made subservient. The section relating to the Slide-rule has been entirely re- written ; and, in this portion of the work, the Author flatters himself there will be found much PKEFACE. 5 that is perfectly new, and many remarks calculated to awaken and stimulate the youthful mind to think for itself; a habit of the utmost value in mathema- tical science, which, being based on truth, courts in- vestigation, and requires that we shall never assent till we can comprehend. In this part, the formulae for surfaces and solids have been so modified as to embrace almost every species of mensuration under the simplest form ; questions for practice are inter- spersed throughout, that the student may test his proficiency, and acquire facility in the use of the rule; and tables are inserted at every step, for the purposes of computation ; a practice in all cases ad- visable, as the instrumental operation and numerical calculation necessarily check and illustrate each other. The reciprocals of divisors, employed as factors, are convenient in practice; but it was deemed ad- visable, upon the whole, to omit them, as the formulae for numerical computation would have then been different from those suited to the Slide-rule, which would have tended to perplex the mind of the learner ; whereas, by retaining the same form for both operations, it is obvious that to understand one is to understand the other ; and the student, instead of coming to regard the instrumental mode of solu- tion as something entirely distinct from the nume- rical, and looking upon the agreement of the two rather as a coincidence than a consequence, as is too often the case, will see that, in fact, they are identical, and 'cannot fail, in a short time, of having i* 6 PREFACE. the very clearest conception of the whole of the subjects treated of. It is somewhat surprising, that, after the lapse of two hundred years, so excellent an instrument as the Slide-rule should he so little known and appreciated by mathematical students in general. To the engi- neer and the excise officer it is perfectly familiar, and of daily utility; but, from its having been al- most exclusively confined to them, there is an idea prevalent among gentlemen engaged in education, that it can neither be understood by their pupils nor be of any utility to them. A more erroneous conception, on both accounts, cannot be formed; for a knowledge of the instrument is acquired with little or no effort., and it may be truly stated, that it is the most valuable adjunct to mathematical study that can possibly be desired. Nothing im- prints a fact so firmly on the mind as repeated exercise. As Demosthenes, when asked the three principal requisites in oratory, summed them up in the word action ; so may we say of learning, that the three great essentials to its success are contained in the word repetition. Dexterity in every art, and skill in every science, must be acquired by this means, and by this alone. But, in the solution of questions that are necessarily laborious, every one feels a great disinclination to work through many examples, much less to repeat them ; the consequence is, so little impression is made on the memory, that the knowledge is, in many instances, forgotten as soon as acquired. Now, by the Slide-rule, the most PREFACE. 7 tedious calculations are effected nearly as easily as the most simple. The student can, therefore, after accompanying them the first time with the numerical solution, go over the operations again and again with the rule, with the greatest ease and rapidity, deepening the impression each succeeding time, and rendering the knowledge obtained distinct and per- manent. In the truth of this, the Author is not only borne out by his own experience, but he can refer, with pleasure, to schools in which they have been adopted, and in which they have proved of the greatest assist- ance; and no one, really fond of knowledge, who may give them a trial, will regret the little extra trouble they may cause, but will rejoice in having found so excellent an aid to study. Mathematical science is of such extensive utility that it ought to be universally understood ; and it is impossible to go five or six times through the present work, which, after the first, may be done in a very few days, with- out being as familiar with the Surfaces and Solids, and with Trigonometry and Navigation, as with the multiplication table; and this is the great object to be attained. To be barely acquainted with them is not sufficient ; knowledge, to be useful, must be at the moment accessible, so that we may be enabled to proceed without error or hesitation ; and that the most intimate familiarity with the above-mentioned studies will be obtained by the method here pointed out, has been again and again tried, and with the happiest results. v 8 PREFACE. The small section allotted to Land Surveying doea not properly come within the design of the work, but it was thought it might prove useful, and has ac- cordingly been inserted. The measuring of a field, which is all that can at all be consistently aimed at here, is so very simple, that one example was deemed sufficient as a guide; but, in teaching the subject, more is necessary; and a very efficient method is to draw on a piece of paper a sketch of a field, which, with the help of a feather-edged plotting scale, or a diagonal scale and a pair of compasses, the pupil should measure, and enter his notes in a field-book, or slate, ruled for the purpose. The sketch should now be handed to the tutor. The learner, then, from his notes, is to construct another, upon paper, from the same scale. When finished, its correctness can be readily ascertained by laying it upon the original, and holding them up to the light, when, if accurately laid down, the lines will, of course, cor- respond. This plan has been tried for many years, and found to convey a very good idea to the mind of the learner. A little occasional field-practice, which is indispensably requisite, soon renders the study pleasant, and the progress certain. The chapter on Cask Gauging will, it is humbly hoped, prove a valuable acquisition to the gauger. The great uncertainty and inconvenience of the four varieties render it extremely desirable to have some general, and, at the same time, easy, and easily re- membered rule of approximation; and from the method employed in making casks, it is obvious that PREFACE. 9 the exact agreement of their shape with any definite geometrical solid must be perfectly fortuitous. The four varieties, however, are exemplified, together with the general rule for frustums; which latter, though rather tedious, would soon become familiar if once adopted. In Navigation, for working a day's reckoning, the rule is peculiarly convenient, and sufficient for all practical purposes; superseding the incessant turn- ing over and transcribing from tables ; which, though in themselves they are one of the most splendid in- ventions of all time, and, in elaborate calculations requiring minute exactness, indispensable, are yet, in their application, as perfectly mechanical as the instrumental operation itself; so that no reasonable objection can be urged against the adoption of the Gunter, that does not apply, with equal force, to the use of Logarithms altogether. For gentlemen, however, who may not desire to use the Slide-Rule, it may be here stated, that the work by no means absolutely requires it ; it is equally available as a Treatise on Mensuration, Trigonome- try, and Navigation. For the purposes of calcula- tion, it would be found a great convenience to copy out, upon a sheet of Bristol board, the tables at pages 115, 116, 118, 123, 126, 129, 136, 137, 138, 150, 182, and 198, as it would save much needless turning over of the pages ; and if each were enclosed in borders, and slightly washed over with different colours, it would make them of easier reference. In studying Trigonometi y, Wallace's Practical 10 PREFACE. Mathematician's Pocket Guide will be found a con- venient set of Logarithmic Tables ; their cost is a mere trifle. Barlow's and Galbraith's Tables are extremely useful. The latter contains the secants and cosecants, which, as complemental to the cosines and sines, offer great facilities in calculation. Kentish's Slide-Rules, furnished with triple slides, are sold by Messrs. Relfe Brothers, 150 Aldergate Street. London price, carefully mounted and var- nished, 5s. each, or beautifully cut in box, 10s. each. Messrs. J. "W. Queen & Co., Importers, Manufactu- rers and Dealers in Mathematical Instruments, 924 Chestnut Street, Philadelphia, will import them to order. CONTENTS. A Box OP ISSTBUXXSTS 13 PRACTICAL GEOMETRY Definitions 15 The Compasses 20 The Parallel Ruler 27 The Protractor 36 The Plain Scale 40 TRIGONOMETRY 48 The Sector 54 LOGARITHM? 81 The Slide-role 101 Ratios and Gauge Points Ill TABLE I. The Circle 112 II. Polygons, linear dimensions of 115 IIL Polygons, areas of 117 IV. Falling Bodies 119 Pendulums 120 Areas of Circles and Surfaces of Spheres.... 120 Diagonals of Squares and Cubes. 122 Velocity of Sound 123 V. Surfaces 124 Formulae for ditto 126 11 12 CONTENTS. PAGl TABLE VI. Solids 131 General Rule for Frustums 133 Formulae for Solids 136 VII. Weight of Spheres 154 VIII. Solidity of Spheres 155 Solar System 156 Miscellaneous Questions 158 Cask Gauging 163 Land Surveying 177 TEIGOSOMETBT ASD NAVIGATION 180 Plane Sailing 183 Traverse sailing 186 Parallel Sailing 188 Middle Latitude Sailing 191 Mercator's Sailing 197 Oblique Sailing 204 Windward Sailing 206 Current Sailing 209 Of a Ship's Journal 211 RECAPITULATION 216 APPENDIX , J23 A TREATISE A BOX OF INSTRUMENTS A BOX OF INSTRUMENTS. THE contents of a case of Mathematical Instruments are, generally, a pair of plain compasses, a pair of bow compasses, a pair of drawing compasses, and a drawing pen ; a parallel ruler, a protractor, a plain scale, and a sector. The plain compasses consist of two inflexible rods of brass, revolving upon an axis at the vertex, and fur- nished with steel points. The bow compasses are a smaller pair, provided with a pen for describing small circles in ink. The sides of the pen are opened or closed with an adjusting screw, that the line may be drawn fine or coarse as required. The drawing compasses are the largest of the three ; one of the legs is furnished with a socket for the reception of either of the four following pieces, as occasion may require : 1. A steel point, which, being fixed in the socket, makes the compasses a plain pair, like the other ; 2. A port-crayon, for the purpose of carrying a piece of blacklead, or slate-pencil, according as paper or slate is used for drawing upon ; 3. A steel pen, like the one attached to the bow compasses, but larger, for the purpose of describing circles of greater diameter ; 4. A 14 A TREATISE ON A BOX OF rowel, or spur wheel, with a brass pen above it, for the re- ception of ink, which the spur, in its circuit, distributes in dots upon the paper. The pen, the port-crayon, and the dotting wheel, are each furnished with a joint, that, when fixed in the compasses, they may be set perpendicular to the paper. The drawing pen is the same as the steel pen of the compasses, only that it is screwed upon a brass rod, of a convenient length for the hand, and into the rod itself is inserted a fine steel point for pricking. The parallel ruler consists of two flat pieces of ebony or ivory, connected together by brass bars, having their ex- tremities equidistant, by which contrivance, when the ruler is opened, the sides necessarily move in parallel lines. The protractor is a semicircular piece of brass, di- vided into 180 degrees, and numbered each way, from end to end. In some boxes this is omitted, and the de- grees are transferred to the border of the plain scale. The plain scale is a flat piece of box or ivory, and is so called from containing a number of lines divided into plain or equal parts. A scale of chords, of a fixed radius, is also graduated upon it. The sector is a foot rule, divided into equal portions, movable upon a brass joint, or axis, from the centre of which are drawn various lines through the whole length of the ruler. The legs represent the radii of a circle, and the middle of the joint expresses the centre. The lines upon the sector are of two sorts, single and double : the single lines run along the margin and the edges ; the double lines radiate from the centre to the extremities of the legs, and are marked twice upon the same face of the instrument, in order that distances may be taken upon them crosswise, when they arc opened to an angular position. INSTRUMENTS AND THE SLIDE-RULE. 15 PRACTICAL GEOMETRY. DEFINITIONS. A POINT is that which has position, without length, breadth, or thickness. A line is length, without breadth, or thickness. A superficies is length and breadth, without thickness. A solid is that which has length, breadth, and thick- ness. An angle is the opening of two straight lines meeting in a point, as RAE. Lines which run side by side, and are always equidis- tant, are called parallels, as SD, KL. A line is perpendicular to another when the angles on both sides of it are equal j and each of these angles i? 36 A TREATISE ON A BOX OF called a right angle ; thus RA is perpendicular to WS f and the angles RAW, RAS, are right angles. An acute angle is less than a right angle, as EAS. An obtuse angle is greater than a right angle, as TAS A figure containing three sides is called a triangle. The boundary of a right-lined figure is termed its perimeter. A triangular figure containing three equal sides is an equilateral triangle, as KBC. If two of its sides only are equal, it is an isosceles triangle, as BHC, in which BH = HC. If the three sides are unequal, it is a scalene triangle, as KBH. A triangle containing a right angle is called a right- angled triangle, as ABC. A triangle containing an obtuse angle is an obtuse- angled triangle, as KHB. An acute-angled triangle contains three acute angles, asNMC. A figure containing four sides is called a quadrilateral. A parallelogram is a quadrilateral whose opposite sides are parallel, as SKLD, KPOL. A rectangle is a parallelogram whose angles are righ* angles, as KMCL. A square is a rectangle whose sides are equal, as ABCD. A rhomboid has its opposite sides equal, but two of its angles are obtuse, and two acute, and these are opposit- to each other, and equal, each to each, as KBML. A rhombus, like a square, has all its sides equal, but two of its angles are obtuse, and two acute, and these are opposite to each other, and equal, each to each, ay PBMO. INSTRUMENTS AND THE SLIDE-RULE. 17 When a quadrilateral has only one pair of its sides parallel, it is called a trapezoid, as PBCO. When none of its sides are parallel it is a trapezium, as KHNL. A line crossing a quadrilateral from opposite angles, is termed a diagonal ; thus AC, BD, are the diagonals of the square ABCD. Figures of more sides than four are called polygons. If all the sides and angles are equal, it is a regular polygon ; if unequal, an irregular polygon. A polygon of five sides is termed a pentagon ; of six, a hexagon ; of seven, a heptagon ; of eight, an octagon ; of nine, a nonagon ; of ten, a decagon ; of eleven, an un- decagon ; and of twelve, a dodecagon. A triangle is sometimes called a trigon j and a quadri- lateral, a tetragon. A circle is a plane figure bounded by a curved line called the circumference, which is everywhere equidistant from the centre. A right line passing through the centre, and meeting the circumference at each extremity, is called the diameter, asSM. A right line reaching from the centre to the circumfe- rence is termed the radius, as HM. An arc of a circle is any part of the circumference, as the curve from S to X. A chord is a right line joining the extremities of an arc, as the straight line from S to X. A segment is a space included between an arc and its chord, as SZXS. A sector is a part of a circle contained by two radii and *he arc between them, as SHXZ. 18 A TREATISE ON A BOX OF Hence a sector is made up of a triangle and a segment A semicircle is half a c.rcle; a quadrant, the fourth part; a sextant, the sixth part; and an octant, the eighth part. A rectilineal solid whose ends are equal, similar, and parallel, and whose sides are parallelograms, is called a prism. If the ends also of the prism are parallelograms, it is a parallelepiped : if all the sides are square, it is a cube : if the ends are unequal and dissimilar, it is a prismoid. A cylinder is a round solid, of uniform thickness, hav- ing circular ends. A pyramid is a solid which has a rectilineal base, and triangular sides meeting in a point called the vertex. A cone is a round solid tapering uniformly to a point. A sphere is a solid every way round. A segment of a solid is the part cut off the top by a plane parallel to its base. A frustum is the part left at the bottom, after the seg- ment has been cut off. Prisms, cylinders, pyramids, and cones are said to be right or oblique according as the base is cut perpendicu- larly or obliquely to the axis. Plain figures formed by the cutting of a cone are called conic sections. A cone may be cut five ways. If the cutting plane passes through the vertex of a right cone and any part of the -base, the section is an isosceles tri- angle ; if through the sides, parallel to the base, a circle ; if obliquely through the sides, an ellipse ; if through one side and parallel to the other, a parabola ; if in any other way, the cutting plane will run beyond into a similar cone inverted over the other, and then the section is termed ac LNslRUMEXTS AND THE SLIDE-RULE. 19 hjperbola. (JVI B. An ellipse may be considered iu an elongated circle, and may be described by driving in tiro pins as centres, passing over them a string with a loop at each end, and working a pencil round within the string, keeping it stretched to its limits.) The centres are usually called foci : the diameter passing through them is termed the transverse ; the short one at right angles to it, the con- jugate diameter. A line perpendicular to either of the diameters is called an ordinate ; and the sections of the diameter met by it are termed abscissas. The vertex of a conic section is the point where the cutting plane meets the opposite sides of the cone. The axis of a parabola or hyperbola is a right line drawn from the vertex to the middle of the base. All round solids may be conceived to be described by the rotation of planes on their sides, or diameters, as axes. A right-angled triangle rotating on its perpendicular, forms a cone ; a parallelogram, revolving on its side, pro- duces a cylinder ; if a circle turn upon its diameter, it shapes out a sphere ; and the revolution of an ellipse gene- rates a spheroid. If the ellipse revolves on the transverse diameter, the spheroid is called prolate ; if on the conju- gate, an oblate spheroid. The figure formed by the revo- lution of a parabola about its axis is termed a paraboloid, or parabolic conoid ; the solid formed in the same way by an hyperbola, an hyperboloid, or hyperbolic concoid. If a section of a curve revolve on a double ordinate as axis, it will generate a spindle ; and this will be circular, elliptic, parabolic, or hyperbolic, according to the curve from which the section is taken. A regular body id a solid contained under a certain 20 A TREATISE ON A BOX OP number of similar and equal plane figures. There are but five kinds, which are, the tetraedron, Laving four equal triangular faces ; the hexaedron, or cube, which has six equal square faces ; the octaedron, which has eight equal triangular faces ; the dodecaedron, which has twelve equal pentagonal faces; and the icosaediwn, which has twenty equal triangular faces. THE COMPASSES 1. To bisect a given line AB. From A and B as centres, with any radius greater than half AB, describe arcs cutting each other in C and D. Join the points C and D, by drawing the straight line CD; this will be perpendicular to AB, which it will bisect in the point E. 2. To bisect a given angle ABC. (See p. 21.) From B as a centre, with any radius, describe the arc DE. From D and E, with the same, or any other radius, draw arcs cutting each other in F. Join BF, and it will bisect the angle as required. INSTRUMENTS AND THE SLIDE-RULE. 21 3. To erect a perpendicular to a straight line AB, from a given point C within it. DOE When the point C is near the middle of the line, on each side of it set off any two equal distances CD, CE. From D and E as centres, with any radius greater than CE or CD, describe arcs cutting each other in F. Join FC, and it will be perpendicular to AB. D C When the point C is at or near the end of the line, from C, with any radius, describe the j?l \! i QI tily opposite the middle ol the line, from C, wili wy sonvenient radius, describe the arc DFE crossing Aft in X> tod E. From D and E, with the same or any t tf .* radius, describe arcs cutting each other in G. Draw '.I, uid it will be perpendicular to Alt When the point C is nearly opposite the end of the line, from C draw any line CD. Bisect CD in E, and from E, with the radius EC, cross AB in F. Draw CF, and it will h*> wpoMioi^ar to AT* INSTRUMENTS AND THE SLIDE-RULE. 23 5. To make an angle equal to a given angle ABO. From B, with any radius, draw the arc GH; and from E, with the same radius, describe the arc KL. Make KL equal to GH, and through K draw the straight line ED. The L DBF = L ABC. 6. To describe a circle through three given points A., B, an4C. From B, with any radius less than BA, describe the aro alcd; and from A and C, with the same radius, cross it in a and I, c and d. Draw straight lines through the points of intersection to moot in D, which will be the centre of the circb required. 24 A TREATISE ON A BOX OF To find the centre of a given circle, take any three points in the circumference, and proceed in like manner. To describe a circle about a triangle, select the three angular points, and proceed in like manner. 7. To construct a triangle of which the three sides A, B, and C are given. Draw the line DE equal to A. From D, with for < radius, describe an arc at F ; and from E, with B for a radius, cross it at F. Draw the lines DF, EF ; and DEF will be the triangle required. To construct an equilateral triangle proceed in the same manner, taking the base each time as radius. 8. To construct a rectangle, whose length A B and breadth C are giveli. At A erect a perron Ocular 41), equal to C. From \) t INSTRUMENTS AND THE SLIDE-RULE. 25 with the distance AB, describe an arc at E ; and from B, with the radius C, cross it at E. Draw DE, EB. A DEB is the parallelogram required. To construct a square, make the perpendicular equal to the base, and proceed in like manner. To construct a rhombus and rhomboid, determine the angle by problem 5, and then proceed in like manner. 9. To draw a line parallel to a given line AB, at a given distance. G H Take any two points E and F in the line AB, and, with the given distance, describe the arcs G and H. Draw the line CD touching the arcs, and it will be parallel to AB. When the line is to pass through a given point C. In AB take any point G, and with the distance GO describe the arc CH ; from C, with the same radius, de- scribe the arc GF, and make FG equal to CH. Through F and C draw the straight line DE, and it will be paral- lel to AB, as required. 