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TREATISE ON PRACTICAL MENSURATION, IN TEN PARTS; CONTAINING The most approved lVIethods of drawing Geometrical Figures; Mensuration of Superficies ; Land~ Surveying; lifensuration of Solids; the Use of'the Carpenter’s Rule; Timber Measure, in which is shown the Method of measur- , ing and valuing standing Timber; Artificers’ Works, illustrated by the Dimensions and Contents of a House; Mensuration of Hay-A stacks, D133... , Canals, Marlpits, Ponds, lVIill-dams, Embank- ments, Quarries, Coal-heaps, and Clay—heaps; Conic Sections and their Solids; the most useful Problems in Gauging, according to the New Imperial lVIeasures 3 Plane Trigonometry with its application to the Mensuration of Heights and Distances; Trigonometrical Surveys; anda Dictionary of the Terms used in Architecture. BY A. NESBIT, BIASTER OF THE CLASSICAL, COMMERCIAL, AND MATHEMATICAL, ACADEIVIY, KENNINGTON LANE, LABIBETH, LONDON; AND AUTHOR or “ A Treatise on Practical Arithmetic; ” “ A complete Treatise on Practical Land-Surveying ; " “ A Treatise on Practical Gauging ; ” “ Kc s to the Arithmetic, Mcnsuration, and Gauging; ” “ An Introduction to Parsmg, adapted to Murray’s Grammar ;” “ An Essay on Education ; ” &c. &c. NE‘V EDITION, ENLARGED, AND GREATLY IMPROVED. TO WHICH IS ADDED, A TREATISE 0N LEVELLING. The whole illustrated by nearly seven hundred Practical Examples, and nearly three hundred Woodcuts. LONDON: LONGMAN. BROWN, GREEN, AND LONGMANS, PATERNOSTER-ROW. 1850. Price Six Shillings, bound. LONDON : SPO'X'TISWOODES and SHAW, hew-streebSquare. T0 ELLIS CUNLIFFE LISTER, ESQ, MJE’. 0F MANNINGHAM-IIOUSE, Near Bradford, AS AN ACKNOIVLEDGMENT OF GRATITUDE fOR NUMEROUS FAVOURS RECEIVED, AND AS A TOKEN 0F ESTEEM FOR HIS GENERAL KNOWLEDGE 0F SCIENTIFIC SUBJECTS, THIS TREATISE 0N PRACTICAL MENSURATION, ‘ In all its Departments, IS MOST HUMBLY AND RESPECTFULLY DEDICATED, BY HIS MUCH OBLIGED, AND MOST OBEDIENT SERVANT, A. NE SBIT. AZ ADVERTISEMENT TO THE TWELFTH EDITION. IN revising the present Edition of this work, the Proprietors have spared neither labour nor expense in order to render it as valuable as possible; and for this purpose they have engaged competent persons, who have made several important and necessary im— provements. They have commenced with the Geo- metrical Definitions, Problems, and Theorems, by arranging the subjects in '1 more concise, elegant, and systematic order, the Geometrical Theorems being much improved and rendered more complete by prefixing a description to the respective theorems. In the Mensuration of Superficies the same order of arrangement has been strictly regarded, and several additional and important notes have been added to the respective problems. On the subject of Land Surveying, great improvements have been made, by introducing a simple method of surveying Without the use of the cross-staff. Similar arrangements and improvements have been observed throughout the Mensuration of Solids, Conic Sections, and their Solids, and the other parts of the book. Thus a large A 3 vi ADVERTISEMENT. saving of space has been effected, enabling the Pro- prietors to add a complete Treatise on Trigonometry without increasing the size of the book. They have, therefore, great pleasure in introducing the present Edition, believing it to be the cheapest and the best school-book on the subject that has ever been pre- sented to the Public. In this Edition two new Parts have been intro- duced. Part the Ninth treats of Plane and Oblique Trigonometry, with a variety of illustrations and practical examples. It will be seen, this subject is treated, upon an entirely new, and improved method, by substituting Arithmetical Complements in the Logarithmic Numbers. Part the Tenth treats of the Application of Plane Trigonometry to the Mensuration of Heights and Distances in a variety of useful questions; the me- thod of finding the Heights of Mountains by means of the Barometer and Thermometer; also numerous examples for finding the magnitude and dimensions of the earth; the vibration of pendulums in different latitudes, on the length of arcs of different meridians, Aid a great variety of other important matter. Lennon, 1844. PREFACE. VARIOUS have been the conjectures concerning the origin of Geometry or Mensuration; but as it is a Science of general utility, there can, I think, be little doubt that its existence is nearly coeval with the creation of Man. Indeed I can see no reasonable objection why we may not attribute its invention to our first parent Adam; espe- cially as we are informed in Holy lVrit, that his son Cain built a city; to do which, it is evident, would require some knowledge of a measuring unit, which is the first principle of Mensuration. By the same infallible testi- mony, we find that the Arts and Sciences were cultivated to a consider. ble extent long before the Flood. J ubal was the father of all such as handled the harp and organ ; and Tubal-cain, an instructor of every artificer‘ in brass and iron. It is also more than probable that Noah was well acquainted with the Art of Mensuration, as practised in his day; for it does not appear that he found any dif- ficulty in building the Ark, which consisted of three stories, according to certain dimensions given him by the Lord of the Universe. Diodorus, Herodotus, and Strabo, are of opinion that the Science of Mensuration had its rise among the Egyp- tians ; whom they represent as constrained, on account of the removal or defacing of the land-marks by the annuab inundation of the Nile, to devise some method of ascer- taining the ancient boundaries, after the waters had re— tired. By Josephus, however, the invention is ascribed to the Hebrews. This writer asserts that the Arts and Sciences, for which Egypt was so long famous, were carried into that country, by the Patriarch Abraham, from Ur of the Chaldees; but as Egypt was peopled by the descendants of Ham, is it not more probable that they had the rudiments of their Sciences originally from their father ? Be this as it may, it is well known that Egypt was, for A 4 viii PREFACE. many ages, the mother and nurse of the Arts and Sciences. From this country they were conveyed into Greece by ' Thales, about 600 years before the Christian JEra. This Philosopher, after travelling into Egypt, and studying, under its sages, Astronomy, Geometry, and other branches of the Mathematics, returned to his own country, and em- ployed himself in communicating the knowledge which he had acquired. The great utility of Mensuration, without which it is im- possible rightly to conduct the affairs of civilized life, in- duced many of the most celebrated Philosophers and Mathematicians of antiquity to study its principles ; and to Thales, Anaxagoras, Pythagoras, Socrates, Plato, Apollonius, Philo, Ptolemy, Aristotle, Euclid, Archi- medes, 850. we are indebted for many substantial im— provements. The moderns, likewise, have not been less solicitous to enrich this Science, than the ancients; ac- cordingly Huygens, Wallis, Gregory, Halley, Euler, Leibnitz, the Bernouilles, Vieta, Metius, Van Ceulen, Barrow, Newton, Sharp, Machin, Moss, Leadbetter, Simpson, Emerson, I-Iolliday, Fletcher, Robertson, Hutton, Bonnycastle, Keith, Beckett, 8w. have greatly improved it by their labours. After so many eminent men have written more or less upon this Science, it may perhaps be thought presumption in me to attempt to add any thing to its stores; but as I can say, Without arrogance, that I have had considerable experience in the Practical Part of Mensuration, in all its Departments, I am persuaded that this work will be found to contain many things both new and useful. With regard to the Rules, indeed, nothing new can be expected ; as they are to be found, with very little varia- tion, in every modern Treatise on l‘vIensuration; but the Questions, which amount to five hundred and eighty, are almost wholly new ones; and a great number of them have been made from actual admeasurements. I have likewise given copious directions for taking dimensions, which art certainly forms a very essential part of Mensur- ation; for if the dimensions be improperly taken, the re- sults must, of course, be incorrect. The work is divided into Ten Parts ; and some of these Parts are again subdivided into Sections. Prat the First contains Practical Geometry, and a few of the most useful Geometrical Theorems; most of which PREFACE. ix are employed in solving Questions in this Work. The Theorems are not demonstrated; but references are given to the elements of Euclid, Simpson, and Emerson, where their demonstrations may be found. Part the Second contains Mensuration of Superficies. In the last Problem of this Part, I have given the inva- luable Rule for finding the areas of curvilineal figures, by means of equi-distant ordinates. This Rule was first de- monstrated by the illustrious Newton; but it is to Mr. Thomas Moss that we owe its present simplicity. Part the Third is divided into two Sections. In the first are given the Methods of surveying and planning single Fields, Woods, Roads, and Rivers ; and also Rules and Directions for Parting off, and Dividing Land. It likewise exhibits four of the most approved Methods of surveying large Estates, illustrated by three distinct Plans, and Field-Books. . Sometimes one of these Methods claims the preference, and sometimes another, according to the different forms of Estates; but they are all approved of by our best Land Surveyors ; and are more accurate and practical than any others, that have come to my knowledge. This Section will be found to contain every thing that is necessary for persons, in general, to know of Surveying. In order to become a complete Surveyor, it is requisite not only to study Works written professedly on the sub- ject, but also to have a considerable portion of Field-prac- tice, under the direction of an able Tutor. The second Section contains a collection of useful Questions concerning Superficial Mensuration, which will serve to exercise the ingenuity of the Learner; and to prove his knowledge of the Theorems and Rules given in the first three Parts. Part the Fourth is divided into four Sections. The first contains the Mensuration of Solids; the second, the De- scription and Use of the Carpenter’s Rule; and the third, Timber Measure. The last Problem of this Section contains Rules and Directions for measuring and valuing standing Timber; many of which were never before published. In this part of the Work, I have been assisted by Mr. Joseph Webster, of Farnley, near Leeds; who has been many years very extensively employed, as a valuer of tim- ber, by the Earl of Cardigan, Lord Mexborough, 85c. 8m. A 5 X PREFACE. In order to render this Problem as useful as possible, I have given a description of Timber Trees, and pointed out the purposes for which their wood is best adapted; for it is impossible to become a valuer of timber without being made acquainted with the properties of trees. The fourth Section contains Miscellaneous Questions concerning the Mensuration of Solids. Part the Fifth treats of the Method of measuring the Works of Artificers ; viz. Bricklayers, Masons, Carpen- ters and J oiners, Slaters and Tilers, Plasterers, Painters, Glaziers, Plumbers, and Pavers. The directions in this Part, for taking the dimensions, making the deductions, 8:0. will be found to be very co- pious. For some of these I am indebted to Mr. Benjamin Jackson, senior, an able and experienced Architect in Leeds ; and to Mr. Joseph Brooke, Teacher of the Mathe- matics, at lVortley, near Leeds, who has had much ex- perience in measuring the VVOI‘kS 0f Artificers. This Part is concluded with a general Illustration, con- taining the dimensions of a House; and exhibiting the methods of ruling the Book, entering the dimensions with the contents ; and forming the bills for workmanship and materials. Part the Sixth treats of the Mensuration of Haystacks, Drains, Canals, Marlpits, Embankments, Ponds, Mill- dams, Quarries, Coal—heaps, Clay-heaps, and other irre- gular figures, by means of equi-distant, parallel sections, founded upon the method of cqui-distant ordinates. This method of finding the contents of irregular figures is pointed out by Dr. Hutton, in his valuable Treatise on Mensuration, octavoie, page 375 ; but Mr. Joseph Beckett appears to have been the first who has applied it with any degree of success. This Part also contains the method of measuring the circular Ponds made upon the W'olds in Yorkshire. This was communicated to me by the Rev. W. Putsey, Master of the Classical, Commercial, and Mathematical Academy, Pickering. In order to give the young Reader an idea of the great improvements made in Agriculture and Commerce, by means of Drains and Canals; and also to make him ac- quainted with some of the stupendous “forks which have been accomplished by the ingenuity, perseverance, and "‘ Price 183. in boards. PREFACE. 11 combined efforts of men; I have concluded this Part with a description of a few of the principal Canals in England, Scotland, France, and China; and with an account of some of the chief Drainages which have been executed in the counties of York and Lincoln. Part the Seventh treats of Conic Sections and their Solids. It also contains a few of the leading properties of the Ellipse, the Parabola, and the Hyperbola. Those who desire more information on this subject, may consult Simpson’s, Emerson’s, and Hutton’s Conic Sections. Part the Eighth displays the method of gauging all kinds of open vessels used by Maltsters, Brewers, Dis- tillers, Wine Merchants, Victuallers, 8m. 8m. In this Part I have applied the method of equi-distant ordinates or sections, to the gauging of vessels whose sides are curved ; such as coppers, stills, 8m. 8m. This I have not seen in any other Work. I have also given the process of gauging and inching a guile-tun, according to the method practised in the Excise. Malt-gauging, and Cask-gauging are likewise treated of, in this Part; and it is concluded with a few Miscellaneous Examples. Part ninth treats of Plain and Oblique Trigonometry. Part the tenth treats on the Application of Trigono- metry to Heights and Distances, with a great variety of useful Problems, and subjects on the Nature, Figure, and Magnitude of the Earth, and of Levelling. In order '15 render the Treatise as useful as possible, I have given a Dictionary containing an explanation of the most general terms made use of in Architecture. This will be found a valuable addition by Bricklayers, Masons, Joiners, and every other person concerned in measuring or building. The Work is brought to a conclusion by a number of questions to be answered verbally by the Pupil; for nothing will tend more to make him an adopt in lVIen— suration, than committing the Definitions and Rules to Memory. Nearly the whole of the Rules upon which Mensuration depends, require Algebra or Fluxions to demonstrate them ; but as these Sciences are too abstruse and sublime to be comprehended by the generality of those who are concerned in Mensuration, it has been thought advisable to give the Rules without the Demonstrations. Those A 6 xii PREFACE. who desire to penetrate the deepest recesses of scientific knowledge, are referred to Simpson’s, Emerson’s and H01- liday’s Fluxions, and to Doctor Hutton’s Mensuration; in one or other of which Works, all the Rules given in this Treatise are demonstrated. The Demonstrations would also have swelled the Work very considerably, both in size and price; but by with- holding them, space has been obtained for a greater body of useful matter, on the subject of Mensuration, than is to be found in any other Work, of equal size, with which I am acquainted. I should be wanting in respect, if I concluded this Pre- face without paying a tribute to the memory of that pro- found and indefatigable Mathematician, the late Mr. John Ryley of Leeds ;—— a man who has enriched almost every periodical, mathematical publication, for nearly half a century, with problems and demonstrations in science, which have struck many of his contemporaries with un- mixed admiration of his genius and attainments. To him I submitted the Plan of this Work, which, I am happy to say, received his entire approbation ; and I flatter myself that it will be generally approved by the Public. It now only remains for me to solicit the indulgence of the candid Reader; for in solving so large a number of new questions as this Work contains, it is almost impos- sible to avoid errors. These, however, it is hoped will be found to be few, and of little moment ; for much care has been employed in working the questions, and correcting the press; and I may add, that neither labour nor ex- pense has been withheld in order to produCe a Work of general Utility. ANTHONY NESBIT. LONDON, Nov. 1841. P. S. A KEY to the Mensuration, containing Solutions to all the Questions which are not solved in that Work, has been published for the use of Teachers and private Students. This will be found of great advantage, as it will render it quite unnecessary for the Tutor to inspect every operation of his Pupil; and will enable those who may not have had an opportunity of acquiring a knowledge of llIensuration in their Youth, and who cannot now have the advantage of a living Instructor, to pursue their pri- vate studies with ultimate success, if they apply them« selves with diligence and perseverance. ADVERTISEMENT TO THE FOURTH EDITION. THE Author has carefully revised and corrected this Edition, and made, in different parts of the Work, several important and useful Additions and Improvements. Part the Eighth, on Gauging, has been entirely re- written, and much improved by the addition of New Matter; and all the Rules and Examples have been adapted to the New Imperial Standard Gallon and Bushel. . The Author has calculated Three New Tables, which exhibit the number of Cubic Inches in the different de— nominations of Wine and Spirit Measure, Ale and Beer Measure, and Corn or Dry Measure, according to the New Imperial Standard Gallon and Bushel; and also given Rules and Directions for reducing the Old English, Irish, and Scotch Measures to the New Imperial Measures; and vice versé. He has also shewn the method of finding the New Im- perial Multipliers, Divisors, and Gauge-Points ; and given .a Table in which they are all arranged in a convenient order. ‘ He has likewise described the New Imperial Sliding Rule, and Diagonal Rod; given particular and general Rules for Cask Gauging; and also calculated a New Table of Multipliers, by which the Content of any Cask may be easily found, in Imperial Gallons. These Alterations and Additions will be found very great improvements to the ‘Nork ; and will tend to secure and increase that popularity and patronage which it has already obtained among Teachers and the Public; as it now contains such a body of Practical Information, on Mensuration in all its Departments, as cannot be found in any other Work on the same Subject. A. NESBIT. ADVERTISEMENT. Ma. NESBIT most respectfully informs his Friends and the Public, that he has removed from Manchester; and in Conjunction with Two of his SONS, has opened a School in Lambeth, near London. —-T hey have Excellent Accommodation for Twenty Boarders; and have erected a Large and Commodious School Room, which is well lighted and ventilated, and fitted up in the most convenient Manner. The Premises are in an airy and a healthful Situation, at the Dis- tance of One Mile from Westminster Bridge, and a Mile and a half from London Bridge; and the Pupils haye the Benefit of an Ex- cellent Play—ground, for their Recreation. The Vicinity of London offers many Advantages for the Education of Youth. -——VVe have the Parks and other Pleasure Grounds, for Recreation; the British Museum, the Adelaide Gallery, the Poly- technic Institution, the Royal Academy of Arts, the National Gal- lery, the Royal Institution of Great Britain, &c. &c., for Inspection and Improvement; and at no great Distance, we have Greenwich, VVoolwich, Hampton Court, Richmond, Windsor, &c. &c. for Holi- day Excursions. Indeed, it is impossible to mention all the Sources of Delight, Amusement, and Improvement, that are connected with London and its Vicinity. The Terms of the Establishment, and other Particulars, may be known on Application ; and the System of Instruction is fully deve- loped in “ An Essay on Education,” lately published by Mr. Nesbit and Sons. fig? IVIit. NESBIT has been succeeded in Manchester, by his Eldest Son, Mr. Jacob Nesbit, who has Excellent Accommodation for a limited Number of Young Ladies, as Boarders. CONTENTS. PART I. GEOMETRICAL DEFINITIONS GEOMETRICAL PROBLEMS - - - - To bisect a given line - - - To bisect a given angle - ~ To draw a line parallel to a given line -- - To erect a perpendicular from a given point, in a given line - — - - - From a given point, to let fall a perpendicular upon a given line - - - - - To find the centre of a given circle, or one already described - - — - — To make a triangle with three given lines - - Having given the base, the perpendicular, and the place of the perpendicular upon the base, to con- struct a triangle - - — - - To describe a square, whose side shall be equal to a given line - - - - - To describe a rectangle, whose length and breadth shall be equal to two given lines - - - Upon a given right line to construct a regular rhom- bus - - — - — - :To construct an irregular rhombus, having given its side and perpendicular height - - - Having any two right lines given, to construct a regular rhomboid - — - - Having given the base, the perpendicular, and the place of the perpendicular upon the base, to con- struct an irregular rhomboid — - - Having the transverse and conjugate diameters given, to construct an ellipse - - - - Upon a given line, to make a regular pentagon - Upon a given line, to make a regular hexagon - u I - 1 *‘1 W com 00‘ co quash-d3 10 10 11 ll 11 12 xvi CONTENTS. Page Upon a given line, to construct a regular octagon - 14 In a given triangle, to inscribe a circle - - 14 About a given triangle, to circumscribe a circle; or to describe the circumference of a circle through three given points - - - - 14 To make a triangle equal to a given trapezium - 15 To make a right angle by the line of chords on the plane-scale - - - - - 15 To make an acute angle - - - - 15 To make an obtuse angle - - - - 15 To find the number of degrees contained in any given angle - - - - - - 16 Upon a given line, to make a regular polygon of any proposed number of sides - - - 16 In any given circle, to inscribe a regular polygon of any proposed number of sides; or to divide the circumference into any number of equal parts - 16 To find a mean proportional between two given lines 17 To raise a perpendicular from any point, in a given line, by a scale of equal parts - - - 17 Given the span or chord line, and height or versed sine of the arch of a bridge or cellar, to find the radius of the circle'that will strike the arch - 17 GEOMETRICAL THEOREMS - - - - 18 An explanation of the principal Mathematical Cha— racters - - - — - - 23 PART II. MENSURATION OF SUPERFICIES - — - 25 A Table of lineal measure - - — — 25 A Table of Square measure - - - 26 To find the area of a square - - - 26 of a rectangle - - - 27 of a rhombus or rhomboides - 29 of a triangle, when the base and perpendicular are given - _ - — 30 of a triangle, the three sides only of which are given - - - - - 31 Any two sides of a right-angled triangle being given, to find the third side - - — - 33 To find the area of a trapezium - — ‘ - ~35 of a trapezoid - - - 36 CONTENTS. To find the a1 ea of an irregular polygon of any number of sides - - - Given the side of a regular polygon, to find the radius of its inscribed 01 cir cumsmibing ch 010 - To find the area of a regular polygon - - —-———— of a regular polygon, when the side only is given - - - - - The diameter of a circle being given, to find the cir- cumference ; or, the circumference being given, to find the diameter - - - - To find the length of any are of a circle - - To find the area of a circle - - - - ———-— of a sector of a circle - - of a segment of a circle . of a circular zone - - ‘ - of a circular ring - - - -—-————-— of a lune — - - - —— of an ellipse — - - of an elliptical segment - - of any curvilineal figure, by means of equi-distant ordinates - - — - A Table of circular segments - - - PART III. SECTION I. LAND-SURVEYING - - - - - Miscellaneous Instructions - - - To measure a field of three sides ~ - - —— a field of four sides — - - a field of more than four sides — an irregular, narrow piece of land - a mere or wood - - - To measure and plan roads, rivers, canals, 8:0. - To find the breadth of a river - i- - To measure a line upon which there is an impedi- ment - - - - - - To part from a rectangular field, any proposed quan- tity of land, by a line parallel to one of its sides - To part from a trapezium or any irregular polygon Whatever, any proposed quantity of land - - To reduce statute-measure to customal y, and vice cmsa - - - - - - xviii CONTENTS. Page To survey and plan Estates or Lordships - - 94 Method I. - — - - - - 94 II. - - - - - - 99 III. - - - - - 100 IV. - - - - - - 100 Miscellaneous Instructions - - - - 101 A Book of Dimensions, Castings, 8w. belonging to Plate III. .- '- — — - 104 Method of surveying single fields without the cross- stafi‘ - — - - - - 109 SECTION II. Miscellaneous Questions concerning Superficial Men- suration - - - - - - 117 PART IV. SECTION I. MENSURATION or SOLIDS - - - - 121 Definitions - - - — - - 121 A Table of solid measure - - - - 124 T 0 find the solidity and surface of a cube - - 124 the solidity and surface of a parallelopipedon 126 the solidity and surface of a prism - - 127 the solidity and surface of a cylinder - 129 the solidity and surface of a pyramid - 130 the solidity and surface of the frustum of a pyramid - - - ‘- — - 131 the solidity and surface of a cone - - 132 the solidity and surface of the frustum of a cone — - - - - - 133 the solidity and surface of a cuneus or wedge 135 the solidity and surface of a prismoid - 136 the solidity and convex surface of a sphere or globe - - - - - - 138 the solidity and surface of the segment of a sphere - - - - - - 138 the solidity and convex surface of the frus- tum of a sphere - - - - - 139 the solidity and surface of a circular spindle 140 the solidity of the middle frustum or zone- of a circular spindle - - - - - 141 the solidity and surface of a cylindrical ring 142 the solidity or superficies of any regular body 143 CONTENTS. To find the solidity of an irregular solid - - the magnitude or solidity of a body from its weight - - ‘ - - - - A Table of the specific gravities of bodies - - To find the weight of a body from its magnitude or solidity - - - - - - SECTION II. Description of the Carpenter’s Rule - - The Use of the Sliding Rule - - - Multiplication by the sliding rule — - — Division by the sliding rule - : - To square any number - - — -. To extract the square root of any number - - To find a mean proportional between two numbers - a fourth proportional to three numbers; or to perform the Rule of Three — - - the areas of plane figures by the sliding rule TIMBER MEASURE — - - - - To find the superficial content of a board or plank - the solidity of squared or four-sided timber SECTION 111. To find the solidity of round or unsquared timber - A Table for measuring timber - - - To measure and value standing timber - - Miscellaneous Instructions for finding the dimensions of standing timber, and for setting—out woods, and valuing them previously to a fall — - A Description of Timber Trees - - - The Oak - - - — - - The Ash - . - The Beech and Elm - — - - The Sycamore and Black Poplar - - . - The Alder, VVeymouth Pine, and Fir - - The Birch - - - - _ The Yew - - - - - - The Holly and BOX - - - - SECTION IV. Miscellaneous Questions concerning Solids the contents of solids - - - ‘ xix Page 145 147 147 148 149 150 150 151 151 151 152 152 ' 156 160 160 166 166 167 167 167 168 169 169 169 171 XX CONTENTS. 1 if t1 H PART V. ‘ ' Page i, ARTIFICERS’ 'WORK _ - _ 175 pg Work—Bricklayers - - — - 176 ‘ Masons - — - - - 178 if > A Description of Portland, Yorkshire, and Purbeck 1 Stone, and of various kinds of Marble - - 181 i; Work—Carpenters’ and Joiners’ - - - 182 53* Slaters’ and Tilers’ - - - 188 i Plasterers’ - — - - 189 Painters’ - - - - - 190 : Glaziers’ - - — - _ 192 l, Plumbers’ - ' — - . .. 193 3' PaverS’ ,- - _ - - 194 3-! VAULTED AND ARCHED Roors - - - 195 3, Prob. I. To find the content of the vacuity of a cir- 1, cular, an elliptical, or a gothic vaulted roof - 196 i To find the concave surface of a circular, an elliptic, ‘ or gothic vaulted roOf - - - - 197 the solid content of a dome ; its height, and ‘ the dimensions of its base being given — - 198 the superficial content of a spherical dome - 198 the solid content of a saloon - - 199 the superficial content of a saloon - - 200 the solid content Of the vacuity formed by 2 groin arch, either circular or elliptical — - 200 the concave surface of a circular or an ellip- tical groin - - - - - 201 GENERAL ILLUSTRATION OR THE WORKS or ARTI- FICERS, FROM THE DIMENSIONS OF A HOUSE - 202 THE BRICKLAYER’S YVORH - - - 202 Abstract of the Brick Work - — - 204 The Bricklayer’s Bill - — - - 205 THE MASON’S WORK - - - - 206 Abstract of the Masonry - - - - 208 The Mason’s Bill - - - - - 210 THE CARPENTER’S AND JOINER’s VVORK - - 213 Abstract of the Carpenter’s and J oiner’s Work '- 213 The Carpenter’s and Joiner’s Bill - — — 215 THE SLATER’s NVORK AND BILL - - - 215 THE PLAs-TERER’S WORK — - - 216 Abstract of the Plastering — — — 217 The Plasterer’s Bill - - - - 218 Useful Works, on Architecture, recommended - 218 CONTENTS. PART VI. The Mensu1ation of Hay- Stacks, D1 ains, Canals, :Marl- -pits, Ponds, Mill- dams, Embankments, Qua1- ries, Coal- -heaps, and Clay- heaps - - To measure a hay- -stack, having a circular base - a hay- -stack, having a rectangular base — To find the number of cubic yai ds, which have been dug out of a drain or canal - - - how many cubic ya1ds have been dug out of a marl-pit - - — - - Directions for measuring ponds and mill-dams - for measuring embankments - - for measuring quarries - - . - for measuring coal—heaps - i - for measuring clay-heaps _ - - A description of some of the principal canals in England, Scotland, France, andChina - — j An account of some of the principal drains-in the ' county of York, and the agricultural improve- ments made by them - — - - An account of agricultural improvements made in the county of Lincoln, by means of drains — - xxi Page 220 220 “324 228 233 237 238 239 240 241 249 Mr. Joseph Elkington’s method of draining land - 250 PART VII. Conic Sections and their Solids - - - 252 Definitions — - — - - 252 In an ellipse, to find the transverse, or conjugate, or ordinate, or absciss, having the other three given - 254 Properties of the ellipse - - - - 256 To describe a parabola . - - - - 257 To find any parabolic absciss or ordinate - -257 the length of the arc of a parabola, cut oil" by a double ordinate - - -258 the area of a parabola — - - 258 the area of a parabolic frustum - - 259 Properties of the parabola - - — - 259 To construct a hyperbola - - - - 259 In a hyperbola, to find the transverse, or conjugate, or ordinate, or absciss - - ~ - To find the length of a hyperbolic curve - - 1.3mm 1“ uqlfiw.4ngus_w ‘Aflnm'alrli 4 sl ’1' l V 3‘2, “wk—s- 1 ., n .f‘o «2M ,_ 1 .W*_~M “an“.-.canumm xxii CONTENTS. To find the area of a hyperbola - — - Properties of the hyperbola - »- - To find the solidity of a spheroid - - - the solidityr of the segment of a spheroid - the content of the middle frustum of a spheroid - - - the solidity of an elliptic spindle - - the solidity of a parabolic conoid - - the solidity of the frustum of a paraboloid - the solidit)r of a parabolic spindle - - the solidity of the middle frustum of a para- bolic spindle - - — the solidity of a hype1boloid - - the solidity of the fi ustum of a hyperboloid the solidity of a hyperbolic spindle - - the solidity of the middle frustum of an elliptic or a hyperbolic spindle - - - l l PART VIII. GAUG IN G. GAUGING - - - u - New Imperial Tables - - - - Tables of Wine and Spirit Measure - - Tables of Ale and Beer Measure - - - Tables of Corn or Dry Measure — - - Rules for reducing the Old Measures to the New Standard; and vice versd - - - Multipliers for performing the same - - for reducing imperial gallons and bushels to old gallons and bushels - for reducing Irish bushels and Scotch bolls, to imperial bushels , and vice versa - A description of the sliding rule — - - The method of finding the multipliers, divisors, and gauge-points - - - - - A table of multipliers, divisors, and gauge-points - To find the content of a vessel in the form of a cube the content of a vessel in the form of a parallelopipedon - - — - - the content of a vessel in the form of a cylinder - - - - - Page 263 264 265 266 ' 268 269 269 270 271 271 272 273 273 275 275 276 276 277 278 279 -279 280 281 284 285 286 287 288 CONTENTS. xxiii Page To find the content of a vessel in the form of a'pris- moid, or the frustum of a square pyramid, or a cylindroid - - - - - 289 the content of a vessel, in the form of the frustum of a cone - - - - 290 the content of a circular vessel, With curved sides - - - - - - 291 To gauge and inch a round guile—tun, according to the method practised in the Excise - - 294 a back or cooler - - - - 299 a cistern, couch, or floor of malt - - 301 CASK GAUGING - - - - - 304 Of the four varieties of casks - - - 304 To take the dimensions of a standing cask - - 304 To take the dimensions of a lying cask - --305 To find the content of a cask of any of the four va1ieties - - - - - - 306 General rules for finding the content of any cask, Without regard to its variety - - 309 To find the content of a cask by the diagonal rod - 312 To ullage a standing cask - - - - 313 a lying cask - - - - 315 Miscellaneous Questions concerning Gauging ~ 316 The method of computing distances by the velocity of sound - - - - - - 3,18 Methods of ascertaining the tonnage of ships - 320 PART IX. TRIGONOMETRY. Explanations of the most useful properties - - 323 The doctrine of right-angled plane triangles, demon- strations of different problems, and a variety of practical examples - - - - 328 The doctrine of oblique-angled plane triangles, de- monstrations, and practical questions - - 335 PART X. Trigonometry applied 1n ascertaining the heights and distances of inaccessible objects 1n a great variety of new and practical questions - - - 344 xxiv CONTENTS. The method of finding the heights of mountains by means of the barometer and thermometer - On the figure of the earth — - - — Properties of the pendulum - - - On the figure and magnitude of the earth - - On the true and elliptic latitude - - - The are of the meridian and the elliptic latitude - The are of the meridian and the true latitude — The are of the meridian between two latitudes - The measure of a degree of the meridian - - The length of a meridional degree - - - The length of a mean degree - — - To find distances on the earth’s surface - - The length of a degree of longitude - - - The measurement of an arc of the meridian - ~ The admeasurement of a base line - - - Errors in taking angles - - — — Of terrestrial refraction - - - - Of the reduction of angles — - - - Of the reduction of angles, from one plane to another Trigonometrical questions respecting the earth’s curvature — ., - - — The dip or depression of the horizon at sea - - LEVELLING - - - - - Tm; BUILDER’S DICTIONARY - .. - QUESTIONS FOR THE EXAMINATION OF THE PUPIL - Page 358 369 369 372 372 373 374 374 374 374 375 37 5 376 378 380 382 385 387 388 389 391 409 434 A TREATISE 0N PRACTICAL MENSURATION. PART I. DEFINITIONS, PROBLEMS, AND THEOREMS, IN GEOMETRY. GEOMETI Y originally signified the Art of measming the Earth, or any distance or dimensions upon or within ;but it is now used for the Science of Quantity, E1;- tension, 01 Magnitude, abst1actedly considered. GEOMETRICAL DEFINITIONS. 1. A point is considered as having neither length, bieadth, nor thickness. 2. A line has length, but 1s conside1 ed as having 11cithe1 b1 eadth nor thickness. 3. Lines are either right, curved, or parallel. 4. A right or straight line lies wholly in the same direction between its extremities, and is the shortest distance between two points. 5. A curved line continually /”/"‘\\\ changes its direction between its extremities. l 6. Parallel lines always remain at “the same distance from each other, ,/ '\ though infinitely produced. ///\\ I . . may. ”Ava-4- mama, (ha—.aawag.m,gm:gmih :4 mm 2 GEOMETRICAL PART 1. 7. A superficies has length and breadth, but is considered as having no thickness. 8. A superficies may be contained within one curved line, but cannot be contained within fewer than three straight lines. 9. A solid is a figure of three dimensions 5. namely, length, breadth, and thickness. 10. An angle is the inclination or opening of two lines, having different directions, and meet— ing in a point, which is called the angular point, as at A; and when A three letters are used, the middle one *7“— denotes that point. ' ll. Angles are of three kinds; viz. right, acute, and obtuse. 12. A right angle is made by two right lines which are perpendicular to each other. 13. An acute angle is less than a right angle, as CAB. ” 14. The complement of an angle is what it wants to complete a right B angle, as the angle DAB is the com- plement of the angle CAB. {K/ M . C 15. An obtuse angle is greater than L a right angle, as BOD. I?“ C .h‘ 16. The supplement of an angle is what it wants of¥wo right angles, as the angle ACD is the supplement of the angle BCD. 17. A triangle is a figure or superficies bounded by three right- lines, and admits of three varieties; viz. equi- lateral, isosceles, and scalene. PART 1. DEFINITIONS. 3 18. An equilateral triangle has all its sides equal. 19. An isosceles triangle has only two of its sides equal. 20. A scalene triangle has all its sides unequal. 21. Triangles are also right-angled, acute-angled, and obtuse-angled. 22. A right—angled triangle has one (:1 right angle, the side opposite to Which ' / p is called the hypothenuse, the other / l two being termed legs, or one the perpendicular and the other the base; / thus AC is the hypothenuse, BC the w I perpendicular, and AB the base. A “ 23. An acute—angled triangle has all its angles acute. 24. An obtuse-angled triangle has one of its angles obtuse, as ACB. 25. The base of any figure is that side upon which it is- supposed to stand, or upon Which a perpendicular is let fall from the vertex or opposite angle; and the altitude of a figure is its perpendicular height. In the last figure, AB is the base, and CD the perpendicular. 26. A figure of four sides and angles is denominated a quadrangle or quadrilateral figure. 27. A parallelogram is a quadrilateral figure, having its ' B 2 4 GEOBIETRICAL PART I. opposite sides pa‘rallel and equal, and admits of four varieties; viz. the square, the rectangle, the rhombus, and the rhomboid. 28. A square is an equilateral paral- lelogram, having all its angles right angles. 29. A rectangle is a parallelogram, having its opposite sides equal, and all its angles right angles. 30. A rhombus is an equilateral g. parallelogram, having its opposite angles equal, two of which are / acute, and two obtuse; which, in a , / regular rhombus, are 60 and 120 / degrees. 31. A rhomboid or rhomboides is a parallelogram, having its opposite sides and angles ,,._-._A_--__-_,._., equal, two of its angles being acute, / /‘ and two obtuse ; and when the figure / is regular, the angles are 60 and 120 4/ / degrees. "WWW-w—WHM 32. A trapezium is a quadrilateral figure, whose opposite sides are not / parallel to each other. ’ A. _._._..~.¥....,-._.- _._ ,h A 33. A diagonal is a right line joining the opposite angles of a quadrilateral figure, as AB. ,.,,. «I 1‘ 34. A trapezoid is a quadrilateral ’"' figure, having two of its opposite sides r, i arallel. . ' 1’ LAW ml 35. Plane figures having more than four sides are generally called polygons; and receive their particular denominations from the number of their sides or angles. mm 1. DEFINITIONS. 5 36. A pentagon is a polygon of five sides; a hexagon of six ; a heptagon of seven; an octagon of eight; 21 nonagon of nine; a decagon of ten; an undecagon of eleven ; and a duodecagon of twelve sides. 37. A regular polygon has all its sides and angles equal ; when they are unequal, the polygon is irregular. 38. A circle is a plane figure, bounded by a curved line, called the circumference, which is every where A equidistant from a certain point within it, called the centre. D 39. The diameter of a. circle is a right line drawn through the centre, and terminating in the circumference on each side, as AB. 40. The radius of a circle is half the diameter; or it is a right. line drawn from the centre to the circumference, as CD. , 41. An arc of a circle is any part of the circumference, as the arc EHF. 42. A chord is a right line joining the extremities of an are, as the line EF ; and the versed sine is part of the diameter cut off by the chord, as GH. 43. A segment is any part of a circle, bounded by an arc and its chord. 44. A semicircle is half of a circle, or a segment cut of? by the diameter, as ADB. 45. A quadrant is the fourth part of a circle, as ADC. 46. A sector is any part of a circle, bounded by an arc and two radii. q 47. The circumference of every circle is supposed to be divided into 360 equal parts, called degrees; each degree into 60 equal parts, called minutes; and each minute into 60 equal parts, called seconds. 48. The are of a quadrant contains 90 degrees, which is the measure of a right angle. 133 6 GEOMETRICAL rim"; 1. ~19. An ellipse is a plane figure f1’\ bounded by a curved line, called the l \ circumference; but as the figure is l \ not a circle, it is described from two A I» 23 points in the longest diameter, called kl/ the foei, or focuses. - l. 50. The longest diameter that can be drawn within an ellipse is called the transverse diameter, as AB; and the shortest the conjugate diameter, as CD. Sometimes these diameters are termed axes. 51. Identical or similar figures, are such as have all the sides and angles of the one equal to all the sides and angles of the other, each to each; so that if the one figure were applied or laid upon the other, all the parts would coincide. 52. The perimeter of a figure is the sum of all its sides taken together. 53. A proposition is something proposed to be done, or to be demonstrated; and is either a problem or a theorem. 54. A problem is something proposed to be done, and requires construction. 55. A theorem is something proposed to be demon— strated; or it is a truth stated, and requires proof or de- monstration. 56. A corollary is a consequent truth, gained from some preceding truth or demonstration. 57. Things which are equal to the same thing, are equal to each other. 58. WVhen equals are added to equals, the wholes are equal. 59. When equals are taken from equals, the remains are equal. ' 60. “Then equals are added to unequals, the wholes are unequal. Gl. \Vhen equals are taken from unequals, the remains are unequal. 62. Things which are halves of the same thing, are equal. 63. The Whole is equal to all its parts taken together. 64. Things which coincide or fill the same space, are identical or mutually equal in all their parts. PART I. PROBLEMS. 7 GEOMETRICAL PROBLEMS. PROBLEM I. To bisect a gwen line AB. From A and B as centres, with any radius, greater than half AB, in your compasses, describe ares cutting each . . A other in m and 11. Draw the line 972C72, and it will bisect AB in C. , I, \ PROBLEM II. To bisect a given angle ABC. From the point B, with any radius, describe the arc AC. From A and C, with the same, or any other radius, make the intersection m. Draw the line Em, and it will bisect the angle ABC, as required. PROBLEBI III. T 0 draw a line parallel to a given line AB. CASE 1. When the parallel line is to pass through a given point C. From any point 772, in the line AB, with the radius 772G describe the are On. From the centre O, with the same radius, de- a scribe the are mr. Take, the dis- 1: ,J tance C71 in the compasses, and apply ,I‘ ,' it from m to 7'. Through C and r g" ,/ draw the line DE, and it will be the [Ti—“*WT‘L“ 0 parallel required. 13 4 3' GEOMETRICAL PART I. CASE 2. When the parallel line is to be at a given dis- tance from AB. From any two points m and 72, in c o ) the line AB, with a radius equal to Trim—Pt—‘T‘? the given distance, describe the arcs 7- and 0. Draw the line CD, to touch these arcs, without cutting ——',,.———~——fi—J them, and it will be the parallel required. NOTE.— This Problem may be more easily performed by means of a parallel ruler, which may also be used to advantage in several operations in Practical Geometry. PROBLEM IV. To erect a perpendicularfrom a given point C, in a given [me AB. CASE 1. then the point is near the middle of the line. On each side of the point C, take ' any two equal distances, Cm and ~\_ C72. From m and n, as centres, , with any radius greater than Cm or C72, describe two arcs cutting each other in 7'. Draw the line Cr, and it will be the perpendicular required. A "‘ c :— i3 CASE 2. When the point C is at or near the end of the given line. From any point m, as a centre, with the radius or distance Cm, de- ‘2; scribe the are 7207', cutting the given ’ line in n and C. Through n and m draw a line cutting the arc in 1‘. / 1. Draw the line Cr, and it will be the perpendicular required. like “x” PROBLEM V. From a given point C, to letfall a perpendicular upon a given line AB. CASE 1. When the point is nearly opposite to the middle of the line. PART I. PROBLEMS. 9 \Vith C as a. centre, and any radius a little exceeding the distance of the given line, describe an are cutting AB in m and 91. VVith‘the centres m and 72, and the same, or any radius T773?” exceeding half their distance, describe ares intersecting each other in ,,._ Draw the line Cr, and CD will be the perpendicular required. CASE 2. “Then the given point C is nearly opposite to the end of the given line. From C draw the line CM to meet AB, in any point l\I. Bisect the line CM in the point N; and with the centre N, and radius ON or MN, de- scribe an are cutting AB in D. Draw the line CD, and it will be the per- pendicular required. NOTE. —-— Perpendiculars may be more easily raised or let fall by means of a. square, or of a feather-edged scale or ruler, with a line across it perpendicular to its edges. PROBLEM VI. To find the centre of a given circle, or one already described. (. / R Draw any chord AB; and bisect it perpendicularly with CD, which \ will be a diameter. Bisect CD in the ._._._°_____, point 0, which will be the centre re— quired. A \ /.7é \ PROBLEM VII. To mahe a triangle with three given lines, any two of which, when together, must be greater than the third. (Euclid, I. 22.) Let the given lines be AB = 12, AC = 10, and BC 2 8. ‘ B5 10 GEOMETRICAL PART I. From any scale of equal parts (which is to be under- stood as employed likewise in the following Problems) lay off the base \ 0 AB. With the centre A, and radius ‘ AC, describe an are. With the cen- tre B, and radius BC, describe another are, cutting the former in C. Draw / the lines AC and BC, and the triangle ix will be completed. / / B Noma— A trapezium may be constructed in the same manner; having the four sides and one of the diagonals. PROBLEBI VIII. flaring given the base, the perpendicular, and the place of the perpendicular upon the base, to construct a triangle. Let the base AB = 12, the perpendicular CD :: 6, and the distance AD 2: 7. Make AB equal to 12, and AD ,0 equal to 7. At D erect the perpen- /l dicular DC, which make equal to 6. / E Join AC and BC, and the figure will i 5 be completed. A ——D n Norm—A trapezium may be constructed in a similar manner, by having one of the diagonals, the two perpendiculars let fall thereon from the opposite angles, and the places of these perpen- diculars upon the diagonal; and a trapezoid may be constructed by drawing the two parallel sides perpendicularly to their base or given distance. PROBLEM IX. To describe a square, whose side shall be equal to a given line. Let the given line AB = 8. n Upon one extremity )5, of the given l line, erect the perpendicular BC, which make equal to AB. 1With A and C as centres, and the radius AB, describe arcs cutting each other in D. Join AD and CD, and the square will be completed. PART 1. PROBLEMS. 11 PROBLEM X. To describe a rectangular parallelogram, whose lengt/i and breadth shall be eqaul to two given lines. Let the length AB = 12, and the breadth BC =‘ 6. At B erect the perpendicular BC, which make equal to 6. With A as a centre, and the radius BC, describe an arc; and I); 0 with C as a centre, and the radius ‘ ‘ AB, describe another arc, cutting ‘ the former in D. Draw the lines L ' ‘_________._ B AD and CD,‘ and the rectangle will :15 be completed. PROBLEM XI. Upon a given right line to construct a regular r/iombus. Let the given line AB = 8. Draw the line AB equal to 8. lVith A and B as centres, and the radius AB, de- _ scribe arcs cutting each other in D ; bf “/12: then with B and D as centres, and / i ’ the same radius, make the inter— / / section C. Draw the lines AD, / DC, and BC, and the rhombus will Lam—w... be completed. PROBLEM XII. To construct an irregular rlzombus, having git-en its side and perpendicular heigizt. Let the side 2 8, and the perpendicular = G. Draw AB equal to 8; at A erect the perpendicular AE, which make equal to 6; and draw EC parallel to AB. With ., p g ,_ the radius AB, and A as a centre, r-—-~—r~><:—————+; make the intersection D; and with / ‘ the same radius, and B as a centre, make the intersection C. Join AD, -———’-.,/ DC, and CB, and the figure will be A “ completed. 136 12 GEOMETRICAL PART I. PROBLEM XIII. IIaving any two rig/at lines given, to coazstract a regular rlwmboid. Let the given lines be AB = 12, and BC = 6. Draw the line AB equal to 12. Take in your com- passes the line BC, and lay it from A to E. WVith A and E as centres, and the radius ‘1), \; AE, make the intersection D. Then f"? g \ with B as a centre, and the same / 7 radius, describe an arc; and with D [AA as a centre, and the radius AB, A I“; B describe another are, cutting the former in C. Draw the lines AD, DC, and BC, and the rhomboid will be completed. PROBLEM XIV.- IIaving yiven the base, the perpendicular, and the place of the perpendicular upon t/ze base, to construct an irre- gular rlwmboz'd. Let the base AB :15, the perpendicular ED :6, and the distance AE = 5. Make AB equal to 15, and AE equal to 5. At E erect the perpendicular ED, which make equal to 6; and join AD. With the radius AB, and D as a centre, describe an arc ; n’ ‘ and with B as a centre, and the / //-’ radius AD, describe another are, 4 ‘ cutting the former in C. Draw the lines DC and BC, and the figure will be completed. \\ I‘ \ A147“; Norm. — The sum of all the interior angles of any quadrilateral figure is equal to four right angles. PROBLEM XV. [lacing tlze transverse and conjugate diameters given, to construct an ellipse. Let the transverse diameter AB : l4, and the cen- jugate diameter CD 2 8. Draw the two diameters to bisect each other perpen- dicularly in the centre 0. IVith the radius A0, and the centre D, intersect AB in F and f: these two PART I. PROBLEMS. 13 points will be the foci of the ellipse. Take any point 122, in the t‘ansverse diameter, and with F and f as centres, and the radius Am, describe the arcs; Gr, Gr, ’ g, {7. Then with the same centres, ‘ and the radius Bm, describe arcs cutting the former in the points _, G, G, g, 9; thus you will. have ; . four points in the circumference F m n 0 f of the ellipse. Again, take avse- ’ ». PL, cond point n, in the transverse aka-7’69 diameter, and proceeding as before, you will determine other four points. .By the same method you may determine as many more as you please ; through all of which, with a steady hand, draw the cir— cumference of the ellipse. V PR ELEM XVI. Moon (1 given line AB, to make a regular pentagon. Make Bm perpendicular to AB, and equal to half of it. Draw Am, and produce it till mn be equal to Bm. ~With the radius Bn, and A and B as centres, describe arcs in- tersecting each other in 0, which will be the centre of the circum~ scribing circle. From the point 0, with the same radius, describe the circle ABCDE; and apply the line AB five times to the circumference, marking the angular points, which connect with right lines, and the figure will be completed. PROBLEM XVII. Upon a given line AB, to make a regular hexagon. WVith A and B as centres, and the radius AB, describe arcs inter- secting each other in 0 ; and with o as a centre, and the same radius, describe the circle ABCDEF. Ap- ply the line AB six times to the circumference, and it will form the hexagon required. 14 GEOMETRICAL PART I. PROBLEM XVIII. Ulnon a given line AB, to construct a regular octagon. On the extremities of the given line AB, erect the in- definite perpendiculars AF and BE. Produce the line AB, both ways to m and n; and with the lines AH and BC, each equal to AB, bisect the angles mAF and nBE. Draw CD and HG parallel to AF or BE, and each equal to AB. With D and G as centres, and the radius AB, describe arcs intersecting AF and BE, in the points F and E. Join GF, FE, and El), and the figure Will be completed. PROBLEM XIX. In a given triangle ABC, to inscribe a circle. Bisect the angles A and B, with J the lines, A0, B0, and 0 will be the ‘ centre of the required circle; and . its radius will be the nearest distance , to anyone of the sides; hence the circle may be described. {b C. PROBLEM XX. About a given triangle ABC, to circumscribe a circle ; or to describe the circumference of a circle through three given points, A, B, C. Bisect the sides AB, BC, with the perpendiculars mo and no ; and 0 will be the centre of the circle, and its radius will be Ac, Bo, or C0. PART I. PROBLEMS. 15 PROBLEM XXI. To make a triangle equal to a given trapezium ABCD. Draw the diagonal DB, and parallel to it draw CE meeting AB produced in E. Join the points DE ; so shall the triangle ADE be equal to the trapezium ABCD. PROBLEM XXII. To make a right angle by the line of chords on the plane scale. Draw the unlimited line AB; then take in your compasses 60° from the line of chords, and with A as a centre, describe the arc ED. Take 90° from the same scale, and set ott‘ that extent from D to C. Draw the line AC; and CAD will be the angle required. PROBLEM XXIII. To mahe an acute angle that shall contain any number of degrees; suppose 35° 30’. Draw the unlimited line AB ; then / O . a . \‘\‘> take 60 1n your compasses, and With 1.: C / A as a centre, describe the arc ED. / Set of? the angle 35° 30’, from D to C. / \‘x Draw the line AC ; and CAD will “'- be the angle required. A 1) in PROBLEM XXIV. To mahe an obtuse angle that shall contain any number of degrees; suppose 128° 35’. Draw the indefinite line AB ; and with the chord of 60° in your com— o , ........... passes, describe the are DE. Set oft )q 90° from D to C; and from C to Gr ‘5 \ | set off the excess above 90°, which \\ I‘.‘ is 38° 35’. Draw the line AG; and \A-«—————<’- GAD will be the required angle. 16 GEOMETRICAL PART I. PROBLEM XXV. To find the number of degrees contained in any given angle BAC. With the chord of 60, and A as a centre, describe the are mn. Take the // b distance mn in your compasses, and mg‘ apply it to the line of chords ; and it i“. will show the number of degrees re- quired. 7—3;“‘6 NOTE 1. ——- The measure of an angle is an arc of any circle eun- taiued between the two lines which form that angle, the angular point being the centre of'the Circle ; and it is estimated by the number of degrees that are contained in that are. NOTE 2.——Angles may be more expeditiously laid down or measured, by means of a semicircle of brass, culled “ a Protractor,” the are of which is divided into 180 degrees. PROBLEBI XXVI. 010072 a given line AB, to make a regular polygon of any proposed number of sides. Divide 360° by the number of sides, subtract the quotient from 180°, and divide the dif- ference by two. Make the angles ABC and BAC each equal to the quotient last found ; and the point of intersec— tion C, will be the centre of the circum- scribing circle. With the radius AC or BC, describe the circle; and apply the chord AB to the circumference the proposed number of times, and it will form the polygon required. PROBLEBI XXVII. In any given circle to inscribe a regular polygon of any proposed number ofsides ; or to divide t/ze circumference into any number of equal parts. Divide 360° by the number of sides, and make the angle ACB, at the centre, equal to the number of degrees con- tained in the quotient ; and the are AB will be one of the equal parts of the circumference ; hence the polygon may be formed. PART I. PROBLEMS. 17 NOTE. —— The sum of all the interior angles of any polygon, whether regular or irregular, is equal to twice as many right angles, wanting four, as the figure has sides. PROBLEM XXVIII. ’ Vofind a mean preportional between two given lines. Let the given lines be AB = 32, and BC 2 18. Make AC = 50, the sum of the given lines ; and with the radius A0 :2 25, and 0 as a centre, describe the semicircle ADC. From the point B, erect the perpendicular BD, and it Will measure 24, the mean propor- tional sought. NOTE. —- A mean proportional between any two numbers may also be found, by multiplying them together, and extracting the square root of their product. PROBLEM XXIX. To raise a perpendicular from any point D, in a giren [me AB, by a scale (3f equal parts. 'Make Dm = 3; and from the ., in points D and m, with the distances 4 and 5, describe arcs intersecting ' each other in 72. From D, through the point 72, draw the line DC, and it will be the perpendicular required. :O‘Tll 5 i) 1? NOTE. — This Problem may be performed by any other numbers in the same proportion ; but 3, 4, and 5 are the least whole numbers that will form a right-angled triangle. PROBLEM XXX. Given the span 0r chord line, and height or versed sine (if the arch of a bridge or cellar, tofind the radius of the circle that will strike the arch. Draw the unlimited line CE ; and take the versed sine 1 8 GEOMETRICAL PART I. from a scale of equal parts, and set it from C to I). Through the point D draw the line (, AB perpendicularly to CE, and make 7 37\\ AD and BD each equal to half the /: ~. given chord; also draw the chords y—flfi— AC, BC. Bisect either of these 5 23%, chords perpendicularly, with a line \ meeting CE in 0, which will be the , centre of the circle ; hence the arch ‘x.-. V ACB may be described. Or, divide E the square of half the chord by the vrsed sine ; to the quotient add the versed sine, and the sum will be the diameter of the circle. NOTE. —- This Problem is extremely useful to Joiners in striking circular arcs, forming centres for bridges, cellars, &c.; and also to Masons and Bricklayers, in describing circular pediments and other ornamental arches. GEOMETRICAL THEOREMS, TIIE DEMONSTRATIONS OF \VIIICH MAY BE SEEN IN THE ELEMENTS OF EUCLID, SIMPSON, AND EMERSON. THEOREM I. “THEN two straight lines intersect each other, the opposite angles are equal. If two straight lines AB, CD, out (-\ each other in the point E, the angle \\ AEC will be equal to the angle DEB, 1““’—,§"<*'*“'*B' and CEB-to AED. (Euclid 1. 15. \ Simp. I. 3. Em. I. 2.) \n THEOREM H. The greatest side of ever;r triangle is opcosite to the greatest angle. (Euc. I. 18. Simp. I. 13. Im. II. 4.) PART I.’ THEOREMS. 19 THEOREM III. When a line intersects two parallel lines, it ‘makes the alternate angles equal to each other. . Let the right line EF fall upon the parallel right lines ‘AB, CD ; the alternate angles AGH, GHD are equal to each other; and the exterior angle EGB is equal to the interior and opposite, upon the same side GED; and the two interior , angles BGH,'GHD, upon the same C side, are together equal to two right angles. (Euc. I. 29. Sémp. I. 7. Em. I. 4.) i l". \\ THEOREM IV. ~\Vhen one side of a triangle is produced, the outward angle is equal to both the inward opposite angles taken together. Let ABC be a triangle, and let one of its sides BC be produced to D; the exterior angle AC1) is equal to the twointerior and Opposite angles CAB, ABC; also the three interior angles of every \\ triangle are together equal to two b w-"U‘w—L‘ right angles. (Em-.1. 32. Simp. I. 9 8c 10. Em. II. 1 8c 2.) THEOREM V. Parallelograms standing on the same base, and between the same parallels, are equal to each other. Let the parallelograms ABCD, ‘Er,-s...,____P 1’- DBCE be upon the same base BC, [/I // / and between the same parallels Alt, // BC; the parallelogram ABCD is // / equal to the parallelogram DBCE. the“? (152662135. Simp.Il.2. 137721116.) ‘5 U 2O GEOMETRICAL PART I. THEOREM V I. Triangles standing on the same base, and between the same parallels are equal to each other. Let the triangles ABC, DBC be ”AA,“ upon the same base BC, and between ,3 \X the same parallels AD, BC; the triangle / ‘\_ \ ABC is equal to the triangle DBC. / / \\ (Euc. I. 37. Simp. II. 2. Em. II. 10.) 54:: __ _ _ A l} U THEOREM VII. In an)r right-angled triangle, the square of the hypo- thenuse is equal to the sum of the squares of the other two sides. Let ABC be a right-angled triangle, having the right angle BAC; the //,i square of the side BC is equal to the // l sum of the squares of the sides AB, /,/' g' AC. (Euc. I. 47. Simp. II. 8. 15m. /,,./ E II. 21.) Lg ”A“?! Norn.~Pythagoras, who was born about 2400 Years ago, dia- eovered this celebrated and useful Theorem; in consequence of which, it is said, he offered a hecatomb to the gods THEOREM VIII. An angle at the centre of a circle is double the angle at the circumference, when both stand on the same are. Let ABC be a circle, and BDC an angle at the centre, and BAC an angle at the circumference, which have the same are BC for their base; the angle BDC is double of the angle BAC. (Euc. III. 20. Simp. III. 10. Em. IV. 12.) THEOREM IX. An angle in a semicircle is a right angle. PART I. THEOREMS. 21 Let ABC be a semicircle ; then the angle ABC in that semicircle is A , a right-angle. (Euc. III. 31. 825772;). III. 13. Em. VI. 14.) THEOREM X. If a line be drawn in a triangle parallel to one of its sides, it Will cut the two other sides proportionally. Let DE be drawn parallel to BC, 1111/ one of the sides of the triangle ABC ; / then ED is to DA, as CE to EA. (Euc. VI. 2. Simp. IV. 12. Em. II. 12.) A D 1‘ THEOREM XI. In the preceding figure, DE being parallel to BC, the triangles ABC, ADE are equi-angular or similar ; there- fore, AB is to BC, as AD to DE; and AB is to AC, as AD to AE. (Euc. VI. 4. Simp. IV. 12. Em. II. 13.) THEOREM XII. In a right-angled triangle, a. perpendicular from the right angle is a mean proportional between the segments of the hypothenuse; and each of the sides, about the right angle, is a mean proportional between the hypothenuse and the adjacent segments. Let ABC be a right-angled triangle, having the right angle BAC; and from the point A let AD be drawn perpendicularly to the base BC ; the triangles ABD, ADC are similar to the whole tri- angle ABC, and to each other. Also the perpendicular AD is a mean pro- portional between the segments of the base; and each of the sides is amean A proportional between the base and its B 7 D 0 segment adjacent to that side ; therefore, ED is to DA, as A 22 GEOMETRICAL PART 1. DA to DC; BC is to BA, as BA to BD; and BC is to CA, as CA to CD. (Euc. VI. 8. Simp. IV. 19. Em. VI. 17.) THEOREM XIII. Similar triangles are to each other as the squares of their like sides. Let ABC, ADE, be similar triangles, having the angle A common to both; then the tri- angle ABC is to the triangle ADE, as the square of BC to the square of DE. That is, similar triangles are to one another in the duplicate ratio of their homologous sides. (Euc. VI. 19. S’imp. IV. 24. Em. II. 18.) THEOREM XIV. In any triangle the double of the square of a line drawn from the vertex to the middle of the base, together with double the square of half the base, is equal to the sum Of the squares of the other sides. In any triangle ABC, double the square of a line CD, drawn from the vertex to the middle of the base AB, // together with double the square of / half the base AD or BD, is equal to A / the sum of the squares of the other A “fif‘j sides AC, BC. (Simp. II. 11. Em. II. 28.) THEOREM XV. In any parallelogram ABCD, the D 9 sum of the squares of the two dia- gonals AC, BD, is equal to the sum of the squares of all the four sides of the parallelogram. ' (Simp. II. 12. Em. III. 9.) A 1‘ THEOREM XVI. All similar figures are in proportion to each other as the squares of their hemologous sides. (Sing). IV. 2-6. Em. III. 20.) PART I. ' THEOREMS. 23 THEOREM XVII. The circumferences of circles, and the arcs and chords of similar segments, are in proportion to each other, as the radii or diameters of the circles. (Em. IV. 8 8c 9.) THEOREM XVIII. Circles are to each other as the squares of their radii, diameters, or circumferences. (Em. IV. 35.) THEOREM XIX. Similar polygons described in circles are to each other, as the circles in which they are inscribed ; or as the squares of the diameters of those circles. (Em. IV. 36.) THEOREM XX. All similar solids are to each other, as the cubes of their like dimensions. (Em. VI. 24.) . § AN EXPLANATION OF THE PRINCIPAL MATHEMATICAL CHARACTERS. THE sign or character : (called equality) denotes that the respective quantities, between which it is placed, are equal; as 4 poles :: 22 yards ; 1 chain :: 100 links. The sign + (called plus, or more) signifies that the numbers, between which it is placed, are to be added to— gether; as 9 + 6 (read 9 plus 6) = 15. Geometrical lines are generally represented by capital letters'; then £5.13 CD01) signifies that the line AB is to be added to the ne 24 MATHEMATICAL CHARACTERS. PART 1. The sign — (called minus, or less) denotes that the quantity, which it precedes, is to be subtracted ; as 15 — 6 (read 15 minus 6) = 9. In geometrical lines also, AB- CD signifies that the line CD is to be subtracted from the line AB. The sign x denotes that the numbers, between which it is placed, are to be multiplied together; as 5 x 3 (read 5 multiplied by 3) :: 15. The sign + signifies division; as 15 -:— 3 (read 15 di- vided by 3) = 5. Numbers placed like a vulgar fraction, also denote division; the upper number being the di- vidend, and the lower the divisor ; as .152 = 5. The signs: 2: :(called proportionals) denote propor- tionality; as 2 : 5 : : 6 : l5, signifying that the number 2 bears the same proportion to 5, as 6 does to 15 : or, in other words, as 2 is to 5, so is 6 to 15. The sign I (called cinculmn) is used to connect several quantities together; as 9 + 3% —— 6? X 2:12—6ix2=6><2=12. l The sign 2, placed above a quantity, represents the square of that quantity; as 5‘ + 372 : 8‘3 = 8 x 8 : 6-1. The sign 3, placed above a quantity, denotes the cube of that quantity ; as §~+173wf ——§l3 2 i217? l 3 = 43 = 4 x 4 x 4 = 64. The sign 3/ or ,/, placed before a quantity, denotes the square root of that quantity; as MWZ : 4/36 = 6. The sign 0’, placed before a quantity, represents the ,eubf root of that quantity; as é/fiTXHét—fgy— 8 : ~3/21 x 31—8 2 9723 = 3/64.:4. 4 Angle; as A A, signifies the angle A; or the angle ABC, B denotes the angular point. 25 PART II. MENSURATION OF SUPERFICIES. THE area of any plane figure is its superficial content, or the measurement of its surface, without any regard to thickness. The dimensions of figures are taken in lineal measure. Sometimes they are taken in inches and tenths; some- times in feet, inches, and parts ; sometimes in feet, tenths, and hundredths ; and. sometimes in chains and links; and the area of any figure is estimated by the number of square inches, square feet, square chains, &c., contained in that figure. Note l..__A measuring tape, (usually called a bar and tape) divided into feet and inches on one side, and links on the other, is admirably calculated for taking di- mensions. Tapes are of various lengths; but those of four poles, or twenty-two yards, are most useful. ‘2. In practice, the dimensions may generally be commodiously entered upon a, rough sketch of the figure. 3. The following tables of lineal and square measures ought to be well under- stood by the learner, before he proceeds further. A TABLE or LINEAL MEASURE. Inches. Link. 7.92 = 1 Foot. 12 1—5151; 1 Yard. w —— _ .. Pole ori 36 4.0404 3 -"—“— Perch. i 198 25 16.5 5 = 1 lEQEi 792 100 66 22 4 =H I __ W ._ _ _ ____ —_ _____,_Ml 7920 1000 660 220 40 l 10 :l 1 i h-d Fur- I 1 ‘rlongs. r 37131;? __.__. ._.l 63360 8000 52801760 320 80 18: 1 l ovate—Two yards make one fathom ; and seven yards one rood of fencina or ditching. O c we“ 26 MENSURATION PART II. A TABLE OF SQUARE MEASURE. Square ' Square i Inches. 1 Link. r~—~--~»———, -—-~<— ~— .A!—#«l Square f 02.7204 =1 1 1 Foot. h —— ~—~-~~—,-~——— Square‘ l44§2.2956=i 1 Yard. l *— i“ E—~» i———~; V» —i Square 1296‘ 20.6611! 9:| l ll—‘erch. —— , 1 | Square 30204 625’ 272.25 30.25:! 1 Chain. -~‘~— ‘ ‘ [ l Square 027204 100001 4356 484} 16: 1 Rood. —— —‘ . — ! ~_4_ Square 1063160 250ml 10890 1210| 40 2.5: 1 Acre. - -7 . 1. S uarer 6272640 100000 43560 4840] 160 10 4 = l lliiilc. 4014489600 64000000 27878400 30976001 102400 6400 2560 010 :2: 1 Nolc.—-Forty-nine square yards make one rood ofdilming. :: o PROBLEM I. To find the area ofa square. RULE . Multiply the length of one of its sides by itself, an": the product will be the area. ' Note 1. — The side of Iquare may be found by extracting the square root of its .rzza. 2. When the area ofa figure is found by duodecimals, it is generally said to be in (ext, inches, parts, &c.: but it is evident that it is in feet, twelfths, inches, 8cm, he- aa 150 144 square inches, and not 12, make a square foot. In this work I have, 31 wevcr, conformed to the general practice. 3. The learner sl'ould carefully examine, war]: over, and put down, all the solutions given in this book, in order that he may the better comprehend the dill ferent rules. The definitions and rules should also be committed to memory. 4. The diagonal ofa square is equal to the square root of twice the square of the side. EXAMPLES. 1) e . ‘ l 1. “That 1s the area of the square I ABCD, the side of which is 8 feet i 6 inches ? f _,.___. __.._._., l A n PART II. or SUPERFICIES. 27 By Decimals. By Dacdecimals. I< eet. ' Feet. Inches. 8.5 8 6 8.5 8 6 425 ‘ G8 0 680 4: 3 72.2.5wa 72 3 Answer. 12 3.00 inches. Answer, 72 feet 3 inches. 2. Required the area of a square floor, Whose side is 18 feet 9 inches. [few 18 feet 9 inches = 18.75; and 18.75 X 18.75 : 351.5625 feet 2 351 feet, 6 in. 9 parts, the area required. 3. that is the area of a square table, the side of which measures 6 feet 8 inches ? Arts. 44ft. 5 2'72. 4 parts. 4. The side of a square court-yard measures 85 feet 3 inches ; what will it cost paving at 23. 9d. per yard? . Ans. £111. 08. 751d. 5. The base of the largest Egyptian pyramid is a square, Whose side is 693 feet ; how many acres of ground does it cover? . Ans. 11a. Or. 4}). 6. What is the side of a square whose area is 132496 square feet? Ans. 364 feet. 7. Requlred the s1de of a square garden that cost W . $3.188. 131d. trenching at 115d. per square yard. Ares. 20 yards. 'ote. __ The pyramids of Egypt are some of the most ancient structures in the world, their antiquity extending beyond the records of all history,and their original uses being entirely unknown. Herodotus, who wrote about 2200 years ago, speaks with as much uncertainty concerning the time of their building as we do at present. Some have conjectured that these majestic monuments were built by the children of lsrael during their bondage in Egypt; and designed as sepnichres for the Egyptian kings. Should this supposition be correct, they must have stood about 3300 years ldieperpendicularlnfightofthelargestofthesestupendousstructures is about 500 feet ; but if measured obliquely, it is 700 feet from the base to the ter- mination at the top. ' PROBLEM II. Tofind the area of a rectangle. RULE. Multiply the length by the breadth, and the product ‘wi11 be the area. 7 ., o 2 " ‘ 28 MENSURATION PART II. Note 1. —— If the length and breadth be in inches, their product must be divided by 144, in order to obtain the area in square feet ; but it' one dimension be in feet, and the other in inches, the product divided by 12 will give the area in square feet. The observation, however, is not applicable when the area is found duodecimnlly. 2. II' the area of a rectangle be divided by one of its sides, the quotient will be the other side; but the area must first be reduced to the same denomination as the given sin e. 3. Acres, mods, and perches may be reduced into square links, by multiplying the 1whole quantity, in perches, by 625, the number of sqhure links in a square perc i. 4. In taking;r dimensions with a chain. it is usual to set down the number of links. In this case, live figures must be cut off, as decimals, towards the right-hand of the area; and those on the left will express the number of acres. The decimals must then be reduced into mods and perches, by multiplying them successively by 4 and by 40, cutting ofi‘five figures, as before, in each product. 5. The diagonal of a rectangle is equal to the square root of the sum of the squares of any two adjacent‘sides. EXAMPLES. n P 1. Required the area of the rcct~ [ angle ABCD, whose length AB is 12.75 chains or 1275 links, and breadth BC, 6.75 chains or 675 links. A n Links. 8.60625 1275 4 'IFF .—- fl? 2.4.2520 6375 ° 40 8925 '_—“—— ? 550 11:999‘1‘1 8.60625 area. A223. 8 . 2:". 171). 2. A door measures 6 feet 3 inches, by 3 feet 6 inches; what did it cost at 13. 8d. per square foot? Ans. £1. 168. 571,11. 3. The length of a mahogany tea-table is 45 inches, and its breadth 42 inches; what is its area in square feet? Ans. 13.125fect. 4. How many square feet are contained in a rectan- gular fir—deal, Whose length is 18 feet, and breadth 11 inches i’ Ans. 16.13feet. 5. Required the area of a granary floor, Whose length is 45 feet 6 inches, and breadth 20 feet 10 inches. Ans. 947fect, 11 2'77. 6. How many yards of carpet, 3 quarters wide, will cover a l'OOnl that measures 28 feet 6 inches, by 18 feet‘ 9 inehesii' * A723. 79 yds. 6 in. wt PART II. OF snrnnmems. ' - 29 7. The area of a rectangular field is 14a. 27'. 111).; what is its length, its breadth being 925 links ? , Ans. 1575 links. 8. A rectangular allotment upon a common, cost £78. 18. 10%(1. digging and levelling, at £7. 10s. per acre; What will be the expense of fencing it half round at 58. 6d. per rood, its length being 1225 links ? Ans. £17. 183. 8d. PROBLEM III. Tofind the area (3f a Mombas 0r rfimnboides RULE. Multiply the length of the base by the perpendicular breadth, and the product Will be the area. Note l.——lf the area of a rhombus or rhomboides be divided by one of its dimen— sions. the quotient will be the other dimension. ‘2. The area of a parallelogram may be found by multiplying the product of any two of its sides by the natural sine of their included angle. - EXA‘ALPLES. 1. Required the area of the irre- gular rhombus A1301), Whose base A13 is 12 feet 10 inches, and perpen- dicular DE 9 feet 8 inches. 57 _. {L 1‘: B [Jere 12ft. 10 in. = 154 inches, and 9ft. 8 in. = 116 inc/res; t/ten (154 X 116) —:— 144 = 178641 —+— 14141 = 121105555 feet = 124ft. 0 in. 8 pm, the area required. D " (7 2. that is the area of the regular [I x rhomboides ABCD, Whose side AB / is 2185, and AD 1426 links? .' / A By 1Vote 2. we have 2185 X 1426 X 3660254 (flee natural sine of 60°, the angle BAD): 2698370 89. links 2 26a. 37'. 371)., tile area required. 3. How many square yards‘oii paving are there in a court- 9} in the form of a, rhoinbus, whdsle base mea- sures 26 feet 9 inches, arid perpendicular, which rises at the fistance of 100 feet from the, end of the 246 feet aginches? Ans. 7271.21; 4. Measuring along the base of a fieldin . C «'5 '3" 3O MENSURATION PART 11. a rhomboides, I found the perpendicular to rise at 678, and its length 1264 links; the remainder of the base measured 2435 links ; what is the area of the field? Ans. 39a. 17'. 15%;). 5. A window contains 36 panes, each in the form of a rhombus, whose base measures 11—2— inches, and perpen- dicular 8% inches; What will it cost glazing at 13. llffd. per square foot? A728. £2. 73. 11-},d. F. A grass-plot, in a gentleman’s pleasure-ground, cost [£33. 1413. 1d. making, at Aid. per square yard ; what is the length of the base, the perpendicular being 40 feet, and the figure a rhombus ? Ans. 50j'ecf. PROBLEM 1V. ’ 'ofiizd' his area (fa triangle, {mien the base and perncm dicular are given. RULE . Multiply the base by the perpendicular, and half the product will be the area. Or, multiply the base by half the perpendicular; or the perpendicular by half the base, and the product will be the area. Note 1._—lfdouble the area of a triangle be divided by one of its dimensions, the quotient will be the other dimension. ‘2. When the base, the perpendicular, and the place ofthe perpendicular upon the base are given, the triangle may be truly constructed; and its other two sides measured by the scale used in the construction. 3. Ifthe area ofa triangle be divided by halfthe perpendicular height. thequotient will be the base; and if the area be divided by half the base, the quotient will be the perpendicular height. 4. in order to find the true place of the perpendicular, in taking the dimensions of a triangle, or any other figure, it will be necessary to use an instrument called a “ cross, or. cross-staff.” which may be made in the following,r mannt'r : Procnre a piece Ot‘board, either of sycamore, box, or mahogany, out of which form the quad- rant 0le circle. of about four inches radius. In it, with a fine saw, make two grooves at right angles to each other, parallel to the radii, and about half an inch deep. Through the centre of the quadrant make a circular hole, large enough to receive a staff of about three quarters of an inch in diameter. The cross or quad- rant must rest and turn upon a shoulder made at a convenient distance from the top of the staff; and it may be prevented from coming ofl‘, by means ot'a screw passing through a hollow cylinder of wood or brass. made to lit the staff imme- diately above the cross. The stafi'must be piked with iron at the bottom, in order to enter the ground readily. It may be about eight feet in length; and it' it be divided into feet and inches, or into links, it will be found very useful in taking di- mensions. Crosses may be constructed in different ways ; but the cross described above I find to be most convenient. EXAMPLES. _ . (l. 1. Reqinred the area of the tri- /.\ angle ABC, whose base All measures / i \ 21.68 feet, the perpendicular DC / g R 9.46 feet, and AD 8.26 feet. .4-.__-4_...._.--..-L;~. PART .II. 0F SUPERFICIES. 31 lleet. . 21.68 Sl46 13008 8672 19512 2)205.0928 102.5464 Answer. 2. \Vhat is the {115111 of a triangular field, the base of which measures 3568 links, the perpendicular 1589 links, and the distance between one end of the base and the place of the erpendicular 1495 links ? Ans.28a.1r. 152p. 3. Required the area of the gable end of a house, the base or distance between the eaves being 22 feet 5 inches, and the 1,101pendicula1‘ from the 1idge to bthe middle of the base 9 feet 4 inches. Ans. 104 ff 7 m. 4 parts. 4. After 111easurintr along the base of a tr1 angle 895 ii -11ks :1 tound the 1)la(e 00f the perpenc licular,‘1 and the per- pendieular itselt_ - 994 links ; the whole base measured 1958 links ; what 1s the area of the tiiangle? . A923. 961. 2r. 37p. 5. How many squ are yards of slating are there in the hipped roof of a squa1e building , the base or length of the eaves from hip to hip being 23 feet 9 inches, and the distance between the middle of the base and the vertex of the roof 9 feet 6 inches ? A723. 50 1/073. 1 ft. 3 in. 6. The area of a triangle is 6 acres 2 mods and 8 perches, and its perpendicular measures 826 links; What will be the expense of, making a ditch, the whole length of its base, at 28. 6d.,pei'~ rood ? Ans. £6. 45. 74d. 51111013111111 15 To find the m ea ofa triangle, the three sides only qf'w/zz'c/z are git e72. RULE. From half the sum of the three sides subtract each side severally , multiply the half sum and the three re1na1 nders ' w continually together, and the square root of the la§t pro— duet will be the area of the t1iangle. » C 4 32 MENSURATION PART 11. No/c l.._lf a triangle be accurately laid down. from a pretty large scale of equal parts, the perpendicular may be measured, and the area found by the last Problem. ‘2. If the rectangle of any two sides ofa triangle be multiplied by the natural sine ()l‘their included angle, the product “ill be double the area of the triangle ; con~ sequently, if double the area of a triangle be divided by the rectangle of any two of its sides, the quotient will be the natural sine of their included angle. 3. Ifthe area ofa triangle be divided by half the sum oi‘the sides, the quotient will be the radius of the inscribed circle. , 4. Many calculations in Arithmetic: Mensuration, 8:0». may'be greatly facilitated by the assistance of logarithms; and it is matter of surprise that their use is not more generally taught in public schools. The method of working by logarithms is extremely simple, and may be applied to almost every kind of calculation ; yet the advantages of this great discovery have hitherto been chiefly enjoyed by those only who have leisure and inclination to col. tivate the higher branches of the Mathematics. The following;r are the most useful linlcs on this subject ; they, however, apply only to numbers greater than unity; for if the numbers be less than unity, the iiidiccs ol‘ their logarithms will be negative, which requires a litt‘r: more ingenuity in the management. illztr’tlplz'cation by Logm'z't/ims. R UL E . Add the logarithms of all the factors together, and their sum will be the logarithm of the product. Division by Logarz‘l/Lms. RULE. From the logarithm of the dividend subtract the logarithm of the divisor; and the remainder will be the logarithm of the quotient. Ride of Three by Logm-z't/nns. IULE. Add the logarithms of the second and third terms together, and from tlieirsuri subtract the logarithm of the first term; and the remainder W111 be the logarithm of the fourth term. Inz'olulz'on, or Raising (if PowL‘I‘s. RULE. Multiply the logarithm of the given number by the index of the proposed power; and the product will be the logarithm of the power sought. Evolution, or Extraction of 110015. RULE. Divide the logarithm of the given number, by the index of the proposed root; and the quotient will be the logarithm of the root required. Further directions on this head appear to be unnecessary, as it is presumed that few Teachers are without a treatise on logarithms ; it may, however, be observed, that Dr. Huttun’s Mathematical Tables are considered the most useful. EXAMPLES. 1. What is the area of the triangle / '\\ ABC, the side AB measuring 25, 1/ / AC 20, and BC 15 chains? /, / E b. v._ [Jere (25 + 20 + 15) —:— 2 =2 60 —:— 2 :- 30 = [luff [/16 sum qfflw sides; {lien 30 —- 2.3 = 5, Met/[mt remainder; 30 —- 20 2 10, the second remainder,- kz72cl 3O —- 15 :: 15, q PART II. ' 0F SUPERFICIES. 33 the third remainder; whence n/ (30 x 5 x 10 x” 15): “22500 = 150 square chains : 15 acres, the area re- quired. By Logarithms. The log. of . . . . 30 . . . . :: 1.4771213 ~—-——- . . . . 5 . . . . :: 0.6989700 . . . . 10 . . . . = 1.0000000 - . . . . 15 . . . . = 1.1760913 Divine by the index of the root . . . . 2)4.35218_2_6 The quotient is the log. of 150, the area . 2.1760913 2. iequired the area of a grass—plot in the form of an equilateral triangle, Whose- side is 36 feet. Ans. 561.18446 feet. 3 What is the area of the gable end of a house form- ing an isosceles triangle, Whose base is 25, and each of its other equal sides 20 feet ? Ans. 195.15618 feet. 4. “that is the area of a triangular field, Whose three sides measure 2564, 2345, and 2189 links? V Ans. 23a. 2r. 0—}, . 5. The thine. sides of g triangular fish—pond measure 29:5, 239, and 185 yards» ,3, what did the ground which it occupies cost, at £185,31wr acre? Ans. £843. 78. 8d. PROBLEM VI. Any two sides of a right-angled triangle being given, to find the third side. ~ RULE. 1. When the two legs are giveéi, to find the hypothennsc. , ‘ Add the square of one of th; legs to the square of the other, and the square root of the sum will be the hypothenuse. 2. When the hypothenuse and one of the legs are given, to find the other leg. From the square of the hypothenuse subtract the square of the given leg, and the square root of the re- mainder will be the required leg. (See Theorem 7.) Note.——Vthen the area ofa right-angled triangle and the hypothenuse are given, the legs may be found by the following General Rule : To the square of the hy— notheuuse add four times the area of the triangle, and the square root of this num- p CO 34 MENSURATION PART 11. ber will be the sum of the legs. From the square of the hypothenuse take four times the area of the triangle. and the square root of the remainder will be the dif- ference of the legs. Add half the difference of the legs to half their sum. and you will obtain the greater leg; but it half the difference of the legs be taken from half their sum, the remainder will be the less leg. EXAMPLES. C 1. In the right-angled triangle ABC, are given the base AB = 36, and the perpendicular BC 2 27, to find the hypothenuse AC. / ti 1; Here 362 + 272 = 1296 -{— 729 = 2025; and V2025 :: 45, tile hypot/zenuse AC. 2. If the hypothenuse AC be 60, and the perpendicular BC 36 ; what is the base AB? IIere 602 — 362 = 3600 — 1296 = 2304 ; and 4/:.304 = 48, the base AB. ' g 3. Required the length of a scaling ladder to reach the top of a wall whose height is 33 feet; the breadth of the moat before it being 44 feet. Ans. SIS-feet. 4. that must be the length of a shore, which, strutting 10 feet 9 inches from the upright of a building, will sup- port a jamb 18 feet 6 inches from the ground? ‘ Ans. 21ft. 4 2'72. Spa/Ms. 5. The distance from the ridge to the eaves of a build- ing is 15 feet, and the perpendicular height of the gable end 9 feet ; what is the breadth of the building ? Ans. 24feet. 6. A ladder 46 feet in length, being placed in a street, reached a window 26 feet from the ground, on one side; and by turning it over, without removing the bottom, it reached another window 35 feet high, on the other side ; What is the breadth of the street? Ans. 67.79695feef. 7. A castle wall there was, whoseheight was found To be 100 feet from th’ top to th’lground ; Against the wall a ladder stood upright, Of the same length the castle was in height : A waggish youngster did the ladder slide (The bottom of it) 10 feet from the side : N ow I would know how far the top did fall, By pulling out the ladder from the wall? Ans. 6 inc/168, nearly PART 11. or SUPERFICIES. 35 PROBLEM VII. T 0 find the area of a trapezium. RULE. Multiply the sum of the two perpendiculars by the diagonal upon which they fall, from the opposite angles, and half the product will be the area. Or, divide the trapezium into two triangles, in the most convenient manner ; and the sum of their areas found by Problem IV., or Problem V., will be the area required. Nate.—If the trapezium can be inscribed in acircle, that is, if the sum of any two of its opposite angles be equal to 180 degrees, its area may be found as follows : .— From half the sum oi‘ the four sides subtract each side severally ; then multiply the , four remainders continually together, and the square root of the last product will be the area. EXAMPLES. 1. It is required to lay down the /D\ trapezium ABCD, and find its area; // AE being 10.26 feet, AF 25.31 feet, /* ________ :----.-._\ AC 40.18 feet, DF 14.32 feet, and A 2 //U BE 12.86 feet. 3 ' 13 IIere (14.32 + 12.86) x 40.18 = 27.18 x 40.18 = 1092.0924 ; and 10920924 —:— 2 :: 546.0462 feet, the area required. 4 By Logarithms. The log. of . . . . . . . 27.18 is . 1.4342495 The log. of . . . . . . . 40.18 is’ . 1.6040099 Their sum . . . . . . . 1s . 3.0382594 The log. of . . . . . . . 2 is . 0.3010300 The diff. is the log. of 546.0462 the area . 2.7372292 2. Lay down the following tra- n H pezium, and find its area; AE mea- suring 1125, AB 3243, 1212,1168, /.- DF 1216, DE 2418, and CF 610 // links. =7 #1.?) Note._-This figure is divided into two triangles, because twoperpendiculars can- not be taken upon either of the diagonals. IIere (3243 x 1168) J; 2 = 3787824 ~.'— 2 = 183912, the area of the triangle ABD. e 6 36 MENSURA TION PART 1:. Also, (2418 >< 610) —:- 2 = 1474930 ,_;_ 2 :: amen, (719 area. of the triangle BCD. T/zen 1893912 + 737490 : 2631402 square lit/rs 7: 26a. 17'. 101)., the area oft/Le trapezium, ABCI). 3. How many square yards of paving are there in a trapezium whose diagonal is found to measure 126 feet 3 inches, and perpendiculars 58 feet 6 inches, and 65 feet 9 inches '3 Ans. 871317569 yard»: 4. In taking the dimensions of a trapezium, I found the first perpendicular to rise at 568, and to measure 835 links; the second at 1865, and to measure 915 links; the whole diagonal measured 2543 links; what is the area of the trapezium ? Am. 22a. 1r. ()1). 5. Lay down a trapezium, and find its area from thi- following dimensions; namely, the side A13 measures 345, B0156, CD 323, DA 192, and the diagonal AC 438 feet. Ans. 52330.33406flicz‘. G. The sides of a trapezium, two of whose opposite. angles are together equal to 180 degrees, measure 39, 32%;, 35, and 37%,- feet; what is its area? Ans. 1131f}. 22712,. 913m. PROBLEM VIII. Tofind tire area {3}" a trapezoid. RULE. llIultiply the sum of the parallel sides by the perpen- dicular distance between them, and half the product will he the are; . Or, half the sum of the sides multiplied by their dis- tance will give the area. EXAMPLES. 1. What is the area of the trape- \ zoid ABCD, the parallel sides AD \0 and BC of which are 25 and 18 ; l S and All, the perpendicular distance . I between them, 38 feet ? l.____.__ y Ifere (25 + 18) X 38 : 43 X 38 = 163-}; and 1634 -—I- 2 = 817 feet, 2716 area required. PART II. or SUPERFICIES. 37 2. Required the area of the trape- zoid ABCD, Whose parallel sides AB ,‘lL-—-—({ and DC measure ~16 feet 10 inches, /§ . and 28 feet 4 inches; DE, the per- / é pendicular distance between them, i 26 feet 9 inches; nd AB 12 feet AL; h 6 inches. Iferc 416 feet 10 in. :: 46.83333, and 28 feet 4 incicex : 98.33333; then (46.83333 + 28.33333) X 26.75 : 75.16666 X 26.75 = 2010708155, half of which is : 100535410775 feet, the area required. 3. The parallel sides of a piece of ground measure 856 and 684 links, and their perpendicular distance 985 links; what is its area? Ans. 7d. Zr. 13%;). 4.. If the parallel sides of a. garden be 65 feet 6 inches, and 49 feet 3 inches, and their perpendicular distance 56 feet 9 inches; What did it cost, at 58325. 10.9. per acre ? Ans. £24. 6.9. 7:}d. 5. The hipped roof of a square building is flat at the top ; the length of the eaves, from hip to hip, is 541 feet (3 inches; the side of the square at the top, is 30 feet 9 inches; and the nearest distance from the top to the caves, is 18 feet 3 inches: how many square yards of slating are contained in the four sides of the roof? Ans. 345.73611 yards. PROBLEM IX. T 0 find the area of an irregular polygon (3f (my numder of sides. RULE. 5? Divide the figure into triangles and trapeziums in the most convenient manner; and find the area of each separately; then the sum of these areas will be the are. of the polygon. Note.‘ln calculating the content of an irregular polygon, it is sometimes more eligible to find the double area of each figure into which it is divided; and half the sum of these double areas will be the area of the whole polygon. EXAMPLES. 1. It is required to lay down the irregular figure ABCDEFGA, and to find its area from the following dimensions. 38 MENSURATION PART II. Diagonals. Perpendiculars. Feet. Feet. Am = 25 Cm = 8 AB = 35.5 G0 = 7.5 A0 = 12 1 C72 2 8.6 An = 24.8 Ep 2 10.4 AD = 36 J Gr 2 9.2 Dr = 23.5 CONSTRUCTION—Draw the line AB, which make 2 35.5 ; and lay off 25 from A to m, at which point erect the perpendicular Cm 2 8 ; join AC and BC, and you will have the triangle ABC. , With C as a centre, and the radius C72, describe an arc; and with A as a centre, and the radius An, describe another are cutting the former in 72. Through n draw the diagonal AD : 36, upon which lay of A0 : 12. At 0 erect the per— pendicular G0 = 7.5 ; join CD, DG, and GA, and the trapezium ACDG will be completed. The trapezium DEFG may be constructed in a similar manner. CALCULATION—Here 35.5 x 8 = 284, double the area of the triangle ABC. Again, (7.5 + 8.6) X 36 = 16.1 X 36 : 579.6, double the area of the trapezium ACDG. Also, (10.4 + 9.2) x 34 = 19.6 X 34: = 666.4, double. the area of the trapezium DEF G. Then, (284 + 579.6 + 666.4) —:—- 2 = 1530 —:- 2 = 765 feet, the area of the irregular poly/gen required.- 2. It is required to lay down a pentangular field, and find its annual value at £2. 53. per acre; the first side measuring 926, the second 536, the third 835, the fourth 628, and the fifth 587 links; and the diagonal from the first angle to the third 1194, and that from the third to the fifth 1223 links. Ans. £18. 108. 7%d. Notch—This field is divided into three triangles, the areas of which maybe found by Problem V. PART 11. or SUPERFICIES. 39 PROBLEM X. Given the side of a regular polygon, tofind the radius of its inscribed or circamscribing circle. RULE. Multiply the given side by the number standing oppo- site to the name of the polygon, in the third or fourth column of the following table, as the case requires ; and the product will be the radius of the inscribed or circum- scribing circle. I. TABLE OF POLYGONS, 8w. No off ~ 'Rad. er the Rad. of the Si des. Names. Inscrlbed Circums. Clrcle. Circle. 3 Trigon, or equil. triangle . 0.288675105773503 4 Tetragon, or square . . 0.500000007071068 5 Pentagon . . . . 10.6881910 0.8506508 6 Hexagon . . . . ‘0’8660254 1.0000000 7 Heptagon . . . . 1.0382617 1.1523825 8 Octagon . . . . 1.2071068 1.3065630 9 Nonagon . . . . 1.3737387 1.4619022 10 Decagon . . . . 1.5388418 1.6186340 11 Undecagon . . . 1.7028437 1.7747329 ' 12 Duodecagon . . 1.8660254 1.9318516 Note 1.——If the radius ofa circle be given, the side of any inscribed polygon may be found by dividing the given radius by the number standing opposite to the name of the polygon, in the fourth column of the preceding table. 2. By the assistance of this Problem, any regular polygon, whose side is given, may be easily constructed in the following manner: Find the radius of the circum- scribing circle; and to the circumference of this circle apply the given side the pro- posed number of times, and you will have the polygon required. Or if the radius gf? circle be given, find the side of the inscribed polygon, with which proceed as e ore. ‘ 3. The perpendicular let fall from the centre of a regular polygon upon one of its sides is equal to the radius of the inscribed circle; and the sum of the sides is equal to the product of one side multiplied by the number of sides. EXAMPLES. l. The side of a regular pentagon is 21.75 ; What are the radii of the inscribed and circumscribing circles ? IIere .688191 X 21.75 = 1496815425 = DC, the radius of the inscribed circle ; and .8506508 X 21.75 =18.5016549 = AC, the radius of the circumscribing circle. 4-0 MENSURATION PART 11. 2. If the side of a regular heptagon be 25.25; What is the radius of the circumscribing circle ? Ans. 29.097658. 3. If the side of an octagonal grass-plot, in a gentle— man’s pleasure ground, measure 86 feet 10 inches; what will be the expense of making a gravel-walk from the middle of one of its sides to the middle of the opposite side, at 2%d. per yard, lineal measure ? A728. 143. 6%(1. 4-. If the radius of a circle be 65 feet; what is the sum of the sides of its inscribed nonagon ? Ans. 400.16356fcet. PROBLEM XI. Tofiiid 016 area of a regular polygon. RHLE. Multiply the sum of the sides, or perimeter of the polygon, by the perpendicular demitted from its centre to one of the sides, and half the product will be the are; . Note 1. — If double the area of a regular polygon be divided by the perpendicular, the quotient will he the sum of the sides. 2. In any figure whatever, the sum of all the inward angles is equal to twice as many right angles, wanting four, as the figure has sides. 3. When every side of any figure is produced out, the sum of all the outward angles is equal to four right angles. EXAMPLES. 1. Required the area of the re- F D gular hexagon ABCDEF, whose side AB is 20 feet 6 inches, and perpendicular P0 is 17 feet 9 inches. / 0 Here 20.5 X 6 X 17.75 =9183.25; F\ /,'[’\\ ,' and 2183.25 -,'~ 2 .2 1091.625 feet \ / l \\ / = 1091ft. 7 in. 6pm, the area re- ; \ quired. A 1’ B 2. that is the area of a court-yard in the form of a regular pentagon, Whose side measures 92 feet 6 inches, and perpendicular 63 feet 8 inches ? A728. 14722.91666fem‘. 3. Required the area of a hcptagonal stone, whose side measures 8 feet 9 inches, and perpendicular 9 feet. Ans. 275ft. 7 in. Gpa. 4. “That will the floor of an octagonal summer-house cost paving with black and white marble, at 48. 6d. per 3 I PART 11. OF SUPERFKCIES. 41 square foot, the side of which measures 9‘ feet 6 inches, and the nearest distance from one of its sides, to the opposite side 22 feet 11 inches? Ans. £97. 198. 4%(17. 5. A hexagonal piece of ground, in a gentleman’s park, cost £29108. 5ft]. planting with trees, at £5. 103. per acre; and a gravel-walk leading from the middle of one of its sides to the middle of the opposite side, cost £2. 38. 3,71%. making, at 3d. per yard, lineal measure; what was the expense of fencing the perimeter of the polygon, at 63. 6d. per rood ? Ans. £27. 173. PROBLEM XII. To find the area of a regular polygon, when the side only is given. RULE. Multiply the square of the given side by the number or area standing opposite to the name of the polygon, in the following table, and the product. will be the area. (This Rule is founded on Theorem XIII.) 11. TABLE OF POLYGONS, Sec. fl\l().of Names 1 Multipliers, or Angle Angle fSides. ‘ ‘ ' 3 Areas. ACB. DAC. 3 Trigon . .l 0.4330127 120° 30° fi 4 T etragon . .l 1.0000000 90° 45° 5 Pentagon . .; 1.7204774 72° 54° 6 Hexagon . ,. 2.5980762 60° 60° 7 Heptagon. .l 3.6339124 513.,0 e .30 8 iOctagon . .; 4.8284271 45° 67%,0 9 Nonagon . .I 6.1818242 40° 70° , 10 i Decagon . . 7.6942088 36° 72° 11 l Undecagon . ‘ 9.3656399 321%O 73T77° ‘, 12 |Duodecagon . 11.1961524 30° 75° .‘w'ote l.-— If the area of a polygon be divided by the number standing opposite to its name, in the foregoing table, the quotient will be the square of the polygon’s side. ‘2. The multipliers in the table of polygons, Problem X., are the radii of the inscribed and circumscribing circles. when the side of the poly on is unity, or 1:, and maybe found by trigonometry, in the following manner: ivide 360 degrees by the number of sides, and the quotient will he the angle ACB at the centre of the polygon, half of which will he the angle ACD ; then say, as the nut. sine of the angle ACD is to AD, so is the nat. co-sine of the angle ACD to CD, the radius of the inscribed circle, or the perpendicular of the polygon: and, as the nat. sine of ACD is to AD, so is the radius (1) to AC, the radius of the circumscribing circle 42 MENSURATION PART 1:. The multipliers in the last table are the areas of their respective polygons, when the side is 1, and may be found thus: Multiply the perpendicular or number in column the third, Table l, by .5 (half of AB), and the product will be the area of the triangle ACB, which being multiplied by the number of sides, we obtain the area or multiplier in Table 2. Or if the mat. tangent of the angle DAC be multiplied by the number of sides, one-fourth of the product will be the multiplier. (See the figure, Problem X.) EXAIIPLES. 1. If the side of a pentagon be 8 feet 4 inches; what is its area ? Here 8feet 4 inches = 8.} feet = 25 —:— 3 ; and (25 ~2— 3)? = 625 —:~ 9, the square of the side; then 1.7204774 X 625 —:- 9 :: 1075298375 —:— 9 = 119.4775972 eet, the area required. 2. What is the area of the base of a hexagonal stone pillar, Whose side measures 1 foot 6 inches ? Ans. 5.8-i567feet. 3. If the side of an octagonal brick pillar measure 1 foot 5 inches ; What is the area of its base? Ans. 9ft. 8 in. 3 pa. 4. Required the area of a decagon Whose side measures 25 feet 9 inches. Ans. 5101ft. 8 in. 1019a. 5. A gardener Wishes to make a hexagonal grass—plot that shall contain 260 square yards; what must be the length of its side ? Ans. 10 yards. PROBLEM XIII. The diameter of a circle being given, tafind the circum- ference,- or, the circumference being given, to find the diameter. RULE I. As 7 is to 22, so is the diameter to the circumference; or, as 22 is to 7, so is the circumference to the diameter. RULE II. As 113 is to 355, so is the diameter to the circum- ference; or, as 355 is to 113, so is the circumference to the diameter. RULE III. thltiply the diameter by 3.1416, and the product will be the circumference ; or, divide the circumference by 3.1416, and the quotient will be the diameter. PART II. or SUPERFICIES. 43 Non: 1....There is no figure that affords a greater variety of useful properties than the circle; nor is there any that contains so large an area within the same perimeter. The ratio of the diameter of a circle to its circumference has never yet been exactly determined ; although this celebrated Problem, called the squaring of the circle, has engaged the attention and exercised the abilities of the ablest mathe- maticians,both ancient and modern. But though the relation between the diameter and circumference cannot be. exactly defined in known numbers ; yet approximating ratiOs have been determined, sufiiciently correct for practical purposes. Archimedes, a native of Syracuse. who flourished about 200 years before the Christian aera, after attempting in vain to determine the true ratio of the diameter to the circumference, found it to be nearly as 7 to 22. The proportion given by Vieta, a Frenchman, and Metius, a Dutchman, about the end of the 16th century, is as 113 to 3-55, which is rather more accurate than the former ; and is a very commodious ratio, for being reduced into decimals, it: agrees with the truth to the sixth figure inclusively. The first, however, who ascertained this ratio to any great degree of exactness, was Ludolph Van Ceulen, a Dutchman. He found that if the diameter of a circle be I, the circumference will be 3.l4l592653589793238462643383279502884 nearly, which is true to 36 places of decimals. This was thought so extraordinary a per- formance, that the numbers were cut on his tomb—stone, in St. Peter’s church-yard, at Leyden. Since the invention of fluxions, by the illustrious Sir Isaac Newton, the squaring of the circle has become more easy g and the late ingenious Mr. Abraham Sharp, of Little Horton, near Bradford, in Yorkshire, has not only confirmed Ceulen’s ratio, but extended it to 72 places of decimals. Mr. John Machin, professor of astronomy in Gresham College, London, has also given us a quadrature of the circle, which is true to 100 places of figures ; and even this has been extended, by the French mathematicians, to 128. 2. The first Rule is the proportion of Archimedes; the second that of Vieta and Metius; and the third is an abridgment of Van Ceulen’s ratio. This Rule is not quite so accurate as the second; but. is most commonly used, as being most con- venient, and, in most cases, correct enough for practice. EXAMPLES. / / 1. If the diameter A8 of a circle ,/ .—-———————————i B \ be 12 ; What is the circumference? \ \j By Rule I. Ah 1 ; 22 1' 12 1 (264 —2— 7) = 3".714285, the cércum- fcreucc requircd. By Rule II. As 113 : 355 :2 12 : 37.699115, tlze circumference o'e- quired. . Bj/ Rule III. IIere 3.1416 x 12 : 37.6992, tlze circumference re- guircd. 2. If the circumference of a circie he 45 ; What is the diameter ? By Rule I. As 22 2 7 :2 45 : 14.318181, the diameter required. 44 MENSURATION PART II. By Rule II. As 355 : 113 :: 45 : 14.328943, the diameter required. By Rule III. Here 45 -:— 3.1416 : 14.323911, llee diah‘zeter re- quired. 3. If the diameter of a. well be 3 feet 9 inches; what is its circumference. Ans. 11ft. 9 in. 4 pa. 4. The diameter of a circular plantation is 100 yards ; what did it cost fencing round, at 68. 9d. per rood ? Ans. £15. 23. llald. 5. that is the diameter of a stone column whose cir- cumference measures 9 feet 6 inches? A728. 3ft. 0 2'72. 3 pa. 6. The circumference of the earth is 25000 miles ; what is its diameter, supposing it a perfect sphere ? A728. 7957.”:28541 miles. 7. The diameter of the sun is 883220 miles; what is his circumference ? Ans. 2774723952 miles. 8. The circumference of the moon is 6850 miles; what is her diameter? Ans. 2180.41762 miles. 9. The diameter of Venus is 7680 miles ; what is her circumference ? Ans. 241227.488 miles. Note. ——-Those who are desirous of making themselves acquainted with the method of finding the distances of the sun, moon. and planets from the earth, and also their diameters, are referred to Martin‘s Trigonometry. vol. i. page 208. ; Fergu— son’s Astronomy, page 100. ; and Bonnycastle’s Astronomy, page 277. PROBLEM XIV. Tofind t/ce length ofcmg/ are qfa circle. RULE I. From 8 times the chord of half the are subtract the chord of the whole are, and g of the remainder will be the length of the are, nearly. Note 1,—Half the chord of the whole are, the chord of half the are, and thc versed sine are sides of a right-angle triangle; any two of which being given, the third may he found by Problem V]. ‘2. The difference of the diameters of any two circles, multiplied by 3.14“}. will give the difference of their circumference; and vice versa. 3. Ha line drawn through or from the centre of :1 circle, hisect a chord,it will he perpendicular to it; or, if it be perpendicular to the chord, it will hisect both the chord and the are of the chord. 4. If more than two equal lines can be drawn from any point within a circle to the circumference, that point will be the centre. 5. Any chords in a circle which are equally distant from the centre are equal to each other. 6. A line perpendicular to the extremity of the radius, is a tangent to the circle. PART 11. or SUPERFICIES. 45 7. An angle formed by a tangent and chord is measured by half the arc of that chord. 8. An angle at the circumference of a circle, is measured by half the are that snbtcnds it. 9. All angles formed in the same segment of a circle, or standing on the same arc. are equal. . 10. An angle at the centre of a circle is double the angle at the circumference, if both stand on the same base. 11. Any angle in a semicircle ls a right angle. _ 12. The angle formed by a tangent to a circle, and a chord drawn from the pornt of contact, is equal to the angle in the alternate segment. ‘ . 13. The sum of any two opposite angles ofa quadrangle, inscribed m a circle, is equal to two right angles. 14. If any side of a quadrangle inscribed in a circle be produced out, the outward angle will be equal to the inward opposite angle. 15. Any two parallel chords in a circle intercept equal arcs. . 16. An angle. formed within a circle, by the intersection of two chords, 1s mea- sured by half the sum of the two intercepted arcs. _ 17. An angle formed without a circle by two secants, is measured by half the dif- ference of the intercepted arcs. 18. The angle formed by two tangents, is measured by half the difference of the two intercepted arcs. EXAMPLE S . l. The chord AB of the whole are is 2—1, and the versed sine CD 9 ; what is the length of the are ACB P i i Here 122+93= 144: +81:225; {D l and V225 = 15 : AC, the chord \ l 1' qf half the are; then (15 X 8 —24) \\ l [,1 +3=(120_24)+3=96+3: \ 3 32, the length of the are required. 2. The chord of the whole are is 45, and the chord of half the are is 25.5 ; what is the length of the arc ? Ans. 53. 3. The chord of half the are is 21.25, and the versed sine 10; What is the length of the are ? Ans. 44.1666. 4. The chord of the Whole are is 30, and the versed sine 8 ; what is the length of the are? A728. 3546—. RULE II. Divide the square of half the chord by the versed sine ; to the quotient add the versed sine, and the sum will be the diameter. 1 Subtract :6 of the versed sine from the diameter; 0 divide g of the versed sine by the remainder; and to the 0 quotient last found, add 1; then this sum being multi— plied by the chord of the whole are, will give the length of the are, nearly. “ l K v n i L. l 46 MENSURATION mm 11. Note—When great accuracy is required, the second Rule should be used, as the first gives the length of the are too little; though near enough for most cases in practice. EXAMPLES. 1. The chord AB is 24, and the versed sine CD 9; what is the length of the are AC1; P Here 123—;9 +9=144~+9+ 9:16+9:25= . _ 2 _ 9 X All the dzameter CE; and 9 X *3: —3- (2o — 50 2 6 25 — 7.38 = 6 —:— 17.62 = 3—1952; 2574672 1 + .34052 X 2—1 2 1.34052 X 24 : 32.17217, the length of Me are required. 2. The upper part or head of a windmv is the segment of a circle, whose chord-line measures 6 feet 9 inches, and versed sine 2 feet 6 inches; what is the length of the arch ? Ans. 8ft. 11 in. 112m. 3. that is the length of the circular arch of a brit ge, the span of which is 15 feet 6 inches, and height above the top of the piers 6 feet 9 inches ? A728. 22ft. 4 in. 9 pa. 4. If the span of the circular roof of a cellar be 21 feet 9 inches ; what is the length of the arch, its height being 6 feet 6 inches? Ans. 26ft. 7 in. 5 pa; 5. If the span of a circular pediment be 18 feet 6 inches; what is the length of the arch, its height above the top of the cntablature being 4 feet 6 inches '3 Ans. 21ft. 3 in. 712a, RULE 11!. As 180 is to the number of degrees in the are, so is 3.1416 times the radius to its length. Or, multiply to— gether the number of degrees in the arc, the radius, and the number .01745329; and the product will be the length of the are. Nona—The length of the are of a‘semicircle, a quadrant, &e. may also be found by taking cue—half, one-fourth, &c. 01 the whole Circumterence. EXAMPLES. 1. Required the length of an arc of 46 degrees 35 minutes, the radius being 12 feet. AS 1800 : 46° 35’ :2 8.1416 x 12 2 9.75641, the length (3f the are required. .s PART 11. or surnnriems. 47 2. What is the length of the semicircular arch of a bridge, the span of which is 18 feet 6 inches ? Ans. 29ft. 0 in. 8 pa. 3. The radius of a cart—wheel, from the centre of the nave to the outside of the felloe, is 2 feet 3 inches ; what is the length of one-sixth of the rim or circumference? Ans. 2ft. 4 in. 3 pa. Note—Those who wish to understand the principles upon which carriage wheels should be made, so that the carriage may be drawn by the least power, may consult Marrat’s Mechanics, article 788. PROBLEM ‘XV. Tofind the area of a circle. . RULES. 1. Multiply half the circumference by half the dia- meter, and the product will be the area. Or, divide the product of the whole circumfe'ence and diameter by 4,’ and the quotient will be the area. 2. Multiply the square of the diameter by .7854, and the product will be the area. 3. Multiply the square of the circumference by .07958, and the product will be the area. Note 1.—.lf the area of a circle be divided by .7854, the quotient will be the square of the diameter. 2. A circle may be considered as a regular polygon of an infinite number of sides, the perimeter of which being equal to the circumference, and the perpendicular equal to the radius: consequently by Problem X1. the area of a circle is equal to half the circumference multiplied by half the diameter. Also, the area ofa circle whose diameter is 1 is .7854 nearly ; and by Theo. 18, circles are to each other as the squares of their diameters: hence we derive the second rule. The area of a circle whose circumference is l, is 07958; and the areas of circles are to each other as the squares of their circumferences : hence our third rule. The following Rules will solve most of the useful Problems relating to the circle and its equal or inscribed square, &c. . Rule l.— The diameter of a circle multiplied by .8862‘269, will give the side of a square equal in area. 2. The circumference of a circle multiplied by .2820948, will give the side of a square equal in area. 3. The diameter of a circle multiplied by .7071068, will give the side of the inscribed square. 4. The circumference of a circle multiplied by .225079l, will give the side of the inscribed square. 5. The area of a circle multiplied by .6360197. and the square root of the product extracted, will give the side of the inscribed square. 6. The side of a square multiplied by 1.414214, will give the diameter of its cir- cumscribing circle. 7.‘ The side ofa square multiplied by 4.442883, will give the circumference of its circumscribing circle. . V 8. The side of a. square multiplied by 1.128379, will give the diameter ofa cirEIe equal in area. 9. The side of a square multiplied by 3.544908, will give the circumference ofa circle equal in area. V 10. When the diameter ofa circle is 1, its area is found to be .78539816339744830Q 9615660845819875721, which is true to 36 places of decimals ; but .7854 is sufiiciently correct for all practical purposes. hll. The areas of circles, are to each other as the squares of their diameters, or t eir radii. 48 MENSURATION PART 11. 12. The area of any circle, is equal to the rectangle of half its diameter and half its circumference. 13. The area of-a circle, is equal to the area of a triangle, the base being equal to the circumference, and the perpendicular equal to the radius. 14. A square circumscribed about a circle is equal to four times the square of the radius. and a square inscribed in a circle to half the circumscribed square. 15. Similar figures inscribed in circles, have their like sides. and also their whole perimetcrs, in the same ratio as the diameters of circles in which they are inscribed. l6. Similar figures inscribed in circles, are to each other as the squares of the diameters of those circles. EXAMPLES. 1. What is the area of a circle whose diameter is 106, and circumference 333 feet ? ' [fare (333 x 106) -;— 4 : 35298 -:— 4 : 8824:} feet, the area required. 2. Required the area of the end or base of a roller, Whose diameter is2 feet 3 inches. Here2 .25 X 2. 25—— — 5. 0625, the square of Me diffihC/Z’C’)’, - and 5.0625 >< .585i— __ 3.97% Sfeet : 8 ji‘. 11 in. 8pm, {lee area required. 3. If the ci1¢cumib1cnce of a cylindrical stone column be 7 feet 9 inches; what is the area of its base. V 4723. zlfzf. 9 in. 4p(/. 4. The diameter of a cylindrical \ (35501 is 3feet inches ; what is the area of its bottom? Ans. 9ft. 7 in. 5 pa. The diameter of a circular building at the iron foundry of Messrs. Fenton, Murray, and lVood, in Leeds measures 73 feet 3 inches; honr many Q‘squarc yards oi Paving are contained in the ground floor? A723. 468.23 5Jards. 6. In the midst of a meadow well stored with0 grass, I engag ’d just one acre to tether my ass , ‘Vhat length must the cord be, that he, feeding all round, May not graze less or more than an acre of ground. ' Ans. 39.250687 yards. PROBLEM XVI. Tofind the area of a sector ({f'a circle. RULE I. Multiply the length of the are by the radius of the sector, and half the product will be the area. Note 1. — The length of the are in ny be found by Problem XIV. 2. If the sector be Dgreater than a St. mic iicle [ind the arc of the remaining sector which subti act lrom the circumference of the \\ hole circle; and the remainder will be the length of the are required. PART II. OF SUPERFICIES. 49 EXAMPLES. 1. The radius AD is 15, the chord BD of the whole are 24, and the versed sine CE 6; what is the area of the sector ABCD? 31 Here (/(122 + 62) = «(1.44 + 36) = V180 = 13.4164 = DC, the chordqfhatf the arc; and (13.4164 x 8—24) + 3 = (107.3312 — 24) —:-3 = 83.3312 + 3 :: 27.77706, the length of the arc ; then (27.77706 X 15) + 2 = 416.6559 —:- 2 2: 208.32795, the area required. 2. What is the area of the sector of a circle whose radius is 15, and the chord of the whole are 18 feet? Ans. 144ft. 8 in. 10 pa. 3. If the radius of a sector be 12 feet 6 inches, and the length of the are 16 feet ; what is its area? Ans. 100 feet. 4. The chord of half the arc is 42 feet 6 inches, and the versed sine 20 feet ; what is the area of the sector? Ans. 1994 ft. 4 in. 9 pa. 5. that is the area of a sector greater than a semi- circle; the chord of the whole are of the remaining sector being 72 feet, the chord of half the are 45 feet, and the radius 37 feet 6 inches? Ans. 2617 ft. 10 in. 6 pa. RULE II. As 360 is to the degrees in the arc of the sector, so is the area of the whole circle to the area of the sector. Note 1. — The area of a semicircle, a quadrant, &c. may be most easily found by taking one-half, one-fourth, &c. of the area of the whole circle. 2. The area of a circle may be found by multiplying the square of the radius by 3.1416; consequently, if the area be divided by 3.1416, the quotient will be the square of the radius. EXAMPLE S. 1. The are of a sector contains 35 degrees, and its radius is 45 feet; what is its area? Here 452 x 3.1416 2 2025 x 3.1416 : 6361.74, the area of the whole circle; then as 3600 .' 35° :2 6361.74 : . :618.5025feet = 618ft. 6 in, the area required. 2. The radius of a sector is 25 feet 9 inches, and the Ilength of its are 218 degrees 54 minutes; what is its . area? Ans. 1266 ft. 7 in. 6 pa. 3. Required the area of the semicircle, the radius of I which is 18 feet 3 inches ? Ans. 523 ft. 2 in. 1 pa. D 50 MENSURATION PART 11. 4. W hat is the a1ea of a quadrant whose radius is 27 feet 6 inches? A723. 593 ft. 11 272. 6 pa. 5. What is the area of a sector whose radius is 50 feet, and the length of its are 60 degrees? Ans. 1309 feet. PROBLEM XVII. Tofind the area of the segment qfa circle. RULE I. Find the area of the secto1, ha Ving the same me as the segment , also, find the arm of the trianple t01med by the chord of the segment and the radii of the secto1; then the difference of these areas, when the segment is less than a semicircle, or their sum, when it is greater, will he the area of the segment. EXAMPLES. (‘ ,45’7\\ -1 .2: i \j" . . ,. 1' \ if" 1/ \ 1. The radius AD is 25, and the / ‘x ; / \ chord AB of the whole are 40; what : 117 2 is the area of the segment ABC? \\ g ,’ \ i II, 117826 ./(259 ~— 20-——) — M<625—400) = M225 :15 = DE, the7ef07'e the versed szne CE = 10. Again, ./(20'— + 102 --) _ /(400 + 100‘)-— __ 4/ 500 = 22. 36968 = AC, the chmd of half the arc; and (22.36068 x 8 —- 40)—:— 3: (178 88544 -— 40) —:~ 3 :: 138. 88544 —:—- 3 2 46.295146. the length of the 9W: A;CB then (46. 295146 x 25): —— 2: 1157 37865 —:— :- 578. 689325, the area of the sector ADBC. Aow,1 AB >< DE: 20 x 15 2 390, the meat of the triangle ADB; hence, 578 689325 — 300“ __ 27 8. 689325, the area of the segment requzred. 2. ‘1‘1 hat is the alea of the segment of a circle; the chord of the whole are being 60 feet, and the chord of half the are 37 feet 6 inches?0 Ans. 987 ft. 6 2n. 3. The c1101d of the whole are is 20, and the versed siueo feet , what 1s the area of the segment? Ans. 69ft. 8 272. 4. “That is the area of a segment, the arc of which is a quadrant, whose radius is 24? Ans. 164.3904. ‘4 PART II. OF SUPERFICIES. 51 5. What is the area of a segment greater than a semi- circle ; the chord of the whole are being 102 feet 6 inches, the chord of half the are 100 feet, the chord of one quarter of the are 57 feet 6 inches, and the diameter of the circle 116 feet 6 inches? Ans. 8408.89399feet. 6. Find the area of a segment whose arc contains 245 degrees 45 minutes; the diameter of the circle being 48 feet 9 inches. Ans. 1545.03853 feet. RULE II. To two-thirds of the product of the chord and height of the segment, add the cube of the height divided by twice the chord, and the sum will be the area of the seg- ment, nearly. Note—\Vhen the segment is greater than a semicircle, find the area of the remaining segment, which subtract from the area of the whole Circle; and the remainder will be the area required. EXAMPLE S. 1. that is the area of a segment of a circle Whose chord is 32, and height or versed sine 8 ? IIere (3'2 x 8) x 3; + 8‘3—z—(32 X 2): 256 x % + 512 —:— 64 2 170.6666 + 8 2 178.6666, the area required. 2. rEhe chord is 65, and versed sine 15; What is the area of the segment? Ans. 675.96153. 8. ‘What is the area of a segment, greater than a semi- circle, whose chord is 30, and height 20? - Ans. 51826171875. RULE III. Divide the height of the segment by the diameter, and find the quotient in the column of heights or versed since, in the Table at the end of Part II. Take out the corresponding area, scg., which multiply by the square of the diameter, and the product will be the area of the segment. ‘ ane l.—1f the quotient of the height by the diameter do not terminate in three places of figures, without a fractional remainder, find the area seg. answering to the first three decimals of the quotient; subtract it from the next greater area srg. ,- multiply the remainder by the fractional part of the quotient. and the product will be the corresponding proportional part to be added to the first area srg. This method ought to be used when accuracy is required ; but for common pur- poses, the fractional remainder may be omitted. 2. \Vhen the area ofa segment greater than a semicircle is required, subtract the quotient of the height by the diameter. from 1: find the arm seg. corresponding to the remainder, which take from .785398, and the difl‘erence will be the area sag. answering to the quotient. 3 The versed sines of similar segments are as the diameters of the circles to which they belong, and the areas of those segments are as the squares of the diameters. D 2 52 MENSURATION PART II. EXAMPLES. 1. What is the area of a segment Whose chord is 32, the versed sine 8, and the diameter of the circle 40? Here 8.0 —:- 40 = .2, the quotient or tabular height; and the corresponding area seg. is .111823 ; hence, .111823 X 402 = .111823 X 1600 = 178.9168, the area of the segment required. 2. What is the area of a segment whose height is 9, and the diameter of the circle 25 ? Ans. 159.09375. 3. Find the area of a segment Whose height is 25, and the diameter of the circle 55. Ans. 1050.60065. 4. that is the area of a segment, greater than a semi- circle, Whose height is 66 feet, and the chord of the whole are 60 feet 10 inches? Ans. 4435 ft. 8 in. 11 pa. 5. The base of a stone column is the greater segment of a circle Whose chord measures 2 feet, and versed sine 1 foot 4 inches; What is its area? Ans. 2.304 feet. PROBLEM XVIII. To find the area of a circular zone, or the space included between any two parallel chords and their intercepted (M‘CS. RULE. Find the area of that part of the zone forming the trapezoid ABCD, to which add twice the area of the segment AED ; and the sum will be the area of the zone required. (See the nextfignre.) Note ].—_When great accuracy is required, the area of the segment should be found by Rule 3, Problem XVII. 2. The chord AD and versed sine Em may be found from the parallel sides AB, DC, and the perpendicular distance DF, by the help of Theorem 12, Part L; much calculation may, however, be saved by measuring the chord and versed sine in taking the dimensions of the zone. (See the Key to the Mensuration, page 22.) 3. When the parallel sides and their perpendicular distance are given, the zone may be constructed in the following manner : draw the side AB ; make AF equal to half the difference between AD and DC, and erect the perpendicular FD. From the point D, draw DC parallel to AB, and. join AD. Bisect AB and AD with the perpendiculars mo, no; and 0 will be the centre of the circle of which the zone is a part; thus you will determine whether the centre of the circle falls within or without the zone. EXAMPLES. l. The greater side AB measures 40 feet, the less DC 30 feet, the per— pendicular FD 35 feet, the chord AD 35 feet 3 inches, and the versed sine Em 7 feet 3 inches; what is the area of the zone ABCD? PART II. OF SUPERFICIES. 53 By Problem VIII. we have (40 + 30) x 35 = 70 X 35 = 2450; and 2450 —:— 2 = 1225, the area of the trape- zoid ABCD. Also, by Problem XVII. Rule 2, we have (35.25 X 7.25) x #31 + 7.253 —:— (35.25 x 2) 2 255.5625 x g 1- 381.078125 +— 70.5 = 170.375 + 5.40536 = 17578036, the area of the segment AED ; hence, 1225 + (175.78036 x 2) = 1225 + 351.56072 = 1576.56072 feet: 1576 ft. 6 in. 8 pm, the area of the zone ABCD, as required. 2. The greater side is 120 feet, the less 75 feet, the chord 39 feet 6 inches, and the versed sine 3 feet 3 inches ; What is the area of the zone? ; ' Ans. 3337ft. 4 in. 1012a. 3. What is the area of a zone Whose greater side mea- sures 72 feet, the less 45 feet, and its breadth 19 feet 6 inches? Ans. 1201 ft. 11 in. 1 pa. 4. The greater side is 80, the less 60, and their dis- tance 70 ; what is the area of the zone ? Ans. 6326.96. PROBLEM XIX. To find the area of a circular ring, or the space included between the circumferences of two concentric circles. RULE. Multiply the sum of the diameters by their difference, and this, product by .7854; and it will give the area re- quired. Or, the difference of the areas of the two circles will be the area of the ring. Note 1. —— The area of a circular ring may also be found by multiplying half the sum of the circumference by half the difference of the diameters. ‘2. The area of part of a ring, or the segment of a sector, may be found by mul- tiplying half the sum of the bounding arcs by the nearest distance between them. EXAMPLES. 1. The diameters AB and CD are 30 and 20; What is the area of A the circular ring ? Here (30 + 20) x (30 —— 20) x .7854 = 50 X 10 X .7854 = 500 X .7854 = 392.7, the area of the ring re- quired. 2. The diameters of two concentric circles are 35 D 3 :L a r. f. r 1 54 MENSURATION PART 11. and 23 ; what is the area of the ring formed by the cir- cumference of those circles? Ans. 546.6384. 3. The inner diameter of a circular building is 73 feet 3 inches, and the thickness of the wall 1 foot 9 inches; how many square feet of ground does the wall occupy? Ans. 412.335feet. 4. “That was the expense of making a moat round a circular island, at 23. 6d. per square yard; the diameter of the island being 525 feet, and the breadth of the moat 21 feet 6 inches? Ans. £512. 133. 7%d. 5. What is the area of the front of a circular arch, built with stones, each 3 feet 6 inches long; the length of the upper bounding are being 35 feet 3 inches, and the length of the lower 24 feet 9 inches? Ans. 105 feet. PROBLEM XX. T 0 find the area of a lune, or the space included between the intersecting arcs of two eccentric circles. RULE. Find the areas of the two segments forming the lune, and their difference will be the area required. (1 Note 1. -—- If ABCD be a square, and circles be ///-\\ \, described from the points Band (I, with the radii Bl) D fl—wrww—«wii and CD; the area of the lune DGEFD will be equal i“ “ to the area of the square ABCD. . . l 1’}. t..__.___.__.l B F’Qi\ \ >392 circles he described on the three sides, as diameters; then will the area of the said triangle be equal to the sum ofthe areas of the two lunes D and E. / ,/ 2. IfABC be a right-angled triangle, and semi- K/T/ \, 4 Several other curious roperties or lunes may be seen in Dr.llutton"s Recrea- tions, and Mathematical ictionary. EXAMPLES. l. The length of the chord AB is 30, the height DC 12, and DE 5 ; what is the area of the lune ACBEA? PART II. or SUPERFICIES. 55- By Rule 2, Problem XVII. we have 30 X 12 X 9; + 123 + (30 x 2) = 360 x g + 1728 —:— 60 = 240 + 28.8. = 968.8, the area ofthe segment ACB. Also 30 x 5 X%+5~"-—:—(30 x 2)=150 X {-;’-+ 125 —2- 60 2 100 + 2.08333 : 102.08333, the area of the segment AEB ; hence, 268.8 — 102.08333 : 166.71667, the area of the lane required. 2. The chord is 24, and the heights of the segments 9 and 4 ; what is the area of the lune? Ans. 93.8542. 3. The length of the chord is 48, and the versed Sines of the segments 18 and 8; what is the area of the lune? Ans. 374.775- Note. _ The answer to this question was obtained by the assistance of Rule 2, Problem XIV., and Rule 3,‘Problem XVII. PROBLEM XXI. Tofind the area of an ellipse. RULE. Multiply continually together the two diameters and the number .7854, and the product will be the area of the ellipse. Note 1. — The area of an elliptical ring, or the space included between the Cir- cumt'ereuce of two concentric similar or dissimilar ellipses, may be found thus: From the product of the two diameters of the greater ellipse subtract the product of the two diameters of the less ; multiply the remainder by .7854, and the product will be the area ofthe ring. Or, subtract the area of the less ellipse from that of the greater, and the remainder will be the area of the ring. 2. If the sum of the two diameters of an ellipse be multiplied by 1.5708. (half of 3.1416,) the product will be the circumference, exact enough for most practical purposes. 3. To halt‘the sum of the two diameters add the square root of half the sum of their squares; multiply the last sum by 1.5708, and the product will be the chem- t'creuce, extremely near. 4. The ellipse is equal to a circle whose diameter is a mean proportional between the two axes; hence we obtain the above Rule. EXAMPLES. C 1. that is the area of the ellipse \ ABCD, whose transverse diameter A 1 AB is 34, and conjugate CD 25? U _ [Jere 34 X 25 x .7854 = 850 x .7854 = 667.59, the area required. 2. If the transverse diameter be 50 feet 9 inches, and the. conjugate 35 feet 6 inches; what is the area of the ellipse? Ans. léléift. 11 in. 11136:. D 4 56 MENSURATION PART II. 3. The transverse diameter of an elliptical bath mea- sures 25 feet, and the conjugate 15 feet; what was the expense of paving a walk round it, with Portland-stone, at 3s. 6d. per square foot; the breadth of the walk being 5 feet 6 inches? Ans. £7 7. 2s. 1—};d. 4. The diameters of an elliptical piece of ground tire 330 and 220 feet; how many quicks will plant the fence forming the circumference, supposing them to be set 5 inches asunder? Ans. 2073. PROBLEM XXII. To find the area qf an elliptical segment, the hase (3f which is parallel to either of the diameters of the ellipse. RULE. Divide the height of the segment by that diameter of the ellipse of which it is a part, and find the area seg. answering to the quotient, in the Table at the end of Part II. Multiply the two diameters of the ellipse and the area seg. thus found continually together, and the product will be the area required. EXAMPLE 8. /i\ 1. What is the area of the ellip- \ tical segment ABC, cut off by the A/Lfi_ml" double ordinate AB ; the height- CG Eli—«L—MF being 12, the diameter CD of the l l ,' whole ellipse 40, and EF 25? \ 1 // Here 12 —:— 40 = .3, the tabular height; and the corn responding area seg. is .198168; then .198168 )1 25 x 40 : 198.168, the area qf the segment required. 2. What is the area of the segment ADI}, the height DG being 28, and the diameters of the ellipse 40 and 25? Ans. 587.23. 3. What are the areas of the two elliptical scgimtnts made by a double ordinate parallel to the conjugate diameter, at the distance of 9 from the centre of the ellipse; the diameters being 40 and GO? Ans. 587.952, and 1297.008. 4. The diameters of an ellipse are 63 feet 9 inches, and PART 11. or snrnnrrcms. a7 95 feet 3 inches, and the height of a segment, cut off by a line parallel to the transverse, is 25—feet 6 inches ; What is its area? Ans. 1781ft. 4 in. 8 pa. PROBLEM XXIII. To find the area of a segment of a circle, or any other carvilinealfigure, by means of equidistant ordinates. RULE . If a right line AN be divided into any even number of equal parts AC, CE, EC, 850.; and at the points of divi- sion be erected perpendicular ordi- . nates AB, CD, EF, 8m; terminated by any curve BDF, 8cc§; and if A be put for the sum of the extreme or first and last ordinates AB, NO; B , for the sum of the even ordinates 3—0 E G I L N CD, GI—I, LM, 860., viz. the second, fourth, sixth, 8m. ; and C for the sum of all the rest EF, IK, 8cc., viz. the third, fifth, 820., or the odd ordinates,‘ wanting the first and last : then the common distance AC, or CE, 8:0. of the ordinates, being multiplied by the sum arising from the addition of A, four times B, and two times C, one-third of the product will be the area ABON, very nearly; that is, (A + 4 B + 2 C) —:— 3 X D = the area, putting D = AC, the common distance of the ordi- nates. (See Prob. 3. Part 6., where the rule is expressed in words.) 'ote l..__When equidistant ordinates or perpendiculars cannot conveniently be taken, in consequence of the bends or corners of the boundary being at unequal distances, you may proceed as in Problem 4, Part ill. 2. By the Rule given in this Problem, the contents of all solids, whether regular or irregular, may be found, by using the areas of the sections perpendicular to the axe, instead of the ordinates ; and it is evident that the greater the number of or- dinates or sections are used, the more accurately will the area or solidity be deter- mined. (See the Sclzolz'um, Prob. 1. Part 6. Case 2.) H K M () EXAMPLES. 1. It is required to find the area of the curved space ABCD, the lengths of the five equi- c distant ordinates being as follow; viz. the first or AD :— 8, the second 10, the third = 12, the fourth 14, and the fifth or last BC 2 l and the length of the base AB : 2 HereA:8+15=23,B=10+ 14:24,C=12 D 5 A B 5; 4. 58 MENSURATION PART 11. [O 4 andD: 1:6; then (A+4B +20) x 4D: 23 + 96 + 24) X ~ = 143 X 2: 286, the area 7eqmred. 2. Given the lengths of seven equidistant 010111.110x of an nTeguku'Inece of ground,, asfbflour;viz.15.19; 23, 25, 30, and 33 fbet; and the length of the ba:e 72fbet:req1ured.the plan and alea. Ans. T he area is 1704 feel. A TABLE OF THE AREAS OF THE SEGMEJTS OF‘A_CIRCLE, WHOSE DIAMETER IS UNITY, AND SUPPOSED TO BE 01110111) INTO 1000 EQUAL PARTS 1 \ Height. A res. H01 ght. . be (Pment. Area Segment. Area 1 1 . 1 ( Segment. 1“; He1ght. .051 .013119 .001 .000042 .002 .000119 .003 .000219 .004 .000337 .005 .000470 1 l f .026 .005546 'f .027 .005867 ', .052 015501 .028 000194 1 .053 010007 .029 .006527 1.054. .016457 .030 .006865 0.55 010911 .006 .000618 .031 .007209 .056 .017369 .007 .000779 .032 .007558 .05 7 .017831 .008 .000951 .033 .007913 .058 .018296 .009 .001135 .034 .008273 .059 .018766 .010 .001329 E .035 .008638 .060 .019239 .011 .001533 E .036 .009008 .061 .019716 .012 .001746 .037 .009383 .062 .020196 .013 .001968 .038 .009763 .063 .020680 .014 .002199 .039 .010148 .064 .021168 .015 .002438 .040 .010537 .065 .021659 .016 .002685 '041 .010931 .066 .022154 .017 .002940 .042 .011330 .067 .022652 .018 .003202 .043 .011734 .068 .023154 .019 .003471 .044 .012142 .069 .023659 .020 .003748 .045 .012554 .070 .024168 .021 .004031 .046 .012971 .071 .024680 '022 .004322 .047 .013392 .072 .025195 .023 .004618 .048 .013818 .073 .025714 .024 .004921 .049 .014247 .074 .026236 .025 .005230 .050 .014681 .075 .026761 PART 11. OF SUPERFICIES. 59 . Area 1 . rca . Area nght' Segment. 1 Helght. Segment. Helght. Segment. .076 .027289 .117 .051446 .158 .079649 .077 .027821 .118 .052090 .159 .080380 .078 .028356 .119 .052736 .160 .081112 .079 .028894 .120 .055385 .161 .081846 .080 .029435 .121 .054036 .162 .082582 .081 .029979 .122 .054689 2163 .083320 .082 .030526 .123 .055345 .164 .084059 .083 .031076 .124 .056003 .165 .084801 .084 .031629 .125 .056663 .166 .085544 .085 .032186 .126 .057326 .167 .086289 .086 .032745 .127' .057991 .168 .087036 .087 .033307 .128 .058658 .169 .087785 .088 .033872 .129 .059327 .170 .088535 .089 .034441 .130 .059999 .171 .089287 .090 .035011 .131 .060672 .172 .090041 .091 .035585 .132 .061348 .173 .090797 .092 .036162 .133 .062026 .174 .091554 .093 .036741 .134 .062707 .175 .092313 .094 .037323 .135 .063389 .176 .093074 .095 .037909 .136 .064074 .177 .093836 .096 .038496 .137 .064760 .178 .094601 .097 .039087 .138 .065449 .179 .095366 .098 .039680 .139 .066140 .180 .096134 .099 .040276 .140 .066833 .181 .096903 .100 .040875 .141 .067528 .182 .097674 .101 .041476 .142 .068225 .183 .098447 .102 .042080 .143 .068924 .184 .099221 .103 .042687 .144 .069625 .185 .099997 .104 .043296 .145 .070328 .186 .100774 .105 .043908 .146 .071033 .187 .101553 .106 .044522 .147 .071741 .188 .102334 .107 .045139 .148 .072450 .189 .103116 .108 .045759 .149 .073161 .190 .103900 .109 .046381 .150 .073874 .191 .104685 .110 .047005 .151 .074589 .192 .105472 .111 .047632 .152 .075306 .193 .106261 .112 .048262 .153 .076026 .194 .107051 , .113 .048894 .154 .076747 .195 .107842 .114 .049528 .155 .077469 .196 .108636 .115 .050165 .156 .078194 .197 .109430 .116 .050804 .157 .078921 .198 .110226 D6 60 MENSURATION PART 11. - Area I - A rea I . Area Helght. Segment. Helght. Segment. Helght. Segment. .199 .111024 .240 .144944 .281 .180918 .200 .111823 .241 .145799 .282 .181817 .201 .112624 .242 .146655 .283 .182718 .202 .113426 .243 .147512 .284 .183619 .203 .114230 .244 .148371 .285 .184521 .204 .115035 .245 .149230 .286 .185425 .205 .115842 ’ .246 .150091 .287 .186329 .206 .116650 .247 .150953 .288 .187234 .207 .117460 .248 .151816 .289 .188140 .208 .118271 .249 .152680 .290 .189047 .209 .119083 .250 .153546 .291 .189955 .210 .119897 .251 .154412 .292 .190864 .211 .120712 .252 .155280 .293 .191775 .212 .121529 .253 .156149 .294 .192684 .213 .122347 .254 .157019 .295 .193596 .214 .123167 .255 .157890 .296 .194509 .215 .123988 .256 .158762 .297 .195422 .216 .124810 .257 .159636 .298 .196337 .217 .125634 .258 .160510 .299 .197252 .218 .126459 .259 .161386 .300 .198168 .219 .127285 .260 .162263 .301 .199085 .220 .128113 .261 .163140 .302 .200003 .221 .128942 .262 .164019 .303 .200922 .222 ,.129773 .263 .164899 .304 .201841 .223 .130605 .264 .165780 J .305 .202761 .224 .131438 .265 .166663 3 .306 .203683 .225 .132272 .266 .167546 1 .307 .204605 .226 .133108 .267 .168430 .308 .205527 .227 .133945 .268 .169315 .309 .206451 .228 .134784 .269 .170202 .310 .207376 .229 .135624 .270 .171089 .311 .208301 .230 .136465 .271 .171978 .312 .209227 .231 .137307 .272 .172867 .313 .210154 .232 .138150 .273 .173758 .314 .211082 .233 .138995 .274 .174649 .315 .212011 .234 .139841 .275 .175542 .316 .212940 .235 .140688 .276 .176435 .317 .213871 .236 .141537 .277 .177330 .318 .214802 .237 .142387 .278 .178225 .319 .215733 .238 .143238 .279 .179122 .320 .216666 .239 .144091 .280 .180019 321 .217599 ] 61 PART II. 01“ SUPERFICIES. . Area . Area - Area Helght. Segment. I-Ielght. Segment. Helght. Segment. .322 .218533 .363 .257433 .404 .297292 .323 .219468 .364 .258395 .405 .298273 .324 .220404 .365 .259357 .406 .299255 .325 .221340 .366 .260320 .407 .300238 .326 .222277 .367 .261284 .408 .301220 .327 .223215 .368 .262248 .409 .302203 .328 .224154 .369 .263213 .410 .303187 .329 .225093 .370 .264178 .411 .304171 .330 .226033 .371 .265144 .412 .305155 .331 .226974 .372 .266111 .413 .306140 .332 .227915 .373 .267078 .414 .307125 .333 .228858 .374 .268045 .415 .308110 .334 .229801 .375 ‘.269013 .416 .309095 .335 .230745 .376 .269982 .417 .310081 .336 .231689 .377 .270951 .418 .311068 .337 .232634 .378 .271920 .419 .312054 .338 .233580 .379 .272890 .420 .313041 .339 .234526 .380 .273861 .421 .314029 .340 .235473 .381 .274832 .422 .315016 .341 .236421 .382 .275803 .423‘ .316004 .342 .237369 .383 .276775 .424 .316992 .343 .238318 .384 .277748 .425 .317981 .344 .239268 .385 .278721 .426 .318970 .345 .240218 .386 .279694 .427 .319959 .346 .241169 .387 .280668 .428 .320948 .347 .242121 .388 .281642 .429 .321938 .348 .243074 .389 .282617 .430 .322928 .349 .244026 ‘ .390 .283592 .431 .323918 .350 .244980 ‘ .391 .284568 .432 .324909 .351 .245934 ‘ .392 .285544 .433 .325900 .352 .246889 % .393 .286521 .434 .326892 .353 .247845 .394 .287498 .435 .327882 .354 .248801 ' .395 .288476 .436 .328874 .355 .249757 .396 .289453 .437 .329866 .356 .250715 .397 .290432 .438 .330858 .357 .251673 .398 .291411 1 .439 .331850 .358 .252631 .399 .292390 .440 .332843 .359 .253590 _ .400 .293369 .441 .333836 .360 .254550 1 .401 .294349 .442 .334829 .361 .255510 .402 .295330 .443 .335822 .362 .256471 .403 .296311 , .444 .336816 3 62 MENSURATION OF SUPERFICIES. PART II. Area Segment. 1 Area Segment. .464 .856730 .483‘1375702! l 1 Area Segment. .445 .337810 .446 .338804 .447 .339798 .448 .340793 ‘ 1 Height. Height. 1 I ! .465 .357727 .484 .376702 ; ‘ .466 .358725 .485 .377701 3 I .467 .359723 .486 .378701 E '.449 .341787 ' .468 .360721 1 .487 .379700 . .450 .342782 N .469 .361719 .488 .380700 .451 .343777 3 .470 .362717 .489 .381699 .452 .344772 ; .471 .363715 .490 .382699 r 1 1 I l V I l 1 J l .453 .345768 . .472 .364713 4 .491 .383699 .454 .346764 .473 .365712 4 .492 5.384699 .455 .347759 .474 .366710 .493 .385699 , .456 .348755 : .475 .367709 .494 .386699 ‘ .457 .349752 .476 .368708 .495 .387699 .458 .350748 . .477 .369707 .496 .388699 .459 .351745 1 .478 .370706 .497 .389699 .460 .352742 ' .479 .371705 .498 .390699 .461 .353739 ‘ .480 .372704 .499 .391699 3 .462 .354736 .481 .3737 3 .500 .392699 1 .463 .355732 . .482 ‘.374702 Note. — The use of the foregoing Table is given in Problem XVII. Rule 3; and. the method of constructing it may be seen in Moss‘s Uaugmg, page 29_ PART III. LAND-SURVEYING. 63 PART III. LAND-SURVEYING, AND MISCELLANEOUS QUESTIONS IN Superficial illensuratz'on. SECTION I. LAND-SURVEYING. LAND is commonly measured with a chain, invented by Mr. Gunter, and known by the name of Gunter’s Chain. ' t is 4. poles, or 22 yards in length, and divided into 100 equal parts, called links ; each link being 7.92 inches. An acre of land is equal to 10 square chains; that is, 10 chains in length and l in breadth; or 220 x 22 = 4840 square yards; or 40 X 4 = 160 square rods, poles, or perches; or 1000 x 100 = 100000 square links. The measurement of land is generally given in acres, roods, and perches; 4 roods being an acre, and 40 perches a rood. A statute-pole or perch is 16%,- feet long; but in dif— ferent parts of the kingdom there are, by custom, poles of different lengths; as 15, 18, 21 feet, 8:0. The Field—600k. The best method of entering the field—notes, is to begin at the bottom of the page and write upward. Each page of the field—book must be divided into three columns. In the middle column must be set down the distances on the chain—line, at which any mark, offset, or other observation is made ; and in the right and left-hand columns respectively, those marks, offsets, and observa- tions must be entered. The crossings of fences, rivers, 860. may be denoted by 64 LAND-SURVEYING. PART IH. lines drawn across the middle column, or part of the right and left-hand columns, opposite the distances on the chain- line, at which they are crossed ; and the corners of fields, and other remarkable turns in the fences, to which ofi’sets are taken, may be defined by lines joining or lying in the same relation to the middle column, as fences, 8w. do to the chain-line. Thus a tolerably accurate representation of the fences, 8:0. may be sketched in the field, which will very much assist the surveyor in drawing the plan. With respect to the characters used to denote stations, the letters of the alphabet will do very well, in small sur— veys; but in those of a large extent, numeral figures must be used, and the sign + (plus) placed before each figure 3 thus, + l, or + 2, which may be read, station first, or cross first, station one, or cross one, 82c. Upon the plan they are generally represented by this (Q) ) mark. Note 1. Many surveyors not only begin at the bottom of the field-book, but also at the right-hand side, and write towards the left, which method I always follow. 2. A straight fence may be denoted in the field-book, by placing S against the line which represents the fence. (See the field-book to Plate III.) 3. In order to assist the memory in planning, a learner sometimes draws a rough sketch of the field or estate he is about to measure, and upon it marks the station in the same manner as they are put down in taking the survey. .Miscellmzeous Instructions. 1. In addition to the chain, the surveyor must provide ten iron arrows, each about 12 or 15 inches in length, and pointed at the bottom, to enter the ground readily; and also bent in a circular form at the top, for the convenience of holding them. To the top of each arrow, 21 piece of red cloth should be attached, to make it more conspicuous among the long grass, Etc. Poles, likewise, generally called “ station stares,” will he wanted as marks or objects of direction, each about 8 or 10 feet in length, pilred with iron at the bottom, and having a red or white flag at the top, that they may be better seen at a distance; and also an offset stafi‘,‘ 10 or 12 links long, to which a cross may be fixed, as described in Prob.4, Part II. 2. In order to measure a line, let your assistant take nine arrows in his left hand, and one end of the chain and an arrow in I’lIS right ; then advancing towards the place to which he is directed, at the end of the chain let him put down the arrow which he holds in his right hand. This the follower must take up with his chain- hand, when he comes to it; the leader, at the same time, putting down another at the other end of the chain. In this manner he must proceed until he has put down his tenth arrow; then advancing a chain further, he must set his foot upon the end of the chain, and call out “change.” The surveyor must then walk up to him, if he have no offsets to take; and the arrows being carefully counted, one must be put down at the end of the chain; then proceed as before, until the whole line be measured. 3. Each change ought to be entered in the field-book, or a mistake of IO chains may happen, when the line is very long. 4. The follower may direct the leader by the motion of his left hand; moving it to the right or left, as circumstances may require. 5. The arrows should always be put down perpendicularly, and in a right line with the object of direction ; Otherwise the line will be made too long. 6. If the ditch be in the field which you are about to measure, both it and the hedge usually belong to the adjoining field. In some places 5, and in others (3 links, from the roots of the quickwood, are allowed for the breadth of the ditches be- tween neighbouring estates ; but for those ditches adjoining roads, commons, and waste lands, 7 links are commonly allowed. 7. The ditches and fences must always be measured with the fields to which they belong, when the whole quantity of land is required ; but in measuring crops SECT. I. LAND-SURVEYING; 65 of corn, turnips, &c. only so much must be measured as is, or has been occupied by the corn, &c. 8. When your view between two stations is obstructed by an intervening hill, the line must be ranged as follows : With your assistant, go to the place whence you can distinctly see both stations; and turning face to face, at the distance of two or three chains, direct each other to the right or left, until you are both in. a right line with the stations ; then, either of you putting down a pole, the line Will be correctly found. 