26 A TREATISE ON A BOX OF 10. To project an ellipse, or oval, the length A B and breadth A C, being given. Bisect AC in D, and upon it describe the semicircle AEG. At C draw a straight line CB, perpendicular to AC ; and from the point A, with the given length as a radius, cross CB in B. From the semicircle AEC and parallel to CB, draw any number of straight lines FGr<7, HK&, NMm, EDrf, &c. On the line AB, at the points of intersection, m, Te, g, &c. erect perpendiculars, and make gf equal to GF, Teh equal to KH, mn equal to MN, &c. Lastly, trace a curve line from B through the points /, h, n,e, &c., and it will give half of the ellipse, from which the other half may readily be constructed. This method is of great utility in describing elliptical arches, stair- cases, &c., and for any purpose in which circular figures, or figures of any shape, require to be elongated without altering the breadth, as in cutting gores for globes, &c., INSTRUMENTS AND THE SLIDE-RULE. 27 in which case AB will be equal to half the circumference, the breadth DE being regulated by the number of gores employed. la instances like these, the length AB will he very great compared with the breadth DE, and AB may then be divided in the same proportion as AC, by other means ; for example, if the length is to be 12 times the breadth, then each of the distances B^r, gk, Jem, will be 12 times the corresponding distances CG-, GK, KM. This method of projecting ellipses is derived from conceiv- ing a right cylinder to be cut by two planes, one parallel, and the other oblique to the base. THE PARALLEL RULER. 1. Through a given point A, to draw a line parallel to a given line DE. Lay the edge of the ruler upon DE, and move it up- wards till it reaches the point A, through which draw BG. BC is parallel to DE. 2. To make an angle equal to a given angle ABC. A Lay the base EF, in a line with BC, and draw ED, ;nrailel to BA. The / DEF = L ABC. A TREATISE ON A BOX OF 3. To find a third proportional to two given straight lines, AB, AC. Place them together so as to form any angle DAE. Take AD = AC. Draw BC, and DE, parallel to it. AB : AC : : AC : AE. 4. To find a fourth proportional to three given straight lines AB. CD, EF. (i K Al Make any angle, LGM. Take GH = AB, GK = CD, and GL = EF Join HK, and draw LM parallel to it AB : CD : : EF : GM. For GH : GK : : GL : GM. 5. Another method. Given AB : CD : : EF : ? K INSTRUMENTS AND THE SLIDE-RULK. 29 Make a rectangle GITLK, of the second and third, CD, EF ; and in one of the sides, produced if necessary, take KM, equal to the first, AB. Draw GM, to meet HL, produced if necessary, in N. AB : CD : : EF : HN. For KM : GH : : KG : HN. And the third problem may be performed in a similar manner, by making a square of the second term 6. Hence to inscribe a square in a given triangle ABC. AD E B K L C Through the vertex A, parallel to BC, draw the straight line AE, and from C raise a perpendicular to meet it in D. Draw DE, equal to DC. Join EB, cutting DC in F. Through F draw FG, parallel to BC, and through H and G draw HL, GK, parallel to DC. GL will be the square required. For FC : CB : : CD : BC + DE rB rTj That is, GH : CB : : CD : BC + CD .-. GH ^ 30 A TREATISE ON A BOX OF That is, the side of the inscribed square is equal to the product of the base and altitude divided by their sum : or the sum is a fourth proportional to the side, base, and altitude. 7. To divide a given line AB, similarly to another CD A- On CD, construct the equilateral triangle CDE, and .'rom the vertex downwards cut off EF AB. Draw FG, parallel to CD, and join EH, EK. Transfer the divisions L and M, to and P. AB is divided, in the points OP, similarly to CD, in H and K, that is, CD : AB : : CK : AP : : CH : AO. 8. Another method. Let AB be the divided line, and AC the line required to be similarly divided. Lay them together, making any angle CAB. Join the extremities CB, and draw GE, FD, parallel to CB. AC is divided similarly to AB ; that is, AB : AC : : AE : AG : : AD : AF. C D E INSTRI'MKNT?! AND THE SLIDE-RULE. 31 9. Hence to divide a line AB, into any number of equal parts, as four. Draw an indefinite line CD, and from C, set off any Distance the intended number of times, in the points 1, 2, 3, 4. On C 4, construct the equilateral triangle EC 4. Make EF = AB, and draw FG, parallel to CD. FG is equal to AB. Join E 1, E 2, E 3 j and FG, that is AB, w'Jl be divided into four equal parts. 10. Or by the other method. E D B From A, draw the straight line AC, making any xngle CAB. From A, set off any distance the intended num- ber of times toward C, in the points 1, 2, 3, 4. Draw the line 4 B, and parallel to it 3 D, 2 E, 1 F. AB will be divided equally into the required number of parts. A TREATISE ON A BOX OP 11. To reduce a trapezium ABCD, to a triangle Draw the diagonal AC, and through B draw BE ; *arallel to it, meeting DC, produced to E. Join AE. The triangle ADE, iy equal to the trapezium ABC. 12. To reduce any polygon ABCDE, to a triangle. C 13 " // \\~~~ D Draw the diagonals CE, CA, and produce AE, both ways, to F and Gr. Draw DG-, parallel to CE, and BF, to CA. Join CF, CG. The triangle CFG, is equal to the polygon ABCDE. 13. To reduce a triangle ABC, to a parallelogram. E F INSTRUMENTS AXD THE SLIDE-RULE. 33 Through the vertex A draw FA, parallel to BC. Bisect BC in D, and raise DE, perpendicular to BC, meeting FA in E. Draw FB parallel to ED. The redangle FD. is equal to the triangle ABC, for the content of a triangle is equal to the product of half the base by the altitude. 14. To reduce a parallelogram ABCD, to a square. A D Bi- Produce BC to E, making CE = DC. Bisect BE in F. Produce DC to K, making FK = FE, and CG = CK. Through K draw KH, parallel to BE, and through G draw GH, parallel to DK. The square GK, is equal to the rectangle AC, and CG is a mean propor- tional between DC and CB j that is, DC : CG : : CG : CB. 15. Hence to make a square equal to any given polygon ALDC. 34 A TREATISE ON A BOX OF B Reduce it to the triangle ACE, and this to the paral- lelogram CHGF, and this to the square NM. 16. To reduce a triangle ABC, to another that shall be of a given altitude AD. Join DC, and through the vertex B draw BE, parallel to the base AC, meeting AD, produced if necessary, in the point E. Through E draw EF, parallel to DC. Join DF. The triangle DAF = ABC. INSTRUMENTS AND THE SLIDE-RULE. 35 THE PROTRACTOR. It is unnecessary to describe the construction of this. It is simply a semicircle, divided into 180 equal parts, termed degrees. As mentioned in the introduction, these degrees are, in some boxes, transferred to the border of the plain scale, which is used precisely as the protractor : it is, however, far from being so convenient as the semicircle. Some protractors are complete circles, and contain, of course 360 degrees. USES. 1. To find the number of degrees contained in any given angle BAG. Lay the central notch of the instrument upon A, and the edge along AB, as in the diagram ; and observe the number cut by the other line AC. 2. To lay down an angle ABC, which shall contain a given number of degrees. 36 A TREATISE ON A BOX OF B C Draw a line BC, of any length. Place the notch against B, and the edge along BC. Prick a point D against the number required, and through it draw the line AB. 3. Through a given point C, to draw a line paralell to a given line AB. D C In AB take any point E, and join CE; and make the angle DCE, equal to the angle CEB, by the line DC. DC is parallel to AB. 4. To divide a given line AB, into any number of equal parts. INSTRUMENTS AND THE SLIDE-RULE. 37 From the extremities of the line AB, draw the lines AC, BD, making equal angles. From A towards C, and from B towards D, set off any distance once less than the intended number of parts. Number one from the extre- mity A, and the other towards the other extremity B, and join the like numbers. AB will be divided as required. 5. To erect a perpendicular to a given line AB, from a point C, within it. D Place the edge along A B, and the notch at C. Then against 90 prick the point D, and draw DC. DC will be the perpendicular required. 6. To let fall a perpendicular upon a straight line AB from a point C. Draw any line CA. Observe the number of degrees contained in the angle CAB. Subtract it from 90, and make the angle ACB, equal to the remainder, by the line CB. CI> will be the perpendicular required. 38 A TREATISE ON A BOX OF 7. To divide a given angle, ABC, into any number of equal parts. Find the number of degrees ; divide it by the required number of parts, and prick off the quotient along the rim, as in D. Join BD. 8. To inscribe a circle in a given triangle, ABC. Bisect any two of the angles ABC, ACB, by the straight lines BD, CD, crossing each other in D, from which let fall the perpendicular DE. DE is the radius of the required circle. INSTRUMENTS AND THE SLIDE-RULE. 9. In a circle to inscribe any regular polygon. 39 Divide 360 by the intended number of sides. Place the instrument with the notch against the centre, and prick off the quotient round the rim. If the number of sides be odd, the instrument will require to be turned round ; if even, half may be pricked off, and lines drawn through the centre, the extremities of which meeting the circle, will give the points required. Connect the points, and the polygon will be completed. ]0. To construct a regular polygon on a given line.AIJ. 40 A llvKATlSE CH A BOX OP Divide 360 by the intended number of sides, subtract the quotient from 180, and halve the remainder. Make the angles CAB, CBA, each equal to this, by lines inter- secting in C. " CA is the radius of a circle, round which the line AB may be carried the required number of times. 11. To describe a circle within or without a regular polygon. Bisect any two angles, ABC, BCF, by the lines BD, CD, and from D let fall DE, perpendicular to the side BC. BD is the radius of the outer, and ED of the inner circle, as required.* THE PLAIN SCALE. THE method of constructing the plain scale is obvious. A number of horizontal lines on one side having been drawn through the whole length of the rule, and a vertical column on the left for the reception of numbers, a distance of two inches is laid down, and divided, upon the top line * Polygons are more expeditiously constructed by means of the Sector, of which hereafter. INSTRUMENTS AND THE SLIDE-RULE. 41 into 9 equal parts, upon tlie next into 8, and so on, down to 3. These are repeated along the rule as often as ita length will allow ; the first portion of each is subdivided into tenths and twelfths ; and numbers are placed in the column, showing how many of the tenths are contained in an inch. A scale of chords, marked C, is graduated along the top ; and at 60 a small r is placed, indicating that distance from the beginning to be the radius of the circle from which they are taken. They are merely the degrees of a quadrant or quarter of a circle, laid down in a straight line ; thus, C B Draw the lines B A, BC, at right angles to each other ; and with any convenient distance, BA, describe the arc AsC ; divide it into 90 equal parts, and join AC. From C as a centre, with the distances C 10, C 20, &c., describe the arcs 10, 10 7 ; 20, 20', &c. meeting the line AC. Fill up the separate degrees, which are not marked in the diagram to prevent confusion, and the scale is complete- 42 A TREATISE ON A BOX OP It is evident, by inspection, that the chord of 60 is equal to the radius, as shown by the letter r upon the rule ; which distance is therefore always to be taken in laying down angles, as will be described when we come to speak of its uses. The other face of the rule is divided along the top into inches, and these into tenths. The next line is divided into 50 equal parts. Under these is what is called the diagonal scale. It consists of 11 equidistant parallel lines, crossed by vertical ones a quarter of an inch apart. By taking these alternately, another scale is obtained, of twice the size of the former. The first of each of these is divided into ten equal parts, above and below ; and oblique lines are drawn from the first perpendicular below to the first division above, and continued parallel, by which con- trivance the first space is divided into 100 equal parts: consequently, if the line contain ten of the large divisions, each of these small spaces is the thousandth part of such line. If, therefore, the large divisions denote hundreds, the first subdivisions will be tens, and the second, units. USES. The plain scale is simply for laying off distances. The manner of using the first side is evident. If the number 47 be required, place one foot of the compasses upon 4, and extend the other to the 7th subdivision of the tenths. If 3 feet 5 inches be required, place one foot on 3, and stretch the other to the 5th division of the twelfths. The following figure is laid down from the scale at the bottom, numbered 15, and may be tested by the pupil. INSTRUMENTS AND THE SLIDE-RULE. . !<),. 43 The scale of chords serves the purposes of the protractor, and is used as follows : 1. To find the number of degrees contained in a given angle, BAG. From A with the radius 60 describe the arc a a. Take the distance from a to a in the compasses, and apply it to the beginning of the scale. The number to which it reaches, shows the degrees contained in the angle. 2. To make an angle ABC, which shall contain a given number of degrees, as 26. From B with the radius GO describe the arc AC. Take 26 from the scale ; place one foot of the compasses in C 44 A TREATISE ON A BOX OP and with the other cross the arc in A Join AB. ABC is the angle required. And in the same manner may be performed all the pro- blems described under the protractor, which need not be repeated here. The diagonal scale is used where more exactness is re- quired, or when a number containing three figures is wanted, as 357, 35.7, 3.57, &c. If the primary divisions denote hundreds, the subdivisions express tens, while the units are counted on the parallels upwards on the left, the quarter-inch scale ; and downwards on the right, the half-inch scale. If the primary divisions (those denoted by the perpendiculars) express tens, the diagonals will be units, and the parallels tenths, and so on ; each smaller division being the tenth of the next larger. In taking off numbers from this scale, proceed in an inverse order to the figures; that is, commence with the units, proceed to the tens, and end with the hundreds; thus, to take off 346 from the larger scale. Place one foot of the compasses upon the sixth parallel where it is crossed by the fourth diagonal, and extend the other to the third perpendicular. To take off 1839 from the smaller scale. On the 9th parallel, where it is crossed by the third diagonal, place one foot of the compasses, and extend the other to the 18th perpendicular. To raise a perpendicular to a given line, AB. Make AC = 4 equal parts. From C, with a distance of 3 from the same scale, make an arc ; and from A with 5 cross it in D. Join DC- -D DC will be the perpendicu- lar required. INSTRUMENTS AND THE SLIDE-RULE. Given two square pieces of deal board ; the side of one 17 inches, of the other 29. It is required to ascertain the side of another that shall be equal to both. Lay down a base, AC = 29, and raise a perpendicular BC = 17. Join AB; apply it to the scale, and it will be found 33.6. For the square of the hypothenuse is equal to the sum of the squares of the base and perpendicular. It is required to find the diameter of a copper, that, being of the same depth as another whose width is 13 inches, may contain thrice as much. Make AB = 13, and raise a perpendicular AC. From B, with twice the distance, cross it in C. Apply CA to the scale; it will be found to be 22 *. For if AB = 1, and BC = 2, then AC = Three men bought a grindstone, 20 inches in diameter ; and agreed that the first should use it till he ground down J of it for his share ; the second to do the same ; and the third to finish the remainder. Ascertain the diameters of the second and third shares. 4G A TREATISE ON A BOX OP Draw the line AB = 20 ; bisect it in C, and on AC describe the semicircle AEDC. Divide AC into three equal parts, in the points FG- ; perpendicular to which draw the lines EF, DGr, to meet the semicircle. Join EC, DC, and produce them till CH be equal to CD, and CK to CE. EK is the diameter for the second person, and DH for the third. By applying them to the scale, EK will be found to be about 16J, and DH rather more than 11}. For AC.CG == CD 3 and AC.CF = CE S .-. GG : CF : : CD a : CE 3 ; and the areas of circles are as the squares of their radii. Four men bought a grindstone of 30 inches in diameter; and agreed that the first should use it till he ground down l-4th of it for his share, deducting 6 inches in the middle for waste ; and then that the second should use it till he ground down l-4th part ; and so on. What part of the diameter must each grind down ? l-5th of the diameter being waste, 125th of the content INSTRUMENTS AND THE SLIDE-RULE. is waste ; therefore, conceiving the whole to contain 25 shares, 1 share will be for waste, and each person will have 6 shares. Hence, draw the line AB = 30, and on AC describe the semicircle AHKLMC. Divide AC into 25 equal parts by problem 10, parallel ruler; and make AG-, GF, FE, ED, each equal to 6 of these parts. From the points G, F, E, D, raise perpendiculars to meet the semicircle in H, K, L, Mj join HC, KG, LC, and MC; and, having drawn circles from H, K, L, M, with C as a centre, produce them to N, 0, P, Q. The diameter AB is 30 : HQ will be found, upon applying it to the scale, to measure about 26; KP, nearly 21 J; LO, about 16; and MN, 6 for the waste. Subtracting these in succession, we have 4 inches for the first, 4$ for the second, 5 for the third, and 10 for the last. The student will find these problems repeated at the end of the Uses of the Slide Rule. 48 A TREATISE ON A BOX OP TRIGONOMETRY. THE circumference of a circle is supposed to be divided into 360 equal parts, called degrees. Each degree is divided into 60 minutes; each minute into 60 seconds; and so on. Degrees are marked with a small at the top of the figure, minutes with ', seconds with ", and so on. Thus, 36 18' 25" 36 degrees, 18 minutes, 25 seconds. The difference of an arc from 90 degrees, or a quarter, is called its complement ; thus FC is the complement of CB. The chord of an arc is a line drawn from one ex- tremity of the arc to the other ; thus CK is the chord of the arc CBK. The sine of an arc is a line drawn from one extremity of the arc perpendicular to the diameter passing through the other extremity ; thus CD is the sine of the arc CB, or angle CAB, which it measures ; and CE is the sine of the arc CF, or angle CAF, which it measures. The sine is half the chord of twice the arc, or angle. INSTRUMENTS AND THE SLIDE-RULE. 49 The tangent of an arc is a line touching the circle in ne extremity of the arc, and meeting a line drawn from the centre through the other extremity ; thus GB is the tangent of the arc CB, or angle CAB; and FH is the tangent of the arc CF, or angle CAF. The secant of an arc is the line meeting the tangent ; thus GA is the secant of the arc CB, or angle CAB j and AH is the secant of the arc CF, or angle CAF. The versed sine of an arc is the part of the diameter intercepted between the arc and its sine : thus DB is the versed sine of CB. The cosine, cotangent, and cosecant of an arc, are the complement's sine, tangent, and secant : co being simply a contraction of complement. Thus CE, or AD, is the cosine of CB, being the sine of the complement CF : so FH is the cotangent of the arc CB, being the tangent of the complement FC; and AH is the cosecant of CB, being the secant of the complement CF. From these definitions it is evident 1st. That when the arc is 0, the sine and tangent are 0, but the secant is then the radius AB. 2d. When the arc is a quadrant, FB, then the sine is the greatest it can be, being the radius of the circle ; aad the tangent and secant are infinite. 3d. The versed sine and cosine together make up the radius. 4th. The radius AB, the tangent BG, and secant AG, form a right-angled triangle. 5th. The cosine AD, the sine DC, and radius AC, also form a right-angled triangle. 6th. The radius AF, the cotangent FH, and cosecant AH, also form a right-angled triangle. And since the 5 50 A TREATISE ON A BOX OF angle FAH = the angle ACD = the angle AGB; these right-angled triangles are similar to each other Hence AD DC AB BG viz. cosine sine radius tangent. . AE EG AF FH viz. sine cosine radius cotangent. AD AC AB AG. viz. cosine radius radius secant. EA AC FA AH. viz. sine radius radius cosecant. GB BA AF FH. viz. tangent radius radius cotangent. So the radius is a mean proportional between the cosine and secant, the sine and cosecant, and the tangent and cotangent. In every case in trigonometry three parts must be given to find the other three ; and one of these, at least, must be a side. The cases in trigonometry are of three varieties : 1st. When a side and its opposite angle are given. 2d. When the two sides and the contained angle are given. 3d. When the three sides are given. When a side and its opposite angle are two of the given parts j then Any side : sine of its opp. angle : : any other side : sine of its opp. angle. INSTRUMENTS AND THE SLIDE-RULE. 51 And, Sine of any Z_ : its opp. side : : sine any othei L. : its opp. side. Make AE = BC, and EF, BD, perpendicular to AC. Then AB : BD : : AE : EF. But AE = BC; .-. AB : BD : : BC : EF. But BD is the sine of C, and EF of A ; .-. AB : sin opp. Z_ C : : BC : sin. its opp. l_ A. n. When two sides and their contained angle are given. Sum of sides : their difference : : tang. \ sum of oppo- site angles : tang, of J their diff. Then \ sum -{- i diff. = greater ; and \ sum \ diff. = less. Let ABC be the proposed triangle, having the two given sides AC, BC, including the given angle C. From C, with radius CA, describe a semicircle, meeting 62 A TREATISE ON A BOX OP BC produced in D and E. Join AE, AD, and draw DF parallel to AE. Then BE = BC + CA, the sum of the sides BC, CA: and BD = BC CA, their difference. Also, CAB + CBA = CAD + CDA = 2CDA ; .-. L CDA =1 1 CAB + CBA. j That is, CDA is half the sum of the unknown angles. Again, L CDA = L DBA + L DAB ; .-. f_ DAB = L CDA f_ DBA = L CAD L CBA, Add to each side DAB ; then, 2 DAB = CAD -f DAB CBA = CAB CBA ; .-. DAB = i | CAB CBA. j That is, DAB is half the difference of the unknown angles. Now EAD being a semicircle, EA is perpendicular to AD, as is also DF; .. AE is the tangent of CDA, and DF the tangent of DAB, to the same radius AD. But, BE : BD : : AE : DF; that is, the sum of the sides : difference of the sides : : tangent of \ sum of opposite angles : tangent of \ their difference. Three angles of a triangle being equal to 180, the sum of the unknown angles is found by taking the given angle from 180. in. When the three sides are given, to find the angles. INSTRUMENTS AND THE SLIDE-RULE, 53 Base : sum of other sides : : difference of those sides : difference of segments of base, made by perpendicular falling from the vertex. Let ABD be the given triangle. From B with the distance BA describe a circle, and produce DB to G-. Then, DG = DB + BA; &nd HD = DB BA, Also, since AE = EF ; . . FD = DE EA. But AD : DG : : HD : FD, (Euclid 3, 46 ;) that is, Base : sum of other sides : : diff. of those sides : diff. of segments of base. Hence, in each of the two right-angled triagles, there will be known two sides, and the right angle opposite to one of them. All cases of plane triangles may be solved by these three problems ; but for right-angled triangles, the follow- ing are more convenient : 54 A TREATISE ON A BOX OF First make the base AB radius ; then, Radius : tang. A : : AB : BC ; and Radius : secant A : : AB : AC. Next make CB radius. Then Radius : tang. C : Radius : secant C : Lastly, make AC radius. : CB : BA; and : CB : CA. \ Then, Radius : sine A : : AC : CB ; and Radius : cos. A : : AC : AB. SECTOR. THE lines on the sector, as before stated, are of two sorts, single and double. Only the double lines are sec- toral ; these proceed from the centre, and are 1st. A scale of equal parts, marked L, and containing '.00 divisions. d. A scale of chords, marked C, running to 60. INSTRUMENTS AND THE SLIDE-RULF.. 55 3d. A line of secants, marked S, running to 75. 