9. 1n surveying hilly ground, the horizontal measure must always be returned, except for paring, reaping, &c., in which cases the hypothenusal measure must 2e given ; to obtain which, it is frequently necessary to divide a hill into various gures. 10. In order to preserve the horizontal line, in ascending or descending a bill, it will be necessary to elevate the chain. If the inclination of the hill be great, the chain must be elevated at several times. King’s Quadrant is admirably adapted for surveying hilly ground. (See my Land-Surveying, page 172.) 11. When a field is bounded by crooked fences, you must measure a line as near to each as the angles or curves will permit; in doing which, you must take the offset perpendicularly to each corner or angle in the fence. Where the fence is curved, those offsets must be so taken, that a right line drawn from the end of any one to the end of the next, on each side, would neither exclude any part of the land to be measured, nor include any of that which is adjacent. Perpendiculars thus erected will divide the whole offset or piece of land contained between the base-line and the fence into triangles and trapezoids. 12. Sometimes it is most convenient to measure a line on the outside of the field, and upon it erect perpendiculars to the crooked fence. These are called “insets ; ” and the area thus included must be deducted from that of the whole figure. (See Ex. 2, Prob. I.) 13. When the fences and ditches are to be measured with the field to which they belong, it is generally most practicable to fix the stations within the fences, at a little distance from the corners, and then to measure to the roots of the quicky’vood: adding or subtracting 5 or 6 links, according to the custom of the place, for the breadth of the ditch. (See Ex. 1, Prob. Il.) ‘ 14. When the offsets are small, their places on the base-line may be determined by laying the ofi‘set~stafi' at right angles across the chain; but when large, they must be found by the cross, and measured with the chain. 15. Any scale of equal parts may be used in laying down figures ; but an ivory feathered-edged plotting-scale is most convenient ; as distances may be pricked off by it, without using the compasses. 16. In laying down an offset, apply the scale to the base-line, and make a small pencil dot at every place where a perpendicular must be erected ; then lay the scale across the base, so that the line which is marked with 00, may coincide with. it, the edge of the scale at the same time touching one of the dots. From the dot, by the edge of the scale, draw a line (which will be perpendicular to the base), and; upon it prick off the offset ; or it maybe pricked off without drawing a line. 17. The base of each triangle and trapezoid, forming an offset, may be found by subtracting the distances on the chain-line from each other. 18. In planning or laying down figures relating to surveying, the upper part of the paper or book used should always, if possible, represent the north : then you“ will have the east on your right hand, and the west on your left. 19. In taking dimensions, it is an admirable method to measure proof-lines, in order,that you may be able to confirm your survey, after it is planned. Some may perhaps deem this tedious and superfluous ; but a person had certainly much better be at the pains of discovering his own errors, than expose himself to ridicule, by suffering them to be detected by some other surveyor. 20. The line in which you have the misfortune to lose an arrow, must be re- measured. , PROBLEM I. T 0 measure afield of three sides. At each corner of the field place a station-staff ; mea- sure along the base till you arrive at the point from which you suppose a perpendicular will rise to the opposite angle; there plant your cross, and turn its index till the mark at each end of the base can be seen through one of 66 LAND~SURVEYING. PART III. the grooves; then apply your eye to the other groove, and if you see the mark at the opposite angle, you are in the right place to measure the perpendicular; if not, more the instrument backward or forward, along the base, till i- you can see the three marks as above directed. Enter in your field-book the distance from the end of the base to the cross; measure the perpendicular, and then the re— mainder of the base. Or, measure the three sides, construct the figure, and determine the length of the perpendicular by the scale ; or the area may be found from the three sides, without the perpendicular. Nola I. Be especially careful, that in measuring the two parts of the base and the perpendicular, no confusion of arrows take place. 2. Triangles, trapeziums, and trapezoids, may be constructed by Problems 7 and 8, Part 1.; and their areas found by the Rules given in Problems 4, .5, 7, and 8, Part II. 3. If a triangle be laid down by a scale of 1 chain or 2 chains to an inch, the perpendicular may be ascertained with considerable accuracy. 4. if the examples in this Problem, or any of the following Problems, be thought too few, more may be easily supplied by the teacher sketching fields, at pleas-arr. with his pen, which the learner may measure by a scale. This method will be found very advantageous ; as it will give the learner a good idea in what manner he must take his dimensions, and enter his notes, when he commences field-practice. 5. The expression. R. oil‘ B, or L. offB, &c. in the field—notes, signifies that you are to turn to the right or left-hand, and measure from B, &c. EXAMPLES. 1. Lay down a field, and find its area from the follmy- ‘ ing notes. 586 ‘ 525 0. Begin at I A, and; go \Vest. Per. on the left JBEZse—Vliae; Per. 07% Me rig/i SECT. I. LAND-SURVEYING. 67 CALCULATION. Triangle ABC. 1253 base AB. 525 perp. Ca. 6265 7 25 double area. if; . O Ofikets taken on the line BC. .30 85 100001 200 64 27000 Double areas of offsets 10000 110 29800 collected. 0 200 15714: 85 2078100 825:1}. Sum. 135 246 2.738% #19685: Double Areas. , ~—= 1476 657825 Triangle ABC. ”5744 ‘ $732544, Ofi‘sets. tr- $740369 3.70184 , 4 E30736 40 321.29qu Area 3a. 27'. 32]). 2. Draw a plan of a field, and find its area. from the fol— lmving notes. CA 0 1252 37 1000 69 824 45 716 72 610 15 424 55 212 O 000 R. off C. LAND-SURVEYING. PART III. AC 683 536 354 0 48 76 56 25 0 Begin at go N. East. ANSXVER. Having constructed the figure, you will find the per- pendicular falling from the angle B, upon the side CA, to measure 528 links. ‘ Double areas. 661056 Triangle ABC. 68543 Ofi‘sets taken on the line AR. 95376 Ditto on CA. 824975 Sum. 50154 Insets on EC. 53)""7774821 Difference. 3874K) =3a. 3r. NEIL, the area, required. 3. Find the area of a field 55mm the following dimen~ sions ' SECT. I. LAND-SURVEYING. 69 AB 1 1946 1000 From A, go East. CALCULATION. IIere, (1946 +1197 +1435)+2=4578—:—2: 2289 :half the sum of the sides ; then, 2289 - 1946:343, the first remainder; 2289 - 1197:1092, the second remainder; and 2289— 1435 = 854, the third remainder ; whence x/(2289 x 343 x 1092 x 854) : «./ 732184316136 2 855677 square 1inks,=8 acres, 2 roads, and 9 perches, the area required. 133/ Logarit/ams. The log. of . . . 2289 = 3.3596458 ———-~——-—- . 343 = 2.5352941 —— 1092 : 3.0382226 . 854 . . . . = 2.9314579 Divide the sum. of the log. by . . . . . 2)11.8646204 The quotient is the log. of 855677, the area 759323102 4. Lay down a field, and find its area from the follow- ing notes. DC proof-line. . 412 Return to D. ' CA 0 638 46 465 32 246 0 000 R. of C. BC 0 462 45 300 32 150 0 000 R ofl‘ B.i .. 70 LAN D-SURVEYING. PART Ill. WABJ 0 725 32 565 40 400 25 250 D, station for a proof-line. 0 000 Begin at A, and go West. ‘ANSWER. Having constructed the figure, you will find the proof- line DC to measure 412 links, as in the field-book; hence, you may conclude there is no error committed in taking or setting down the dimensions. You will also find the perpendicular falling from the angle C, upon the base AB, to measure 402 links; from which, and the ofi‘sets, we obtain the area = 1a. 37'. 2517. PROBLEM II. To measure afield offom' Sides. Measure a diagonal, and find the perpendiculars falling upon it, from the opposite angles, as already directed. Or, if you measure a diagonal and the four sides, the figure may be constructed, and the perpendiculars deter- mined by the scale. Nola 1. Always make choice of the longer diagonal, because the longer the base- line ol‘a triangle, the more obtuse is its subtending angle; consequently the per-- pendiculartwill be shorter, and its place more easily and accurately determined. if you measure the four sides and both the diagonals, one of them will serve as a proof-line, after you have laid down the figure. 2. If neither of the diagonals can be measured, in consequence of obstructions, you must measure tie-lines across the angles of the field. which will enable you to plan it .7 and hence the diagonal and perpendiculars may be found by the scale. 3. When two perpendiculars cannot be taken upon either of the diagonals. such fields must be divided into two triangles, or into two right-angled triangles and a trapezoid. 4. Unskilful surveyors afl‘ect to reduce trapeziums into rectangles, by adding each two opposite sides together, and taking half their sum respectively for a mean length and breadth; but this method leads to very erroneous results. (See my Land-Surveying, page 124.) EXAMPLES. 1. Draw a plan of a field, and find its area. from the following notes. AC Diana 1 155 1000 495 91 5 360 520 From A, go N. W'est. SECT. I. LAND-SURVEYING. 7} To the Fence. 48 660 53 630 to A. ‘10 500 25 380 50 200 62 000 From 1), g9 South. To the Fence. 40 1012 42 95 to 1). 54 600- 33 4-00 12 260 30 150 65 000 From C "0 East. To the Fence. 25 615 35 550 to C. 30 400 10 240 22 1.50 45 000 mm B, go North. To the "1 Fence. 33 1090 45 104-5 to B. 55 800 40 500 48 300 30 000 Begin at A, and gmi “Est. ,;:t:::x.—;:;:-u, :; 17:77:71,: .35, 5C l?‘ 31 I . 5 .‘l ' . [I \‘~\ .‘ t ”l," , .0". .‘ § 72 LAND-SURVEYING. PART III. CALCULATION. Trapezium ABCD. Ofisets taken on the line AB. 520 001‘ 30 56 £5 * p‘ :8 :12 1015 sum. 78 101 1155 diag. ’300 245 5075 23400 ”505 5075 2; 404 48 1015 40 202 1015 g . 24% .1}qu double area. 200 ’_‘_ - T606 ' Ofiets taken on the line AB. 40 45 23400 2% g; égggg Double areas 300 45 24745 collected. 273m 21)?) 341,9 : 312 98055 sum. {59 : Qfl‘lvets taken on the line BC. 45 30 10050 22 35 2880 w ~ Double are as 67 65 64:00 150 150 9750 collected. 37356 T5550 3999 67 65 23%? sum. 10050 9]?) . ‘ 22 ”’35 30 10 25 2% 32— 6—0 90 65 % .583?) 37)?) ~ 3900 6400 SECT. I. ’ LAND-SURVEYING. 73 Ofil‘ets taken on the line CD. 65 30 14250 30 12 4620 9—5 {2 6300 Double areas 150 1 10 17400 collected. 4750 M) 33600 95 42 5084 ilggfi 4620 81254sunL 12 54 33» 32 45 96 140 350 1800 4800 43.: 12% 6300 33600 9 42 £4 :0 87 82 200 62 17400 164 4931 5084 Ofisets taken on the line DA. 62 25 53 50 40 48 112 . 65 101 200 120 30 22400 13 1 3030 ' 65 A .32 7.899 22400 75 40 13388 Double areas .Eg _5_§ 12090 COIlected. 6000 93 3000 75 ”13_0_ 88:0 Em) 27:!0 ‘ 3 ~ ”31' —__ 93 Web 7‘1’ LAND-SURVEYING. PART 111. 1172325 1" I" ' $33518 Whom double areas 81254 collected. 58820 - .86868 40 34.7472©Area 7a. 07'. 342p. To find tke area of the foregoing figure by the met/rod generally called “ Casting.” Having constructed the figure, and taken out the chain- lines, draw with a pencil by A 72 - M the stralght edge of a clear g \. piece of lantern-horn, the four a (/1 ' right lines AB, BC, CD, and DA, in such a manner, that the parts included may be equal to those excluded, as nearly as you can judge by your eye; then the diagonal , AC Will be found to measure 1274, and the perpendicu- lars Ba, 550, and Da, 583 links. ‘ Note. The use of the Parallel Ruler, in straightening crooked fences, is given in the Author’s Land— Surveying, in twelve new problems, comprising every case that can possibly occur in practice. CALCULATION. 550 583 }perp. 1133 sum. 1274 diag. 4532 7931 2266 l l 33 31443442 (£7335? Ans. 7a. 02-. 3422‘!)- SECT. I. LAND-SURVEYLN G. 75 Note. Although the method of finding the area by casting is adopted. by most practical surveyors, it is certainl less correct than that by offsets, &c. ; a learner, therefore, ought to practise bot , until he can habitually come very near to the truth by the former. 2. Lay down a field, and find its area from the follow- ing dimensions. AF 2718 1865 575 000 D B From _.A_,& O 953 E. 1142 C. 0 W’est. Area 20a. 37'. 17%;). 3. Required the plan and area of a. field from the fol- lowing notes. 1646 720 .~. IJ. Off C. 2030 620 A, go B 850 From I. Diag. 385 D :_KC# Base. East. Area 11a. 37". 71). 4. Draw a plan, and find the area of a field, from the" following dimensions. Return Drag. Ding. 76 LAND-SURVEYING. PART III CD 0 1244 47 1000 70 800 85 600 68 400 30 200 0 000 R. of 0 BC 0 720 85 650 112 550 88 450 360 Cross the tencc. 300 83 200 130 100 100 000 O :R.ofldi 1X8 1350 1000 Begin at A, an go West. ANSWER. Having constructed the figure by means of the four sides and the diagonal AC, you will find the diagonal BD to measure 1460 links, as in the field-book; hence, it may be concluded that no error has been made in taking or enter— ing the dimensions. You will also find the perpendicular from the angle A, upon the diagonal BD, to measure 618 links, and the perpendicular from the angle C, upon the same diagonal, 613 links; hence the area is 9a. 27‘. 1113. PROBLEM III. To measure afield of more than four sides. Any piece of land, consisting of more than four sides, ‘ may be surveyed by reducing it into triangles and tra- g peziums ; thus, a field of five sides may be reduced into a t ‘ triangle and a trapezium; of six sides, into two trape- ‘ SECT. I. LAND-SURVEYING. 77 ziums ; of seven, into two trapeziums and a triangle; of elght, mto three trapeziums, 8:0. EXAMPLE S. 1. Required the plan and area of a field, from the fol- lowing dimensions. ~—CE‘ Ding. 1165 912 168 B. D 308 415 R 011' C AC Base. / 1195 V A 293 233 B. From A, go “rest. CALCULATION. Triangle ABC. Trapezz'um BCDE. 1195 base. 308} er 233 perp. 168 p P“ 3585 476 sum. 3585 1165 diag. 2390 2% 278435 2856 476 476 ”4’53?“ @ 554540 2)832975 4.16487 4 .65948 40 2. Lay down a field, and find its area from the fellow- ing notes. {7—138 1792 Diag. E. 562 954 R. 011' D. E3 78 LAND-SURVEYING. PART 111. {fit—:l 2243 Diag. C 728 1785 560 624 13. From . A, 5L“, N. “Test. Area 20a. 07'. 3143]). 3. Required the plan and area of a field from the fol- lowing dimensions. D 1695 B 1265 From 5126 5000 4040 4000 3000 2000 1720 1000 R. off C. 11C 4300 4000 3000 2000 1930 1000 Diag. 1490 Ding. “Test. Area 108a. 3r. 121;). 4. Draw :1- plan of a. field, and find its area from the fo1r lowing notes. BD 1238 1000 Return to 1000 Return to Diag. B. Ding. SECT. I. LAND-SURVEYING. 79 DE , 1500 Diag. 1000 R. of D. FD 1160 1000 R. of F. EF 1260 1000 R. off E. CE 1080 R. of C. —AC”' 2100 Diag. 2000 1340 Diag. \Begin at WAK‘agdr g0 West. ANSWER. ' Having laid down the figure, you will find the perpen- diculars of the trapezium ABCD : 708 and 390, and of DCEF=68O and 808 links; hence the area qf thefield is 25 acres, 2 roads, and 26—17 perches, E 4 80 LAND-SURVEYING. PART III. PROBLEM IV. To measure cm irregular, narrow piece of land. Divide the ground to be surveyed into triangles and trapezoids, by measuring a base-line, in a convenient posi- tion, and upon it erecting perpendiculars t0 the boundaries, on each side. Note 1. The general method of measuring a few lands or ridges together, in common fields, «Ste. is by taking several breadths, dividing their sum by the number for a mean breadth, and multiplying this by the length, for the area; but it is much more accurate to note the place of each breadth upon the base or chain-line, and find the area by the rule for trapezoids. h2. Indorder to obtain the breadths correctly, you must measure directly across t e Ian 5. 3. If the lands be curved, or longer on one side than on the other, by measuring along the middle, you will obtain the mean length. 4. Faring, reaping, &c. should always be surveyed with a slack chain, in order to obtain the measurement of the surface. EXAMPLES. 1. Draw a plan of an irregular piece of land, and find its area from the following dimensions. AB 0 1314 126 234 1005 980 52 785 125 {“3 ; 312 700 1 W\ 555 152 215 460 /\-// *4 335 100 336 260 360 000 232 From A, go East. AN SVVER. Double Area's. 324337 offsets on the right. 65glj6 ditto 0n the left. 2>98081§wm 4962106=4a. 37'. 243p. tile area required. 2. Lay down a field, and find its area from the follow- ing notes. 65 92 94 no lo 0 55 112 175 205 230 250 266 From 3. Find the area of five lands, : mensions. LAND-SURVEYING. BC 526 400 300 200 100 000 R. of B. AB— 1235 l 100 900 650 500 350 200 000 A, go 2436 2000 1 700 1 400 1 000 700 400 000 526 to C. 435 '328 250 224 245 280 365 North. Area 6a. 1r. 2}). from the following di- 205 210 214 218 220 215 209 205 Area 5a. 0r. 29p. 4. Required the area of six lands, from the following , notes. ':§3_’63: 3000 2500 2000 1800 1000 800 000 E5 199 214 236 201 175 Area 7a. Or. 81). 82 LAND-SURVEYING. PART in. Note. The sum of the breadths in the last example is 1025, which being divided by 5, their number, we obtain 205 for the mean breadth ; then 3365 multiplied my 205 is equal to 689825 square links, which are equal to 6 acres, 3 roods, 24 perches. the aria, which is too little by 24 perches, proving the inaccuracy of the common metho . If a piece of ground he narrowest about the middle, the common method will give the area too much. PROBLEM V. To measure a mere or wood. By the help of your cross, fix four station-staves on the outside of the mere or wood, so as to form a rectangle ; then measure each of its sides, taking insets to the boun- daries, which must be treated as before directed. It is, however, a more expeditious method than the former, to measure the four sides of a quadrilateral figure. having one right-angle; then construct the figure, and measure the diagonal and perpendiculars by the scale. Note I. If amere or wood be of a triangular shape, its area may be found bv aegigring the three sides of a. circumscribing triangle; and proceeding as 2. By this Problem you may survey fields into which you are not permitted to enter, or which contain obstructions. EXAMPLE. Let the following figure represent a wood, the are: of which is required. Set up your cross at station A, and let your assistant fix the marks B and D, so that the angle at A may be a right angle; and measure the line AB, taking insets to the fence, as you proceed. Then fix the mark C, as most convenient ; measure the other three lines, and you Will find in yourfield- book the followin dimensions. D ' DA l 0 1550 ' 160 1440 I 50. 1200 . 1000 0 900 t SECT. I. LAND-SURVEYING. 83 CD ’ 0 950 120 500 0 000 L. off C. BC 1340 0 1050 1000 i 50 700 N 60 400 0 000 L. of B. AB 0 1150 1000 50 900 0 . 550 100' 300 110 160 0 000 Begin at A, and i go North. ANS‘VER. Having constructed the figure, you will find the diago- nal BD to measure 1930, the perpendicular Aa 923, and Ca 605 links ; hence the area of the wood is 12a. 3T. 20;). PROBLEM VI. To measure and plan roads, rivers, canals, £30. In measuring roads, rivers, or canals, angles or tie-lines must be taken at the different turns, in order to lay down the chain-lines; and offsets must be taken to the boundaries, as you proceed, to enable you to draw the plan. Note 1. The length of a road is generally returned either in miles, furlongs, and poles, or else in miles and yards. 2. A machine called a “ Perambulator ” is sometimes used to ascertain the lengths of roads. It has a wheel of 8 feet 3 inches, or halfa pole in circumference, which being made to pass over the ground, puts in motion the clockwork within : and the distance measured is pointed out by an index on the outside. This instrument is much more expeditious for measuring the length of a read than the chain; but it is certainly less correct: for by the wheel passing over stones, sinking into holes, &c. the distance is made to appear more than it is in realitv. E6 84 LAND-SURVEYING. PART III. EXAMPLES. 1. Let the following figure represent a serpentine road, a plan of which is required. Begin at + 1, and measure to + 3, taking offsets on both sides, as you proceed. Return to + 2, and measure to + 4, from which run a line to + 3, which will tie the first and second lines. Return to + 4, and continue the line to + 6. From + 6, proceed as before, until you ar- rive at + 14 ; and you will have obtained the following dimensions, from which a plan may be drawn. SECT. I. 58 68 50 Go from 30 7O 86 70 +8 is 200 fr0m+ 10 120 Go from OlH-ROD 10000 50 Go from 20 50 52 20 18 Go from 38 15 28 55 Begin LAND-SURVEYING. to + 14. 350 200 150 100 000 at+l, 85 60 44 +13 is 184 from +11 80 Line 5. to + 12. +11. 33 5O Cross—fence. Line 4. to + 9. Cross-fence. + 8. 84 60 65 + 7 is 160 from+5. Line 3. to + 6. 70 +5. 80 + 4 is 232 from + 3. 93 Line 2. t0+3. Cross-fence. + 2. 95 80 70 Line 1. 2. Let the. foregoing figure represent a river, a plan of which is required. 86 LAND-SURVEYING. PART III. Begin at a, and measure to 0; taking ofi‘sets to the river’s edge, as you proceed. From 0 measure to d; and there take the tie or chord—line db, which will enable you to lay down the first and second lines. Continue the second line to n ; and from m, measure to 7', at which place take the tie-line 1‘72 ; and thus proceed until you come to the end of your survey at x. If the breadth of the river be everywhere nearly the same, its breadth taken in different places, by the next Problem, will suffice ; but if it be very irregular, dimen— sions must be taken on both sides, as above. “Then the area is required, it must be found from the plan, by dividing the river into several parts ; and taking the necessary dimensions by the scale. Note 1. Any bog, marsh, mere, or wood, whatever may be its number of sides. may be measured by this Problem. 2. In taking an angle with the chain, some surveyors never measure more than 100 links from the angular point ; but when the lines, including the angle, are of a considerable length, it l3 much better to measure the chord-line at a greater dis- tance from the angle, than one chain ; because a small inaccuracy in constructing the figure, when the angular distance is short, will throw the lines, when far pro. duced, considerably out of their true position. It sometimes happens, however, in consequence of obstructions, that it is impossible to measure the chord—line at a greater distance from the angular point, than one chain. In such cases, multiply both the angular distance and chord-line by 2, 3, 4, or any larger number, as circumstances may require; and use the products resulting, in laying down the angle. 0r, which is the same thing, lay down the angle by a scale 2 or 3 times as large as that by which you intend to draw the plan. By this means the radius or angular distance be equaled, and the chordJine bd, will be increased in length ; and consequently the lines at: and on, including the angle, will be more correctly laid down in position. PROBLEM VII. Tofind the dreadflz ofa river. EXAMPLE. Let the following figure represent a river, the breadth of which is required. Fix upon any object B, close by the edge of ‘ the river, on the side opposite to which you stand. By the help of your cross, make AD perpendicu- lar to AB; also make AC :: Cl), and erect the perpen- dicular DE; and when you have arrived at the point E, in a direct line with CB, the distance DE will be = AB, “a seer. I. LAND-SURVEYING. 87 the breadth of the river; for by Theo. 1, Part I., the angle AGE 2 DOE, and as AC: CD, and the angles A and D right angles, it is evident that the triangles ABC, CDE, are not only snnllar but equal. Note 1. The distance between A and the edge of the river must be deducted from DE, when it is not convenient to fix A close by the river’s edge. 2. This Problem may also be well applied in measuring the distance of any inac- cessible object: for let AC equal 8, CD equal 2, and DE equal 10 chains; then, by similar triangles, as CD : DE : : AC : AB, equal to 40 chains. PROBLEM VIII. To measure a line upon which them is an impediment. EXAMPLE. ‘ Let CDEF represent the base of a building, through which it is necessary that a straight $3 line should pass from A. Measure from A to m ,- at m, erect the perpendicular ma, which measure ‘ until you are clear of the impediment, b as at e. Erect the perpendicular ce, which measure until you are beyond the building, as at I). Erect the per- pendicular bd; and make 672 equal to a . g m F me, at which point you will be in a b w\ direct line with mA. Erect the per- pendicular nB, which measure; then Am added to the sum of CI) and nB, Will give the length of the whole line , AB. A Note 1. This Problem is very useful when you meet with ponds, bogs, buildings &c. upon a chain.line. 2. When a cross is not at hand. a perpendicular may be erected by the chain, in the following manner : Measure 40 links from m to r,- and let one end of the chain be kept fast at 7', and the eightieth link at m,- take hold of the fiftieth link, and stretch the chain so that the two parts mw, and rw, may be equally tight; then wili mm be perpendicular to 1221' ; for mw, mr, and rw, are in proportion to each other as 3, 4, and 5. (See Problem ‘29, Part I.) PROBLEM IX. To partfrom a rectangularfield, any proposed quantity of land, by a line parallel to one of its sides. RULE. Divide the proposed area, in square links, by the side upon which it is to he parted oil", and the quotient will be the length of the other side of the figure required. 88 LAND-SURVEYING. PART 111‘. Note I. ll‘ there be offsets on the line upon which the proposed quantity is to be parted off, deduct the area of the offsets from it, and proceed with the remainder as above. 2. Acres, reeds. and perches may he reduced into square links, by multiplying the whole quantity, in perches, by (5:25, the number of square links in a perch. EXAMPLES. 1. From the rectangle ABCD part off 2 acres, 1 mod, and 32 perches, by a line parallel to AD=700 links. D F 0 flow 2a. 1r. 32p.:392p., and 392 X 625 2945000 square links; then 245000 2450 . —~-————::—~~: _, { =53: " ‘ 700 7 300 11111 s T A L DF. '2. Part of 2a. 3r. 32p. upon the chain-line. AB, so that the ofisets taken on that line may be included. ’ A 1200 0 —*—*--—-*~—*———7 1000 ' 900 60 600 80 300 50 000 0 From A, g0 East. 2a. 37'. 32p.=295000 square links. ,, , 57000 the area of the offsets. 12,00)‘2380,00 the difference. 198,4 linkszAllzBF. Hence the irregular figure AGBFE, contains 2a. 3r. 32p. Note. When it is required to part from a trapezium, approaching very nearly to a rectangle, any number of acres, &e. by a line parallel to one of its sides ; it may be done by this Problem. PROBLEM X. To part from a trapezium or any irregular polygon what- ever, any proposed quantity of land, by a line drawn parallel to any of the sides, or by a line drawn from any (ft/1e angles, orfrom any assigned point in one of Me sides, to any oft/re opposite sides. RULE. Having surveyed and laid down the Whole figure, draw a guess-line in the direction required, parting off; as nearly SECT. I. LAND-SURVEYING. 89 as can be judged, the proposed quantity; after which, by the scale, measure with the greatest accuracy, the guess- line, and also the quantity thus parted-off. Then, if the guess~line or line of division be drawn from an angle, or from any assigned point in a side, divide the difference between the proposed quantity and the quantity parted off, by half the guess-line ; and the quotient will be the per- pendicular to be set off, on one side, or the other, of the guess-line, accordingly as the quantity parted off is more or less than the quantity proposed. To the end of this perpendicular, from the point assigned, draw a new line of division; and it will part of the quantity required. But if the guess-line be drawn parallel to any of the sides, divide the difference before mentioned, by the whole guess-line, and the quotient will be the perpendicular to be set off from each end of the guess-line, on one side, or the other, as above. EXAMPLES. 1. From a trapezium, whose dimensions arepcontained in the following notes, part off 2a. 27'. 241). by a line parallel to the side AB. .1313" . 1249 Diag 1000 Return to E— AC 1 1 12 Diag 1000 'R. of A: —DA 550 R. off D. #01)” 979 R. of? C. , ———~_._BC_ 557 ,RLOE B. ’73“ 1 l 14 , 1000 Begin I,” air}, and go West. 90 LAND-SURV EYL’G. PART III. Having laid down the figure, draw the guess-line mnparallel to AB ; and from 72, let fall the perpendicular an ; then suppose mn = 1058 links, an will be=230, and Aa21052 ; therefore Ba 2 1114 — 1052 = 62 links. Square links. Then 1055 x230—242650Ithe area of the trapezoid ’ _ L Acmm. And 230X31: 7130 the area (i; the triangle am. The sum is:9 49780 the area oiBthe trapezium 72m. 2a. 2r. 24p.=265000 15220 the difference between the quantity proposed, and the quantity parted of? by the guess-line, which being divided by 1058, we obtain 14.4 links, to be set off perpendicularly from m and 72 towards D and C. Hence EF is the true line of division ; and the trapezium ABEF contains 2a. 27'. 24p. As A is very nearly a right angle, measure, in the field, 230 + 14.4 2244.4 links, from A to F ; and upon any part of the line AB (towards B) as at e, erect the perpen- dicular er, which make = 2444 links; stake out the line ErF, and the work will be completed. Note. In order to prove the operation, find the area DCEF; then if it be equal to the difference between the area. of ABCD and the quantity parted oil; the work is undoubtedly right. 2. From any irregular field, whose dimensions are con— tained in the following notes, part- ofi’ 2a. 37‘. 20p. towards the line AE, by a fence made from the angle D to the side AB. EB 1398 Diag. 12 Diag. o m, proof-line, goes to D, R. off C. and measures 324. SECT. 1. Having laid down the figure, draw the guess—line D72, which suppose=766 links ; then the diagonal AD Will be: 824, the perpendicular Ea = 978, and the perpendicular ar : 372 links; also re will be : 228, and m = 52 links. LAND-SURVEYING. O 35 50 60 50 30 0 Begin at AC 1260 1000 R. off" A. EA * 400 200 000 R. off E. “DE— 600 450 300 000 i}; an‘L Dia a. U go West. . 2 LAND-SURVEYENG. PART HI. Square links. 267800 the area of the trapezium Ar DE. 12000 the area of the offsets 13h 16000 taken on the line A' 12348 ' Ar. The sum is =308148 the area of ANDEA. 2a. 37'. 20p. 2287500 208% the difference between the quantity proposed, and the quantity parted off by the guess-line, which being divided by 383, (half the guess-line,) gives 54 links to be set off from 72 towards A. Hence, DF is the true line of division ; and the irregular figure AFDE contains 2a. 37'. 20p. Now, by the scale, Ac :: 377 links. Measure therefore, in the field, 377 links from A to c ; stake out the line DcF, and the work will be completed. Note 1. The Rules given in this Problem, for parting-off land from irregular fields, are generally adopted by Practical Land—Surveyors; because they may be applied to any irregular figure whatever. Land, however, may sometimes be parted off more directly; for instance, the last example may be performed by finding the area of the irregular figure ADE, and subtracting it from the quantity to be parted ofi‘; then, if the difference be divided by half the line AD, the quotient will be the perpendicular of the triangle ADF ; the side AB being nearly straight from A to F. Now, at the distance of this perpendicular, draw a line parallel to AD ; and it will intersect AB in F, the point to which the division fence must be made. 2. It is not absolutely necessary to survey and plan the whole figure, in order to part a portion from it, as the guess-line and portion parted ofi“ may be measured in the field; but, in my opinion, the former, in general, is a more eligible method than the latter. 3. In order to divide a trapezium or an irregular polygon, among any number of persons, by fences made in a given direction, proceed thus : Part oil" the first per- son’s share, then from the remainder of the figure, part ofi‘ the second person's share; and thus continue, till the whole field be divided. 4. Those who desire to see a greater Variety of Examples in surveying single Fields, and to make themselves fully acquainted with the Methods of Laying out, Parting off, and Dividing Land ; also of Dividing a Common, &c. of variable Value, among any Number of Proprietors, in the Proportion of their respective Interests, may consult my Treatise on Practical Land-Surveying, Seventh Edition, in which I flatter myself they will find these subjects satisfactorily treated. PROBLEM XI. To reduce statute-measure t0 custommy, and vice versa. It has been before observed, that by custom the perch varies in different parts of England ; and With it, conse- quently, varies the acre in proportion. In Devonshire and part of Somersetshire, 15 5 in Corn- wall, 18; in Lancashire, 2i ; and in Cheshire and Staf- fordshire, 24 feet in length, are accounted a customary perch. seer. I. LAND-SURVEYING. 93 GENERAL RULES. I. To reduce statute-measure to customary, multiply the number of perches, statute-measure, by the square feet in a square perch, statute-measure; divide the product by the square feet in a square perch, customary measure, and the quotient will be the answer in square perches. II. To reduce customary measure to statute, multiply the number of perches, customary measure, by the square feet in a square perch, customary measure; divide the product by the square feet in a square perch, statute- measure, and the quotient Will be the answer in square perches. Note I. It is scarcely necessary to remark that the length of any perch mul- tiplied by itself will give the number of square feet, in a square perch of the same measure ; hence we have 16.5 x 16.5: 272.25, the statute perch ; 15x15: 225, the Devonshire and Somersetshire perch ; 18x18=324, the Cornwall perch; 21 x21: 441, the Lancashire perch ; and 24x 24:576, the Cheshire and Staffordshire perch. 2. When it is intended to find the area of an estate in customary measure only, it is generally thought most convenient to take the dimensions by a chain properly adapted for that purpose. The Devonshire and Somerset chain, is 60 feet; the Cornwall chain, 72 feet ; the Lancashire chain. 84 feet ; and the Cheshire and Stafl‘ordshire chain, is 96 feet in length. Each of these chains is divided into 100 equal links, in the same manner as the statute chain ; consequently, the customary measure is found by the same Rules as the statute-measure. 3. It may also be observed, that 4840 square yards make one statute acre; 4000 make one Devonshire or Somersetshire acre; 5760 make one Cornwall acre; 7810 make one Lancashire acre ; and 10240 square yards make one acre of the customary measure of Cheshire and Stafi‘ordshire. (See the Author’s Land~Surueyz‘ng. Part: II. III. and VI.) EXAMPLES. 1. In 25a. 27-. 20p. statute, how many acres, 8m. cus- tomary measure, of 15 feet to a perch ? By Rule I. 25A. 2R. 20?. 4 To? 40 47% 272.25 :16.5 X 16.5 72m 8200 8200 28700 8200 4,0 15 x 15:225)’1_1“162‘2m(496,1 ©1772—th erasing. Ans. 94 LAND-SURVEYING. PART III. 2. In 31a. 0r. 1}). customary measure, of 15 feet to a perch, how many acres, &c. statute-measure? By Rule II. 31A. OR. 1r. 4: E1 40 EST 225 24805 9922 9922 4,0 272.25fi’i‘i‘6’fi576d41ao 19152—26 725727277® Ans. 3. Reduce 56a. 372 36p. statute, to customary measure, of 18 feet to a perch. Ans. 47a. 37220112. 4. In 47a. 37‘. 2017. customary measure, at 18 feet to a perch, how many acres, &e. statute-measure ? Ans. 565:. 3r. 36p. PROBLEM XII. T0 survey and plan Estates 0r Lordships. Various methods are adopted by different surveyors, in taking the dimensions of Estates or Lordships; I shall, however, describe only four which I conceive to be the most accurate and practical. METHOD I. Having made yourself acquainted with the form of the estate, either by actual examination, or by the assistance of a previous plan, select two suitable places, at the greatest convenient distance from each other, as grand stations; and measure a principal base, or what is generally called a “main-line,” from one to the other, noting every hedge, brook, or other remarkable object, as you cross or pass it; taking offsets likewise to the bends or corners of the hedges that are near you. Next, fix upon some other suitable place, towards the outside of the estate, as a third grand station; to which, sncr. I. LAND-SURVEYING. 95 from each extremity of the diagonal or main—line, or from two convenient points in it, lines must also be run. These three lines being laid down, will form one large triangle; and in a similar manner, if necessary, on the other side of the diagonal or main-line, a second triangle may be formed. The survey must then be completed by forming smaller triangles, on the sides of the former; and measuring such lines as will enable you to obtain the fences of each in- closure, and prove the whole of your work. Arote l. The method of measuring estates by dividing them into triangles, is particularly described in my Survey- ing ; and illustrated with several rough and finished Plans, and an engraved field-book. It is also exemplified in this Work by Plate 1., which is a rough plan of an Estate exhibiting the chain-lines and stations used in taking the survey. The field-notes are not given, as they would have occupied too many pages of copper-plate; it may, however, be observed that the method of entering the notes, sketching the fences, 8:0. is precisely the same as in the field-book belonging to Plate 111.; and the directions of all the lines may be easily as» certained from the following particulars. The first or main-line leads from + l to + 8; the sc- cond line from + 8 to + 10; and the third from + 10 to —‘;~ 1 ; which three lines form the first large triangle. The fourth line extends from + 2 to + 15; and the fifth from + 15 to + 8; which two lines and part of the main-line form the second large triangle. The sixth line leads from + 9 to + 11; the seventh from + 20 to + G; the eighth from + 7 to + 22; the ninth from + 21 to + 4 ; the tenth from + 24 to + 13 ; and the eleventh from + 12 to + 23; which complete the survey of the first triangle. The twelfth line extends from + 5 to + 17; the thir- teenth from + 25 to the main—line, southward of + 3; the fourteenth from + l to + 14; the fifteenth from + M to + 26; the sixteenth from + 27 to + 16 ; the seven- teenth from + 18 to + 28; and the eighteenth from + 28 to + 19 ; which finish the whole survey. 2. The content of the Estate may be found in the fol- lowing manner : Measure the lines upon the Plan, and take the necessary offsets, by a scale of 8 chains to an inch; and v-Xsbm 96 LAND-SURVEYING . PART III. enter the dimensions in a field-book. From the dimen- sions thus obtained, draw a plan by a scale of 2 chains to an inch; then straighten the fences as directed in Pro- blem II. ; and measure diagonals, perpendiculars, 8:0. from which compute the content of each field. The diagonals, perpendiculars, and contents may be entered in a Book of Castings, similar to that belonging to Plate 111.; and if you should not have a scale of 8 chains to an inch, any other scale will do just the same for practice. 3. Taking the dimensions, 850. as directed in the last note, will be found of infinite service to the learner; as it will tend to make him very expert in entering the field- notes, laying down the lines, and casting the contents, what are no small acquisitions towards becoming a com- plete Land-Surveyor. SECT. 'I. O—-~-----‘--------- Scale 8 Chains to an I B‘ LAND-SURVEYL'G. PLAN 1. __-- ..-__°. ‘..-.--. ncn. Ft... Kr! s _.|I-.9|.IJI.¢[!EI! nvofovk. F- 97 m. i i I l 1 PART x t I l I ’7 I ”J & ‘ ‘ . / Q)» _ , . » .UJ , .1». .......... 9, rm .......... 20/ , , K ‘ )4 i‘.\ mi] . ~ 7 7 _ / 'D-SURVEYING. PLAN II. Scale 8 Chains to an Inch. \ , \ . \ x _ x _ \ . ~ \ 7 x A \ . 11 .2\ L , ;Ilc|ush0 l... ‘Ow‘xlnnxllyl l. . 10/ ‘ _ vl‘\l¥’n _ _l — \ _ \\ .. \ d~/ \ .1 ‘ . \ _ \\ . K _ xx _ \ . \ Z \ ....................... O\-||xlllxnlununv 98 SECT. I. LAND-SURVEYING. 99 METHOD II. Measure a main—line as nearly to one of the out-boun— daries of the estate as the curves in the hedges will per- mit ; noting the crossings of fences, and taking offsets as before directed. At a convenient distance, measure another main-line parallel or nearly parallel to the first line, so that a num- ber of fences running in that direction may be obtained; and from any two stations in the first line measure lines to any station in the second main-line, forming a triangle; so will a station in the second main-line become deter— mined or fixed. From the first main-line to the second, or from the se- cond to the first, measure lines in order to obtain all those fences which run in that direction. The remainder of the fences of the inclosures contained between the first and second main-lines being obtained by running lines in the most convenient manner; you will have completed the dimensions of a portion of the estate, which may then be laid down. Parallel or nearly parallel to the second main-line, and at a proper distance from it, measure a third; and pro- ceed with the internal lines as before, and you will obtain the dimensions of another portion of the estate, which may also be laid down; and thus go forward until you finish the survey. ZVote 1. This method is illustrated by Plate II., which displays the chain-lines and stations used in taking the survey. The first main-line leads from + 1 to + 6; the second line from + 6 to + 7; and the third line, or second main- line, from + 7 to + 16. The fourth line extends from + 16 to + 1; the fifth, or tie—line from + 16 to + 2; the sixth from + 2 to + 14; the seventh from + 17 to + 18; the eighth from + 12, through + 18, to + 3; the ninth from + 4 to + 10; and the tenth line leads from + 8 to + 5; thus all the fences between the first and second main-lines are obtained. The eleventh line, or the third main-line, leads from + 19 to + 29; and the directions in which all the rest of the lines were run, may be easily ascertained by inspecting the plan. 2. The content of this Estate may be found in the same manner as directed in Note 2, Method I. F 2 LAND-SURVEYING. PART III. METHOD HI. An estate of four sides may frequently be conveniently surveyed as follows : Measure four lines in such a manner that offsets or insets may be taken to the four out-boun- daries of the estate ; and tie the first and fourth lines to- gether by a diagonal or tie-line measured from one to the other, at the distance of five, six, or more chains from the angular point, according to the extent of the survey; thus you will be enabled to lay down the first four lines, and also the out-boundaries of the estate. Next proceed to obtain the internal fences, by measur- ing lines in the most convenient manner; some of which must be run from the first to the third, or from the second to the fourth line, or in some other more proper direction, so that they may become proofs, and fast-lines, into which other lines may be run with propriety. In thus proceeding, it is evident that a great deal will always depend upon the dexterity and ingenuity of the surveyor; as no directions can be given that will suit every particular case to be met with in practice. Note. —— This method is illustrated in my Surveying, by Plate VII. ; and also in this Work, by Plate 111., which is :1 Rough Plan of an Estate in the Township of Farnley, in the parish of Leeds. METHOD IV. The'method which I here intend to describe is a com- pound of all the methods of surveying with the chain; for as there are never two estates to be met with which are exactly alike, sometimes one method claims the preference, and sometimes another; but a skilful surveyor will always adopt that by which he can take his dimensions and proofs with the greatest accuracy by the fewest lines. If an estate be in the form of an irregular polygon of five, six, or more sides, and the fences very crooked, such an estate may generally be most easily surveyed by divid- ing it into triangles, as in Plate I.; but if many of the fences of the different inclosures run a considerable way in the same direction, and the fields in general be pretty neat trapeziums, it is commonly more eligible to proceed as directed in Method II. Sometimes an estate varies so much in its shape that all the methods before described may be used with propriety and advantage ; and it frequently happens that an in- genious surveyor adopts methods, in particular cases. en- error. I. LAND-SURVEYING. 101 tirely new to himself; care, however, must always be taken to make one line depend upon another, throughout the whole of the survey, so that when you come to lay it down, you may find no lines whose positions are undetermined. Nata—For the method of measuring Hilly_Ground, see my Land-Surveying, Seventh Edition, page 172. MISCELLANEOUS INSTRUCTIONS. l. I}: ranging a line, you must be extremely careful. to keep the poles in a direct line with each other; which you may accurately effect by always having, at least, two behind you ; and if your sight be obstructed by hedges, you must cut down the intervening parts. 2. In measuring your main, or any other chain-line, put down stations at every place to which you apprehend it may be necessary to run lines, in order to complete the survey; and if it happen that you put down more than are wanted, it will be immaterial. At each station you must put down a small stake, called a “ station~stake," with the number of the station upon it; and also cut a mark in the ground, which may be easily done by having a small spade upon the bottom of the offset-staff. 3. In measuring your internal lines, it will give you the least trouble to run them from one station to another, if you can make it convenient; if not, you must run them from, and continue them to some chain-line, and measure the distance upon that line, to the nearest station. The place where you run upon, or cross a chain- line, may be easily ascertained by setting up poles, at two of the neareststations in that line; and the crossing will be at the place where you are in a direct line with these poles. 4. The angle which the first line makes with the meridian line must be taken with a compass; and in doing this, an allowance must be made for the variation of the needle, which is about 24 degrees westerly. 5. In order to plan a large survey, provide a sheet of drawing- paper, ofa proper size for use; and trace with a pencil, a meridian or north and south line, in such a position, that your first station may be in some convenient point in this line. Then, from your first station, draw your first or main-line, making its proper angle with the meridian, which you may then take out with Indian rubber. Further directions appear to be unnecessary; as any person who is acquainted with the methods of laying down triangles, trapeziums, and trapezoids, will find no difficulty in planning an estate. 6. The most expeditious method of laying down crooked fences, is by means of an offset-scale, which must be used with the plot- ting-scale in the following manner: Lay one edge of the plotting- scalc close by the base-line, and bring the end of the offset-scale in contact with the edge of the plotting-scale, so that the edges of the scales may form a right angle; then by the edge of the offset-scale, prick off, in its proper situation, the first olfsct, with a pencil finely pointed. Keep the plotting—scale firm, and slide the offset-scale to F 3 102 LAND-SURVEYIN G. PART 111. the place of the next perpendicular, which prick oil' as hef‘re; and thus proceed until all the offsets are finished. 7. In order to find the area of an estate, practical surveyors gene- rally straighten the crooked fences of each field, as directed in Pro- blem II. ; and then divide the fields into trapeziums and triangles, and take such dimensions, by the scale, as are necessary to find the area of each field. They then collect all the areas into one sum; afterward find the area of the whole survey, as if it were a single field; and if it be equal or nearly equal to the sum of the separate areas, previously found, they justly infer that their survey is correct. 8. Those who do not approve of finding the area by the method of casting, may make use of the offsets taken in the survey, where con- venient; and if more he wanted, they may be measured by the scale; for in measuring a number of small parts by it, some will probably be taken a little too large, and others a little too small, so that, in the end, they will nearly counterbalance each other. 9. Practical surveyors generally lay down their lines by a scale of 4 chains to an inch, when their surveys are very large; and in com- puting the contents, they measure the bases and diagonals by the same scale, but. the perpendiculars by a scale of 2 chains to an inch; consequently the product of the bar“, and perpendicular ofa triangle, will be its area. To treat small surveys, in a similar manner, by a scale of 2 chains and of 1 chain to an inch, must, of course be very correct. 10. Rivers, large brooks, public roads, and common sewers, should not be included in the area, but. only delineated upon the plan; and marshes, bogs, heaths, rocks, &c. should also be represented or spe- cified, and their measurements separately returned. 11. Sometimes the content of each field is entered within the field itself; sometimes the fields are numbered, and their areas set down, one after another, in some convenient part of the plan; and some- times they are entered in a book of particulars. 12. When you wish to transfer a rough plan to a fresh sheet of paper, or to a skin of parchment or vellum, in order to make a finished plan, proceed thus: Take a sheet of writing—paper of the same size as the rough plan, and rub one side of it with black-lead powder; then lay it upon the sheet which you intend for your new plan, with the black side downward; upon both lay the rough plan; and upon the whole place weights or books, to keep them from moving. Next, run your tracer gently over all the boundaries upon the rough plan, so that the black lead under them may be transferred to the fresh sheet. Separate the papers and trace the lines thus transferred with a fine pen and Indian ink; as common ink ought never to be used in plan- ning. You must then proceed to enter such names, remarks, or ex- planations, as may be judged necessary; drawing by the pen, the representations of hedges, bushes, trees, woods, hills, gates, stiles, bridges, roads, the bases of buildings, &c. in their proper places; run- ning a single dotted line for a foot-path, a double one for a carriage- road, &c. Rivers, brooks, lakes, 8w. should be shaded with crooked or waved lines, bold at the edges, and fainter towards the middle. Hills. may also be represented by crooked lines, hold about the middle of the hill, but fainter towards the top and bottom ; and the bases of seer. I. LAND-SURVEYING. 103 buildings must be shaded by straight, diagonal lines. Roads should be shaded with a brownish colour, laid on with a camel-hair pencil; and fields in a variety of forms, with a fine pen and Indian ink. In some convenient part of the Plan, you may write, in conspicuous cha- racters, the title of the Estate, ornamented with a compartment. In another vacancy introduce the Scale by which the plan has been laid down; and also a meridian-line, with the compass or fiower-de-luca pointing north. The whole may then be bordered with black lines at a convenient distance from each other; and the space between them shaded by a hair-pencil, with Indian ink. 13. A plan well finished with Indian ink, as directed above, has a very elegant appearance, and is considered by most persons to excel those done in colours; but the work is very tedious, and requires much time to do it well, in consequence of which most surveyors prefer finishing their plans with colours. Some, however, not only embellish them with Indian ink, but also wash the different fields with various shades of colouring. 14. In colouring a plan, meadow and pasture ground should be washed with a transparent green, the pasture rather lighter than the meadow; arable land with a mixture of red and yellow, or of red, brown, and yellow, of various shades, so that too many fields may not appear exactly alike; and some surveyors use both light blue and lake, in colouring plans. An excellent green, of various shades, may be made by mixing light blue and gamboge. Rivers, brooks, lakes, &0. should be coloured with Prussian blue mixed with a little Indian ink. The hedges must be done with a strong shade, which should be softened off towards the middle with a lighter one. If the quickwood hedges be not done with a pen and Indian ink, they may be represented by running strong narrow shades, of various colours, round the boundaries of the different fields. 15. Reeves’s and Newman’s water-colours are entitled to recom- mendation. They must. be prepared for use in the following manner: Dip one end of the cake in clear water, and rub a little of it upon a clean earthen plate; then mix it with water, by your hair-pencil, until you have brought it to any consistence you please. Indian ink must be prepared for use in the same manner. 16. When parchment is used in planning, it must first be rubbed with clean flannel dipped in the best Paris whiting. This operation clears its surface from grease, and makes the pen slide more freely. 17. The learner will fully comprehend what has been said on the sub- ject of embellishing plans, by examining a well-finished coloured map. 18. The method of transferring plans by means ofa sheet of paper, rubbed with black lead, will do very well for small surveys ; but when the survey is large, it is necessary, not only to transfer the original plan, but also to reduce it to a smaller scale, in order that the finished plan may be ofa convenient size. This may be effected in different ways; as by squares, by proportional compasses, 8w. ; but the most expeditious and accurate method is by an instrument called a Pentagraph. Those who desire to make themselves acquainted with the method F4 104 LAND-SURVEYING. PART 111. of using this instrument, are referred to my Land-Surveying, Seventh Edition ; also to Adam's Geometrical and Graphical Essays, in which valuable work they will likewise find the description and use of the Plane-Table, Theodolite, Spirit-Level, and every other Mathe- matical Instrument. Note. — The area of the following Estate, Plate 111., was found from a Plan laid down by a scale of 2 chains to an inch. The crooked fences were straightened as directed in Problem 11., and the diagonals measured by the scale used in planning; but the perpendicular; by a scale ofl chain to an inch. The diagonals, perpendiculars, &c. are contained in the following Book of Castings. A Book qf Dimensions, Castings, cS'c Belonging to Plate III. c: 5' E. 3"“ 3 j 8* S p: Quantit ’ uantitv in 2 g c: ,2 35 in ) QA. RIP. . in E 8 E A. Dec. 23 t3 2; 3 .3 1 1375 143 227 370 5.08750 5 0 14 2 1542 280 298 578 8.91276 8 3 26 3 1120 264 208 472 5.28640 5 1 . 5g— 4 1226 217 213 430 5.27180 5 1 34 5 1225 220 194 414 5.07150 5 0 11; 6 1552 374 252 626 9.71552 9 2 344 7 1088 210 168 378 4.11264 4 0 18~ , 8 874 130 _ 180 1.57320 1 2 11;} ‘ 9 { 1262 254 76 330 4.16460 } 5 0 R 1030 84 _ 84 .86520 I 10 790 136 190 y 326 2.57540 2 2 12 Whole Quantity — 52.63652 ‘52 2 21;. LAND-SURVEYING. 1 05 FINIS ‘ [o-r/J’. $35 4029 /;/3/2'r§6' / .50061 _, 63%: / , ,1 %m0+/*/0 3/0? a '4K 02+J’. 662 5002‘? (J 1/0026f Cr/k/flf/Z'f/ly/O C/V /10, .00+//0. 1/102 204/100 1000 WWW—— . 24 (5’00 ' '46} <7.th (95/ 40 .300 , ,1 ‘flffi— 6%07/0 +1/r/0 %/ / ‘9 ' [0+6'. 409/ 434600 504200 +//J’ 400/015 y 1000 ,900 +// 501900 7 6’0 .400 4.0 400 C . c/émm+//.70 (1/7 “fine-d”. ”—‘_ (av/a. 20/070 20° 00 3d” €00 2.50 106 ¢%%9 @9250 l Jc/ /&&7)2/ + y /w77y + {4/ 0 WI. LAND-SURVEYING. /n + /CL , ”z // 37.30 730800 OIfiflloo .fiOQQW 5034/ 09 000/ 006 000 10 00 0000 000 000 v , W W% E Kama mojlowl / arm“ W I. + mL #+ I .I o .7 % /007’ 94 1470 \50 6 7", +//‘ 006 #20080 / ~~/\6)\~9‘6’\}\ 0% _9 0‘ 7 02 ,1 0 4 0. ff 8 0 3 0 100 _Z/f’_ J 12 0’0 +76. ~o 24 c: +IJ $36 6: [0+2 Z; V, a? m PART III y SECT. I. LAND-SURVEYING. 107 50 I m 674 §\\ Q’AN was gas 1% QQQ 94 d’2 ($0 3. [0+9. ”ch QQQ § \§ K K\ Q9: 03% QQ N) 6‘ %; 2d) \ Q Q Q \x E :36" 40 30 ($0 9. _/0 + a“. K i» Q Q QC: QQ QQ E5 \§ § 3 0 6' 0 62 3 0 0 0 2700dv 230090 e7 4/70 + 4 2 0 0 0 #0 /600M‘ + 3 1 2 0 0 6d? 7 0 0 0 + 2 d’ 0 0 63 4 0 0 63’ .0 0 0 40 . o/gM/v +70 Waite I April 18th, 1814. The 1Votes qf an Estate situate in 1 the Township 0f Farnley, in the Parish 0;“ Leeds. FIELD BOOK. PART III. LAND-SURVEYING. Plan from tlzcjbreyoing Field Book. Scale 6 Chains to an Inch. .108 seer. I. LAND-SURVEYING. 109 THE METHOD OF SURVEYING WITHOUT THE CROSS-STAFF. IT has been suggested, and deemed advisable in revising this work, to add a few more examples on the subject of land-surveying, where it will be observed that we have varied the method in a slight degree from the former, chiefly in that of forming and keeping the field-book, as also without the aid of the cross-staff; in taking dimen- sions, the stations being represented by the numbers 1, 2, 3, 4, &c., which are read from station first to station second, from second to third, from third to fourth, &c. ; inserting three cyphers at the commencement of each station, and by keeping the fence on the right hand, the offsets will always appear on the right-hand side of the field-book, unless you cross the fence, when they will be- come insets: the learner is, however, strongly recom- mended to make himself fully acquainted with the former part on this subject. PROBLEM XIII. To measure afield of four sides. EXAMPLES. 1. Let the following figure represent a field to be measured without the aid of the cross-staff. Begin at station 1, and measure the first side to station 2, from 2 to 3, and so on until you get round the field; taking special care to take and enter all the offsets in a book, as the irregularity of the fence may require; then measure the diagonal from 1 to 3, and the field-book will be completed. CONSTRUCTION. Draw the first side, and set off from 1 to 2, 680 from a scale of equal parts, then take the second side 720, and the diagonal 1080 in your compasses, and the point of in tersection will be the third station ; in like manner, 800 and 515 taken from the same scale, and the point of inter- section will be the fourth station; the diagonal being drawn, and the perpendiculars measured on the same scale, will be 7722, 450, and 724, 360 PART III LAND-SURVEYING. 110 «flaw .04 4 3 2 0 1+ 2 24 ~0 Ono/filao 00500 OAMOW 33005 50000.00000 00000 ,100000 10860 02060 20840 18010 54210 87520 76420 N 76430 7 11080 I» 3 2 1 FIELD BOOK. \‘a , SECT. I. LAND-SURVEYING. 111 CALCULATIONS. Area of the trapezium - 874800 Area of the offsets, first side - 27000 Area of the offsets, second side 24480 Area of the offsets, third side - 24840 Area of the offsets, fourth side 30420 Double areas 2)981540 4,90770=4a. 3r. 25p. 2. Let the following figure represent a field of five sides, it is required to form the field-book, and find- its area. 4 540 7 7 00 3 32001 .5 0000 30005 20020 1‘000 10022 A 0000 30004 34010 30022 210/0 1600 3 0000 33003 22040 44010 1003 2 0000 35002 21040 12020 00 A I Q Q Having measured the field, formed the field-book, and constructed the figure, we find the perpendiculars to be, m2, 160, 724, 322, and 05, 146 links. 112 LAND-SURVEYIN". PART 11]. j , A 3 Area of the trapezium - - 289200 1 ‘ it ‘1 Area of the triangle - - 78840 * ~' & Area of the first side offsets - 13400 . i 3 : Area of the second side offsets - 13840 '~ ' .; Area of the third side offsets - 5220 if ., H ‘ Area of the fourth side offsets - 6320 ; i a ” Double areas 2)406820 : , ‘ 2,034fi) ,’ _%_ . r . 9 '1 ; .7 ‘ Ans. -a. Or. OEP. .- ." 3. Lay down a field, and find its area, from the follow- ing notes. 3 1 l LAND-SURVEYING. SECT. I. ,5 1/31 A. 3 M 2 ”0 /4 [04+ ) 3 OMOO 002M00 MOO 0W0u0n 0000 2000_000000 00000 000 000 5040 8110 2509/40 07460 263nm0¢u0 321.0 2210 322 0 311 0 3291110 Cu .0 I4 3 2 Qt. 7 O FIELD BO 0L t3 . LAND-SURVEYING. PART III. Having laid down the figure from a scale of equal parts, we find the perpendiculars as follows : m2, 184, 724, 296, and 05, 200, consequently the respective areas will be as below :— Square links. Area of the trapezium - - 240000 Area of the triangle - - 96000 Area of the first side - - 2340 Area of the second side - - 6040 Area of the third side - - 7040 Area of the fourth side ~ - 4472 Area of the fifth side - - 3720 359612 Deduct inset 4000 Double areas 2)%56T2 mm 4: 3,11224 40 4,4896?) Ans. 1a. 3}. 4721—! , sncr. I. LAND-SURVEYING. 1 15 4. Let the following figure represent a field of six sides it is required to be measured and planed. CONSTRUCTION. Suppose the field-book to be formed as below, draw the first side, and set off 700 from your scale of equal parts; take 600, the second side in your compasses, which inter< sect with the diagonal 1, 3, 1000, and the point of inter- section will be the third station; with the side 1, 6, and the diagonal 3, 6, intersect again, and you have the sixth station. Then take the side 3, 4, and the diagonal 6, 4, and the point of intersection is the fourth station ; again, take the two sides 4, 5, and 6, 5, and the point of intersection is the fifth station : thus the field will be divided into two trapeziums, and the perpendiculars will be 712, 340, m6, 220, p3, 310, 05, 470. 6' 6‘20! 3 4400‘ 710003 6800/ xe—iré—eo {7C 31 0 w; :zre 6‘ 0000 #479006“ 510000 68005 4700 32024 21020 1,. 0000 50004 34030 1800 3 0000 60003 3,200 2 00030' ,70002 55044 3000 18050 I 0000 1 16 LAND-SURVEYING. PART III. Area of the trepeziums - - - 1043600 Area of offsets on the 1st, 2d, 3d, and 4th sides 66360 11709960 Area of the inset on the 6th side ~ - 224100 Double areas mfl’fécfi)‘ $743780 4 177571020 40 301047800 Ans. 5a. 1r. 30p. Note I. It must always be remembered that when the cross-stafi‘is not used in the field in measuring the perpendiculars from the diagonals &c., great care should be taken in laying down the field accurately from the scale of equal parts; and, for this purpose, it will be necessary that: you use as large a scale as possible in order to get the true dimensions, and when due attention is paid to these observations, it is a matter of doubt, whether the perpendiculars can be measured truer by using the cross in the field, or by measuring them on the figure by the same scale with which the field was constructed. 2. The plan of a field may be very readily taken by laying the rough draft upon a sheet of paper, then with a fine pricker or needle begin atone corner of the figure, and prick holes through the paper so as to leave an impression on the paper below : this must be continued at all angles, at every offset, and where every oilSet begins and ends until you get round the field, then draw the lines with a fine pencil ; it may then be coloured if required. SECT. n. MISCELLANEOUS QUESTIONS. 117 SECTION II. MISCELLANEOUS QUESTIONS CONCERNING SUPERFICIAL MENSURATION. 1. THE length of a floor is 35 feet, and its breadth 24 feet ; what is the area of three such floors, deducting for two well-holes, one of which measures 10 feet 6 inches by 4 feet 3 inches, and the other 8 feet 9 inches by 3 feet 6 inches? Ans. 244443 feet. 2. From a mahogany plank 16 inches in breadth, 6 square feet are to be sawn off ; at what distance from the and must the line be struck? Ans. 4% feet. 3. What must be the length of a chord which will strike the circumference of a circular plantation that shall contain just an acre and a half of ground? Ans. 48.072 yards. 4. A gentleman has ordered a rectangular court-yard to be paved, which measures 45 feet 9 inches in front, and 32 feet 6 inches broad. A foot-path 5 feet 6 inches in breadth, leading to the front door of his house, is to be laid with Portland-stone, at 33. 4d. per square foot, and the rest With Purbeck-stone, at 2s. 3d. per square foot; what will be the expense of the Whole? Ans. £176. 193. lid. 5. The annual rent of a triangular field is £43. 158.; its base measures 25, and perpendicular l4 chains; what is it let for per acre ? Ans. £2. 10s. 6. The transverse diameter of the ellipse in Grosvenor Square measures 840 links, and the conjugate 612, within the wall; the wall is 14 inches thick; what quantity of ground does it enclose, and how much does it occupy? Ans. The wall encloses 4a. 07'. 610., and occupies 1760.531 square feet. 7. The area of a right-angled triangle is 60, and the hypothenuse 17; required the two legs. Ans. 15 and 8. 8. Two sides of an obtuse-angled triangle are 5 and 10 chains ; what must be the length of the third side, that the triangle may contain just 2 acres of ground? Ans. 8.06225 or 13.60147 chains. 118 MISCELLANEOUS QUESTIONS mar 111. 9. Two travellers, A and B, departed from an inn, at the hour of eight in the morning; A proceeded north-west, at the rate of 6 miles an hour, and B north-east, at the rate of 8 miles an hour; how far were they distant from each other, at twelve o’clock of the same day? Ans. 40 miles. 10. T wo boys amusing themselves at a game called snatch-apple, in a room 13 feet high, find that by standing 12 feet from each other, the apple, which is suspended from the ceiling by a string, and in a right line between them, when put in motion, just touches each of their mouths. Required the area of the sector described by the string and apple; the perpendicular height of each boy’s mouth, from the ground, being 5 feet. Ans. 64.35726feet. 11. “That is the area of an isosceles triangle inscribed in a circle whose diameter is 24; the angle included by the equal sides of the triangle being 30 degrees? Ans. 134.3538. 12. A maltster has a kiln that is 18 feet square ; but he intends to pull it down, and build a new one, that may dry three times as much at once as the old one; what must be its length, if its breadth be 24~~feet? Ans. 40% feet. 13. The side AB of a triangular field is 40, BC 30, and CA 25 chains; required the sides of a triangle parted off by a division-fence made parallel to AB, and proceeding from a point in CA, at the distance of 9 chains from the angle A. Ans. 16, 19.2, and 25.6 chains. 14. A field in the form of a right-angled triangle is to be divided between two persons, by a fence made from the right angle meeting the hypothcnuse perpendicularly, at the distance of 880 links from one end ; required the area of each person’s share, the length of the division-fence being 660 links. Ans. 2a. 37‘. 24—37. and 1a. 27*. 21% . 15. It is required to part from a triangular field whose three sides measure 1200, 1000, and 800 links respectively, 1 acre, 2 roods, and 16 perches, by a line parallel to the longest side. Ans. The sides of the remaining triangle are 927, 772%, and 618 links respectively. 16. The perambulator, or surveying wheel, is so con- trived, as to turn just twice in the length of a pole, or 16-;- feet ; what is its diameter? Ans. 2.626feet. 17. Required the side of an equilateral triangle whose area is just two acres. Ans. 6.79617 chains. SECT. II. 1N SUPERFICIES. 119 18. The sides of a triangle are 20, 30, and 40 respec- tively; what is the area of its inscribed circle? Ans. 130.8999. 19. In an isosceles triangle, two circles are inscribed touching each other and the sides of the triangle; the diameters of the circles are 9 and 25; required the sides of the triangle. Ans. 44.27083, 44.27083, and 41.66666. 20. The base of a field, in the form of a trapezoid, is 30, and the two perpendiculars are 28 and 16 chains re- spectively; it is required to divide it equally between two persons, by a fence parallel to the perpendiculars. Ans. The divisionfence is 22.8035 chains, and it di- vides the base into two parts, whose lengths are 17.0087, and 12.9913 chains respectively. 21. A field in the form of an equilateral triangle, con- tains just half an acre; what must be the length of a tether, fixed at one of its angles, and to a horse’s nose, to enable him to graze exactly half of it? Ans. 48.072 yards. 22. The diameter of a circular estate is 25 chains; what is the length of the chord which divides it into two segments whose areas age to each other as 2 to 1 ? ' Ans. 24.1062 chains. 23. In turning a one—horse chaise within a ring of a certain diameter, it was observed that the outer wheel made three turns while the inner wheel made two. The wheels were of equal height; and supposing them fixed at the statutable distance of 5 feet asunder on the axle-tree; what was the circumference of the track described by each wheel ? Ans. The length of the track described by the outer wheel is 94.248 feet, and that described by the inner wheel 62.832 feet. 24. If the frustrum of a cone whose diameters are 8 and 12 feet respectively, be made to.revolve with its slant side upon a horizontal plane, until it returns to its first posi- tion; what will be the area of the space passed over by the slant side, the length of which is 10 feet? ' Ans. 1570.8feet. 25. Being desirous of finding the height of a steeple, I placed a looking-glass at the distance of 100 feet from its base, on the horizontal plane, and walking backward 5 feet, I saw the top of the steeple appear in the centre of the glass; required the steeple’s height, my eye being 5 feet 6 inches from the ground. Ans. 110 feet. 120 MISCELLANEOUS QUESTIONS. PART III. 26. Three men bought a grinding-stone of 50 inches diameter, each paying :i‘ part of the expense; what part of the diameter must each person grind down for his share? Ans. The first must grind down 9.1752, the second 11.9537, and the third 28.8675 inches. 27. Wanting to know the height of the cathedral at York, I measured the length of its shadow, and found it to be 200 feet. At the same time, a staff 5 feet long, cast a shadow of 4 feet; required hence the height of that elegant and magnificent structure. Ans. 250feet. 28. At Matlock, near the Peak, in Derbyshire, where there are many surprising curiosities of nature, is a rock by the side of the river Derwent, rising perpendicularly to a wonderful height, which, being inaccessible, I endea- voured to measure in the following manner: At the dis- tance of 340 feet from the bottom of the rock, I fixed a staff, 8 feet in length, perpendicularly to the horizontal plane ; and at a convenient distance from this, I fixed an- other, 3 feet long, so that by looking over both their tops, I could just see the summit of the rock. The distance between the staves I found to be 4 feet 6 inches. From these data I determined the perpendicular altitude of the rock ; and you are requested to repeat the process. Ans. The height of the rock is 128.5925 yards. 29. There are two columns in the ruins of Persepolis, left standing upright; one of which is 64: feet above the horizontal plane, and the other 50. Between these, in a right line, stands a small statue, the head of which is 97 feet from the summit of the higher, and 86 feet from the top of the lower column, whose base is just 76 feet from the centre of the figure’s base; required hence the distance between the tops of the two columns. Ans. 157.03687feet. 30. A gentleman a garden had, Five score feet long and four score broad; A walk of equal width half round He made, that took up half the ground: Ye skilful in geometry, Tell us how wide the walk must be. Ans. 25.96876 feet. Note 1. Persepolis, mentioned in the 29th question, was the ancient capital of the Persian empire. This immense and renowned city was taken by Alexander the Great, about 300 years before the Christian era, who at the instigation of the de- praved courtezan Thais, ordered it to be set on fire. Its magnificent ruins are about 50 miles N. E. of Schiras, the present capital of Persia. 2. The 30th question was first proposed in the Ladies‘ Diary for the year 1709. The poetry is not very elegant; but the question is more original in its present form than it would be in prose. SEGT. L MENSU‘RATION or sonnet ‘ 121 PART IV. MENSURATION or SOLIDS. \ SECTION I. DEFINITIONS. 1. A SOLID is a figure which generally consists of three dimensions; viz. length, breadth, and thickness. . 2. The measurement of a solid is called its solidity, capacity, or content. 3. The contents of solids are estimated by a cube, whose side is one inch, one foot, one yard, 8m. ; hence the solidity of a body is said to be so many cubic inches, feet, yards, Ste. as are contained in that body. 4. A cube is a solid having six equal square sides. 5. A parallelopipedon is a solid having six rectangular sides, every 1 - opposite two of which are equal and ‘ parallel. 6. A prism is a solid whose ends are two equal, parallel, and similar plane figures; and its sides rectangles. It. is called a triangular prism when its ends are triangles; a square "3% prism, when its ends are squares; a pentagon§~ 2) arm, when its ends are pentagons, 8m. ‘1' ,h’ , G g _ 122 MENSURATION PART IV. 7. A cylinder is a solid conceived to be described by the revolution of a right- angled parallelogram about one of its sides, which 1emains fixed, and is called the axis of the cylinder; or it is a solid whose ends a1e parallel circles, and its sides right lines. 8. A pyramid is a solid the base of which is any plane figure what— ever, and its sides are triangles, meeting in a point, called the vertex of the pyramid. 9. A cone is a solid conceived to be described by the revolution of a right-angled triangle about one of its legs, which remains fixed,’ and is called the axis of the cone; or it is a pyramid of an infinite number of sides, having a circle for its base. 10. The Ifrustmn of a pyramid o1 cone is that part 11 Inch remains, when the top is cut off by a pl:1 ne pa1 allel to the base. The pa1t cut off the top is called a segment. 11.A wedge is a solid whose base is arectangle, its two ends plane triangles, and its two opposite sides terminate in an edge. 12. A pfismoid is a solid whose bases or ends are two right-angled parallelograms, being parallel but not similar to each other; and its sides four plane trapezoids. sucr. I. or SOLIDS. 123 by the rotation of a semi-circle about its diameter, which remains fixed, and is called the axis or diameter of the sphere; or it is a solid bounded by one continued convex surface, every part of which is equally distant from a point within, called the centre. 14. The segment of a sphere is any part of it cut off by a plane. If the plane pass through the centre, it will divide the sphere into two equal parts called hemispheres. 15. The zone of a sphere is a part intercepted between two parallel planes; and if these planes be equally distant from the centre, it is called the middle zone of the sphere. 16. A circular spindle is a solid conceived to be formed by the revolu- tion of a circular segment about its chord, which remains fixed. 17. A cylindrical ring is a round solid, or a cylinde bent into a ring. 18. A regular body is a solid contained under a certai; number of similar and equal plane figures. 19. The faces of the solid are the plane figures under which it is contained; and the linear sides or edges are the sides of the plane faces. 20. The greatest number of regular bodies which can be formed is five; viz. lst, the tetraedron, or regular pyramid, which has four triangular faces; 2nd, the hex- aedron, or cube, which has six square faces; 3rd, the octaedron, which has eight triangular faces; 4th, the do- decaedron, which has twelve pentagonal faces; and 5th, the icosaedron, which has twenty triangular faces. Nole. If the following figures be made of pasteboard, and the lines cut halt through, so that the parts may be turned up and glued together, they will represent the five regular bodies; namely, figure 1 the tetraedron, figure 2 the hexaedron, figure 3 the octaedron, figure 4 the dodecaedron, and figure 5 the icosaedron. G2 . . g. 1 L MENSURATION PART 1 '. \ 4 A/\ 9&9: \/\/\/\/ OH“ 3\/ \ A TaZJIe of Solid fiIeasur° e. Cubic Inches. Cubic Foot. I 1728 = 1 . Cubic Yard. 1 46656 27 z: 1 ( c.1161.- 7762892 44922;- 16ggl = IV *0 Fur! 496793088000 52874196000l 10648000 1 I I l ( 64.000 1358061056000I47191952000i5451776000 82768000 J J __ (1-1;- I’ROBLEM I. Tofind the solidity ofa cube. RULE . Ifiultiply the side of the cube by itself, and that product again by the side; and the 121st product will be the 60111111] required. Iwz‘el. The surface of a cube may he f1 unzi by multiplying 111-11 square of 115 5111‘ by six. ’lhe s1de ofa cube ma3 be f011n11‘o3 ext; ".r‘ting the (111111. mot of its 5011 11. :3. .The 1(st method 01 ghing bo3s a comm tent knowh dge ()1 the (1111'.le 3 19 ms in M1 2737511110071 0]” Solids, is to 111w 1111. 113mm {01.11111011300161‘11i3 wi. tau 1391”} 1.113.111t21g10u5; as the 16.111101 11.111. 1. be tau: .1: both the 1736075,! and [1:1 ,1; .150 .1: the 5.1m: time. SECT 1. OF ’somns. 125 EXAMPLE 8. 1. The side AB or BC of the cube ABCDEFGH is 5 feet 6 inches ; what is its solidity? By becimals. By Duodece'mafs. Feet Ft. In 5 5 5 6 5.5 5 6 . 275 27—6 275 2 9 30.25 30 3 5 5 5 6 15125 151 3 pa 15125 15 1 6 166.375 ' 66 4 ’5 Am. . 12 1 4.506 12 . Ans. 166]}. 4 in. 82m. 6 000 o By Logarithms. The log. of -. . . . 5.5 is . . . . . 0.7403527 flfultiplg/ by the index of the power . . . 3 The product is the log. cf 166.375, the ans. . 2.2210881 2. “That is the solidity of a Cube, whose side measures 3 feet 6 inches? Ifere 3feet 6 inches=3.5; and 3.53:3.5 x35 X 3.5: 12.25 x 3.5 242.875 feet =42ft. 10 in. Spa. (07' 6” se conds) the solidity required. 8. If the side of a cubical piece of timber measures 2 feet 9 inches; What is its solid content? . Ans. 20 ft. 9 in. 6:11pm 4. “That is the solidity of a cubical stone, Whose side. measures 2 feet 10 inches? Ans. 22ft. 8 in. 11” 4’”, G 3 126 MENSURATION PART IV. 5. that did the inside of a cubical bin, that holds exactly 5 Winchester bushels, cost painting, at l%ld. per square foot? Ans. 28. 61nd. PROBLEM II. To find the solidity of a parallelopipedon. RULE. Multiply the length by the breadth, and that product by the depth or altitude, and it will give the solidity required. Note 1. In order to find the surface of a parallelopipedon, multiply the periphery of its end by its length, and the product will be the area of the sides ; of which add twice the area of its end, and the sum will be the whole surface. 2. The surfaces of all similar solids are to each other as the squares of their like lineal dimensions. This is evident from Theorem 16, Part I. 3. If the content of a parallelopipedon be divided by the product of any two of its dimensions, the quotient will be the other dimension. 4. Parallelopipedons of equal altitudes are to each other as their bases; and if equal bases they are to each other as their altitudes. EXAMPLES. 1. What is the solidity of the pa- rallelopipedon ABCDEFGH, Whose length AB is 12 feet 9 inches, breadth BC 8 feet 6 inches, and depth or alti- tude AR 6 feet 3 inches? By Decimals. By Duodecimals. Feet. Feet. In. 12.75 12 9 8.5 8 6 6875 T0270 10200 6 4 6” 1087373 108 4 6 6.25 6 3 0 541875 650 3 0 216750 27 1 1 6’” 650250 67774—1_¥6 A225. 67734373 __1,__, ~24) 12 113—2560 12 1.500% Ans. 677ft. 4m.1” 6’”. 666656 sncr I. or soups. 127 By Logarithms. The log.‘0f . . . . 12.75 is . . . . 1.1055102 . . . . 8.5 is . . . . 0.9294189 . . . . 6.25 is . . . . 0.7958800 Their sum is the log. of 677.34375, the ans. 233368001 2. The length of a piece of timber is 24.65 feet, the breadth 2.82 feet, and the depth 1.56 feet; what is its solidity? Ans. 108.44028feet. 3. In a quarry near Balbec, is a stone whose length is 70 feet, its breadth 14 feet, and its depth 14 feet 5 inches; what is its solidity? Ans. 14128 ft. 4 in. 4. The length of a cellar is 25 feet, its breadth 15 feet, and its depth 7 feet; what did it cost digging, at 3%d. per solid yard? Ans. £ 1. 103. 411,—d. 5. A brick-kiln measures 27 feet 6 inches in length, 18 feet 4 inches in breadth, and 10 feet 9 inches in height; how many bricks, each 10 inches long, 5 inches broad, and 3 inches thick, will it contain at once, admitting them to be close piled? Ans. 62436. 6. A parallelopipedon measures 18 feet in length, 12 feet in breadth, and 8 feet 6 inches in depth; what did it cost painting, at 9%d. per square yard? Ans. £4. 23. lOid. PROBLEM III. Tofind the solidity of a. prism. RULE . Multiply the area of the base by the perpendicular height of the prism, and the product will be the solidity. Note I. If the area of the base in inches, be multiplied by the height in inches, the product must be divided by I728. in order to obtain the content in cubic feet; but when the area of the base in inches, is multiplied by the height in feet, the product divided by 144, will give the solidity in feet. 2. The surface of a prism may be found thus: Multiply the area of one side by the number of sides ; to the product add twice the area of the base, and the sum will be the surface required. Or, if the circumference of a prism be multiplied by its length, the product will be the surface of all the sides; to which add twice the area of its base, and the sum will be the whole surface. h3.bThe altitude of a prism may be found by dividing the solidity by the area of t c ase. 4. The area of the base may be found by Problems 11 and 12, Part II. 5. All prisms and cylinders of equal bases and altitudes are equal to each other. EXAMPLES. 1. Required the solidity of the triangular prism ABC DEF, whose length AB is 14 feet 8 inches, and each of G 4 128 MENSURATION PART IV. the equal sides of the base ADE 6 feet 4 inches. Here 14ft. 8 in. : 176 inches, and 6ft. 4in.=76inches; then by Prob. 12, PartII., we have .4330127 x 763 : .4330127 X 5776 : 2501.0813552, the area of the base ADE. Or, [23/ PIOI). 5, Part II, we have (76 +76+76)—:-2 :228—2—2:114,Izalfthe sum of the sides; and 114— 76: 38: each 9 emamder, - whence ./ (114 X 38 X 38 X 38) : M625 5408: 250108136, the area of the base as before , then (2501'08136 X9 176) — 1728 : 44019031936 — 17 2:8 254373976 feet :254ft. 8 2'72.10pa., the solidity required. ‘ 2. The perpendicular altitude of a stone pillar is 8 fe ct 9 inches, and the side of its square base 2 1eet3 inches; What is its solidity? Ans. 44ft 3 271.61Pa. 3. The length of a square bee m 01 timber is 32 feet 6 inches, and its circumterence 01 gi1t 5 feet; "$11 at is its solidity? Ans. 50ft. 9 2'22. 44 pa. 4. The side of a pentagonal stone measures 10 inches, and its length 9 feet 4 inches ; what is its solidity? A71s.11.15124feet. 5. A hexagonal prism measures .1 2fcet 6 inches across the centre 01 aits end, from corner to corner, and 25 feet 3 inches 111 length; What 1s its solidity? [1223. 102. 502225 f et. 6. 111 the walls of Balbcc is a stone in the form 01 square prism, whose length measures 63 feet, and its breadth and depth each 12 feet ; What is its solid content? Ans. 9072feet. 7. The gallery of an assembly-room is supported by 16 octagonal prisms of wood, each of which measures 4 feet 8 inches in circumference, a d 12 feet 111 height- equired the solidity of the Whole, and what they cost pailnting at 9d. per square yard? Ans. The solidity Q)" the 16 prisms is 315.4572368feet, 72d they cost £3. 148. 8d. painting. Note. Balbec, mentioned in the sixth question, is a town of Syria. situated about ".7 miles noith of Damascus. Its remains 01 antiquity display, according to the best judgts the boldest plan that ever was attempted. The inhabitants of Asia consider Solomon as its founder; and the nobleness of the architecture, the beauty of the ornaments, and the stupendous execution of the whole, seem to fix its foundation to a period belore the Christian era; but. in all probability, the Jens knew little of the Grecian stile of building and ornamenting 1n the time of Solomon. gsncr I. OF somns. .129 PROBLEM IV. Tofind the solidity of a cylinder. RULE. Multiply the area of the base by the perpendicular alti- stude of the cylinder, and the product will be the solidity. Kale 1. If the circumference of a cylinder be multiplied by its height, the product will be the convex surface; to which add twice the area of the base, and the sum will be the whole surface of'the cylinder. 2. The altitude of a cylinder may be found by dividing the solidity by the area of the base. 3. The area of the base of a cylinder may be found by Problem 15. Part ll. 4. Similar prisms and cylinders are to each other, as the cubes of their altitudes. or of any other like dimensions. EXAMPLE S. 1. Required the solidity of the cylinder ‘ABCD, the diameter of the base AB of which is 2.76 feet, and the altitude BC 4.58 feet. Here 2.762 X .7854 r: 2.76 X 2.76 X .7854: 7.6176 x .7854 : 5.98286304, tile area of the base; and 5.98286304 X 4.58 = 274015127232 feel, the solidity 7‘6- gulred. 2. The altitude of a cylindrical stone column is 8 fee- 3 inches, and its circumference 5 feet 6 inches ; what is its solidity? Ans. 19.86018 feel. 3. The circumference of a cylindrical piece of timber is 6 feet 8 inches, and its length 24 feet; what is its solid content? Ans. 84.88533 fleet. 4. The diameter of a rolling—stone is 18.7 inches, and its length 4 feet 9 inches; required its solidity. Ans. 9.05952feel‘. 5. The diameter of a well is 3 feet 9 inches, and its depth 45 feet ; what did it cost sinking at 73. 3d. per cubic yard? Ans. £6. 13s. 5451’. 6. The greater diameter of a hollow iron roller is 1 foot 9 inches, the thickness of the metal 1% inch, and the length of the roller 5 feet ; now supposing a cubic foot of cast iron to weigh 464 pounds avoirdupois; what did the roller cost at 193. 9d. per hundred weight, and how many times will it turn round in rolling 5 acres of land? Ans. The roller cost £13. 15. 043d” and it will turn round 7923.16926 times in rolling 5 acres of land. G o ~ I“-_>.V\~r,’_»:i“1 J30 ' MENSURATION PART Iv. PROBLEM V. Tofind the solidity ofa pyramid. RULE. Multiply the area of the base by .3. of the perpendicular height, and the product will be the solidity. Note 1. The surface ofa pyramid may be found thus: Multiply the perimeter of the base by the length of the side, or slant height, and half the product will be the surface of the sides; to which add the area of the base, and the sum will be the whole surface. ‘2. The perpendicular height of a pyramid or cone is a line drawn from the vertex to the middle or centre of the base; and the slant height is the distance between the middle of one side of the base and the vertex. 3. In order to find the slant height of a pyramid from the perpendicular height, or the perpendicular height from the slant height, it will be necessary to find the distance between the centre of the base and the middle of one of its sides, which may be done by the following Rule: Multiply the number answering to the poly- gonal base, in the third column of the table of polygons, Problem 10, Part 11., by the side of the base; and the product will be the distance required. 4. If three times the solidity of a pyramid be divided by the area of the base, the quotient will be the perpendicular altitude. 5. The area of the base of a pyramid may be found by Problems 11 and 12, Part II. 6. Every triangular pyramid, is the third part of a prism of the same base and altitude. 7. In any pyramid, a section parallel to the base is similar to the base ; and those two sections are to each other as the squares of their distances from the vertex. EXAMPLES. 1. Required the solidity of the hexa- E gonal pyramid ABCDE, each side of its base being 3 feet 9 inches, and its per- pendicular altitude EF 15 feet 9 inches. By Prob. 12, Part II., we have 2.5980762 x 3.752: 2.5980762 X 14.0625 = 36.5354465625, the area of tile base; and 365354465625 x 15.75 —:— 3 = 36. 5354465625 x 5.25 = 191.811094453125 feet, the solidity required. 2. The three sides of a triangular pyramid are 6, 7, and 8 feet, and its per- pendicular altitude 18 feet; what is its solidity? Ans. 121.998975 feet. 3. The altitude of a square pyramid is 15 feet 9 inches, and the side of its base 2 feet 6 inches; required its so- lidity. Ans. 32ft. 9 2'72. Qpa. 4. The spire of a church is a regular hexagonal py- ramid, whose perpendicular height measures 48 feet, the side of its base 10 feet, the perpendicular height of the cavity or hollow part 42 feet, and its side at the base 8 feet; how many cubic feet of stone are contained in the spire? Ans. 1829.0456448 feet. . sncr. I. or SOLIDS. 131 5. The slant height of a regular, pentagonal, stone py- ramid, measures 18 feet 6 inches, and the side of its base 13 feet; what was its Whole expense; 63. 9d. being paid for each cubic foot, and 631,-d. per square foot, for polishing the sides? Ans. £35. 158. Hid. PROBLEM VI. T 0 find the solidity of the frustum qfa pyramid. GENERAL RULE. To the areas of the two ends of the frustum add the square root of their product; and their sum being mul- tiplied by % of the perpendicular height, will give the solidity. Note 1. This rule will also give the solidity of the frustum of a right cone; and likewise that of the frustum of an elliptical cone, generally called a cylindroid. (See Prob. 4. Part VIII. Gauging.) 2. To find the perpendicular altitude of the frustum of a pyramid: To the areas of the two ends add the square mm of their product, for a. divisor; then, divide three times the solidity or content, by this divisor, and the quotient will be the altitude required. _ This rule will hold good, whatever may be the form of the ends of the frustum. RULE II. If the ends be regular polygons. Add together the square of a side of each end, and the product of those sides ; multiply the sum by g of the height, and this product by the tabular area belonging to the polygon, (Prob. 12, Part II.) and the last product will be the solidity. Note. The surface of the frustum of a pyramid maybe found thus: Multiply half the sum of the perimeters of the two ends, by the slant height, and the product will be the surface of the sides; to which add the areas of the ends, and the sum will he the whole surface of the frustum. EXAMPLES. 1. ”What is the solidity of the frustum ABCDEF, of a square pyramid; the side AB of the greater end being 5 feet, the side DE of the less end 2feet, and the perpen- dicular height GH 12 feet? By the General Rule. IIere 5 X 5 = 25, the area of the greater end; and 2 x 2 = 4, the area of the less,- also, ,/ (25 x 4) = \/ 100 = 10, the square root of their product; then (25 + 4 + 10) X 4 = 39 X 4 = 156feet, the solidity re- quired. G 6 MENSURATION Pill 1’1" IV. By Rule II. H'Jrc 51’ + 22 + (5 x 2): 25 + 4 +10 2 39; time 39 X 4 x 1 : 156fect, the solidity as bejbre. 2. Each side of the greater end of a piece of squared timber is 28 inches, each side of the less end 14 inches, and its length 18 feet 9 inches ; how many solid feet does it contain? A725. 59.548611fcet. 8. The length of a piece of timber is 17 feet 3 inches, and its ends are similar rectangles; the length of the greater being 36, and its breadth 20 inches; the length of the less 18, and its breadth 10 inches, how many solid feet are contained in the frustum? A725. 50.8125feezf. 4. What is the solidity of an octagonal stone pillar; each side of the greater end being 12 inches, each side of the less end 8 inches, and the perpendicular altitude 9 feet? A733. 30.58003fleet. 5. The roof of a portico is supported by four regular, hexagonal, marble pillars, whose perpendicular altitude is i5 feet, the side of the greater end 8 inches, and the side of the less end 5 inches 5 what did they cost at it}. 105. per solid foot ? A728. £209. 98. 43:36]. 6. The perpendicular altitude of a square chimney, in Leeds, is 120 feet 9 inches, the side of its base 10 feet 9 inches, and the side of its top 5 feet 9 inches. The cavity or hollow part is a square prism, whose side is 2 feet 6 inches ; how many solid feet are contained in the chimney, and what is the area of its four sides? Am. 2726 solidity {3}" Me ciu'nnzeg/ 2'8 7715421875 cat/lie .f'eez‘, mad #26 area (fate/bur sides is 39856039421 square feet. PROBLEM VII. Tofind tile solidify (3f a cone. RULE. Multiply the area of the base by the perpendicular height, and 1} 0f the product will be the solidity. Note 1. The altitude of an oblique cone, or one whose axis does not make a right angle with the plane of a base, may be found bydemitting a perpendicular from the vertex of the cone, upon the plane. 2. If the circumference of the base or" a cone be multiplied by half its slant height, the product will be the convex surface; to which add the area of the base, and the sum will be the whole surface of the cone. 3. The solidity of every cone is equal to one third of the solidity of its circum- scribing cylinder. 4. All prisms, cylinders, pyramids, and cones, of the same altitude, are to each ; quired. A SECT. 1. OF SOLIDS. 133 other as their bases ; consequently, prisms, cylinders, pyramids, and cones of equal bases, are to each other as their a titudes. 5. The perpendicular altitude of a cone may be found by dividing three times the solidity by the area of the base. 6. The area of the base of a cone may be found by Problem 15, Part II. 7. The surfaces of similar right cones are as the squares of their axes, and their solid contents as the cubes of their axes. EXAMPLE S. 1. Required the solidity of the cone ABC; the diameter AB of the base being 12 feet, and the perpendicular altitude DC 18 feet 6 inches. Hem .7854 X 122 = .7854: X 144; 2 113.0976, the area of the base; and (113.0976 X 18.5) —:- 3 : 9092.30.56 —:- 3 = 697.4352 feet, the solidity re- 2. The diameter of the base of a conical piece of timber is 5 feet 6 ' inches, and the length of its slant side 25 feet 9 inches; “ what is its solidity? Ans. 202.75954feet. 3. The circumference of the base of a conical stone is 12 feet 6 inches, and its perpendicular altitude 8 feet S) inches ; what is its solidity? A723. 36.266927 feet. 4. “That will be the expense of tooling or dressing a conical spire, at STE-d. per square foot; the circumference of the base being 30, and the slant height 45 feet? Ans. £ 18. 5S. 7—.Bd. PROBLEM VIII. Tofind the solidity of the frustum of a. cone. RULE I. To the product of the diameters of the two ends, add the sum of their squares ; then this sum being multiplied by the height, and again by .2618, (one-third of .7854,) will give the solidity. RULE II. Divide the difference of the cubes of the diameters of the two ends, by the difference of the diameters ; then this quotient being multiplied by .7854, and the product by :1,— of the height, will give the solidity. Nate 1. The solidity of the frustum of a cone may also be found by the genera. rule, given in Problem 6. 134 MENSURATION PART 1v. 2. The surface of the frustum of a cone may be found thus: Multiply half the sum of the perimeters of the two ends, by the slant height, and the product will be the convex surface; to which add the areas of the ends, and the sum will be the whole surface of the frustum. 3. To find the perpendicular altitude of the frustum of a cone: To the product of the diameters of the two ends add the sum of their squares, and multiply this result by .2618 for a divisor ; then, divide the solidity of the frustum by this divisor, and the quotient will be the perpendicular altitude. EXADH’LES. 1. What is the solidity of the frustum ABCD of a cone; the diameter AB of the greater end M‘ being 7 feet, the diameter CD of the less end 5 feet, and the pcrpeidicular altitude EF 8 feet 3 inches ? By Rule I. Here (7 X 5) + 72 + 53:35-1- 49+ 25 , ‘ :109 ; and 109 X 8.25 X 2618289925 A W X .2618 = 23542365 feet, tlze solidity required. By Rule II. Here (73 —53)—:- (7 —5) x .7854 x 8.25 +3: (343 — 125) +9 x .7854 x 2.75 : 218+2 x .7854 >< 2.75:109 x .7854 >< 2.75—85.6086 x 275223542365 feet, the solidity as bgf’ore. 2. The greater diameter of a piece of timber is 2 feet 6 inches, the less 1 foot 3 inches, and the length 25 feet 9 inches; what is its solidity? Ans. 73.73351feet. 3. The circumference of a piece of round timber is 65 inches at the greater end, 35 inches at the less end. and its length 33 feet 9 inches ; what is its solid content? A728. 48.027773feet. 4. The pediment in the front of the court-house, in Leeds, is supported by four stone columns, whose perpen- dicular altitude is 22 feet 9 inches, the diameter at the bottom 2 feet 9 inches, and at the top 2 feet 3 inches; re- quired their solidity. Ans. 448.185237 feet. 5. The slant height of the hovel or chimney of a pot- kiln is 45 feet, the circumference at the bottom 113 feet 9 inches, and at the top 31 feet 3 inches; required its convex surface. Ans. 3262.5 feet. 6. The top diameter of a lime-kiln is 15 feet 10 inches, the bottom diameter 3 feet 8 inches, and the perpendicu- lar depth 21 feet 4 inches ; how many bushels of lime will it burn at once ; allowing five peeks, or 2688 cubic inches to a bushel? Ans. 115680693 bushels. seer. I. or SOLIDS. 135 7. The shaft of Pompey’s pillar is a single stone of granite whose perpendicular altitude is 90 feet, the diameter at the base 9 feet, and at the top 7 feet 6 inches; required its solidity. Ans. 4824.3195 feet. Note. Pompey’s Pillar is situated near Alexandria, in Egypt. It is a fine regular column, of the Corinthian order; and its whole height, including the capital and pedestal, is 114 feet. Upon its top has been a statue, which must have been of a gigantic stature, to have appeared of the ordinary size of a man, to a spectator at the bottom. The remains of this statue, consisting of a foot and an ancle, were only dis- covered about 30 years ago, by some jolly sons of Neptune, that ascended the pillar, and drank a bowl of punch upon the top of it, amidst the shouts and ac» clamations of the natives, who were astonished at the address and boldness of the British tars. This object they accomplished by means ofa rope, which they contrived to draw over the top of the pillar. by the assistance of a paper kite. One of the company then ascended; and a kind of shroud being constructed, seven persons more went up; and the whole descended again, without the least accident, except the falling of a portion of the capital, which they carried off in triumph. Learned men and travellers have made many fruitless attempts to discover in honour of whom this majestic monument was erected. The best informed have concluded that it could not be in honour of Pompey, since neither Strabo nor Diodorus Siculus has mentioned it. The Arabian historian Abulfeda, in his de- scription of Egypt, calls it “ The Pillar of Severus ; ” hence Mr. Savary conjectures that it was erected to the honour of the emperor Severus by the inhabitants of Alexandria, to whom, as history informs us, he granted several favours. Severus died at York, A.D. 211 ; and Pompey’s Pillar did not receive that name until the fifteenth century. PROBLEM IX. Tofind the solidity of a cuneus 09‘ wedge. RULE . To twice the length of the base, add the length of the edge, multiply the sum by the breadth of the base, and this product by the perpendicular height of the wedge ; and ,1, of the last product will be the solidity. Note 1. The surface ofa wedge may be obtained by finding the area of each of its sides or parts separately, and adding them into one sum. 2. When the length of the edge is equal to the length of the base, the solidity of the wedge is evidently equal to half the solidity of a prism of the same base and altitude; but if the edge be longer or shorter than the base, the wedge is manifestly greater, or less than halfa prism. - EXAMPLE S. 1. Required the solidity of the wedge ABCDEF, the length AB of its base being 26 inches, the breadth BC 18 inches, the length of the edge EF 15 inches, and the per- pendicular height PE 28 inches. Here (26x 2)+15=52+15=67; . .‘ and(67x18x28)-:—6=67X3x28 \ __________ = 5628 cubic inc/res, the solidity re- ‘ _‘ quired. A n T.‘ A. 136 MENSURATION rant 1V. 2. tequired the solidity of a wedge, the length and breadth of the base being 82 and 35 inches, the length of the edge 115 inches, and the perpendicular height 8 feet. Ans. 90.41666feef. 3. What is the soldity of a wedge, the length and breadth of Whose base is 5 feet 9 inches and 3 feet 6 inches, the length of the edge 4 feet 3 inches, and the perpendicular height 9 feet? Ans. 82.6875fecf. PROBLEM X. Tofind t/te solidity of a prismoid. iULE. To the sum of the areas of the two ends add four times the area of a section parallel to and equally distant from both ends ; multiply this sum by the perpendicular height, and (1) of the product will be the solidity. Note 1. The length of the middle section is equal to half the sum of the lengths of the two ends; and its breadth is equal to half the sum of their breadths. 2. The surface of a prismoid may he obtained by adding the areas 01' its four sides and two ends into one sum. 3. As the ends of a. prismoid are rectangles, their areas may be found by 13:0— ‘nlem 2, Part II. 4. To find the perpendicular height of a prismoid: To the sum of the areas of the two ends add {our times the are: of the middle section. for a divisor; then, divide six times the content by this divisor, and the quotient will he the perpen- dicular altitude. This Rule will hold good for cylindroids, whatever mw be the form of their ends ; and also for all frustums and solids whose parallel sections are similar figures. REMARK. ' he last Rule, and the similar Rules given in the preceding Problems. will he found useful when the content or capai'ity of a solid, and all the horizontal dimen— sions are given, to find the perpendicular altitude. This is often the Case in ‘ 0 orders for the making of cisterns, couches. tuns, vats, tubs, and other Vt various descriptions. (See Gauging, Parr V111.) SCHOLIUM. The Rule given in this Problem for a prismoid, is very clearly demonstrated in Simpson’s Fluxions, page 178, first edition; and also in a similar manner at page 302, of Holliday’s Fluxions. There is likewise a very elegant demonstration give: in Proposi- tion 111., Section 1., Part IV., of Dr. Hutton’s l‘vIensuration, third edition, in which it is shown, to be true for all frustums whatever, and for all solids whose parallel sections are similar figures; and Eir. Fletcher, in his second edition of his Universal lVIeasurcr, Part 111., page 54, and llIr. iVIoss, in his Gauging, page 175, third edition, say that it is nearly true for any other solid, whatever 111:3:7l)’: its form. EXABIPLES. l. ‘vVhat is the solidity of the prismoid ABCDEF iii, the length AB of the greater end being 18, and its d" seer. I. OF SOLIDS. 137 breadth BC 8 inches; the length EF of the less end 12, and its breadth FG 6 inches ; and the perpendicular height PE 18 inches? Here 18 X 8 +12>< 6:144 + 72 = 216, the sum of the areas of the two ends. Also, (18+ 12)-:—2 = 30 _:._ 2 = 15, the length of the middle section ; and (8 + 6 —:— 2 = 14 —:— 2 :7, the breadth of the middle sec~ tion ; then 15 x 7 X 4: 105 X 4 2 420, four times the area (f the middle section ; whence (216 + 420) X 18 -2— 6 :2 636 x 3 = 1908 cubic inches, the solidity required. 2. How many solid feet of timber are contained in a beam whose ends are rectangles; the length and breadth of the greater being 32 and 20 inches ; the length and breadth of the less 16 and 8 inches ; and the perpendicular length 25 feet ? Ans. 61»},»feet. 8. The perpendicular altitude of a stone pillar is 8 feet 6 inches; the length and breadth of the greater end are 26 and 16 inches ; and the length and breadth of the less end 18 and 10 inches ; required the solidity of the stone. Ans. 17.118055 feet. 4. The length and breadth of a fish—pond at the top are 182 and 64 yards ; the length and breadth at the bottom 116 and 48 yards; and the perpendicular depth 12 feet 9 inches ; what did it cost digging, at 64§d. per cubic yard? Ans. £832. lls. 6d. 5. lWhat is the capacity of a coal waggon, the inside dimensions of which are as follow : at the top, the length is 80%, and breadth 56 inches ; at the bottom, the length is 41-17, and the breadth 32 inches; and the perpendicular depth 47 inches? Ans. 129814 cubic inches. Note. Coals are frequently carried from pits in prismoidal waggons, having four cast iron wheels, which run upon a railway. They are prevented from leaving the railway by a particular construction of the wheels; the inner edge of the rim of each wheel being made to project. Coals are conveyed, in waggons of the above description, from the pits in the vicinity of New- castle, to the river Tyne ; from those near Sunderland, to the river Weir ; and from Middleton Colliery. to the town of Leeds. In the last question are given the dimensions of a Leeds coal waggon, which differs but little from those of other places. Instead of rails, the waggon-road between Middleton and Leeds, is made with bars of cast-iron; and the waggons are drawn by a machine which moves along the road before them, by the force of steam. This machine was invented by Mr. Blen— kinsop, of Middleton; and is called “ Blenkinsop’s Patent Steam Carriage.” It is considered to be one of the most curious pieces of mechanism that ever appeared in England, or perhaps in the world. It moves at the rate of about three miles and a half in an hour ; and is capable of drawing 30 waggons at once, upon level ground, which are computed to weigh 105 tons 138 MENSURATION PART IV. PROBLEM XI. Tofind the solidity qfa sphere or globe. RULE. Multiply the cube of the diameter by .5236, or the cube :1" the circumference by .016887, and the product will be the solidity. Note 1. The convex surface of a sphere may be found by multiplying the circum- "erence by the diameter. Or, multiply the square of the diameter by 3.1416, and :L'ne product will be the convex surface. The solidity ofa sphere is equal to two thirds of the solidity of its circumscribing a 'linder. 13. The surface of a sphere is equal to 4 times the area of a circle of the same liamoter as the sphere; or to the area of a circle whose diameter is double that of he sphere; or to the convex surface of the circumscribing cylinder. EXAMPLE S . 1. What is the solidity of a globe whose diameter AB is 25 inches? IIere 253 X .5236 :25 X 25 X 25 x .5236 = 625 x 25 x .5236: 15625 x .5236 :2 8181.25 cubic in— ches, the solidity required. 2. “That is the solidity of a spherical stone whose circumfer- ence measures 5 feet 6 inches? Ans. 2.80957fcct. 3. If the circumference of a cannon-ball be 18.6 in- ches; what is its solidity ? Ans. 108.66541 inches. 4. The diameter of the moon is 2180 miles; what is her solidity in cubic miles ? Ans. 54246174752 miles. 5. The diameter of the earth is 79.57% miles; what is His solidity in cubic miles, and its convex surface in square miles ? Ans. The solidity is 263858149120 cubic miles, and the convex surface is 198944286 square miles. PROBLEM XII. Tofind the solidity of a segment of a sphere. RULE I. To three times the square of the radius of the segment’s base, add the square of its height; and this sum multi- plied by the height, and the product again by .5286, will give the solidity. seer. I. or SOLIDS. 139 RULE II. From three times the diameter of the sphere subtract twice the height of the segment; multiply the remainder by the square of the height, and that product by .52361 for the solidity. Note 1. The surface of the segment ofa sphere may be found thus: Multiply the circumference of the whole sphere by the height of the segment, and the product will he the convex surface; to which add the area of the base, and the sum will he the whole surface. 2. The area ofa spherical triangle, or the spherical surface included by the arcs of three great circles of the sphere, intersecting each other, may be found by the following rule ; As two right angles, or 180 degrees, 'Are to the area ofa great circle of the sphere; So is the excess of the three angles above two right angles, To the area of the triangle. EXAMPLES. 1. The hadius An of the base of c the segment ABC is 9 inches, and -- the height C72 7 inches; what is the solidity of the segment ? By Rule I., we have 92 X 3 + ”3:81 X 3+49: 24134-492292; and 292 x 7 x 5236:2044 X .5236 210702384 inches, the solidity re- quired. 2. If the diameter of a sphere be 3 feet 6 inches; What is the solidity of a segment Whose height is 1 foot 3 inches ? Ans. 6.545 . feet. 3. What is the solidity of a spherical segment Whose height is 24.8 inches, and the radius of its base 30.6 inches ? Ans. 25.73099 feet. 4. Required the solidity and convex surface of each of the frigid zones, Whose height is 330.0075 miles, and the diameter of the earth 7957:} miles. Ans. The solidity is 13236797534398 cubic miles, and? the convex smface 2's 82502097425 square miles. PROBLEM XIII. Tofz'ml the solidity of thefrustum or zone qf a sphere. RULE . To the sum of the squares of the radii of the two ends, add i of the spuare of their distance, or the height of the 140 MENSURATION PART iv. zone; and this sum being multiplied by the said height, and the product again by 1.5708, will give the solidity. Note. The convex surface of a zone may be found by multiplying the circumfer- ence of the whole sphere by the height of the zone. EXAMPLES. 1. Required the solidity of the zone ABCD; the greater diameter AB being 28, the less dia- meter CD 20, and the height 72m 15 inches. Here 142 + 102 + 152 + 3 = 196 +100+225+3=296 + 75: 371; and 371 X 15 X 1.5708 : 5565 x 1.5708 = 8741.502 inches, the solidity required. 2. What is the solidity of a zone whose greater diameter is 0 feet 3 inches, less diameter 6 feet Qinches, and height 5 feet 6 inches ? Ans. 370.32425 feet. 3. Required the solidity and convex surface of the torrid zone, the top and bottom diameters of which are each 7297.735 miles, its height 3173.14565 miles, and the circumference of the earth 25000 miles? Aus. The solidity is 14941555316338.46698 cubic miles, and the convex surface is 7932864125 square miles. PROBLEM XIV. Tofiud the solidity ofa circular spindle. tULE. Find the area of the revolving segment ACBEA, which multiply by half the central distance OE. Subtract the. product from f, of the cube of AE, half the length of the spindle; and multiply the remainder by 12.5664, and the product will be the solidity. Note 1. The surface of a circular spindle may be found by the following rule: Multiply the length AB of the spindle by the radius OC of the “evolving arc ACB. Multiply also the length of the said are by the central distance ()E,br distance between the centre of the spindle and the centre of the revolving are. Su'itract the latter product from the former ; and the remainder being multiplied by 5283‘), will give the surface required. 2. The convex surface of any segment or zone of a circular spindle, cut oil" per- pendicularly to the chord of the revolving arc, may be found in the same manner; using the length of the solid, and the part of the are which describes it, instead of the length of the whole spindle and whole are, seer. 1. or somns. 141 EXAMPLE S. 1. Required the solidity of the cir- cular spindle ABCD, whose length AB is 30, and middle diameter CD 16 inches. By Rule 2, Prol). 14, Part II, we have152—1-8 +8:225—:—8+8 :- 28.125 +8236.125, the diameter CF; and 8 x g —.‘— (36.125 ——(8 X 41) )+50 = 5.3333 + (36.125 —6.56) = 5.3333 "“3“” — 29.565 = .18039; then 1 + .18039 X 30 = 1.18039 X 30 = 35.4117, the length ofthe revolv- ing arc ACB ; and by Rulel. Prob. 16, we have 35.4117 x 18.0625 2 639.6238, half of which is 319.8119, the area of the sector AOBC. Again, 00— CE = 18.0625 —8 = 10.0625 : OE; and OE >< AE = 10.0625 X 15 2 150.9375, the area of the triangle AOB ; then 319.8119 — 150.9375 = 168.8744, the area of the revolving segment ACB. A7020, 63/ theRule, we have (153+3-— 168.8744 X 10.0625 -:— 2) x 12.5664 : (3375 + 3~168.8744 >< 5.03125) X 12.5664 : (1125 — 849.6493) x 12.5664 2 275.3507 X 12.5664 : 3460.16703648 inches, the solidity required. 2. that is the solidity and superficial content of a cir- cular spindle, Whose length is 96, and middle diameter 72 inches ? r" Ans Solidity, 138.51593 feet. ' Superficies, 130.82761 feet. PROBLEM XV. Tofind the SOl‘let'Z/ of the middle frustum or zone of a circular spindle. RULE . From the square of half the length of the Whole spindle, take 7} of the square of half the length of the frustum; and multiply the remainder by the said half length of the frustum. 1V1ultiply the revolving area which generates the frustum, by the central distance, and subtract this pro- duct from the former; then the remainder being multi- plied by 6.2832, will give the solidity required. 142 MENSUBATION PART IV. EXAMPLES. . 1. ‘vVhat is the solidity of the frustum ABCD, whose length mn is 40, its greater diameter EF 32, and its least diameter AD or BC 24 inches ? Draw DGr parallel to mn, then we have DGr : ; mn=20. Also, 4 EF — %, AD = 16 —- 12 H =2 4 2 EG; and DG2 + EG") \‘\ i :202 + 42 2400 +16 2416, the square of the chord DE; then by Theo. 12, Part 1., DE2 —2— EG=416 -:— 4 = 104, the diameter of the generating circle, half of which is 52, the radius OE or OH ; consequently OE—E1252—16 = 36, the central distance OI. Again OH2 — 012 : ”23—4369 2 2704 — 1296 =1408, the square of HI, half the length of the whole spindle. By Rule 3, Prob. 17, Part 11., we have 4 —:— 104 = .03846, the tabular height; and the corresponding Area Seg. is .00994; then .00994 x 1042 = .00994 X 10816 = 107.51104, the area of the segment DEC. Also, mn x Dm = 40 x 12 = 480, the area of the rectangle mDCn; and 107.51104 + 480 :2 587.51104 the generating area mDECn. Alon), by the Rule, we have (1408 —203—-f— 3) X 20 = (1408 —- 400 —:— 3) X 20 : (1408 —— 13333333) x 20 = 127466667 X 20 = 2549333334, the first product,- and 587.51104 X 36 = 21150.39744, the second product, which tahcnfrom thefirst, leaves 4342.9359; then 43429359 X 6.2832:27287.53484688 inches, the solidity required. 2. “lhat is the solidity of the middle frustum of a cir- cular spindle, whose length is 50, greatest diameter 40, and least diameter 30 inches? Ans. 30.84257 feet. PROBLEM XVI. To find the solidity of a cylindrical ring. RULE. To the thickness of the ring, add the inner diameter; and this sum being multiplied by the square of the thick- ness, and the product again by 2.4674, will give the? solidity. ‘ snow. I. or soups. 143 Note. The surface of a cylindrical ring may be found by the following rule: fi'o the thickness of the ring, and the inner diameter; and this sum being multiplied by the thickness, and the product again by 9.8606, will gun: the surface required. EXAMPLE S. 1. What is the solidity of a cylin- drical ring, whose thickness AB or CD is 6, and the inner diameter BC 20 inches ? IIcre (20 + 6) X 69 x 2.4674 :26 X 36 X 2.4674 : 936 x 2.4674 2 2309.4864 inches, the solidity re- quired. 2. that is the solidity of an anchor-ring, the inner dia- meter of which is 24.6 inches, and its thickness 6.4 inches ? Ans. 3133.005824 inches. 3. Required the solid and superficial contents of a cy- lindrical ring, whose thickness is 9, and inner diameter 32 inches. Ans Solidity, 8194.2354 inclzes. ' Szwerflcz’cs, 8641.8824 inclzes. PROBLEM XVII. Tofind the solidity or superficieg of (my regular body. RULES. 1. Multiply the tabular solidity, in the following table, by the cube of the linear edge of the body, and the pro- duct will be the solidity. 2. Multiply the tabular area, by the square of the linear edge, and the product will be the superflcies. Solidities and Surfaces of Regular Bodies. , r l 351$ Names. Solidities. Surfaces. 4 Tetraedron . . . .1178511 1.7320508 6 Hexaedron . . . 1.0000000 6.0000000 8 Octaedron . . . .4714045 3.4641016 12 Dodecaedron . . 7.6631189 20.64.57.288.3 20 IIcosaedron . . . 2.1816949 8.6602540 : Note. The above table shows the solidity and superflcies of the five Regular Bodies, when the linear edge of each is' l, or unity. 1 44 MENSURATION EXAMPLES. 1. “That is the solidity and super- flcies of the tetraedron ABCD, Whose linear edge is 8 inches? Here .1178511 x 83 = .1178511 x 512 = 603397632 inches, the ' m solidity ; and 1.7320508 X 82 = 1.7320508 x 64 = 1108512512 inches, the superficies. 2. “That is the solidity 0f the hexaedron ABCD, Whose side AB is 6 feet? A723. 216fcef. . ‘ 3. Required the solidity of the ~ octaedron ABCDE, whose 1inear -_ . “ ‘ side is 22 inches. ‘ 3 ' A728. 5019.515116 inches. ' 4. The linear side of the dodecae- .. ’ dron ABCDE is 5.68 feet; what is - r. its solidity ? " ' . ‘ Ans. 1404.2698-1fiet. " 5. What is the solidity and super- ‘- ficies of the icosaedron ABCDEF, Ck. ' Whose linear edge is 15 ? '1 . . "r Solzdzty, 7363.220? . ' r. Ans. ‘ . ”.0 --.. , - ‘ Supmfimes, 10-i:).00.‘ la. PART IV. sncr. I. or SOLIDS. 145 Note. The five Regular Bodies are sometimes called Platonic Bodies. It appears from an ancient Greek Epigram, quoted by Scarburgh in his “ Eng- lish Euclid,” (Oxford. 1705,) that “the five Platonic Bodies, which the wise Pythagoras found out, were indeed discovered by him ,7 but Plato elucidated and taught them in the clearest manner; and Euclid took them as the foundation of his own imperishable renown.” Pythagoras was a native of the island of Samos, and was born about 590 years before Christ; Plato was born at Athens 13.0. 429; and Euclid flourished at Alexandria, about 300 years before the Christian aera. ‘ PROBLEM XVIII. Tofind the solidity of an irregular solid. RULES. 1. Divide the irregular solid into different figures; and the sum of their solidities, found by the preceding Pro- blems, will be the solidity required. 2. If the figure be a compound solid, whose two ends are equal plane figures; the solidity may be found by multiplying the area of one end by the length. 3. To find the solidity of a piece of wood or stone, that is craggy or uneven, put it into a tub or cistern, and pour in as much water as will just cover it; then take it out, and find the content of that part of the vessel through which the water has descended, and it will be the solidity required. 4. If a solid be large and very irregular, so that it can- not be measured by any of the above Rules, the general method is to take lengths, in two or three different places; and their sum divided by their number, is considered as a mean length. A mean breadth and a mean depth are found by similar processes. Sometimes the length, breadth, and depth, taken in the middle, are considered as mean dimensions. Note. The unhewn blocks, in the freestone quarries, in the vicinity of Leeds, are generally measured by the method described in the last Rule. The dimensions, however, are not, in general, taken to the extremities of the stones, in order to make an allowance for the waste in hewing. EXAMPLE S. 1. The lower part of a stone is a parallelopipedon, the breadth of whose end is 7, and its depth 5 feet. The upper part is a triangular prism, the perpendicular of whose end is 4 feet; required the solidity of the stone, its length being 18 foot. II 146 MENSURATION. PART Iv. By Rule I. Here 7 X 5 X 18 = 35 x 18 = 630, the solidity of the lower part,- and 7 X 2 X 18 : 14 x 18 =252, the solidity ofthe upper part; then 630 + 252 = 882feet, the solidity of the whole stone. By Rule II. Ilerc 7 X 5 + 7 X 2: 35 +14 :2 49, the area If the end; and 49 X 18 = 882 feet, the solidity as before. 2. Being desirous of finding the solidity of an irregular piece of wood, I immersed it in a cubical vessel of water; and when it was taken out, the water descended 6 inches ; required the solidity of the wood, the side of the vessel being 30 inches. Ry Rule III. [[ere 30 X 30 X 6: 900 X 6 = 54-00 inches, the solidity required. 3. The lengths of an irregular block of marble, taken in different places, are 8 feet 6 inches, 8 feet 10 inches, and 9 feet 2 inches; the breadths 4 feet 7 inches, 4 feet 2 1“ inches, and 4 feet 9 inches ; the depths 3 feet 2 inches, 0 feet 5 inches, and 3 feet 11 inches; required its solidity. By Rule IV. LENGTHS. BKEADTHS. DEPTIIS. feet in. feet in. feet in. 8 6 4 7 3 2, 8 10 4 2 3 5 9 2 4 9 3 11 3)26 6 3)l3 6 3310 6 8 10 mean. "4 6 mean. 3 6 mean. feet in. 8 10 mean length. 4 6 mean breadth. 3 4 4 5 3 9 3 6 mean depth. T117— 3 19 10 6” 139 l (37’ Am. SECT. I. OF SOLIDS. ‘ 147 4. W'anting to know the solidity of an irregular block of marble, I immersed it in a cylindrical tub of water, whose diameter was 34.8 inches ; and on taking it out, I found the fall of the water to be 12.6 inches ; what was the solidity of the marble ? Ans. 11984.50028 inches. 5. Required the solidity of an irregular block of York- shire-stone, whose dimensions are as follow : viz. lengths taken in different places, 11 feet 3 inches, and 11 feet 9 inches ; breadths, 5 feet 5 inches, 5 feet 9 inches, and 6 feet 4 inches; depths, 4 feet 5 inches, 4 feet 8 inches, and 5 feet 2 inches. Ans. 318 ft. 7 in. 9 pa. PROBLEM XIX. Tofind tile magnitude or solidz'tyof a body from its weight. RULE. As the tabular specific gravity of the body, Is to its weight in avoirdupois ounces, So is one cubic foot, or 1728 cubic inches, To its content in feet, or inches, respectively. A TABLE or THE SPECIFIC GRAVITIES OF BODIES. Fine Gold . . . . 19640 Bristol-stone . . . 2510 _ Standard Gold . . 18888 Portland-stone . . . 2496 Quick Silver . . . 14000 Mill-stone . . . . 2484 Lead . . . . . 11325 Yorkshire-stone . . 2442 Fine Silver . . . 11091 Clay . . . . . . 2160 Standard Silver . . 10535 Grind-stone. . . . 2143 Copper . . . . . 9000 Burford-stone . . . 2049 Gun Metal . . . 8784 Brick . . . . . . 2000 Cast Brass . . . 8000 Light Earth . . . 1984 Steel . . . . . 7850 Ivory . . . . . . 1825 Iron. . . . . . 7645 Chalk . . . . . 1793 Cast Iron . . . . 7425 Sand . . . . . . 1520 Tin . . . . . . 7320 Lignum-vita3 . . . 1333 Load-stone . . . 4930 Ebony . . . . .1331 Granite. . . . . 3000 Coal . . . . . . 1250 Marble . . . . . 2700 Pitch . . . . . . 1150 Glass . . . . . 2642 Mahogany . . . . 1063 Flint . . . . . 2570 Box-wood . . . . 1030 H 2 148 MENSURATION 01‘ SOLIDS. 111111 IV.‘ Sea—water . 1026 01‘ 1030 Maple . . . . . . 7551 Common water . . 1000 Cherry-tree . . . . 715 Oak . . . . . . 925 Pear—tree . . . . . 661 Gunpowder, shaken . 922 1 Cedar of Lebanon . . 613 Logwood . . . . 913 E1111 . . . . . . 600 13136011 . . . . . 852 Willow . . . . . 585 ASh......800Fir.......550 Yew......797Poplar......383 Apple-tree . . . . 7'93 Cork. . . . . 24:0 Plum-tree . . . . 785 Atmospheric Air . . '12 Note 1. The spemfic gravities of bodies are their relative weights contained under the same [11101111111111tu1lo,as (11‘11111c foot,:11‘11b1(‘1n111&c. 2. As acubic foot of n .1111 weighs just 111(‘0 011111‘1‘5 211011111111015, t‘11‘ 1111111111315 in the 10191301111: ’lnble expres s not only thcs specific grm‘itics ot the several bodies, but also the weight ofu cubic foot of each in '111)1r1111po1s ounces 1 '5. The several sorts of wood, mentioned i111hcprcccdi11g 1211111321 are supposed to 101111: 4. In some tables, sea-water is 1026, and in others 1030. EXAMPLES. 1. VVlmt is the solidity of :1 marble chimney—piece, whose weight is 210 II). 15 oz. avoirdupms? oz. 11). 02. ft. ft. As 2700 ; 210 15 z: 1 : 12.21223 16 1275 210 3375 1 10013375 1.25 2. What is the solidity of an irregular YOi‘hslni‘e—stonc, whose weight is2 28 {11.15 oz.:1voi1‘1l1mozs? 11223.1»:— foot. 3. A piece of carved 111ahog: my we 10115 4915.140... avoirdupois , v 11:11; 15 its solidity? A223. 19"2 . 19192 22297268. PROBLEM XX. 3'0 find the weight of a body flom zz‘s magnitude 02 30l2d22‘3/. RULE. As one cubic foot or 1128 cubic 111 1.0 cs, Is to the content 01 the body, So is its tabular specific p‘1‘11Vity, '10 the weiWht of the ho; SECT. II. CARPENTER’S RULE. 149 EXAMPLES. 1. The solidity of a grindstone is 3 feet; what is its weight? ft. ft. oz. lb. oz. A81 : 3 3: 2143 : 401 13.A7zs. 3 In??? 16)§g9(401 2. The solidity of a beam of dry oak is 25 feet; what is its weight? Ans. 1445 lb. 5 oz. 3. Required the weight of a block of marble, whose length is 63 feet, and its breadth and thickness each 12 feet; these being the dimensions of one of the stones in the walls of Balbec. Ans. 683.4375 tons, which is nearly equal to 1726 bur- tlten of an East India, skip. SECTION II. DESCRIPTION AND USE OF THE CARPENTER’S RULE. THIS instrument is commonly called Cages/milk Slid- ing Rule, and is much used in measuring timber and arti- ficers’ work ; not only in taking the dimensions, but also in casting up the contents. It consists of two pieces of box, each one foot in length; and connected together by a folding joint. One side or face of the rule is divided into inches and half—quarters, or eighths; and on the same face there are also several plane scales, divided into twelfth parts, which are designed for planning such dimensions as are taken in feet and inches. On one part of the other face are four lines marked A, B, C, D; the two middle ones, B and C, being upon a. slider. Three of these "lines, viz. A, B, C, are exactly alike, and are called double lines; because they proceed from 1 to 10, twice over. The fourth line, 1), is a single one, pro- ceeding from 4 to 40, and is called the gin? line. H 3 _. .3_:,;.:12~,";;._«91_7,; -Au .. _, r 150 CARPENTERS RULE. PART iv. The use of the double lines, A and B, is for working proportions, and finding the areas of plane figures ; and the use of the girt-line D, and the other double line C, for finding the contents of solids. When 1 at the beginning of any line is accounted 1, then the 1 in the middle will be 10, and the 10 at the end 100; and when 1 at the end is accounted 10, then the l in the middle is 100, and the 10 at the end 1000, &c.; and all the smaller divisions are altered in value accord- ingly. Upon the girt-line are also marked W G at 17.15, and A G at 18.95, the wine and ale gauge—points, to make the instrument serve the purpose of a Gauging Rule. On the other part of this face there is commonly either a table of the value of a load, or 50 cubic feet of timber, at all prices, from 6d. to 24d. per foot ; or else several plane scales divided into twelfth parts, and marked 1, 43, %, and 21-, signifying that the inch, ginch, 8m”. are each divided into 12 equal parts. The edge of the rule is generally divided decimally, or into tenths; namely, each foot into 10 equal parts, and each of those into ten other equal parts. By this scale dimensions may be taken in feet,tenths, and hundredths of a foot, which is a very commodious method of taking di- mensions, when the contents are to be cast up decimally. THE USE OF THE SLIDING RULE. PROBLEM I. Multiplication by the Sliding rule. RULE. Set 1 upon A, to the multiplier upon 13 ; then against the multiplicand on A, will be found the product upon B. Note. When the third term runs beyond the end of the line, seek it on the other radius or part of the line; and increase the product 10 times, or 100 times, &c. as the case requires. EXAMPLES. 1. What is the product of 24 multiplied by 12 ? As 1 on A : 12 on B :1 24 on A: 288 on B, the A223. 2. Required the product of 36 muitiplied by 18. A725. 648. SECT. II. CARPENTER’S RULE. 151 3. Multiply 12.8 by 6.5. Ans. 83.2. 4. “That is the product of 68 multiplied by 35 ? Ans. 2380. PROBLEM II. Division by the sliding rule. RULE. Set the divisor on A, to 1 upon B; then against the dividend on A, is the quotient on B. Note. When the dividend runs beyond the end ofthe line, diminish it by 10 or 100 times, in order to make it. fail upon A ; and increase the quotient in the same pro- portion. EXAMPLES. ]. What is the quotient of 432 divided by 12 ? As 12 onA: 1 onB :: 432 onA : 36 072B, tkeAns. 2. What is the quotient of 9752 divided by 46 ? Ans. 212. 3. If a board be 8 inches broad ; what length must be sawn of}; to make a foot square ? Ans. 18 inches. PROBLEM III. T0 square any number. RULE. Set 1 upon D, to 1 upon C; then against the number upon D, will be found the square upon 0. EXAMPLES. 1. What is the square of 25? AslonD110n C :2 25 onD : 625 on C, t/zeAns. 2. Required the square of 36. Ans. 1296. 3. What is the square of 58 ? Ans. 3364. PROBLEM IV. To extract the square root of any number. RULE. Set 1 upon C, to 1 upon D; then against the number on C, is the root on D. H 4 152 CARPENTER’S RULE. PART Iv. Note. In order to value this rightly, you must suppose the 1 on C to be some of these squares. l, 100, 10000, &c. which is nearest to the given number ; and then the root corresponding will be the value of the 1 upon D. EXAMPLES. 1. What is the square root of 144 ? As 1 on C I 1 on D :1 144 on C: 12 on D, the Ans. 2. Required the square root of 900. A715. 30. 3. ~What is the square root of 9586 ? Ans. 97.9. PROBLEM V. T0fl7ld a mean proportional between two numbers. RULE. Set one of the numbers on C, to the some on D ; then against the other number on C, will be the mean on 1). EXAMPLES. 1. What is the mean proportional between 9 and 16? As 9 on C19 on D I: 16 on C 1 12 on 13,1118 A723. 2. Required a mean proportional between 15 and ‘2 ' Ans. 2J..1 3. The segments of the hypothenuse of a right-angled triangle made by a perpendicular from the right-angle, :u' 1 18 and 32 ; what is the perpendicular ? Ans. 21. H l ‘7 PROBLEM VI. "ofind afourth proportional to t/zree numbers ; or to perform the Rule of Three. RULE. Set the first term on A, to the second on B ; then against the third term on A, stands the fourth on B. Note. The finding of a third proportional is exactly the some; the second num- her being twice repeated. ’Qhus, suppose a third proportional was required to 80 and 60. As 80 on A160 on B : : 60 on A : 45 on B, the third proportional sought. EXAMPLES. 1. What is the fourth proportional to the three num— bers, 12, 24, 36 ? As 12 on A : 24 on B1186 on A I 72 on B, the A728. seer. II. CARPENTER’S RULE. 153 2. If 1 foot of timber cost 35.; What will 180 feet cost? Ans. 5403. 3. If 50 feet of timber cost 7L; what will 2500 feet cost? Ans. £ 350. PROBLEM VII. T 0 find the areas of plane figures by the sliding rule. RULE S. l. T 0 find the area of a rectangle or rhonzboz'des. As 1 upon A, is to the perpendicular breadth upon B ; so is the length upon A, to the area upon B. 2. To find the area qfa triangle. As 2 upon A, is to the perpendicular upon B ; so is the base upon A, to the area upon B. 3. Tofind the area of a trapezium. As 2 upon A, is to the sum of the two perpendiculars upon B ; so is the diagonal upon A, to the area upon B. 4. To find the area qfa regular polygon. As 2 upon A, is to the perpendicular upon B ; so is the sum of the sides upon A, to the area upon B. 5. ”0 find the diameter and circumference of a circle, the one from the other. As 7 upon A, is to 22 upon B; so is the diameter upon A, to the circumference upon B ; and vice versa. 6. ’I'ofind the area of a circle. As 4 upon A, is to the diameter upon B; so is the cir- cumference upon A, to the area upon B. Or, as 1 upon D, is to .7854 upon C; so is the diameter upon D, to the area upon C. Note. In order to exemplify the foregoing Rules, the learner may work the ques- tions in the respective Problems of Part 11. PROBLEM VIII. To find the contents of solids by the sliding rule. RULES. 1. Toflnd the solidity of a cube. As 1 upon D, is to the side upon C; so is the side upon D, to the solidity upon C. 2. Tajincl the solidity of a parallelopipedon. As 1 upon A, is to the breadth of the base or end upon 11 5 154 TIMBER MEASURE. PART IV B ; so is the length of the base upon A, to the area of the base upon B; and, as 1 upon A, is to the length of the solid upon B; so is the area of the base upon A, to the solidity upon B. 3. Tofind t/ce solidity of a prism, 07’ a cylinder. Find the area of the base by the last Problem, with which proceed as in the last Rule. 4. Tofind the solidity qfa cone, or a pyramid. Find the area of the base by the last Problem; then, as 3 upon A, is to the length of the solid upon B; so is the area of the base upon A, to the solidity upon B. Note. Examples for practice may be found in Section I. SECTION III. TIMBER MEASURE. PROBLEM I. Toflnd the supmficial content of a board 07- plank. RULE. Multiply the length by the breadth, and the product will be the superficial content. Note. If the board taper, add the breadth of the two ends together; and half their ‘ sum will be a mean breadth. By the Sliding Rule. As 12 upon B, is to the breadth in inches upon A; so is the length in feet upon B, to the content upon A, in feet and fractional parts. EXAMPLES. 1. If the length of a plank be 12 feet 6 inches, and its breadth 1 foot 3 inches ; What is its superficial content? By Decimals. By Cross Zlfultzplicatz'on. feet. feet in. 12.5 12 6 1.25 1 3 6‘23“ 1ft; 250 3 1 6/ 12L 1T756‘ Ans. 15'625 Ans. h SECT. 111. TIMBER MEASURE. 155 By the Sliding Rule. As 12 on B: 15 on A1: 12.5 on Bn15.6 on A. 2. Required the content of a board whose length is 25 feet 8 inches, and breadth 1 foot 7 inches. Ans. 40ft. 7 2'72. 8 pa. 3. What is the content of a board whose length is 18 feet 10 inches, and breadth at the broader end 2 feet 3 inches, and at the narrower end 1 foot 7 inches? Ans. 36ft. 1 in. 2 pa. 4. If the length of a mahogany plank be 35 feet 9 inches, and its breadth 3 feet 6 inches ; what is the value of three such planks at 23. 9d. per square foot ? Ans. £51. 12s. 3%d. Note. Mahogany is a production of the warmest parts of America. It is also found plentifully in the islands of (Tuba, Jamaica, and 11ispaniola. This tree grows tall and straight, rising often sixty, and sometimes a hundred. feet from the ground to the arms ; and is about 4 feet in diameter. The foliage is a beautiful d :ep green 5 and the appearance of the whole tree is very elegant. The mahogany brought from Jamaica is most valued, in consequence of its firmness, durability, and beauty of colour. This wood has been used in medicine with the same effect as Peruvian bark. PROBLEM II. To find the solidity of squared or fOur-sided timber. RULE . Multiply the mean breadth by the mean thickness, and this product by the length, and it, will give the solidity, according to the customary method. By the Sliding Rule. As 1 upon A, is to the mean breadth upon B ; so is the mean thickness upon A, to the mean area upon B, or the area of the middle section ; and, as 1 upon A, is to the length upon B ; so is the mean area upon A, to the solidity upon B. Note 1. 1f the tree be equally broad and thick throughout, the breadth and thick- ness, taken in any part, will be the mean breadth and thickness; but if it taper re- gularly from one end to the other, the breadth and thickness must be taken in the middle. Or, take the dimensions of the two ends, and half their sum respectively will be the mean breadth and thickness. 2. Some measurers multiply the square of one-fourth of the circumference, or quarter girt, taken in the middle, by the length, for the solidity; but when the ends are not equal squares, this method always gives too much ; and the more the breadth and thickness differ, the greater will be the error. 3. The Rule given in this Problem. is generally used in pros/ice ; and when the tree is equally broad and thick throughout, it gives the true content; but in taper- ing timber, it always gives too little; consequently, the method mentioned in the last note is sometimes nearer to the truth. When the'true content of a tapering tree is wanted, it must be found by the General Rule, for the frustum of apyramid, given in Problem Vl., Section 1., or by the Rule for a prismoid, given in Problem X. 116 106 TIMBER MEASURE. PART IV. EXAMPLES. 1. If the length of a piece of timber be 9.8 feet, its breadth 2.6 feet, and its thickness 1.5 foot; what is its solidity ? IIere 2.6 X 1.5 X 9.8 = 3.90 x 9.8 a 38.220. Ans. By t/ze Sliding Rule. As 1 on A 1 2.6 on 13 :2 1.5 on A ; 3.9 on B, the mean area. As 1 on A : 9.8 on B I I 3.9 on A : 38.22 on B, the solidity. 2. The mean breadth of a piece of timber is 2.86 feet, its mean thickness 1.93 foot, and its length 18.64 feet; what is its solidity? Ans. 102889072 feet. 3. Each side of the greater end of a piece of squared timber is 28 inches, each side of the less end 14 inches, and its length 18 feet 9 inches ; how many solid feet does it contain ? Ans. 57.421875jeet. Note. The true content measured as the {rustum of a pyramid, is 59.5486 feet. See Example 2, Problem Vl., Section I. 4. The length of a piece of timber is 17 feet 3 inches ; at the greater end, the breadth is 36 inches, and the thickness 20 inches; and at the less end, the breadth is 18 inches, and the thickness 10 inches; what is the solidity? A723. 48.515625 feet. Note. The true content. measured as the frustum of a pyramid, is 50.3125 feet. See Example 3, Problem Vl., Section I. 5. A piece of timber is 32 inches broad, and 20 inches deep, at the greater end; 16 inches broad, and 8 inches deep, at the less end; and its length is 25 feet ; what is its solidity? Ans. 58.01.» feet. Note. The true content, measured as a prismoid, is (31‘g feet. See Example 2, Problem X., Section 1. PROBLEM III. Tofind the solidity qf'romzd or unsquared timber. RULE 1. Multiply the square of 1 of the circumference, or quarter girt, by the length, and the product will be the content, according to the common practice. SECT. III. TIMBER MEASURE. 157 By the Sliding Rule. As the length upon C, is to 12 upon D; so is 71 of the girt in inches, upon D, to the content upon C. RULE II. Multiply the square of :1; of the girt by twice the length, and the product will be the solidity nearly. By the Sliding” Rule. As twice the length upon C, is to 12 upon D; so is 1:1;- of‘the girt upon D, to the content upon C. Note 1. The girt must be taken in the middle of the tree, ifit taper regularly ; if not, take several girts, and their sum, divided by their number, will give something near a mean girt. 1f the tree he very irregular, divide it into two or more parts, and find the content of each separately. ‘2. The girt of a tree is generally taken with a string, which being folded into four equal parts, and applied to a carpenter‘s rule, gives the quarter girt; but it is more expeditious to take the girt with a tape divided into equal parts of 4 inches each; and numbered at 4 inches from the end with l, at 8 inches with 2, at 12 inches with 3, &c., by which means the quarter girt can be obtained by merely girting the tree. If the second Rule be used, the girt may be taken with a tape divided into equal parts of 5 inches each. Each division of 4 or 5 inches, may also be subdivided into halves and quarters, and numbered successively with %, g, %; hence the quarter girt may be obtained with accuracy. The other side of the tape may be divided into feet and inches. 3. 1n measuring a tree which has its bark on, an allowance is generally made, by deducting so much from the girt as is judged sufficient to reduce the tree to such a circumference as it would have without its bark. This deduction may he most easily made from the quarter girt; and should be proportioned according to the thickness of the bark. For ash, beech, clm, young oak, &c. g or g of an inch for every foot of the quarter girt, will generally be sufficient; but for old oak, whose bark is much thicker, it will be found necessary to allow an inch, an inch and a. half, and sometimes two inches, for every foot of the quarter girt. 4. That part of the boughs, or of the trunk ofa tree, which is less than 24 inches in circumference, or 6 inches quarter girt, is generally cut off, and sold at an inferior price ;' not being considered timber. 5. ‘he first ltule in this Problem, gives the solidity about % part less than the true quantity, or nearly what the quantity would be if the tree were squared ; so that it seems intended to make an allowance for the squaring of the tree. 6. The second Rule gives nearly the true content, whether the tree be acylinder, or the frustum of a cone; and as it is full as easy in practice as the first, it ought always to be used when accuracy is required; if the Rules for a cylinder, the frus- tum of a cone, &c. be thought too prolix. (See Examples 3 and 4.) 7. If a piece of round tapering timber be cut through exactly in the middle, the two parts will measure to the most possible ; and to more than the whole by either of the Rules. 8. To find where a round tapering tree must be cut, so that the part next the greater end may measure the most possible: From the greatest girt take 3 times the least; then, as the difference of these girts, is to the remainder, so is one-third of the whole length, to the length to be cut off from the less end. 0r, cut the tree where the girt is one-third of the greatest girt. Wllien the greatest girt does not exceed 3 times the least, this Rule is imprac» ticab e. 9. Forty feet of round timber, measured by the quarter-girl: method, are called a load ; but when the content of timber is correctly found, fifty feet are accounted a load. Fifty feet of dry oak weighs rather more than a ton and a quarter. 10. ’l he content of a piece of timber, according to the quarter-girt method, may be readily found from the following Table, thus : Multiply the area corresponding to the quarter girt, in inches, by the length of the timber, in feet; and the product will be the solidity in feet and decimal parts. .éq...._s.é an. i 158 TIMBER MEASURE. PART IV. A TABLE FOR MEASURING TIMBER. . 1 I ' Qélgtfer Area. ! Q'Guiit'er Area. (15:11:01 Area. 1' Inches. Feet. I‘I Inches. Feet. 1 Inches. Feet. 1 6 .250 12 1.000 1 18 2.250 I 1 64 .272 124 1.042 1 184 2.376 I 1 64 .294 124 1.085 1 19 2.506 1 64 .317 | 124 1.129 1 194- 2.640 1 _._.~___ I I “ I 7 .340 13 1.174 1 20 2.777 1 74 .364 34 1.219 204 2.917 I 74 .390 1 134 1.265 1 21 . 3.062 I 74 .417 1 134 1.313 1 214 3.209 1_ L __.~_ 8 .444 14 1.361 1 22 3.362 84 .472 144 1.410 1 224 3.516 84 .501 144 1.460 1 23 3.673 1 84 .531 144 1.511 234 3.835 I 9 .562 I 15 1.562 I 24 4.000 I 94 .594 154 1.615 1 244 4.168 94 .626 154 1.668 1 25 4.340 94 .659 1 154 1.722 1 254 4.516 I I I' A . 10 .694 16 1 1.777 1 26 4.6.94 1 1 104 .730 164 1 1.833 264 4.876 1 104 .766 164 I 1.890 1 27 5.062 I 104 .803 164 1 1.948 I 274 5.252 1 _ I1 I 11 .840 17 2.006 I 28 5.444 1 114 .878 174 2.066 1 284 5.640 . 114 .918 174 2.126 11 29 5.840 1 114 .959 174 2.187 1 294 6.044 1‘ 1 30 6.250 I EXAMPLE S. 1. “That is the solidity of :1 tree Whose girt in the middle is 100 inches, and length 18 feet P SECT. m. TIMBER MEASURE. 159 By Rule I. In. In. Ft. In. 25:71f of 100. 2 1 25 2 1 W 21—2 50 2 1" 623 1 1 4 _1’ 18 18 5000 781—— 1 6’ Ans. 625 w 144)11235(78.125ft. Ans. By the Table. 4.340 18 @727) 4340 78.120 Ans. By the Sliding Rule. As 18 upon C : 12 upon D :1 25 upon D 2 78 upon C. By Rule II. V In. In. Ft. In. 20:31oflOO. 1 8 2o 1 8 if) 1 1 4’ 36 1 8 .144)1~1400(100 ft. Ans. 2 9 4' 144 36 00 100 0 0 By the Sliding Rule. As 36 upon C : 12 upon D I I 20 upon D : 100 upon C. 2. The length of a tree is 32 feet 6 inches, and its girt in the middle, after allowing for the bark, is 60 inches; What is its content? Ans By the first Rule, 50.78125 feet. ' By the second Rule, 65 feet. 3. The circumference of a cylindrical piece of timber is 6 feet 8 inches, and its length 24 feet; what is its solidity ? Ans {By the/inst Rule, 66ft. 8 in. ' ' By the second Rule, 85ft. 4 in. e an 160 TIMBER MEASURE. 121111 .13 Note. The true content measured as a cylinder, is 84 feet 10 inches. See Ex- ample 3, Problem lV., Section I. 4. The circumference of a piece of round tapering timber is 65 inches at the greater end, 35 inches at the less end, and its length 33 feet 9 inches; what is its solidity? Bz/ tlzefirst Rule, 36. 621 feet. Bu the second Rule, 46. 875 feet. Nola The true content measured as the frustum of a cone, is 84 feet. See Ex- ample 3, Problem Vlll., Section I. 5. Vflmt is the content of an 0:11: tree Whose dimensions are as follow: the length of the trunk is 45 feet, and its (eiu .‘1ter girt, in the middle, 30 inches; the length of :1. 1.11 We boubgh 16 feet, and its quarter {Iii‘t 141 inches; and the Clengtli of another bough 12 feet, and its qu1rt1 girt 10 inches? A215. 311. J6111J'eet. Ans. PROBLEM IV. To measure and value standing timber. Tofind the content. RULE. Obtain the dimensions by some of the methods oesc1ibcd in the following pages, and 13nd the solidity by th 0 Rules given in the List I’ioblem. The firs 1: Rule is gone} lull 1y adopted, and the content found by the Sliding Rule. MISCELLANEOUS INSTRUCTIONS FOR. FINDING THE DIMENSIONS OF STANDING TIMBE; AND FOR SETTING OUT ‘vVOODS AND VALUING THEM PREVIOUSLY TO A FALL. ]. V ARIOUS methods are used in Older to obtain the dimensions of standing trees ,bnt their flirt :111d ulti tudcs may be found most cor- rectly by means of :1 laodei, and a long staff dixided into feet and inches, and numbered from the top to the bottom, on one side, and from the bottom to the top on the opposite side, for the convenience of taking dimensions. SECT. 111. TIMBER MEASURE. 161 The ladder will enable you to take the girt at or near the middle of the tree; and with the staff you may measure the boughs and the upper part of the trunk. The lower part of the trunk may generally be measured by a tape. 2. Divide the arc of the quadrant of a circle of any convenient radius, into 90 equal parts, or degrees; and figure them at every 10 degrees thus; 10, ‘20, 30, &c. to 90. Upon that radius which is contiguous to 90 degrees, place two.small brass sights; and from the centre or angular point suspend a plummet. Fix the quadrant to a square statl', of a convenient length for use, by means ofa nail passing through both; and upon the end of this nail screwa small nut, so that the quadrant may be made fast to the stafl“ at pleasure. The neck of the nail should be round, so that the quadrant may be readily turned upon it; but that part which is contained within the staff should be square, to prevent the nail from turning round with the quadrant. Then, in order to find the height of a tree, screw the quadrant fast to the staff, so that the plummet may hang exactly at 45 degrees, when the staff is perpendicular to the horizon; move the staff backward or forward, always keep it perpendicular, until you can see the top of the tree through both the sights; measure the distance be- tween the bottom of the staff and the bottom of the tree, to which add the height of your eye ; and the sum will be the height of the tree, supposing the ground to be horizontal. If, by reason of impediments, you cannot retire so far from the tree as directed above, make the quadrant fast, so that the plummet may hang at 63!,- degrecs, when you view the top of the tree through both the sights; then twice your distance from the tree added to the height of your eye, will give the height of the tree. If this also be impracticable, take any angle of altitude, and measure the distance to the bottom of the tree; then, by a scale of equal parts, draw a line equal to the mea- sured distance; and at one end of this line erect the per- pendicular, and at the other end, by Problem XXIII., Part L, make an angle equal to the angle of altitude. :Measurc the perpendicular of the right-angled triangle thus formed, by the same scale from which the base was taken, to which add the height of your eye; and you will obtain the height of the tree. 3. Divide a square statf, AB, of about 7 or 8 feet in length, into feet and inches, for the convenience of mea- suring the distance between the place of observation and the tree, or taking any other dimensions. Upon one side of this staff, at a commodious distance from the bottom, fix a rectangular board, CDE F, whose length DE is exactly equal to twice its breadth CD, which breadth may be about 4 or 5 inches. At C and D tix sights, or small iron pins ; and also at G and E; making DG and GE each equal to CD. Then, when the top of a tree is seen through the 162 TIMBER MEASURE. PART Iv. sights at C and G, the tree’s height is equal to your distance from its bottom added to the height of your eye; but if seen through the sights at C and E, its height is equal to twice your distance from its bottom, adding the same height as before. In making an observation with this instrument, it ought to be fixed perpendicularly to the horizon, which may be done by means of a plummet suspended from n. In taking the altitude of a tree growing upon an inclined plane, you must endeavour to make your observations from a place upon a level with the bottom of the tree. If this cannot be done, direct the horizontal sights at C and D, towards the lower part of the tree, and let your assistant make a mark upon it; then find the height of the tree above this mark, as before, to which add the distance of the mark from the ground, which must, in this case, be considered the height of the eye; and the sum will be the height of the tree. The simplicity of this instrument seems to give it a preference to the quadrant; practice will, however, soon enable a person to judge pretty correctly of the heights of trees in general, independently of any instrument. 