4th. A line of polygons, marked POL. These are num- bered backwards from 4 to 12. Upon the other face, the sectoral lines are, 1-:. A line of sines, marked S, running to 90. The sines may be easily distinguished from the secants, though marked the same ; as the distances of the sines diminish towards the end, while the secants increase. 2d. A line of tangents, marked T, running to 45. 3d. Between the line of tangents and sines there is another line of tangents, beginning at a quarter of the length of the former, to supply their defect, and extend- ing from 45 to 75, marked t or T. The distance from T to T, from S to S, from C to C, and from L to L, is the same ; so that at whatever distance the sector may be opened, the angles formed by those lines will always be equal. The polygons are laid down to a shorter radius, for the sake of including the pentagon and square. The radius, the chord of 60, the sine of 90, the tangent of 45, the secant of 0, all are equal. The method of constructing the sectoral lines is exhi- bited in Fig. 1, fronting the title-page. From the point A with any convenient radius describe the circle BODE, and draw the diameters BD, CE, cross- ing each other at right angles in the centre. Produce C to F, and through B draw BG, a tangent to AB. Join EB, BC, CD. In the construction of the following Scales, only the primary dii-isions are drawn, the smaller ones being omitted to prevent confusion : Divide AD into ten equal parts, and these again into tenths ; so shall AD be a line of equal parts. 66 A TREATISE ON A BOX OP Divide the arc BC into 9 equal parts, and these again into tenths, and with the compasses from B, as a centre, transfer the divisions to the straight line BC; so shall BG be a line of chords. From each of the divisions of the arc BC let fall perpen- diculars upon AB, and number them backwards ; so shall AB be a line of sines. From the point A, through the divisions of the arc BC, draw straight lines to meet BG; so shall BG be a line of tangents. And from D to the same divisions of the arc BC, draw straight lines cutting AC ; so shall AC be a line of half-tangents. The distances from the centre A to the divisions on the line of tangents being transferred to AF, AF will be a line of secants. From B, in the arc BE, cut off the fifth part of the circumference, 72, and transfer it with the compasses to the straight line BE. Do the same with the sixth, seventh, eighth, &c. ; so shall BE be a line of polygons. Transfer these distances to the lines upon the rule, and the sectoral part is complete.* From the property of similar triangles, AB : BC : : Ab : be; hence, if AB were divided into ten equal parts, and A.b contained four of those parts, then BC being di- vided into ten equal parts, be would contain four of those parts. And if AB were the sine of 90, and Ab the sine of * These lines are best laid down by the help of tables of natural sines, tangents, gJ-r} } ri*'ch contains a portion of the line L. When the distance is taken from the centre 1 , along 58 A TREATISE ON A BOX OP either of the legs to any point, BC. or D, it is called a lateral distance ; when it is taken from any point of one leg to the corresponding point of the other leg, as from B to B, from C to C, from D to D, or from L to L, it is Called a parallel distance. To make any length then, aa for instance an inch, a parallel distance from 3 to 3, or D to D, the points of the compasses are to be opened aa inch ; then placing one point against the 3 of one leg of the rule, open the rule till the other point will fall upon the 3 of the other leg. Observe, the points of the com- passes must always be placed upon the lines nearest the opening of the rule upon those which would, if continued, meet in the centre. As distances are taken from all the sectoral lines in the same manner, and as they are of most extensive utility when used conjointly, it will not be necessary to treat of them separately. USES OF THE SECTOR. To divide a given line AB into any number of equal parts, as 6. A- ... . , , _. ,B 12 34 56 Take the distance from A to B in the compasses, and make it a parallel distance from 6 to 6 on the line L; then the parallel distance from 1 to 1 will be the sixth part of the line AB, as required. To find a third proportional to two numbers, 4 and 6. Take 6 laterally, and make it a parallel distance from 4 to 4 on L : then take the parallel distance from 6 to 6, and apply it laterally : it will be found to measure to 9, the third proportional required. Or ir ake the lateral dis- INSTRUMENTS AND THE SLIDE-RULE. 59 taiice 4, a parallel distance between 6 and 6 ; then will the lateral distance 6 be found a parallel distance between 9 and 9. To find a fourth proportional to three given numbers, 3, 4, and (7. Take the lateral distance 3, and stretch it as a parallel from 4 to 4 on L; then take the lateral dis- tance 6, it will be found to extend as a parallel from 8 to 8. Or make 4 a parallel distance from 3 to 3j then will the parallel from 6 to 6 measure laterally to 8. The reason is obvious ; for in all cases, Any lateral distance : its parallel distance : : any other lateral distance : its parallel distance. And conversely, Any parallel distance : its lateral distance : : any other parallel distance : its lateral distance. To measure the lines of the peri- c meter of any figure, one of which, as AB, contains a given number of equal parts, as 4. Make AB a pa- rallel distance from 4 to 4 on L. Take CB, and it will be found a parallel distance from 3 to 3 ; take AC, and it will be found a parallel distance from 5 to 5. To multiply any number, 3, by another, 7. Make the lateral distance 3 a parallel distance from 1 to 1 on L; then take the parallel distance from 7 to 7 ; it will be found the lateral distance of 21. To divide any number, 40, by another, 5. Make the parallel distance of 40 the lateral distance of 5 on L' 60 A TREATISE ON A BOX OF then will the parallel distance of 1 be the lateral distance of 8. To divide a line in any required proportion, as 2, 3, and 4. Take the length and make it a parallel distance front 9 to 9, their sum, on the line L ; then will the parallel distance from 2 to 2, 3 to 3, and 4 to 4, be the parts re- quired. If the number is greater than 100, take some aliquot part of it, and then multiply the result by the number by which it was divided. To measure an angle ABC with the line of chords, C. With any radius BA, describe the arc AC ; make BA a parallel distance from 60 to 60 on C ; then take AC, and moving it along, find the numbers to which it will apply as a parallel distance. At a given point B, in a given line AB, to make an angle containing any number of degrees, as 30. Open the sector to any distance, and take the parallel distance from 60 to 60, with which describe the arc DE. Take the parallel distance from 30 to 30 ; and setting one INSTRUMENTS AND THE SLIDE-RULE. 61 foot of tht compasses on D, prick off the distance to E ; through E draw BC. ABC is the angle required. And so of all the problems given under the protractor. To measure any angle, ABC, with the line of sines. From any point, A, let fall the line AD, perpendicular to BC ; then in AD the sine of B, the radius being BA Therefore, take BA, and make it a parallel distance from 90 to 90, on the line S; then take AD, and moving it along, find the numbers to which it will apply as a pa- rallel distance At a given point A, in a given line AB, to make an angle containing any number of degrees, as 30. ,C Make AB a parallel distance from 90 to 90 on S ; take the parallel distance from 30 to 30, and from B describe an arc. Draw AC touching the arc. CAB is the angle required. And so of all the problems under the protractor. 6 62 A TREATISE ON A BOX OP To measure any angle, ABC, with the line of tangents, T. ,A At any distance,!), raise the perpendicular DE, to meet AB; then is DE the tangent of B, the radius being BD. Make BD a parallel distance from 45 to 45 on T ; then take DE, and moving it along, find the numbers to which it will apply as a parallel distance. At a given point A, in a given line AB, to make an angle containing any number of degrees, a,* 30 D Take any point, C, at which erect the perpendicular CD, Make AC a parallel distance from 45 to 45. Take the parallel distance from 30 to 30, and from C cut off CE equal to it. Join AE. EAB is the angle required. And so of the rest under the protractor. To measure any angle, BAG, with the line of secants. INSTRUMENTS AND \HE SLIDE-RULE. 03 'lake any point, D, and raise the perpendicular DE. Makc'AD a parallel distance from to on the line of secants ; then take AE, and moving it along, find the numbers to which it will apply as a parallel distance. At a given point, A, in a given line AB, to make an angle containing any number of degrees, as 40 C Kaise BC at right angles to AB. Make AB a parallel distance from to ; then take the parallel distance from 40 to 40, and with it, from the point A, cross BC in D. Join AD. DAB will be the angle required. And so of the rest under the protractor. The line of secants cannot be employed to much advantage when the number of degrees is under 30, nor the line of sines when above 70, as is evident from an inspection of the rule. To find the chord, sine, tangent, and secant of 30 de- grees, to a radius of two inches, AB. (See diagram p. 64.) Take 2 inches in the compasses, and make it a parallel distance from 60 to 60 on the scale of chords ; it will also be a parallel distance from 45 to 45 on the tangents ; and from 90 to 90 on the sines. Therefore, taking the parallel distance from 30 to 30 on the line C, it will give the chord 64 A TREATISE ON A BOX OF of 30 degrees, BB; from 30 to 30 on the line S, it will be the sine of 30 degrees, BC ; from 30 to 30 on the line T, it will be the tangent of 30 degrees, BD. When the number of degrees is above 45, the same opening will not 6uffi.ce for the tangents ; it will then be necessary to set the radius to the 45 of the smaller tangents, and take the aperture as usual. So for the secants, make 2 inches a parallel distance from to ; then will the parallel dis- tance of 30 be the secant AD. When the rule is opened to the radius of the smaller tangents, it is also opened to the radius of the secants. All the problems given under the protractor may be performed by any of these lines ; but polygons are most conveniently constructed by the lines POL. ; and in taking distances from these, the points of the compasses are to be placed on the same line as that from which the chords are taken. The side of the hexagon, 6, does not reach to the chord of 60, as has been mentioned; a smaller radius being chosen for the purpose of including the 5 and 4, the pentagon and square. Since the radius of a circle is equal to the side of an inscribed hexagon, the radius is always to be made a parallel distance from 6 to 6. An example or two will suffice. INSTRUMENTS AND THE SLIDE-RULJ5. 65 To construct a pentagon on a given line AB. Take AB, and make it a parallel distance from 5 to 5. Take the parallel distance from 6 to 6 for a radius, and with it from A and B describe arcs crossing each other in C, and from C describe a circle. The distance AB will run 5 times round it, and form the pentagon required. And so of any regular polygon. In a given circle, to inscribe a regular heptagon. Find the centre of the circle, and make the radius a parallel distance from 6 to 6 ; take the parallel distance from 7 to 7, and it will run 7 times round the circle as required. And so of any regular polygon. 6 66 A TREATISE ON A BOX OF APPLICATION OF THE SECTOR TO TRIGONOMETRY. llequired the height of a tower, AB, the angle of ele- vation, ACB, 200 feet distant, being 47 1'. The angle B will be 90 47 = 42 }. Taking AC radius, AB is the tangent of C, or 47 i. Hence, Rad. : CA : : tang. 47 J : AB. Take 200 from a scale of equal parts, (the diagonal scale is the best,) and make it a parallel distance from 45 to 45 on the smaller line of tangents ; take the parallel distance from 47 J to 47 i, and apply it to the same scale of equal parts; it will be found to measure about 218 feet, the height of the tower. Otherwise, Sine B : CA : : sine C : AB. On the line of sines stretch 200 from 42 J to 42 J ; take the parallel distance from 47 J to 47J, it will measure 218 as before. What was the }v.;rpendicular height of a balloon, when its angles of elev; I* on were 35 and 64, C.K taken by two observers at the iiiiu time, bc'n on the rime side of it, INSTRUMENTS AND THE Sl.irE-RDLE. 67 and in the same vertical line, their distance beinp 880 yards ? The external angle DEC = DAB -f ADK ; .-. ADB = 64 35^.29. Then, Sine ADB : AB : : sine A : BD. Take 880 from a scale of equal parts, and make it a parallel distance from 29 to 29 on the line of sines ; then will the parallel distance from 35 to 35 measure 1011 for the line BD. Again, Sine C : BD : : sine DEC : DC. Take 1041 and make it a parallel distance from 90 to 90, the angle C being a right angle ; then take the paral- lel distance from 64 to 64, and it will measure 936 for CD, the perpendicular height of the balloon. Wanting to know the distance between two inaccessible trees from the top of a tower, 120 feet high, which lay in the same right line with the two objects I took the angles formed by the perpendicular wall and lines con- ceived to be drawn from the top of the tower to the bottom of each tree, and found them to be 33 and 64$. What 1.4 the li-l 9 _U v v 398^ Find the value of - - ; . Over 33 of D oo 2 set 40, then Over 24 is 21.2 Over 32 is 37.6 Ditto 37.6 96.4 , 40 (24 a + 32 3 + 59.2 2 ) Find the value of 4f -. Over 46 of 4o 9 D set 40, then Over 24 is 10.9 32 is 19.35 59.2 is 66.2 96.45 It sometimes happens that we require to multiply three numbers together. This cannot be done by the kind of rule we have been considering in one operation, but it may be effected by dividing by the reciprocal of one of the numbers. Thus, let it be required to find the product of 4X7X8. The reciprocal of 4 is T = 25 j hence we have to divide by .25, 7 times 8. To .25 on A set 7 of the slide, under 8 on A is 224. By inverting the slide, and pushing it evenly in, that is, until the end is under the beginning of A, it will be seen that the numbers on B are the reciprocals of the corresponding ones on A ; hence if instead of the D line, a line similar to A were laid down under the slide, in an inverted position, it would furnish a series of reciprocals, and then three numbers might at once . IXSTKUMENTS AND THE SLIDE-RULE. lO'A be multiplied together, by taking one of them on this in- verted line, one on the slide, and the third on the A line, under which would be the product. Moreover, if, instead of laying it down so as to make the commencement of it fall under the end of the slide, it were drawn out toward the right hand till some other number than unity stood under the end of the A line, then the product of the three numbers would be divided by this constant number. For instance, supposed we wished to divide by 2, 7 times IT \s 8^6 8 times 6 ; that is, to find the value of - . Let a the inverted line be placed so that the 2 shall fall under the end of the A line ; then over the 7 of this inverted line place 8 of the slide, and under the 6 of A will be 168. Hence, if a person pursued an occupation in which his calculations required to be divided by a constant number, he might have a rule constructed to suit himself for that particular number. A few such rules are in use. The officers of the customs have frequently to measure pieces of timber, the length of which is taken in feet, and the breadth and thickness in inches. Now, multiplying these three dimensions together, and dividing by 144, gives the solidity in cubic feet. Hence let the A, B, and C lines be laid down as Us'ial, and instead of D substitute an inverted A line, so placed that 144 shall fall under the end of the slide. Then, if a piece of timber measures 55 feet long, 24 inches broad, and 9 thick ; under 55 of A place 24 of the slide, and over 9 of the inverted line is 82^ cubic feet, the content. In malt guaging again, the number of cubic indies in a bushel is 2218.19. Hence, taking the di- mensions in inches, let the inverted line be so placed that the nuiii'.c;- ^>^1>. 1'J shall fall under the end of the slide; 110 A TREATISE ON A BOX OP then if a cistern of malt measures 30 inches long, 16 broad, and 12 deep ; to 30 on A set 16 of the slide, and over 12 of the inverted line is 2.6 nearly, the content in bushels. If we wish to obtain the result in gallons, (as 8 gallons make a bushel,) take 8 times one of the dimen- sions : for instance, to 240 on A set 16 of the slide, and over 12 of the inverted line is 20.78 gallons. In practice these rules are of the utmost convenience possible, and the principle might be carried out with ad- vantage to a much greater extent than it yet has been. OBSERVATIONS. There are three kinds of measure lineal, superficial, and solid : lineal, for such things as have length only ; superficial, for those that have length and breadth ; and solid, where there are length, breadth, and thickness. When lines vary proportionally they vary simply as then measures ; when surfaces vary proportionally they vary as the squares of their like measures ; and when solids varj proportionally they vary as the cubes of their like measures. Thus, let there be two similar funnels, or cones, A and B ; and let A be filled with water to the depth of 1 foot, and B to the depth of 2 feet ; then the circumference of the top of the water in B will be twice that of A, both being lines; the area of the top of the water in B will be 4 times that of A, or 2 a , both being surfaces: and the weight or quantity of the water in B will be 8 times that of A, or 2 3 , both being solids : and so of all surfaces and polids that vary proportionally. If a number of regular polygons have oqual periinetero. INSTRUMENTS AND THE SLIDE-RULE. Ill tfiut contains the greatest amount of surface in which the perimeter is distributed among the greatest number of sides ; and, as a circle may be conceived to be a polygon of an infinite number of sides, it therefore contains the greatest quantity of space within the shortest bounding line. A regular polygon contains more than an irregular polygon of the same number of sides, their perimeters being equal ; thus, an equilateral triangle has a greater area than any other triangle of equal ambit ; and a square is the largest quadrilateral that can be constructed with sections of the same line. In the same way as the circle contains the largest sur- face within the least compass, so the sphere contains the greatest bulk within the smallest space. RATIOS AND GAUGE POINTS. At the back of the rule will be found a quantity of tabular work, adapted to various kinds of calculation : these consist of ratios and gauge points. Ratios express the proportions existing between certain lines, or num- bers ; thus, if the diameter of a circle be 113, its circum- ference will be 355; and, as the circumference varies as the diameter, therefore 113 : 355 expresses the ratio of the diameter of any circle to its circumference. Gauge points are the square roots of divisors ; thus, if we require to reduce square inches to square feet, we must divide by 144, which number may be chosen on A; if instead of this we divide by 12 a , we take 12 upon the D line, and, 112 A TREATISE ON A BOX OP for the sake of distinction, 12 is called a gauge point In rules having the D line commencing with unity, when the slide is set to any gauge point, it is also set to the cor- responding divisor, the one standing under the slide, the other above it ; and therefore, with such rules, it would be immaterial whether we used divisors or gauge points; as however, the formulae for many surfaces, and almost all solids, require the use of the D line, it is far more conve- nient for valuing the numbers, to make use of gauge- points, and therefore the tabular work is so constructed. TABLE I. contains a list of ratios belonging to the circle, commencing A B 113 diameter = 355 circumference. 44 diameter = 39 side of equal square. That is, under 113 on A set 355 on B, then the num- bers on A will be a series of diameters, and the numbe, 3 beneath them on B will be their corresponding circumfe- rences ; and so of all the rest. EXAMPLES. 1. If the diameter of a circle is 8 inches, what is its circumference ? Set the rule as directed, then under 8 is 25.13 inches. 2. The diameter of a circle is 9 inches, what is the side of an equal square ? Under 44 of A set 39 of B ; under 9 is 7.97 inches. INSTRUMENTS AND THE SLIDE-RULE. 113 S. The radius of a circle is 6 inches, what is the leijgth of an arc of it containing 31 J degrees? Under 57.3 de- grees on A set 6 inches on B ; under 31 degrees on A is 3.3 inches. 4. The circumference of a circle is 75, what is the di- ameter ? Ans. 23.87. 5. The diameter is 7, what is the circumference ? Ans. 22 nearly. 6. The diameter is 17, what is the circumference ? Ans. 53.4. 7. Suppose the diameter of the earth to be 7960 miles, what is its circumference ? Ans. 25,000 miles. 8. The diameter of a circle is 6 inches, what is the side of a square inscribed within it ? Ans. 4.24. 9. The circumference of a circle is 12 feet, what is the side of an equal square ? Ans. 3.38. 10. The circumference of a circle is 15 inches, what is the side of its inscribed square ? Ans. 3.375. 11. The side of a square is 10 inches, what is the diameter of an equal circle ? Ans. 11.28. 12. The side of a square is 20 yards, what is the cir- cumference of an equal circle ? Ans. 70.83. 13. The side of a square is 19 inches, what is the side of an equal equilateral triangle ? Ans. 28.88. 14. The area of a circle is 27, what is the area of a square inscribed in it? Ans. 17.18. lu* 114 A TREATISE ON A BOX OF 15. An arc of 38 degrees measures 5 inches, irO' iJ the radius of the circle of which it is a part? Aa "?.(><>. 16. I have a circular piece of wood, whose diameter ig 15 inches, and wish to cut the largest square out of it ; what will be the length of each side ? Ans. 10.6 inches. The method of obtaining these ratios in whole numbers is a beautiful exemplification of the abridgment of labour effected by the slide-rule ; and of performing, with the utmost facility, operations that would require considerable time and trouble by any other means. Archimedes dis- covered that the ratio of 7 to 22 nearly expressed that of the diameter of a circle to its circumference. Purbachius, in the fifteenth century, making the diameter 120, reck- oned the circumference at 377. Metius, two centuries later, subtracted the 7 and 22 from the 120 and 377, and obtained the numbers 113 and 355. This last ratio is easily remembered, from its containing the first three odd numbers in pairs, and it is remarkably accurate, the quotient of 355 by 113 being true to the sixth place of decimals. The obtaining of these ratios in integers, how- ever, must have been a task of considerable labour. To determine them by the slide-rule is the work of a moment. By various modes of computation it may be shown that if the diameter be 1, the circumference will be nearly 3.1416 ; therefore, under 1 of A set 3.1416, as nearly as possible, and run the eye along until you find two num- bers coinciding : such will be 113 and 355, which will be the ratio required. The advantage of having the ratios in whole numbers, for the purposes of the slide-rule, is obvious, as they can be set with greater rapidity and exactness than decimals. INSTRUMENTS AND THE SLIDE-RULE. 115 The following table will enable the student to solve the pr- questions numerically : Diameter 1, circumference 3.1416.' side of equal square, .8862. side of inscribed square .7071. Circumference 1, diameter .3183. side of equal square .2821. side of inscribed square .2251. Side of square 1, diameter of equal circle 1.128. circumference of equal circle 3.545. side of equal equilateral triangle 1.5196. Area of circle 1, area of inscribed square .6366. Area of square 1, area of inscribed circle .7854. area of inscribed octagon .8284. The length of an arc of 57.2957795 degrees = radius of circle Solution of question 8: .7071 X 6 = 4.2426. TABLE II. contains the Linear Dimensions of Polygons described within and without Circles, and commences thus : No. of Sides. Inscribed Polygon. Circumscribed Polygon. A. Diam. B. Side. A. Diam. B. Side. 3 4 15 9.9 13 7 15 1 26 1 That is, if the diameter of a circle be 15, the side of an equilateral triangle inscribed within it will be 13 : hence, under 15 of A set 13 of B; then the numbers on A will be a series of diameters, and the numbers beneath them on B will be the sides of the corresponding triangles; and so of the rest. The method of obtaining them is first by computation, and then as for the ratios before described. 