4. In order to obtain the girt ofa tree, at or near the middle, pro- ceed thus: If the tree taper pretty regularly, take the girt. about two feet from the bottom, to which add 24 inches, the supposed girt at the top, and half the sum will be nearly the girt at the middle; as the height of the trunk ought only to be taken to that part which will measure 24 inches in circumference. Or, take the quarter girt about 2 feet from the bottom, to which add 6 inches; and half the sum will be the quarter girt at the middle nearly. If the upper part of the trunk be nearly as thick as the bottom, which is sometimes the case; take the girt as far from the ground as you can reach, from which girt make a small deduction, at discretion, in order to reduce it as nearly as you can to the girt at the middle. 5. In measuring timber in a wood, you will frequently meet with 8, 10, 12, or more trees. nearly of the same dimensions, so that by measuring one of a medium size, you may, by it, estimate the contents of the rest, without measuring them; and thus be enabled to proceed with considerable expedition. 6. When afall of wood takes place, it is often necessary to cut down many young trees, some of which will not contain more than a foot of wood, in order to make room for others to come to a greater maturity. If the quarter girt of these be less than 4 inches, their contents cannot be found by the Sliding Rule, consequently they must be computed by the Pen; practice will, however, soon enable a person to estimate the contents of small trees with considerable ac- curacy, independently of calculation. Young trees are called pole-wood, and valued at a less price than larger trees, or such as are called timber. 7. In valuing a wood, the branches, the bark of oak trees, and the under-wood must be taken into consideration, as well as the timber and pole-wood. The strongest of the under-wood, and the boughs or branches which are not timber, are called cord-wood, and valued in Yorkshire and other northern counties, by the statute cord of 128 cubic feet; viz. 4 feet in breadth, 4 feet in height, and 8 feet in length. SECT. III. TIMBER MEASURE. 163 Cord-wood is sometimes piled up and sold in parcels of those dimen- sions; and sometimes it is made into piles much larger, and the num- ber of cords they contain found by admeasurement. In Sussex, and several other southern counties, a pile of wood con- taining 126 cubic feet, viz. 3 feet broad, 3 feet high, and 14 feet long, is accounted a cord of wood. The strongest of the cord-wood is made into props for the roofs of coal-mines, posts, rails, &c.; and the other into charcoal, which is used in manufacturing gunpowder, polishing copper or brass, making powerful fires for melting metals, &c. The price of cord-wood is at present from nine to fifteen shillings per cord. Bark is valued by the ton; and the price fluctuates in proportion as we have supplies from foreign countries. The price is, at present, from 12!. to 151., according to the quality of the article. Bark is a very strong astringent, and is chiefly used in tanning the hides of cattle into leather. Under-wood is valued by the acre, if the number of acres contained in the wood be known; if not, it must be estimated by the great, at the discretion of the valuer. The price of under-wood is regulated by the quantity, quality, and demand there is for it. In some woods it is worth 51. per acre, and in others it is scarcely worth 55. Part of the under-wood is used for wicker-work, such as panniers, coal-baskets, &c. ; and some of it is only fit for hedging materials. 8. When a fall of wood is about to take place, the Valuer or VVood- man must first mark all the trees which are intended to stand, by causing a ring of white or red paint to be made quite round the trunk of each tree. This is called “ setting-out.” The trees that are permitted to stand should be at proper distances from each other; of different ages; and such as are likely to make the greatest improvement. Those which are straight and elegant, and have few branches, are generally preferred; this part of the busi- ness, however, depends very much upon the skill and experience of the woodman. When the wood has all been set out, those trees which are not marked must be valued; and in doing this, attention must be paid to the sizes, qualities, and species of the trees. Large trees, if sound, are always worth more per foot, than small ones; and if the crooked oak trees, and large crooked boughs, be particularly adapted for ship timbers, they are worth more per foot than straight trees. In measuring and valuing, some persons number every tree, by a scribe-iron, or with paint; others only count them, and mark each tree with a cross, to show that it has been measured; and others number the large trees, and count the small ones, setting down the contents of 10, 15, or 20 of these in one sum. 9. No regular rules can be given by which the quantity of bark, and the cord-wood a tree contains, may be computed; as the bark of trees differs very much in thickness, and the cord-wood depends entirely upon the quantity of top. Sometimes there is a great deal more bark upon the top of a tree, than upon the bole. If the bole of a tree measures 30 feet or 360 inches in length, 60 inches in circumference in the middle, and the bark be half an inch thick ; we have 860 multiplied by 60 multiplied by 11;, equal to 10800 cubic inches, the solidity of the bark, which being divided 164 TIMBER MEASURE. PART 1v. by 2218, the cubic inches in an imperial bushel, we obtain 5 bushels nearly, the quantity of bark upon the trunk, to which must be added the quantity estimated to be on the top; practitioners, however, are never at the trouble of calculating the quantity of bark upon the trunk, but make an estimate of what they suppose to be upon the whole tree. Besides, when bark is chopped, it will measure to a great deal more than in a solid state, as found by the above calcula- tion. In most places in the West Riding of Yorkshire, 6 pecks are ac- counted a bushel of bark, and 9 such bushels a quarter; and it has been found by experiment, that about 8 quarters of the bark of pretty large trees, or about 12 of that of small trees, or pole-wood, will weigh a ton; consequently, we may reckon about 10 quarters, upon a me- dium, to make a ton. 10. When an estate, containing a wood, or a great number of trees in the hedge-rows, is intended to be sold, all the trees should be valued previously to the sale; as there have been instances where Estates have been sold, without considering the woods; and the pur- chaser has immediately disposed of the timber for nearly the price of “the Estate. 11. “'hen a rough calculation of a large wood is wanted, or time will not permit you to be very accurate, let an acre be measured oti‘ in that part of the wood which appears to be of a mean value, and measure and estimate the timber, &c. which it contains; then this estimation being multiplied by the number of acres in the wood, will. give the value of the wood nearly. Or measure and estimate an acre of the best and an acre of the worst; and multiply half the sum of the two estimations by the number of acres in the wood, for the whole value. 12. In measuring and valuing standing timber, various methods are adopted, by different persons, in setting down the dimensions, &c. Some enter the oaks, elms, &c. separately in different parts of the book; and others enter them promiseuously, having a column in which they specify the name of each tree. Some valuers estimate the quantity of cord-wood in the top of each tree, or in the tops of two or three trees taken together; and others value the top ofcach tree. The two following examples will serve to illustrate what has been said on the subject of measuring and valuing standing timber. The content of each tree was found by the Sliding Rule; and the learner ought to repeat the process, in order to make himself expert at casting by the Rule. EXAMPLES. 1. Find the contents of six oak trees, and the value of the timber contained in each tree, from the following dimensions and values. SECT. In. TIMBER MEASURE. 165 Number Bush. Cords Length Quarter Contents Values of of of in flirts in in per Value of each Trees. Burk. Wood. Feet. Inches. Feet. Foot. Tree. 8. d. £ 5. d. 1 1% o 12 8% 6 . o 012 o 2 3.;— o 163.. 1231. 17% 2 6 2 3 9 3 32 .5. 39% 18 89 4 3 1818 3 4 15 :1. 23 15g 39g 3 0 519 3_ 5 2.1. o 15 9.1,— 957 2 3 1 1 4% G 80 :13- 44 28% 248 6 0 74: 8 0 Total l344i ’ 1% —— —-. —— -—— 103 2 7% 2. Required the contents of ten trees, and the value of the timber in each tree, from the following dimensions and values. l i iLengths Quarter Contents Value No. of Namesofthe Value ofl in Girts in in per Valueof Trees. Trees. lthe'l‘ops.£ Feet. Inches. Feet. Foot. each’l'ree. ‘1 s. d.‘ s. d. 39 s. d. 1 Elm. .10 5 6 42.;- 12.41. ‘14-}: 3 0 6 129 2Ash. J‘o 503841—1172— 3633 305 96 3 Pine 'i0 3 O: 36 14—12~ ’2—12 3 6 9 3 9 411%. .10 1628 10 194: 19 113811- 5 Beech .;o 4 6; 39% 16% 5 2 6 9 7 6 6 SycamoreO 3 6 26;}— 9f 17'1 2 6 2 31%- 7 Poplar 0 2 6348-511—‘3 543 3 0 8 4 3 8 Yew. .1 2 6; 9% 38% 961g 4 92218 4.12. 9 Fir. .p 2 0132»; 13—1 392—; 2 0‘ 3 19 0 10 Larch .0 2 6l 30;;— 151E 5111— 3 6 819 4;. _ 1 1 Total ‘2 12 6: — —— —— —— 178 11 3;}l Core. Having described the methods of measuring and valuing standing timber; and. in the lust Example, mentioned a variety of trees, it is presumed that this section cannot be better concluded than by a short description of the natures, pro- perties, and uses of these trees. Besides, it is absolutely necessary, in order to become a valuer of timber, to be made acquainted with a few of the leading properties of trees, and their comparative usefulness. 166 A DESCRIPTION PART iv A DESCRIPTION OF THE NATURE AND PROPERTIES OF SOME OF THE MOST USEFUL TIMBER TREES. THE OAK. THE OAK stands at the head of all British timber trees, as well for its utility, as for its majestic appearance. It is slow of growth; but if permitted to stand, it arrives at a size equal, if not. superior, to that of any other tree of the forest; and by the vast arms which it throws out on every side, it forms a mass which fills the eye of the spectator, and impresses him with gigantic ideas of its masculine strength. The oak delights most in a rich, strong soil, in which it strikes its root to a great depth. It loves hilly, better than boggy ground, and thrives best in large plantations. The uses to which oak is applied, are numerous. It is a firm wood, and will endure all weathers, climates, and seasons; hence it is used for posts, rails, window-frames, casks, water-pails, carts, wag— gons, wheel-spokes, &c. In machinery, no other wood is equal to it, where a great stress is to be borne, as in mill-work, 8w. ; and in water- works it is inferior to none. It is also used for household furni- ture, such as tables, bedsteads, chests of drawers, 8w. ; but it has ac- quired its chief fame, in this country, by its use in ship-building, being much superior to foreign oak; and has, no doubt, contributed very materially to the naval glory of OLD ENGLAND. Some oak trees have arrived at an enormous size. In Dr. Hunter's edition of Evelyn's Sylva, is given a. figure of the old oak of Cow- thorpe, in Yorkshire, which measures 48 feet in circumference, at a yard from the ground. About a mile and a half from Shrewsbury, there is an oak whose girt is 44 feet at the bottom, 27 at the distance of 8 feet from the ground; and it is 41}, feet in height. In Hainault Forest, near Barking, in Essex, there is an oak which measures 36 feet in circumference. The tree has been known through many cen— turies by the name of FAIRLOP. Mr. Gilpin, in his “ Remarks on Forest Scenery and other Woodland Views,” says that the tradition of the country traces this tree half-way up the Christian aara. Some of our best poets have noticed the usefulness and remarkable longevity of the oak. “ Let India boast her plants, nor envy we The weeping amber and the balmy tree, While by our oaks the precious loads are borne. And realms commanded which those trees adorn.” _ Para. “ The monarch oak, the patriarch of the trees, Shoots rising up, and spreads by slow degrees : Three centuries he grows, and three he stays Supreme in state, and in three more decays." _. Diu‘nm‘ SECT. In. or TIMBER TREES. 167 , THE ASH. This tree generally grows tall and elegant, and makes a graceful appearance, when contrasted with trees of greater bulk. It flourishes most in woods, but will also thrive well in good soils, upon open grounds. There are few trees which excel the ash in utility; for its wood, next to that of" oak, is employed for the greatest variety of purposes. It may be peculiarly termed the husbandman’s tree,- for it is one of the principal materials in making ploughs, barrows, carts, waggons, spokes and felloes for wheels, and various other implements for rustic use. It is also employed by the turner for dairy utensils; and at sea it is used for oars and hand-spikes. The toughness of its wood rendered it a favourite with the heroes of old, for the shafts of their potent spears; hence it is poetically termed “ the martial ash.” Homer arms his heroes with spears of ash. “ From Pelion’s cloudy top an ash entire, 01d Chiron fell’d, and sharp’d it for his sire.” — POPE's HOMER. THE BEECH. This is one of the most stately of timber trees; large woods are wholly composed of it in some parts of this country. It particularly delights in a chalky soil, where it will flourish and arrive at a great size, although the land may have all the appearance of barrenness. The wood of the beech is brittle, and apt to decay; but being of a fine grain, and easily wrought, it is used for a variety of domestic purposes. It is employed by the turner and cabinet-maker; the former using it for his larger ware, and the latter for common chairs and other articles of furniture. It can be split so thin, that it is used for band-boxes, hat-cases, book-covers, and scabbards of swords. THE ELM. The common elm is a large timber tree, of great beauty and utility. It grows to a. great height, and at the same time, if permitted, ex- pands some massy arms. It loves an open situation, and a black clayey soil. The wood of the elm is hard and tough; and is used for a great variety of purposes. Is is particularly serviceable in situations where it is kept constantly wet ; and is therefore used for ship-planks, water- pipes, pumps, mill-wheels, &c. It is also employed for axle-trees, wheel-naves, gate-posts, chopping-blocks, 8:13. THE SYCAMORE. This tree is of quick growth, arrives at a large size, and flourishes best in open places and sandy ground. Its wood is soft and very white; hence it is proper for the use or the turner, who makes it into bowls, trenchers, and other domestic utensils. In consequence of its lightness, it is also occasionally wrought up as cart and plough timber. 168 A DESCRIPTION PART Iv. THE BLACK POPLAR. The name of black seems given to this tree, in order to distinguish it from the while poplar; for its leaves are a beautiful green, and the tree has nothing dark in its appearance. It loves a rich and moist soil; it arrives at a great size; and is one of the tallest and most stately to be seen when full grown The wood of the poplar is tougher and harder than fir; and is frequently used instead of it, for laths, packing-boxes, rooting, floor- ing, &c. The white poplar does not arrive at so large a 5120 as the black. It grows best in moist situations; and is very conspicuous from the whiteness of its foliage. TIIE ALDER. The alder flourishes best in boggy situations, and by the sides of rivers. &c. It sometimes arrives at a large size; it is, however, chiefly used in this country for poles. Virgil, in his Georgics, tells us tha , by hollowing its trunk, it was aneiently made into boats or canoes. The wood of the alder will remain long sound under water. It is employed for pumps, water-pipes, piles, &c. It is likewise used for shoe-heels, clogs, and turners’ work. THE W'EYMOUTH PINE. This tree is a native of North America, where it is called the “ white pine.” It frequently arrives at the height of 100 feet; and is therefore preferred to the rest of the pine tribe for the masts of large ships. Our men of war are generally furnished with masts of this species of tree, from the timber-yards of Nova Scotia. Lord VVeymouth introduced it into this country ; on which account it is generally known by his name; and has, for some time, been a great favourite with planters. THE FIR. Linnmus considers all the fir tribe to be only different species of the pine. Some of them grow best on mountainous situations ; others prefer bogs and swamps; and they alone often compose exte isive woods, clothing barren and desolate regions, unfit for human culture. Many species of this tribe, are now become'common in our planta- tions. Of these I shall only mention a few of the principal. The Scotch Fir grows naturally in some parts of the Highlands of Scotland; and also upon the mountainous parts of Norway, Sweden, and Russia. The fir grown in the native forests of Scotland is good, but affords a very scanty supply; consequently, the greatest part of what is consumed by us, is brought from Norway, and the countries border- ing upon the Baltic. No wood is at present used amongst us in such quantities as the fir, which, under the name of deal, is employed about buildings, for boards, planks, beams, rafters, joists, Sac. Fir is also used for the masts and yards of ships; and from it we obtain the important productions of turpentine, tar, and pitch. It SECT. III. or TIMBER TREES. 169 \ may, therefore, be called the sailors’ tree, with as much propriety as the oak. The Spruce Fir is a native of Norway; and is said to afford the white deal. From the green tops of this tree is made the antiscor- butic beverage called “ Spruce-beer.” ‘ The Silver Fir grows in Norway and Germany, and from its trunk the yellow deal is said to be procured. It also yields great abundance of tar. The Larch Fir is a native of the Alps and Apennines; and for beauty and durability of wood, it greatly surpasses the Scotch fir. It is now become a great favourite in this country, and thrives well on barren and sandy soils. In some countries ships are built of this wood, which are represented as proving very durable. The larch is remark- ably resinous, and from it is obtained Venice turpentine. THE BIRCH. There are four species of birch. The common birch-tree may be cultivated upon barren land, where better trees will not grow; for there is no ground so bad as not to allow this to thrive in it. It will grow in moist springy land, or in dry gravel, or sand, where there is little surface; hence persons who are in possession of poor land, can— not employ it better than by planting it with these trees. They have sometimes been planted upon ground that produced nothing but, moss ; and in nine or ten years after planting, have been sold for 101. per acre; the after-produce greatly increased. The wood of this tree is used by broom-makers, hoop-benders, and turners. THE YEW. The yew is a native of this country, and is found in rocky and mountainous situations. It is frequently planted in church-yards, probably on account of its being an e‘ver-green. The leaves are of a poisonous nature; both horses, cows, and children have died in con- sequence of eating them. This tree is remarkable for its toughness and elasticity; and is celebrated for the purposes to which its wood was anciently applied in making that most formidable weapon of our ancestors,—the long- bow. ' “ Th’ elastic yew, whose distant wound With England’s rivals heap’d the ground.” -—VViLLiA.\i’s MISCELLANIES. The wood of the yew is at present valued by cabinet-makers and inlayers on account of its beautiful red veins; and it is also a good material for flood-gates, axles, cogs for mill-wheels, and other works where strength and durability are required. The yew seldom grows to a great height; but sometimes arrives at an amazing thickness of trunk. In Darnley church-yard, near lVIatlock, in Derbyshire, is a yCWo tree whose circumference is 33 feet. THE HOLLY. The holly is an ever-green, and a native of the woods in this coun- try, where it sometimes arrives to the height of 20 or 30 feet. It is, I 170 A DESCRIPTION OF TDIBER TREES. PART IV. however, more usually seen in gardens and hedgerows, in the state of a shrub. The wood of a full-grown holly is valuable. It is the whitest of all our hard woods, and is therefore used by inlayers; and it is some- times stained black, to imitate ebony. It is also excellent for the uses of the turner, carver, and mill-wright; being extremely firm and durable. THE BOX. The box is another ever-green tree or shrub, which is sometimes met with here in a wild state; but more commonly in gardens and plantations. A strong shallow soil, of the limestone kind, seems to suit it best. The wood of the box is of a pale-yellow colour; and being very hard, smooth, and solid, it is much used for various purposes. Mathe- matical and musical instruments are made of it; also knife-handles, combs, shuttles, &c. It is the only tree whose wood is serviceable to the engraver. Having described the natures, properties, and uses of most of the trees found in our woods and plantations, I shall conclude with the following extract from l\Ir. John Tuke’s Survey of the Agriculture of the North Riding of Yorkshire; a work that abounds with useful and just observations: — “ Most people, I think, concur in this point, that for the last half century, the wood in this kingdom has been terribly on the decline. “ That gloomy prospect is now become tremendous, and sufficiently visible to awaken the fears of every thinking person. The axe is often heard, but the planter is seldom seen. Let us cast our thoughts towards the future support and welfare of our navy—our sole pro- tection l—and we must tremble at the continual disappearance of our oak. Some speedy method must be adopted to remedy this great national evil; or, besides the danger from fierce external foes, we must determine to go barefoot: we should never think of looking to foreign countries for a constant supply of oak—bark to tan our leather. t Britain help herself! “ Each nobleman and gentleman should insert, in the agreement with his tenants, a clause to compel them to plant and protect, in the corners of their fields, and upon pieces of waste ground. a certain number of good oak, elm, and ash trees, annually. These trees should be found by the landlord; and he should enforce the perform~ ance of this clause as rigidly as the payment of the rent; then will the rising generation have cause to bless the wisdom and policy of the present age.” Note. It has, however, been found within the last few years, that England can have a sufficient supply from her own colonies. It is now ascertained that the timber in the fertile islands of New Zealanti far exceeds that of the English, both in size and variety; the covvdie tree is a fine specimen; it possesses a high degree of flexibility, as well as strength: the wood is fine grained, and is of sufficien length and strength to serve for the main and tore top masts of the largest three- decker; the Board of Admiralty being now procuring supplies of this timber for the use of the Royal Navy. SECT. 1v. MISCELLANEOUS QUESTIONS. 171 SECTION IV. MISCELLANEOUS QUESTIONS CONCERNING SOLIDS. 1. WHAT is the solidity of a cubical stone, whose dia- gonal is 30 inches? Ans. 9545.94154 inches. 2. The length of a parallelopipedon is 10, its breadth 6, and its depth 4 feet; what is the length of a similar parallelopipedon, whose solidity is three times as much ? Ans. 14.42249feet. 3. The altitude of a cylinder is 20, and its diameter 10 inches; what is the altitude of another cylinder whose solidity is twice as much, its diameter being 30 inches? Ans. 4% inches. 4. The diameter of the old Winchester bushel is 18% inches, and its depth 8 inches; what is the diameter of that bushel whose depth is 7 inches ? Ans. 19.77733 inches. 5. Two men bought a conical piece of timber, which is to be equally divided between them, by a plane parallel to the base; what will be the altitude of each part, the height of the cone being 21 feet, and the diameter of its base 3 feet 6 inches ? Ans. The altitude qf the upper part is 16.66771 feet; and that of the lower part 4.33229 feet. 6. Each side of the greater end of a piece of timber, in the form of the frustum of a square pyramid, is 5 feet, each side of the less end 2 feet, and its perpendicular alti— tude 15 feet ; it is required to divide it equally between two persons, by a plane parallel to the ends. , Ans. The height of the part mat the less end is 10.25709, and the height of the part next the greater end is 4.74291 feet. 7. A gentleman has a bowling-green, 500 feet in length, and 300 feet in breadth, which he would raise 1 foot higher, by means of the earth to be dug out of a ditch with which he intends to surround it; What will be the depth of the ditch, if its breadth be every where 9 feet ? Ans. 10.18744 feet. 1 2 172 MISCELLANEOUS QUESTIONS 1111111 1v. 8. What must be the diameter of the bore of a cannon, which is cast for an iron ball of 36 lb. in weight, so that the diameter of the bore may be i of an inch more than that of the ball? Ans. 6.4747 inc/res. 9. If a heavy sphere whose diameter is 8 inches, be put . into a conical vessel, full of water, whose diameter is 10, and altitude 12 inches; it is required to determine how many cubic inches of water will run over. Ans. 210.17722 cubic incfies. 10. A cubical foot of brass is to be drawn into wire 31-0- of an inch in diameter; what will be the length of the wire, allowing no waste in the metal ? Ans. 31.25217 miles. 11. How many 4-inch cubes can be cut out of a 12- inch cube ? Am. 27. 12. The length of a piece of square timber is 12 feet, and its solidity 48 feet; what will be its solidity when it is made into the greatest cylinder possible ? Ans. 37.6992feet. 13. A farmer borrowed part of a haystack of his neigh» bour, which measured 8 feet every way, and paid him back by two equal cubical pieces, each side of which mea- sured 4 feet: Query, was the lender fully paid? . Ans. He was paid 0721117111017. 14. A piece of timber 1s 20 inches broad, and 16 inches 1 thick; at what distance from the end must a section be 1 made, so that the part sawn off may measure exactly5 cubic feet ? Ans. 27 inches. 1 15. A ship’s hold measures 120 feet in length, 33 in ‘ breadth, and 6 1n depth , how many bales of goods 6 feet l long, 4 feet broad, and 3 feet deep, may be stowed the1 ein, ‘ leav mg a gangwa J, 3 feet b1 oad, the whole length of the i hold? " A723. 300. 16. The diameter of a cilcular island is 300 feet, and it is surrounded by an oblique- -sided moat; how many cubic yards of earth were dug out of it, the breadth at the top being 10 feet, the breadth at the bottom 4 feet, and the ! pe1pendicular depth 8 feet? Ans. 2019. 93244 02112207 ymds. l 17. Suppose it be found, by measurement, that a man | of wa1, with its ordnance, rigging, and appointments, ‘d1aws so much water as to displace 55000 cubic feet 01 sea- 11 atcr; requil ed the weight of the vessel. Ans. 157 2—33 tom. 18. The length of a cylindrical piece of timber is 15 H seer. 1v. CONCERNiNG soups. 1 {3 feet, and its diameter 4 feet; what will be its solidity when hewn into a square prism? Ans. 120 feet. 19. How many bricks, each 8 inches long, 4 inches broad, and 3 inches thick, will build a wall 200 feet long, 10 feet high, and 1 foot thick? Ans. 36000. 20. ‘Vhat is the weight of a cast iron cannon-ball whose diameter is 4 inches ? Ans. 9 pounds. 21. The ball upon the top of St. Paul’s church, in Leeds, is 3 feet in diameter, and it is covered with copper 7,15 of an inch thick; required the value of the copper, at 25. 39,117. , per pound avoirdupois. Ans. £7 125. 3igd. 22. What is the weight of a bomb-shell, or hollow sphere of cast iron, whose inside diameter is 9 inches and the thickness of the metal an inch and a half ? Ans. 140.4746 lb. 23. The length and breadth of the greater end of a mill-hopper, in the form of a prismoid, are 40 and 30 inches; the length and breadth of the less end 10 and 6 inches; and its perpendicular altitude 36 inches; how many imperial bushels will it hold ? Ans. 8.277 bushels. 24. A tree whose length is 28 feet, girt 45 inches in the middle, with a rope 1 inch in diameter; required its true girt, and likewise its solidity, according to the quarter- girt method of measuring timber; the length of the rope, when extended in a right-line, being equal to the circum- ference of its centre, when formed into a circle, in taking the girt of the tree. _Ans. The true girt 2's 41.858364 inches, and the solidity 21.293157 feet; hence appears the propriety of girting trees with a small string, or with a mea- sunny-tape. 25. A leaden pipe, 12 feet in length, weighs 100 lb. avoirdupois ; required the thickness of the lead, the dia- meter ot' the bore being 2 inches. Ans. .2408 qfan inch. 26. Required the thickness of the shell of a hollow sphere of copper, weighing 3 lb. avoirdupois, so that being put into common water, it may be just immersed by its own weight ? Ans. .1042 of an inch. 27. The bore of a syringe which holds one pint, wine measure, is 1% inch in diameter; what is the length of the piston? Ans. 16.33986 inches. 28. If a balloon contains 1000 yards of silk, {2- of a yard I 3 ' 174 MISCELLANEOUS QUESTIONS. PART 1v. wide; What is its diameter, admitting it to be a perfect sphere? Ans. 15.45095 yards. 29. A hollow sphere of ebony, whose diameter is 18 inches, is found to sink just 6 inches in sea-water; re— quired the thickness of the shell. Ans. .647333 of an inc/z. 30. One ev’ning I chanc’d with a tinker to sit, Whose tongue ran a great deal too fast forhis wit; He talk’d of his art with abundance of mettle; So I ask’d him to make me a flat-bottom’d kettle: Let the top and the bottom diameters be, In just such proportion as five is to three; Twelve inches the depth I proposed, and no more; And to hold in ale-gallons seven less than a score. He promis’d to do it, and straight to work went; But when he had done it, he found it too scant ; He alter’d it then, but too big he now made it; For though it held right, the diameters fail’d it : Thus making it often too big, and too little, The tinker at last quite spoiled his kettle; But he vows he will bring his said promise to pass, Or he’ll utterly spoil every ounce of his brass : Now, to save him from ruin, I pray find him out The diameter’s length, for he’ll ne’er do’t I doubt. Ans. The bottom diameter is 14.640098, and the top diameter, 24.400163 inches. Note. This ingenious question was, I believe, first proposed in the. Ladies’ Diary, for the year 1711. PART v. ARTIFICERS’ WORK. 175 PART V. ARTIFICERS’ WORK. The Artificers whose Works are here to be treated of, are Bricklayers, Masons, Carpenters and Joiners, Slaters and Tilers, Plasterers, Painters, Glaziers, Plumbers, and 1Pavers. THE contents of the works of all artificers, whether super- ficial or solid, must be found by the Rules given in the foregoing Problems, for the respective figures. This ought to be particularly attended to in taking the dimen- sions. The following Rule, which should be got by heait, is bette1 adapted for Cross illultiplication, than any other I have seen; and as 12 fourths make 1 third, 12 thirds 1 second, 12 seconds 1 inch, and 12 inchesl foot, these numbe1 s, when not too large, may be“ more expeditiously turned into the higher denominations by the Pence Table, than by dividing them by 12. RULE. Feet multiplied into feet give feet. Feet multiplied into inches give inches. Feet multiplied into seconds give seconds. Inches multiplied into inches give seconds. Inches multiplied into seconds give thirds. Seconds multiplied into seconds give fourths. The measures chiefly used by Artificers, are contained in the following table :— 12 inches , 1 lineal foot. 144- square inches . 1 square foot. 9 square feet 1 square yard. 63 square feet make 1 square rood. 100 square feet 1 square. 272},- square feet, or} 1 square rod, pole, or 301 square yards perch. Note. As the number 2721 is rather troublesome to divide by it is customary, in practice, to divide by 272, omitting the ‘; but if this be not thought correct enough, convert the i into decimals; 01 re uce the di1isor to the improper frac- 14 ’15 ‘ if r P 9 176 BRICKLAYERS’ WORK. PART V. tion 138:), and multiply the content in feet by 4, and divide the product by 1089; and the quotient will be the number of rods sought. Divide the remainder, if any, by 4, and the quotient will be the feet. The rod of 272% square feet, is the square of 16% feet ; but in some places the cus- tomary rod is the square of 18 feet, viz. 324 square feet. BRICKLAYER 8’ WORK. Bricklayers generally compute their work by the rod of 2721,, square feet, and at the rate of a brick and a half thick ; consequently, if a wall be more or less in thickness than this standard, it must be reduced to it, in the follow- ing manner : Multiply the superficial content of the wall, in feet, by the number of half bricks which it is in thick- ness; and E of the product will be the content sought. In some places, however, brickwork is measured by the rod of 63 square feet; that is, 21 feet in length, and 3 feet high; and then no regard is paid to the thickness of the wall, in measuring; as the price of the workmanship is regulated according to the thickness. late 1. In taking the dimensions of a building, measure half round the outside and half round the inside; and the sum of these two will be the true compass of the building. Or, 4 times the thickness of the wall taken from the whole compass, on the outside, or added to the whole compass within, will give the true compass; which multiply by the height, and the product will be the superficial content of the walls; and if the breadth of the building be multiplied by the height of one gable- end, you will obtain the content of both the gable-ends. In some places it is customary to take the whole compass, on the outside, in. order to make an allowance to the workmen, for the trouble of turning the corners; but this should not be done for both workmanship and materials, except specified in the agreement; but for workmanship only. If the workmen be allowed so much per yard, lineal measure. for the corners, which is the custom in some places. the true compass of the building ought then to be taken for both workmanship and materials. 2. When the height of a building is unequal, measure a piece round the bottom, so as to make the upper part all of one height ; and in doing this, holes must be dug in the ground, to enable you to take your dimensions to the foundation. Several altitudes of the bottom part must be taken ; and their sum, divided by their number, may be considered as a mean altitude. 3. In most buildings of two or more stories, the walls decrease in thickness, to- wards the top; and this diminution generally consists of half a brick, in each story. The thickness is set OFF on the inside, and commonly in a place where the floor will be laid; a contrivance by which the set-o is concealed. The stories that are of different thicknesses must be measured separately; except by agreement, one price he allowed for the whole, which is not often the case. When the walls of a building are of ditl‘crent thicknesses, the upper rooms are broader than the lower ones ; hence a set—rfl'may be discovered, although the walls are plastered and the floors laid. 4. Doors and windows must always be deducted for materials, if there be no stipulation to the contrary; but for workmanship, these deductions are seldom made, except the doors and windows he very numerous, as in workshops, &c. ; or larger than the usual size, as in shop—fronts, &c. In these cases the Surveyor must exercise a distretionary judgment; for the windows of buildings vary so much in size and number, that no special Ilule can be given. 5. If a chimney stand by itself, without any party wall being.' joined to it, take the girt. in the middle, for the length, and the height of the story for the breadth; but if the chimney-back be a parryuwall, and the wall be measured by itsell, you must girt the chimney round, to the wall, on each side, for the length ; and take the breadth the same as before. When a clu'nmcy is wrought upright from the mantcl-trec to the ceiling, the PART v. BRICKLAYERS’ wonr. 177 thickness of the whole, is generally considered the same as that of the jambs ,- and no deduction is ever made for the vacancy between the floor and the mantel-tree, because of the gathering of the breast and wings, to make room for the hearth in the next story. Chimney shafts above the roof, are measured by girting them, in the middle, for the length, and taking the height for the breadth. Their thickness is generally accounted half a brick more than it is in reality, in consideration of the plastering and scaffolding. (S. In some places, double measure, for workmanship, is allowed for chimney/s, in consequence of their heii g more troublesome to be made than the other parts of the building; and in others they are done at so much per yard, lineal measure, or at so much per piece. It is also customary, in most places, for bricklayers, to charge so much extra for every arch they turn; and this charge is regulated by the size of the arch. They also make a difference in the price between an inside and an outside arch, charging less for the former than the latter. EXAMPLE S. l. The length of a wall is 86 feet 9 inches, its height 12 feet 6 inches, and its thickness 3 bricks; how many standard rods of brick—work does it contain? By Decimals. By Cross flfultiplication. Feet. Feet. Inches. 86.75 length. 86 9 12.5 height. 12 6 43375 1041 0 17350 43 4 6” 5 867,571 fi84‘r6 1084.375 6 ,,__~_~6 @6506 3 0 7'. fl. 2'27. 3)6506.250 7'. fz‘. in. 272)2168 9 0(7 1264 9 272)?!§§§7§9(7 264 9 Ans. 2. In taking the dimensions of a cottage, I find that half the compass, on the outside, is 40 feet, half the corn- pass within 37 feet, the height from the foundation to the eaves 9 feet, the height of the gable—end 5 feet, and its breadth 18 feet; how many rods of brickwork are con- tained in the building, deducting for a door, which mea- sures 6 feet by 3, and a window whose height is 5, and breadth 4 feet? A723. 11 roads, 52fcct. 3. The circumference of a circular building at the iron foundry of hiessrs. Fenton, lilurray, and Wrood, in Leeds, is 241 feet, and its height, from the bottom to the eaves, 49 feet 6 inches; how many rods are contained in the wall; deducting for a door whose breadth is 12, and height 141 feet; for 96 windows, each of which measures I 5 178 MASONS’ WORK. PART v. 6 feet 7inches by 4 feet; and for 14 Windows, whose heights are 9 feet 6 inches, and breadths 5 feet 4 inches? Ans. 135 rods, 19 feet. 4. The true compass of a building, two stories high, is 120 feet 6 inches ; the height of the lower story is 10 feet 6 inches, and the thickness of the wall 2 bricks; the height of the upper story is 8 feet 3 inches, and the thick- ness of the wall 1?7 brick ; the gable-end measures 20 feet in breadth, and 8 feet in height, and is 1% brick thick; what did the brick—work cost at 14L 123. per standard rod ; deducting for 6 windows in the lower story, each of which measures 5 feet 6 inches by 4 feet 3 inches, and a door whose height is 6 feet 6 inches, and breadth 3 feet 9 inches, and four Windows in the upper story, Whose dimensions are 4 feet 9 inches by 4 feet 3 inches ? A723. £ 136 68. 9d. MASONS’ WORK. All kinds of stone-work belong to Masonry; and the measures generally used are the lineal foot, the square foot, the square yard, the square rod of 63 feet, and the cubic foot. Carved mouldings, &c. are generally measured by the lineal foot; and ornamental work, such as arches, archi- traves, friezes, cornices, chimney-pieces, &c. by the square foot. Also all tooled or cleansed work is measured by the square foot ; viz. door-posts, window-jambs, flags, steps, 8:0. ; but mug/l flagging is generally measured by the square yard. TValls are sometimes measured by the square yard, and sometimes by the rod of 63 square feet. Columns, pillars, blocks of marble or stone, &c. are measured by the solid foot; and sometimes the contents of walls are found in the same measure. Solid measure is chiefly used for materials, and super- ficial for workmanship; in some places, however, masons are paid so much per rod, for workmanship and ma- terials, and the price is regulated by the thickness of the wall. Note 1. The dimensions of stone buildings are taken in the same manner as in Bricklayers’ Work; and deductions must be made for doors, windows, &c., except the agreement prohibits it. These deductions. however, ought only to be made for materials; as the workmen are fully entitled to receive pay for the whole as walling, in consequence of the trouble of fixing the windowjambs, &c. PART v. MASONS’ WORK. 179 In measuring tooled or cleansedfronts, the doors and windows must be deducted ; as the pdrice of the workmanship, in these cases, is too considerable for them to be include . 2. The walls of the upper stories of buildings are, in general, not so thick as those of the lower stories; but the price for the workmanship is commonly the same, in the consideration of the trouble of scaffolding, and the labour of carrying up the materials. 3. In some places it is customary to measure door-posts, window-jambs, steps. &c. by the cubic foot for the materials, and the superficial foot for the workman- ship ; and in others, so much per superficial foot, is charged for workmanship and materials; and in taking the breadth, 0r girt, the tape is made to ply close over every part of the stone that has been tooled, except itappear that the workmen have intenlz'onally tooled more than is necessary. The length of a circular window—head, or door-head, is found by taking half the sum of the greater and less arches. 4. In making the notch in the window-jambs, for the frame of the window, some workmen are in the habit of tooling or chiseling the jambs further on the inside, than the window-frame requires, in order to make the work measure as much as possible: few architects will, however, allow more than 3 or 4 inches for both sides of the notch. 5. In measuring a flight of steps make the tape ply close over them, in the middle, from the top to the bottom, for the length ; and take the length of a step for the breadth. Or, if the steps be all of one size. multiply the area of one step by the number of steps. The ends and the landings ought always to be measured by themselves. 6. All the parts of ornamental frontispieces must be measured separately: viz. plinths, dados, columns, pz'lasters, archz'lrlwes, frz’czes, corm'ces, pedz'menls, &c. &c. 7. It is customary, in some places, to allow double measure for all kinds of circular work, such as cylindrical or conical columns, circular pedz'menls, arched door or window—heads, &c. ; and also for cornz'ces. feathered gables, &c. : in other places, only the area and half the area are allowed; it is much better, however, to proportion the price to the workmanship, and take the true measurement. It is also customary, in most places, to pay the workmen so much per yard lineal, for the trouble of turning the corners of buildings, and hewing the stones in a proper manner to form these corners. ' If the corners be formed by tooled or cleansed coins, they must be measured separately, and added to the tooled or cleansed work. In this case no other charge ought to be made for the corners. . It may likewise be observed, that masons are generally allowed something extra for the trouble of turning arches. EXAMPLE S. 1. The length of a wall is 86 feet 9 inches, its height 10 feet 6 inches, and its thickness 2 feet 3 inches; re- quired its superficies and solidity. By Decimals. By Cross illultiplication. Feet. Feet. Inches. 86.75 length. 86 9 length. 10.5 height. 10 6 height. 43375 867 6 8675 43 4 6” 910.875 superficies. 910 10 6 superficies. 2.25 thickness. 2 3 0 thickness. 4554375 1821 9 0 1821750 227 8 7 6’” 1821750 204.9 5 7"6solidity. W solidity. 24- 180 MASONS’ wonK. PART v 2. The length of a flight of Yorkshire-stone steps is 10 feet 6 inches, and their breadth 3 feet 9 inches; the landing measures 5 feet 3 inches by 4 feet 6 inches; to what do they amount, at 18. 9d. per square foot, for work- manship and materials? Ans. £5 105. 3d. 3. The side of a square pillar, of Portland-stone, me' - sures 1 foot 9 inches, and its height 7 feet 6 inches ; what is the value of 6 pillars of the same dimensions, at 53. 9d. per cubic foot? Ans. £39 128. 5d. 4. The circumference of the base of a stone column measures 6 feet 10 inches, the circumference at the top 3 feet 8 inches, and the slant height 9 feet 3 inches; re— quired the expense of cleansing or polishing the convex surfaces of 4 such columns, at ls. 9d. per square foot. ANS. 5816 198. 117131. 5. The threshold of a door measures 4 feet 6 inches in length, ll inches in breadth, and 5 inches in thickness; the head is 4 feet ll inches long, 10 inches broad, and 7 inches thick; each jamb is 6 feet 5 inches in height, 10 inches broad, and 7 inches thick; how many cubic feet of stone do they all contain ? Ans. 10ft. 4 in. 213a,. 6. A window-sole measures 5 feet 10 inches in length, and 15 inches in girt; the head is 5 feet 8 inches long, and girts 11 inches; the height of each jamb is 6 feet 4: inches, and the girt, including 3 inches for the notch, 1 foot 2 inches ; what will be the expense of tooling the soles, heads, and jambs of 8 such windows, at 7d. per square foot? Ans. £6 Ts. 23:617. 7. The mantel-tree of a fireplace measures 6 feet 3 inches in length, 9 inches in breadth, and 6 inches in thickness; the height of each jamb is 4 feet 9 inches, the breadth 9 inches, and the thickness 6 inches; the length of the slab is 6 feet 3 inches, and its breadth 2 feet (5 inches; the two coves are each 4 feet 9 inches high, and 9 inches broad ; the chimney—piece measures 6 feet 9 inches by 6 inches; what did the marble cost, the jainbs and mantel being at Ill. 105. per cubic foot, and the coves, slab, and Chimney-piece at 125. 6d. per square foot .9 Ans. £42 183. 1%d. 8. Farnlcy Chapel, in the parish of Leeds, measures 66 feet 8 inches in length, and 32 feet 7 inches in breadth, on the outside; its perpendicular height from the foundation to the ridge is 32 feet, and from the foundation to the caves, 22 feet 6 inches; and the thickness of the wall is 2 feet; how many rods of materials are contained PART v. A DESCRIPTION, ETC. 18} in the building; taking its true compass, and deducting for 7 arched windows, whose heights, from the sole to the crown or middle of the arch are 11 feet 831,— inches, from the sole to the spring of the arch, or where it begins to turn, 9 feet 4 inches, and breadths 4 feet 9 inches; and for an arched door, whose height, from the threshold to the crown of the arch, is 9 feet 4% inches, from the thresh- old to the spring of the arch, 7 feet, and breadth 4 feet 9 inches ? A728. 66 roads, 23 feet. 9. A cloth-mill at Armley, in the parish of Leeds, mea- sures 192 feet in length, and 32 in breadth, on the out- side ; its perpendicular height, from the foundation to the caves, is 35 feet, the height of the gable-end 8 feet, and the thickness of the wall 2 feet: how many roods does the. building contain; taking the true compass, and making deductions for 2 doors, one of which measures 9 feet 6 inches by 6 feet, and the other 9 feet 6 inches by 5 feet ; for 69 windows, whose dimensions are 7 feet 6 inches by 4 feet, and 78 windows, each of which is 7 feet in height, and 4 feet in breadth ? Ans. 179 mods, 2033 feet A DESCRIPTION OF PORTLAND, YORKSHIRE, AND PURBECK STONE; AND ALSO OF VARIOUS KINDS OF MARBLE. THE learned and ingenious Mr. Robert Boyle observes, that a com- petent knowledge of the stones used in building, is of the greatest importance ; one stone dug out of a quarry being found to moulder away in a few winters, while another will brave the weather for many ages. The same author adds, that although some stones will decay in a few years, others will not have attained their full hardness in half a century. FREE-STONE is very much used in building. It is of a sandy nature, and may be cut freely in any direction; hence it derives its name. It is generally of reddish, yellowish, whitish, or greyish colour; and. is sometimes mixed with small particles of mica, or vestiges of shells. There are many excellent quarries of free-stone in England. Those in the peninsula of Portland, in Dorsetshire, and in the vicinity of Leeds, in Yorkshire, are the most celebrated. 182 CARPENTERS, AND PART v. PORTLAND-STONE is of a soft nature, when dug out of the quarry; but in process of time becomes harder. Some of the finest structures in London are built with this stone, as the piers and arches of West- minster Bridge, the magnificent cathedral of St. Paul’s, &c. &c. YoRKstE-STONE is harder when dug out of the quarry, than Port- land; and it becomes still more hard by being exposed to the weather. Large quantities of this stone are sent to different parts of England. It is noted not only for its use in building, but also for its durability when laid under water; hence it is much used for bridges, docks, 8w. Kirkstall Abbey, situated about three miles from Leeds, is built of this stone; and most of the stones of which the outer walls are com— posed, are not only perfectly sound, but extremely hard. A few are in a state of decay ; which prove the justness of Mr. Boyle’s observ- ations before quoted. It is 667 years since this Abbey was built. Part of it was un- roofed, at the dissolution of the house, in the reign of Henry VIII. ; and the whole is now completely in ruins. PURBECK—STONE is dug out of quarries in the isle of Purbeck, in Dorsetshire; and is of a much harder nature than either Portland or Yorkshire-stone. There are numerous sorts of this stone, the finest of which take a good polish; and are used for chimney-pieces, hearths, grave-stones, &c. The coarser kinds make excellent paving-stones. MARBLE is a fine valuable stone, extremely hard and compact; and capable of receiving a very beautiful polish. It is chiefly used in works of ornament, as columns, statues, altars, tombs, chimney-pieces, &c. &c. Of this stone there are many varieties, as white, black, purple, 8w. English white marble is generally veined with red. Derbyshire marble is variously clouded and diversified with brown, red, and yel- low. That of Devonshire is either black with white veins, or red variegated with grey and orange. iMarbZe of Auzsergne, in France, is of a pale red, mingled with violet, green, and yellow. Various other kinds are denominated by the places from which they are brought, as Liege and Namur, in the Netherlands; Languedoc. in France; Savoy, Sicily, Spain, &c. 8:0. CARPENTERS’ AND JOINERS’ WORK. To this branch belongs all the wood-work of a house; viz. flooring, partitioning, roofing, wainscotting, &c. &c. Flooring, partitioning, and roofing are generally com~ puted by the square of 100 feet. In some places, how- ever, naked flooring and roofing are measured by the cubic foot for materials, and by the square of 100 feet, for workmanship. \Vainscotting, doors, window-shutters, &c. are com- monly measured by the square foot; enriched mouldings PART v. JOINERS’ WORK. 183 and several other articles, by the lineal foot; and some things are done at so much per piece. Note 1. In measuring naked flooring, take the length of the room added to the bearing of thejoz'sts, or what they are let into the wall, at each end, for one dimen- sion; and the breadth of the room added to the bearing of the girders, for the other dimension. It'there be no girders, the length of the room must be taken for one dimension; and its breadth added to the bearing of the joists for the other. The girders andjoz'sts of floors intended to bear great weights, ought to be let into the wall, at each end, g of the wall’s thickness; but in common flooring, the girtlilers have seldom more than 9 or 10 inches bearing, and the joists about 5 or 6 inc es. For boarded flooring. take the length and breadth of the room; and deductions must always be made for hearths and well-holes. In naked flooring, no deductions are made for hearths, in consequence of the additional trouble and waste of materials ; but well-holes must always be de- noted. If the materials of the naked flooring be charged by the cubic foot, multiply the solidity of one joist by the. number of joists of the same dimensions; and when the girders are all of the same size, proceed with them in a similar manner; if not, find the solidity of each separately. 2. Partitions are measured from wall to wall for one dimension, and from floor to floor, or as far as they extend, for the other dimension; and deductions must be made for doors and windows, except the agreement includes them. Strong partitions, made with framed timber, are generally measured in the same manner as flooring. Weather-boarding is sometimes computed by the square of 100 feet, the same as partitioning ; sometimes by the Square foot, the same as wainscotting; and some- times it is measured by the lineal foot. 3. In roofing, the length of the house in the inside, added to g of the thickness of each gable, is to be taken for the length; because the purlz'ns ought to be let into the wall, at each end, 3,— of its thickness; and the breadth or girt will be equal to twice the distance measured from the ridge, down the principal rafter, over the end of the tic-beam, to the wall. If the roof be covered with tiles or slates before the Joiners’ Work be measured off, which is very often the case, you must endeavour to ascertain the true girt as nearly as possible. In general twice the distance between the top of the ridge and the extremity of the eaves, will be rather too much ; this, however, depends very much upon the depth of the tie-beam. In measuring roofing, whether for workmanship or materials, no deductions ought to be made for the holes of chimney-shafts, sky-lights, or luthern-lights, on account of their additional trouble, and the waste of materials. When angles, running from the ridge to the eaves, are formed in a roof, that angle which bends outward is called a hip, and that which bends inward a valley,- and in roofing, it is customary to allow the Joiner some additional measurement, in consideration of the waste of timber, and extra trouble. In most places the length of the hip or valley is multiplied by 2feet, and the product added to the content of the roof. The sides of the hipped roof of a rectangular building are trapezoids, and the ends triangles; hence, it is evident that if the contents of these figures be found separately. their sum will be the content of the whole roof. Or, if the length of one side and the breadth of one end, taken half way between the ridge and the eaves, be multiplied by the girt of the roof, the product will be the content. When the solidity of the tie-beams, rafters, king-posts, purlins, &c. in the roof of a building, are required; they must be found in the same manner as directed for girders and joists, in naked flooring. 4. The pitches of roofs are frequently made according to the fancy of the de- signer. Formerly they were made much higher than Architects, in general, make them at present ; some regard, however, ought always to be paid to the covering they are to receive. When the length of the rafters is g of the breadth of the building, within the walls, the roof is said to be of a true pitch. In this case, the angle formed at the ridge, by the rafters, is 830 37' ; and the perpendicular height of the gable 3 of the breadth of the building, in the inside. Some Architects use this pitch when the coveringis pantiles; and others take no more than half the breadth of the building. When the covering is slate, some persons take % of the breadth of the building for the height of the gable; and others consider this pitch too low, and take § of the breadth. When the height of the gable is equal to 3» the breadth of the building, the angle 184 CARPENTERS’ AND PART v. at the ridge is 90°; when it is equal to §, the angle is 112° 37’ ; and when equal to i, the angle is 1260 52’. 5. Waz‘nscotiing is measured by taking the compass of the room, as it is upon the floor, for the length, and the height from the floor to the ceiling, or as far as the wainscot extends, for the breadth ; and in doing this, you must gird over the swelling panels, 85c. ; making the tape ply close into all the mouldings. Cornz'ces are sometimes done by the lineal foot, and sometimes by the square foot ; they, however, must always be measured separately from the wainscotting. Chimneys, window-seats, check-boards, sofiits, casings, architruves. Etc. must be measured by themselves; and deductions ought always to be made for doors, win- dows. fire-places, and other openings. 6. In tileasuring a door, it is usual to take its length, measured along the straight edge, and once its thickness for the length; and its breadth, pressing the tape into all the mouldings of the panels, and once its thickness for the breadth. If a door he panelled on both sides, it is customary, in some places, to allow double measure for workmanship; but if one side only be panelled, the area and half the area are taken for workmanship. It is much preferable, however, to take the real measurement, and regulate the price accordingto the thickness of the door, the number of the panels, &c. This is the method followed in London, and ought to be adopted in every other place. (See Crosby‘s builder’s Price Book, by Mr. John Phillips, Surveyor, age 178.) 1'3 For the archz'trm'e of a door, measure round the outermost edge for the length ; and take the girt of its front and two edges. for the. breadth. xlrcln'travcs are sometimes executed by the lineal foot. 7. Windows are measured by taking the distance between the under side of the sill and the upper side of the top-rail, for the height; and the distance between out- side and outside of the jambs, for the breadth. Sometimes windows are made at so much per piece. Window-shutters and the archz‘travcs of windows are measured in the same manner as doors, and their architraves. 8. In measuring stair-cases, make the tape ply close over the steps, from the top to the bottom, for the length ; and take the length of a step for the breadth. ()r, if the steps be all of the same size, multiply the area of the riser and tread by the number of steps. The ends and landings must be taken by them selves. T/zc strings are measured either by the lineal or cubic foot ; the string boards by the square foot; and the brackets are generally charged at so much per piece. For l/It.’ balustradc, take the whole length ol'the hand-rail for the length ; and the height of the baluster and hand-rail, upon the landing, for the breadth. Formerly there was much more carved work in balusters than there is at present, and it was then customary to allow double. measure ; but now the true measure is taken, and the price is proportioned according to the workmanship. Hand-rails are sometimes executed by the lineal foot, and the balusters charged at so much a piece; and in some places both hand—rails and balusters are computed by lineal measure. In this case they must be taken separately. 9. Cdntrz‘mzfor vaults is measured by taking its length for one dimension, and girting over the arch for the other ; but in groin-centrmg, it is customary, in some, places, to allow double measure, on account of the additional trouble. It is a more proper method, however, to proportion the price to the workmanship. (See Crosby‘s Builder‘s Price Book, page 165.) 10. The following articles are generally done by lineal measure. ; viz. boring to windows, skirting boards, beads, fillets, stops, Clippings, uslragals, and water trunks. Examples in Flooring. 1. The length of a floor is 45 feet 7 inches, and the bearing of the joists is 6 inches at each end; its breadth is 22 feet 5 inches, and the bearing of the girders is 9 inches at each end; required the number of squares of naked and boarded flooring. PART v. Jomens’ WORK. 185 By Cross jlfzcltiplication. NAKED FLOORING. feet. inches. feet. inches. 22 5 45 7 1%? J___9 23 ll breadth. 4-6 7 length. 2;: :23 ll breadth. l 51 5 92 0 42 i 8 5” l ,00)l 1,14 1 5 Ans. ll sq—zarcs mm4feet. BOARDED FLOORING. feet. inches. 45 7 length. 22 5 breadth. T62 10 90 o 18 11 11” L00)ni21 9 11 Ans. 10 squares and 21 feet. 2. In a common naked floor, there are 11 girders, whose lengths are 20 feet 9 inches, and scantlings (viz. breadths and depths), 1 foot by 7 inches; 144 joists, each 8 feet 6 inches in length, and their scantlings 5} inches by 4 inches; required the solidity of the whole. Ans. 320ft. l in. Qpa. 3. In a. naked floor, the girder is 1 foot 2 inches deep. 1 foot broad, and 25 feet long; there are 10 bridging joists, whose scantlings are 5 inches by 3 inches, and lengths 24 feet; 10 binding joists, whose scentlings are 10 inches by 5% inches, and lengths 9 feet 6 inches. The ceiling joists are 24 in number, each 5 feet 9 inches in length, and their scantlings 3 inches by 21,— inches ; What is the value of the whole, at 43. 9d. per cubic foot ? Ans. £23. 33. 91-61. 4. A house of two stories, besides the ground floor, measures 45 feet 9 inches in length, and 21 feet 6 inches in breadth. There are 6 hem-tbs whose dimensions are, 186 CARPENTERS, AND PART v. two of 6 feet by 4 feet 6 inches ; two of 6 feet by 4 feet 3 inches ; and two of 5 feet 9 inches by 4 feet. The well- hole measures 12 feet 8 inches by 4 feet 3 inches; re- quired the expense of the boarded flooring, at 61. 5s. per square. ' Ans. £ 168. SS. Examples in Partitioning. 1. The length of a partition between two rooms is 25 feet 6 inches, and its height 8 feet 10 inches ; how many squares does it contain? Ans. 2 squares, 25 feet. 2. A partition measures 36 feet 4 inches in length, and 12 feet 3 inches in height; what did it cost, at 61. 15s per square? Ans. £30 Os. 9d Examples in Roofing. 1. How many squares are in the roof of a building whose length is 86 feet 10 inches, and girt 28 feet 6 inches? ‘ Ans. 24 squares, 74:} feet. 2. The length of a house, within the walls, is 32 feet 3 inches, and the purlins have 9 inches bearing at each end; the girt of the roof is 23 feet 4 inches; how many squares does it contain ? Ans. 7 squares, 87:}‘fieet. 3. The length of a hipped roof, at the eaves, is 46 feet 8 inches, and the breadth of the end 25 feet; the length of the ridge is 21 feet 8 inches, and its distance from the eaves, 18 feet 9 inches; each of the 4 hips measures 22 feet 6 inches in length; how many squares are contained in the roof, allowing 2 feet in breadth, for the hips ? Ans. 19 squares, 30 feet. 4. Let the following figure represent a truss for the roof of a building; the tie-beam AB, is 28 feet long, 7 inches broad, and 12 inches deep; the principal rafters AC, BC, measure 16 feet 9 inches in length, 6 inches in breadth, and 10 inches in depth; the king-post CD, is 9 feet 3 inches in height, and its seantling at the bottom, 1 foot 4 inches by 6 inches; the braces DE, DF, are 5 feet 3 inches long, and their scantlings 6 inches by 4 inches; the punchins GH, KL, are 3 feet 3 inches in height, and their seantlings 6 inches by 4 inches; how many cubic feet are contained in the whole, deducting the two pieces cut out of the king-post, whose lengths are 6 feet, and breadths 37} inches? Ans. 37ft. 6 in. 6”. PART v JOINERs’ WORK. 187 NIH- u l‘ l “WWW. ‘ , Note. If the pieces, sawn out of the king-post, in order to give the ends of the braces what Joiners calla “ Square butment," be 2% inches or more, in breadth, and exceed 2 feet in length, they must be deducted from the solidity of the king- post ; but if they be of smaller dimensions than these, no deduction must be made, as the pieces cut out are considered to be of little or no value. in measuring these pieces, the shortest lengths ought always to be taken ; because the ends, in general, must be cut square, to render the pieces fit for use. Examples in Wainscotting, <30. 1. A room of wainscot, being girt downwards over the mouldings, is 12 feet 10 inches in height, and 98 feet 6 inches in compass, as upon the floor; what did it cost, at ls. 3d. per square foot? Ans. £79 Os. lid. 2. A door measures 7 feet 2%, inches by 3 feet 10 inches; there are two architraves, one on each side of the door, whose lengths are 19 feet 7% inches, and breadths 9% inches; the lining-boards or casings round the door—way measure 17 feet 7 inches in length, and 11 inches in breadth. Now there are five doors with architraves and linings of the above dimensions, in the Court-House, in Leeds; required the content of these doors, in square feet, and also the contents of their architraves and linings. Ans. The contents qfthe doors are 138ft. l in. 11” ; the architmves, 155 ft. 4 in. 4” 6’” ,- and the linings 80 feet, 7 inches, 1”. 3. The pulpit of a church measures 12 feet 6 inches by 4 feet 10 inches; the minister’s reading-desk 12 feet 4 inches by 4 feet 3 inches; and the clerk’s reading-desk 10 feet 8 inches by 3 feet 6 inches. The divisions between the pews are 72 in number, each of which is 10 feet 6 inches in length, and 3 feet 8 inches in height; the doors and ends, taken together, measure 332 feet 3 inches in length, and 3 feet 8 inches in height ; what did the whole cost, at ls. 6d. per square foot? Ans. £310 10s. 7%,d. 4. A room of wainscot is 135 feet 6 inches in compass, and 14 feet 10 inches in height; the cornice girts 10 inches, and its length is equal to the compass of the room; 188 SLATERS’ AND TILERS’ WORK. PART v. the door is mahogany, and measures 7 feet 4 inches by 3 feet 9 inches; the length of the surrounding architrave is 20 feet 3 inches, and its girts 9 inches; the :asings round the door-way measure 17 feet 11 inches in length, and 1 foot 2 inches in breadth; there are three pair of window-shutters, each of which measures 7 feet 3 inches by 4 feet 9 inches; the soffit and cheek-boards of each window, are 21 feet 9 inches in length, and 15 inches in breadth; the fire-place measures 6 feet 3 inches by 4 feet 9 inches ; what did the whole cost, the door being charged 18s. 6d. the cornice 25. 3d. the architrave 15. 9d. the window-shutters ls. 6d. and all the other articles Is. 2 ’. per square foot? Ans. £ 161 ls. 61d. SLATERS’ AND TILERS’ WVORK. SLATING and T ILING are generally computed either by the square yard, or by the square of 100 feet ; and the length of the ridge is taken for one dimension, and the girt of the roof, from caves to eaves. for the other dimen- sion. Note 1. In measuring slatz‘ng, the tape is made to ply over the eaves, and returned up the under side till it. meets the wall or eaves—board, in order to make an allow— ance for the double row of slates at the bottom ; but in tiling, the distance between the extremities of the eaves is taken for the girt of the roof. 2. It is customary. in some places, instead of returning the tape to the wall or eaves-board, to add 3 feet to the girt of the roof, viz. 18 inches for each side; and when [zips or valleys occur, either in slating or tiling, their length is generally added to the content of the roof; that is, an allowance of one foot in breadth, the whole length of the hips and valleys, is made for the waste of materials. 3. Sky-lights (Ind clzmmey—slzafls are generally deducted, if they be large 2 but no deductions ought to be made for common lutherndights, garret windows on the roof, and small chimney-shafts. EXAMPLES. 1. How many square yards of slating are there in a roof whose length is 65 feet 10 inches, and breadth or girt 29 feet 6 inches ? By Cross fllultiplication. feet. inches. 65 10 29 6 (30'9““2 130 0 32 11 mmfifi ”ii—5‘7“? Ans. 215 yards, and 7 feet. PART v. PLASTERERS’ WORK. 189 2. WVhat was the expense of covering the roof of a barn with pantiles, at 21. 33. 6d. per square; the length being 37 feet 8 inches, and the girt 25 feet 6 inches ? Ans. £20 173. 9%6]. 3. The roof a school measures 56 feet 9 inches in length, and 32 feet 4 inches in girt; what did it cost covering with Tavistock slates, at 2!. 153. per square? Ans. £50 9.9. 2—1d. 4. A roof measures 28 feet 4 inches in length, and 24: feet 6 inches in girt, exclusive of the allowance at the eaves, which is 18 inches on each side; what did it cost covering with Yorkshire slates, at 28. 9d. per square yard; deducting for two sky-lights, one of which measures 10 feet 8 inches by 6 feet 9 inches, and the ether 6 feet 10 inches by 5 feet 6 inches ? Ans. £10 43. 7d. PLASTERER 8’ WORK. PLASTERERS’ work is principally of two kinds; namely, plastering upon laths, called ceiling; and plastering upon walls or partitions made of framed timber, called render— mg. These different kinds must be measured separately, ex- cept they be done at the same price, which is not often the case. The contents of ceiling and rendering are estimated either by the square yard, or by the square of 100 feet; and if cornices do not exceed 9 inches in girt, they are rated at so much per foot, lineal measure; but all above this girt are computed by the square foot. Enriched mouldings are measured by the lineal foot. Notc 1. When Plasterers find materials, deductions must always be made for fire-places, doors, and windows,- and the returns at the tops and sides of doors and windows must be measured separately; but for workmanship only, these de- ductions are generally omitted, the plastered returns being allowed to counter- balance them. 2. Deductions are never made for corm'ces, enriched mouldings, festoons, or other ornaments; because the spaces occupied by them are always plastered previously to such ornaments being made; and the length ofa cornice, running round the top of a room, is always taken equal to the compass of the room, as upon the floor. It is likewise customary to allow so much a piece extra, for every corner above four, in the cornice. 3. In measuring plastered timber partitions, in large warehouses, &c. where the quarters and braces project from the plastering, % part of the whole area is gene . rally deducted ; because the projecting timbers are not plastered. This work is commonly called “ rendering between quarters.” 4. .leflewashing and colouring are measured in the same manner as plastering: and in timber partitions, 7} of the whole area is commonly added, for the sides of the quarters and braces. 190 PAINTERS’ WORK. PART v. EXAMPLES. 1. A ceiling measures 43 feet 10 inches in length, and 25 feet 6 inches in breadth; how many square yards does it contain? By Cross Multzplz'cation. feet. inches. 43 10 25 6 EEKTO 86 o 21 11 9)1117 9 7.5.21 1 9 Ans. 124 yards, and 1g foot. 2. The compass of a room is 138 feet 8 inches, and its height 10 feet 3 inches; what will be the expense of plastering the wall with stucco, at 9d. per square yard, workmanship only? Ans. £5 18s. 5%d. 3. A partition which measures 85 feet 10 inches in length, and 15 feet 6 inches in height, is rendered be- tween quarters on both sides, and whitewashed; what did the Whole cost, workmanship and materials; the lathing and plastering being charged ls. 6d. per yard, and the whitewashing 2%d. per yard? Ans. £21 lls. 9d. 4. A room measures 32 feet 10 inches in length, 21 feet 6 inches in breadth, and 12 feet 3 inches in height; and the girt of the cornice is 11 inches. The ceiling cost Is. 9d. per yard, the rendering 10d. per yard, and the cornice 1s. 4d. per foot; what was the expense of the whole? Ans. £ 19 13s. 3%d. PAINTER 8’ WORK. PAINTERS generally compute the contents of all large articles, such as wainscotting, doors, window-shutters, 8w. by the square yard; and every part is measured upon which the colour is laid. Cornices and enriched mouldings are estimated by the lineal foot; window-frames at so much a piece; and Window- squares at so much per dozen, according to their size. PART v. PAINTERs’ WORK. 191 Deductions must always be made for fire-places and other openings. Note 1. In measuring u'afnscottz'ng, doors, window-shutters, tg’b‘. painters always gird over the swelling panels, in taking both the length and breadth; and press the tape into all the mouldings. ‘2. Balustradcs are generally measured by taking the length of the hand-rail, for one dimension, and twice the height of the baluster, upon the landing added to the girt of the hand-rail, for the other dimension. 3. For lattice-work, double the area of one side is generally taken for the mea- surement of both sides ; and the area and half the area of one side, for pzzlisading; but no general rules can be given for these works, as they vary so much in the dis- tance of their perpendiculars and horizontal parts. 4. Painters proportion their prices to the nature of the colouring, the number of coats the work receives, &c. &c. EXAMPLES. 1. If a room be painted, Whose height is 15 feet 6 inches, and compass 98 feet 9 inches; how many yards does it contain ? By Decimals. By Cross JWaltz'plication. feet. feet. inches. 98.75 98 9 15.5 15 6 11937—5 5—07 ' 49375 98 0 9875 _ 49 6 6” 9)1530.625 Q 9)1530 7 6 ”Those Ans. 170 6“”? 6 yds. 2. A door measures 7 feet 4% inches by 3 feet 8% inches; what did 4 such doors cost painting on both sides, with 3 coats, at 10d. per yard? Ans. £1 Os. 4%d. 3. The length of the hand-rail of a staircase is 21 feet 9 inches, and the girt of the hand-rail, and twice the height of the baluster, upon the landing, 7 feet 10 inches; how many yards of painting are contained in the balustrade? Ans. 18 yards, 8 feet, 4% in. 4. The lattice Window of a dairy measures 4 feet 10 inches by 3 feet 6 inches; how many square yards of painting does it contain, taking the area of both sides ? Ans. 3 yards, 6 feet, 10 in. 5. The compass of a wainscotted room is 118 feet 6 inches, and its height, to the under side of the cornice, 12 feet 8 inches; the door measures 7 feet 2 inches by 3 feet 6 inches; the lining round the door-way is 17 feet 4 inches long, and 9 inches broad; there are four pair of window-shutters, each of which measures 6 feet 10 inches £92 GLAZIERs’ WORK. PART v. by 4 feet 6 inches; the sofi‘it and check-boards of each window, are 19 feet 6 inches in length, and H inches in breadth; the fire-place, which is to be deducted, measures 6 feet 6 inches, by 5 feet 2 inches; the door znd window- shutters are painted on both sides; required the expense of giving the whole 4 coats ; the cornice, which is 117 feet in length, being charged 8d. per lineal foot, and all the other articles ls. 10d. per square yard. Ans. £21 83. 3d. GLAZIER C’ W'ORK. GLAZIERS take their dimensions either in feet, inches. and parts, or in feet, tenths, and hundredths ; and estimate their work by the square foot. Note 1. The most general method of measuring: a window is by taking: the length and breadth, without makingr any deduction for the cross-bars between the panes ; sometimes, however. the measurement of a window is found by multiplying the area of one pane by the number of panes. 2. If windows be circular or elliptical, they must be measured as if they were squares or rectangles; the greatest lengths and breadtns always being taken. in order to make a compensation for the waste of glass in cutting the panes into proper shapes. EXAMPLES. 1. How many feet of glass artfontained in a window which measures 7 feet 10,—:1 inches )y 4 feet 7.}; inches? B}; Cross ilIultz'plication. ft. in. pa. 7 10 9 4 7 6 31 7 O 4 7 3 3’” O 311 4 6”” 36 6 2 7 6/1723. 2. A pane of plate glass measures 2 feet 8 inches by 1 foot 6 inches; what did 8 such panes cost, at 143.6(1. per square foot? Ans. £23 4.9. 3. The diameter of a circular window is 2 feet 9:} inches ; "for how many feet must the glazier be paid, taking the window as a square? Ans. 7ft. 9 in. 6% pa. 4. The base of a triangular sky-light measures 10 feet 6 inches, and the perpendicular 5 feet 8 inches; what did it cost glazing, at ls. 10d. per square foot? Ans. £2 14s. 6%(1. PART V. PLUMBERS’ WORK. 193 5. There is a house with three tiers of windows, four in a tier ; the height of the first tier is 7 feet 2 inches, of the second 6 feet 4 inches, and of the third 5 feet 8 inches; and the breadth of each Window is 4 feet 6 inches. The height of a semi-circular window, above the door, is 2 feet ; what did the Whole cost glazing, at 28. 6d. per square foot Ans. £44 28. 6d. PLUMBER 8’ WORK. PLUMBERS’ work is generally done at so much per pound, or else by the hundred weight of 112 pounds; and the price is regulated according to the value of the lead at the time when the work is performed. Note. Sheet lead. used in roofing, guttering, &c. commonly weighs from 7 to 12 pounds per square foot; and leaden pipe varies in weight, per yard, according to the diameter of its bore in inches. The first of the following Tables shows the weight ofa square foot of sheet lead. in pounds, to each of the subjoined thicknesses, in tenths and hundredths of an inch ; and the second exhibits the general weight of a yard of leaden pipe, accord- ing to the diameter of its bore. TABLE IR TABLE II. 0 OJ PFC" LT" i U a) '2 {A '5‘ E 4.1 $95 534.1%??? 5.32 $53 :44: E 8 :4 “ S % ,5 8 A 2 m2: :4 V‘ B '5 m3 " 5 55‘ .3" TI, 5.899 3 .15 8.848 4} 10 .11 6.489 ! .16 9.438 1 12 4 6.554 4 9.831 14 16 .12 7.078 1 .17 10.028 14 18 4 7.373 1 .18 10.618 14;. 21 .13 7.668 1 .19 11.207 2 24 .14 8.258 ; 4 11.797 —— 4 8.427 g .21 12.387 EXAMPLES. 1. A sheet of lead measures 18 feet 10 inches in length, and 5 feet 6 inches in breadth; required its weight, at 9; lb. to a square foot. K 194 ravnns’ WORK. PART v By Cross fifulaplication. Ft. In. 18 10 5 6 94: 2 9 5 103*7 content. As 1ft. 1 911'). 8 012: l03ft. 7 in. 1 8 cwz‘. 3 ya. 415. 0 oz. 10 637:, the weigfit required. 2. What is the weight of a sheet of lead, Whose length is 15 feet 10 inches, breadth 4 feet 6 inches, and thickness i- or .2 of an inch? Ans. 84053625 pounds. 3. If I buy 150 yards of leaden pipe, Whose bore is 1%; inch; What Will it cost me at BZ-d. per pound, admitting each yard to weigh 21 pounds? Ans. £49 48. ALE—d. 4. What cost the covering and guttering of the roof of a church with lead, at £1 188. per cwt. ; the length of the roof being 82 feet 9 inches, and its girt 65 feet 3 inches; the length of the guttering 165 feet 6 inches, and its breadth 1 foot 9 inches; admitting the thickness of the lead to be 52 of an inch? A728. £948 153. llgd. PAVERS’ WORK. PAVERS estimate their work by the square yard; and such dimensions must always be taken as will give the true area. _ EXAMPLES. l. A rectangle measures 85 feet 9 inches in length, and 43 feet 6 inches in breadth; how many square yards of paving does it contain? By Decimals. By Cross Multiplication. Feet Feet. In. 85.75 length. 85 9 length. “43.5 breadth. . §3_-_6 breadth. 42875 287 3 25725 340 0 -diEiiliL 42 {L97 9)37eo.125 9)3730‘ 1 e i£4583 3453., {he Ans. ii—léfld8_:1fl./lm. PART V. VAULTED AND ARCHED ROOFS. 190 2. The base of a triangle measures 76 feet 8 inches, and the perpendicular 42 feet 3 inches ; what did it cost paving with Aberdeen granite, at 10s. 6d. per square yard? Ans. £94 9s. 6d. 3. If the parallel sides of a trapezoid be 68 feet 7 inches, and 45 feet 3 inches, and their perpendicular distance 98 feet 6 inches; what will it cost paving with Guernsey pebbles, at 63. 6d. per square yard ? Ans. £202 8.9. 11 id. 4. How many yards of paving does the trapezium con- tain, whose diagonal measures 136 feet 8 inches, and per- pendicular 68 feet 2 inches, and 56 feet 4 inches? Ans. 945 yds. 21ft. 5. The length of a street is 538 feet 6 inches, and its breadth 65 feet 8 inches; what did it cost paving with Purbcck-stone, at 53. 6d. per square yard? . Ans. £1080 93. 9%,al. 6. A rectangular court—yard measures 96 feet 9 inches in length, and 74 feet 6 inches in breadth. Across the middle and round the extremities of the yard, is a foot- way, 5 feet 3 inches broad; and paved with Guernsey granite, at 93. 6d. per square yard. The rest is paved with Jersey pebbles, at 53. 9d. per square yard; required the expense of the whole. Ans. £272 8.9. 32d. VAULTED AND ARCHED ROOFS. ARCHED Boo/3‘ are either vaults, domes, saloons, or grains. Vaulted roofs are formed by arches springing from the opposite walls, and meeting in a line at the top. Domes are made by arches springing from a circular or polygonal base, and meeting in a point at the top. Saloons are formed by arches connecting the side walls to a flat roof, or ceiling, in the middle. Grains are formed by the intersection of vaults with each other. Vaulted roofs are commonly of the three following sorts : 1. Circular roofs, are those whose arch ‘is some part of the circumference of a circle. 2. Elliptical or oval roofs, or those whose arch is an oval, or some part of the circumference of an ellipsis. 3. Got/tic roofs, or those which are formed by two cir- K 2 196 VAULTED AND ARCHED ROOFS. PART V. cular arcs, struck from different centres, and meeting in a point over the middle of the breadth or span of the arch. Note. Domes and saloons are of various figures; they, however, seldom occur in the practice of measuring; but most cellars are covered either with vaults or groins. PROBLEM I. To fiml the content of the vacuity of a circular, an elliptic, or a Gothic vaulted roof. RULE. Multiply the area of one end by the length of the roof or unit, and the product will be the content required. Note 1. If the arch be the segment ofa circle, the area of the end may be found by Problem 17, Part II. ; ifit be elliptical, multiply the span by the height, and the product by.7854, for the area of the end ; but if it be a Gothic arch, the area ot‘the end must be obtained by finding the areas of the two circular segments and the triangle of which the end is composed. 2. The upper sides of all arches, whether vaults or groins, are built up solid, above the haunches. to the same height as the crown of the arch. 3. The solidity of the materials in any arched roof, may be found thus: find the content of the whole, considered as solid, from the spring of the arch to the upper side of the crown; find also the contents of the vacuity; then the difference of these two contents will be the solidity required. 4. The whole arch, considered as a solid, will be a parallelopipedon, the content of which may be found by Problem ‘2, Part IV. EXAMPLES. 1. Required the content of the vacuity of a semi-circular vault, the span 01' diameter of which is 20 feet, and its length 60 feet. .7854 400:.the square of 20. 157.08Vthe area of the end. 60 the length. 9424.80 Ans. 2. The span of an elliptical vault is 30 feet, its height 10 feet, and its length 50 feet ; What is the content of the vacuity? Ans. 11781 feet. 8. Required the content of the vacuity of a Gothic vault, whose span is 30 feet, the chord of each arch 32 feet, the versed sine, or distance of each arch from the middle of these chords, 8 feet, and the length of the vault 35 feet 6 inches. Ans. 27755.0502fcct. PART V. VAULTED AND ARCIIED ROOFS. 197 4. Let ABCD denote the end or D H C upright section of a semi-circular ' roof. The span EF is 18 feet, the thickness of the wall AE or FB, at the spring of the arch, 3 feet, the thickness GH, at the crown of the arch, 2 feet, and the length of the vault 56 feet 9 inches ; how many solid feet are contained in the roof? Ans. 7761.425 feet. PROBLEM II. To find the concave or convex surface of a circular, an elliptic, or a Got/tic vaulted roof. RULE. Multiply the length of the arch by the length of the vault, and the product will be the superficies required. Note. The convex length of an arch may be easily found by making a line ply close over it; but for the concave length, this method is not quite so applicable; for if care be not taken, the dimension will be made too short. If the arch be the segment of a circle, its true length may be found by Problem 14, Part II. EX MIPLE S. l. The span of a semi-circular vault is 30 feet, and its length 40 feet; What is its concave surface? 3.1416 30 2)94.2480 “477.124_= length of the arch. 40 Ans. £64960 square feet. 2. The length of a vault is 62 feet 9 inches, and that of the arch 54 feet 6 inches; howr many square yards are contained in the roof? Ans. 379.986 yards. 3. Required the concave surface of a bridge consisting of 5 circular arches ; the span of each arch being 96 feet, the height, above the top of the piers, 36 feet, and the length 45 feet. Ans. 28955.232feet. Note. Those who desire to make themselves acquainted with the essential pro- perties, dimensions, proportions, and other relations of the various parts of a bridge, are referred to Dr. Ilutton’s Principles of Bridges. in this valuable little work, the learned Doctor proves, that the equilibrial arch, described in Problem V., is the most proper for a bridge of several arches. Next to it, the elliptical arch claims the preference; after it the cycloidal arch; and lastly the arch of a circle. K 3 198 VAULTED AND ARCHED ROOFS. PART v. As far parabolic, hyperbolic. and catanariau arches, they might never to be ad- mitted into A bridge consisting of several arches; but may, in some cases, be used for a bridge of one arch, which is to rise an unusual height. PROBLEM III. Tofind the solid content qfa dome ; its height, and the dimensions (gfz'ts base being given. RULE. Multiply the area of the base by the height, and 3 of the product will be the solidity. EXAMPLES. 1. “That is the solid content of :1, hemispheriea dome , the diameter of the base being 40 feet? .7 54: 1600 :2 < % AB : 3.163858 x 1.625.0621728, the area of the triangle ABC; K 4 200 VAULTED AND ARCHED ROOFS. PART V. consequently 5.0621728—2.296875225652978, the area of the transverse section, AECBA. Now, 2.7652978X50:l38.26489 feet, the content of the solid part, which being tahen from the whole upright space, will leave the content of the vacuity within the room. 2. What is the solid content of a saloon with a circular quadrantal arch of 2 feet radius, springing over a rectan- gular room of 20 feet long, and 16 feet broad ? Ans. 580.2065 cubicfeet. 3. A circular building of 40 feet diameter, and 25 feet high to the ceiling, is covered with a saloon, the circular quadrantal arch of which is 5 feet radius; required the capacity of the room in cubic feet? Ans. 30766.496 cubicfeet. PROBLEM VI. Tofind the superficial content of a saloon. RULE . Find its breadth by applying a. line close to it across the surface; and its length by measuring along the middle of it, quite round the room; then the product of these two dimensions will be the surface required. Note. The area of the flat ceiling must be added to the area found by the above Rule, in order to obtain the whole surface of the saloon. EXAMPLE S. l. The girt across the face of a saloon is 5 feet 3 inches, and its mean compass 94 feet 6 inches; what is the area of its surface? Here 94.5 X 5.25 :: 496.125feet, the area required. 2. The mean compass of a saloon is 126 feet 10 inches, and the girt across its face 8 feet 6 inches; what is the area of its surface? Ans. 1078 ft. 1 in. PROBLEM VII. Tofind the solid content of the vacuity formed by a groin arch, either circular or elliptical. RULE. Multiply the area of the base by the height, and the product by .904, and it will give the solidity required. PART v VAULTED AND ARCHED ROOFS. 201 Note. Groins are sometimes measured as if they were solid, in consideration of the great trouble and waste of materials in forming the arches and mtersectlons. EXAMPLES. 1. What is the content of the vacuity formed by a cir- cular groin, springing from the sides of a square base, each side of which is 14 feet? 14 Elizarea of the base. 7=height, or radius. 1372 .904 548—8 12348 @7268 = solidity required. 2. What is the solid content of the vacuity formed by an elliptical groin; the side of its square base being 24 feet 6 inches, and its height 8 feet 3 inches? - Ans. 4476.6645 feet. PROBLEM VIII. To find the concave surface of a circular or an elliptical groin. RULE . Multiply the area of the base by 1.1416, and the pro- duct Will be the superficies required. Note. In measuring works where there are many groins‘ in a range, the cylin- drical pieces between the groins, and on their sides, must be taken separately. EXAMPLE S . 1. What is the concave surface of a circular groin arch; the side of its square base being 15 feet 6 inches ? Here 15.5 x 15.5 = 240.25, area of the base; and 240.25 x 1.1416 : 274.2694 feet, the answer required. 2. The base of a groin is a rectangle whose sides are 20 and 26 feet ; required the concave surface of the arch. Ans. 593.632 feet. K 5 202 GENERAL ILLUSTRATION. mar v GENERAL ILLUSTRATION. HAVING gone through the Works of‘ Artificers,‘ and noted the methods of measuring buildings, and computingr their contents, I now proceed to give a gene ‘al illustra~ tion of the whole, by assigning the dimensions of a house, and from thence computing the contents of the works of the different Artificers employed in building it. In performing this task, are shown the methods of ruling the book, entering the dimensions, with the con- tents ; then the method of abstracting the contents ; and lastly, of forming the bills of expenses of the work and materials. The building of which I have made choice, for the general illustration, consists of two stories, beside the cellars, and of two rooms upon each floor; which will be found quite sufficient to exemplify the methods of mea— suring the works of Artificers. The whole length of the building, on the outside, is 53 feet 6 inches, and its breadth 24 feet. The valls of the cellars are 6 half bricks, or 24 inches in thickness ; those of the lower story 20 inches; and those of" the upper story 16 inches. One of the lower rooms is considered to be the kitchen, and the other the parlour; and they are frequently dis- tinguished by these denominations, in the following notes. The plans of the different stories, and the elevation, could not be given without a folding plate ; but it is pre- sumed that the reader will find no difficulty in compre- hending the dimensions without them. Note. The columns of numbers,in the following forms, are sufficientlyexplained by the titles at the tops of them; excepting the figures 2. 3, &c. in the first column; which figures signify that there are more than one article of the same dimensions; consequently the contents arising from the dimensions to which these figures are prefixed, must be multiplied by 2, 3, &c.; and the products entered in the column of contents. THE BRICKLAYER’S IVORK. K 6 420 O lIiddle walls of ditto. Half Dimensions. Bricks Contents. Titles. thick. 4 Ft. In. Ft. In. “‘7 0 6 1543 6 The cellar walls. 10 6 l' 907 PART v. GENERAL ILLUSTRATION. 203 Hal‘ Dimensions. Bricks Contents. Titles. thick. Ft. In. Ft. In. 20 0 , Common arches over the two 2 -39 0* o 760 0 cellars. 2T) 0 , . 10 O o 200 0 Ditto under the passage. 1L%Z*2— 5 1911 0 Outer walls of the ground w story. 1%; (6) 4 1837 6 Ditto of the upper story. i224 (T 3 6 9 162 0 Gable-ends of the outer walls. :2] 0— . ~ 2 25 6 3 10! 1 0 Partition walls. Til—‘5” 6 9 3 141 9 Gable-ends of ditto. 2 —T3 0 5 247 O Chimney shafts of the ground _ 9j_ story. 2 12 6 . 8 2 4 204 2 Ditto of the upper story. 2 g g 3 90 0 Ditto of the gable-ends. Gmfoa The deductions are as fallen.” viz. 2 3 6 6 47 10 Cellar doors. 5 g 5 44 0 Front door. 2 g 5 35 0 Back door. 2 g g 5 65 0 Windows of the lower rooms. ”*5? . 3 4 6 4 76 6 Ditto of the upper rooms. :3 18 4 34 2 Staircase window. 6 8 ,, . .- . 4 3 6 o 93 4 Doors 1n partition walls. AK6 204 GENERAL ILLUSTRATION. PART v. In order to abstract the foregoing contents ; that is, to collect them into one sum, make the deductions, and re- duce the neat contents to the standard thickness of one brick and a half, proceed thus : Make only two columns for the whole contents, and two for the deductions of the same thickness; viz. one column for the contents that are one brick in thickness, and the other for the contents that are one brick and a half in thickness ; and dispose of the superior denominations in one or both of these columns, by entering them more than once; thus, the contents that are 2 bricks in thickness must be set down twice in the one-brick column ; those that are 2—;— bricks in thickness, once in the one-brick column, and once in the brick-and- half column ; and those contents whose thickness is 3 bricks, must be entered twice in the brick-and-half column; then add up all the columns, and reduce the sums in the one brick columns to the standard thickness of one brick and a half, which add to the respective sums in the brick-and—half columns. Lastly, take one of the sums thus obtained, from the other, and the remainder will be the whole reduced content of the brick-work. Abstract oft/1e Brick- Wbrk. ! Contents. Contents. Deductions. Deductions. i 112 brick thick. 1 brick thick. 1% brick thick. 1 brick thick. Ft. In. Ft. In. Ft. In. Ft. In. 1543 6 760 0 47 10 44 0 1543 6 200 0 47 10 35 O 420 0 1911 0 44 0 65 0 420 0 1837 6 35 0 76 6 760 0 1837 6 65 0 76 6 200 0 247 0 93 4 34 2 1911 0 204 2 333 0 34 2 162 O 204 2 243 6 “3&7 1071 0 7201 4 576 6 2 141 9 27‘, 3)730 8 247 0 W8 ~243 6 90 (L Wit? 8509 9 m 4800 10 13310 7 576 6 12734 1 PART v GENERAL ILLUSTRATION. 205 Then 12734 feet being divided by 272, we obtain 46 rods 222 feet of brick-and-half wall, for workmanship and materials. Next find the contents of the corners or coins of the different stories of the house, to which add the deductions for doors and windows; and you will obtain the quantity to be charged for workmanship only. Cellar walls. Walls of the ground story. Feet. In. Feet. In. 2 0 thickness. 1 8 thickness. 4 multiply. 4 multiply. 8-71 breadth. 6’8 breadth. 10 6 height. 13 0 height. 80 0 78—7) 4 0 8 8 84-73 content. 86 8 content. 6 half bricks thick. 5 half bricks thick. 3)5o4 o 3)433 4 168 0 reduced content. 144 5 reduced content. Walls of the upper story. Feet. In. Feet. In. 1 4 thickness. 66 8 content. 4 multiply. 4 multiply. - 5“? breadth. 3)266 8 1a height. 88 10 reduced content. 64 O 168 0 ditto. 2 8 144 5 ditto. 66 8 content. 401 3 total of the coins. _“‘ 576 6 deductions. 272)977 9(3 rods 161 feet for 816 workmanship only. 161 rem. THE BRICKLAYER’S BILL. Rods. Feet. .58 s. d. 46 222 of brick-and-half wall, workmanship and materials, at per rod . -—- —- —— 3 161 of ditto, for workmanship only, at per rod . . . . . . . .—..——— £____ 206 [0 l0 l3 Dimensions. Ft. In. 155 0 2 5: w53 6" _.2_9__ 14 6’“ “3‘0 5 6 2 O _5,_6~ 1 3 _6fi__8.,, 1 6 i 3; i5_h6_ 1 8 ~§HOH _2,_é- h_ 72767 * 3 6 “10*? “if; —5T 1 8‘ 5 (T l 4 big—Z” 1 3 _}_§, _7IT GENERAL ILLUSTRATION. PART V. THE MASON’S WORK. Contents. Titles. Ft. In. M 348 9 Stone base or plinth. 40 1 Faeia to the front of the house. 29 O J ambs of the front door. 11 0 Head of ditto. 6 10 Frize. 10 0 Cornice. 15 0 Threshold. 36 8 Steps to the front door. 37 4: Ends of ditto. 26 3 Landing of ditto. 32 0 Front Window jambs. 18 4 Soles of ditto. 13 4 Heads of ditto. 35 0 Upper window jumbo. 22 6 Soles of ditto. 13 6 Heads of ditto. PART V. GENERAL“ ILLUSTRATION. 207 Dimensions. Contents. Titles. / _ Ft. In. Ft. In. _. 18 0 30 o B k d v 1 1 8 ac 001 Jambs and lead. 5 6‘ T1 ‘ 1d 1’ ' 1 6 8 3 LI‘thO o ditto. 1?? 14 7 St . . . 1 3 aircase W1ndow Jambs. 5‘6‘ . 1 4* 7 4 Sole of ditto. '15 (2) 5 10 Head of ditto. “8‘6; Jambs to the fire-places in the 2 1 8 _ 28 4 lower rooms. 2 f g 16 6 Mantels to,ditto. 2 3 g 25 8 Slabs or hearths to ditto. 4 If i 22 8 Coves to ditto. 4 “f g 18 4 Back cover to ditto. 2 “5"0‘ 11 8 Chimney-pieces or cornices to 1 2 ditto. 2 “SWCT 21 4 Jambs to the fire-places in the 1 4 upper rooms. 2 All 2 12 O Mantels to ditto. 2 4 0 8 O Chimney-pieces or cornices to 1 O ditto. 2 g (6) 18 o Slabs to ditto. 4 ‘11 (2) 18 8 Coves to ditto. :72),— . ,. 4 1 12 0 Back coves to ditto. 208 GENERAL ILLUSTRATION. PART v. ‘1 1 - | Dimensions. Contents. 3 Titles. _. Ft. In. Ft. In. ' J / ”F 14: 6 ‘ . 2 3 4 96 8 ‘ Chimney—tops of stone. 2 1‘: 12 47 6 Bases and facias to ditto. 58 0 H C ' ‘ h f f i 1 10 106 4 ormce to t e ront o the house. g 55 e B . .. 1 6 83 3 locking course to ditto. 4 g g 82 0 Tabling. 3 6 15 1 6 78 9 Steps down to the cellars. 3 6 L . . 2 6 8 9 andlng of ditto. ’20 o . . 2 18 O 720 O Tooled flagging 1n the cellars. 28 g 190 0 Ditto between the cellars. 16 6— 36 9 Well-hole of the cellar—stairs to be 3 #6” deducted. 20 8 385 9 P lished firm in in the kitchen 18 8 0 ‘og g ' ' 2g g 196 4 Ditto in the passage. 2 2 23 10 Kitchen fire-place to be deducted. ABSTRACT OF THE MASONRY. In making this abstract, all the articles of the same price must be collected together: thus, the polished work must be brought into one sum, the tooled work into another; the cornices into another, &c. &c. PART v. GENERAL ILLUSTRATION. ' 209 Polished Wbrk. Feet. In. 348 9 stone base or plinth. 40 1 facia to the front. 29 0 jambs to the front door. 11 0 ‘head of ditto. ‘ 6 10 frize. 15 0 threshold. 36 8 steps to the front door. 37 4 ends of ditto. , 26 3 landing of ditto. 32 0 front window jambs. l8 4 soles of ditto. 13 4 heads of ditto. 35 0 upper Window jambs. 22 6 soles of ditto. 13 6 heads of ditto. 28 4 jambs to the fire-places in the lower rooms. 16 6 mantels to ditto. 21 4 jambs to the fire-places in the upper rooms. 12 0 mantels to ditto. 96 8 chimney tops. 47 6 bases and facias to ditto. 83 3 blocking course to the front. 82 O tabling. 1073 2 sum. 1 Tooled Work. Feet. In. 30 0 back door jambs and head. 8 3 threshold of ditto. l4 7 staircase Window jambs. 7 4 sole of ditto. 5 10 head of ditto. 78 9 steps down to the cellars. 8 9 landing of ditto 153 6 sum. Cornices. Feet. In. 10 0 cornice of front door. 11 chimney-pieces in the low rooms. 8 O ditto in the upper rooms. 106 4 cornice to the front of the house. 136 0 sum. 00 210 GENERAL ILLUSTRATION. PART v. Slabs and Caves Feet In. 25 8 slabs of the low room fire-paces. 22 8 coves of ditto. 18 4 back coves of ditto. 18 O slabs of the upper room fire-plates. 18 8 coves of ditto. 12 0 back coves of ditto. 115 4 sum. ’ holed Flagging Feet. In. 720 O flagging in the cellars. 1970 O ditto between the cellars. 910 0 sum. £36 9 deduction. 9)8j3 3 9? square yards, Fall's/zed Flagging. Feet. 11!. 385 .9 flagging in the kitchen. 23 10 deduction. SEQ—3 62 0 square yards. THE MASON’S BILL. Feet. In. £ 5. (I. 1073 2 of polished work . at per foot, —- — —- 153 6 of tooled ditto . . . at —— per foot, - — —- 136 0 of cornices . . . . at per foot, — —- — 115 4 of slabs and coves . at per foot, — -— -— 97 yards of tooled flagging . at per yard, - — ~— 62 ditto of polished ditto . . at per yard, — - - PART v. GENERAL ILLUSTRATION. 211 THE CARPENTERS AND J OINER’S WORK. Dimensions. Contents. Titles. Ft. In. Ft. In. ?g g 445 9 Naked flooring of the parlour; ,, _ viz. girders, and binding and bridging joists. fig 3 385 9 Boarded flooring of ditto. _5 E— 23 10 Hearth to be deducted from the 4 4 boarded floorino. o . 2 33 :1; 948 10 Naked flooring of the upper rooms. u ”ii":— 824 10 Boarded flooring of ditto. 9 4—6— 34 6 Hearths to be deducted from the ‘ 3 10 boarded flooring. 1,17 18 90 1 Naked flooring of the staircase. g 187 74 5 Boarded flooring of ditto. 10 if g 60 0 Deal steps to staircase, first flight. 10 8 1(7) 4 10 End of ditto. 2 8 38 O Foot-pace. g g 5 6 Face of ditto. _;——(T f 1 (1 fl' ht 10 1 6 60 0 Steps 0 tie secon 1g .. 10 8 if; 4 10 Ends of ditto. 76—7? 92 9 Balustrade of the staircase. 39 6 I 25 4 i String-board of dittO. 212 GENERAL ILLUSTRATION. PART v. Dimensions. Contents. fines. — Ft. In. Ft. In. C“? 23 4 Ceiling—joists over the upper 2 20 4 948 10 rooms. i1) g 224 0 Ceiling-joists over the staircase. 9 3; Z 1441 10 Roofing. 1 g 35 0 Front door, 6 panelled. 1‘3) g l 14 4 Casings t‘ ditto. 1‘ Gil "0 5 Arehitrave “o ditto o 10 I ‘ ‘ ' 6 9 3O 4 Back door, A; panelled. 4 6 . 19 O 11 l Casing to ditto. 0 7 g g ‘ 21 0 Cellar door, 4 panelled. ' l g g 20 7 Kitchen door, 6 panelled. 16 10 - ‘2 " Casinws to ditto e19”: ’ ° ‘ 18 10 i 12 6 Arohitrave to ditto. 0 8 l g g E 20 7 Parlour door, 6 panelled. O .4 18 1 i ‘2 7 Casings to parlour door. Tim . V . 9 l 410 Arelntrm e“ to mt!“ . ~ 0 8 s l 3 :2 g g 41 2 Upper room doors, €panelled. 9 16 18 l 25 3 1 Casings to ditto. PART v. GENERAL ILLUSTRATION. 213 Dimensions. Contents. . Titles. # Ft. In. Ft. In. '- 4 1?) lg 50 2 Architraves to ditto. 2 _. 5—170. 50 6 Window shutters to lower win- 4 4 dows. 2 ‘33—3 0— 46 O Casings to ditto, viz. cheek-boards, 1 0 window-boards, and soffits. 2(7) (7) 15 9 Architrave to parlour window. 2‘? (8) 23 8 Casings to staircase window. 2;; g 16 1 Architrave to ditto. 3 2(1) 3 61 0 Casings to upper windows. 3 2g 3 42 7 Architraves to ditto. 2 Z 3 47 2 Sashes of the lower windows. 2 Z 26 8 Ditto of the staircase window. 3 :3, 18 57 6 Ditto of the upper windows. ABSTRACT OF THE CARPENTER’S AND JOINER’S VVO’RK. This abstract must be made in the same manner as that of the Masonry ; viz. by collecting ‘all the articles of the same price into one sum, making the proper deductions, &c. &c. Naked Flooring. Feet. In. 445 9 parlour floor. 948 10 upper floors. 90 l staircase. 1,00)14,84 8 sum. l4squares, 84 feet, 8 inches. 214 GENERAL ILLUSTRATION. PART V. Bearded Flooring. Feet. In. 385 9 parlour floor. 824 10 upper floors. 74 5 staircase. 1285 0 sum. 58 4: hearths to be deducted. hmflgfifé 12 squares. 26 feet, 8 inches. Doors, 4 panelled. Feet. In. 30 4 ba cl; door. 21 0 cellar door. 51 4 sum. Ceiling Joists. Feet. In. 948 10 upper rooms. 224 O staircase. 1,00)ll,72 10 sum. 11 squares, 72 feet, 10 inches. iS'tairs Steps. Doors, 6 panelled. Feet. In. Feet. In. 60 0 first flight 35 0 front door. 4 10 ends of ditto. 20 7 kitchen door. 38 0 foot—pace. 20 7 parlour door. 5 6 face of ditto. 41 2 upper room door:. 60 0 second flight. 1717777*1 sum. 43: 10 ends of ditto. 4- 2‘6 4 string-board. l99 77 6 sum. Door and W'imlow Casings. Feet. In. 1-4: 4 front door. 11 1 back door. .12 7 kitchen door. 12 7 parlour door. 25 3 upper room doors. 46 0 lower rooni windows. 23 8 staircase window. 61 0 upper room windows. 206 6 sum. PART v. GENERAL ILLUSTRATION. 215 Arcflitraves. Feet. In. 20 5 front door. 12 6 kitchen door 25 1 parlour door. 50 2 upper room doors. 15 9 parlour window. 16 1 staircase window. '12 7 upper windows. 182 7 sum. Window Sa sites. Feet. In. 47 2 lower windows. 26 8 staircase Window. 57 6 upper windows. 131 _4 sum. THE CARPENTER’S AND JOINER’S BILL. Sqrs. Ft. In. :6 s. d. 14 84 8 of naked flooring, . at ——~ per square, 12 25 8 of boarded flooring, at — per square, 11 72 10 of ceiling joists, . . at — per square, 14 41 10 of roofing, . . . at -—- per square, 1 I l l l l | l I 199 6 of stair-steps, . . at —— per foot, —- — .— 92 9 of balustrade, . . at —— per foot, —— — .1 . 117 4 of doors 6 panelled, at —- per foot, -— -— -- 51 4 of doors 4 panelled, at —-— per foot, -- —- ~ 206 6 of doors and Window casings, . . . at —- per foot, — —— —- 182 7 of architraves, . . at -—- per foot, -— - ~— 131 4 of window sashes, . at —— per foot, — —- — THE SLATER’S WORK. The length of the roof, for the slating, is 52 feet 6 inches, and its girt t, allowing 18 inches at each of the eaves, is 29 feetc 10 inches , hence the content is 15661- square feet: 15 squares 661-feet, at per square . . . £_ a 216 GENERAL ILLUSTRATION. PART V- THE PLASTERER’S WORK. Dimensions. Contents. Titles. Ft. In. Ft. In. 18 8 771 6 Ceilings of the lower rooms, 20 8 3 coats. 72—8— . . 0 10 137 9 Cornices of ditto. 82 8— 82 8 Enriched mouldings in the parlour, _ llneal measure. 9 6 74 5 Ceiling in the lower part of the 7 lg staircase, 3 coats. 3: 18 28 10 Cornice of ditto. 82 8 868 0 “Tans of the parlour, hard-finish- 10 6 mg. g 2 23 4 Door of ditto to be deducted. g g 32 6 Window to be deducted. g g 28 10 Fire-place to be deducted. 82 8 868 0 \Valls of the kitchen, 2 coat; 10 6 7 g g 23 4 Door to be deducted. 2 8 35 0 Back door to be deducted. g g 32 6 Window to be deducted. f; (3): 28 10 Fire-place to be deducted. 7 6(5 1: 633 6 1Walls of the lower part of (he 10 6 staircase, hard-finishing. ()‘( 4‘ 558 1 Ditto of the upper part. 217 PART V. GENERAL ILLUSTRATION. Dimensions. Contents. Titles. _ Ft. In. Ft. In. 2 , g 44 0 Front door to be deducted. 6 8 Kitchen and parlour doors to be 2 3 6 46 8 deducted. 13 0 39 0 Space occupied by the staircase, __ 3 O to be deducted. g 18 34 2 Staircase Window to be deducted. 5 8 i 25 6 W’indow at the top of the stair- 4 6 case, to be deducted. 2 g g 46 8 Upper room doors to be deducted. L60 4* 573 2 CT ’ h ' 3 i 9 6 e1 111g over 1: e staircase, coats. 60 4— . . 0 10 50 3 Cormce of ditto. 21 4 - - , ~ ~ 2 19 4 824 10 Ceillng of the uppe1 rooms, 3 coats. 8'5 0 . . 2 0 10 141 8 Cornlces of dltto. 85 o ' fl 1 9 2 9 3 1572 6 Walls of the upper looms, .. coats. 2 g g 46 8 Doors to be deducted. 2 Z 2 51 0 Windows to be deducted. 5 0~ . t _ 2 4 6 45 0 Fire-places to be deducted. ABSTRACT OF‘ THE PLASTERING. This abstract must be made by collecting all the ceiling of 3 coats, into one sum; the hard-finishing into another, 650. &c. ; and by making the proper deductions. ‘ L 218 ~1l;:\'|'.i.‘AT. ILL’USTI‘AI‘ION. PART V l Ceiling. " Uni-«h ring. Deductions. l i j""‘——_'_'““’"“ '_"—' “Wm Corniccs. i ' .. 11;; ~.‘_ ‘ V . Hard , ‘ l i .5 Loats Finisl . 2, : ( oats Finishing. 2 Coats. 1 3 Ft. In 1 H. In. Ft In. Ft. In. Ft. in. l 771 (r U 968 O 23 4 23 4 137 i”; i 74 5 r 157; G :“2 c 35 o 28 10 i l 573 0 I 9140 ‘5 28 10 32 6 5 a J l 824 10 2003‘. T 1:62 4 44 0 28 10 141 8 , ,——._—*# phi, __,,_,,_ . , ,7, ‘E 199243 11 sec 5: Wing 3, :16 2 3(1) 3 35,8 (:1 r “t"“‘*"“ ”’23.; ‘ ~~~ ”W a a “'*""" E ids. F‘t. In, 9?l1.6:flll - 311$ Ft. In. 34 2 45 0 -‘ 249 ‘2 11 Yds. 13, in. ‘ 942 O 2 27 6 ”—2" 193 i H "”—““ " 969 4 ,r-__,,i . 46 8 ' "“h’fl rm: rmsrnmcn’s BILL. Yds. Ft. In. 249 2 11 of ceiling, :3 coats, at — per yard, . 193 1 ll of rendering, hard-finishing, at — per-yard, . . . . . . . — — — Le w ~ f. l I 1 2-12 0 2 of ditto, '2 coats, at —~ per yard,. . — — — 358 6 of ecfri-riic-«".~‘, at —— per foot, . . . — — —— 82 8 of enriched mouldings, at —— per foot... .......—~- Note. The follOnii'y ‘v'orks will he found extremely useful to Bricklayers, Masons. Joiners, or any other persons who are desirous of obtaining a knowledge of Architecture. Jililler’s Designs for (lottages, Farm-houses, Country-houses, Villas, Lodges for Park or Garden Entrances, with Plans of the Offices belonging to each Design, on 32 Quarto Plates, Price, sewed, 103.607. Rawlin’s Desigm of Houses for Gentlemen and Tradesmen, Par— sonages and Summer Retreats; with Back-fronts, Sections, &c.; together with Bangui-Hing Rooms and Churches; to which is added the lVIasonry of semicircular and elliptical Arches, on 51 Royal Quarto Plates, Price 1/. is. The Student’s Imtrmtm. in drawing and working the Five Orders or“ x‘u’chitecture; ruliy cxlrlsaining the best Methods of striking re— gular and quirked mouldings, for diminishing and glueing of Columns and Capitals, for finding the true diameter of an Order to any giren Height7 for strikingr the Ionic Volute circular and elliptical, with finished Examples. on a large Scale. of the Orders, their Planceers, &c. on 33 Octavo Hutu. liy I’cter Nicholson, 1’rice,l)ound, 103. CI]. PART V. GENERAL ILLUSTRA'I‘IOJ. 2’19 Pain’s Builder’s Assistant, demonstrating in the most easy and prac- tical manner, all the principal Rules of Architecture, from the Ground Plan to the ornamental Finish, illustrated with several new and use— ful Designs of Houses, with their Plans, Elevations, and Sections, on (12 Folio Plates, Price, bound, 165. Pain’s Practical IIouse Carpenter, containing a great Variety of use- ful Designs in Carpentry and Architecture: as Centering for Groins, Niches, (SEC. Examples for Roofs, Sky-lights, &c. The Five Orders laid down by a New Scale, llouldings, Sac. at large, with their Enrich— ments. Plans, Elevations, and Sections of Houses for Town and Country; Lodges, Hot-houses, Green-houses, Stables, 8w. Design for a Church, with Plan, Elevation, and two Sections; an Altar- piece, and Pulpit. Designs for Chimney-pieces, Shop Fronts, and Door Cases. Section of a Dining-room, and Library. Variety of Staircases, with many other important Articles, and useful Embel- lishments, on 146 Quarto Plates, Price, bound, 188. This is Pain’s last Work. 220 MENSL’RATION or HAY-STACKS. PART VI. PART VI. THE METHOD OF MEASURING HAY-STACKS, BRAINS, CANALS, MARL-PITS, PONDS, MILL-DAMS, EMBANK- MENTS, QUARRIES, COAL-IIEAPS, AND CLAY-HEAPS. HAY-STACKS. THE contents of hay—stacks are found in order to ascer- tain their weights ; which must, of course, vary according to the density of the hay. ‘ Some stacks will not weigh more than 8 or 10 stones per cubic yard, and others will weigh 15 or 16 stones ; it is not, however, the measurer’s province to determine the weight, but only the number of cubic yards which the stack contains; and leave the buyer and seller to settle about the weight as they think proper. PROBLEM I. To measure a hay-stack, [lacing a circular base. CASE I. PV/zen the stack is straight from the bottom to the eaves, andfrom the eaves t0 the top, as in the following figure,- t/te upper part may be taken as a cone, and the lower part as a conical frustam. RULE. hiultiply the square of the circumference at the bottom AB, by .07958, or for general practice, by .08 ; and the product will be the area of the base. Find the area of a section at the eaves DC, in the same manner. To the sum of these areas, add the square root of their product; multiply this sum by the perpendicular height GF : HD, and g, of the product will be the soli- dity of the frustum ABCD. PART VI. MENSURATION or HAY-STACKS. 221 Multiply the area of the section at the eaves, by the perpendicular height FE 2 DK, and :1) of the product will be the solidity of the. cone DCE. . To the solidity of the frustum add that of the cone; and the sum will, be the content of the whole solid ABCEDA. Note 1. Some measurers take the dimensions of ha -stacks, canals, marl-pits, &c. with a tape divided into yards, tenths, and hundre ths ; but one divided into feet and tenths, is considered, by most practitioners, to be much preferable. 2. When the dimensions are taken in feet, the content must be divided by 27, in order to reduce it to cubic yards. ' EXAJVIPLE S. 1. The circumference at the base K E AB, of the following figure, is 40, "‘“"7 the circumference at the eaves DC 60, the perpendicular height GF=HD 15, and FE 2 DK 16 feet ; how many solid yards does the stack con- tain ? CALCULATION. ' IIere 40 X 40 X .082128, the area of the base ; and 60 X 60 X .082288, 113-- the area of a section at the eaves ; A B also, ,/ (128 X288): ,,/36864=192, the square root of their product; then (128 +288+ 192) X 15—2—32608 X 5 23040 cubic feet, the content of the frastam ABCD. Again, (288 X lG)+3=4608—:-3=1536 cubic feet, the content of the cone CDE. ~ Lastly, (3040+1536)+27=4576+972169 yards, 13 feet, the answer required. 2. The circumference of the base of a hay-stack is 52 feet 9inches, the circumference at the eaves 75 feet 6 inches, the perpendicular height of the lower part 16 feet 3 inches, and that of the upper part 18 feet 6 inches ; how many tons are contained in the stack; admitting each cubic yard to weigh 15 stones or 210 pounds P Ans. 28 tons, 10 cwt. 1 gr. 17 lb. CASE :1. When the stack is bulged from the bottom to the top, as in the following figure ; the equidistant ordinate method, described in Problem 28, Part II, must be adopted. ' I ' L 3 222 MIZNSLRATION or IlAY-STACKS. 11.111'1'1'1. RULE. l. I? ind the areas of as many circular sections, taken at equal perpendicular distances from the bottom, as you iudqe sufficient. bv multiplving the squaie oi the (111 cu .11— terL 11ce of each section IN .08. I 1ocecd \1' itl 1 these areas in the same. manner as it they were cipiidistant ordinates , and the result will be the soliditv of the stack, from the bottom to the uppermost 01 last section. 2. l\Iultiply the area oi the base 01 the remaining part, at the top, which may be considered as a cone, by its per- pendicular height; and § of the product will be the solidity. Add these two soliditics together, and the sum will be the content of the whole stack. Fair 1. Always make. choice of an 0le 1111 Tiber of sections, in order that the num— ber of parts into which the solid is divided. may be equal. I’ivc or seven will, in general, be sufficient; of which one must be at the bottom, and another at the cares, 01' as near to them as possible. Great care must be taken to obtain the dimensions of the sections at equal perpendicular distances; for it the slanting distances be taken, it is evident that the content will be made too much. SCHOLIUM. The method of finding the areas of curvilineal figures, by means of equi-distant, perpendicular ordinates, was first demonstrated by the illustrious Sir Isaac Newton. IlIr. Robert Shirtcliii'e, in his Theory and Practice of Gauging; ap- pears to have been the first who applied it to finding the aicas of" curvilincal vessels used by brewers, Distillers, &c , and after him ilI1. Samuel 11"”a11e1, in the Appendix to Ovcrley s Gauging Ihcii Rules, however, were extremely tedious; and, though true to de- 111011st1ation, were not general, but particular, according to the num- ber of ordinates used. To obviate this inconvenience, the general rule given in Problem 2.3-, Part II. of this ‘Vork, was deduced from Sinrsox’s Dissertations, page 109, by l\Ir. Thomas bloss; and demonstrated in his valuable 'Ireatise of Gauging, page“ 2.35 D1 Hutton, in his Mensuration, Proposition I , Section II., Part IV., has also given an elegant demonstration of the same Rule, and adds, 111 a (To10llary, that it will obtain for the czo itents of all solids, by using the areas of the sections perpendicular to the are, instead oi the ordinates. The Doctor particularly recommends it for the puipose of gauging and u llaging casks; hence, it is evident that it may be applied 11 1th propriety and success to the IVIensuration of Hay- stacks, Canals, DIarl— —pits, and other irregular figures, as being the. best approxima- tion that has yet been, or zaperhaps ever can be given , for by taking an indctinite number of sect 'ions, the content of an irregulai solid may be obtained to any degree of accuracy. N) [O 00 PART VI. MENSURATION OF HAY-STACKS. EXAMPLES. 1. Let the annexed figure re- present a hay-stack, whose di- mensions are as follow; Viz. the . girt at the bottom A13236, at CD254, at EFZGG, at GE: 58, and at KL:37 feet. The perpendicular distance between each section is 5 feet, and the perpendicular height of the co- nical part KLM 4 feet 6 inches ; how many cubic yards are con- tained in the stack? CALCULATION. IIere 36 X 36 x .08:103.68 the area of the bottom or first section; 54 x 54 x .08=233.28, the area of the second section ; 66 X 66 X .08 = 348.48, the area ofthe thirdsection; 58 X 58 X 082269.12, the area of the fourth section ; and 37 x 37 X .08: 109.52, the area of theflfth or last section; then by proceeding according to the Bulefor eqai-distant ordinates, ’we have A: 103.68 4— 109.52 2213.2, 13 .2 233.28 +269.12_—:502.4, 02348.48, and D25 ; consequently (A+4B + 2C)+3 x D : (213.2 + 2009.6+696.96)—:—3 x 5 rs (2919.76 x 5)+3=14598.8—:—3=4866.26 feet, the so- lidity of the part ABLK. Again, (109.52 x 4.5)+3=492.84—:—3=164.28 feet, the solidity of the conical part KLM. , Lastly, (4866.26 + 164.28)—:—97 : 5030.54+27=186.31 cubic yards, the content required. 2. The dimensions of a hay-stack are as follow; Viz. the girt of the bottom or first section 2 127.2, the girt of the second = 145.4, the girt of the third 2 156.5, thegirt of the fourth _—; 168.7, the girt of the fifth :: 48.3, the girt of the sixth = 121.8, and the girt of the seventh or last section :2 68.6 feet. The perpendicu1ar distance be- tween each section is 8 feet, and the perpendicular height of the conical part, at the top, 7 .4 feet ; how many cubic yards are contained in the stack? Ans. 2970.49641 yards. 224 MENSURATION or IIAY-STACKS. PART VI. PROBLEM II. To measure a hay-stack, having a rectangular base. CASE I. I'Vhen the stack is straight from the bottom to the eaves, andfrom the eaves t0 the top, as in the flillowingflgure; the lower part may be taken as a prismoid, and the upper part as a triangular prism. RULE. 1. Multiply the mean length of the bottom by the mcar breadth, and the product will be the area of the bottom. Find the area of a section at the caves in the same manner. llIultiply half the sum of the lengths of the bottom and eaves, by half the sum of the breadths ; and the product will be the area of a section equally distant from the bottom and caves. To the area of the bottom add the area of the section at the eaves, and four times the area of the middle section; multiply this sum by the pcrpendiculan height, from the bottom to the eaves; and s, of the pro- duct will be the solidity of the lower part. 2. Multiply the breadth at the eaves by the perpen- dicular height from the caves to the top; and half the product will be the area of the end; which being multi- plied by the mean length, will give the solidity of the upper part. 3. Add these two solidities together, and the sum will be the content of the whole stack. Note 1. Sometimes stacks are longer on one side than the other, and broader at one end than the other; in such cases, take half the sum of the lengths for a mean length, and half the sum ot‘the breadths for a mean breadth. 2. Some stacks are higher and broader at the ends than in the middle; when this is the case, a proper allowance must be made in taking the dimensions. Allowance also ought to be made for the thatch. 3. It' the ends 01' the upper part he not equal. find the area of both ends; and take halfthcir sum for a mean area. which multiply by the length for the solidity. 4. Ifthe length of the top or ridge GH, be more or less than the length of the base or cave liF, it is evident that the upper part of the stack is in the form ot‘a cunens or wedge; hence its true content may be found by Problem 9, Section I. Part; IV. EXAMPLES. 1. Let the annexed figure represent a hay-stack, the dimensions of which are as follow; viz. the mean length at the bottom 36.8, and the mean breadth 18.3; the mean length at the eaves 44.6, and the mean breadth 25.9 ; the perpendicular height from the bottom to the caves 18.6, and from the caves to the top 15.5 feet; how many cubic yards does the stack contain ; the mean length of the upper part being 43.7 feet? PART VI. MENSURATION 0F IIAY-STACKS. 225 CALCULATION. Here 36.8 X 183267344, the area of the base ; and 44.6 X 25.9:1155.14, the area of the section at the eaves. Also, (36.8 +44.6)~I—2=81.4—:—2=40.7, the length of the middle section; and (18.3+25.9)—:-2=44.2—:—2:22.1, the breadth of the middle section; then 40.7 X 22.1 X 4:: 3597.88, four times the area of the middle section; whence (673.44+ 1155.14 +3597.88) X 18.6—z—6 = 5426.46 X 3.1 :2 16822.026 feet, the solidity of the prismoid ABCDEF. Again, (25.9 X 15.5)—:—2=401.45—:—-2: 200.725, the area ofthe end EDG; and 200.725 X43.7=8771.6825 feet, the solidity of the prism FEDGH. Lastly, (16822.026 + 877l.6825)-—:—97 225593.7085—z-27 =947.915l cubic yards, the answer required. 2. How many cubic yards are contained in a hay—stack of the following dimensions; viz. the length at the bottom 49.4, and the breadth 24.6; the length at the eaves 58.8, and the breadth 34.2; the perpendicular height of the lower part 25.8; the perpendicular height of the upper part 21.7 feet, and its length the same as that at the eaves? Ans. 2335.143 yards. CASE II. When a stack is bulged from the bottom to the eaves, and from the caves to the top, as in the following figure ; recourse must be had to the equi-distant ordinate method. RULE. 1. Find the areas of as many equidistant, parallel sec- tions as you think sufficient, with which proceed asif they were the equi-distant ordinates; and the result will L 5 22f) MENSURATION OF HAY-STACKS. PART VI. be the solidity of the stack, from the bottom to the upper- i’nost section. , 2. Multiply the area of the end of the remaining piece, at the top, which may be considered as a triangular prism, by its length; and the product will be its solidity. 3. Add these two solidifies together, and the sum will be the content of the whole stack. Note 1. The weight of a stack maybe ascertained to a considerable degree of accuracy, in the following manner: Cut out a portion extending to the centre of the stack, from the top to the bottom. and weigh it ; and also measure the vacuity from which it is taken: then say, as the content of this part is to its weight, so is the content ofthe whole stack 0 its weight. 2. it the stack have a Net 1211' base, the portion to be weighed should not he cut off the end, but taken ( ihout half way between the end and the middle; v. here it may be supposvd the hay is ofa medium density. E XAM PLE s . l. Let the annexed figure represent a hay—stack, the dimensions of which are as follow; viz. the length Ali 38 feet, and breadth BC 13 feet; the length DE 41 feet, and breadth EF 18 feet; the length GH 45 feet, and breadth HI 22 feet; the length KL 44: feet, and breadth LM 19 feet; the length NO 42 feet, and breadth Ol’ 12 feet; the mean length of the triangular prism, at the top, 41 feet, and the perpendicular height of its end 5 feet; how many cubic yards are contained in the stack _: the perpendicular distance between each parallel section being 6 feet ? CALC ULA'IION. IIH'e 38 X 13:45)}, [271,6 arm of the first section; '41 X 182738, {/18 arm of 5/13 second section; 45 X222990, the area of the (lain! section; 4H X 192836, t/ze area of t/mfourt/z section; rind 42 X 122504, the area oft/aeflfl/I, 07' {(1st section; Men 63/ proccrdinf/ according to tile Role PART VI. MENSURATION or HAY-STACKS. 227 for equi-distant ordinates, we have'..:\::494+504=998, B:738+836:1574, 0:990, and D26; consequently, (Ari—4B +2C)-I—3 >< D=(998 + 6296 r~1980)><-§—:9274: >< 22185-18 feet, the solidity oftke part ABCPON. Again, (12 X 5)~:—2:60—1—2:30, the area of the end CPR; and 41X 80:1280feet,‘t/m solidity of the prism NOPRS. Lastly, (18548121230)—:—-27=1,9778--21—27=732 yards, 14: feet, the answer required. 2. T he dimensions of a hay-stack are as follow; Viz. the length of the bottom or first section 70.8, and its breadth 20.2 ; the length of the second 7 5.4., and its breadth 28.7; the length of the third 81.2, and its breadth 32.4; the length of the fourth 86.7, and its breadth 36.8; the length of the fifth 8429, and its breadth 34.6 ; the length of the sixth 83.4, and its breadth 30.9 ; the length of the seventh 82.7, and its breadth 16.5 ; the mean length of the trian— gular prism, at the top, 81.5, and the perpendicular height of its end 6.4 feet’; how many cubic yards are contained in the stack; the perpendicular distance between each section being 8 feet? Ans. 4668.8 yards. 3. \Vhat is the weight of a liay-stzu-i; which measures 425 cubic yards 18 feet; the weigl‘u' of a. piece cut out of the stack from the top to the bottom, being 1 ton, 3 cwt. 2 qrs. 15 1b., and the vacuity from which it is taken, 16 cubic yards 12 feet? Ans. 30 fans, 11 cwt. 3 qrs. 1 ll). REMARK. Sometimes hay is stored up in barns or shades. ‘When this is the case, the lower part of the mow, from the bottom to the eaves, will be a parallelopipedon, the content of VVlllL’ll may be found by Prob- lem 2; and the upper part, from the eaves to the ridge of the build- ing, Will be a. triangular prism, the content of which may be Obtained by Problem 3, Part IV. ; then the sum of these two contents will be the whole content of the mow. If the hay does not extend to the ridge, the upper part will be a trapezoidal prism, the content of which may he obtained by finding the area of the end by Problem 8, Part 11.; and multiplying this area by the length of toe prism. (See Problem 18, Part IV., on the method of measuring compound and irregular :oiids.) Nate. \Vheu hay-mows are irregular, mean Jimmsims must be taken. L6 228 MENSURATION OF PART VI. DRAIN S AND CANAL S. A DRAIN is an artificial channel made to convey water from marshes, bogs, and other low grounds, for the pur— pose of making ii'nprovements in agriculture, &c. A CANAL is an artificial, navigable river, most commonly made for the purpose of transporting goods from one place to another, by water—carriage. Note 1. Drains and Canals are made with sloping sides, and are almost invariably dug by the cubic yard; hence, it is of the utmost moment to ascertain their con- tents with accuracy. The common method of measuring them is to take the breadths of the top and bottom, in dilferent places, and their sum being divided by their number, the quotient is considered as a mean breadth. Several depths are likewise measured, in various places, and their sum divided by their number, is taken for a mean depth ; then the length, breadth, and depth being multiplied continually together, the last product is taken {or the content ; but it is evident that this process must lead to very erroneous results. ‘2. In this country, many thousands of acres of bogs, marshes, and fans have of late been made fit for the purposes ofagriculture, by means of drains, particularly in the counties of York and Lincoln; and improvements of this kind are still car- ried on in the latter county, with great spirit. PROBLEM III. Tofind the number of cubic yards wlcz'c/z have been dug out qfa drain or canal. RULE. Find the areas of as many equi-distant perpendicular transverse sections, as you judge sufiicient, with which proceed as if they were equi-distant ordinates; and the result will be the content between the first and last sec- tions. (See Problem 23, Part 11.) Alote l. The distance between the sections must be measured along the middle of the bottom of the drain or canal ; and when the ground is very uneven, it will be necessary to take the sections nearer to each other, than when it is pretty level. 2. When a drain or canal is very long, divide it into several parts or lengths, and find the content of each separately by the above Rule; and the sum of these contents will be the content of the whole. 3. Let the annexed figure repre- A ‘ m n B sent a perpendicular, transverse sec- : ,_ , tion of a drain or canal, where the i 5 ground is even; then it is evident i u: that if a line AB, be stretched across 1) (i the top, it will be parallel to the bottom DC, admitting it to be level ; and the perpendicular Dm will be equal to the perpendicular Cn, which perpendiculars may be easily found by erecting a straight stati‘ perpendicularly to the bottom DC, and so as to touch the line AB, at m and n. PART VI. BRAINS AND CANALS. 229 In this case, the section ABCD is a. trapezoid, the area of which may be found by Problem 8, Part II. 4. Let the subjoined figure de- note a perpendicular, transverse sec- tion of a drain or canal, where the ground is uneven; then it is evident that the section is a trapezium, be- cause AB is not parallel to DC. A Now, as two perpendiculars cannot be taken upon either of the diagonals, the figure must be divided into two triangles in the follow- ing manner: Measure the diagonal BD, and at right- angles to it, the perpendicular Cn , measure also the line AB, and at right angles to it, the perpendicular Dm; hence, the area of the section may be found by Problem 7, Part II. Or, the section may be divided A into two right- angled triangles, and a trapezoid, as in the annexed figure, by measuring the line AB, and at right-angles to it, the two perpen- diculars Dm and Cu; and hence 1)“— C the area may be found. Or, the area of the section may B ---—--~ 3 be obtained by dividing it into the A m {L _____ B two triangles ADm, BCn, and the "“l:,- in ‘ trapezoid DmnC. I f In this case, Dm and C71, which i\/ are at right-angles to DC, are the D 0 parallel sides of the trapezoid, and also the bases of the two triangles; and their perpendiculars are Ar and Ba. 5. If the bottom of a drain or canal be a curve line, as in the sub- joined section; then the area of the quadrilateral figure ABCD, must be found by some of the foregoing methods; and the area of the part DEC, by the method of equi-distant ordinates, described in Pro- blem 23, Part II. Note. The Rule given in Problem 23, Part 11., being expressed in an algebraic form. is seldom perfectly comprehended by learners , but the following one may be easily understood, and committed to memory. RULE. To the sum of the first and last ordinates, add four times the sum of all the even ordinates, and tv ice the sum of all the odd ordinates, not including the first and last; multiply this. sum by the common distance of the ordinates, divide the product by 3, and the quotient will be the area required. 230 MENSURATION OF ”3 V 5'1 r-E M EXAMPLES. 1. Let the subgeqnent figure represent part of a (train or canal; required the number of cubic yards that were dug out of it; the distance between each perpendicular. transverse section being 60 feet, and the dimensinns nt' these sections, as follow: First Section ABCD. ,7, ‘ Feet. A I; "w" "_ ”fie—(fr.— 7! AB 2: {50.5 ‘ 3 E 1 : DC 2 28. 7 1 ~>~~—~=~» '1 Dm:Cn :15.4 i I Sécoml Scolion EFC H. 3 \: Feet ' ’5 Hm: 14.8 rt r‘ ? Tiaird Section KLMN. : Feet. ‘ 1 ' _ Km:12.2 ? ‘ Nm:13.5 I . U f.' 22272 227.4 ' U ‘ '! 31522183 Kit/VX/ E L72. i ' H k ' Fourli} Section PQRS. 0‘) V 1‘ , “L.- H K. 10:09?“ Fifi/z Satin); '1‘ UVKV’. , : t 1 Feet. I ’ 3 : TU 261.5 I J ‘ 1;..1 WV:38.4 ; ; 7 “7,721:sz 210,2 1‘ J .. . ‘1‘ *___| ~._-_._., , , (t :: ' _ - « .' //J " qr 9- 3 l\ ' 7 1' 4/ P M = 1.4 I / . ’ ,« , __ ," z, I ' 5/“: ‘ *1: [1) .._ .34 7\.\; 1 I PART VI. BRAINS AND CANALS. 231 CALCULATION. First Section. Here (60.6 +28.7) X 15.4:89.3 X15.4=1375.22; and 1375.22+22687.61, the area of the trapezoid ABCD. Second Section. Here 52.3 .\< 135270605, double the area of the triangle FGH; and 63.4 X 14.8 2 938.32, double the area of the triangle EFH; then (706.05 + 938.32) +2: 1644.37 +22822.185, the area of the trapezium EFGH. Third Section. Here 12.2 X 13.521647, double the area of the triangle KmN; also (13.5 +18.3) x 27.4 2 31.8 X 27.4 2 871.32, double the area of the trapezoid NmnM; and 23.8 x 18.3 2435.54, double the area of the triangle LMn ; then (164.7 4» 871.32 +435.54)+2 =1471.56+2 2 735.78, the area of the trapezium KLMN. ' Fourth Section. IIere 14.6 x 152222192, double the area of the tri- angle SPm; also (14.6 +20.5) X 29.72 35.1 x 29.7: 1042.47, double the area of the trapezoid SmnR ; and 20.5 X 173235465, double the area of the triangle QRn ; then (221.92 + 1042.47 + 354.65)—:-221619.04—z—22809.52, the area of the trapezium PQRS. Fifth Section. Here (61.5+38.4) X 10.22999 X10.221018.98; and 10189842250949, the area of the trapezoid TUVVV; and by the Rulefor equi-distant ordinates, we have A2 O, B 2 5.3 +5.8 211.1, 0:751, and D:38.4—:—4=9.6; then (A+ 413 + 2 C) x 4 D2(44.4+ 14.8) X 9.6—2-3 2 59.2 X 3.2- 218944, the area of the part VXWV; then 509.49+ 18944269893, the area of the whole section TUVXW. Tofind the solid content. Here A2687.61 + 698.93 2 1386.54, 13 = 822.185 + 8095221631705, 02735.78, and D260; then, (A+ 413 +20) >< 7}D2(1386.54 +6526.82 + 1471.56) >< 60—2—3: 9384.92 >< 2021876984 cubicfeet; and 187698.4—:—272 6951.792 cubic yards, the answer required. 2. Required the number of cubic yards dug out of part of a. drain or canal, from the following dimensions ; each 232 MENSURATION on MARLPITS. PART VI. perpendicular, transverse section being divided into two triangles in the same manner as the second section EFGH, in the foregoing figure; and the common distance of the sections 100 feet: First Section. Second Section. Feet. Feet. HF : 35.9 HF 236.1 Gn :: 8.2 Gn : 8.9 EF 241.7 EF 239.8 Hm: 9.6 Hm: 9.1 Third Section. Fourth Section. Feet. Feet. HF 237.1 HF 236.5 Gn = 9.8 Gn = 9.2 EF 242.2 EF 241.5 Hm: 9.4 Hm=10.2 Fifth Section. Sixth Section. Feet. Feet. HF : 37.1 HF =38.6 Gn = 9.5 Gn = 9.4 EF 242.8 EF =43.2 Hm:lO.2 Hm: 9.8 Seventh Section. Feet. HF =37.6 G71 = 9.8 EF 242.6 Hm: 10.3 Ans. The area of the first section=347.35; the area ofthe second=341.735; the area of the third=380.l3; the area of the fourth: 379.55 ; the area of the fifth: 394.505; the area of the siccth:393.1; the area of the seventh=403.63; and the content:225259% cubic feet: 8342 cubic yards, 25% feet, the answer required. MARLPITS. MARI. is a kind of rich clay, and is used as manure for land, in Lancashire, Cheshire, Derbyshire, and other counties in England ; and is commonly dug by the cubic yard. PART. VI. MENSURATION or MARLPITS. 233 Marlpits are of various forms. Sometimes they are laid out in the shape of a rectangle, sometimes in that of a trapezium, and sometimes as an irregular polygon. In digging a marlpit the sides are always sloped, in order to prevent the upper edges from slipping in ; and as the bottom is seldom perfectly level, there is generally a little variation in the depths. PROBLEM IV. To find how many Cubic Yards have been dug out of a Marlpit. CASE I. W'lzen the top and bottom of a marlpit are rectangles, the pit may be considered as a prismoid. RULE. To the sum of the areas of the top and bottom, add four times the area of a section half-way between them ; multiply this sum by the mean perpendicular depth, and -3, of the product will be the solidity. Note 1. ‘When the sides and ends are regularly sloped, the length of the middle section will be equal to half the sum of the lengths of the top and bottom, and its breadth equal to half the sum of their breadths ; but if the inclination of the sides and ends be not regular, the length and breadth of the middle section must be found by actual admeasurement. 2. A mean depth must be found by taking several depths, at equal distances from each other, and dividing their sum by their number. EXAMPLES. 1. Let the subsequent figure represent a marlpit, the dimensions of which are as folloxr : viz. the length AB of the top =58.6 feet, and the breadth BC 2 36.2 feet; the length EF of the middle sec- tion :55.7 feet, and its breadth FG = 33.4 feet; the length KL of the bottom = 52.8 feet, and breadth LM : 30.4 feet; how many cubic yards of marl were dug out of the pit, its mean perpendicular depth being 8.6 feet ? A it 234 MENSURATIOX or MARLPITS. PART VI. CALCULATION. Here 58.6 X 36.2:2121.32, the area of the top; 55.7 V I\ 33.4 x 421860.38 >< 427441.52, four times the area of the middle section; and 52.8 X 30.4: 21605.12, the area of the bottom; then (2121.32+7441.52+ 1605.12) X 8.6.4; 11167.96 X8.6=9604-’i.456, which being divided by 6, um obtain 16007409 cubic feet:592.867 cubic yards, (he answer required. 2. A marlpit measures 86.4 feet in length, and 36.8 feet in breadth at the top ; the length of the middle sec- ‘ "7 tion is 82.2 feet, and its breadth 32.4 feet; the bottom i< 78.3 feet in length, and 28.6 feet in breadth; how many cubic yards of marl were dug out of the pit, its mean perpendicular depth being 10.5 feet ? Ans. 1048.098 yards. CASE II. Tofi'izd the number of cubic yards which have been dug out of a marlpit, the top and bottom ofwhich are triangles, trapeziums, polygons, or any otherfigurc whatever. RUL E . Take such dimensions as will give you the area of the top, the area of the middle section, and the area of the bottom; proceed with these areas in the same manner as directed for a prismoid, and the result will be the content. nearly. (See the SCHOLIUJI, Prob. 10, Part I V. loci. 1.) Note 1. The above Rule will give the content of the generality of marlpits very near the truth; if, however, a pit exceed 10 or 12 feet in depth, and the sidts b" not pretty regularly sloped, it will be more accurate to find the areas of :3 (our. distant, horizontal sections ; and proceed with them in the. same manner as it they were equi-distant ordinates. (See Me SCHO LI U31, Problem l, Case 2. Port 1'1.) 2. When the top and bottom of a maripit are trapeziums approaching nearly to rectangles, and the inclination of the bottom from the plane of the horizon very considerable, so as to make the pit much deeper on one side than on the other. the content may be most correctly obtained by finding the areas of several equi—distant, perpendicular sections; and proceeding with them as directed in the last Problem. The contents of the two pieces at the ends, resembling wedges, may be found by Problem 9, Part IV., Section I. 3. The content of the space or road leading into the pit, may be obtained by the last Problem. The area of the first section, on that where the road commences, will evidently be nothing; and the distance between the first section and the middle of the slant height of the pit must be taken for the length of the space. 4. When the edges of a pit give way, and slide in, these slips must be measured. if they be thrown out or removed by the labourers. The general method of doing this is by multiplying the mean length, the mean breadth, and the mean depth continually together. for the content. This method is certainly inaccurate ; yet in such cases as these, it may, in general, suffice; and sometimes it is the only one that can be adopted with any probability of success. N) 03 C7 l PART VI. MENSURATION OF MARLPITS. EXAMPLES. 1. Let the following figure denote the uppermost, horizontal section of a inarlpit, the diagonal AC of which mea- sures 86.5, the perpendicular Da 26.8, and the perpendicular Ba 32.6 feet. The diagonal of the middle section is found to be 80.6, and one of the perpen- ‘ (liculars 23.9, and the other 29.3 feet. The diagonal of the bottom measures 74.3, and the perpendiculars 20.7, and 26.5 feet; how many cubic yards were dug out of the , pit, its mean perpendicular depth being 10.8 feet ? CALCULATION. [Jere (26.8 +32.6) >< 86.52594 X 86.5 2 5138.1 ; and 5138.1—1—22256905, the area ofthe top. Also, (23.9 +29.3) X 80.62532 x 80.6 : 4287.92 ; and 4287.92 + 2 X 4 2 2143.96 X 4 = 8575.84, four times the area of the middle section. Again, (20.7 + 26.5) x 74.3 247.2 x 74.3 : 3506.96; and 3506.96—2—221753.48, the area ofthe bottom. Lastly, (2569.05 + 8575.84 +'1753.48) X 10.8 +6 2 12898.37 X 1.8 2 23217.066 cubic feet: 859.8913 cubic yards, the answer required. 2. Admit the following figure ABCDE to denote three horizontal sections of a marl- pit; also the figure FLMG to ----- ----------J ........................... ? g represent the space or road “ / leading into the pit; required the number of cubic yards that were dug out of each ; the mean depth of the pit being. 11.4 feet; the equi-distant, H K perpendicular sections at HK, LM, trapezoids; the whole length of the space 43.8 feet; and the other dimensions ' as below. M B F G 236 MENSURATION OF MARLPITS. PART VI. IIorizontal Sections of the Pit. Uppermost Section. flfiddle Section. Bottom Section. Feet. Feet. Feet. AC=118.6 AC=111.7 AC=104.8 Ea = 43.4} Ea = 39.8} Ea = 36.4} Ba 2 35.2 Ba = 31.6 Ba = 28.3 CE =102.8 CE = 95.9 CE :- 88.6 Da = 26.4 Da = 22.6 Da = 19.5 Perpendicular Sections of the Space. Section at HK. Section at LM. Feet. Feet. AB=12.4 AB:12.8 DC : 8.5 DC 2 7.9 szCn = 5.7 Dm:Cn 211.2 (See thefigare in the last Problem, ZVOL‘e 3.) CALCULATION OF THE PIT. Uppermost Section. Here (43.4 + 35.2) X 118.6 : 78.6 X 118.6 2 9321.96, double the area of the trapezium ABCE; and 102.8 x 26.4 =27l3.92, double the area of the triangle ODE; then (9321.96 +2713.92)+2=12035.88+2=6017.94, the area of the whole section ABCDE. 11liddle Section. Here (39.8 + 31.6)x111.7= 71.4 X 111.7 : 7975.38, double the area of the trapezium ABCE; and 95.9 X 22.6 22167.34, double the area of the triangle CDE ; then (7975.38 +2167.34) + 2 X 4 : 10142.72 —:—— 2 X 4 25071.36 x4=20285.44, four times the area of the whole section ABCDE. Bottom Section. Here (36.4+28.3) X 104.8 = 64.7 X 104.8 2 6780.56, double the area ofthe trapezium ABCE; and 88.6 x 19.5 = 1727.7, double the area of the triangle CDE ; then (6780.56 +1727.7)-:~ 228508.26 —2— 2 = 4254.13, the area of the zchole section ABCDE. PART VI. MENSURATION OF PONDS, ETC. 237 Tofind the solid content. Here (6017.94 + 20285.44 + 4254.13) X 11.4 + 6 2 30557.51 X 1.9 :2 58059.269 cubic feet=2150.343 cubic yards, the content of the pit. CALCULATION OF THE SPACE. Section at HK. Here (12.4 + 8.5) X 5.7 = 20.9 X 5.7: 119.13; and 119.13 +2:59.565, the area of the trapezoid ABCD. Section at LM. erre (12.8 + 7.9) x 11.2 = 20.7 x 11.2 = 931.84.; and 231.84+2=115.92, the area ofthe trapezoid ABCD. 1 Tofind the solid content. ‘ IIere A=115.92, B = 59.565, C = o, and D=43.8 + 2 1291.9; then (A+4 B + 2 C) X % D=(115.92 +238.26) X 21.9 ~2- 3 : 354.18 x 7.3 2:: 3585.514 cubicleetrz 95.759 :cubic yards, the content of the space. 3. The diagonal of the first or uppermost horizontal section of a marlpit is 118.6, and the perpendiculars are 64.8, and 43.5 feet; the diagonal of the second section 112.4, and the perpendiculars 61.7, and 40.6 feet; the diagonal of the third section 106.3, and the perpendiculars 58.5, and 37.7 feet; the diagonal of the fourth section 100.8, and the perpendiculars 55.9, and 34.6 feet; the diagonal of the fifth or bottom section 94.3, and the per- pendiculars 52.4, and 31.5 feet; required the number of cubic yards that were dug out of the pit, the sections being taken at equal distances from each other, and the mean perpendicular depth of the pit 16.8 feet. Ans. The area of the first section:6422.19; the area of the second:5749.26 ; the area of the third=5113.03; the area of thefourth = 4561.2; the area of the fifth : 3955.885; and the content : 86584.365 cubic feet : 3206.828 cubic yards, the answer required. DIRECTIONS FOR hiEASURINGr PONDS AND hflLL-DAMS. PONDS and mill-dams are commonly dug by the cubic yard; and assume a. variety of shapes. 238 MENSURATION OF EMBANKMENTS. PART VI. If the top and bottom of a pond or dam be rectangles, it may be treated as directed in the last Problem, Case I. ; if the top and bottom be triangles, trapeziums, or polygons, the content may generally be found by Case 11., of the last Problem; but if both these methods fail, you must proceed as directed in Problem 3, for drains and canals. In this case, the contents of the pieces at the ends, re- sembling wedges, must be found by Problem 9, Part IV, Sect. I. hIr. W. Putsey, Teacher of the Mathematics, at Picker- ing, informs me that the ponds made upon the wolds in Yorkshire, are generally of a conical shape; hence their contents may be found by Problem 7, Part IV, Sect. I. Mr. P. who is a practical measurer, has also very kindly communicated the following method of taking the dimensions of a conical pond in which there is water. EXAMPLE. Let ABC denote a perpen- dicular section of a conical pond, whose dimensions are required. Extend a cord over the pond, with which take the diameter AB; make a ring fast to the middle of the cord or diameter, as at D, through which put the end of the plumb-line ADC; let your assistant keep one end of the diameter at B ; while you hold the other, and the plumb- line at A ; permit the plummet to descend to the bottom of the pond, as at C ; then draw back both the cords, and measure CD, which will be the perpendicular depth of the pond. Note 1. When the top of a pond is not a perfect circle, measure two diameters at right angles to each other; and take half their sum for a mean diameter. 2. Cellars generally form parallelopipedons ; and when they are dug by the cubic yard, their contents may be found by Problem 2, Section 1., Part IV. DIRECTIONS FOR MEASURING EMBANKMENTS. THE most correct method of measuring embankmcnts, is to proceed in the same manner as directed in Pro- blem 3, for drains and canals; and if the following ob- servations be well understood, no difiiculties will arise in PART VI. MENSURATION or QUARRIES'. 239 taking the dimensions, and finding the areas of the sections. Let ABCD represent a trans- verse perpendicular section of an embankment, made upon level ground; then if DC be parallel to AB, it is evident that the section is a trapezoid, the area of which may be found A by P1 oblem 8, Part II. The distance of the parallel sides, or height of the sec- tion, may be obtained by placing a staff perpendicular to the horizon, at A or B, and producing DC to E or F; then will AE or BF be the perpendicular distance of the parallel sides AB, DC; and AB, the breadth of the bottom of the embankment, is evidently equal to ED +DC + CF. Again, let GHKL, denote a perpendicular, transverse section of an embankment, made upon uneven ground ; ,9 ........... then it is evident that the sec- tion is a trapezium; and as the diagonals and perpendiculars cannot be measured, the con- tent must be found in the fol- lowing manner: Produce LK, H the horizontal line ot the top, both ways, to M and N, and let fall the perpendiculars MG and NH; then by Problem 8, Part II., half the sum of these parallel sides multiplied by their perpendicular distance MN, will give the area of the trapezoid GHNM, from which sub- tract the sum of the areas of the two right-angled triangles GML, HNK, and the remainder will be the area of the section GHKL. Note. Unskilful measurers affect to determine the contents of embankm ents by finding what they call mean breadths and thicknesses , but no person who has a scientific knowledge of mensuration, will have recourse to such an eironeous pro— cess, if the foregoing method can by any means he adopted. See Diains and Canals, Note 1. B .--.._..,...--..._.z 7; DIRECTIONS FOR MEASURING QUARRIE S. THE baring of quarries is generally done by the cubic yard ; and sometimes the stones themselves are got by the same measurement. 240 LUSNSURATION or COAL-HEAPS. PART VI. Quarries, in general, are very irregular ; they may however, commonly be measured by some of the methods already described for canals, marlpits, ponds, &c. If a. quarry be so irregular that none of these methods can be adopted, the general method is to take such dimen- sions as will give the area of a mean horizontal section; then this area being multiplied by the mean depth of the quarr ; the product is taken for the content. Note 1. Sometimes a quarry may be most correctly measured by dividing it into several parts ; and taking the dimensions of each part separately. 