116 A TREATISE ON A BOX OF EXAMPLES. 17. The diameter of a circle is 12 inches, what will bo the side of an equilateral triangle inscribed therein ? Under 15 of A set 13 of B; under 12 of A is 10.4 inches. 18. The diameter of a circle is 11 J inches, what is the side of a regular pentagon inscribed within it ? Ans. 6.76. 19. A circle whose diameter is 9i inches has a regular hexagon surrounding it, what is the length of each side ? Ans. 5.33. 20. A person having a circular piece of ground 37 yards in diameter, wishes to make within it a flower-bed of a heptagonal form, whose area shall be a maximum ; what will be the length of each side ? Ans. 16. 21. If I make the diameter of a circle a parallel dis- tance on the line L of the sector from 100 to 100, what parallel distance must I take off as the side of an undoca- gon inscribable therein ? Ans. 28.1. The following table will enable the student to solve the pre ceding questions numerically. The diameter of the circle being unity, No. of Sides. Side of Inscribed Polygon. Side of Circumscribed Polygon. 3 .8660254 1.7320508 4 .7071068 1.0000000 5 .5877853 .7265425 6 .5000000 .5773503 7 .4338837 .4815745 8 .3826834 .4142136 9 .3420201 .3639702 10 .3090170 .3249197 11 .2817325 .2936264 12 .2588190 .2679492 Solution of question 20: .4338837 X 37 = 16.0536969. INSTRUMENTS AND THE SLIDE-RULE. 117 TABLE III. contains the Areas of Polygons, commenc- ing thus : No. of Sides. C. Area. D. Side. 3 5 3.9 43 3 5 That is, if 3 be the side of an equilateral triangle, its area will be 3.9, and, as similar surfaces vary as the squares of their like measures, if over 3 of D we set 3.9 on C, then the numbers on D will be a series of sides, and the numbers over them on C their corresponding areas. EXAMPLES. 22. The side of an equilateral triangle is 2, what is its area ? Over 3 of D set 3.9 on C ; over 2 of D is 1.732, the area required. 23. Required the area of a regular nonagon having a side of 7.3 yards. Ans. 329 yds. 24. What is the area of an undecagon whose side measures 6.4 feet ? Ans. 383.6 feet. 25. The side of an octagon is 4.9 feet, what is its area ? Ans. 116 nearly. The side being given in inches, to find the area in square feet, take 12 times the number on D, for the number of inches in a square foot is 12 2 . EXAMPLES. 26. The side of an equilateral triangle is 19 inches, how many square feet does it contain ? Over 36 of D set 3.9 of C j over 19 of D is 1.086 feet. 118 A TREATISE ON A BOX OF 27. The side of a regular pentagon is 53 inches, how many square feet does it contain? Ans. 33.55. 28. What is the area of a nonagon, each of whose sides measures 27 inches? Ans. 31.29. The side being given in feet, to find the area in square yards, take 3 times the number on D, for the number of feet in a square yard is 3 a . EXAMPLES. 29. The side of a regular pentagon is 7 feet, how many square yards does it contain ? Over 15 of D set 43 of C; over 7 of D is 9.36 yards. 30. What is the area of a heptagon whose side measures 17 feet? Ans. 116.7 yards. 31. A decagon measures 20.2 feet along each side, what is the area? Ans. 348.8 yds. The following table will enable the student to solve the pre- ceding questions numerically. Multiply the subjoined numbers by the square of the side. Equilateral Triangle 4330127 Pentagon 1.7204774 Hexagon. 2.5980762 Heptagon 3.6339124 Octagon 4.8284272 Nonagon 6.1818242 Decagon 7.6942088 Undecagon 9.3656411 Dodecagon 11.1961524 Solution of Question 22: .4330127 X 2 s = 1.7320508. INSTRUMENTS AND THE SLIDE-RULE. 119 FALLING BODIES. TABLE IV. is of a miscellaneous nature, and will be understood by inspection. It is found that a heavy body, in the latitude of London, falls 579 feet, or 193 yards in 6 seconds ; and the spaces descended by falling are as the squares of the times ; hence, as directed by the table, over 6 seconds on D set 579 feet on C, (or 193 yards, if the distance be required in yards,) then the numbers on G will be a series of distances fallen, and the numbers be- neath them on D the seconds elapsed in falling. The same law applies to bodies projected directly upwards, the retardation corresponding with the acceleration in an inverse order. EXAMPLES. 32. How many feet will a body fall in 1 second ? Over 6 of D set 579 on C; over 1 of D is 16^ feet. 33. If a ball is propelled straight upwards, and is found to be 18 seconds before it again falls to the earth, how many yards has it ranged ? 9 seconds occupied in ascend- ing, 9 in descending ; over 6 of D set 193 on C ; over 9 is 434 yards. 34. Standing at the mouth of a well, which, by means of reflecting the sun's rays into it with a mirror, I per- ceived to be of considerable depth, I dropped a stone into it, and found it reached the water in 3J seconds; what was its depth ? Ans. 197 feet. 35. How long would a cannon-ball, fired perpendicu- larly upwards, be in rising a mile, if it went no higher ? Ans. 18.12 seconds. A TREATISE ON A BOX 01 PENDULUMS. A pendulum 22 inches long, as shown by the table, makes 80 vibrations, or 40 revolutions per minute, and their lengths vary reciprocally as the squares of their times, their velocity being regulated by the force of gravity, like that of falling bodies. Hence, invert the slide, and set 22 inches on B over 80 vibrations, or 40 revolutions on D ; then the numbers on the inverted line will be a series of lengths, and the numbers beneath them on D the corresponding number of vibrations or revolu- tions. EXAMPLES. 36. What is the length of a pendulum vibrating 60 times per minute ? Over 60 is 39.2 inches. 37. What is the length of a pendulum vibrating 64 times per minute ? Ans. 34.4 inches. 38. What is the length of a pendulum making 29 re- volutions per minute ? Ans. 42 inches. EXPERIMENT. Suspend from a hook in the ceiling a string with a bullet at the endj set it vibrating, or swing it round so as to cause it to revolve, and compare its motions with a watch. The next part of the tabular work relates to the areas rf circles and surfaces of spheres, and is as follows : Circle 43 area C = 7.4 diameter D. 23 area C ^ 17 circumference D. Sphere 172 surface C = 7.4 diameter D. 92 surface C = 17 circumference. INSTRUMENTS AND THE SLIDE-RULE. 121 The student will perceive from this that the surface of a sphere is 4 times the area of a circle of equal diameter That is, if an orange were perfectly round, and cut into two equal parts, then the external surface of the rind in each half would be just double the surface of the part cut by the knife. Similar surfaces varying as the squares of their like measures, the dimensions being taken on D the areas will be on C. EXAMPLES. 39. The diameter of a circle is 5 inches, what is its aiea? Over 7.4 of D set 43 of C; over 5 is 19.63 square inches. 40. What is the area of a circle whose circumference is 12 inches? Ans. 11.46 inches. 41. The circumference of a sphere is 12 inches, what is the surface ? Ans. 45.8 square inches. If the dimensions are given in inches and the area is required in feet, take 12 times the number on D; and if the dimensions are in feet, and the area is required in square yards, take 3 times the number on D. EXAMPLES. 42. The diameter of a circle is 19 inches, what is its area in square feet? 12 times 7.4 = 88.8; over 88.8 on D set 43 on C ; over 19 is 1.97 square feet. 43. The circumference of a circle is 43 inches, how many square feet does it contain? Ans. 1.021 square feet. 44. The diameter of a sphere is 17 feet; what is its surface in square yards? Ans. 100.8 square yards. 122 A TREATISE ON A BOX OP The nezt part of the table is 1 C = side of square or cube D. 2 G = diagonal of square or diameter of circumscribing circle D. 8 C = diagonal of cube or diameter of circumscribing sphere D. That is, set 1 of C over the side of a given square on "D, then under 2 of C will be its diagonal, or the diameter of its circumscribing circle. EXAMPLES. 45. A circle 12 inches in diameter has a square in- scribed within it, what is the length of each side ? Over 12 of D set 2 of C; under 1 of C is 8.48 inches. 46. A cube measures 7 inches along the side, what will be the diagonal of the face, and what of the cube? Ans. 9.9 diagonal of the face. 12.12 diagonal of the cube. 47. What is the longest line that can be taken in a cubical box whose sides measure 19.4 feet? Ans. 33.6 feet. 48. A square inscribed in a circle measures 43 inches, what is the diameter of the circle? Ans. 60.8 inches. 49. The diameter of a sphere is 26.7 inches; what will be the side of the largest cube that can be cut from it ? Ans. 15.41. 50. Standing within a cubical room I found that the distance from one of the top corners to the opposite cor- ner at bottom was 23.3 feet; what was the distance of the ceiling from the floor? Ans. 13.45 feet. INSTRUMENTS AND THE SLIDE-RULE. 123 VELOCITY OF SOUND. THE flight of sound is uniformly proportional to the time; hence use the A and B lines as directed in the table. EXAMPLES. 51. I observed the flash of a gun 12 seconds before hearing the report; how far was it distant from me? To 65 seconds on A set 14 miles on B ; under 12 of A is 2.59 miles on B. 52. I observed a flash of lightning, and 7 seconds after- wards heard the thunder ; how far distant was the electric cloud? Ans. Ij mile. 53. A person standing on the bank of a river, heard the echo of his voice reflected from a rock on the opposite bank in 5 seconds after; what was the breadth of the river? Ans. 950 yards. The subjoined table will enable the student to solve the ques- tions by computation : A body falls 16 2 ' T feet in the first second. A pendulum vibrating seconds in the latitude of London, is 89.1*396 inches. In a pendulum describing a conical surface, the time of revo- lution is equal to the time of two oscillations of a simple pen- dulum, equal to the height of the cone: that is, a pendulum takes the same time in going half round a circle as it does in falling across it. Putting d diameter, c circumference, Area of Circle = .7854 d* or .07958 c. Surface of Sphere = 3.1416 d or .31832 c'. 124 A TREATISE ON A BOX OP Putting s side of square or cube, then diagonal of square, or diameter of circumscribing circle = a v /2 = X 1.4142136. Diagonal of cube, or diameter of circumscribing sphere = s ^/ 3 = s X 1.7320508. Sound flies about 380 yards per second. Solution of Question 33 .16 T ' ? X 9 a = 1302f feet = 434j yards. SURFACES. WE now come to Table 5, which consists of a number of gauge points for the mensuration of surfaces, quadri- laterals, triangles, parabolas, circles, cycloids, and ellipses ; and the surfaces of prisms, cylinders, pyramids, cones, and spheres. The area of a rectangle is equal to the product of the length and breadth. The area of a trapezoid is found by multiplying half the sum of the parallel sides by the perpendicular distance between them. A triangle is half a rectangle, and therefore its area is half the product of the height and base.* The areas of trapeziums and multilaterals are found by dividing them into triangles. A parabola is equal to 3 of its circumscribing parallelo- * If a quadrilateral can be inscribed in a circle, its area will be, (putting s semiperimeter, and a, b, c, d, the sides,) = \/ s a sb. s c. s^d. If one f th e sides, as d, is supposed to vanish, the figure merges into a triangle, and the formula becomes V^H^ g^b. T^c. a. That is, for the quadrilateral, from half the sum of the four sides subtract each side separately: multi- ply the four remainders together; the square root will be the area. For the triangle, from half the sum of the three sides subtract each side separately; multiply the three remainders find the half sum together ; the square root \vill be the area. INSTRUMENTS AND THE SLIDE-RULE. 125 gram, and therefore its area is found by taking of the product of the height and base. A circle may bo con- ceived to be a polygon of an infinite number of sides. Now, a polygon is made up of as many triangles as the figure has sides, and the area of each triangle is found by taking the product of the height and half the basej there- fore the area of the whole polygon will be equal to the perpendicular multiplied by half the perimeter; this, when the figure merges into a circle, becomes the radius multiplied by half the circumference j or, which is equiva- lent, the diameter multiplied by i of the circumference. Now, when the diameter is 1, the circumference is 3.1416, hence 1 X .7854 = the area of a circle whose diameter is unity. And since similar surfaces are to each other as the squares of their like dimensions, the area of any circle will be equal to the square of its diameter multiplied by .7854. The area of the sector of a circle, in like manner, will be found by multiplying the radius by J the length of the arc.* The area of a cycloid is 3 times that of its generating circle. From the method described in page 12, of projecting the circle into an ellipse, it is obvious that the area will be in proportion to the elongation, that is, equal to the product of the axes multiplied by .7854. The sides of a prism being parallelograms, it follows that the perimetrical surface will be equal to the product of * To find the length of the arc, from 8 times the chord of J the arc subtract the chord of the whole arc, and divide by 3 ; the quotient is the length, nearly. To find the length of the chord of J the arc, add together the square of the versed sine, or height of the segment, and the square of J the chord ; the square root is the length of the chord of .} the arc. 11* 126 A TREATISE ON A BOX OF the perimeter and height. The same rule applies to the cylinder, which is a round prism. The sides of a pyramid being triangles, the area of each of which is found by multiplying J the base by the height, the sloping surface will be found by multiplying J the perimeter by the slant height. The same will be the case with the cone, which is a round pyramid. The surface of a sphere is equal to the convex surface of its circumscribing cylinder, which surface, as above shown, is equal to the circum- ference multiplied by the depth ; that is, in this case, the circumference multiplied by the diameter. The same holds good for the surface of any part of the sphere, it still being equal to the surface of the corresponding paral- lel section of the cylinder. We may now refer to Table 5, at the back of the rule, the gauge points of which are determined as follows. To reduce square inches to square feet, we divide by 144, and -j/ 144 = 12, the gauge point for squares ; 144 -j- .7854 == 183.3462, and j/ 183.3462 = 13.54 the gauge point for circles, when the dimensions are taken in inches, and the area is required in square feet ; and so of the rest. Putting s side, b base,^? perimeter, h perpendicular height, H slant height, d diameter, c conjugate axis, t transverse; then Area of square = Parallelogram square. square. }U square. square. Surface of prism ) ph Surface of pyra- ) _ _ \pll and cylinder ) square, mid and cone ) square. INSTRUMENTS AND THE SLIDE-RULE. 127 Area of circle ; Ellipse = -: r circle. circle. Area of cycloid = -^ Up- Surface of sphere = -. circle. circle. Surface of spherical zone = circle. That is, to obtain the area of a square, over the square gauge point on D, set 1 of the slide, then over the side on D is the area; for the parallelogram over the square gauge point on D, set the base, then under the height on A is the area ; or find a mean proportional between the base and height, then over the square gauge point on D set 1, and over the mean proportional on D will be the area ; for the parabola, over the square gauge point on D set f of the base, then under the height on A is the area; for the surface of a sphere, over the circular gauge point on D set 4 of the slide, then over the diameter on D is the surface ; and so of the rest. EXAMPLES. 54. The diameter of a circle is 17 inches; how many square feet doe.- it contain ? Over 13.54 on D set 1 of the slide, over 17 Is 1.576 feet. 55. The base or double ordinate of a parabola is 39 inches, the height or absciss 11.1 inches; what is tho area in feet? f of 39 = 26. Over 12 of D set 26 of the slide, under 11.1 of A is 2 feet. 56. The diameters of an ellipse are 20 and 17 feet, what is the area in square yards? Over 3.385 on D set 20 on the slide, under 17 on A is 29.67 yards; or, to 17 128 A TREATISE ON A BOX OP of 1) set 17 of the slide, under 20 is 18.44, the mean pro- portional : then, over 3.385 on D set 1, over 18.44 is 29.67, as before. 57. The side of a square measures 17 inches; required the area in square feet. Ans. 2 square feet. 58. As a wheel, 5 feet in diameter, is rolled along by the side of a wall, a nail, bent sideways over the tire, scratches it, and marks out a succession of curves, termed cycloids ; what is the area of each in square yards ? Ans. 6.545 yards. 59. A circular field measures 283 yards in diameter; how many acres does it contain ? Ans. 13 acres nearly. 60. A globe is 7 feet in diameter ; what is the extent of its surface in square yards? Ans. 17.1. 61. A piece of land measures 95 links by 74; how many square perches does it contain? Ans. 11.24. 62. How many square yards of canvas will be required to construct a conical tent, 57 feet round the bottom, the slant height of which is to be 22 feet ? Ans. 69.7. 63. Kequired the surface, in square feet, of a pentago- nal prism, the length 168 inches, and each side of the base 33 inches. Ans. 192.5. 64. How many square yards are contained in a para- bola, of which the base is 126, and the height 210 inches? Ans 13.6. The following Table of Divisors will enable the student to solve the preceding questions numerically. The same formulae apply. INSTRUMENTS AND THE SLIDE-RULE. 129 Dimensions in Area in Square. Cii^le. Inches Sq. Inches 1 1.2732 Feet 144 183.3462 Feet Yards Yards 1296 9 1650.1164 11.4591 Rods 272.25 346.639 Links Perches.... 625 795.7737 Yards Perches.... 30.25 38.5154 Acres 4840 6162.4719 Chains Acres 10 12.7323 Solution of Question 58 : 3X5 = 6.545 square yards. 11.4591 The following is not adapted for the slide-rule ; but as it is an excellent method, and requisite to complete the mensuration of surfaces, it is accordingly inserted. To find the areas of plain figures by an odd number of equidistant ordinates. Find the centre of the figure w, and draw the diameters rip, dd. On each side of w set off any equal distances ws, sv, vo, ?rr, rt, tm, as often as may be deemed necessary, and through the points m, t, r, &c., draw the ordinates aa, bb, cc, &c., and measure their lengths ; also the distance nm, or op, which are equal to each other. Place in a line the letters x 4e 2o (Contractions for the words, extreme, four times even, twice odd.) Sot the first ordinate, aa, under x ; the second, bb, un- J30 A TREATISE ON A BOX OP der 4 e ; the third, cc, under 2 o ; the fourth under 4 e ; the fifth under 2 o ; the sixth under 4 e ; and so on, alter- nately, to the last, which set under x. Add up the three columns separately.. Multiply the one under 4 e by 4 ; and the one under 2 o by 2. Add the three together, and multiply by the common distance ws. For the end areas multiply the sum of the extreme ordinates, standing under x, by twice the height nm ; that is, the sum of the bases by the sum of the heights. Add this product to the other, and divide by 3, gives the area. EXAMPLE. 65. In a curvilinear figure, 7 ordinates were taken in the following order, 20, 32, 38, 41, 39, 33, 22; the common distance ws was 8 ; the distance nm 3 : required the area x 4 e 2 o "20 ~32 ~3i 22 41 39 ~42 J 77 _6 106 252 __f 154 424 154 42 620 _8 4960 252 1737i area. . INSTRUMENTS AND THE SLIDE-RULE, 131 The greater the number of ordinates taken, the more correct will be the area found ; and when the curve ana is abrupt, the distance nm should be small. If the curve taper gradually the distance nm may be taken equal to mt, and then the extreme ordinates will be 0. The num- ber of ordinates must always be odd. Beginning with one in the middle insures this. 66. In a curvilinear figure 5 ordinates were taken, 70, 79, 80, 78.6,69; their common distance was 24; the height of each of the end areas 8 : required the area. Ans. 8176.53. 67. In a triangle the ordinates were 0, 2, 4, 6, 8 ; the common distance 3; required the area. Ans. 48. 68. The ordinates in a triangle are 0, 3, 6, 9, 12, 9, 6, 3, 0; the common distance is 4; what is the area? Ans. 192. SOLIDS. THE next part on the Slide Rule is Table 6, which consists of a number of gauge-points for the mensuration of Solids. The content of a prism, or cylinder, is found by multiplying the area of the base by the height. Pyra- mids and cones are J of their circumscribing prisms and cylinders, and therefore their content will be equal to the pro- duct of the area of the base, and of their height. A globe is f of its circumscribing cylinder, and therefore its content is equal to the area of one of its great circles multiplied by f of its diameter. The number of cubic inches in a gallon, 132 A TREATISE ON A BOX OF is 277.274 ; hence, if the dimensions of a square prism are taken in inches, and the content is required in gallons, it will be (putting I the length, and s the side of the prism) Zs 277 274 "^ ^ 277.274 = 16.65, the gauge point for square prisms. Since the pyramid is J of the prism, we may multiply by the whole heiyht, and divide by three Is 9 times 277.274 : that is, the content will be QQI 090 an< * I/ 831.822 = 28.84, the gauge-point for square pyramids. A gallon of water weighs exactly 10 Ibs. ; .-. 