2. When the stones that are left jutting out of the sides of the quarry, are mea- sured with the vacuity, mean dimensions of these stones must be taken, and their contents subtracted, in order to obtain the true content of the vacuity. DIRECTIONS FOR BIEASURING COAL-HEAPS. WVHEN the stock of a collicry is to be taken, or an ex- change of tenants takes place, it becomes necessary to ascertain the quantity of coals which are laid unsold at the pits; and as pit-heaps are generally very irregular, both in extent and thickness, it is no easy task to find their contents with a tolerable degree of accuracy. The method generally adopted, is to take such dimen- sions as will give the area of a mean horizontal section of the heap; and to multiply this area by the mean thick- ness, for the content. Note I. The perpendicular height of any point at the extremity of a coal-heap, may be obtained in the following manner: Place a spirit or water—level upon the top of the heap. close by the edge; let your assistant hold a pole upon the ground, in a perpendicular direction, at the bottom of the heap: direct the level towards the pole, and note that point in it which is seen through the level; measure the distance between the bottom of the pole and this point; subtract the height of the level from the said distance; and the remainder will be the height of the heap, at the place where the observation is made. 2. When a coal-heap is upon level ground, the height or thickness of any place between the pit and the extremity of the heap, may be found thus: By Note 1, find the height of the heap where it is most elevated ; then let your assistant place the pole upon that part of the heap the height of which you wish to obtain; and without removing the level from the highest point of the heap, direct it towards the pole ; measure the distance between the bottom of the pole and the point seen through the level; from this distance take the height of the level; and if the re- mainder be subtracted from the greatest height of the heap, before found, you will obtain the height required. 3. If the ground on which the coal-heap is laid be level, and the upper surface of the heap a regular inclined plane, rising gradually from the pit to the utmost extremity of the heap. which is sometimes the case, it is evident that the mean thickness of the heap will be equal to half the sum of the two heights taken at that plarthof the heap adjoining the pit, and at the utmost extremity or highest point of e cap. 4. W'hen a coal-heap is extremely irregular, it is generally necessary to divide it into several parts, and take the dimensions of each part separately; in this case, heights must be taken in such places as are most likely to give the mean thickness of each part ; and when the ground is not level, a proper allowance must be made for this circumstance. PART VI. MENSURATION or CLAY-HEAPS. 241 a? 5. In some parts of England, 5 peeks, \Ninchester measure, or 2688 cubic inches, make a bushel of coals, and 36 bushels a chaldron ; therefore, if the cubic :inches in a coal-heap be divided by 2688, or the cubic feet by 1.555, the quotient will be the number of bushels contained in the heap; but as this measure is not general, every person who measures a coal-heap ought to make himselfacquainted with the customary measure of the place. — The content of the new imperial coal-bushel is 2815 cubic inches. (See the Author’s Arithmetic, pp. 53. 353. For the New Im- perial Measures, see Part VIII. of [his Work.) 6. Notwithstanding what has been advanced on the subject of measuring quarries and coal-heaps, agreat deal will always depend upon the ingenuity of the measurer; for it is impossible to give directions that will suit every particular case to be met with in the practice of measuring these irregular figures. CLAY-HEAPS. CLAY is frequently dug out of pits, and laid upon the surface of the ground, generally in very irregular heaps ; and then sold by the cubic yard or by the ton, for the purpose of making bricks. When this is the case, it becomes necessary to ascer- tain its solid content, as nearly as possible; and this may be approximated by the following methods : Find the area of the base of the heap, and also the area of its top, in the most convenient manner, according to the rules and directions given in the different Problems of Part 11.; and then take half the sum of these areas, for the mean area. -—Also, take several depths, and divide their sum by their number, for a mean depth. Multiply the mean area by the mean depth, and the product will be the solid con- tent; and, when this is in cubic feet, divide it by 27, to reduce it to cubic yards. Note 1. When the base of a clay-heap is very irregular, its area may be found by Problem 5, Part III. 2. When the dimensions are taken in yards, tenths, and hundredths, the soli-s content will be in cubic yards, and decimals of a yard; but if the dimensions b' taken in feet, tenths, and hundredths, the content will be in cubic feet, and decimals of a foot. 3. When a clay-heap is not very irregular, the mean area may sometimes be found by taking the dimensions ofthe top of the heap ; and increasing each dimen- sion, as you take it, so as to give (as nearly as you can judge) the dimensions of the middle horizontal section. —This method of measuring is generally called “giving and taking,- " as the dimensions thus obtained are smaller than those of the base of the heap, but larger than those of the top, 4. When a clay-heap assumes something of the form of a parallelopipedon (Pro- blem 2, Part IV.), some measurers take lengths, in two or three different places ; and divide their sum by their number, for a mean length. A mean breadth and a mean depth are found in the same manner; and the continued product of these three dimensions is taken for the solid content. 5. When a heap is very large and irregular, it is frequently advisable to divide it into two or three parts ; and take the dimensions, and tind the content of each part separately. 6. When a clay-heap resembles a. prismoid, or the frustum of a cone. the rule given in Problem 10, Part IV., may be applied with propriety. For other useful Rules and Observations, the reader is referred to Problem 18, Part IV. ; and also to the methods of measuring marl-pits and coal—heaps. 7. A cubic foot of clay, according to the table of specific gravities. page 147., weighs 2160 ounces avoirdupois ; consequently, a cubic yard will weigh 1.627 tons; therefore, if we multiply the number of cubic yards, in a clay-heap, by 1.627, the product will be the weight in tons. M 242 :11msn:1;.\'nox 01‘ CLAY- 11111115 11.1111 V1. 8. By experience, it has ban mund. that “hen cla} is ver3 compact a cubic 33rd will weigh 15‘- tons; consequently. the multiplier will be 1.75; but when it is less dense, a cubic yard will no: 111111111 more than ll or 1} tons , in these cases, the multiplier \\ ill be 1.2501 1. :'1. to b11113 cubic yards into tons 9. A measuring-tape divide: into feet tenths, and hall'- tenths, is best adapted for taking the dimensions; as we then have no trouble in reducing the inches to the decimals of a foot; for, in this case. the tenths are given by the tape; and the half-tenths are 5 hundredths of a foot; hence we have always the dimensions in feet and tenths, or in feet, tenths. and hundredths of a foot; thus, 28.6% feet =28.Gb feet, 81c. 10. The rules and directions gixen for clay—heaps will apply to all compost and manure heaps; when it is r1 gum-1'1 to determine their solid contents for the pur— poses of sale, &c. EXAMPLE. Required the 11.1.1.1 .11 1- of cubic yards, in two clay—heaps, resembling paralieiopipedons, ironi the following dimen— sions; and also the weight oi the clay, supposincre each cubic yard to we} eh 1.627 tons. 1 1 1 ‘ No. ileng'ms. Breadths. Depths. Feet. Feet. Feet 1 93.85 56.55 9.65 86.75 48.45 8.85 74.65 39.35 7.70 2 112.25 43.25 8.35 58.210 39.70 46.4.3 ‘ Center)?! 133/ [/n" 1/1in (mdfourz‘la 1V0tes. No. 1. Length. Breadth. Dep th. Feet. Feet. I1 cut. 93.85 56.55 9.65 86.75 48.45 8.85 74. 65 39.35 £0 3)E):55.25 :7. 1—4155 3EE) 85. 083 mean £8. 116 mean. 8.666 meme. Then, 85.083 x 15.116124095853628 squme' feet, the mean area; and 4.05.3.8 833628x 8.666— -3547 72335540258 cubicfeef, {/ze 501517 ronhnt 0f1\o.I PART VI. A DESCRIPTION or CANALS. 243 O .. No. II. Length. Breadth. Depth. Feet. Feet. Feet. 62.25 43.25 8.35 58.30 $9.70 ‘27 464:5 2)82.95 3)167-00 41.476 mean. R666 mean. ~-—* Then, 55.666 >< 41.475 =2308.74735 square feet, the mean area; and 2308.74735 x 8.35 =19278.0403725 cubic feet, the solid content of lVo. If. Lastly, 35477.335 +19278.040 = 54755.375 cubic feet, the solid content of both heaps; then, 541755.375 +27 2 2027.9768 cubic yards, the whole content. IV eight by the seventh Alote. Here, 2027.9768 >< 1.627=3299.5182536 tons: 3299!; tons, the weight required. - - REMARK. Having, in the preceding pages, treated of the Mensu- ration of Drains and Canals, it is presumed that a short account of a few of the principal ones, will not be un— acceptable to the young reader; as it will tend to give him some idea of the great Improvements that Agriculture and Commerce have received, and are daily receiving by the use of them; and Will inform him What stupendous ‘Works have been effected by the Ingenuity, Perseverance, and united Efforts of men. A DESCRIPTION or some on THE PRINCIPAL CANALS IN ENGLAND, SCOTLAND, FRANCE, AND CHINA. CANALS are to be met with in every civilised country; and perhaps it will not be going too far to say, that the internal commerce of no nation has received greater improvements by them, than that of Great Britain. The the of Bridgewater’s Canal, is a work that begins at VVorsley, seven miles from Manchester; where, at the foot of a mountain, com- M 2 244 A DESCRIPTION or CANALS. PART W. O posed of coal, a basin is cut, containing a great body of water, which serves as a reservoir to the navigation. The canal runs under a hill, by a subterraneous passage from this basin, nearly three quarters ofa mile, to the Duke’s coal—works. In some places the passage is cut through the solid rock, and in others arched over with brick; and at certain distances, air funnels, some of which are thirty-seven yards in perpendicular height, are cut through the rock to the top of the hill. At Barton-bridge, three miles from the basin, is an aqueduct, which, for upwards of 200 yards, conveys the canal across a valley and the navigable river Irwell. There are three arches over this river; the middle one is 60 feet wide, and 38 feet above the water in the Irwell; and will admit the largest barges that navigate that river, to pass under with masts and sails standing. At Longford bridge, the canal turns to the right, and crossing the DIersey, passes near Altringham, Dunham, Grappenhall, and Kaulton, into the tideway of the Mersey, at Runcorn Gap, where barges can come into the canal from Liver- pool at low water. This canal is 29 miles in length. It was begun in 1758, and finished in 1763, under the direction of the great mechanic and en- gineer, Brindley. Note. From this canal, at \Vorsley, there is a branch cut to the town of Leigh; and thence to Wigan, where it communicates with the Leeds and Liverpool canal.—From the Duke’s canal, there is also another branch which terminates at Manchester; and thus unites this great emporium with the maritime town of Liverpool. The Grand Junction Canal, is a work that joins several other canals in the centre of the country, which thence form a communication between the rivers Thames, Severn, iVIersey, and Trent; and, conse- ‘quently, an inland navigation to the four principal sea-ports, London, Bristol, Liverpool, and Hull. This canal commences at Braunston, on the west-borders of Nor- thamptonshire, passes by Daventry, to Stony Stratford. in Bucking- hamshire, thence on the confines of Bedfordshire, west of Leighton Buzzard, to Tring, Berkhamstead, and Rickmansworth, in Herts, and through Middlesex, by Uxbridge, to Brentford, where it enters the Thames, 12 miles, by that river, above London. Its whole length is upwards of 90 miles. The Grand Trunk Canal, is a. work that forms a communication between the rivers Mersey and Trent, and, of course, between the Irish sea and the German ocean. Its length is 92 miles, from the Duke of Bridgewater’s canal at Preston on the Hill, in Cheshire, to VVildon-ferry, in Derbyshire, where it communicates with the Trent. The canal is carried over the river Dove, in an aqueduct of 23 arches, and over the Trent, by an aqueduct of 6 arches. At Preston on the Hill, it passes under ground 1241 yards; at Barton, and in the neighbourhood, it has two subterrancous passages; and at Hardcastle Hill, in Stafiordshire, it is conveyed under ground 2880 yards. In the neighbourhood of Stafford, a branch is made from this canal, to run near Wolverhampton, and to join the Severn near Bewdley: from this again other branches cross VVarwickshire to Braunston, where commences the Grand Junction Canal to the Thames at Brentford. PART VI. A DESCRIPTION or CANALS. . 245 This canal was begun in 1766, by Brindley, who died before its completion; but it was finished by Mr. Heushall, his brother-iu-law, in 1777. The Leeds and Liverpool Canal. “ The first act for this canal' passed in 1770, by which the undertakers, a body corporate, by the name of “ The Company of Proprietors of the Canal Navigation from Leeds to Liverpool,” were empowered to raise the sum of 320,0001. in shares, deemed personal estate. The work was prosecuted with spirit; and in a few years there were finished 83%, miles on the Yorkshire side, viz. from Leeds to Holme Bridge, near Gargrave; and 28 miles on the Lancashirc side, viz. from Liverpool to Newbrough ; but the money subscribed being all expended e works were discontinued for several years. In 1790 another act was obtained, to enable the company to vary the line from the north to the south side of the river Calder; and to raise a further sum of money: in consequence of which the works were again resumed. In 1794 another act passed, to enable the company to make a con- siderable alteration in the line, so as to pass by the market-towns of Burnley, Blackburn, Chorley, and W'igan, where it joins the head of the Douglas Navigation ; by means of which it unites the Lancashire side of this canal at Newbrough. The whole length of the canal, according to this arrangement, is 129 miles. The summit of this canal is at a village, called Foulridge, near Colne, where it passes under ground 1680 yards, by a tunnel, which is 18 feet high, and 17 feet wide. The fall, eastward, from the sum- mit to the navigable river Aire, at Leeds, is 409% feet; and westward to the basin, at Liverpool, 431 feet; which basin is about 52 feet. above the river Mersey, at low water. Between the summit and Leeds are 44 locks, viz. 15 to Holme Bridge, (from which place to near Bingley, there is a level pool of 17 miles in length,) 11 from Bingley to the Junction with the Bradford canal, and 18 from thence to Leeds. Between the summit and Liverpool are 47 locks. There are several large aqueduct bridges on this canal, particularly those over the river Aire below Bingley, and above Gargrave. It has a communication with the town of Bradford, by means of a branch, about three miles in length, called “ Bradford Canal.” The general advantages which the public will derive from this canal, are more than the limits of this paper will allow to be ex- plained. It will tend greatly to encourage trade and manufactures; will bring into extensive use inexhaustible rocks of lime-stone, slate, flags, and free-stone; excellent mines of coal ; and great quantities of timber for building ships, houses, &c. This canal is 42 feet wide at the top, with 5 feet depth of water; and the dimensions of the locks upon it, are about 70 feet in length, and 15% feet in width. It was completed in the latter part of the year 1816; and now forms a most eligible communication between the port of Liverpool and the town of Leeds; and from thence, by the river Aire, and other canals and river navigations, to the port of Hull. Thus the object of M 3 246 A DESCRIPTION or emu-us. PART vr. the original promoters of this noble undertaking is obtained, viz. “ A lVavz'gable Communication between the East and ll'cst Seas. K353. Mr. Thomas Beeston, ofthe Canal office, Leeds, and Mr. John Bottomley, Cf the Canal ofli‘eeullradford, very kindly favoured me with the above description of the Leeds and Liverpool Canal. The Great Canal, in Scotland, is a work that forms a junction be- tween the Forth and Clyde. Its length is 85 miles from the influx of the Carton, at Grangemouth, t0 the Junction with the Clyde, six miles above Dumbarton. In the course of this navigation, the vessels are raised to the height of 155 feet above the level of the sea; and passing afterward upon the summit of the country for 18 miles, they descend into the river Clyde, and have from thence free access to the Atlantic ocean. This canal is carried over 36 rivers and rivulets, and :2 great roads, by 38 aqueducts of hewn stone. In some places it passes through mossy ground, and in others through solid rock. The road from Edinburgh to Glasgow passes under it near Falkirk, and over it, by means of a drawbridge, six miles from Glasgow. In the course of this inland navigation are many striking scenes; par- ticularly the romantic situation of the stupendous aqueduct over the Kelvin, near Glasgow, 420 feet in length, carrying a great artificial river over a natural one, where large vessels sail at the height of 65 feet above the bed of the river below. The utility of this communi- cation between the German and Atlantic oceans, to the commerce of Great Britain and Ireland, in their trade with Norway, Sweden, and the Baltic, must be strikingly evident; as it shortens the nautical distance in some instances 800, and in others 1000 miles. This canal cost about 800,000l. It was begun in 1768, under the direction of Mr. Smeaton, and finished in 1790, by Mr. ‘Whitworth; Blr. Smeaton having resigned, in consequence of the bad state ofhis health. The Royal Canal of Languedoc, in France, is a work that eli‘ccts an inland communication between the Mediterranean and Atlantic. From the port of Cette, in the Mediterranean, it crosses the lake of Thau; and, below Toulouse, is conveyed by three sluices into the Garonne. At St. Ferreol, near Revel, between two rocky hills, is a grand basin, above 1000 feet in diameter, into which the rivulet Laudot is received; and hence three large cocks of east brass open, and discharge the water, which then goes under the name of the river Laudot, and continues its course to the canal called Rigole de la Plaine. Thence it is conveyed to another reservoir near Naurouse, out of which it is conveyed by sluices, both to the lVIediterranean and Atlantic, as the c' nal requires it; this being the highest point be- tween the two seas. Near Beziers are eight sluices, which form a regular and grand cascade, 986 feet long, and 66 feet high, by which vessels cross the river Orb, and continue their voyage on the canal. Above it, between Beziers and Capestan, is the Mal-Pas, where the canal is conveyed. for the length of 720 feet, under a mountain. At Adge is a round sluice, with three openings, three different depths of water meeting here ; and the gates are so contrived, that vessels may pass through by opening which sluice the master pleases. The canal has 37 aqueducts, and its length from Toulouse to Beziers, where it joins the river Orb, is 152 miles. PART VI. AN ACCOUNT OF BRAINS. 247 This canal was begun in 1666, and completed in 1681, under the direction of the engineer Riquet. The Imperial Canal, in China, is a stupendous work, extending nearly through the whole empire, from north to south; and serves to convey goods between Pekin and Canton; being only interrupted by a mountain, in the province of Kiang-si, of about one day’s journey. This canal was finished about the year 980; and 30,000 men were employed 43 years in completing it; its whole length being 825 miles. The trafiic upon it is exceedingly great; and it is, in various other respects, an object of wonder and admiration to Europeans. There are in England, as well as in other nations, many more Canals far too numerous to be mentioned here; those who desire to see a more circumstantial account, may consult Mr. Phillips’s General History of Inland Navigation, which is a work replete with useful in- formation. AN ACCOUNT or some or THE PRINCIPAL DRAINS IN THE COUNTY OF YORK, AND THE AG RIC ULTU HAL IMPROVE M ENT S MAD B BY T HE M . EXTENSIVE Drainages have of late years been executed in this county; and little land, which is capable of improvement by such means, now remains undrained. The Drainage known by the name of the “ Holderness Drainage,” lies chiefly adjoining to and on the east side of the river Hull. It extends from north to south about eleven miles; and contains, by survey, 1121] acres. Previously to the execution of this undertaking, much of the level was usually covered with water for above half of the year ; and on some parts were extensive cars or lakes. This Drainage, though still imperfect, has made a great improve- ment in the adjoining lands, many of which from having been of little or no value, are now let from fifteen to thirty shillings per acre; and this has been accomplished at an expense of somewhat less than 51. per acre. The Act of Parliament for executing this Drainage, was obtained in 1762; but it was not until the year 1775, that an assessment was levied for 24,000]. to carry it into ei’l‘ect. Since that time, engineers have been consulted on the further improvement of this Drainage; and the consent of the proprietors has been obtained for raising fur— ther supplies; in consequence of which the sum of 80,0001. has been advanced, in order to complete the work. The winter of 1763-4 being unusually wet, the water on the west- side of the river Hull was higher than common, and occasioned a breach in the bank of the river, near the upper end of this level. by which the whole of it was overflowed, together with part of the town— ships of Stonet‘erry, l\'Iarfleet, Bilton, Preston, Southcoats, and Dry- fool. The turnpike road between Hull and Ileadon, for nearly M 4 248 AN ACCOUNT OF BRAINS. PART VI. four miles, stood from two to four feet deep in water, as did likewise that leading from VVhitecross to Beverly; and persons going to market were obliged to pass in boats; but it is now believed that these low-lands will never be flooded again, if the banks of the river Hull can be preserved strong enough to resist the water of floods. The Beverley and Barmston Drainage lies parallel to the Holderness Drainage, but on the contrary side of the river Hull; and extends from Barmston, along the course of that river, nearly to Kingston- upon-Hull, a distance of about twenty-four miles. The Act of Parliament for executing this Drainage, was obtained about the year 1792. It is divided into two parts, the upper or more northern of which, containing 2,136 acres, extends from the township of Foston to Barmston, where it has an outfall into the German Ocean. The lower or southern part commences at the same point, and has its outlet into the river Hull, at a place called Wincolmlee, near the town of Hull. It contains 10,432 acres. This Drainage is very effectual, as the water in the drains com- monly stands three or four feet below the surface of the land. The expenses hitherto incurred, amount to 125,0001. or about ten pounds fifteen shillings per acre; and the average improvement of the land is about fourteen shillings per acre per annum ; but as the an- nual expense of keeping the drains and works in repair, amounts to between twelve and thirteen hundred pounds, the clear interest upon the capital expended, is reduced to little more than 51. per cent. per annum; hence it appears not to have been a very profitable under- taking to the proprietors. The public, however, has been benefited by land being brought into cultivation, which before was of little value. The Keyingham Drainage is divided into two parts, having distinct outlets. The eastern part commences at Sand-le-Mere, and dis- charges itself into the Humber, at a place called “ Stone-Creek ;” and is in length about fourteen miles. The other part lying a little to the north and west of the former, commences at Aldborough, and empties itself also into the Humber at a place called “ Headon Haven ; " being in length about eleven miles. The Act of Parliament for executing this Drainage was obtained in 1802 ; and the level contains 5,505 acres of land. The whole expenses of the act and the completion of the works, amounted to 30,000l. or about five pounds nine shillings per acre. Prior to the commencement of this Drainage, the Commissioners made a valuation of the low grounds; and subsequently to the com- pletion of it, they made an estimate of the improvements which had been effected, amounting to 2,475l. per annum, which is rather more than SI. per cent.; hence it appears that this has been a profitable undertaking. The Harford and Derwent Drainage contains upwards of 10,500 acres, of which nearly 4,500 are situated in the East Riding, and the remainder in the North Riding. The Act of Parliament for this Drainage, was obtained in 1800; and it has been completed at a total expense of 41,6121. which is PART VI. AN ACCOUNT or DRAINS. 249 nearly at the rate of 41. per acre; but as the land has been benefited to the amount of 17s. per acre; after paying the annual assessment, the disbursements are paying an interest, to the proprietors of about 211. per cent. per annum ; hence it appears that this has been a very advantageous undertaking. Sparkling-Moor and Walling-Fen Drainage was completed about thirty years ago; and has greatly improved about 20,000 acres of land which were before subject to be flooded. It commences at I‘VIarket \Veighton, and enters the H umber near a place called “ Foss- dyke Clough,” situated upon the banks of that river. This Drain also serves as a navigable Canal; and turns out greatly to the advantage of the land-owners and proprietors, not only in draining the lands, but by facilitating the carriage of the produce and manufactures of a large extent of country, which before were con- veyed at a very great expense by land-carriage. i Yore. Those who desire to see more on this subject are referred to the excellent Agricultural Survey of the East Riding of Yorkshire, by H. E. Strickland, Esq. AN ACCOUNT OF AGRICULTURAL IMPROVEMENTS MADE IN THE COUNTY OF LINCOLN, BY MEANS OF DRAINS. THE extensive Drainages which have been executed in different parts of the county of Lincoln are far too many to be enumerated here; but the following extract from Mr. Young’s Agricultural Survey of that County, will give the reader some idea of the great improve- ments that have been made by them: “ It is very probable that no county in the kingdom has made equal exertions in this very im- portant work of draining. The quantity of land thus added to the kingdom has been great; fens of water, mud, wild fowl, frogs, and agues, have been converted to rich pasture and arable ground, worth from 208. to 403. an acre: health improved, morals corrected, and the community enriched. These, when carried to such an extent, are great works; and reflect the highest credit on the good. sense and energy of the proprietors. “ Without going back to very remote periods, there cannot have been less than 150,000 acres drained and improved, on an average, from 5s. an acre to 258. ; or a rental created of 150,0001. a year. But suppose it only 100,000!. and that the profit has, on on average, been received during the period of thirty years; the rental has in that time amounted to three millions, and the produce to nearly ten; and when with the views of a political arithmetician, we reflect on the circulation that has attended this creation of wealth through industry; the number of people supported ; the consumption of manufactures; the shipping employed; the taxes levied by the State; and all the classes of the community benefited; the magnitude and importance M " 250 ELKINGTON’S MODE or DRAINING. PART W. of such works will be seen; and the propriety well understood or giving all imaginable encouragement and facility to their execution. “ Early in the days of republican France, decrees were issued for draining marshes: I do not ask what progress has been made; but I would demand, if any drainages equal to this have been executed in that kingdom during a century? From Bourdeaux to Bayonne, in one of the finest climates in Europe, nearly all is marsh. \Vhat Frenchman has been so actuated by the blessings of republican security, as to lay out one louis on that or any other marsh or bog. “ Undertakings of this kind prove the reliance of the people on the secure possession of what their industry creates; and an English- man must examine his native land with a cold heart, who does not pray for the continuance of a system of legislation which has tended so powerfully to adorn, improve, and cultivate the country, and to diffuse prosperity and happiness through the whole of society. " MR. JOSEPH ELKINGTON‘S METHOD OF DRAINING LAND, APPROVED OF, AND RECOMMENDED BY THE BOARD OF AGRICULTURE. SEVERAL gentlemen have favoured the public with Treatises on the subject of draining land; but Dir. Elkington’s Method appears to be now almost universally adopted. Mr. Elkington was a VVarwickshire farmer, and practised the art of draining land with unexampled suc- cess. The publication of this art was represented, to the Board 01' Agriculture, to be one of the greatest means of promoting the im- provement of this country that could be suggested. In consequence of this, the House of Commons, on the 10th of June, 1795, voted an Address, “ That His Majesty would be graciously pleased to give directions for issuing to Mr. Joseph Elkington, as an inducement to discover his Mode of Draining, such sum as his Majesty, in his Wl5< dom, shall think proper, not exceeding the sum of lOOOl. sterling.” l‘vIr. Elkington’s health being extremely precarious, there was a risk that the public might lose the benefit of the knowledge he had acquired, by the experience of above thirty years, in a species of im- provement, which, in these kingdoms, ought to be considered as the basis of every other. T o prevent so unfortunate a circumstance, the Board resolved to send B‘Ir. John Johnstone, to visit, in company with Dir. Elkington, the principal Drainages he was then executing; and to take Drawings and Sections of the same. The result of that journey was drawn up by 311'. Johnstone; and has been published under the title of “ An Account of the Mode of Draining Land, according to the System practised by Mr. Joseph Elkington.” This Work has been well received by the public; and has already been of great service in promoting agricultural improve- merits. It not only points out the best mode of draining bogs and marshes,- but also treats very c-xttnsively of the most approved me- PART VI. ELKINGTON’S moon or DRAINING. ‘ .251 thods of making under—drains; hence it will be found of infinite ser- vice to every person who is troubled with springy grounds. The following extracts from some of the Agricultural Reports of those counties in England, where Mr. Elkington executed the most remarkable Drainages, will serve to shew the superiority of his method over every other. County of Warwick, by illr. Jo/m Wedge. “ Draining is, without doubt, the first step towards the improve- ment of all wet land. It has been practised with much success in this county for several years; but more particularly so, since Mr. Elkington, a farmer of this district, introduced a method of draining boggy lands, by making deep drains, and boring at the bottom or sides of them, through the difl'erent under-strata, so as to tap the springs, and thereby, in many instances, cure large tracts of land with very few drains.” County of Leicester, by M: John Elan/r. “ The most capital improvements have been made under the direc- tion of a Mr. Elkington, who is suppdsed to be the first in that line in the world. After forming the drain, by beginning at the fall, and working upwards, he makes use of a borer to find the spring, with which he generally succeeds, which has a wonderful effect in draining the land.” County of Worcester, by ill/z W. J. Pomcroy. “ In speaking of under-drains, it may be thought right to mention, that various experiments have been made at Ewell-Grange, the seat of the Earl of Plymouth, and in that neighbourhood; but that by boring, after iVIr. Elkington’s method, deserves to be most particularly noticed.” CounIy of Derby, by BI)‘. Thomas Brown. “'Everykother method seems to bend to that practised by Mr. Elkington, whose practice is becoming every day more extensive, and seems to me the most effectual of all others, for carrying olt'subter- raneous waters. He lays a stone drain from three to six feet. below the surface, in such a direction as to cut the source of the spring, and with such a declivity as to scour itself. Wherever he finds the source of the spring below the level of his drain, he bores, and with such judgment, that, to a stranger, his anger scents possessed of the virtue of that rod with which l‘vloses struck the l‘UL'l-I ; for the water imme- diately gushes out, and perhaps lays land th , before was too wet to carry a sheep, sufficiently dry to carry the hunk-st ox. This method is certainly efi‘ectual against springs.” Some readers may perhaps think that I have expatiated too much on the excellency of Mr. Elkington‘s mode of Draining; but I am of opinion that this science, which tends so powerfully to promote the general welfare, cannot be too much ret-mmrwmled. MG CON IC SECTIONS PART VII. PART VII. CONIC SECTIONS AND THEIR SOLIDS. DEFINITIONS. I. Come Sections are plane figures formed by cutting a cone. According to the different positions of the cutting plane, there will arise five dif- ‘, ferent figures or sections. " 2. If the cutting plane pass through the vertex of the cone, and any part of the base, the section will be a triangle. 3. If a cone be cut into two parts, by a plane parallel to the base, the section will be a circle. 4. Ifa cone be cut by a plane passing through its two slant sides, in an oblique . direction, the section will be an ellipse, or 3: . , ellipsis. 5. If a cone be cut by a plane, which is parallel to either of its slant sides, the section will be a parabola. PART VII. ' AND THEIR SOLIDS. 253 6. If a cone be cut into two parts, by a plane, which, if continued, would meet the opposite cone, the section is called a hyperbola. Note. All the figures that can possibly be formed by cutting a cone, are mentioned in the preceding defi- nitions; and are five in number, viz. a triangle, a circle, an ellipse, a parabola, and a hyper-bola ; but the last three only, are usually denominated (IONIC SECTIONS. 7. The vertex of a conic section, is that point where the cutting plane meets the opposite sides of the cone, as at A and B ; hence the ellipse and the hyperbola have, each of them, two vertices ; but the parabola has only one vertex. 8. The axis or transverse diameter of an ellipse or a hyperbola, is the distance AB between the vertices; and the axis of a parabola is a right line drawn from the verv tex, so as to divide the figure into two equal parts, as Mn. 9. The conjugate diameter of an ellipse or hyperbola is a right line drawn through the centre of the figure, perpendicularly to the transverse; hence the semi-con- jugate diameter of a hyperbola is the perpendicular dis- tance from the centre of the figure, or middle of the transverse diameter AB, to the verticcs of the opposite cones. 10. An ordinate is a right line drawn from any point in the curve, perpendicularly to the axis. 11. An absciss is that part of the axis which is con- tained between the vertex and the ordinate. 12. The parameter of any diameter is a third propor- tional to that diameter and its conjugate. The parameter is sometimes called the latus rectum. 13. The focus is a point in the axis where the ordinate is equal to half the parameter. 14. The ellipse and hyperbola have each two foot: but the parabola has only one focus. 15. A spheroid or ellipsoid is a solid generated by the revolution of an ellipse about one of its diameters. If the revolution be made about the transverse diameter, the solid is called a prolate spheroid ; but if about the conjugate diameter, an oblate spheroid. 25-1 CONIC SECTIONS PART VII. 16. A conoia’ is a solid formed by the revolution of a parabola, or hyperbola, about its axis; and is accordingl‘,r called parabolic, or hyperbolic. — i The parabolic conoid is also called a paraboloid ; and the hyperbolic conoid, a hyperboloid. 17. An elliptic, a parabolic, or a hyperbolic spindle, is a solid formed by the revolution of a segment of an ellipse, a parabola, or a hyperbola, about its double ordinate, which remains fixed. PROBLEM I. In an ellipse, to find tile transverse, or conjugate, or ordinate, or absciss ; having the other t/zree given. CASE I. FVleen the transverse, conjugate, and (1630533 are giren, (4) find the ordinate. NILE. As the transverse diameter is to the conjugate, so is the square root of the product of" the two abscisses, to the, ordinate. Nate 1. An ellipse may be constructed by Problem 15, Part I. ; its area found by Problem ‘21, Part IL; and its circumference by Note 3, of the same Problem. Also, the area of an elliptical segment may be obtained by Problem ‘22, Part II. 2. Au ellipse may also be constructed in the following manner: Haringr found the loci F, f. as directed in Problem 15, Part 1., take a thread equal in length to the transverse diameter AB, and fasten its ends, with two pins, in the points i“, t; then stretch the thread to its greatest extent; and by moving a pencil round. within the thread, keeping it always tight, you will trace out the curve of the ellipse. This construction is founded on the second pmpcrty of the ellipse, given at the end ct this Problem. EXAMPLES. 1. In the ellipse ABCD, the trans- C ‘ verse diameter AB is 60, the con— P jugate diameter CD 20, the absciss A ‘39}; BE 12, and the absciss AB 48 ; “DJ/V what is tne length of the ordinate 32F .9 As 60 : 20:: V(48 X12) .' 20+60,\//(48 X12)::;0M57i_i =1— >< 24:8:EF, tlze ordinate required. 2. If the transverse diameter be 100, the conjugate 7.3, and the less absciss 2O ; what is the ordinate ? Ans. 3t). PART vn. AND THEIR SOLIDS. 255 CASE. II. When the transverse, conjugate, and ordinate are given, to find the absciss. :ULE. As the conjugate diameter is to the transverse, so is the square root of the difference of the squares of the semi-conjugate and ordinate, to the distance of that ordi- nate from the centre. Then this distance being added to the semi-transverse, will give the greater absciss; but if subtracted from it, the remainder will be the less absciss. EXAMPLES. l. The transverse diameter AB is 60, the conjugate diameter CD 20, and the ordinate EF 8; What are the abscisses AE and BE? As 20: 60 I: 4/ (102— 89) : 60 + 20 ¢(109—8)9: 3 ,._/ (100—64) =3 4/ 86 = 3 x 6 = 18.: EG, the distance from the centre. Then 30+18=482AE, the greater absciss; and 30 ——18=l2:BE, the less absciss. 2. What are the two abscisses to the ordinate 30, the diameters being 100 and 75 ? Ans. 20 and 80. CASE III. PV hen the conjugate, ordinate, and absciss are given, to find the transverse. RULE . From the square of the semi-conjugate subtract the square of the ordinate; and extract the square root of the remainder. Add this root to the semi-conjugate, if the less absciss be given; but when the greater absciss is given, subtract this root from the semi-conjugate; and reserve the sum or difference. Then say, as the square of the ordinate is to the rectangle of the conjugate and absciss, so is the reserved sum or diti‘erence to the trans- verse diameter. EXAMPLES. l. The conjugate diameter CD is 20, the ordinate EF 8, and the absciss BE 12 ; required the transverse diameter AB. 256 CONIC SECTIONS PART VII. Here “002—89): ¢(100 — 64): V3626, and 10 + 6:16; then, as 82 : 20x12:: 16 ; (20x12x16)—:-83= (20 X 12 X 2)+8:(5 X12 X 2)+2=:5 x 12:60, the trans- verse diameter required. 2. If the conjugate diameter be 75, the ordinate 30, and the greater abseiss 80; what is the transverse ? Ans. 100. CASE IV. When the transverse, ordinate, and absciss are give I, to find the conjugate. RULE. As the square root of the product of the two abscisses is to the ordinate, so is the transverse diameter to the conjugate. Note. The transverse diameter may be found from the conjugate, in the same manner; using the abscisses of the conjugate, and their ordinate perpendicular to the conjugate. - EXAMPLES. 1. The transverse diameter AB is 60, the ordinate EF 8, and the absciss BE 12; required the conjugate diameter CD. Here 60~12=48=AE, the greater absciss, and ,/(48 x12)= «576224; then, as 24 : 8: :60 : (8 x 60)—:—24: 60+3=20, the conjugate diameter required. 2. If the transverse diameter be 100, the ordinate 30, and the absciss 20; what is the conjugate diameter ? Ans. 75. PROPERTIES OF THE ELLIPSE. 1. The square of the distance of the focus from the centre, is equal to the differ- ence of the square of the semi-axes. 2. The sum of two lines drawn from the foci, to meet at any point at the curve, is equal to the transverse axis. 3. The parameter, or double ordinate at the focus, is equal to the square of the conjugate diameter divided by the transverse. 4. The rectangles of the segments of any diameter, are as the squares of their or- dinates. 5. All the parallelograms circumscribed about an ellipse, are equal to one an- other, and each equal to the rectangles of the two axes. 6. An ellipse is to the rectangle of the two axes, as any circle is to the square of its diameter. 7. The areas of ellipses are to one another, as the rectangles of their transverse and conjugate axes. 8. As the transverse axis of an ellipse is to the conjugate, so is the area of a circle whose diameter is the transverse. to the area of the ellipse. 9. The square of the transverse diameter is to the square of the conjugate, as the rectangle of the abscissa is to the square of the ordinate. 10. The area of an ellipse is a geometrical mean proportional number between the area of two circles whose diameters are equal to the transverse and conjugate diameters of the ellipse. PART v11. AND THEIR SOLIDS. 257 PROBLEM II. To describe a parabola, the absciss CD, and the ordinate BD being-r given. Construction. Bisect the ordinate BC in a; join Ca, and draw a n per— pendicular to it, meeting CD pro- duced in n. Make CE and CF each equal to Dn ; and F will be the focus of the parabola. Take any number of points, r, r,. &c. in the axis, through which draw the indefinite, double ordinates, srs, srs, 8:0. perpendicularly to CD. With the radii EF, Er, 8:0. and the centre F, describe arcs cutting the corresponding ordinates in the points, 3, s, &c. ; then draw the curve through all the points of inter— section, and ACB will be the parabola required. Note. Various methods of constructing parabolas and hyperbolas, may be seen in Emerson's Conic Sections. PROBLEM III. Toflnd any parabolic absciss or ordinate. RULE. As any absciss is to the square of its ordinate, so is any other absciss to the square of its ordinate. Or, as the square root of any absciss is to its ordinate, so is the square root of any other absciss to its ordinate; and vice versa. EXAM PLES. 1. The absciss CE is 9, and its ordinate DE 6 ; what is the ordinate AG, whose absciss CG is 16? As 9: 62:: 16: (36x16)+9= 5764—9264 ; and4/64282AG, the D E F ordinate required. Or, as“ 9 Z 6:: M16 1 (6X V16) +4/9=(6x4)—:—3=24-:—3=8= AG, the same as before. A G O C. . 258 001110 SECTIONS PART VII. 2. The ordinate AG is 8, the ordinate DE 6, and the absciss CG 16; what is the abseiss CE? As 8'3: 16::6‘3 : (16 X 6'3) +83=(16 >< 36)-—I—6~l:576+ 64:9:CE, the absciss required. 07“, as 8: M16226 2 («/16x6)—:—8=(4 x6) +8221 +823, - and 3 X 32920143, the same as before. If an absciss be 12 , and its 01 dlnate lo , 1equired tin ordinate Whose absciss 1s 27. A718. 22' 4. If an absciss be 18, and its ordinate 12, what is the abseiss Whose ordinate is 16? Ans. 32. PROBL CM IV. Tofind the length of the are (3)“ a parabola, cut qfi' by a douhle ordinate. tULE. To the square of the ordinate add f of the square ol’ the absciss ; multiply the square root of the sum by 2, and the product will be the length of the curve required. Note. This rule will not hold good. when the absciss is greater than half the or- dinette , that is, when the ordinate falls below the focus. EXAMPLES. 1. If the absciss CE be 6, and the ordinate DE 12 ; .hut is the length of the curve DCF? Ifere 1‘23—l—6- ><. 1:1/‘4—1—(36 X4)—j—3:l H—l—Hia— 1-14+48:l92 , aid 4/ i913 9213.8 )\ ix 2 I: 27. the length of the cane 7egz/i1ed. 71:23 .Vtht is the length of a parabolic curve .‘ihoi'e absciss 1s 7, and ordinate 15? Am. 31.071" 3. PROB EM Y. Tofind the area ofa parabola, its base and height being given. RULE. Multiply the base bV the in-ig; 1t and - ot the product will 1 e tl 1 n ‘2 .3 It. It 1. fl\ EXAMPLES. I /-m1 emfi 1. “That is the area of the pam— / bola, ABC, whose hnght 11D is l‘o’ __ and base or double ordinate AB 21 P 1‘ -’ PART VII. AND THEIR SOLIDS. 259 Here 24x 18 32(24. X 18 x2)—:—3= 8.x 18 x2: 288, z’lae area required. 2. The absciss is 24, and the double ordinate 72 ; What is the area of the parabola? Ans. 1152. PROBLEM VI. To find the area (y‘a parabolic frustam RULE . Divide the difference of the cubes of the two ends of the frustum, by the difference of their squares; multiply the quotient by 3.— of the altitude, and the product will be the area EXAMPLES. 1. What is the area of a parabolic frustum AEFC, the end AC being 2—1, EF 16, and the altitude DG 10? Here (243 —.163) + (24:2 —— 163)=(13824 - 4096) —:- (576 ~~256) = 9728 + 3202304; and 30.4 X (10 x2)—=.— 3: (30.4 X 10 x 2) +32608+3=202§, the area required. 2. Required the area of a parabolic frustuin, the greater end of which is 20, the less end 12, and the height 6. Ans. 9Q. PROPERTIES OF THE PARABOLA. l. The distance of the focus from the vertex is equal to the square of any ordi- nate divided by 4 times its absciss. 2. The parameter, latus rectum, or double ordinate at the focus, is equal to the square of any ordinate divided by its absciss. :5. The distance between the focus and the vertex is equal to % of the parameter, or g. of the ordinate at the focus. 4. As the parameter is to the sum of any two ordinates, so is the difference of those ordinates to the difference of their abscisses. 5. All parabolas are similar to each other. #5. The area of a parabola is equal to g of the area of the circumscribing parallel- “gram. 7. The square of any two ordinates are to each other as their corresponding ab. susses. S. Abscisses are to each other as the squares of their ordinates. PROBLEM ‘VII. To construct a leg/perbola, having the transverse and con- jugate diameter given. Construction. Draw CD equal to the conjugate dia- meter; and bisect it perpendicularly with a transverse diameter AB. From E, the middle of the transverse 260 CONIC SECTIONS rAnr VII diameter, with the radius EC or ED, describe the circle CfDF cutting AB produced in F and f, which points will be the foci of the liyperbola. In AB produced, take any number of points 772, m, &c.; and from F and f, as centres, with the distances Bm, Am, as radii, de- scribe arcs cutting each other in the points .9, s, &0. Through these points draw the curve sBs, and it Will be the hyperbola required. Nole. If the straight lines EC. ED, be drawn from the point E, through the ex- tremities of the conjugate axis CD, they will he the asymptotes of the hyperbola; the property of which is to approach continually to the curve, but not to meet it, until they be infinitely produced. (See Emerson's Conic Sections, Book II. Prob. 75.) PROBLEM VIII. In a hyperbola, to find the transverse, or conjugate, or ordinate, or absciss. CASE I. The transverse and conjugate diameters, and the two abscisses being given, to find the ordinate. RULE. As the transverse diameter is to the conjugate, so is the square root of the product of the two abseisses to the ordinate. Note. The less absciss added to the transverse diameter, will give the greater absciss. EXAMPLES. 1. In the hyperbola ABC, the B transverse diameter is 30, the eon- / juvate 18, and the absciss BD 10; \ what is the ordinate AD ? 2 As 30:18::‘/(40x10);18+ A U U 30¢ (40x 10) = (3 x w 400) + 5 = (3 X 20)+5 :3 X 4212, the ordinate required. 2. The transverse diameter is 48, the conjugate 42, and the less absciss 16; What is the ordinate? Ans. 28 i’ART VII. AND THEIR SOLIDS. 0 261 CASE II. The transverse and corg'agate diameters, and an ordinate being given, to find the abscisses. RULE . To the square of half the conjugate, add the square of the ordinate; and extract the square root of the sum. Then say, as the conjugate diameter is to the transverse, so is the said square root to half the sum of the abscisses, or the distance between the ordinate and the centre. Then this distance being added to half the transverse diameter, will give the greater absciss; and their differ- ence will be the less absciss. EXAMPLES. 1. The transverse diameter is 30, the conjugate 18, and the ordinate 12; What are the two abscisses? Here x/(9?+122)= «(81+ 144): v225=15; then, as 18 I 30::10 2 (30x15) —:- 18:(5X15)—:-3=5X5=25, half the sum of the abscisses ; hence, 25 + 15:40, the greater absciss ; and 25-15::10, the less abseiss. 2. If the transverse diameter be 48, the conjugate 42, and the ordinate 28; What are the two abscisses? Ans. 64 and 16. CASE III. The transverse diameter, the two abscisses, and the ordi- nate being given, to find the conjugate. RULE. As the square root of the product of the two abscisses is to the ordinate, so is the transverse diameter to the conjugate. EXAMPLE S. 1. The transverse diameter is 30, the ordinate is 12, and the two abscisses are 40 and 10; required the con- jugate diameter. Here «(40X IO): M400=9O, the square root of the product of the two absez'sses; then, as 20 : 12::30 : (12 262 * CONIC SECTIONS PART VII. x 30) —:- 202(12 X G) -—:— 4:3 X 6218, the conjugate re— quired. 2. The transverse diameter is 48, the ordinate 28, and the abseisses are 64 and 16 ; required the conjugate diameter. Ans. 42. CASE IV. The conjugate diameter, the ordinate, and the two abscisses being given, to find the transverse. RULE. To the square of half the conjugate add the square of the ordinate, and extract the square root of the sum. To this root add half the eonj ugate, when the less absciss is used, but subtract it when the greater absciss is used; and ‘reserve the sum or difference. Then say, as the square of the ordinate is to the product of the abseiss and conjugate, so is the sum or difference, above found, to the transverse required. I EXAMPLES. 1. The conjugate diameter is 18, the ordinate is 12, and the less absciss 10 ; required the transverse diameter. Here x/(92+122)= «(814-144): ,/225=15; and 15 +9224. Also, 12 X 12: 144, the square of the ordinate,- and 18x10=180, the product of the absciss and con- iugate; then, as 144: 1801224 : (180 X24)+l44=(15 >< 24)+l2 :_ 15 x 2:30, the transverse required. 2. The conjugate diameter is 42, the greater absciss is 64, and the ordinate 28 ; what is the transverse diameter? Ans. 48. PROBLEM IX. To find the length of a hyperbolic carve. RULE. To 21 times the square of the conjugate, add 9 times the square of the transverse; and to the said 21 times the square of the conjugate, add 19 times the square of the transverse; and multiply each sum by the absciss. T o PART VII. AND THEIR SOLIDS. 263 each of these two products add 15 times the product of the transverse multiplied by the square of the conjugate ; then say, as the less of these sums is to the greater, so is the double ordinate to the length of the curve, nearly. EXAMPLES. 1. The transverse diameter of the hyperbola ABC, is 30, the conjugate 18, the absciss BD 10, and the double ordinate AC 24; what is the length / _: of the curve ABC ? 31““1) c erre (181' x 21) + (303 x 9) = (324 X 21) + (900x 9): 6804+8100 214904291 times the square (3)" the cory'ugate, added to 9 times the square of the transverse ; and 6804 +(900 X 19) =6804+ 17100: 23904221 times the square of the conjugate, added to 19 times the square of the transverse; then 14904 X 10: 149040, and 23904 X 102939040. Again, 324 x 30 x 1 72145800215 times the product of the transverse multiplied by the square of the conjugate; and 149040+145800=294840, and 239040+145800= 384-840; then, as 294840 3 3848402224 1 31.326007, the length of the curve required. 2. The transverse diameter of a hyperbola is 45, the conjugate 27, the absciss 15, and the double ordinate 36 3 what- is the length of the curve ? Ans. 46.989. PROBLEM X. Tofind the area of a hgperhola, the transverse, conjugate, * and absciss being given. RULE. Multiply the transverse by the absciss ; to the product add 4} of the square'of the abseiss; and multiply the ‘square root of the sum by 21. Add the product, last found, to 4 times the square root of the product of the transverse and absciss; and divide the sum by 75. Di- vide 4 times the product of the conjugate and absciss by the transverse; then this quotient being multiplied by the former, will give the area, nearly. 264 come SECTIONS PART VII. EXAMPLE S. 1. Required the area of a hyperbola, whose transverse diameter is 30, the conjugate 18, and the absciss 10. Here 214/(30 X 10 +103 x?) 2 21 1/ (300 + 100 X i) 2 21 4/(300-1- 500 —:— 7): 21 4/(300 + 71.428571): 214/ 371.428571221x19.2722404.712. Again, (4 4/ (30 x 10) + 404.712) —-:— 752(4 M 300 + 404.712) + 752(4 >< 17.3205 +404.712) —:— 752 (69.282 + 404.712)—:—752473.994+7526.3199. Lastly, (18 X10 x 4) —:— 30 X 6.3199 2 (18 X 4) —I— 3 x 6.319926 X 4 X 6.3199224 x 6319921516776, the area required. 2. The transverse diameter of a hyperbola is 50, the conjugate 30, and the absciss 25 ; what is its area? Ans. 805.0902. PROPERTIES OF THE HYPERBOLA. 1. 'lhe square root of the sum of the squares of the semLaxes, is equal to the distance of the focus from the middle of the transverse axis. 2. If half the transverse axis be subtracted from the distance between the focus and the middle of the said axis, the remainder will be the distance of the focus from the vertex. 3. The difference of two lines drawn from the two foci to any point in the curve, is equal to the transverse axis. 4. The parameter, latus rectum, or double ordinate at the focus, is equal to the square of the conjugate divided by the transverse. Note. Those who desire to obtain a complete knowledge of the properties of (ionic Sections, are referred to the works of Emerson, Hutton, and Simsou, on that subject. PROBLEM XI. Tofind the solidity qfa spheroid. RULE. Multiply the square of the revolving axis by the fixed axis, and the product thence arising by ." 236 ; and the last product will be the solidity. Note 1. The multiplier .5236 is 3‘; of 3.1416. 2. A semi—spheroid is equal to § of a cylinder, or to double a cone of the same base and altitude. EXAMPLES. 1. What is the solidity of the pro- late spheroid ABCD, whose trans- verse or fixed axis AC is 50, and the conjugate or revolving axis BD 30? 11mm 30'} X 50 X .5236 = 900 X 50 X .5236245000 >< .5236293562, tile solidity required. 0 PART v11. AND THEIR series. 265 2. \Vhat is the content of an oblate spheroid, Whose axes me 50 and 30. 9 Am. 39270. 3. Requhed the solidity of 21 11101.1 te s'pheroid, whose axes ale 45 and 95. Ans.100727.5 PROBLEM X11. T 0 find the solidity oft/1e segment Igfa spheroid. ‘ CASE 1. ”71622 the base is circular, or parallel to the revolving axis. RULE. From triple the fixed axe, take double the height of the segment, multiply the difi‘e1ence by the square of the height, and the p1 oduet thence arising by .5236; then say, as the square of the fixed axe is To othe square of the levolving axe, so is the last produrt to the solidity 1e— quiied. EXAMPLES. 1. Required the solidity of HM segment ABC of 21 p1 olate sphe1oit1. the height Bm being 5, the fixed axe, BD 50. and theo reV 01V 111g a\~: EF 30. Here (50 x 3) —(5 x 2) X 5'-‘ x .5236 = (150 —- 10) X 25 X .5236 = 110 X 25 X .5236: 3500 x .5236: 1832.6 , then, as 2500: 900 :: 1832.6 2 65 9. 736, the solidity required The axes of 21 pl olate sphetoid an: 100 and 60; what is the solidity of that segment Whose 11.1,:igl1t is 10, and its base parallel to the revolving axe ‘r Ans. 5277.888. 3. The axes of an oblate spheroid are 50 and 30; what is the solidity of that segment whose height is 6, and its base parallel to the revolving axe I" Ans. 4084.08. o l I : I CASE 1L When the base is elliptical, or perpendicular to the re- rolving are. RULE. From triple the revolving axe, take double the height of the segment ; multiply the di-flierence by the square of N 266 come SECTIONS PART vm the height, and the product thence arising by .5236; then say, as the revolving axe is to the fixed axe, so is the last product to the solidity required. EXAMPLES. 1. What is the solidity of the segment ABC of a prolate spheroid, the height Bm being 6, the fixed axe EF 50, and the revolving axe BD 30 P Here (30 X 3)—(6 X 2») X 62 X .5236: (90—12) x 36 x .5236 = 78 x 36 X .5236 = 2808 X .5236 :14702688 ; then, as 30 : 00 :2 14702688 2 2450.448, the solidity required. 2. The axes of an oblate splicroid are 50 and 30; what is the solidit}r of that segment whose height is 6, and its base perpendicular to the revolving axe ? Ans. 156054688. PROBLEM XIII. 'ofind the content of the middlefi'ustmn ofa spheroicl. CASE I. Ithn the ends are circular, or parallel to the revolving awe. RULE. To twice the square of the middle diameter, add the. square of the diameter of one end ; multiply the sum by the length of the frustum, and the product thence arising by .2618, for the solidity. Nate 1. A cask in the form of the middle frustum of a prolate spheroid is by Gaugers called a cask of thefirsl rarl'rty ; hence this llule is useful in cask gauging. 2. The old ale gallon contains 282.'the old wine gallon 231, and the old Win— chester bushel 2150.4 cubic inches—The new imperial gallon contains 277.274. and the new imperial bushel 2218.192 cubic inches. (See the New Imperial Tables, Part VIII.) 3. 1f the content of any vessel, in cubic inches. be divided by 277.274, and 2218.192; the respective quotients will be the content in imperial gallons and bushels. (Sec Gauging, Prob. 1, Part VIII.) ' EXAMPLES. I. “That is the solidity of the middle frustum EGHF of a prolate spheroid; the middle diameter CD being 30, PART v11. AND THEIR SOLIDS. 267 the diameter of each end EF or GH 18, and the length AB 40 ? * IIere ((302 X 2) + 182) X 40 X .2618 = ((900 x 2) + 324) x 40 x A 6—- .2618:(1800+324)><40><.2618= _ 2124 x 40 x .2618—:84960 X .2618: 122242.528, the solidity required. 2. XVhat is the solidity of the middle frustum of an oblate spheroid; the diameter of each end being 40, the ‘ middle diameter 50, and the length of the frustum 18 ? Ans. 31101.84. 3. The length of a spheroidal cask is 30, the bung dia- meter 24, and the head diameter 18 inches; What is its content in imperial gallons? Ans. 41.088 gallons. CASE II. When the ends are ellipticdl, or perpendicular to the ' revolving axe. RULE. To twice the product of the transverse and conjugate diameters of the middle section, and the product of the transverse and conjugate of one end; multiply the sum by the height of the frustum, and the product thence arising by .2618, for the solidity. EXAMPLES. 1. In the middle frustum EFGH, \ F of an oblate spheroid, the diameters ________ of the middle section AB are 50 A --------- B and 30; those of the end EF or GH 40 and 24; and its height nm 18; what is the solidity ? IIere (50 X 30 X 2) + (40 x 24) X 18 x .2618 = (3000+ 960) x18 x .2618 = 3960 X18 x .2618: 71280 x .2618 2 18661.104, the solidity required. 2. What is the solidity of the middle frustum of a pro- 1ate spheroid; the diameters of the middle section being 100 and 60, those of each end 80 and 48, and its height 36? Ans. 149288.832. N2 268 CONIC SECTIONS PART v11. PROBLEM XIV. To find the solidity of an elliptic spindle. RULE. From 3 times the square of the middle or greatest dia- meter take 4 times the square of the diameter at 4 of the length, or equally distant between the middle and one end ; and from 4 times this last diameter take 3 times the middle diameter; and i of the quotient arising from dividing the former difference by the latter, will give the central distance. Find the axes of the ellipse ; and also the area of the ge nerating segment. Divide b3 times this alea by the length of the spindle; 1111111 the quotient subt1act the middle diameter; and multiply the 1e1nainde1 by 4 times the cent1a1 distance, before found. Subtract this ploduct from the squaie of the middl- diameter; multiply the remaindei 1)] the length of the spindle, and the p1oduct thus obtained b f 52116, 101 the solidity. EXAMPLES. 1. “711% is the solidity of the 7. elliptic spindle ACBDA, the length ‘ , AB being 40, the middle diameter CD 12, and the diameter EF, ‘at_1 1 “1 of the length, 9. 49546. '9 Here (122 X 3— 9.495463 X 4) ~1— (9.49546 X 4 —— 12 X 3) X ,1, : (432 — 360.0546) —:— (37.98184 — 36) X::(71.345’1—2—1.98184>X Jr: 36 X 1:9, the central distance 0G. Also, (9+ 12—:— 2) X 2:(9+6) X 2:15 X ”=30, the con— jugate diameter CH. By Problem 1, Case 4, Part VII, we have 30— 6:24 =GH, the greater absciss; and ¢(24 X 6): M144:l2; then, as 12 : 20 (AG) 2230 (CH) : 50, he transze'lse diameter KL. Again, by Problem 22, Part II, we have (Se—30:2, the tabular height ; and the corresponding Area Seg. 2'3 .111823; then. 111823 X 50 X 30 :5. 591 15 X 30: 167.7345, the area of the generating segment ACB. 1‘17010, (167 7345 X 3 -+-40——12) X 9 ,X 4: (503. 2035+ 40 PART VII. AND THEIR SOLIDS. 269 - 12) x 36 = (12.5800875 —l2) X 36 = .5800875 X 36 =20.88315; Vt/zen (123—2088315) X 40 X .5236: (144 —- 20.88315) X 40 X .5236 : 123.11685 x 40 X .5236 :2 4928.674 X .5236:2578.5593064, the solidity required. 2. The length of an elliptic spindle is 80, the middle diameter 24, and the diameter at i- of the length 18.99094; what is the solidity P Ans. 206284744512. PROBLEM XV. To find the solidity of a parabolic conoz'a’. RULE. Multiply the square of the diameter of the base by the altitude, and the product thence arising by .3927, (one half of .7854 g) and the last product will be the solidity. Note 1. This Rule will give the solidity of any segment of a paraboloid, whose base is a circle. ‘2. A paraboloid is equal to half of its circumscribing cylinder. EXAMPLES. 1. What is the solidity of the para— boloid ADB ; the height Dm being 42, and the diameter of the base AB 24 ? flare 242 x 42 X .3927 = 576 X42 X .3927 : 24192 X.3927:9500.1984, the solidity required. 2. What is the solidity of a para- boloid, the height of which is 30, and the diameter of its base 50 ? Ans. 29452.5. 3. Required the solidity of the segment of a paraboloid, the height of which is 9, and the diameter of its base 12. Ans. 508.9392. PROBLEM XVI. Tofind the solidity of t/zefrustum 0f aparaboloid, whey: its ends are pewendz'czdar to the axis of tile solid. RULE. Multiply the sum of the squares of the diameters of the W0 ends by the height of the frustum, and the product N 3 270 come SECTIONS PART v11. thus obtained by .3927 ; and the last product will be the solidity. Note. If the lower frustums of two equal paraboloids be joined together at their greater ends, they will form a figure which, by Gaugers, is called a cask of the third vane/y,- hence this Rule may be applied in Gauging. EXAMPLE S. 1. What is the content of the parabolic frustum ABCD; the dia- meter AB of the greater end being 30, that of the less end DC 24, and the height mn 9 ? Here (309+242) X 9 X .3927=(900 A m +576) x9 x.3927=1476 x9x.3927 \ , M, 213284. X .3927: ’2166268, file content of thefrustmn. 2. What is the solidity of the frustum of a paraboloid; the diameter of the greater end being 80, that of the less end 40, and the altitude 45 ? Ans. 141372. 3. The length of a cask, in the form of two equal frustums of a parabolic conoid, is 30, the bung diameter 24, and the head diameter 18 inches; what is its content in imperial gallons ? Ans. 38.2397 gallons. PROBLEM XVII. To find the solidity ofa parabolic spindle. RULE. Multiply the square of the middle diameter by the length of the spindle, and the product thence arising by A1888, and it will give the solidity. Note. A parabolic spindle is equal to 183 of its circumscribing cylinder ; hence we obtain the multiplier .41888, which is 133 of .7854. EXAMPLES. 1. The length AB of the parabolic Spindle ACBD is 81, and the middle diameter CD 27; what is its solidity? Here 272 x 81 x .41888 : 729 X 81 X .41888 = 59049 X .41888 = 24734.44512, the solidity required. 2. What is the solidity of a parabolic spindle, whose length is 20, and middle diameter 8? Ans. 536.1664. ram VII. AND THEIR SOLIDS. 271 PROBLEM XVIII. To find the solidity 0f the middle frustum of a parabolic spindle. RULE. Add 8 times the square of the middle diameter, 3 times the square of the end diameter, and 4 times the pro- duct of those diameters into one sum; then this sum being multiplied by the length, and the product thence arising by .05236, (one—fifteenth of .7854) will give the solidity. Note. A cask in the form of the middle frustum ofa parabolic spindle, is called a cask of the second variety,- and is the most common of any of the varieties ; hence the above Rule is very useful in Cask Gauging. EXAMPLE S. 1. What is the solidity of the middle frustum ABCD, of a para- bolic spindle; the diameter of the end AB being 12, the middle dia- meter EF 16, and the length mn 20? Here (162 X 8)+(l2‘-’ X 3)+(16 X 12 x 4)=256 X 8 +144 X 3+192 X4: 2048+432+768 :2 3248 ; and 3248 X 20 X .05236: 64960 x .05236: 3401.3056, the solidity required. 2. What'is the solidity of the frustum of a parabolic Spindle, Whose length is 25, middle diameter 20, and end diameter 15? Ans. 6643.175. 3. The length of a cask, in the form of the middle frustum of a parabolic spindle, is 30, the bung diameter 24, and the head diameter 18 inches; What is its content in imperial gallons? Ans. 41.4009 gallons. PROBLEM XIX. Tofind the solidity of a hyperboloid. RULE. To the square of the radius of the base, and the square of the diameter in the middle between the base and the vertex; then this sum being multiplied by the altitude, N 4 - 'SM'QZLZ‘E” 'SE‘V 51(11p1103 0113, 31 11311111 f 871 0p1111111; 0111 pm: ‘179 10101111111) 0111111111 0111, ‘175 p110 $301 0111 10 110101111111) 0111 ‘0} s1 11111181111 011011.101LC11 :3 ‘10 11110 10180113 0113, }0 110101111311) 01111‘ ”5, 'umgsauf ml} 011) 523329103 «9212 ‘z;‘0199=0039' X 00501 29852“ X 15 X 9517 “8712 3951:6135 +08 + 001:2“ + z9 + :01 WE . (113 3.5 0131111310 0111 13111; 1L 1 um 10101111111) 0113101111 0113, ‘51 p110 $901 0111 30 (10 10101111311) 011], ‘Qg: 33111011 11110 101130.115 0111 ‘10 EIV .1010111 #12111 0111 ‘GOEIV 1111113111} 0110111011101 0111 10 1€11p1103 0111 81 11311111 '1 'SIi'IJNVXCI 111111109 0111 0113 111111 ‘gggg' £01 311131.11: 00110111 1011mm 0113, 13111; ‘01)1131111; 011], A11 1)011d11111111 3111011 11mg 3111; 11011; €.101011.11211) 0113mm 0111 J10 0.1311103 0113, ppm ‘1.103,0111131p-111108 1131101 13111: 3501130153” 011; 30 samnbs 0113, 10 11mg 0111 OJ. "in: 21 'p_zozoq.cadfiy 7) J10 111229572.1f3711J1’0 xfigpggos 01/2 pug/'01]: 'KX I‘i'zirIEIOUJ ‘QL'EEQH‘FQ 'WV M1111 —11OS 0111 S1 11211111 E501 10101111211) 0113mm 0111 1111!; ‘SL assu<1 0111 30 su1p1211 0111 ‘0; S1 14010111011161 1; ‘10 01311111112 0111‘ -5 WWW”. ”72221208 9712 ‘88'08683=9959' X 00891 = 9869' X 95 X 588[ mm 55881 = 9911 + 949201: + ,-95 ME £178 mu 10101111211) 0111111111 0113, pm: ‘95 031111 0113, '10 .11] 811113131 0111 ‘95 8111011 .10 0pm1111; 0111 CEIQV 131010q.10(LC11 9111 30 X41011” 0111 8'1 111311111 '1 'SFI’IJI‘EVXEI 11.110103 0r“ 0111?? 11111 ‘9959' £01 130111131110 511111 qonpmd 0111 p111: '1111 11111111 sxomoas 01x00 2.12; PART VII. AND THEIR SOLIDS. 273 PROBLEM XXI. Tofind the solidity of a hyperbolic spindle. RULE . To the square of the greatest diameter, add the square of double the diameter taken at 41 of the length ; then this sum being multiplied by the length, and the product thus obtained by .1309, will give the solidity, nearly. Note VS hen great accuracy is not lequired, this Rule may be used for any spindle formed b1 the 1'e1olution of a tonic section, or part of a conic section, about its axiS' , as it 11 ill always "he nearly the true solidity. EXAMPLES. 1. What is the solidity of a hyperbolic spindle, Whose length is 36, the greatest diameter 24, and the diameter at 10f the length 16.73338 ? Here 242 + (16.73338 x 2)?= 249+ 33.466762=576 + 11200240249 : 16960240249 ; then 16960240249 x 36 x .1309 = 610568648964 X .1309: 7992.34361493876, the answer required. .The 1e 1gth of a hyperbolic suindle is 30, the gre gatest diameter 20, and the diameter at 1_ of the length 13.94449; What is the solidity? Ans. 4625.20177. PROBLEM XXII. Toff/2d the solidin of the middle frustum of an elliptic or a layperbolic spindle. 11111.1“ To the sum of the squares of the greatestand least dia— meters, add the square of double the diameter taken exactly in the middle between them; and this sum being multiplied by the length, and the product again by .1309, will give the solidity. \11te. This Rule 11ill also give nearly the content of am frustum or segment 11 1"1ned by the revolution of 1 (101110 section. 01' part or a CW 110 seition, either about 1'1 10 axis of the section, or about (1111 other line EX AMPLE S. l. The length of the middle frustum of an elliptic spindle is 60, the greatest diameter 48, the least dia- meter 36, and the diameter in the middle between them 45.23635; what is the solidity? N 1) 274 CONIC SECTIONS AND THEIR SOLIDS. PART VH. Ifere 482+ 369 + (45.23635 x 2)2= 4894—3694—9047273 : 2304 +1296 +8185.30944529=1178530944529 ; i/m/ 1178530944529 X 60 x .1309: 7071185667174 x .1309 292561.82038330766, tile solidity required. 2. The length of the middle frustum of a hyperbolic spindle is 40, the greatest diameter 32, the least diameter 24, and the diameter in the middle between them 29; what is the solidity? A728. 25991.504. 3. What is the content of the middle frustum of any spindle; the length being 50, the greatest diameter 40, the least diameter 30, and the diameter in the middle be- tween them 37.69696? Ans. 53565871567. 4. What is the content of the segment of any spindle; the length being 15, the greatest diameter 12, and the middle diameter 9? Am. 918.918. \ Cu PART VIII. GAUGING. ' 27 PART VIII. GAUGING. GAUGING is the art of finding the capacities or contents of all kinds of vessels used by Maltsters, Brewers, Distillers, Wine-Merchants, Victuallers, 8:0. 820.; such as cisterns, couches, coppcrs, coolers, tuns, vats, stills, casks, 860. &c. REMARK. The Art of Gauging is of such general utility, in the common affairs of life, that there are few persons who do not frequently want its assistance.——To Candidates for the Excise and Customs, it is indispensable; and to Victuallers, Common Brewers, Maltsters, Distillers, Rectifiers, \Vine-Mcrchunts, Spirit—Merchants, Oil- Merchants, Cider and Perry Makers, Vinegar Makers, (we. &c., it is of considerable moment; for without its aid they could not ascertain, with accuracy, the quantity of those articles which they buy, sell, or manufacture. NEW IMPERIAL TABLES. Few Acts of Parliament immediately interest so many individuals, as that passed on June 17th, 1824, for esta- blishing uniformity of Weights and Measures. Every person that has any dealings, either in buying or selling, particularly in commodities that are measured, must feel desirous to know how he is affected by the change that has taken place ; in order that he may neither be the ob— ‘ject of intentional fraud himself, nor take, inadvertently, any undue advantage in his transactions with others. i The content of the New Imperial Standard Gallon is fixed at 277.274 cubic inches ; and this is now the stan- dard measure of capacity, to be used throughout the United Kingdom of Great Britain and Ireland, for both Spirits, Wine, Ale, Beer, and all other liquids ; and also for every kind of grain, and all other dry commodities. But as the capacity of the gallon is changed, so likewise is the capacity of the tierce, the hogshead, the puneheon, the pipe, and the tun, in Wine and Spirit Measure ; the firkin, the kilderkin, the barrel, the hogshead, the pun- cheon, the butt, and the tun, in Ale and Beer Measure; and the peck, the bushel, the strike, the coom, the quarter, the wey, and the last, in Dry Measure. This being the N 6 276 «‘jiAUGING. PART V111. case, the author has «Moulated Tables, which exhibit the capacities of cacl; ol' Ilwsr- denominations, in cubic inches. "ABLES OF ‘W [N I?) AND SPIRIT MEASURE. TABLE I. Ezrltilu'h'ug tlns dffi'én Ht dcwonmzaa‘ions in, flux meet. me. 4 gills . . . . . . , . make 1 111111, Inf. L pints. . . . . . . . . . 1 (111211t,{//. 4 quarts . . . . . . . . . 1511111111,:70]. 42 gallons . . . . ' . . . . l tierce, tic). 63 gallons . . . . . . . . . 1 l1ogsl1o:1 As 277.274 3 2821154 : 54.92040 new gallons:54 gal. 3 qt. 1 pt. 173.9271, very nearly. 4. How many old gallons of ale. are there in 1 hogs- hcad, or 54 gallons of imperial measure ? As 282 : 277.274: 204 : 53.09502 old gallons : 53 gal. 0 qt. Opt. 3 gills, very nearly. 5. How many new bushels of corn are there in 1 last, or 80 old bushels ? PART VIII. GAUGING. 