1 Ib. occupies 27.7274 cubic inches; dividing this by the specific gravity of any metal, and taking the square root of the quotient, gives the gauge point for such metal. The gauge points for polygonal prisms are obtained by dividing the number of cubic inches in a gallon by the polygonal numbers given at page 118, and taking the square root. In treating of surfaces, it was seen that the area of a circle, inscribed in a square whose side is unity, is .7854. Now, 277.274 -j- .7854 = 353.0353 ; consequently, the dimensions being taken in inches, as before, the content of a cylinder will be I d a (putting I length and (^diameter) ~ H-JKU andy'SSS.OSSS = 18.78, the gauge point for cylinders. In the same way as the pyramid was determined by taking 3 times the prismatic divisor, so the content of the cone will be found by taking 3 times the cylindrical divisor; and 3 X 353.0353 Id* = 1059.106 ; consequently, content = -inrq ing? an< i I/ 1059.106 = 32.54 the conical gauge point. The globe, being $ of the cylinder, will be twice the cone ; hence the divisor will be the half of 1059.106, namely, 529.553; INSTRUMENTS AND THE SLIDE-RULE. 133 therefore, putting d the diameter, the content of the globe dd* will be 590 553 and -j/ 529.553 = 23, the gauge point for globes. By having divisors and gauge points thus prepared, round solids are reduced to square ones, by which means their contents are determined with the greatest ease, as they all come under the general formula Is" -p-j in which I represents the length, s the side or diame- ter, as the case may be, and G the prepared divisor, or, for the purposes of the Slide-rule, its square root, the gauge point. For finding the solidities of frustums the followifig is an invaluable rule, and of general applicability : Find the area of the top, the area of the bottom, and four times the middle area ; their sum is six times a mean area, which, being multiplied by one-sixth of the depth, gives the con- tent. Now, since by the above-mentioned divisors we have reduced round solids to square ones, the rule be- comes : Add together the square of the top, the square of the bottom, and four times the square of the middle, and multiply the sum by one-sixth of the depth. But four times the square of a number is equal to the square of twice that number ; therefore the rule becomes still easier. Add together the square of the top, the square of the bot- tom, and the square of twice tJie middle, and multiply by one-sixth of the depth. Moreover, when solids do not bulge in the middle, like globes and spindles, but taper regularly like cones and pyramids, then the sum of the top and bottom will l>e twice the middle diameter. There- fore, for all regularly tapering frustums the above given rule becomes still more concise, viz. : Add together the 12 134 A TREATISE ON A BOX Of square of the top, the square of the bottom, and the square of their sum, and multiply by one-sixth of the depth. In the same way as for cones, we multiplied by the whole of the height, and took three times the divisor, so for frus- tums we may take the whole of the height, and divide by six times the divisor. Now, 6 times 277.274 = 1663.644, whose square root = 40.78, the gauge point for square frustums. Also, 6 times 353.0358 =2118.2148, whose square root = 46, the gauge point for round frustums. Moreover, as a rule that applies to frustums, applies also to the complete solids themselves, and as this is of such general utility, we shall illustrate it by a few examples. EXAMPLES. 69. A carpenter has a block of wood 11 inches square at top, 13 inches square at bottom, and 12 inches deep : does it contain an exact cubic foot, or more, or less ? Top 11" == 121 Bottom 13 a = 169 Sum 24 9 = 576 866 2 = f of depth. 1732 = 4 cubic inches more than a solid foot. 70. A prismoid, 24 inches deep, measures 12 inches by 10 at top, and 16 by 12 at the bottom; what is the jontent in cubic inches ? INSTRUMENTS AND THE SLIDE-RULE. 135 Top 12 X 10 = 120 Bottom 16 X 12 = 192 Sum 28 X 22 = 616 928 X 4 = 3712 cubic inches. 71. A wedge measures 8 inches along the edge; the base is 12 inches long, and 4 thick, and the perpendicular height 18 inches; what is the solidity? Top 8X0=0 Bottom 12 X 4 = 48 Sum 20 X 4 = 80 128 X 3 = 384 cubic inches. 72. A cylindroid, or solid bounded at one end by a circle 6 inches diameter, and at the other by an ellipse whose axes are 12 and 10, is 24 inches deep; how many gallons will it contain ? Top 6 X 6 = 36 Bottom 12 X 10 = 120 Sum 18 X 16 = 288 444X4=1776; aod = 5 ' 03 alloPS - 73. What is the solidity of a globe whose diameter is 1 ? See diagram page 48, and suppose E to be halfway between A and F, and then the diameter being 1, A C will be 1, and AE = i 136 A TREATISE ON A BOX OP .. twice EC = ~ v/3 the middle diameter; then Square of top = Square of bottom = 1 Square of twice middle = 3 4X ^+.7854 = .5236, the content as determined by other modes. To return to the slide. Putting I or Ti, length, height, Dr depth, according as the solid is considered lying or standing ; d and D } less and greater diameter ; m, middle diameter, taken halfway between them ; r and R, less and greater radius ; s and S, less and greater side ; g, square root of product, or mean proportional between two dimen- sions; f } fixed axis; and v, revolving axis; then the capa- cities of solids will be denoted by the following formulae : 1. Prism = prism ; 2. Pyramid = 5. Sphere = globe fv* 6. Spheroid = -4-r globe INSTRUMENTS AND THE SLIDE-RULE. 137 7_9 7. Rectangular Prism = 8 Rectangular Pyramid = square pnsrn square pyramid 2AJ 8 9. Parabolic Prism = rj- square pyramid 10. Elliptic Cylinder = .ft*.. . cylinder 11. Elliptic Cone = - cone 17 T^i 12. Paraboloid, or Parabolic Conoid = ,. , , cylinder or -^-5 cylinder 13. Hyperboloid, or Hyperbolic Conoid h(D + 2m|) ~~ round firustum' globe ?/> 14. Parabolic Spindle = ^^ X -8 15. Spindles in general = or globe l(^+ ^ 9 +*+ S\ 16. Fmstum of Pyramid - ^ frustum I (d a + D" + d + Z>|) 17. Frustum of Cone = - round frustum 12* 138 A TREATISE ON A BOX OP 18. Frustum of Paraboloid =- , cylinder or cylinder 19. Frustum of Hyperboloid = ^^ round frustum globe 20. Middle Frustum of Parabolic Spindle _ Z( 9 ^ (of 2 diff.) 8 ) cone 21. Middle Frustum of Spindles in General _ 7(

we have \Ji for h or I. The parabolic spindle being T 8 5 of the cylinder, and the cylinder | of the globe, multiply- ing these together we have .8. The difference between an oblate and prolate spheroid will be best understood by considering the revolutions of a parallelogram. Suppose a parallelogram 12 inches by 6 to be divided by two lines across the middle, at right angles to each other, so as to cut it into 4 equal portions, each 6 by 3. Then, if the parallelogram revulve on the short axis, it will generate a cylinder 6 inches deep, and having a diameter of 12 inches; consequently, its content will be 6 X 12 s X -7854. If it revolve on the long axis the cylinder produced will be 12 inches deep, and 6 diameter, and its content 12 X 6 a X -7854 ; that is, in each casefv* .7854.* Two- * In short, all the formulae for round solids are but modifica- /w 2 tions of the general expression ~- ; and even angular solids 140 A TREATISE ON A BOX OF /* thirds of this, or fv* .5236, gives the spheroid = - The earth is a spheroid slightly oblate, the polar diameter, as determined by careful measurements of a degree at different parts of its surface, being about 26 miles less than the equatorial, the prominence of the torrid zone having, it is presumed, been acquired, at the commence- ment, from the operation of centrifugal force : it being supposed that the earth was formed from matter in a semi- fluid state, and set rotating on its axis before the parts had been allowed time to consolidate. ILLUSTRATION OF FORMULA. 74. Formula 1. What is the content, in gallons, of a vessel in the shape of a square prism, 1 inch deep, and 29 inches along each of the sides ? Referring to the back of the rule, 16.65 will be found the gauge point for square prisms ; therefore over 16.65 of D set 1 ; over 29 is 3.03 gallons, the content.* 75. Formula 2. Each side of aii hexagonal pyramid is 46 inches, its perpendicular depth 90 inches; what is the content in gallons? Over 17.9 set 90; over 46 is 594.8 gallons. 76. Formula 3. The depth of a cylinder is 40 inches, may come under the same form, if we conceive them to be described by the rotation of planes, and the generated surfaces subsequently shaped into polygons by lateral compression. * Finding the content of solids whose depth or thickness is unity is generally termed " gauging areas," because, in such cases, the surface and solidity are both represented by the same Dumber, Is 2 and 2 being equivalent when I becomes 1. INSTRUMENTS AND THE SLIDE-RULE. 141 its diameter 21.5; how many gallons will it contain? Over 18.78 set 40; over 21.5 is 52.37 gallons. 77. Formula 4. The depth of a cone is 24 inches, its diameter 17 ; how many Ibs. of tallow will it hold ? Over 10.75 set 24; over 17 is 60 Ibs. 78. Formula 5. "What is the weight of a globe of brass 8 inches in diameter? Over 2.51 set 8; over 8 is 81.2 Ibs. 79. Formula 6. The fixed, or transverse axis, of a prolate spheroid is 54 inches, its conjugate 33 ; how many bushels will it contain? Over 65.08 set 54; over 33 is 13.88 bushels. 80. The fixed, or conjugate diameter of an oblate spheroid is 33 inches, its transverse 54 ; how many bushels will it contain? Over 65.08 set 33; over 54 is 22.7 bushels. 81 . Formula 7. A cistern in the shape of a rectangulai prism, or parallelepiped, is 82 inches long, 54 broad, and 37.5 deep ; how many gallons will it contain ? To 54 on D set 54 of the slide, then under 37.5 is 45, a mean pro- portional. Over 16.65 set 82; over 45 is 598 J gallons. 82. Formula 8. A vessel oblong at top, and tapering downward to a point, measures 48 inches by 75 ; its depth is 63 inches ; how many Ibs. of hot hard soap will it hold? Over 48 of D set 48, then under 75 is 60, a mean proportional. Over 9.16 set 63; over 60 is 2700 Ibs. 83. Formula 9. A prismatic vessel, 10 inches deep, whose ends are in the shape of a parabola, measures 80 142 A TREATISE ON A BOX OP inches along the straight side or double ordinate, from the middle of which to the vertex is 60 inches ; how many gallons will it contain? Over 60 set 60, under 80 ig 69.28, a mean. Over 28.84 set 20 (twice depth;) over 69.28 is 115.4 gallons. 84. Formula 10. The axes of an elliptic cylinder are 67 and 52, its depth 50 inches ; how many bushels will it contain ? Over 52 set 52 ; under 67 is 59, a mean- Over 53.14 set 50; over 59 is 61.6 bushels. 85. Formula 11. The axes of an inverted elliptic cone are 16 and 9, the depth 19 inches ; how many pints will it hold ? Over 16 set 16 ; under 9 is 12, a mean. Over 11.5 set 19 ; over 12 is 20.6 pints. 86. Formula 12. A vessel in the shape of a parabolic conoid is 42 inches deep, and the diameter of the top is 24 inches; what is the content in gallons ? Over 18.78 set 21 (half 42;) over 24 is 34.25 gallons: or by the second, over 18.78 set 84 (twice 42); then over 12 is 34.25, as before. 87. Formula 13. "What is the content, in gallons, of a hyperbolic conoid, the diameter at top being 52 inches, the diameter in the middle 34, and the depth 25 inches ? Over 46 set 25, then- Over 52 is 31.9 Over 68 is 54.6 86.5 gallons. 88. Formula 14. What is the content, in gallons, of a parabolic spindle, the diameter of which is 28 inches, INSTRUMENTS AND THE SLIDE-RULE. 143 and length 70 inches? Over 23 set 70; over 28 is 103.7. Then to 103.7 on A set commencement of slide, and over .8 is 82.96 gallons. 89. Formula 15. The length cf a spindle is 20 inches, the greatest diameter 6 inches, and the diameter halfway between it and the point 4.74 inches ; what is the content in cubic inches ? Over 2. 76 set 20, then Over 6. is 94. . Over 9.48 is 235.5 329.5 cubic inches. 90. Formula 16. How many gallons will be contained in the frustum of an octagonal pyramid, each side of the greater base being 17.5 inches, of the less 14 inches, and the perpendicular depth 47 inches? Over 18.55 set 47, then Over 14. is 26.8 Over 17.5 is 41.8 Over 31.5 is 135.2 203.8 gallons. 91. How many Ibs. of hot hard soap will the above contain ? As polygonal pyramids are not figures of frequent oc- currence, it was not deemed necessary to insert gauge points for any other quantities than gallons and cubic feet, the weight therefore must be determined by a second pro- cess, which, since a gallon of water weighs 10 Ibs., is ef- fected by multiplying the content in gallons by 10 times 144 A TREATISE ON A BOX Of the specific gravity. Now the specific gravity of hot hard soap is shown on the rule to be .99, ten times which :=9.9. Therefore, to 203.8 on A set commencement of slide, then over 9.9 is 2017 Ibs. 92. Formula 17. The frustum of a cone is 43 inches deep, the diameter at one end 36, at the other 20 inches how many bushels will it contain ? Over 130.17 set 43 ; then Over 20 is 1.02 Over 36 is 3.28 Over 56 is 7.96 12.26 bushels.* 93. Formula 18. The diameters of the frustum of a paraboloid are 30 and 40 inches, the depth 1 8 inches ; how many gallons will it contain ? Over 18.78 set 9 (half 18,) then- Over 30 is 23. Over 40 is 40.7 63.7 gallons. f 94. Formula 19. How many bushels will be contained in the frustum of a hyperbolic conoid, the top and bottom * If the frustums of two equal cones be joined together at their greater ends they form a figure called by gaugers a cask of the 4th variety. flf the frustums of two equal paraboloids be joined together at their greater ends they form a figure called by gaugers a cask of the 3d variety. INSTRUMENTS AND THE SLIDE-RULE. 145 diameters of which are 23 and 40 inches, the middle 36 inches, and depth 20 inches? Over 130.17 set 20, then Over 23 is .62 Over 40 is 1.88 Over 72 is 6.13 8.63 bushels. 95. Formula 20. The length of a vessel in the form of the middle frustum of a parabolic spindle is 20, the greatest diameter 16, and least 12 inches; what is the content in gallons ? Here twice the difference of the diameters = 8 ; therefore, over 32.54 set 20, then Over 12 is 2.72 Over 16 is 4.83 4.83 12.38 Over 8 is 1.2, one-tenth of which is .12 12.26 gallons.* 96. Formula 21. The bung diameter of a vessel is 36 inches, the head 30, twice the diameter taken midway between them 67.8 inches, and the length 40 inches ; how many gallons will it contain ? Over 46 set 40, then Over 30 is 17. Over 36 is 24.5 Over 67.8 is 86.9 128.4 gallons. * A cask in the form of the middle frustum of a parabolic spindle is termed by gangers a cask of the 2d variety. 13 146 A TREATISE ON A BOX OP 97. Formula 22. A vessel in the form of the middle frustum of a prolate spheroid is 40 inches long, the bung diameter is 36, and the head 27 inches ; what is the con- tent in gallons ? Over 32.54 set 40, then Over 27 is 27.4 Over 36 is 49. 49. 125.4 gallons.* 98. Formula 23. How many cubic feet are contained in the middle zone of a sphere, the axis of which is 44 inches, and the height of the zone 14 inches? Over 46.9 set 14, then Over 44 is 12.32 Over 14 is 1.26, one-third of which = .42 11.9 bushels. 99. Formula 24. What is the content in gallons of the shoulder of a still in the form of the frustum of a sphere, the top and bottom diameters being 42 and 36 inches, and the height 30 inches? Over 18.78 set 15 (half 30,) then Over 36 is 55 Over 42 is 75 Over 30 is 38.3 + * = 12.7 181. gallons. * A cask in the form of the middle frustum of a prolate sphe- roid is termed by gaugers a cask of the 1st variety INSTRUMENTS AND THE SLIDE-RULE. 147 100. Formula 25. A copper basin in the form of the segment of a sphere is 18 inches deep, the diameter across the top 40 inches ; how many gallons will it contain f Over 23 set 18, then Over 18 is 11 Over 20 is 13.6 + twice ditto = 27.2 51.8 gallons. The content of cylindroids, prismoids, and wedges, is found by taking the mean proportionals of the products of the top and bottom dimensions, and of the product of their sums, making use of the round frustum gauge points for the cylindroid, and the square frustum for the prismoid and wedge. ILLUSTRATION. 101. The perpendicular depth of a cylindroid is 52 inches, the diameters at top 60 and 46, at bottom 42 inches ; what is the content in bushels ? Bottom 42 X 42, mean proportional between which =42 Top 60x46 " " =52.54 Sum 102X88 " *=94.74 Over 130.17 set 52, then over 42 = 5.4 52.54= 8.5 94.74 = 27.5 41.4 bushels. Referring to the Table on page 150, the round frustum divisor for bushels is 16945.74. 148 A TREATISE ON A BOX OP The numerical solution of this question, therefore, will be as follows : Bottom 42 X 42 = 1764 Top 60 X 46 = 2760 Sum 102 X 88 =8976 13500 52 27000 67500 16945.74 ) 702000.00 ( 41.426 bushels. 6778296 2417040 1694574 7224660 6778296 4463640 3389148 577476 For questions 102 and 103 the divisor, as shown on page 150, will be 13309.15. 102. The length and breadth of a coal wagon at top are 81 and 55 inches, at bottom 41 and 29 inches; the depth is 47 inches ; how many bushels will it contain ? Top 81 X55, mean proportional between which = 66.8 Bottom 41x29 " " = 35.5 Sum 122X84 " =101.2* * It must be observed that, in irregular solids, the menn pro- portional of the sum is not the sum of the mean proportionals ; INSTRUMENTS AND THE SLIDE-RULE. 149 Over 115.3 set 47, then over 66.8 = 15.6 35.5= 4.4 101.2 = 36.1 56.1 bushels. 103. A heap of malt is piled into the form of a wedge, 24 inches deep, the base is 40 inches long, and 20 broad, the edge 20 inches long; how many bushels does it contain ? Top 20 X Bottom 30x20, mean proportional between which = 28. 28 Sum 60X20, " " =34.37 Over 115.3 set 24, then over 28.28 = 1.46 34.36=2.14 3.6 bushels. the former mast be taken, not the latter. In prismoids, also, attention must be paid to the position of the sides ; for the top and bottom areas of two prismoids may be the same, and yet their middle area, and consequently their content, different. For, suppose a prismoid to be 12 inches by 10 at top, and 9 inches by 6 at bottom, the 12 falling over the 9, and the 10 over the 6 : if now we shift the position of the top parallelogram so as to bring the short side over the long one of the bottom, then the figure becomes distorted, and the content altogether altered. In the first it will be In the second Bottom 12 X 10 = 120 Bottom 12 x 10 = 120 Top 9 X 6 = 54 Top 6 X 9 = 54 Sum 21 X 16 = 336 Sum 18 X 19 = 342 510 XL 516 X fa o 13* 150 A TREATISE ON A BOX OF The following Table of Divisors will enable the Student to solve the preceding Questions numerically. The same formulae apply. DIMENSIONS IN INCHES. V SQUARE SOLIDS. ROUND SOLIDS. Content in & Prism. Pyramid. Frustum. Cylinder. Globe. Cone. Frustum. Cubic Inches . 1. 3. 6. 1.2732 1.90985 3.8197 7.6394 Cubic Feet . 1728. 5184. 10368. 2200.16 3301 .24 6600.4 8 13200.96 Pints . 34.659 103.978 207.956 44.129 66. 194 132.38 3 264.776 Gallons . 277.274 831.82 1663.64 353.03(5 529. 554 1059.1 )8 2118.216 Bushels . 2218.19 6654.57 13309.15 2824.29 4236 435 8472.8 7 16945.74 Hot hard \ Soap, Ibs. ' J .99 28. 84. 168. 35.65 53.475 106.95 213.9 Cold ditto . 1.02 27.14 81.42 162.84 3455 51.8 32 103.66 4 207.327 Tallow . .915 30.28 90.84 181.68 3855 57. 83 115.65 231.3 Flint Glass 3.21 8.64 25.92 51.84 11 16. 5 33. 66. Plate ditto . 2.418 11.3 33.9 67.8 14.4 21. 6 43.2 86.4 Platinum . 21.45 1.29 3.87 7.74 1.642 2.4 63 4.927 9.854 Gold . 19.25 1.44 4.32 8.64 18 2.7 5 5.5 11. Mercury . 13.61 2.03 6.09 12.18 2.584 3.8 77 7-753 15.5 Lead 11.35 2.44 7.32 14.64 3.11 4.6 7 9.34 18.68 Silver 10.53 2.64 7.92 15.84 3.36 5.0 4 10.08 20.16 Copper . 8.81 3.155 9.46 18.92 4. 6. 12. 24. Brass . 8.41 3.3 9.91 19.82 4.2 6.3 12.6 25.2 Wt. Iron & Steel 7.82 3.54 10.62 21.24 4.5 6.7 5 13.5 27. Ct. Iron, Tin, 1 & Zinc . J 7.24 3.8 11.41 22.82 4.8 7.2 14.44 28.8 Tee & Gunpowd. .93 29.81 89.43 179.86 37.94 56.91 113.82 227.64 POLYGONAL SOLIDS. CONTENT IN GALLONS. CONTENT IN CUBIC FEET. DIMENSIONS IN INCHES. Priam. Pyramid. Frustum. Prism. Pyramid. Frustum. Trigonal 640.34 1921.01 3842.02 3090.64 11971.93 23943.87 Tetragonal 277.274 831.82 1683.64 1728. 5184. 10368. Pentagonal . 161.161 483.48 966.96 1004.37 3013.12 6026.23 Hexagonal . . 106.72 320.16 640.32 665.11 1995.32 3990.64 Heptagonal . 76.39 229.17 458.34 475.52 1426.56 2853.12 Octagonal . 57.42 172.27 344.54 357.88 1073.64 2147.28 Nonagonal . 44.85 134.56 269.12 279.53 838.58 1677.17 Decagonal . 36.03 108.11 216.22 224.58 67375 1347.50 Undecagonal . 29.6 88.81 177.63 184.5 553.51 1107.02 Dodecagonal 24.76 74.30 148.6 154.34 463.02 926.03 8* Numerical solution of Question 78. =81.2 Ibs. 6.3 Numerical solution of Question 90. Top 14.8 196 Bottom 17.52 = 306.25 Sum 31.52 _ 992.25 70241.5 1494.5 X 47 70241.5 ; and = 344.54 203.8 gallons. INSTRUMENTS AND THE SLIDE-RULE. 151 EXAMPLES FOR PRACTICE. 104. What is the weight of a prism of steel 7 inches square, 15 inches long? Ans. 20T Ibs. 105. What would be the weight of a pyramid of ice, 8 inches square at bottom, and 13.8 inches high ? Ans. 9.87 Ibs. 106. A cylindrical glass-pot, 24 inches diameter, is charged with flint glass to the depth of 15 inches : what is its weight ? Ans. 785 Ibs. 107. An inverted cone is 23 inches deep, its diameter at top 10 inches : what quantity of tallow will it contain ? Ans. 19.8 Ibs. 108. What quantity of gunpowder, shaken down, will fill a shell whose internal diameter is 9 inches? Ans. 12.8 Ibs. 109. The axes of an oblong or prolate spheroid are 6 and 8 inches : what quantity of mercury will it contain ? Ans. 74.2 Ibs. 110. The axes of an oblate spheroid are 6 and 8 inches : what quantity of mercury will it contain ? : Ans. 99 Ibs. 111. What is the weight of a rectangular block of ice, 12 inches by 10 thick, and 30 inches long ? Ans. 121 Ibs. 112. The top of an inverted rectangular pyramid mea- sures 17 inches by 13 ; its depth is 44 inches : how many gallons of water will it contain, and how many Ibs. ? Ans. 11.69 gallons, 116.9 Ibs. 152 A TREATISE ON A BOX OP 113. The base of a parabola is 32 inches, its absciss 24 inches ; the depth b 1 inch : how many gallons will it hold ? Ans. 1.84 gallons. 114. The diameters of an elliptic cylinder are 25 and 20 inches, the depth 13 inches : how many gallons will it contain? Ans. 18.4 gallons. 115. An elliptic cone of silver is 10 inches high, the diameters at bottom 5 inches by 4 : what is its weight in Ibs. avoirdupois? Ans. 19.8 Ibs. 116. A paraboloid of copper is 12 inches high, the diameter of the base 8 inches : what is its weight ? Ans. 96 Ibs. 117. A vessel in the shape of an hyperboloid is 25 inches deep, the radius of the top 26, and the middle diameter 34 inches : what quantity of cold hard soap will it hold? Ans. 883 Ibs. 118. The length of a parabolic spindle is 82 inches, its diameter 10 inches : required the content in gallons Ans. 4.83 gallons. 119. The length of a cast-iron spindle is 20 inches, its greatest diameter 9 inches, and the diameter halfway be- tween that and the point 6 inches : what is its weight ? Ans. 155.8 Ibs. 120. The frustum of a nonagonal pyramid, 25 inches deep, measures 9 inches along each side at top, and 12 at bottom: how many gallons will it contain? Ans. 61.8 gallons. 121. Suppose a cask to consist of two equal frustums of a cone, the length of which is 40 inches, the bung dia- INSTRUMENTS AND THE SLIDE-RULE. 153 meter 32, and the head 24 : what is the content in gallons? By formula 17. Ans. 89.4 gallons, 4th variety. 122. Suppose a cask, of the same dimensions, to be composed of two equal frustums of a paraboloid : re- quired the content. By formula 18. Ans. 90.6 gallons, 3d variety. 123. Let the cask be the middle frustum of a para- bolic spindle, and the dimensions remain the same : what is the content? By formula 20. Ans. 98.2 gallons, 2d variety. 124. Let the cask be the middle frustum of a prolate spheroid, the dimensions continuing the same : what is the content? By formula 22. Ans. 99.1 gallons, 1st variety. 125. What will be the content of the middle frustum of a spindle having the same dimensions, and also the diameter halfway between the head and bung 29.6 inches? By formula 21. Ans. 96.45 gallons, true con- tent. 126. Required the content in cubic feet of the middle frustum of a sphere, the height of which is 24, and the least diameter 18 inches. Ans. 7.72 feet. 127. Find the content in gallons of the frustum of a sphere, the height of which is 9 inches, and the radii at its ends 14 and 10 inches. Ans. 16.47 gallons. 128. What is the weight of the segment of a globe of leal, the height of which is 6 inches, and the radius of the base 8 inches ? Ans. 293 Ibs. 129. The depth of a cylindroid is 50 inches, the diame- 154 A TREATISE ON A BOX OP ters of the elliptic base are 60 and 44 inches, the diameter of the circular top 40 inches : required the content in gallons. Ans. 298.3 gallons. 130. The depth of a prismoid is 50 inches ; the base is a parallelogram 60 inches long, 44 broad ; the top is a square, the sides of which are each 40 inches : what is the content in gallons? Ans. 379.8 gallons. 131. The frustum of a square pyramid is 30 inches deep, each of the sides at bottom 36, and at top 25 inches : what is the content of each of the wedges into which a diagonal plane, passing through its extremities, divides it? Ans. 62.97 gallons, lower hoof or wedge; 38.77 ditto, upper ditto. TABLES VII. and VIII., at thi back of the rule, are adapted for the use of the E slide. Table VII. exhibits the weight of metallic spheres, commencing thus : D diameter. E weight. Platinum 4 inches = 26 Ibs. avoirdupois. Gk)ld 6* inches = 100 Ibs. That is, over 4 on D, set 26 Ibs. on E, then the numbers on D will be a series of diameters, and the numbers over them on E their corresponding weights. EXAMPLES. 132. A sphere of platinum weighs 51 Ibs. : what is its diameter? Over 4 set 26 Ibs.; under 51 Ibs. is 5 inches. 133. A sphere of silver weighs 7 Ibs.: what is its diameter? Ans. 3.284 inches. INSTRUMENTS AND THE SLIDE-RULE. 155 134. Rockets receive their names from a comparison of the external diameters of their cases with leaden balls : what, then, is the diameter of a 5-pound rocket ? Ana. 2.86 inches. 135. A globe of wrought iron weighs 19.7 Ibs. : what is its diameter? Ans. 5.1 inches. 136. A spherical vessel, filled with mercury, holds 258 Ibs. : what is its diameter ? Ans. 10 inches. 137. Thirteen Ibs. of gunpowder fill a shell: what is its diameter? Ans. 9.04 inches. 138. A sphere of brass weighs 81.2 Ibs. : what is its diameter ? Ans. 8 inches. Table VIII. is used precisely like Table VII., and is for finding the diameters and circumferences of spheres from their solidities; and also the solidities of regular bodies, the tetrahedron, &c. EXAMPLES. 139. The solidity of a sphere is 33.6 : what is its diameter ? Over 4.6 of D set 51 of E; under 33.6 is 4. 140. A globe contains 98.5 solid feet : what is its cir- cumference? Ans. 18 feet. 141. The side of a tetrahedron measures 2.2 inches: how many cubic inches does it contain? Ans. 1.25 cubic inches. 142. The side of an octahedron measures 3.3 inches : how many cubic inches does it contain? Ans. 16.875 solid inches. 156 A TREATISE ON A BOX OF 143. A dodecahedron contains 15 cubic feet: what is the length of each of its sides ? Ans. 1.25 feet. 144. The solidity of an icosahedron is 162 : what is the length of each of its sides ? Ans. 4.2. SOLAR SYSTEM. THE concluding part of the tabular work on the rule is for the use of the line A in conjunction with that of E. According to Kepler's famous discovery, the squares of the periodic times of the planets are proportional to the cubes of their mean distances. Now, since the line A is laid down twice, and the line E thrice, in the same spa.ce, when the slide E is laid evenly in, the cubes of the num- bers on A will be equal to the squares of the numbers on E ; when in any other position, the cubes of the numbers on A will be proportional to the squares of the numbers on E. Hence, if under 95 millions of miles on A we set 365 days, or 52 weeks, or 13 lunar months, or 1 year, on E ; then the numbers on A will be a series of planetary distances, and the numbers beneath them on E their periods of revolution, in days, weeks, months, or years, according as 365, 52, 13, or 1, is selected. EXAMPLES. 145. The distance of Mercury from the sun is 37 mil- lions of miles; what is the length of his year? Under 95 set 365 ; under 37 is 88 days. 146. Mars is about 687 days in revolving round the sun ; what is his distance ? Ans. 144 millions. 147. Herschel's mean distance is about 1823 millions of miles ; how many years does he consume in traversing his orbit? Ans. 83.8 years. INSTRUMENTS AND THE SLIDE-RULE. 157 148. If a planet revolved in an orbit 20 million miles from the sun ; how long would it take in passing round him? Ans. 35 J days. 149. Suppose the recently discovered planet to be 3,000 millions of miles distant from the sun ; how many years does it take in traversing its orbit ? Ans. 178 years. 150. The nearest of Saturn's moons is 108 thousand miles distant from him, and the time of its periodic revo- lution about 22 J hours; the second is distant 140 thou- sand : what is its periodic revolution ? Under 108 of A set 22 J of E; under 140 is 34 hours nearly. 151. The fourth satellite of Saturn spends 65 hours in passing round its primary; required its distance from ' him. Ans. 217,000 miles. The following table will enable the student to solve the previous questions numerically : WEIGHT OF METALLIC SPHERES. Platinum 4 inches diam. = 26 Ibs. av. Gold 6.5 ... =100 Mercury 3 ... = 7 Lead 7.5 ... = 90 Silver 4.6 ... = 18 Copper 6 ... = 36 Brass 5.4 ... = 25 Wt Iron and Steel 3 ... = 4 Ct. Iron, Tin, and Zinc 6 ... = 30 Ice and Gunpowder ....7 ... = 6 Solidity of Sphere = .5236 d* = .01688 e. Solidity of Tetrahedron = .11785 s 1 . " Octahedron = .47140 a 1 . " Dodecahedron = 7.66312 . ' Icosahedron = 2.18169 * 158 A TREATISE ON A BOX OF Numerical Solution of Question 133. lb. 3 lb. 18 : 4.5 : : 7 : d 3 ; or taking the j 126.06 126.86 46 41.5 30. 37. 69.6 130LM 136.72 138.75 47 51. 23.7 31.9 59.4 UBJfl 117.73 121.72 48 48. 25.2 33.4 61.8 126^2 122.78 126.4 49 56.4 32.3 41.2 76.4 Muaa 225.21 230.08 50 45.3 28.2 34.9 66.2 134.12 13255 134.52 Total 7415.12 7312.05 7423.04 168 A TREATISE ON A BOX OV In 47 out of the 50, the rule I have proposed agree? nearest with the truth; in the three marked with an asterisk, Dr. Hutton's comes nearer. The total error by his is 103 gallons ; by mine, 8 gallons. In setting down the dimensions, the most concise way will be to place the length on the left hand, with a brace between it and the diameters, recollecting that 33 is the gauge point, when three dimensions are used; and 46, when four dimen- sions are taken. EXAMPLE. 177. The length of a cask is 40 inches, the head 30, the bung 36, and twice the middle 67.8 inches : required the content. By Proposed Eule for 3 dimensions. 30=33.05 40 { 36 =47.6 47.7* 128.35 gallons. By General Rule for 4 dimensions. f30 =17. 40^36 =24.5 (67.8 = 86.9 128.4 gallons. The dimensions of casks are taken most readily with the long and cross callipers, and the bung and head rods. * If, as in this case, the number found standing over the bung diameter appear to be more than 47.6, and less than 47.7, set down both, as above. INSTRUMENTS AND THE SLIDE-RULE. 169 When these are not at hand, their place may be supplied as follows : Procure a straight piece of deal, about | of an inch square, and 6 feet long, for a measuring-rod ; and, with a camel's-hair pencil and Indian ink, divide it into inches and tenths. Take another piece, AB, an inch square, and about 4 feet long ; and near one end, as at a, cut a notch, and 2 inches from it, b, make a mark, and place a cipher 0. Then divide the distance from b to the end B into inches and tenths. Also procure two pieces of string, each with loops at one end, and heavy plummets of lead at the other. Before tying the loop, ou one of the strings slip 3 pieces of cork, e, v, -z, about i of an inch thick, and } of an inch square. Then, To take the Dimensions of a standing Cask. With a piece of string and chalk, by problem 2, page 20, strike a line across the middle of the head of the cask ; lay the rod AB over this line, and bring the plummet depending from a up to the bulge of the cask. Then slip the other plummet along to c, till it touches the cask in like manner. The number now cut by c will be the intemal bung diameter C, the distance ab, of 2 inches, being an allowance for twice the thickness of the staves. With the measuring rod take the distance from y (the under side of the rod AB) to the ground f. Also the distance from o (the upper side of the rod) to n } the head of the barrel. Then yf minus twice on, will be the internal length of the cask, the thickness of the square rod, AB, being supposed 15 170 A TREATISE ON A BOX OF equal to the thickness of the head of the cask, which is generally 1 inch. To take the middle diameter D, slip the top cork up to e, till the distance ye is equal to on. The length of the cask being known, slip the second cork down to v, the distance ev being J of the length; in the same manner adjust the cork z, if deemed necessary. Then add together the distances vv, zz, and subtract their sum from the bung diameter, or deduct twice the dis- tance fl.w, if the curve of the cask be uniform; the remainder will be the middle diameter, D. In the same way might a diameter be taken halfway between D and C. The oblique line sx, measured from the inside of the chimb to the outermost sloped edge of the opposite stave, will be the internal head diameter ; or twice the distance at e may be deducted from the bung. If only three dimensions are taken, the corks may be dispensed with ; but in ullaging standing casks, they will be found ex- tremely convenient. For taking the dimensions of lying casks, a common pair of callipers may be made by any carpenter, as annexed. kbc, efq, are precisely like a carpenter's square. The arms Ik, ef, may be an inch .3 square, and 2 feet 6 inches long ; the blades be, fq, about f of an inch thick, and an inch broad. At c and q two pieces are fixed at right angles, the distance cd being 4 inches. In the face of the arm We, let a groove be ploughed and worked under with a side tool, to a dove-tailed shape, like the section shown at m. The under side of the arm ef is to be cut to match it like the section shown at n. The arm cf will now slide INSTRUMENTS AND THE SLIDE-RULE. 171 along the arm Nc ; in fact, it would be preferable if it were cut like a slide-rule, but carpenters have not tools for effecting this. One inch from a (which is opposite to d) make a mark, and place a cipher 0. Then from to k will be 25 inches ; divide this into inches and tenths, and number it from towards k. On the arm ef at the point opposite to h make a mark, and 1 inch from it toward e place 25 ; then divide the space from 25 to e into inches and tenths, and number them backward. When this arm is made to slide in the other, and drawn out to measure the length or bung diameter, the number standing oppo- site the end k will denote such length or bung diameter. To find the content of a large circular vessel, that ap- pears to bulge irregularly, by an odd number of equidis- tant diameters. 178. Let the vessel be the cask on page 169, and let there be taken 9 diameters, commencing with the head, level with 3 -f c 3 ; c 9 -f 4d a -f- e 3 ; e 3 -f 4/ -f- g a , &c. j that is, a 8 -f- 4Z* 9 -f 2d> -f- 4d* + 2^ + 4/ a -\-g a ', namely, the square of the extremes, plus 4 times the square of the even diameters, plus twice the square of the remaining odd diameters. By the slide-rule the con- tent may be found by taking it as three successive frus- tums. The same rule obviously applies to the ullaging of a standing cask. EXAMPLE. 179. The depth of liquor, in a cask partly filled, is 20 inches ; five oquidistant diameters, measured from the sur- face downward, are 28, 27, 26, 24, and 22 inches : rt quired the content. x* 4e 2o* 784 ^729 676 484 576 2 1268 1305 1352 4 "5220 1268 1352 7840 X 5 -~- 1059.108 = 37 gallons. For ullaging a lying cask, the following role may be employed. From 10 times the wet inches, subtract the bnng ; mul- tiply the remainder by the content, and divide by 8 times the bang; the quotient gives the liquor in the cask ; i. e. (IQW-B)C SB To find the content of vessels whose bases are nearly of an elliptical form, proceed for the area of the base as di- rected on pa^e 1*29, and (after multiplying by the com- mon distance of the ordinates,) instead of dividing by 3, multiply by the depth of the vessel, and divide by the pyramid divisors, these being equal to 3 times the prism divisors. If the vessel also bulge up the sides, take an odd number of equidistant areas, and proceed as in the last example. And thus may any solid be measured ; always observing, that when equidistant area* are taken, the^ro- i/ 93 A Set up poles at A, B, C, and D, so that, standing at A, you can see B and D, the end of the sides whose meeting forms the angle : and so of the others. And suppose the hedge to run on straight, or nearly so, from A to a, then to bend and run on straight to c, and so on. Form a field- book, as on the right of the diagram, by ruling two lines down the middle of a page; and, in using this, begin at the bottom and write upward, placing the main lines in the middle, and the offsets right or left, as they are on the right or left of the line measured. Then, suppose you commence surveying at A ; let your attendant lead the chain toward B ; and when he gets it extended, see that he is in a straight line between yourself and B, directing 178 A TREATISE ON A BOX OF him by a wave of the hand, right or left, according as you wish him to move to one side, or the other. His position being correct, he is to place an arrow in the ground, and walk on till the chain is again extended, when he places another arrow, while you take up the first; and so proceed. But when you arrive at 5, opposite to a, measure ab, (at right angles to AB,) with an offset staff, which may be a thin piece of deal 10 links long. Now, suppose from A to b, 92 links, and ai32 links; place the 92 in the middle column over A, and 32 on the left hand of it ; and so pro- ceed till you arrive at the end of B, which suppose 384 ; set this down in the field-book, and over it place the letter B, and above this draw a line across the page. The A being placed at the bottom, and B at the top, shows that the intervening numbers are the measures of the AB line. Proceed with the rest in like manner, making the circuit of the field, and returning to A. Then from A, measure the diagonal AC. With these dimensions plot the field from a scale of equal parts, (feather-edged plotting scales are best for this purpose,) and drop the perpendiculars Dm, B, and from the scale ascertain their lengths. These are set down underneath the diagram, as they are not sup- posed to have been .measured in the field; but if a cross staff, or theodolite, be employed, they are to be taken while proceeding along the diagonal, and set down in the field-book, like offsets,- and then the sides AD, DC will not require measuring, supposing there are no offsets on them. To find the area. Add together Dm, B?*, and multiply their sum by AC. For the offsets ; for the tri- angle A ab, multiply Ab by ba. For the succeeding trapezoid, add together ab, cd, and multiply by bd ; and go proceed. Then, as, in all these cases, this gives double INSTRUMENTS AND THE SLIDE-RULE. 179 the area, add the whole together, halve the sum, and di- vide by the number of links in an acre, viz. 100,000, (that is, point off the 5 right-hand figures,) and the content is in acres and decimal parts; the latter of which being multiplied by 4 and 40, (which need not be set down,) gives the roods and poles. In taking the distances from the field-book, the numbers up the middle column, 92, 208, &c., have to be subtracted from each succeeding, and the offsets, 0, 32, 36, &c., to be added together in pairs. The work will stand thus : 180 92 116 176 136 210684 208 32 68 36 33 2944 7888 388 194 928 1056 408 6336 543 276 696 528 408 4488 1164 2944 7888 .6336 4488 2)232340 1652 1940 1.16170 210684 25.8720 Content, 1 acre 25 poles. 180 A TREATISE ON A BOX Off TRIGONOMETRY AND NAVIGATION. THE trigonometrical slide is a slide containing the loga- rithmic sines and tangents, the former of which work to the line D, and the latter to the line A, which lines, as before explained, are also logarithmic. But it is to be recollected, that it is the distances only that are loga- rithmic, not the numbers ; hence, when the slide is laid evenly in, then the numbers on A are the natural tan- gents, and the numbers on D the natural sines of the degrees marked on the slide : when in any other position they are proportional to the natural sines and tangents of those degrees; and, therefore, if we set the first term of a proportion over or under the second, then the third will stand over or under the fourth, the fifth over or under the sixth, and so on. In making use of the tangent line, three points may be taken as radius, either the beginning, middle, or end of the slide ; but the middle point, marked 45, will in practice be most convenient. In using the sine slide, two points may be taken as radius, either the beginning or the end of the slide, as may be found necessary for preventing the numbers from over- running. Having already given several questions under the sector, and Navigation being only an application of Trigonometry, it will be sufficient here to shew the mode INSTRUMENTS AND THE SLIDE-RULE. 181 of working an example or two with the slide-rule, which the student will find infinitely superior for the purpose. Take the tower at page 68. By the sine line, sin. 42 J : 200 : : sin. 47 J : height, sin. 42 J sin. 47 \ ., , or onn = . . , therefore 200 height ; Under 42 1 of the sines bring 200; then under 471 ia 218.2, the height. By the tangent line, making AC radius Rad. or tan. 45 : 200 : : tan. 47 J : height; 200 height tan. 45 tan. 47 J" Over 45 of the tangents set 200 ; then over 47 J is 218.2, the height, (as before.) 182 A TREATISE ON A BOX O? NAVIGATION. fc ^ g K ^ ^ P* M H EH 8 W PS 53 PJ pj H O .0 pi g 02 te N ^ 02 a ^" ^ P% m H r* P* cd 02 H p O 02 02 H ri pq 02 .! 02 ^Q 02 pq W p4 CO ja . _^ l*-M-fcH BH-HN-4* KHMN-4* g "o I~ [~ l~ 1~ CO CO O CO . U3 O O O ^ ^ "^ ^ '3 CH CH 2- - ** rH CM Tj< lO CO t~ C35 O I-H CM * O co r^ Ci o 5 r-* CM CO "^ IO rH CO CM CO "* S 5] GO QO OO l^ 10 co o i^- t-- t~ t^- CO TH i i C75 CO CO CO O >O CO O t~- xffl IO US Tt< Tft o O HN HN f~4M 1 fc) O . 9 O5 t~ CO >O Tfl CO CM i-H O * CO 01 1^ co ia rH IO Tt< * CM rH O CO CM rH l-'a i O O CM >O OO I-H * CO O5 (M O CO O CO C ^ r?5 B ^3 ^ ' ti jj H H Pi ^* ^~* PH PS O K H pq r5 S O & INSTRUMENTS AND THE SLIDE-RULE. 1P3 PLANE SAILING. Plane sailing supposes the earth to be a plane, the meri- dians parallel to each other, and the lengths of degrees everywhere equal ; and involves the consideration of four quantities, difference of latitude, nautical distance, de- parture, and course. Let K (diagram p. 48, disregarding the circle,) denote a point on the earth's surface, and KG its meridian. Draw a line from K to A, and suppose a ship to sail along it from K till she arrive at A ; then KA will be the distance sailed ; DA, the departure from the meridian ; KD, the difference of latitude ; and the angle DKA, contained between the meridian and the rhumb sailed on, the course. The difference of latitude is thus represented by a vertical line, the departure by a horizontal one, the distance by the hypothenusal line forming with the other two a right-angled triangle, and the course by the angle included between the difference of latitude and the distance. Then, if we make distance radius, the de- parture becomes the sine, and the difference of latitude the cosine, of the course ; or, if diff. lat. be made radius, de- parture becomes the tangent of the course. EXAMPLES. 1. A ship sails 38 S., 255 miles W. : required the diff of lat. and departure. Complement of 38 = 52, then Sin. 90 : 255 : : sin. 52 : diff. lat. : : sin. 38 : dep Under 90 of the sines set 255 ; then Under 52 is 201 miles, diff. lat. and under 38 is 157 miles, departure. Ib4 A TREATISE ON A BOX OF 2. A ship sails from lat. 44 50' N. between S. and E. till she has made 64 miles of easting, and is then found to be in lat. 42 56' N. : required the course and distance. 44 50' 42 56 1 54 = 114 miles, diff. lat. As 114 : rad. : : 64 : tan. of course. Under 114 set 45 of the tangents, then under 64 is 29 20', the course. Again, sin. 29 20' : 64 : : sin. 90 : dist. Under 29 20' set 64, then under 90 is 130.6 miles, distance. 3. A ship in lat. 45 25' N. sails N.E.b.N. 1 E. till she comes to 46 55' N. : required the distance and departure. N.E.b.N. J E. = 39 22 J', comp. of which = 50 371'. 4 6 55' 45 25 1 30 = 90 miles, diff. lat. Sin. &0 371' : 90 miles : : sin. 39 221' : 73.8 miles de- parture : : sin. 90 : 116.4, distance.* * As the learner is supposed by this time to be familiar with the mode of operation, it will be sufficient for the future to indi- cate the proportion, without repeating the directions for setting the slide. Thus, in the above instance, under 50 37' set 90 miles, then under 39 22J' will be 73.8, and under 90 will be 1 J 6 4 miles ; and so of all others. When the word rad. occurs as the first or second term, before adjusting the slide run the eye along the proportion to see if the word sin. or tan. follows, INSTRUMENTS AND THE SLIDE-RULE. 185 EXAMPLES FOR PRACTICE. 4. A ship sails from lat. 56 50' N. on a rhumb be- tween S. and S. W. 126 miles, and is then found to be in lat. 55 40' : required the course she sailed, and her de- parture from the meridian. Ans. Course, 56 15'; departure, 104.8 miles. 5. A ship in lat. 44 50' N. sails S. 29 20' E. 130.8 miles : required diff. lat. and departure. Ans. 64 dep. : 114 diff. lat. 6. A ship in lat. 45 25' N. sails N.E.b.N. * E. 116.4 miles : required dep., diff. lat., and latitude come to. Ans. 74 dep. ; 90 miles, or 1 30' diff. lat ; and 46 55' N. lat. come to. 7. A ship at sea sails from lat. 34 24' N. between N. and W. 124 miles, and is found to have made 86 miles of westing : required the course steered, and diff. of lat., or northing made good. Ans. Course, 43 54'; diff. lat. 129'; 35 53' N. lat. come to. 8. A ship in lat. 24 30' S. sails S.E.b.S. till she has made 96 miles of easting : required the distance sailed, and diff. of lat. made good. Ans. Diff. lat. 143.7; dis- tance, 172.8 ; lat. come to, 26 54' S. and use the sine or tangent line accordingly. And in every case it will be advisable for the beginner to construct a diagram, as nothing tends so much to make the operation perfectly un- derstood ; and what is thoroughly understood at the commence- ment is seldom afterwards forgotten. 16* 186 A TREATISE ON A BOX OF TRAVERSE SAILING. When a ship sails upon several courses, the zigzag lina fche describes is called a traverse ; and the reducing the courses into one, and thereby finding the course and dis- tance made good upon the whole, is called the resolving of the traverse. For this purpose, construct a table of six columns, in the first of which is the course, and in the second the distance ; then find the diff. lat. and dep. for each course, and enter it N. or S., E. or W., as it may be. Add up the columns separately ; the difference of the third and fourth will give the diff. of lat., and the diff. of the fifth and sixth, the departure. Then, having obtained the total diff. lat. and dep. which the ship has made, find the corresponding course and distance. EXAMPLE. 9. A ship from the equator sails N. 48, W. 37, N.W. 18, N.E. 70, N.N.E. 24, and E. 32 miles: required her course, distance, and latitude reached. Course. Dist. Diff. Lat. Dep. N. S. E. W. N. W. N.W. N.E. NN.E. E. 48 37 18 70 , 24 32 48 12.72 49.5 22.18 49.5 9.18 32. 37 13.72 132.4 90.68 49.72 49.72 40.96 INSTRUMENTS AND THE SLIDE-RULE. 187 First she sails due N., and so will have no departure : therefore place 48 under N. Her second course is due W., and so she will have no diff. of lat. ; therefore place 37 under W. Her third course is 45, and therefore her departure and diff. of lat. will be equal. Over 18 place 90 of the sines, and under 45 is 12.72, which place under N. and W. Her fourth course is also 45, and therefore her dep. and diff. of lat. will be equal. Over 70 place 90, and under 45 is 49.5, which place under N. and E. Her fifth course is 22 J, cos. of which = sin. 67 J. Over 24 place 90, then under 22 J is 9.18, her departure, which place under E. ; and under 67 J is 22.18, her diff. lat., which place under N. Her last course is 32 due E., and so she will have no diff. of lat. : therefore place 32 under E. Add up the three columns. As there is no number standing under S. the diff. of lat. is 132.4 = 2 12' N. Subtract the W. from the E., and the remainder is 40.96 E. for the total departure : then As 132.4 : rad. : : 40.96 : tan. 17 12', the course. Again, sin. 17 12 f : 40.96 : : sin. 90 : 138.6 wiles, the distance. EXAMPLES FOR PRACTICE. 10. A ship from Cape Clear, lat. 51 25' N., sails SS.E. J E. 16, E.S.E. 23, S.W.b.W.J W. 36, W.t N. 12, and S.E.b.E. JE. 41 miles : required the equivalent course and distance, and the latitude of the place which the ship has arrived at. Ans. Course, 18 12' ; distance 62.75 miles ; lat. in, 50 25' N. 188 A TREATISE ON A BOX OF 11. From Cape St. Vincent, in lat. 37 2' N., a ship sailed S.W.b.S. 49, S.b.E. 56, S.E.b.E. 38, S.W. 84, NN.W. 72, and E.N.E 24 miles : required the course, distance, and latitude come to. Ans. Course, 26 15' ; distance, 112 ; lat. in, 35 22' N. PARALLEL SAILING. Since the meridians meet at the poles, it follows that the length of a degree on any parallel of latitude dimi- nishes as it recedes from the equator. To ascertain this diminution, when a vessel sails on a parallel of latitude, or changes her longitude only, is the object of parallel sailing. Let FA (diagram, page 48,) represent the earth's semi-axis ; FCB, a quadrant of a meridian; B, a point on the equator; C, a point on the meridian, and conse- quently the arc CB, or angle CAB, the latitude of C ; and let the quadrant revolve on AF; then the circles described by the points C, B, or similar parts of them, will be proportional to their radii EC, AB. Now AB, or AC : EC, or AD : : rad. : cos. CAB ; that is, difference of longitude, or distance between any two meridians on the equator, or parallel described by B : the distance between those meridians on the parallel described by C : : radius : the cosine of the latitude ; or the lengths of degrees on different parallels vary as the cosines of the latitudes. Hence, if in any right-angled triangle ADC, the acute angle at the base CAD, be made equal to the latitude, and the length of the base AD equal to the departure, or meridian distance, or distance be- tween any two meridians on a parallel of that latitude ; INSTRUMENTS AXD THE SLIDE-RULE. 189 then the hypothenuse AC will be equal to the arc of the equator, or the difference of longitude corresponding to that meridian distance. EXAMPLES. 12. Required the number of miles contained in a degree of longitude, in lat.- 55 N. cos. 55 = sin. 35. sin. 90 : 60 miles : : sin. 35 : 34.4 miles. 13. A ship from lat. 42 52' N. in long. 9 17' W. sails due W. 342 miles : required the longitude come to. cos. 42 52' = sin. 4 7 8'. sin. 47 8' : 342 : : sin. 90 : 467, diff. long. 467 = 7 47' 9 17 17 4 W. long, come to. 14. A ship sailed 224 miles upon a due "W. course, and by observation found she had differed her longitude 6 18', or 378 miles : required latitude. 378 : sin. 90 : : 224 : sin. 3620 / ; and sin. 36 20' = cos. 53 40', the latitude required. 15. Two ships in lat. 46 30' N., distant asunder 654 miles, sail both directly N. 256 miles : required their distance. 256 = 4 16' 46 30 50 46 N., lat. reached. 190 A TREATISE ON A BOX OP Then cos. 46 30' : 654 miles : : cos. 50 46'; or sin. 43 30' : 654 : : sin. 39 14' : 601 miles, the dis- tance. 16. Two ships in lat. 45 44' N., distant 846 miles, sail directly N. till the distance between them is 624 miles : required the lat. reached and dist. sailed. Cos. 45 44' = sin. 44 16'; then 846 : sin. 44 16' : : 624 : sin. 31; and sin. 31 = cos. 39, lat. come to. Then 59 0' 45 44 13 16 = 796 miles, dist. sailed. EXAMPLES FOR PRACTICE. 17. A ship in lat. 54 20' N. sails directly W. on that parallel till she has differed her longitude 12 45' : re- quired the distance sailed. Ans. 446 miles. 18. A ship from Cape Finisterre, lat. 42 52' N., long. 9 17' W., sailed due "W. 342 miles : required the longi- tude come to. Ans. 17 4' W. 19. A ship sails on a certain parallel directly W. 624 miles, and has then differed her longitude 18 46', or 1126 miles : required the latitude of the parallel sailed on. Ans. 56 20'. 20. A ship from a port in lat. 54 N. sailed due E. 200 miles ; then, having run due S. an unknown number of miles, sailed W. 250 miles, and, by observation, found she had arrived at the meridian of the port she sailed from : required the lat. come to, and distance run in tho S. direction. Ans. 42 43' lat. come to; 677 miles run INSTRUMENTS AND THE SLIDE-RULE. 191 MIDDLE LATITUDE SAILING. Middle Latitude Sailing is a composition of plane and parallel sailing, and is used for reducing the departure to miles of longitude. Now, when two places lie not on the same parallel, their difference of longitude, reduced to miles of easting, or westing, if reckoned on the higher parallel, would be too small, and if on the lower parallel, too great. The common way of reducing it is, by taking it, as the name implies, on a parallel midway between the two; which, though not strictly correct, is sufficiently so for most nautical purposes. For the solution of ques- tions of this kind, we have only to place together the two triangles treated of under Plane and Parallel Sailing, and resolve them separately, observing to begin with that in which two parts are given, and then the unknown parts of the other triangle will be easily obtained. See triangle ACK, (diagram, page 48.) By plane sailing, the angle at K is the course ; KD, the difference of latitude ; DA, the departure ; and KA, the distance sailed. By parallel sailing AD is still the departure, or meridian distance, on the parallel midway between the latitude left and latitude reached ; CAD, the angle of the middle latitude ; and AC, the difference of longitude. The following examples will illustrate the modes of solution. EXAMPLES. 21. Required the course and distance from the east point of St. Michael's, in lat. 37 49' N., long. 25 11' W., to Start Point, in lat. 50 13' N., long. 3 38' W. 192 A TREATISE ON A BOX OP 50 13' N. 25 11' W. 37 49 N. 3 38 W J)12 24 = 744, diff. lat. 21 33 = 1293, diff. long. 6 12 37 49 44 1 mid. lat., complement of which = 45 59'. Then sin. 90 : 1293 : : sin. 45 59' : 930, the departure. 744 : rad. : : 930 : tan. 51 20', the course = N. 51 20' E. sin. 51 20' : 930 : : sin. 90 : 1191 miles, the distance. 22. A ship from Brest, in lat. 48 23' N., long. 4 30' W., sailed S.W. f W. 238 miles : required the lat. and long, come to. S.W. i W. == 53 26', comp. of which = 36 34' j Then sin. 90 : 238 : : sin. 36 34' : 141.8 diff. lat. == 2 22'. 48 23' N. 2 22 46 1 lat. come to. 1 11 = * diff. lat. 47 12 mid. lat., comp. of which 42 48'. Then sin. 42 48' : 238 : : sin. 53 26' : 282 diff, long. = 4 42' W. 4 30 W. 9 12 long, come to. INSTRUMENTS AND THE SLIDE-RULE. 193 23. A ship from lat. 17 N., long. 24 25' W., sailed N.W. | N. till, by observation, her lat. is found to be 28 34' N. : required the distance sailed and long, come to. N.W. * N. = 36 34', comp. of which = 53 26'. 28 34' N. 17 N. 34 = 694 diff. lat. 5 47 17 22 47 mid. lat., comp. of which = 67 13'. Then sin. 53 26' : 694 : : sin. 90 : 864 miles, the distance; and sin. 67 13' : 864 : : sin. 36 34' : 558, diff. long. = 9 18' W. 24 25 W. 33 43 long, come to. EXAMPLES FOR PRACTICR 24. A ship from lat. 26 30' N., long. 45 30' W., sailed N.E. N. till her departure was 216 miles: re- quired the distance run, and lat. and long, come to. Ans. Dist. 341 miles; lat. come to, 30 53' N.; long. 41 24' W. 25. From lat. 43 24' N., long. 65 39' W., a ship sailed 246 miles, on a direct course between S. and E., and was then, by observation, in lat. 40 48' N. ; required 194 A TREATISE ON A BOX OF the course, and long. in. Ans. Course, 50 40'; long come to, 61 23' W. 26. A ship from Cape St. Vincent, lat. 37 2' N., long. 9 2' W., sails between S. and W.; the lat. come to ia 18 16 r N., and departure 838 miles : required the course, distance run, and long, come to. Ans. Course, 36 40' j dist. 1403 miles ; long, come to, 24 48' W. 27. A ship from Bordeaux, in lat. 44 5(K N., 35' W., sails between the N. and W. 374 miles, and makes 210 miles of easting : required the course, and lat. and long, come to. Ans. Course, 34 10' ; lat. come to, 49 59' N., long. 5 45' W. 28. A ship from lat. 54 56' N., long. 1 10' W., sailed between N. and E. till, by observation, she was found to be in long. 5 26' E., and had made 220 miles of easting: required the lat. come to, and course and distance run. Ans. Lat. come to, 57 34' N. ; course, 54 20'; distance, 271 miles. 29. A ship from a port in N. lat. sailed S. E. i S. 438 miles, and differed her long. 7 28': required the lat. sailed from and come to. Ans. Lat. sailed from, 51 40' j come to, 46 16 . INS1RUMENTS AND THE SLIDE-RULE. 195 TO DETERMINE THE DIFFERENCE OF LONGITUDE MADE GOOD UPON COMPOUND COURSES, BY MIDDLE LATITUDE SAILING. WITH the several courses and distances find the latitude and departure made good and the ship's present latitude, as in Traverse Sailing. Take the middle latitude between the latitude left and latitude arrived at; then with the departure made by the traverse table, and the middle lati- tude, find the difference of longitude by Middle Latitude Sailing. In high latitudes this method will be somewhat incorrect, and therefore it will be advisable to employ the more tedious mode of computing the difference of longi- tude for every separate course, which is most readily done as follows : Complete the traverse table, as before, to which annex five columns : in the first put the several lati- tudes the ship is in at the end of each course and distance ; in the second, the sums of each consecutive pair of lati- tudes ; and in the third, half the sums, or middle latitude ; then find the difference of longitude answering to each separate middle latitude, and its corresponding departure, and place it in the fourth or fifth (namely the east or west) difference of longitude columns, according as the de- parture is east or west : then the difference of the sums of the east and west columns will be the difference of longi- tude made good, of the same name as the greater. 196 A TREATISE ON A BOX OF EXAMPLE. 30. A ship from lat. 66 14' N., long. 3 12' E., sails NN. E. * E. 46, N. E. i E. 28, N. I W. 52, N. E. b. E. i E. 57, and E. S. E. 24 miles : required her course, and longi- tude in. to 1 * MSN 8 j | = < COiO 3' O t o ' co 3 S-- - I * ^ ^a 3 8[pp PT !I CO COrH 2 36C5 62 4770 72 6335 82 9145 3 180 13 787 23 1419 33 2100 43 2863 58 3764 334905 73 6534 S3 9606 4 240 14 848 24 1484 34 2171 44 2946 54 3865 645039 74 6746 84 10137 5 300 In 910 26 155035 2244 45 3030 .1.1 396S 65 '5179 75 6970 80 10765 6 361 If, 973 26 161636 2318i 46 31 1C, .ifi 4074 66 5324 76 721^ 80 11532 7 421 17 10*5 27 168437 2393; 47 3203 n 4183 07 5474 1 1 7467 87 12522 8 482 18 1098 28 1751 :38 2468 48 3292 58 4294 68 5631 78 7745 88 13916 9 542 I'l llfil 29 1819 39 2545 49 33S2 .V.I 4409 Ii9 5795 79 8046 88 16300 To return to the diagram. Let the angle DAC be the course; AD the difference of latitude; AC the distance; and DC the departure; then AB being the elongated, or meridional, difference of latitude, AG will be the elon- INSTRUMENTS AND THE SLIDE-RULE. 199 gated distance, and BG the elongated departure, that is the real difference of longitude. Now, AD : DC : : AB : BG; that is, Diff. Lat. : Dep. : : Merid. Diff. Lat. : Diff Long. And AB : BG : : Had. : tan. CAD ; that is, Merid. Diff. Lat. : Diff. Long. : : Rad. : tan. course. To find the meridional parts answering to any number of degrees and minutes, take proportional parts of the dif- ferences found by subtraction. EXAMPLES. 33. Required the meridional parts answering to 37 43', viz. 37g. 38 = 2468 37 = 2393 75 Then 2393 43 + 54 225 = 2447 merid. parts for 37 43'. 300 6,0 ) 322,5 54 nearly. 34. Required the meridional parts for 27 58'. 28 = 1751 27 = 1684 Then 1751 2 67 = 1749 merid. parts for 27 58'. 6,0 ) 13,4 200 A TREATISE ON A BOX OF 35. Required the number of degrees answering to 3625 meridional parts. 3665 = 52 3625 3569 = 51 3569 96= Ior60' 56=?' 96 : 60' : : 56 : 35'. .-. 3625 = 51 35'. EXAMPLES IN MERCATOB/S SAILING. 36. Required the course and distance from the east point of the Azores, lat. 37 49' N., long. 25' 11' W., to Start Point in lat. 50 13' N. long. 3 38' W. 25 11' W, 60 13' N. Meridional parts = 3495 3 38 W. 37 49 N. " = 2454 21 33=1293 miles 12 24 = 744 miles diff. lat. 1041 merid. diff. long diff. lat. Then 1041 : rad. : : 1293 : tan. 51 10' course, whose comp. =38 50'; and sin. 38 50 7 : 744 : : sin. 90 : 1186, the distance. Compare this with Example 21. 37. A ship sails from lat. 38 47' N., long. 75 4^ W., 267 miles N. E. b. N. : required the ship's present place. N. E. b. N. = 33 45' course : comp. of which=5615' Bin. 91: 267 : : sin. 56 15' : 222 diff. lat.= 342'N. 38 47 N. 42 29 laL come to 42 29' merid. parts = 2821 38 47' " 2528 293 merid. diff. lat INSTRUMENTS AND THE SLIDE-RULE. 201 Rad. : 293 : : tan. 33 45' : 196 diff. long. =3 16' E. 75 4'W. 3 16 E. 71 48 long. in. 38. A ship from Nova Scotia, in lat. 45 20' N., long. 60 55' W., sailed S. E. | S., and, by observation, was found to be in lat. 41 14' N. : required the distance sailed, and long, come to. S. E. i S. = 42 11' course, whose comp. =47 49'. 45 20' merid. parts = 3058 41 14 do. =2720 4 6 = diff. lat. 338 merid. diff. lat. liad. : 338 : : tan. 42 11' : 306 diff. long. = 5 6' 60 55' W. 5 6 E. 55 49 W. long. in. Sin. 47 49' : 246 : : sin. 90 : 332, distance. EXAMPLES FOR PRACTICE. 39. Required the direct course and distance between the Lizard in lat. 50 0' N., and Port Royal, in Jamaica, in lat. 17 40' N., differing in long. 70 46', Port Royal lying so far to the W. of the Lizard. Ans. Course, 60 33'; distance, 3645 miles. 40. Suppose a ship from the Lizard, in lat. 50 N., sails S. 35 40' W. 156 miles : required lat. come to, and how much she has altered her longitude. Ans. Lat. come to, 4753'N.; diff. long. 2 19'. 202 A TREATISE ON A BOX OP 41. A ship in lat. 54 20' N. sails S. 33 45' E., until, by observation, she is found to be in lat. 51 45' N. : required the distance sailed, and the diff. long. Ans. Distance, 186.4 miles : diff. long. 2 52' E. 42. A ship from lat. 45 26' N. sails between N. and E. 195 miles, and then, by observation, is found to be in lat. 48 6' N. : required the direct course, and diff. long. Ans. Course, N. 34 52' E., or N. E. b. N. 1 7' E. ; diff. long. 2 43' E. 43. A ship from lat. 48 50' N. sails S. 34 40' E., till her diff. long, is 2 44' : required lat. come to, and dis- tance sailed. Ans. Diff. lat. 2 41' ; distance 196 miles. 44. A ship from 54 36' N. sails S. 42 33' W., until she has made 116 miles of departure : required the lat. she is in, her direct distance sailed, and how much she has altered her longitude. Ans. Lat. come to 52 30' ; distance, 171.5 miles; diff. long. 3 15'. INSTRUMENTS AND THE SLIDE-RULE. 203 TO DETERMINE THE DIFFERENCE OF LONGITUDE MADE GOOD UPON COMPOUND COURSES, BY MERCATOR'S SAILING. With the several courses and distances, find the latitude and departure made good, and the ship's present latitude, as in traverse sailing. Take the meridional difference of latitude between the latitude left and latitude arrived at. Then, with the course made good by the traverse table, and the meridional difference of latitude, find the difference of longitude by Mercator's Sailing. In high latitudes, this method will be somewhat incorrect; and, therefore, it will be advisable to employ the more tedious mode of computing the difference of longitude for every separate course, which is most readily done as follows : Complete the traverse table as before, to which annex five columns. In the first, put the several latitudes the ship is in at the end of each course ; in the second, the meridional parts corresponding to each latitude; and in the third, the dif- ference of each consecutive pair of meridional parts. Then find the difference of longitude answering to each separate course, and its corresponding meridional differ- ence of latitude, and place it in the fourth or fifth (viz. the east or west) difference of longitude columns, according as the course is east or west; then the difference of the sums of the east and west columns will be the difference of longitude made good, of the same name as the greater. EXAMPLE. 45. A ship from lat. 66 14' N., long. 3 12' E., sails NN. E. i E. 46, N. E. } E. 28, N. } W. 52, N. E. b. E. t 204 A TREATISE ON A BOX OP E. 57, and E. S. E. 24 miles : required her course and longitude in. SO rHlH tHTjl ss is &1(N \ f 1-HT* O &<* | t- o^t 511 s a tiw 3 ^ Illl /; w QJ S5 .ja 1 I I O OO ^ CO 3538 -* I t i i * S II 8," g ^ If*" EXAMPLES FOR PRACTICE. 46. A ship from lat. 5730'N., long. 147'W., sailed SS. E. 48, S. W.b. S. 54, E. b. S. 71, N. E. 63, and W. N. W 50 miles : required the lat. and long, of the place come to. Ans. By 1st rule, lat. come to, 56 50' N., long 2' W. INSTRUMENTS AND THE SLIDE-RULE. 205 47. Four days ago, we took our departure from Faro- head, in lat. 58 4(X N., and long. 4 50' W., and since have sailed as follows : N.W. 32, W. 69, W. N. W. 93, W. b. S. 77, S.W. 58, and W. f S. 49 miles : required our present lat. and long. Ans. By Rule 2, lat. come to, 58 35'; long. 15 54' W. OBLIQUE SAILING. Oblique Sailing is the application of oblique angled plane triangles to the solution of problems at sea ; and is particularly useful in going along shore, and surveying coasts and harbours. EXAMPLES. 48. Coasting along the shore, I saw a cape bear from me NX. E. ; then I stood away X.W. b. W. 20 miles, and observed the same cape to bear from me N. E. b. E. : required the distance of the ship from the cape at her last station. See figure, page 72. Sin. 33 45' : 20 :: sin. 78 45' : 35.3 miles. 49. A point of land was observed, by a ship at sea, to bear E. b. S. ; and after sailing N. E. 12 miles, it was found to bear S. E. b. E. It is required to determine the place of that headland, and the ship's distance from it at the last observation. See figure, page 78. Sin. 22 30 7 : 12 : : sin. 56 15' : 26.1. 50. At noon, Dungeness bore N. b. W., distance 5 leagues ; and having run N. W. b. W. 7 knots an hour, at 5 P. M. we were up with Beachy Head : required the dis- tance of Beachy Head from Dungeness. Ans. 26.6 mile* u 206 A TREATISE ON A BOX OF WINDWARD SAILING. Windward Sailing is the method of gaining an intended port by the shortest and most direct method possible, when the wind is in a direction unfavourable to the course the ship ought to steer for that port. In order to attain this point, it is evident that the ship must sail on different tacks; and, therefore, the object of this sailing is, to fiud the proper courses to be steered on each board, that the vessel may arrive at the intended port with the least delay possible. By the term board is to be understood the shifting of the direction, or alteration of the course. Thus, if a vessel sails on two boards, she shapes out the letter V ; if on three boards, the letter N ; and so on. EXAMPLES. 51. A ship is bound to a port 48 miles directly to the windward, the wind being SS.W., which it is intended to reach on two boards ; and the ship can lie within 6 points of the wind ; required the course and distance on each tack. Describe a circle, and from the centre, which call A, draw a line in a SS.W. direction, to represent the direction of the wind, and call the lower extremity of this line B, and let it represent the port intended to be reached. Then the wind blowing from B to A, and A being the position of the ship ; from A, to the left of the line BA, draw a line, making with it an angle of 6 points, or 67 30' ; this will, of course, be due W. From the centre of the circle A, to the right of the line AB, draw another line, making with AB an angle, like the other, of 67 30'. This line will be south-east. From the point B, paralle INSTRUMENTS AND THE SLIDE-RULE. 207 to this last line, draw a line, cutting the one running west, in a point, which call C. Then AC will be the course of the ship on the first board, and CB that on the second. Now, the angles at A and B will be each 67 J, and at C 45, opposite which is the line AB, 48 miles. Then, sin. 45 : 48 miles : : sin. 67 J : 62.7 miles, the distance to be sailed on each board ; so that she will have to sail 125.4 miles to make 48. 52. The wind at N. W., a ship bound to a port 64 miles to the windward, proposes to reach it on three boards, two on the starboard, and one on the larboard tack, and each within 5 points of the wind : required the course and dis- tance on each tack. Describe a circle, and from its centre, which call A, draw a line in a N. W. direction, to represent the direc- tion of the wind, and let its upper extremity denote the port intended to be reached, which call B. From A draw two lines, one to the left and the other to the right of the line B A, each making with it an angle of 5 points ; con- sequently, the first will pass through the W. b. S. rhumb, and the second through the N. b. E. Call the lower ex- tremity of the line passing through the S. b. W. rhumb, C ; the upper extremity of the other, D. From B draw a line to the right of the line BA, parallel with CA. Bi- sect BA, in a point, which call E. Draw a line from E to C, parallel with the line DA, and prolong it upward till it cuts the line running right of B, in a point, which call F. Then, in the triangle EAC, the angles at A and C are each 56 15', and the angle at E 67 30', and the line EA is 32 miles. Therefore, sin. 56 15' : 32 : : sin. 67 30' : 36.25 miles = AC, BF, CK, or EF, and 208 A TREATISE ON A BOX OP twice 36.25 = 72.5. Hence, she must first sail W b. S. 36J miles, then N. b. E. 72 miles, then W. b. S. 36* miles. It may be here observed, that whatever number of boards it may be found expedient a ship should make, the sum of the distances on each tack will be the same as if the place had been reached on two boards only. 53. A ship is bound to a port 26 miles directly to wind- ward (the wind being N. E.,) which it is intended to reach on two boards, the first being on the larboard tack, and the ship can lie within 6 points of the wind : required the course and distance on each tack. Ans. Course on the larboard tack, E. S. E. ; on the starboard, NN. W. ; dis- tance on each board, 34 miles, nearly. 54. The wind at N. E., a ship is bound to a port bearing NN. E., distance 68 miles, which it is proposed to make at four boards ; the coast, which is to westward, trends NN. E. also ; so that the ship must go about as soon as she reaches the straight line joining the ports : required the course and distance on each board, the ship making her way good within 6 points of the wind. Ans. Course on the larboard tack, E. N. E. E. ; on the star- board, N. W. b. W. J W. ; first and third distances, 47.8 miles; second and fourth distances, 37.2 miles. 55. A ship close hauled within 5 points of the wind, and making 1 point of leeway, is bound to a port bearing SS. W., distant 54 miles, the wind being S. b. E. ; it is intended to make the port at three boards, the first of which must be on the larboard tack, in order to avoid a reef of rocks : required the course and distance on each INSTRUMENTS AND THE SLIDE-RULE. 209 tack. Ans. Course on the larboard tack, S. W. b. TV. ; on the starboard, E. b. S. ; distances on the larboard tack, each 37.45 miles; distance on the starboard tack, 42.4 miles. CURRENT SAILING. When a ship sails exactly with the current, her velocity will, of course, be accelerated; and, when in due opposi- tion to the current, it will be retarded by the difference of the velocities of the wind and stream. When she is urged by the wind in one direction, aud by the current in another, her course, agreeably to the law influencing all bodies acted upon simultaneously by two forces, will lie in the diagonal of the parallelogram formed by those forces; that is, will be the third side of a triangle of which the drift of the current and the action of the wind form the other two, the angle between them being known. N.B. That point of the compass to which a current runs is called its setting, and its rate per hour is called its drift. EXAMPLES. 56. A ship sails by the compass directly S. 96 miles, in a current that sets E. 45 miles in the same time : re- quired the ship's true course and distance. Describe a circle, and from its centre, which call A, draw a line in a south direction, and make it equal to 96 from a scale of equal parts, and call the lower extremity B. From the point B, in an easterly direction, draw a line equal to 45, from the same scale, and call its extremity C. Join AC. The angle BAG will be the course, and C the point at which the ship will have arrived. Then, 18* 210 A TREATISE ON A BOX OF 96 : rad. : : 45 : tan. 25 7', the ship's course = SS. E. 2 6', easterly. And, sin. 25 T : 45 : : sin. 90 : 105.9 miles, distance sailed. 57. A ship has made by the reckoning N. J W. 20 miles, but, by observation, it is found that, owing to a current, she has actually gone NN. E. 28 miles : required the setting and drift of the current in the time which the ship had been running. Ans. Setting, N. 64 48' E., drift, 14.1 miles. 58. A ship from a port in lat. 42 52' N., sailed S. b. W. W. 17 miles in 7 hours, in a current setting be- tween the N. and W. ; and then the same port bore E. N. E., and the ship's latitude, by observation, was 42 42' N. : required the setting and drift of the current. Ans. Setting, 71 55', drift, 2.9 knots an hour. 59. A ship, bound from Dover to Calais, lying 21 miles to the S. E. b. E. E., and the flood-tide setting N. E. E. 2 1 miles an hour : required the course she must steer, and the distance run by the log, at 6 knots an hour, to reach her port. Ans. Course, 39 14'. Distance to be run 19.4 miles. 60. From a ship, in a current, steering W. S. W. 6 miles an hour by the log, a rock was seen at 6 in the evening, bearing S. W. i S. 20 miles. The ship was lost on the rock at 11 p. M. : required the setting and drift of the current. Ans. Setting, S. 75 10' E., drift 3.11 miles per hour. INSTRUMENTS AND THE SLIDE-RULE. 211 OF A SHIP'S JOURNAL. A journal is a register of transactions occurring on board a ship, and should contain a particular detail of every thing relative to the navigation of the vessel as the courses, winds, currents, &c. that her situation may be known at any instant at which it may be required. The computations made to determine the place of a ship from the courses and distances run in 24 hours, are called a day's work ; and the latitude and longitude of a ship deduced therefrom, are called the latitude and longitude in, by account, or, by dead reckoning, in contradistinction to the latitude and longitude as determined by observation. At the time of leaving land, the bearing of some known place is to be observed, and its distance found, either by observation, or by taking its bearing at two different times, from two different places, and determining its dis- tance accordingly. The log-book, which is to contain a daily transcript from the log-board, is to be divided into 7 columns. In the first, put the hours; in the second and third, the knots and fathoms sailed per hour ; in the fourth, the courses ; in the fifth, the winds ; in the sixth, the leeway; and in the seventh, any remark that may be thought necessary. It is better, however, to omit the leeway column, and, on transcribing from the log-board, to make the proper allowance, and to enter the amended courses only, in the log-book. After this, allow for the variation, and bring them into a traverse table. Find the ship's distance, difference of latitude, and departure, by plane sailing; then, by Mercator's, or middle latitude 212 A TREATISE ON A BOX OF sailing, find the difference of longitude, and enter it ac- cordingly. The following specimen will furnish an idea ; but the present work being intended principally to show the instrumental modes of computation, the student is referred to works exclusively on Navigation, for more com- plete information upon the subject. .a " 02 S = y> M o .-*?. * Obser- vation. (M ^ CD QO O C Of this class are the formulae for determining the di- mensions of spheres from their weight, or solidity; and for exhibiting the relations of similar solids to each other. A 3 (6.) a 3 : e* : : A 3 : E 2 , whence E = ej/ 3 - The formula for determining the distance of a planet from its periodical revolution, and conversely. The principle of rules containing inverted lines is shown as follows ; Jjet a, b, and c, denote any logarithmic INSTRUMENTS AND THE SLIDE-RULE.- 225 distance on the A, B, and C lines ; and, in lieu of D, let an inverted line A be laid down, so that unity upon it coincides with the extremity of the C line. Then the value of the same distance upon this line will be -j ; but if it be drawn aside until some other number r fall under T the extremity, then its value will be ; and .-. we shall have; : c : : a : Z>, whence b = - > where r is a constant A r divisor, and A, c, a, any three numbers. Solutions of the more difficult Questions. Example. 161, page 160. 1 : 6 : : 3 : 18, the depth of 2 /2V the entire cone ; hence $ of the depth is cut off; .-. ( ^ ) O N>' 8 or = of the solidity is cut off, and the remaining frustum 19 4 76 8 116 18 07 > 5 O* this is yrr^j which added to ^= = TTT-T ; hence 135 : 18 s : : 116 : ? 3 135 E : 18 D : : 116 E : 17.12 D, the distance from the surface of the water to the bottom of the cone : hence 17.12 12 = 5.12, the depth of the water. Ex. 162. 2 : 18 : : 3 : 27, the height of the entire pyramid; hence - of the height is cut off; .-. {=J or^y 26 of the whole is cut off, and the remaining frustum is ~ 1 . ,. . . 26 26 , 3 ' - oi this is ; so each person will have TT=- and will o ol ol ol be for waste. The various bulks will therefore be aa 3, 29, 55, and 81. 226 A TREATISE ON A BOX OP Hence 81 : 27 3 : : 55 : ? 3 : : 29 : ? 3 : : 3 : 9 3 j 81 E : 27 D : : 55J5J : 23.73 D : : 29 E: 19. 17 D : : 3 E: 9 D; then 27 23.73 == 3.27 ; 23.73 19.17 = 4.56 ; 19.17 9=10.17. Ex. 172. 4 Ib. : 3 3 : : 108 Ib. : ? 3 4 E : 3 D : : 108 E : 9 inches D, the diameter of the globe ; to find the content of which in gallons, the globe gauge point is 23 ; divide, then, by 23 a , 9 times 9 a . 23 D : 9 C : : 9 D : 1.37 gallons C; 12 + 1.37 = 13.37 gallons, the quantity virtually put into the vessel. Again, 5 : 20 : : 15 : 60, the height of the entire pyramid ; to find the content of which, in gallons, the pentagonal pyramid gauge point is 21.98 ; divide, then, by 21.98 3 , 60 times 15 3 . 21.98 D : 60 C : : 15 D : 27.92 gallons (7; 27.92 13.37 == 14.55, the content of the pyramidal segment above the surface of the water; then 27.92 : 60 3 : : 14.55 : ? 3 27.92 E : 60 D:: 14.55 E : 48.28 inches D, the height of the segmental pyramid above the water; .-. 48.28 40 = 8.28, the depth of the vessel unoccupied. 1 /1\ 3 Ex. 173. As - of the diameter is to be left, ( = ) or 5 V 5/ 1 124 - of the solidity will be left, and j^ will be turned down; .-. each will turn down - ; the various bulks L2o will therefore be as 1, 32, 63, 94, and 125. Hence 125 : 10 3 : : 94 : ? 3 : : 63 : ? 3 : : 32 : ? 3 : : 1 : 2 3 ; 125 E : 10 D : : 94 E : 9.09 D : : 63 E : 7.96 D-..32E : 6.35 D::IE:2D. INSTRUMENTS AND THE SLIDE-RULE. '2'27 Ex. 174. To find, in pints, the contents of a globe whose diameter is 3.6 inches. The pint gauge point for globes is 8.13 ; divide, then, by 8.13 a , 3.6 times 3.6 3 . 8.13 D : 3.6 C:: 3.6 D : . 703(7; .703 + I = .703 + .777 = 1.48; y then .703 : 3.6 3 : : 1.48 : ? 3 .703 E : 3.6 D :: 1.48 E : 4.615 D, the diameter; and 113 A : 355 B : : 4.6154 : 14.49.5, the circumference. Ex. 175. Let the diameters be 30 and 50. The round or conic gauge point for gallons is 46 ; the con- tent, therefore, by formula 17, page 137, is D 30 =? viz. 50 = 14.2 46 55.5 gallons, the content of a vessel whose depth is 12 inches, and diameters 30 and 50. Then, since the depth remains unaltered, the content will vary as the squares of the diameters ; hence 55.5 gals. : j ^ 1 : : 14 gals. : 1* 55.5 C : 30 D : : 14 G : 15.06 D, bottom diameter; 55.5 C : 50 D : : 14 C : 25 1 Z>, top diameter. Round Timber. Instead of using the quarter girt, as mentioned at page 176, it will be preferable to take the tchole girt, andybwr times the divisor; that -is, putting L length in feet, y girt in inches, then the content by the common method will be !j thus, in question 183, 48 D : 48 C : : 39 D : 31.7 228 TREATISE ON A BOX OP INSTRUMENTS. feet C. To find the true content the formula will be ~^-' } thus, in question 184, 42.53 D : 48 O : : 39 D 4.4. oo : 40 J cubic feet a Casks. The following exhibits the formulae for the four varie- ties under the simplest form : 1st var. (#' + 2.1*) 2d var. (g'+2.l' ^of (2diff.) Fr.Pro. Sphd. g2 54 a Fr. Par. Spin. g2 54* 3d Tar . 4th var. Fr.TwoParb. 26.6 a Fr.TwoCo. It will be found a great improvement to the rule to copy the formulae at pp. 136, 137, 138, on the back of one of the slides. June, 1848. THE END. STEREOTYPED BY I. JOHNSON A Ofil CATALOGUE OP PRACTICAL AXD SCIENTIFIC BOOKS, PUBLISHED BT HENRY CAREY BAIRD, INDUSTRIAL PUBLISHER, 3STo- 40O -W-A-IalSrTJT STREET, PHILADKLPHIA. Any of the Books comprised in this Catalogue will be sent by mail, free of postage, at the publication price. r IfEW ASD ENLARGED CATALOGUE, 95 pages 8vo., with full descriptions of Books, -will be sent, free of postage, to any one who will favor me with his address. A RMENGATTD, AMOTTBOTIX, AND JOHNSON. 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A PRACTICAL TEEATISE ON STREET OR HORSE- ** POWER RAILWAYS : Their Location, Construction, and Management ; with General Plans and Rules for their Organization and Operation; toge- ther with Examinations as to their Comparative Advantages over the Omnibus System, and Inquiries as to their Value for Investment ; including Copies of Municipal Ordinances relat- ing thereto. By ALEXANDER EASTON, C. E. Illustrated by 23 plates, 8vo., cloth $2 00 pDRSYTH. BOOK OF DESIGNS FOR HEAD-STONES, MURAL, L AND OTHER MONUMENTS : Containing 78 Elaborate and Exquisite Designs. By FORSYTE. 4to., cloth $5 00 *#* This volume, for the beauty and variety of its designs, has never been surpassed by any publication of the kind, and should be in the hands of every marble-worker who does fine monumental work. pAIRBAIRN. 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A TREATISE OH A BOX OF INSTRUMENTS, And the Slide Rule ; with the Theory of Trigonometry and Lo- garithms, including Practical Geometry, Surveying, Measur- ing of Timber, Cask and Malt Gauging, Heights, and Distances. By THOMAS KESTISH. In one volume. 12mo. . . $1 25 T7-OBELL. ERNI. MINERALOGY A thorf method of Determining and Classifying Minerals, by mean*^ of simple Chemical Experiments in the Wet Way. Translated from the last German Edition of F. Vox KOBELL, with an Introduction to Blowpipe Analysis and other addi- tions. By HEXBI ERXI, M. D., Chief Chemist, Department of Agriculture, author of "Coal Oil and Petroleum." In one volume. 12mo. ... . -50 T ANDRTN. A TREATISE ON STEEL : Comprising its Theory, Metallurgy, Properties, Practical Work- ing, and Use. By M. H. C. LAXDRIX, Jr., Civil Engineer. Translated from the French, with Notes, by A. A. FESQCET, Chemist and Engineer. 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ASSAYER'S GUIDE; Or, Practical Directions to Assayers, Miners, and Smelters, for the Tests and Assays, by Heat and by Wet Processes, for the Ores of all the principal Metals, of Gold and Silver Coins and Alloys, and of Coal, etc. By OSCAR M. LIEBEB. 12mo., cloth $1 25 T OVE. THE ART OF DYEING, CLEANING, SCOURING, AND *-* FINISHING : On the most approved English and French methods ; being Practical Instructions in Dyeing Silks, Woollens, and Cottons, Feathers, Chips, Straw, etc.; Scouring and Cleaning Bed and Window Curtains, Carpets, Rugs, etc.; French and English Cleaning, etc. By THOMAS LOVE. Second American Edition, to which are added General Instructions for the Use of Aniline Colors. 8vo. . 5 00 M M HENRY CAREY BAIRD'S CATALOGUE. 17 AIN AND BROWN. QUESTIONS ON SUBJECTS CONNECTED WITH THE MARINE STEAM-ENGINE : And Examination Papers; with Hints for their Solution. By THOMAS J. MAI.V, Professor of Mathematics, Royal Naval College, and THOMAS BROWN, Chief Engineer, R. 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Illustrated. 12mo. cloth .... ..... $2 25 jtfORRIS. A HAND-BOOK FOB LOCOMOTIVE ENGINEERS AND - 1 ' MACHINISTS: Comprising the Proportions and Calculations for Constructing Locomotives ; Manner of Setting Valves ; Tables of Squares, Cubes, Areas, etc. etc. By SEPTIMUS NORRIS, Civil and Me- chanical Engineer. New edition. Illustrated, 12mo., cloth $2 00 MTSTROM. ON TECHNOLOGICAL EDUCATION AND THE iN CONSTRUCTION OF SHIPS AND SCREW PROPELLERS: For Naval and Marine Engineers. By JOHN W. NTSTROM, late Acting Chief Engineer U. S. N. Second edition, revised with additional matter. Illustrated by seven engravings. 12mo. $2 50 Q'NEILL. A DICTIONARY OF DYEING AND CALICO PRINT- U ING: Containing a brief account of all the Substances and Processes in use in the Art of Dyeing and Printing Textile Fabrics : with Prac- tical Receipts and Scientific Information. By CHAHLES O'NEILL, Analytical Chemist ; Fellow of the Chemical Society of London ; Member of the Literary and Philosophical Society of Manchester ; Author of " Chemistry of Calico Printing and Dyeing." To which is added An Essay on Coal Tar Colors and their Application To HENRY CAREY BAIRD'S CATALOGUE. 19 Dyeing and Calico Printing. By A. A. FESQCET, Chemist and Engineer. With an Appendix on Dyeing and Calico Printing, as shown at the Exposition of 1S67, from the Reports of the Interna. tional Jury, etc. In one volume Svo., 491 pages . . $6 00 QSBOSN. THE METALLURGY OF ISON AND STEEL : Theoretical and Practical : In all its Branches ; With Special Re- ference to American Materials and Processes. By H. S. OSBORN, LL. D., Professor of Mining and Metallurgy in Lafayette College, Easton, Pa. Illustrated by 230 Engravings on Wood, and 6 Folding Plates. 8vo., 972 pages $10 00 QSBOEN. AMEEICAN MINES AND MINING : U Theoretically and Practically Considered. By Prof. H. S. Os- BORX, Illustrated by numerous engravings. Svo. (In. preparation.) pAINTER, GILDES, AND VAENISHEE'S COMPANION : Containing Rules and Regulations in everything relating to the Arts of Painting, Gilding, Varnishing, and Glass Staining, with numerous useful and valuable Receipts; Tests for the Detection of Adulterations in Oils and Colors, and a statement of the Dis- eases and Accidents to which Painters, Gilders, and Varnishers are particularly liable, with the simplest methods of Prevention and Remedy. With Directions for Graining, Marbling, Sign Writ- ing, and Gilding on Glass. To which are added COMPLETE INSTRUC- TIONS FOR COACU PAINTING AND VARNISHING. 12mo., cloth, $1 50 pALLETT. THE MILLEE'S, MLLLWEIGHT'S, AND ENGI- * NEES'S GUIDE. By HENRY- PALLETT. Illustrated. In one vol. 12mo. . $3 00 pEBKINS. GAS AND VENTILATION. Practical Treatise on Gas and Ventilation. With Special Relation to Illuminating, Heating, and Cooking by Gas." Including Scien- tific Helps to Engineer-students and others. With illustrated Diagrams. Bv E. E. PERKINS. 12mo., cloth . . . $1 25 "DEEXINS AND STOWE. A NEW GUIDE TO THE SHEET-LEON r AND BOILEE PLATE SOLLEE: Containing a Series of Tables showing the Weight of Slabs and Piles to Produce Boiler Plates, and of the Weight of Piles and the Sizes of Bars to Produce Sheet-iron ; the Thickness of the Bar Gauge in Decimals ; the Weight per foot, and the Thickness on the Bar or Wire Gauge of the fractional parti of an inch; the Weight per sheet, and the Thickness on the Wire Gauge of Sheet- iron of various dimensions to weigh 112 Ibs. per bundle; and the conversion of Short Weight into Long Weight, and Long Weight into Short. Estimated and collected by G. H. PERKINS and J. G- STOWK . . .... $259 20 HENRY CAREY BAIRD'S CATALOGUE. pHILLIPS AND DARLINGTON. RECORDS OF MINING AND * METALLURGY : Or, Facts and Memoranda for the use of the Mine Agent and Smelter. By J. ARTHUR PHILLIPS, Mining Engineer, Graduate of the Imperial School of Mines, France, etc., and JOIIN DARLINGTON. Illustrated by numerous engravings. In one vol. 12mo. . $2 00 pRADAL, MALEPEYRE, AND DUSSAUCE. A COMPLETE * TREATISE ON PERFUMERY: Containing notices of the Raw Material used in the Ait, and the Best Formulae. According to the most approved Methods followed in France, England, and the United States. By M. P. PRADAL, Perfumer-Chemist, and M. F. MALEPEYRE. Translated from the French, with extensive additions, by Prof. H. DUSSAUCE. 8vo. $10 pROTEAUX. PRACTICAL GUIDE FOR THE MANUFACTURE * OF PAPER AND BOARDS. By A. PROTEAUX, Civil Engineer, and Graduate of the School of Arts and Manufactures, Director of Thiers's Paper Mill, 'Puy-de- Dome. With additions, by L. S. LE NORMAND. Translated from the French, with Notes, by HORATIO PAINE, A. B., M. D. To which is added a Chapter on the Manufacture of Paper from Wood in the United States, by HENRY T. BROWN, of the "American Artisan." Illustrated by six plates, containing Drawings of Raw Materials, Machinery, Plans of Paper-Mills, etc. etc. 8vo. $5 00 DEGNAULT. ELEMENTS OF CHEMISTRY. By M. V. REGNAULT. Translated from the French by T. FOR- REST BENTON, M. L., and edited, with notes, by JAMES C. BOOTH, Melter and Refiner U. S. Mint, and WM. L. FABER, Metallurgist and Mining Engineer. Illustrated by nearly 700 wood engravings. Comprising nearly 1500 pages. In two vols. 8vo., cloth $10 00 TDEID. A PRACTICAL TREATISE ON THE MANUFACIURE OF " PORTLAND CEMENT: By HENRY REID, C. E. To which is added a Translation of M. A. Lipowitz's Work, describing anew method adopted in Germany of Manufacturing that Cement. By W. F. REID. Illustrated by plates and wood engravings. 8vo. . . . . . $7 00 TDIFFAULT, VERGNAUD, AND TOUSSAINT. A PRACTICAL 11 TREATISE ON THE MANUFACTURE OF COLORS FOR PAINTING: Containing the best Formulae and the Processes the Newest and in most General Use. By MM. RIFFAULT, VERGXAUD, and TOUS- SAINT. Revised and Edited by M. F. MALEPEYRE and Dr. E>rn, WINCKLER. Illustrated by Engravings. In one vol. 8vo. (In 1 ffepa ration.) HENRY CAREY BAIRD'S CATALOGUE. 21 TDIFFAULF, VERSNAUD, AND TOUSSAINT. A PRACTICAL TREATISE ON THE MANUFACTURE OF VARNISHES: By MM. RIFFACLT, VERGXAUD, and TOUSSAIXT. Revised and Edited by M. F. MALEPEYRE and Dr. EMIL WINCKLER. Illus- trated. In one vol. 8vo. (In preparation.) gHUNK. A PRACTICAL TREATISE ON RAILWAY CUEVES AND LOCATION, FOR YOUNG ENGINEERS. By WH. F. SHTJSK, Civil Engineer. 12mo., tucks . . $2 00 OMEATON. BUILDEE'S POCKET COMPANION: Containing the Elements of Building, Surveying, and Architec. ture ; with Practical Rules and Instructions connected with the suh- ject. By A. C. SMEATOX, Civil Engineer, etc. In one volume, 12mo ........ .. . . $1 50 THE DYER'S INSTEUCTOR: Comprising Practical Instructions in the Art of Dyeing Silk, Cot- ton, Wool, and Worsted, and Woollen Goods: containing nearly 800 Receipts. To which is added a Treatise on the Art of Pad- ding; and the Printing of Silk Warps, Skeins, and Handkerchiefs, and the various Mordants and Colors for the different styles of such work. By DAVID SMITH, Pattern Dyer, 12mo., cloth $3 00 OMITH. THE PRACTICAL DYEE'S GUIDE: Comprising Practical Instructions in the Dyeing of Shot Cobourgs, Silk Striped Orleans, Colored Orleans from Black Warps, ditto from White Warps, Colored Cobourgs from White Warps, Merinos, Yarns, Woollen Cloths, etc. Containing nearly 300 Receipts, to most of which a Dyed Pattern is annexed. Also, a Treatise on the Art of Padding. By DAVID SMITH. In one vol. 8vo. $25 00 QIHAW. CIVIL AECHITECTUEE : Being a Complete Theoretical and Practical System of Building, containing the Fundamental Principles of the Art. By EDTV vnn SHAW, Architect. To which is added a Treatise on Gothic Ar hi- tectnre, Ac. By THOMAS W. SILLOWAY and GEORGE M. HARD- ING , Architects. The whole illustrated by 102 quarto plates finely engraved on copper. Eleventh Edition. 4to. Cloth. $10 00 OLOAN AMEEICAN HOUSES: A variety of Original Designs for Rural Buildings. Illustrated by 26 colored Engravings, with Descriptive References. By SAMUEL SLOAN, Architect, author of the " Model Architect," etc. etc. 8vo. $2 50 OCHINZ. RESEAECHES ON THE ACTION OF THE BLAST. FUENACE. By CHAS. SCHINZ. Seven plates. 12mo. . . . $4 25 22 HENRY CAREY BAIRD'S CATALOGUE. OlMITH. PARKS AND PLEASURE GROUNDS : Or, Practical Notes on Country Residences, Villas, Public Parks, and Gardens. By CHARLES H. J. SMITH, Landscape Gardener and Garden Architect, etc. etc. 12mo. . . . . $2 25 STOKES. 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In 1 vol. quarto . $10 00 OULLIVAN. PROTECTION TO NATIVE INDUSTRY. By Sir EDWARD SULLIVAN, Baronet. (1870.) 8vo. . $1 50 mABLES SHOWING THE WEIGHT OF ROUND, SQUARE, AND A FLAT BAR IRON, STEEL, ETC. By Measurement. Cloth ...... 63 rPAYLOR. STATISTICS OF COAL: Including Mineral Bituminous Substances employe! in Arts and Manufactures ; with their Geographical, Geological, and Commer- cial Distribution and amount of Production and Consumption on the American Continent. With Incidental Statistics of the Iron Manufacture. By R. C. TAYLOR. Second edition, revised by S. S. HALDEMAN. Illustrated by five Maps and many wood engrav- ings. 8vo., cloth $6 00 rpEMPLETON. THE PRACTICAL EXAMINATOR ON STEAM * AND THE STEAM-ENGINE : With Instructive References relative thereto, for the Use of Engi- neers, Students, and others. By WM. TEMPLETON, Engineer ]2mo. $1 25 HENRY CARET BAIRD'S CATALOGUE. 23 rpHOMAS. THE MODERN PRACTICE OF PHOTOGRAPHY. * By R. W. THOMAS, F. C. S. 8vo., cloth . ... 75 fPHOMSON. FREIGHT CHARGES CALCULATOR. By ANDREW THOMSON, Freight Agent . . . . $1 25 JIURNING : SPECIMENS OF FANCY TURNING EXECUTED ON * THE HAND OR FOOT LATHE : With Geometric, Oval, and Eccentric Chucks, and Elliptical Cut- ting Frame. By an Amateur. Illustrated by 30 exquisite Pho- tographs. 4to $3 00 BURNER'S (THE) COMPANION: Containing Instructions in Concentric, Elliptic, and Eccentric Turning; also various Plates of Chucks, Tools, and Instru- ments ; and Directions for using the Eccentric Cutter, Drill, Vertical Cutter, and Circular Rest; with Patterns and Instruc- tions for working them. A new edition in 1 vol. ]2mo. $1 50 TTRBIN BRULL. A PRACTICAL GUIDE FOR PUDDLING U IRON AND STEEL. By ED. URBIN, Engineer of Arts and Manufactures. A Prize Essny read before the Association of Engineers, Graduate of the School of Mines, of Liege, Belgium, at the Meeting of 1865-6. To which is added a COMPARISON OF THE RESISTING PROPERTIES OF IRON AND STEEL. By A. BRULL. Translated from the French by A. 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In which are ex- plained, in an easy and familiar manner, the various Methods of Manufacturing every description of Raw and Refined Sugar Goods, as sold by Confectioners and others . . . $2 00 XL. TABLES FOR QUALITATIVE CHEMICAL ANALYSIS. By Prof. HEINRICH WILL, of Giessen, Germany. Seventh edi- tion. Translated by CHARLES F. HIMES, Ph. D., Professor of Natural Science, Dickinson College, Carlisle, Pa. . . $1 25 -TTrriLLIAMS. ON HEAT AND STEAM : Embracing New Views of Vaporization, Condensation, and Expan- sion. By CHARLES WYE WILLIAMS, A. I. C. E. Illustrated. 8vo. $3 50 yyORSSAM. ON MECHANICAL SAWS: From the Transactions of the Society of Engineers, 1867. By 6. W. WORSSAM, Jr. Illustrated by 18 large folding plates. 8vo. $5 00 WOHLER. A HAND-BOOK OF MINERAL ANALYSIS. " By F. WOHLKR. Edited by II. B. NASON, Professor of Chemistry, Rensselaer Institute, Troy, N. Y. With numerous Illustrations. 12mo. ...... $3 00 W 1 LlbKAKJL A 000 661 934