279 As 2218.192 2 2150.4 22 80 1 77.55505 new bushels =2 77 bush. 2 pk. 0 gal. 1 qt. 1% pint, very nearly. 6. How many old bushels of corn are there in 1 last, or 80 new bushels ? As 2150.4 3 2218.192: I 80 : 82.52202 old bushels = 82 bush. 2 pk. 0 gal. 0 qt. 1 pt. 1} gill, very nearly. 2d. Multipliers, by which old gallons and bushels may be reduced to imperial gallons and bushels. Multiply .83311 by the content of any vessel, in old Wine gallons ; and the product will be imperial gallons. Multiply 1.01705 by old ale gallons; and the product will be imperial gallons. Multiply .96945 by old corn bushels ; and the product will be imperial bushels. EXAMPLES. 1. If a cask contain 1 hogshead, or 63 old gallons of wine ; how many imperial gallons Will it hold? Here .83311 X 6325248593 imperial gallons, the same as obtained in the first example, by the preceding rules. 2. If the content of a vessel be 1 hogshead, or 54 old gallons of ale; What is its content in imperial gallons ? Here 1.01705 >< 54:54.9207 imperial gallons, the same as obtained in the third example, by the preceding rules. 3. How many imperial bushels are contained in 1 last, or 80 old bushels ? Here .96945 x 80 = 77.556 imperial bushels, the same as found in the fifth example, by the preceding rules. 3d. Multipliers, by which imperial gallons and bushels may be reduced to old gallons and bushels. Multiply 1.20033 by the content of any vessel, in im- perial gallons ; and the product will be old Wine gallons. Multiply .98325 by imperial gallons; and the product Will be old ale gallons. Multiply 1.03153 by imperial bushels ; and the product will be old Winchester bushels. EXAMPLE S 1. If a cask contain 52. 48593 imperial gallons , how many old wine gallons will it hold? Here 52.48593 X1. 20033: 63 old wine gallons. 2. If the content of a vessel be 54 9207 imperial gal- lons; What IS the content in old ale gallons? Here 54. 9207 x .98325=54 old ale gallons. \ 980 GAL’GING. PART mu. 3. How many old Winchester bushels are contained in 77.556 imperial bushels? Here 77.556 x 1.03153280 old bushels. (See the pre- ceding Examples.) 41h. Multipliers, for reducing Irish bushels and Scotch bolls t0 imperial bushels; and rice rersé. Illultiply .98188 by Irish bushels, and the product will be imperial bushels ; or multiply 1.01846 by imperial bushels, and the product will be Irish bushels. Multiply 3.96239 by Scotch wheat bolls, and the pro— duct Will be imperial bushels; or multiply 25233 by im— perial bushels, and the product will be Scotch wheat bolls. Multiply 5.78013 by Scotch oat bolls, and the product will be imperial bushels; or multiply imperial bushels by .173, and the product will be Scotch oat bolls. EXAMPLES. 1. How many imperial bushels are there in 1 last, or 80 Irish bushels? Here .98188 X 80278.5504 imperial bushels. 2. How many Irish bushels are there in 78.5504- im- perial bushels ? Ilere 78.5504 X 101846230 Irish bushels. 3. How many imperial bushels are there in 10 Scotch bolls of wheat? Herc 3.96239 >< 1002396239 imperial bushels. 4. How many Scotch bolls of wheat are there in 396.239 imperial bushels? flare 396.239 >< 252382100 Scotch balls of wheat. 5. HOW many imperial bushels are there in 100 Scotch bolls of oats ? flare 5.78013 >< 1002578043 imperial bushels. (5. How many Scotch bolls of oats are there in 578.0413 imperial bushels ? Here 578.043 >< .1732100 Scorch balls (3)" oats. Irish liquid gallons may be reduced to imperial gallons, by mulii— plying by .78478; and imperial gallons may be reduced to Irish liquid gallons, by multiplying by 1.27425. 1. The new imperial gallon is greater than the old wine gallon, by 46.274» cubic inches, which is nearly (1) of 277.274, or % of 231 ; hence, 6 old gallons are equal to 5 new gallons. 9. The new gallon is less than the 01d ale gallon, by 4.726 cubic inches, which is nearly 3‘5 of 282, or 519 of 277.974; hence 60 new gallons are nearly equal to 59 old gallons. PART Vin. GAUGING. 281 3, The new gallon is greater than the old corn gallon, by 8.474 cubic inches, which is nearly equal to 315 of 268.8 ; and the new bushel is greater than the old bushel, by 67.792 cubic inches; hence 82 new bushels are nearly equal to 83 old bushels. 4. Old wine gallons may be reduced into new gallons, by subtract- ing (1); and new gallons may be brought into old wine gallons, by adding 1. 5. Old ale gallons may be reduced into new gallons, by adding 3%; and new gallons may be brought into old ale gallons, by subtract- . ‘ L 111g 39. 6. Old bushels may be reduced into new bushels, by subtracting 3%,; and new bushels may be brought into old bushels, by adding :37. OBSE RVATION. Those who wish to see more on the subject (if the IVew Imperial Weights and fifeasnres, are referred to the Author’s Treatise on Practical Arithmetic, Third Edi- tion, in which he has given extensive Tables, comparing the old and new measures; and also a copious abstract of the Act of Parliament, for establishing uniformity of Weights and 1Weasures throughout the united kingdom (3)" Great Britain and Ireland. A DESCRIPTION OF THE SLIDING RULE. THIS instrument is in the form of a parallelopipedou; has four sliding pieces, which run in grooves, and is com- monly made of box. It was invented by Mr. Thomas Everand, about the year 1683, and is generally called Everand’s Sliding Rule ; but it has since been much improved by Mr. Verie ; and is now adapted to the new imperial measure. It is of various lengths, as 6, 9, 12, 18 inches, 8L0. ; but 12 inches is the most common length. T his rule is much used in Gauging, in consequence of the ease and expedition With which calculations may be made by it. Upon the first face of the Rule, is a line marked A, which is called Gunter’s Line, from its inventor lllr. Edmund Gunter; and is numbered from the left to the right With the figures 1, 2, 3, 4, 5, 6, 7, 8, 9, 10; and the space between each of these figures are graduated into subdivisions. At 2218.2, is a brass pin, marked 1MB, signifying the cubic inches in the imperial bushel; and at 277.3, is another brass pin, marked IMG, denoting the number of cubic inches in the imperial gallon. 282 GAUGING. PART VIII- The second line on this face is upon the slide, and marked B. It is divided exactly in the same manner as that marked A. There is also another slide B, which is used with the former; the two brass ends are then placed together, in which position they form a double radius. At 277.3, on the second radius, is a brass pin marked G or IG, denoting the number of cubic inches in the im- perial gallon. By these lines, multiplication, division, proportion, &c. may be performed; and the manner of reading and using them is precisely the same as the lines A and B, upon the Carpenter’s Rule. (See Section 2, Part IV.) Upon the same face of the Rule, is another line marked BID, signifying malt-depth; and is numbered from the right to the left, with the figures, 3, 4, 5, 6, 7, 8, 9, 10, 2. The l, or 10 on this line is placed directly opposite to MB, on the line A; and if it be called 1000, the last di- vision at the left end of the Rule, will denote 2218.2, the cubic inches in the imperial bushel—This line is used with the lines A and B, in malt gauging. Upon the second face of the Rule, or that opposite to the one already described, is a line marked D. This line begins on the upper edge of the Rule, and is numbered" from the left to the right with the figures 1, 2, 3, 31, 32, which is at the right end of the Rule. The line is then continued from the left end of the other edge, 32, 4, 5, 6, 7, 8, 9, 10. At 18.79 is a brass pin, marked 1G, signifying the cir- cular gauge-point for imperial gallons ; at 47.1 is another brass pin, marked BIS, which is the square gauge-point for imperial malt bushels; and at 53.1, is a third pin, marked MR, denoting the round or circular gauge-point for imperial malt bushels. Upon this face of the Rule, is a line marked C. This line is upon the slide; and is numbered and divided in the same way as the lines A and B. Belonging to the Rule, is also another slide C, which is used with the former, in the same manner as the two slides marked B. The line D, on this Rule, is similar to that marked 1), on the Carpenter’s Rule ; and is used with the two slides C, just described ; in finding the contents of vessels whose form is that of a cube, a parallelopipedon, a cy- linder, &c. &c. By these lines the square root of any number may also PART VIII. GAUGING. 283 be readily extracted ; for if 1 on C be set to 1 on D, we have against any proposed number on C, its root on D. The back of the first slide, marked C, is divided, next the upper edge, into inches and tenths; and numbered from the left to the right, with l, 2, 3, 4, 5, &c. ; the se- cond line is marked spheroid, and the third second variety ; and both are numbered from the left to the right with the figures 1, 2, 3, 4, &c.; and the spaces between each of these figures are divided into ten equal parts. These three lines are used for finding a mean diameter between the bung and the head of casks of the first and second varieties ; which is performed in the following manner : Find the difference between the bung and head diameters, on the first line, or line of inches; then against it, for each variety, is a number, which being added to the head, will give the mean diameter sought; hence the cask is reduced to a cylinder, Whose content may be found by Problem III. The back of the second slide C, contains the gauge- points, divisors, and factors or multipliers, for imperial gallons and bushels. The third face of the Rule contains a line marked Serf. St. or SS, signifying segments standing. This line is numbered from the left to the right, on the upper edge of the Rule, with l, 2, 3, 4, 5, 6, 7, 8. It is then continued on the other edge, 8, 9, 10, 20, 860. to 100. This line is used with the two slides, marked C, in finding the ullage of a standing cask, or the quantity of liquor it contains when it is not full. Upon the fourth face of the Rule, or the one opposite to that last described, is a line marked Seg. Ly. or SL, denoting segments lying. This line is numbered nearly in the same manner as the last; and is used with the slides C, in ullaging lying casks. Note l. There are various kinds of Sliding Rules, some of which have a line marked E, for extracting the cube roots of numbers; and others a line for the third variety of casks ; they are, however, all upon the same principle, and may be easily understood from the foregoing description. 2. Very neat and commodious Sliding Rules, have lately been made flat, with only two slides. These slides are the lull thickness of the rule; and being divided on each side, answer every purpose of the four sides ; and the rule is considerably reduced both in size and weight. 3. It may not be improper to observe, that in the preceding description of the Sliding Rule. the figures mentioned, refer to the larger figures placed at the primary divisions of the rule; most of the subdivisions being marked with smaller figures—For the method of estimating the values of the divisions and subdivisions, gie lievader is referred to the Description of the Carpenter’s Rule, Section 2, art . . 284 GAL-'6 ING. PART VIU. THE METHODS OF FINDING THE MULTIPLIERS, DIVISORS, AND GAUGE- POINTS, CONTAINED IN THE SUBSEQU EN" TABLE. 111:; multipliers and (11113013 arc chiefly obtainc l by (111215111, 111 the 101101111 111;; manner. - Divisors for Multipliers for squa1es.circlcs )77 .L)7~1)l. 0000 850. (. (11"60721212] eizalgrzll01’1‘. ilk 2218.19L’fl.0000 &c. (.OOO151 272231672th 6215 7261.9. Divisors for Blultiplicrs for squares. circles. 277.274).785o988 :.C \009830mnpe22m callous. 2218.192).785398 82C. (.000351 2222;; 2621:1230: 8.2655. Divisms for ci1clcs. .785398) 277 7.27 4(353. O1 2122pe22algal/aim. .785398)2 218 8.192(2821. 29 2222136? rial bus/2:12.» GAUGE-POINTS. The Ojauge- -11oi11ts ale 101111d by extractin 112111.119 square mots 0f the b(111130139. Divisors for Gauge-points for 591111] US. Squ '11 L‘S. 277 .2711) (/2622 Sf/‘llfné’ (16. 65 222212222le 1,124.09}; 2218.192) 7100219 axe (—17.10 Impenal 62231212" .1. Divisors for Gauge—points for circles. circles. 353. 01) 2/102)" square (18. 79 2221219622621." 9152 Gib 2824:. 29) *oois are (53.14 2222per2al.2 " 297:: 12 "“11113 11 the 1111111015 111 the following 21111113119111 0b~ 13.111 ed. :3 5011 9 PAR No A Table q)" .llzcltzplim-s, Divisors, and Gauge-Pointsfor Squares and Circles. ta. 0 :3 ’3' an "d -1 o 0" ... S 5 5,. (D I: m m 0 n ('V :1‘ O :5 H S ,.. '3. '6 5:: r9 -: ‘11) £2 < E . O '1 U} a: D :2. Q L: 5 Ha ‘I’ "d O E . rr .1] v >1- U] u :3" C: S D H- 5 n ..... .. (D u NOTE. —— The Areas, «Sac. are all in inches. The side or diameter 1 . 11 superficialfbot . . . 14 solid fioot . . . . . Imperial gallon . . . . Imperial bushel . . . . A pound of hard soap cold A pound of hot soap . . A pound of green soap . 1£(finvhucsofi&map A pound of tallow not . A pound of green starch . A pound of dry starch . BIultiplicrs for Divisors for Gauge-points for Sq ua res. .006944 .000578 .003607 .000451 .03 845 .035714 .033956 .039123 .031844 .028736 .024813 Circles. .785398 .005454 .000454 .002833 .000354 .028939 .028050 .0306 .030731 .025101 .022565 .019491 Squares. 1. 144. 1728. 277.274 2218.192 27.14 28. 25.67 25.56 31.4 34.8 40.3 Circles. 1.27324 183.34 2200.16 353.04 2824.29 34.56 35.65 32.68 32.54 39.98 44.32 51.3 Squares. 1. 12. 41.57 16.65 47.10 5.21 5.29 5.06 5.05 5.6 5.9 .35 Circles. 1.128 13.54 46.91 18.79 53.14 5.97 5.72 5.7 6.32 6.66 7.16 7 T VIII. GAUGIN G. 286 GAUGING. PART vin- PROBLEM I. The side ofa vessel in theform of'a cube, being given in inches; to find its content in imperial gallons and bushels. RULE. Multiply the side by itself, and that product again by the side ; and the last product will be the content of the vessel in cubic inches ; which being multiplied by .003607, and .000451 ; or divided by 277.3, and 2218.2, the respec- tive products or quotients will be the content of the vessel in imperial gallons and bushels. Note 1. The definitions of the cube, the parallelopipedon, the cylinder, &c. &c. may be seen in Part 1V.; and if the content ofany vessel be found in cubic inches, by the rules given in that Part, and then divided by 292, 231, and 2150.4, the respec- tive quotients will be the content of the vessel in old ale and wine gallons, and malt bushels; but the Rules and Examples given in the following Problems will be found more particularly adapted to Gauging, than those in Part IV. 2. If the content of any vessel, in cubic inches, be divided by 277.3, and 2218.2; the respective uotients will be the content in imperial gallons and bushels. 3. Gaugers ‘ ways take their dimensions in inches or in inches and tenths ; and when we say the Side or diameter of a vessel measures so many inches, we mean the internal, not the external dimensions- EXAMPLES. 1. The side of a cubical vessel measures 46 inches; What is its content in imperial gallons and bushels ? [lore 46 X 46 x 4622116 X 46 =97336, the content in cubic inches ; then 97336 X 00360723510909 imperial gallons ; and 97336 X 0004512438985 imperial bushels. 0r, 97336 +277.3:351.0133 imperial gallons ; and 97336+2218.2:43.8806 imperial bushels. Note. The divisors are used in all the following solutions; the work, however, may be proved by the multipliers given in the preceding Table- Bg the Sliding Rule. In this operation, the square gauge-points 16.65, and 47.10, upon the line D, must be used. 072 D. On C. 07% D. 072 C. 16.65 . , . . , 351.0 imperial gallons. AS{47.10 ’ 46 ' ' 46 44.0 inlzperial bushels. Note. New Sliding Rules have not always the square guage-points muriged upon them ; but small brass pins may be easily inserted at those points. 2. The side of a cubical cistern is 134 inches; what is its content in imperial gallons and bushels? Ans. 86768986 imperial gallons, and 1084.710'1 im- perial bushels. PART vm. GAUGING. 287 3. The side of a cubical Wine-vat measures 135.6 in- ches ; how much wine will it contain at once ? Ans. 8991.4389 gallons = 8991 gal. 1 qt. 1% pt. PROBLEM II. The length, breadth, and depth of a vessel, in the form of a parallelopipedon, being given in inches ,- to find its content in imperial gallons and bushels. RULE. Multiply the length by the breadth, and the product thence arising by the depth ; and the last product will be the content in cubic inches; which being divided by 277.3 and 2218.2, will give the content in imperial gal— lons and bushels. Note. As the sides of vessels, in the form of a parallelopipedon, are seldom per- fectly regular and parallel. it is best to take several lengths, in different places ; and divide their sum by their number for 'a mean length. A mean breadth and depth may be found in the same manner EXAMPLE S. l. A cistern in the form of a parallelopipedon, mea- sures 96 inches in length, 65 in breadth, and 48 in depth; what is its content in imperial gallons and bushels ? Here 96 X 65 X 48:299520, the content in cubic inches ; then 299520+277.321080.l298 gallons; and 299520—;- 2218.2=135.0284 bushels. Bg the Sliding Rule. As 96 on C: 96 on DZ: 65 on C: 79 on D, which is a mean proportional between the length and breadth. (See Problem 5, page 152.) Then, 0n D. On C. 0n D. - 0n C 16.65 , , _ , 1080.0 gallons. AS{47.10 ' 48 -- 79 ' {135.0 bushels. 2. The length of a vessel in the form of a parallelo- pipedon, is 136, its breadth 94, and its depth 62 inches ; What is its content in imperial gallons and bushels? Ans. 2858,3050 gallons, and 357.3203 bushels. 3. A water-trough measures 85.3 inches in length, 54.7 in breadth, and 32.9 in depth ; how many imperial gallons will it contain? Ans. 553.5825 gallons = 553 gal. 2 qt. 0:;— pt. 288 GAUGING. PART VIII. 4. A maltster has a cistern whose length is 132, breadth 118, and depth 46 inches; how much barley can he steep at a time, admitting the water to occupy ,~ of the cistern? .‘1 Ans. 193.80l7 bushels : 2i qr. l bus/1. 311171;}?5. PROBLEM III. The dimnyeter and dept/i of a vessel, in the form of (1. cylinder being given,- I‘o find its content in imperial gallows and bushels. RULE. Multiply the square of the diameter by the depth; and divide the product by 353, for imperial gallons ; and bf 2824.3, for imperial bushels. Noll? 1. As cylindrical vessels are seldom perfectly round, it is best to mount» cross diameters, in diifcrent parts; and divide their sum by their number, for it mean diameter. 2. Ifa cylindrical vessel be placed in an inclining position, so that the liqum‘ forms an elliptical surface, the content may be found by the following Rule: Muh tiply the square of the diameter of the vessel by half the sum of the greatest an l least depths of the liquor; and divide the product by 3:33, for imperial gallo:a.~.. (See Notes, Prob. 5.) EXAMPLES. l. The diameter of a cylindrical vessel is 3—1, and it: depth 45 inches; what is its content in gallons and bushels? Ifere 34: X 34 x 45 = 52020 ; tlzen 5202()—:~ 353 = 147.3654 gallons; and52020—z—982l3:18.4187 bushels; By tllc Sliding Rule. Here the circular gauge—points 18.79, and 53.14, uptn the line D, must be used. On D. 071 C. 022, D. 022, C. 18.79 . ,. _, 1 _ 147.4 gallons. AS{03.14 ' 4’) '- 34 ' 18.4 bushels. 2. The diameter of a cylindrical vessel is 416.7, and its depth 68.4 inches ; What is its content in gallons and bushels? Ans. 422.5860 gallons, and 52.8176 lugs/2.079. 3. A cylindrical mash—tun measures 94- inches in dia- meter, and 82 in depth; how many bushels of malt will it contain at once ? A723. 256.5421 bushels. 4. At Heidelberg, in Germany, is a cylindrical wine cask, PART VIII. GAUGING. 289 whose depth is 27, and diameter 21 feet; how many gal- lons Will it contain, imperial measure? . Ans. 582869575 gallons. Note. The convivial monument of ancient hospitality, mentioned in the 4th Ex- ample, was formerly kept full of the best Rhenish wine, and the electors have given many entertainments on its platform ; but it now only serves as a melancholy instance of the extinction of that hospitality ; for it is suffered to moulder in a damp vault, quite empty. Although this vessel is of an extraordinary magnitude, yet it is much inferior in capacity, to many of the London porter-vats PROBLEM IV. Given the dimensions of a vessel in the form of a prismoz'd, or the frustum of a square pyramid, of a cylindroid ; to find its content in imperial gallons and bushels. . RULE. To the sum of the areas of the two ends, add four times the area of the middle section parallel to them; multiply :this sum by the perpendicular depth, and % of the product .will be the content in cubic inches ; which divide by 277.3 for imperial gallons, and 2218.2 for imperial bushels. (See the Scholz'ztm, Prob. 10. Sect. I. Part I V) Note 1. A cylindroid is a figure resembling the frustum of a cone; but having elliptical instead of circular ends. Sometimes one end is circular, and the other e i tical. 2? When the vessel is a prismoid, the length of the middle section is equal to ;half the sum of the lengths of the two ends ; and its breadth is equal to half the sum of their breadths. 3. If the ends be elliptical, the transverse diameter of the middle section will be equal to halfthe sum of the transverse diameters of the two ends; and the conjugate Idiameter equal to half the sum of the conjugate diameters of the two ends. 4. If one end be an ellipse and the other a circle, add the transverse diameter of the elliptical end to the diameter of the circular end ; and take half the sum for the transverse diameter of the middle section. The conjugate diameter of the middle section may be found in a similar manner; it is better, however, in all cases of .p-ractz'ce, to take the real dimensions of the sections. 5. W’hen the ends are rectangles, their areas may be found by Problem 2; when they are circles, we may obtain their areas by Problem 15 ; and when they are ellipses, we can find their areas by Problem 21, Part II. EXAMPLES. l. The length and breadth of the bottom of a vessel in .the form of a prismoid measures 72 and 64, the length and breadth of the top 96 and 82, and the perpendicular ldepth 65 inches ; What is its content in gallons and bushels ? Here (72 X 64) + (96 X 82) 24608 +7872=l2480, the larea of the two ends. _ Also, (72+96)—:—2=l68 +2284, the length of the middle section ; and (6%: + 82) —:— 2 = 146 —:— 2:73, the 0 290 GAUGING. PART Vin. breadth of the middle section ; then 84 X 73 x 4:6132 x 4 =24528,fou7~ times the area of the middle section ,- whence (12480 + 24528) X 65+6=(37008 >< 65)-+—6: 2405520—:~6 2400920, the content in cubic inches ; then 400920—z—277.3 = 1445.7987 gallons ; and 400920 —:— 2218.2: 180.7411 bushels. Note. This and some of the following Problems may be performed by the Sliding Rule ; but as the operations are too tedious for practice, they are omitted. 2. Each side of the bottom of a cistern, in the form of the frustum of a square pyramid, measures 86, each side of the top 78, and the perpendicular depth 74 inches; what is its content in gallons and bushels ? Ans. 1795.7831 gallons, and 224.4931 bushels. 3. The perpendicular depth of a vessel, in the form of the frustum of an elliptical cone, is 46.8 inches ; the transverse and conjugate diameters of the bottom measure 49.6 and 37.8 inches; the transverse and conjugate dia— meters of the top, 67.2 and 50.4 inches; required the content in gallons and bushels. Ans. 343.8297 gallons, and 42.9825 bushes. 4. The perpendicular depth of a vessel with an elliptical base and a circular top, is 52.6 inches; the transverse and conjugate diameters of the bottom measure 61.6 and 46.2 inches ; and the diameter of the top measures 42.6 inches ; required the content in gallons and bushels. Ans. 345.4747 gallons, and 43.1882 bushels. PROBLEM V. Tofind the content of a vessel in the form of thefrustmn of a cone, in imperial gallons and bushels. RULE. To three times the product of the top and bottom di- ameters, add the square of their difference; multiply the sum by the depth, and divide the product by 1059 for imn perial gallons, and 8472.9 for imperial bushels. Note 1. The general rule given in Problem 6, Section 1., Part IV. will give (Eu- content ofany frustum, whatever may be the form of the two similar ends ; that l:._ whether they be polygons, circles, or ellipses. ‘2. In taking the dimensions of circular vessels, it is best, in all cases. to measure cross diameters, as directed in the second Note, in Case I. of the next Problem. 3. When the frustum of a cone is cut bya diagonal plane passing through the opposite extremities of the ends; or when a vessel of that shape is placed in an in- clining position, so that the liquor just touches the opposite extremities of the top and bottom, the two parts or sections thus formed are called elliptic hoofs; and their contents may be found by the following Rule: Multiply the product of the diameters of the ends of the frustum by a. mean proportional between them; and cube the diameter of the hoof’s base. PART VIII. GAUGING. 291 From the greater of the numbers thus found subtract the less; and divide the remainder by the difference of the diameters ; then the quotient being,r multiplied by the perpendicular height of the hoof, and the product by .2618, will give the copltent in cubic inches ; which divide by 277.3, and you will obtain the content in ga cos. 4. If the greater end of the frustum be considered as the base of the hoof, the product of the diameters by a mean proportional between them, will be less than the cube of the diameter of the base; but the contrary will be the case, when the less end of the frustum represents the base of the hoof. 5. If a vessel be placed in such a position that the liquor just covers the bot- tom, and part of one side, measure the diameter at the bottom, the diameter at the upper extremity of the liquor, and also the liquor’s perpendicular depth, or height of the hoof; then proceed with these dimensions as directed in Note 3, and you will obtain the content. 6. In Nesbit’s and Little’s Practical Gauging, there are Rules given for finding the contents of all cylindrical, pyramidical, and conical hoofs, or ungulas, that can possibly be formed, by placing in various positions vessels containing fluids. EXAIVIPLE S. l. The bottom diameter of a vessel, in the form of the frustum of a cone, is 46, the top diameter 62, and the depth 60 inches ; what is its content in gallons and bushels ? Here 62 X 46 X 3:2852 X 3: 8556, three times the pro- duct of the top and bottom diameters. Also, (62—46)3: 16 X 16:256, the square of their difference; then (8556 +256) x 60:8812 >< 60:528720; and 528720 —:- 1059: 499.2634 gallons ; likewise, 528720 + 847292624013 bushels. -2. The bottom diameter of a guile-tun is 115, the top diameter 98, and the depth 75 inches; how many impe- rial gallons will it contain? Ans. 2414.9433 gallons. 3. The bottom diameter of a Wine-vat is 78.6, the top diameter 64.3, and the depth 72.8 inches; What quantity of Wine will it contain? Ans. 1056.3513 gallons PROBLEM VI. To find the content of a circular vessel, with curved sides. CASE I. When the sides are not much curved, as in the following figure. RULE . To the sum of the squares of the top and bottom diam- eters, add four times the square of the middle diameter; multiply this sum by the depth ; divide the product by 2118.24, and 16945.74 ; and the respective quotients will be the content in imperial gallons and bushels. o 2 292 GAUGING. PART VIII. Note 1. The foregoing divisors are found by multiplying the circular divisors 6 y2. As vessels of this kind are seldom perfectly round, it is best to measure two diameters at right angles to each other; and take half their sum for a mean dia- meter. (Sec Problem 7.) 3. The Rule given in this Problem is the same, in substance, as that given in Problem 10, Section I., Part IV. (See the Scholz'um in that Problem.) EXAMPLES. 1. The diameter AB, of the fol- lowing vessel, measures 50, the dia- meter CD 56, the diameter EF 54, and the depth 77272 46 inches; what is its content in gallons and bushels ? Here 502 +542: 9500 + 2916: 5416, the sum of the squares of the top and bottom diameters ; and 562 x 4:3136 X 4 : 12544, four times the square of the middle diameter. Then, (5416 + 12544) x 46: 17960 X 46 : 826160; and 826160 —:— 2118.24: 390.0219 gallons; also, 826160+16945.74:48.7532 bushels. 2. The bottom diameter of a mash-tun is 75.3, the top diameter 81.6, the diameter taken in the middle between them 84.7, and the depth 56.9 inches; what is its content in imperial bushels? Ans. 137.7527 bushels. 3. If the length of a cash be 30, the bung diameter 24, and the head diameter 18 inches ; what is its content in imperial gallons? Ans. 41.8082 gallons. CASE II. Tofind the content of a circular vessel, when its sides are much curved, by tahingfioe diameters, at equal distances from each other, as in the following figure. RULE. Add into one sum, the squares of the top and bottom diameters, twice the square of the middle diameter, and four times the sum of the squares of the diameters taken at one—fourth and at three-fourths of the depth; multiply this sum by the depth; divide the product by 4236.48, and 33891.48; and the respective quotients will be the content in imperial gallons and bushels. biota The foregoing divisors are found by multiplying the circular divisors by twe ve. PART VIII. GAUGING. , 293 REMARK. ‘ The Author has deduced the preceding Rule, from the method of equidistant ordinates, described in Problem 23, Part II. ; and it may be used for all vessels of an ordinary size; but if a vessel be very deep, its sides very much curved, and great accuracy required, the content must be found by the following Rule : Find the areas of as many equidistant, parallel sections as you think necessary, by multiplying the square of the mean diameter of each sec- tion by .7854. Then to the sum of the areas of the two end sections, add four times the sum of the areas of all the even sections, and twice the sum of the areas of all the odd sections, not including the sections at the ends ; mul- tiply the sum by the common distance of the sections; divide the product by 3; and the quotient will be the content in cubic inches—Divide the content thus found by 277.274, and 2218.192 ; and the respective quotients will be the content in imperial gallons and bushels. Note 1. Always make choice of an odd number of sections, in order that the num~ ber of parts into which the vessel is divided may be equal. Seven or nine will, in general, be sufficient, except when the vessel is very deep, in which case it may be necessary to take eleven, thirteen, &c. as the case may require. 2. The perpendicular depth of the. vessel must first be taken, in order to deter- mine the common distance of the ordinates, which may be found by dividing the whole depth by the number of sections minus one. 3. Great care must be taken to obtain the diameters of the sections at equal per- pendicular distances from each other; for if their distance be measured upon the side of the vessel, it is evident that the process will be incorrect. (See Pro- blem 1, Case 11., Part VI.) EXAMPLES. 1. The diameter AB, of the following vessel, measures 49, CD 62, EF 68, GH 65, KL 56, and the depth mn 48 inches; what is its content in imperial gallons and bushels ? Here 4994-569 = 2401+ 3136 :5537, the sum of the squares of the top and bottom diameters ; 682 x 2 = 4624 x 2:: 9248, twice the square of the middle diameter ; and (622 +- 659) X 4=(3844+4225) X 428069 X 4:32276,f0ur times the sum of the squares of the diameters taken at one- fourth and at threefourths of the depth. Then, (5537+ 9248 +32276) x 48 247061 X 48 22258928 ; and 2258928 —:— 42364825332087, the content in imperial gallons; 0 3 294 GAUGING. PART mt. also 2258928+33891.48=66.6517, the content in impe- rial bushels. 2. The bottom diameter of a guile—tun measures 74.2, and that of the top 84.3 inches ; the diameter taken at 1,; of the depth from the bottom, is 93.6, and that at 5% of the depth 98.7 inches; the middle diameter 102.4, and the depth 60.8 inches ; how much ale will it contain at once? Ans. 1544.1428 gallons. 3. What is the content of a vessel in imperial gallons and bushels, the depth of which is 120 inches, and the mean diameters of seven equidistant, parallel sections, as follow: the diameter of the bottom or first section 124, the second 146, the third 161, the fourth 164, the fifth 166, the sixth 157, and the diameter of the top, or seventh section 144 inches ? Ans. The area of the first section is =12076.8104, the second: 16741.5864, the third:20358.3534, the fourth = 211241184, thefifth:21642.4824, the sixth219359.3‘_46, and the seeenth=16286.0544 ; then by the method of eyesi- distant ordinates, or parallel sections, we find the content = 927509436 cubic inches = 82052206 gallons, and 1025.6525 bushels, the answer required. Note. In order to find the content ofa copper with a convex or aconcave bottom, generally called arising or afalling crown, pour in as much water as will just cover the bottom ; and make marks with chalk upon the sides of the copper, at the surface of the water ; then draw it into a vessel of a known measure, or one that may be easily gauged ; after which find the content of the remainder of the copper, as directed in this Problem, which being added to the number of gallons requisite to cover the bottom, will give the whole content. —- The content of a still may also be found in the same manner. PROBLEM VII. To gauge and inch a round guile-tun according to the method practised in the Excise. RULE. Take cross diameters of the guile-tun, in the middle of every ten inches from the bottom upwards; that is, mea- sure the first diameters at five inches from the bottom, the second at fifteen, the third at twenty-five, 8m; and take half the sum of each two for mean diameters. Di- vide the square of the first mean diameter by 353.04, multiply the quotient by .‘O : and the product will be the content of the first ten inches of the vessel. Find the content of every ten inches of the perpendicular depth in PART Vin. GAUGING. 295 the same manner, which being added together, will give the content of the whole tun in imperial gallons. Note. It must here be observed that the square of the diameter of any section divided by 353.04, will give the area of that section in imperial gallons ; for in finding the areas of figures, Gaugers always consider their depth to be one inch. 2. Mash-tuns are used for the purpose of macerating malt in hot water, to ex- tract from it the saccharine substance ; and guile-tuns, sometimes called working- vats, are vessels in which wort is fermented, in order to convert it into ale or beer. REMARK. The most eligible method of taking the dimensions of circular vessels, is to quarter them, in order to obtain cross diameters as correctly as possible—First, take the diameter of the top of the vessel, which multiply by .707, or .7; and the product will be the side of the in- scribed square. (See Part II. Prob. 15, Rule III. in the IVotes.) . Apply the side of the square thus found, four times to the top of the vessel, by your dimension cane, or tape; and mark, with chalk, each angle of the inscribed square; and thus will the top of the vessel be quartered. —With a steady hand, draw a chalk line, from each quartering point, down the inside of the vessel, to the bottom ; taking care that these lines be equally distant from each other, not only at the bottom of the vessel, but also at every . horizontal section; otherwise the vessel will not be truly quartered in every part. Place your dimension cane, or other instrument, in a perpendicular direction, with its end resting upon the bottom of the vessel, about half way between the centre of the bottom and the side; at the same time laying a rod, or holding a cord diametrically across the top, so as to come in contact With the dimension cane; and thus you will have the perpendicular depth of the vessel, upon the dimension cane, where it intersects the rod or cord at the top. Now, in order to take cross diameters in the middle of every ten inches, mark the dimension cane with chalk, at 5, 15, 25, 35 inches &c.; and place that instrument in the same manner in which it stood in taking the depth; then bring a rod horizontally in contact with the cane, at 5 inches from the bottom, so that the rod and cane may form right angles ; and let the end of the rod, at the same time, touch one of the quartering lines; and there make a mark with chalk, on the inside of the vessel; and you will have one point at which a diameter must be taken. 0 4 296 GAUGING. PART VIII. Do the same at 15, and 25 inches, 8:6. from the bot- tom; and you will obtain all the points in one of the quartering lines, at which diameters must be taken. These points may then be transferred to the other three quartering lines, by a pair of compasses or a dipping piece; and hence you may proceed to take cross diameters, as before directed; always beginning at the bottom. Note. When vessels are small. cross diameters may be most easily taken by an instrument called the Diameter Rod, or Rule. When they are large, they must be measured by the Dimension Cane. EXAMPLES. 1. Let the following figure ABCD represent a guile-tun, in the form of the frustum of a cone, Whose perpendicular depth mn is 30, the mean dia- meters pg‘ 42.2, rs 45.3, and was 48.6 inches; what is its content in imperial gallons? Here (42.2 x 42.2) —:— 353.04 = 1780.84 —:- 350.04 :. 5.044 gallons, the area of the section at pg ; and 5.044 X 10:50.44 gallons, the content of the first 10 inches of the perpendicular depth. Again, (45.3 X 45.3) + 353.04 = 2052.09 + 353.04 = 5.812 gallons, the area ofthe section at rs ; and 5.812 x 10:58.12 gallons, the content of the second 10 inches of the depth. Also, (48.6 x 48.6) + 353.04 = 2361.96 —:— 353.04 : 6.690 gallons, the area of the section at was ; and 6.690 X 10:66.90 gallons, the content of the third and last 10 znches of the depth. Lastly, 50.44 +58.l2+66.90= 175.46 gallons :4 bar- rels, 3 firhins, and 4.46 gallons, the whole content of the gaile-tan. Note. In Practice, when the diameters are taken in the middle of every ten inches, it is not necessary to multiply by 10; for if the decimal dot, in the area, he removed one figure towards the right—hand you will obtain the content the same as if you actually multiplied the area by l0; but in the foregoing.r Erampie I have multiplied the area of each section by 10. in order to show the learner how to pro- ceed when diameters are taken in the middle of every 6 or 8 inches ; or when the vessel is not full of liquor. 2. If the depth of the liquor in the foregoing guile—tun be 26.8 inches ; how many gallons does it contain ? Herc 50.44—l—58.l2=108.56 gallons, the content of the first 20 inches of the depth ; and 6.69 x 6.8245492 gal- [PART VIII. GAUGING. ~ 297 Ions, the content of the remaining6.8 inches; then 108.56 +45.492=154.052 gallons, the content required. 3. If the depth of the liquor in the preceding guile-tun be 18.4 inches; how many gallons does it contain ? Ans. 99.2608 gallons. REMARKS. 1. In order to facilitate the Practice of Gauging, it is necessary to be furnished with a measuring—tape, divided into inches, a gauging-rod, divided into inches and tenths, a chalk-line, plummet, &c. as the case may require; and also a dimension-book ruled in columns preper for the dimensionsintended to be taken. 2. The following Table, in which are entered the di- mensions, areas, and contents of the foregoing guile-tun, may serve as a specimen for mash-tuns, guile-tuns, cop- pers, stills, 8:0. 3. By the solution to the first Example, we find the content of the first 10 inches from the bottom to be 50.44 gallons; and as 9 gallons make 1 firkin, and 4 firkins 1 barrel, we have only to divide 50.44 gallons by 9 and 4, successively; and we obtain 1 barrel, 1 firkin, and 5.44 gallons, for the reduced content of the first 10 inches.— In the same manner we have 58.12 gallons = 1 barrel, 2 firkins, and 4.12 gallons, for the reduced content of the second 10 inches; and likewise 66.90 gallons 2 1 barrel, < 3 firkins, and 3.90 gallons, for the reduced content of the third 10 inches of the depth; the whole being equal to 4 barrels, 3 firkins, and 4.46 gallons, as in the preceding Solution and following Table. A. B.’s Guile-tun, ganged March 10th, 1827. Mean Areas in EDepths. Diagggdrs. Diam. Gallons. Areas. Contents. 9 1 |- bar. firk. gal. arwfirk. gal. 10 41842.6 42.21 5.044 0 0 5.044 1 1 5.44 10 44.71459 45.31 5.812| 0 .0 5.812 1 2 4.12 10 48.5i48.7i 48.6i 6.690‘. 0 0 l 6.690 1 3 3.90 30 Whole depth and content . . . 4 l 3 4.46 l ‘ REMARK. In order to find the content of the foregoing guile- tun, at every inch from the bottom, proceed thus: Add the ,- 0 0 298 GAUG-ING. PART VIII. area of the first section to itself, and the sum will be the content at two inches of the depth; to this add the area, and the sum will be the content at three in and thus continue adding the area of the firs until you come to the tenth inch ; a content at every inch of' the first ten inches. content add the area of the second section, obtain the content at eleven inches from the thus proceed until you come to the 30th in the area at every ten inches; and you will 11 same ches ; t section, nd you will have the To this and you will bottom; and oh, changing ave the con- tent at every inch of the depth, as in the following Table. A Table of the foregoing Guile-tun. l Writ Contents. ‘12)“ Contents. Vile-t Contents. bar. firk.= gall. , WETT‘_‘ WETZET 1 0 0 5.0441 11 1 2 92.252 21 3 I 0 7.250 i 5.044 5.812 I 6.690 ‘2'0T 1.088 T2 T3“8.064 22 TlT4.9‘4“0 5.044 5.812 1 6.690 “T 0 1 16.132 13 T 8 4.876 "$8,212.630 . 5.044 5 ‘812 J§£92 ‘4“? 72.176’W21‘ 0 1688 ‘2? 3 ‘810320 ' 5.044! 5.812 6 690 5 0773213“? 0 7.500 25 ‘3“?7010 _._-J __5-0441 5.812 ,n 6999 6 17)~ 3 1322647? ”2*‘1‘481‘2 W?” 0 4.700 I 5.044 I 5.812 I l6.690 773—5375373687 T7197 21.174 ’WIT 172.390 5.044 I 5.812J {6.690 T T045352 18 ‘2— 2 6.9861 28 T 2 {0.080 5.044 5.812 6.690 TflT T0386 19 2 7313.748. 29 4 I 2 {6.770 1 5.044 5.812; Jp‘fiQO Win—5m 20 "t 0 pm}??? 3 4.186 1 5.812 s I 6.690. 1 . z 4 1 . ,nnnq Note 1. The stars or asterisks, in the preceding Table, tend to prevent mistakes, in tabulatz‘ng other vessels. g Table is evident ; for when a vessel is once gauged ake the depth of the liquor, and against it we find the pth be 23 inches, we have 3 B. 2 F. 2.63 G. for the epth be 23.5 inches, we take one-half of the area of the third one-half of 6.69 G. = 3.345 G., which being added to the content we obtain 3 B. 2 F. 5.975 G. for the content required. Again if the areas; and their use will 2. The use of the foregoin and inched, we have only to t content required: thus if the de content. If the d section; namely, at 23 inches, denote the changes of the PART vnI. GAUGING. 299 depth be 23.8 inches, we take 1% of 6.69 G = 5.352 G ; and this being added to the content at 23 inches, we have 3 B. 2 F. 7.982 G. for the content at %.8 inches of the depth. 3. Any mash-tun, guile-tun, copper, or still, Whose sides are curved, may be gauged and inched in the same manner; but if the sides be much curved, cross- diameters must be taken in the middle of every 4,6, or 8 inches of the perpendicular depth, as the case may require; and the areas of the different sections must be used, as before directed, in tablz'ng the vessels. 4. in order to gauge and inch a mash-tun, or a guile-tun, in the form of a pris- moid, or the frustum of a square pyramid, take lengths and breadths in the middle of every 6, 8, or 10 inches of the perpendicular depth, as the case may require ; and divide the product of each length and breadth by 277.3, and the quotients will be the areas of the sections in imperial gallons; then enter the dimensions, areas, and contents, in Tables, as before directed. 5. Large mash-tuns, and guile-tuns are generally fixed in an inclining position, called the drip or fall of the tun, in order that the liquor may be conveniently drawn ofl'; when this is the case, pour in as many gallons of water as will just cover the bottom ; and make marks upon the sides of the tun, at the surface of the waterd; then draw it off, and proceed with the remainder of the tun, as before di- recte . 6. The water that is used to cover the bottoms of guile-tuns, and of coppers with rising orfallz'ng crowns, must either be measured with a gallon measure, or drawn Qfi‘,‘ into a vessel ofa known capacity, or into one that may be easily gauged ,- and the quantity must be added to the content of the other part of the guile-tun or copper. \ PROBLEM VIII. To gauge a back or cooler. Backs or coolers are vessels which receive the wort, when let out of the copper, in order to be cooled. They are commonly of a rectangular form, and seldom exceed nine or ten inches in depth. RULE. Multiply the mean length by the mean breadth ; divide the product by 277.3, and the quotient will be the area in imperial gallons; which being multiplied by the mean depth, will give the quantity of wort contained in the cooler. Nole 1. In taking the dimensions of a cooler, find the middle of both ends, and also that of the sides, at which take the length and breadth; but as coolers are never fixed in a horizontal position, but inclined a little, in order that the wort may run off, and as their bottoms are frequently warped ; it is necessary to take the depth of the liquor in several places, and divide the sum of these depths by their number, for a mean depth. 2. In order to find a constant dipping-place, which will save much trouble, pro- ceed thus: Find a mean depth, as directed in the last note; then try in different parts of the cooler, until you find a place of the same depth as your mean depth; at which make a mark upon the side of the cooler, for a constant dipping-place. 3. It has been found by experiments that 10 gallons of hot wort will on] mea- sure to 9 gallons when the wort is cold; therefore, an allowance of one ga lon in every ten must be made, when wort is gauged hot. In order to do this, multiply the numlbier of warm gallons by .9, and the product will be the number of gallons when co . EXAMPLES. l. The length AB of a cooler is 125.6 inches, the breadth CD 73.4 inches, and the depth of the wort taken 0 6 300 GAUGING. PART VIII. in ten different places, as below; how many gallons are contalned in the cooler? Depths. Inches. At a 24.3 I) = 4.2 D c = 4.2 d 24.5 6 24.4 9 =4.6 ll 24.5 k 24.9 771:4.8 7?. =4.6 10)45 0 4.5 mean. Inches. 125.6 length. 73.4 breadti' . 5024 3768 8792 277.3)9219.04 (33.24 gallons, (.736 area for one inc]; deep. Gallons. 33.24 area. 4.5 mean dept/z. 16620 13296 ITQTSSO the content 0f the wort in the cooler. By the Sliding Rule. Divisor. Length. Breadth. Area. As 277.3 on A I 125.6 072B I: 73.4 on A : 03.24 on B. And, Unity. Area. Depth. Content. As 1 on A 1 33.24 on B I: 4.5 on A : 149.58 072 B. Note. The constant dipping—place may be either at d or 12 ; but it is preferable to find one nearer the side of the cooler, as at r. 2. If the mean depth of warm wort, in the foregoing cooler, be 5.8 inches ; how many gallons will there be when the wort is cold? PART VIII. GAUGING. 301 Gallons. ’ 33.24 area. 5.8 depth. 26592 16620 pl§277§2 gallons of warm wort. .9 multiplier. 173.5123 gallons when cold. 3. If the mean depth of warm wort in the preceding cooler be 6.4 inches; how many gallons when cold ? Ans. 191.4624 gallons. Note l. The depth of the liquor in a cooler is always taken to the tenth of an inch ; and in order to facilitate the practical part of Gauging, Excise ofl‘icers gene- rally make a Table, exhibiting the content in barrels, firkins, and gallons, at every tenth of an inch. This is called tent/Zing a cooler,- and may be performed in the following manner: Find the area of the cooler. and reduce it to barrels, firkins, and gallons ; which being divided by ten, will give the content at one-tenth of an inch in depth. Add this content to itself, and the sum will be the content at two- tenths. Again, to this content add that of the first tenth, and you will obtain the content at three-tenths of the depth. Continue this operation until you arrive at six or seven inches of the depth, which will generally be found sufficient; as the wort in coolers is not often deeper. 2. It has been before observed that coolers are generally rectangular; but should vou meet with one of any other shape, its area may be found by the Rules given in art 11. PROBLEM IX. To gauge a cistern, couch, or floor of malt. RULE. Take the dimensions as directed in the last Problem; then multiply the mean length by the mean breadth ; di- vide the product by 2218.2, and the quotient will be the area; which being multiplied by the mean depth, will give the content in bushels. Note 1. According to Act of Parliament, barley must lie under water in the cistern, forty hours; in which time it is supposed to swell or increase to one-fourth more ; so that four bushels in twenty are allowed for this increase. From the cistern the barley is removed to the couch ; and after having lain there twenty—four hours, it is deemed a floor. The same allowance is made in the couch as in the cistern; but when the corn has been thrown out of the couch into the floor, and there grown according to the usual custom, it is supposed to increase one-half; consequently an allowance is made of ten bushels in every twenty. ‘2. If cistern or couch-bushels be multiplied by .8. the product will be neat bushels; but floor-bushels must be multiplied by .5, in order to reduce them to neat bushels. 3. The duty is always charged upon the best gauge of the cistern, couch, or floor; and in order to find from which the charge will arise, without reducing them to neat bushels, proceed thus: Multiply the best gauge of the cistern or couch by 1.6; and if the product exceed the floor bushels, the charge must be made from the cistern or couch; but if not, the charge must be made from the floon This multiplier is found by dividing eight-tenths by five-tenths. (See the last Nate.) '302 GAUGING. PART Tm. EXAMPLES. 1. The mean length of a cistern is 96, the mean breadth 6}, and the mean depth 32 inches ; what is the area and content in malt bushels? Tofind tile area. Inches. 96 length 64 breadtiz. 3871 576 2218.2)6144.0(2.769 area in bus/ids. By Me Siiding Rule. M. n. Length. Breadth. Area. AS 2218.2 072 A : 96 on B I I 64 on A 1 2.77 072 B. Tofind the content. Bushels. 2.769 area. 32 dept/e. W38 8307 WOS content. By t/ze Sliding Rule. Unity. Area. Depth. Content. As 1 on A: 2.77 on B :: 32 on A: 88.6 0.72 B. 01', The content may be readily found by means of the line M 1), on the Sliding Rule, Without knowing the area ; thus, Length. Depth. Breadth. Content. As 96 on B: 32 on MD I I 64 on A : 88.6 on B. 2. The mean length of a floor of malt is 115, the mean breadth 112, and the mean depth 4.6 inches; what is its content in floor-«bushels ? Inches. 115 length. 112 breadt/z. 236 115 115 2218.2)12880.0(5.806 area in basaeis. PART VIII. GAUGIN Gr. 303 Bushels. 5.806 area. 4.6 depth. 34836 23224 26.7076 content. By the Sliding Rule. Length. Depth. Breadth. Content. As 115 on B: 4.6 on MD :: 112 on A: 26.7 on B. 3. The mean length of a cistern is 126.4, the mean breadth 62.6, and the mean depth of the barley 32.8 in- ches; how many neat bushels are contained in the cistern? Inches. 126.4 length. 62.6 breadth. m 2528 7584 2218.2)7912.64(3.567 area in bushels. Bushels. 3.567 area. 32.8 depth. 28536 7134 10701 116.9976 content in cistern bushels. .8 multiplier in .Note 2. 93159808 content in neat bushels. 4. If the mean depth of the barley in the foregoing cistern be 38.6 inches; how many neat bushels does it contain? Ans. 110.14896 bushels. 5. The length of a couch is 136.2, the breadth 72.6, and the depth of the barley 42.8 inches ; What is its con- tent in neat bushels ? Ans. 152.60768 bushels. 6. If the depth of the barley in the foregoing couch be 46.3 inches; how many neat bushels does it contain? Ans. 165.08728 bushels. 7. The length of afloor of malt is 236, the breadth 212, and the depth 5.2 inches; What is its content in neat bushels? Ans. 58.643 bushels. 304 GAUGING. PART vm. 8. If' the best cistern-gauge be 68.4, the best conch- gauge 69.8, and the best floor-gauge 109.5 bushels ; from which will the charge of the duty arise? Ans. From the couch. Note. In this Problem, we have supposed the cistern, couch, and floor to be in the form of a parallelopipedon, which is most commonly their shape; but their contents may be obtained by the Rules given in Part IV., Section 1., Whatever form they may assume. PROBLEM X. Cask Gauging. The performance of this part of Gauging is the most difficult that occurs, as no Rules can be given by which the exact form of casks may be ascertained. There are commonly reckoned four forms or varieties of casks, viz. 1. The middle frustum of a spheroid. 2. The middle frustum of a parabolic spindle. 3. The lower frustums of two equal paraboloids. 4. The lower frustums of two equal cones. The Rule for finding the content of the 1st variety, is given in Prob. 13. Part VII. ; that for the 2d in Prob. 18. Part VII. ; that for the 3d in Prob. 16. Part VII. ; and the content of a cask of the 4th variety may be obtained by Prob. 8. Sect. I. Part IV.; but it is very probable that there never was a cask that agreed exactly with any of the varieties; for very few casks are to be met with that will contain so much as the first form, or so little as the third or fourth; so that the second variety is the most general form of casks. Note. Excise officers generally consider all casks as belonging to the first variety, and gauge them as such ; but this practice ought to be abolished, as being injurious to the Trader. (See Nesbz't‘s and Little‘s Practical Gauging.) To take the dimensions of a standing cask. Measure the distance between the inside of the chimb, close to the head, and the outermost sloped edge of the opposite stafi‘, which will be the head diameter within the cask, very nearly. In order to find the bung dia- meter, lay a straight rod AB across the centre of the head, and perpendicular to it, place another straight rod am, so as to touch the bulge of the cask at C; mea- PART VIII. GAUGING. 305 sure the distance between the outer edge of each chimb at b and 0; also measure ab, which should be equal to mn ; then be being added to twice ab, will give the bung diameter CD, including the thickness of the staff on each side of the cask. From this take twice the thickness of the staff, at the bulge, as nearly as your judgment directs, having regard to the size of the cask; and you will obtain the internal bung diameter. It is unnecessary to give any directions for taking the length of a cask in this position, admitting a hole to be in the upper head, which is most commonly the case. Note 1. The external bung diameter ofa standing cask may also be found by dividing the circumference by 3.14“}. (See Prob. XIII. Part 11.) 2. The staves of casks, in general, are thicker at the bulge than at the head; London-made casks, however, have their staves commonly much thicker at the head than at the bulge. The best method of forming a correct judgment is to examine empty casks of the same size and make as those you are about to gauge. By this means, you will come to tolerably correct conclusions relating to the deductions necessary to be made for the thickness of the heads, staves, 8w. (Sc-e the Notes in the next Article.) To take the dimensions of a lying cask. Measure the head diameter in the same manner as directed for a standing cask; and in taking the bung diameter CD, make an allowance for the thickness of the staff at the bung-hole. The length may be most expeditiously found by a pair of callipers, al— lowing for the thickness of the heads, according to the size of the cask; but as it cannot be expected that every person concerned in Gauging is furnished with this instrument, the length may be obtained in the follow- ing manner .—Apply a straight rod AB, to the bulge of the cask; and, at right angles to it, place two others, ma and no, touching the chimbs at each end, and making mr equal to 721'; then measure the distance 71272, from which subtract the depths of the chimbs, together with the thick- ness of the heads, as nearly as you can judge, and the remainder will be the internal length of the cask. Xolc I. In taking the dimensions of a cask, the gauger ought carefully to ob- serve that the bung-hole be in the middle; that the bung-staff be regular and even 306 GAUGING. PART VIII. within ; and that the staff opposite the bung-hole be neither thicker nor thinner than the rest, which he may easily ascertain by the gauging-rod ; and if any impro- priety be discovered, a proper allowance must be made for it, in the dimensions. 2. It is also necessary to observe that the heads of the casks be equal, and truly circular; if not, take cross diameters of each head, and divide their sum by four for a mean diameter. To determine the content of a cash of any of the four varieties, by finding a mean diameter, or reducing it to a cylinder. RULE. Multiply the difference between the bung and head diameters, when it is 6 inches, or less, .68 1 By 2‘: for the g variety, .5 4 Or, if the difi’erence between the bung and head diame- ters exceed 6 inches, .7 11 By 2:; for the 2 variety. 3 .52 4 Add the product to the head diameter, and the sum will be a mean diameter. Square the mean diameter, and’ multiply that square by the length of the cask; then, if the product thence arising be divided by 353.04, you will obtain the content of the cask, in imperial gallons. By the Sliding Rule. Find the difference between the bung and the head diameters, on the inside of the slide marked C; and oppo- site to it, for each variety, is a number, which being added to the head, will give the mean diameter. Then, as the gauge-point on ’D, is to the length of the cask on C, so is the mean diameter on D, to the content on C. Note. Neither of the above methods of finding a mean diameter, is strictly true; but in consequence of their simplicity, they are generally adopted by officers of the Excrse. (See Moss’s Gauging, Section X.) PART VIII. GAUGING .307 EXAMPLES. 1. If the length of a cash be 30, the bung diameter 24, and the head diameter 18 inches; What is its content in imperial gallons, for the first, second, and third ‘varieties? For the middlefrustum of a spheroid, 07' the first variety. Bung diameter 24 inches. Head diameter 18 inches. Difference 6 Multiplier .68 7&8 36 Product ....... 4.08 Head diameter 18. Mean diameter 22.08 Ditto ........... 22.08 17664 4416 4416 487.5264 square. 30 length. Divisor 358.04)14625.7920(41.428 gallons. 3. By the Sliding Rule. The difference between the bung and head diameters is 6 inches, against which, on the line of inches, we find 4.16 on the line marked spheroid, which being added to 18, the head diameter, gives 22.16 inches, for the mean diameter. Then, 09% D. 072 C. 072 D. 0% C. As 18.79 : 3O :: 22.16 : 41.5 gallons Note. The content given by the Rule en. page 267. is 41.8088 gallons. 308 GAUGING. PART VIII. For the mlddlefrustum of a parabolic spindle, or the second variety. Bung diameter 24 inches. Head diameter 18 inches. Difl'erence _6 Multiplier...... .62 12 36 Product 8.77 Head diameter. 18. Mean diameter —2~1.—72 Ditto 21.72 4171.738? square. 30 length. 353.04)14152.7520(40.088 gallant. By the Sliding Rule. Against 6, the difference of the diameters, on the line aof inches, we find 3.8 on the line marked second variety, which being added to the head diameter 18, we obtain 21.8 for the mean diameter. Then, 0n D. On C. 072 D. On C. As 18.79 : 3O :: 21.8 : 40.15 gallons. Note. The content found by the Rule on page 271. is 41.4009 gallons. For the lower frustums of two equal paraboloiols, or the third variety. Bung diameter 24 inches. Head diameter 18 inches. Difference 75 Multiplier .55 *3?) 30 Product 5 Head diameter. 18. Mean diameter #21180 Ditto 21.30 MW square. 30 length. 353.04)13610.7060(38.552 gallons. PART vnr. GAUGING. 309 By the Sliding Rule. The mean diameter, found as above, is 21.3 inches. Then, 07; D. 072 C. 077, D. On C. As 18.79 : 30 :: 21.3 : 38.6 gallons. Note. The content found by the Rule on page 270. is 38.2397 gallons. 2. The length of a cask is 45, the bung diameter 36, and the head diameter 27 inches; What is its content in imperial gallons, for the first, second, and third varieties? Ans. For the first variety, 141.343; for the second, 136.796; (mdfor the third, 131.586 imperial gallons. Note. By the following General Rules, the content of this cask is 133.188 gal— ions, which is between the second and third varieties. (See the Second Example.) GENERAL RULES Forfinding the contents of cashs from the head diameter, bang diameter, and length ; without paying any regard to their variety. RULE I. Add into one sum, 39 times the square of the bung diameter, 25 times the square of the head diameter, and 26 times the product of those diameters ; and multiply the sum by the length of the cask. Then, if the product thus obtained, be multiplied by .000031473, or divided by 3177 3.244; the result will be the content of the cask, in imperial gallons. Note. This Rule is taken from Dr. Hutton’s excellent Mathematical and Philo— sophical Dictionary, vol. i. page 528., and has now been adapted, by the author, to the new imperial gallon. It gives the contents of casks less than the Rule for the second variety, and more than that for the third, being nearly in the middle between them. lut the Dr. observes that, “ It agrees well with the real contents of casks ; as hath been proved by several casks which have actually been filled with a true gallon measure, after their contents had been computed by this method.” RULE II. Divide the head diameter by the bung diameter, to two places of decimals; and find that number in the column of quotients, in the following Table, against which we have a multiplier or factor, for imperial gallons. Multiply this factor by the square of the bung diameter, and the product thence arising by the length of the cash, and the last result will be the content of the cask in imperial gallons. GAUGING. PART VIII: Note I. If the quotient of the head by the bung diameter do not terminate in two places of figures, without a fractional remainder, find the multiplier answering to the first two decimals of the quotient, and subtract it from the next greater multiplier; then if the remainder be multiplied by the fractional part of the quotient, the product will be the corresponding proportional part to be added to the first multiplier. This method ought always to be used when the fractional remainder is large, or accuracy required. 2. This Rule gives the content very nearly the same as Rule 1., and will he found much easier in practice, as it requires a great deal fewer figures in the operation, particularly when the fractional remainder is rejected. A.NEW7TABLE Of multipliers orfactors, forfinding the content of any cask, in imperial gallons. Q‘gggenfiefi Multipliers Qggéienlfeaf’g Multipliers divided by Imggrial [1 dlilvided by Imgogrial 315.133??? Gallone i Slime??? Gallons. i .50 .0018333 .76 .0023038 .51 .0018494 .77 .0023240 .52 .0018657 .78 .0023443 .53 .0018822 .79 .0023649 .54 .0018988 .80 .0023856 .55 .0019155 .81 .0024064 .56 .0019324 .82 .0024274 ’.57 .0019495 .83 .0024486 .58 .0019667 .84 .0024698 .59 .0019842 .85 .0024913 .60 .0020017. .86 .0025131 .61 .0020193 .87 .0025348 .62 .0020372 .88 .0025568 .63 .0020552 .89 .0025788 .64 .0020734 .90 .0026012 .65 .0020917 .91 .0026236 .66 .0021102 .92 .0026462 .67 .0021288 .93 .0026689 .68 .0021476 .94 .0026918 .69 .0021666 .95 .0027149 .70 .0021858 .96 .0027381 .71 .0022050 .97 .0027615 .72 .0022244 .98 .0027850 .73 .0022440 .99 .0028086 .74 .0022638 L00 .0028325 .75 40022837 eARr Vin. GAUGING. 3 1 1: EXAMPLES. . 1. The length‘of a cask is 30, the bung diameter 24, and the head diameter 18 inches; What is’ its content in imperial gallons? By Rule I. Here 24 X 24 X 39 r: 576 X 39 = 22464, thirty~nine times the square of the hung diameter; 18 x 18 X 25 = 324 X 25 = 8100, twenty times the square of the head diameter; and 24 X 18 X 26 = 432 X 26: 11232, twenty- sise times the product of the diameters. Then, (22464 +' 8100 + 11232) X 30 X .000031473 2 41796 X 30 X. .000031473 : 1253880 X .000031473 : 39.46336 im- perial gallons. 0r, 1253880 + 31773.244 2 39.46339 imperial gallons, the same as before. . > By Rule II. Ifere l8 -:— 24 — .75, the quotient of the head divided by the hung diameter. Opposite to this in the preceding 'Tahle, we have the multiplier .0022837; then, .0022837 X 242 X 30 = .0022837 X 576 X 30 = 1.3154112 X 30 = 39.462336 imperial gallons, the same as by the first Rule. late. The content of this cask, found by the foregoing Rules, for the first variety, is 41.428 gallons ; for the second variety, 40.088 gallons; and for the third variety, 38.552 gallons: hence we see that the content given by the General Rules is be- tween the second and third varieties. 2. The length of a cask is 45, the bung diameter 36, and the head diameter 27 inches; What is its content in imperial gallons ? Ans. By the first Rule, 133.188 ; and by the second Rule, 133.185 imperial gallons. 3. What is the content of a cask Whose bung and head diameters are 32 and 24, and length 40 inches? Ans. By the first Rule, 93.5427 gallons. 4. The bung diameter of a cask is 45, the head diameter 34.2, and the length 56.4 inches; what is its content in imperial gallons? , Ans. By the second Rule, 263.1169 gallons. 5. What is the content of a cask Whose bung and head diameters are 37.3 and 28.6, and length 52.8 inches ? Ans. By the second Rule, 170.2364 gallons. 312 GAUGING. ‘ PART VIII. DIAGONAL ROD. To find the content of a cash by the diagonal or gauging TOd. This rod has two lines of numbers graduated upon it. One of them is divided into inches and tenths, for the purpose of taking dimensions; and the other is called a diagonal line, and expresses the content of any cask, in imperial gallons, corresponding to the cask’s diagonal in inches and tenths. then the rod is put into the bung-hole of a cask, so as to meet the head, Where it intersects the staff opposite the bung-hole, the content of the cask is exhibited in imperial gallons on the diagonal line, reckoning from the end of the rod to the centre of the bung-hole. The diagonal rod is very much used in gauging, in con- sequence of the ease and. expedition with which the con— tents of casks may be obtained by it, and it commonly gives nearly the same as the General Rules. Its construction is founded upon Theo. 20. Part 1.; viz. that similar solids are to each other as the cubes of their like dimensions. ' Note 1. Those who have not a diagonal rod, may nevertheless find the content of a cask by that method in the following mannerz..Measure the diagonal of the cask in inches and tenths ; then multiply the cube of this diagonal by .002266, and the product will be the content of the cask in imperial gallons. 2. If the bung and head diameters, and length of a cash be given, the diagonal may be found by the following Rule:_— To the square of half the sum of the diameters, add the square of halfthe length ; and the square root of the sum thence arising will be the diagonal, or distance between the centre of the bung«hole and the point where the middle of the opposite staff and head intersect each other. EXAMPLE S. 1. The diagonal of a cask is found to be 20 inches; What is its content in imperial gallons ? By the Diagonal Line. Opposite to 20 inches, the given diagonal of the cash, we find on the diagonal line, 18 gallons, the content re- gained. By Note I. Here 20 X 20 X 20 :: 8000, the cube of the. diagonal ,- and .002266 X 8000: 18.128 gallons, the same as by the diagonal line. 2. The length of a cask is 80, the bung diameter 24, and the head diameter 18 inches; required the diagonal and content. rim": VIII. GAUGING. 313 By lVote 2, we have (24+18) -.'— 2 = 42 -+— 2:91, half the sum of the diameters; and 30 —:— 2 = 15, half the length of the cash,- then 442134-152) : ~/(441 +225) = 4/666 2 25.8 inches, the diagonal. Again, by Note 1, we have 666 X 25.8 x .002266 = 17182.8 x .002266 = 38.9362248 gallons ; which is nearly the same as given by the General Rules. (See the first Example.) 3. The length of a cask is 45, the bung diameter 36, and the head diameter 27 inches; required the diagonal and the content. Ans. The diagonal is 38.71 inches, and the content 131.4437 gallons; which is the same as found for the thde variety; and difers only 1.74 gallons from the content given by the General Rules. (See the second Example.) PROBLEM XI. To ullage a standing cash. The liquor contained in a cask when it is not full, is called the wet ullage ; the vacuity or space not occupied by the liquor, is termed the di'g nllage ; and the method of finding the content of the liquor, is called ullaging a. cask. RULE. Divide the wet inches by the length of the cask, to three places of decimals; and if the quotient exceed .500, add to the said quotient one-tenth of the excess ; but if it be less than .500, subtract one-tenth of the deficiency: then the whole content of the cask being multiplied by the sum in the first case, or the remainder in the second, will give the ullage required. . ' Note. When the whole content of the cask is not known, it must be found from proper dimensions, before the above Rule can be applied. By the Sliding Rule. Set the length of the cask upon C to 100 on the line marked Segt. St. or SS; then against the wet inches on C, you will have a segment on the line SS, which call a fourth number. Set 100 on the line marked A, to the content of the cask upon B ; then against the fourth number, before found, on A, is the quantity of liquor in the cask, upon B. P 314 GAUGING. PART VIII. EXAMPLE S . 1.. The length of a cask is 30, the bung diameter 24_. the head diameter 18, and the wet inches 22 ; what is the ullage, in imperial gallons? Here 22 —:— 30 = .733, which exceeds .500, by .233, one—tenth part of which is .0233; then .733 + .0233 :- .7563, the multiplier. The whole content of the cash, as found in the General Rules, in the last Problem, is 39.46 imperial gallons; then, 39.46 X .7563 : 29.843598 imperial gallons, the ullage required. By the Sliding Rule. On C. On SS. On C. On SS. As 30 3 100 :1 22 : 75.5, a fourth number. And, On A. On B. On A. On B. As 100 : 39.46 I: 75.5 : 29.6 gallons. 2. Let the dimensions and content be the same as in the last example: what is the ullage in imperial gallons, for 8 wet inches? Here 8 -—:- 30 = .266, which is less than .500 by .234, one-tenth part of which is .0234; then .266 ~— .0234 : .2426, the multiplier; hence, 39.46 x .2426 = 9.572996 gallons, the ullage required. Note. If we add 29.84, the ullage of the first example, to 9.57. the ullage of the second ; we obtain 39.41 gallons, the whole content of the cask, nearly. By the Sliding Rule. On C. On SS. On C. On SS. As 30 : 100 :2 8 : 24.5, a fourth number. And, On A. On B. On A. On B. As 100 2 39.46 22 24.5 ' 9.7 gallons. 3. The length of a cask is 50, the bung diameter 40, the head diameter 30, and the wet inches 20; what are the wet and dry ullages, in imperial gallons? . Ans. By Rule 11. Problem X the content is found to he 182.696 ; and hence the wet ullage 71.25144, and the dry ullage 111.44456 imperial gallon.Q _ PART VIII. GAUGING. V 315 PROBLEM XII. To ullage a lying cash. RULE . Divide the wet inches by the bung diameter, to three places of decimals; and if the quotient exceed .500, add to the said quotient one-fourth of the excess; but if it be less than .500, subtract one-quarter of the deficiency: then multiply the Whole content of the cask by the sum in the first case, or the remainder in the second, and the product will be the ullage required. By the Sliding Rule. Set the bung diameter upon C to 100 on the line marked Seg. Ly. or SL; then against the wet inches on C, you will have a segment on the line SL, which call a fourth number. Set 100 on the line marked A, to the content of the cask upon B; then against the fourth number on A, is the quantity of liquor in the cask, upon B. EXAMPLES. 1. The length of a cask is 30, the bung diameter 24, the head diameter 18, and the wet inches 15 ; what is the ullage, in imperial gallons ? Here 15 —:—- 24 = .625, which exceeds .500 by .125, onefonrth part of which is .03125 3 then .625 + .03125 = .65625, the multiplier. The whole content of the cash, as found by the General Rules, in the tenth Problem, is 39.46 imperial gallons; then, 39.46 X .65625 = 25.895625 imperial gallons, the tillage required. By the Sliding Rule. On C. 0n SL. 0n C. 0n SL. As 24 : 100 :1 l5 3 67, afoarth number. And, On A. On B. On A. On B. As 100 3 39.46 :2 67 l 2 26.4 gallons. 2. Let the dimensions and content be the same as in the last example ; what is the ullage, in imperial gallons, for 9 wet inches? Here 9 + 24 = .375, which is less than .500, by .125, r 2 316 MISCELLANEOUS QUESTIONS PART VIII. onefourth part of which is .03125 ; then .375 — .03125 = .34375, the multiplier; hence, 39.46 X .34375 = 13.564375 imperial gallons, the ullage required. No/e. If we add 25.89, the ullage of the first example, to 13.56, the ullage of the— second; we obtain 39.45 gallons, the whole content of the cask, nearly. By the Sliding Rule. 0n C. ' 0n SL. 0n C. 0n SL. As 24 : 100 :1 9 : 32.5, a fourth number. And, On A. On B. On A. On B. As 100 : 39.46 I: 82.5 1 12.8 gallons. 3. The length of a cask is 32, the bung diameter 25.6, the head diameter 19.2, and the wet inches 15.8 ; What are the wet and dry ullages, in imperial gallons? Ans. By Rule I]. Problem X the content is found to he 47 .892660224: and hence the wet ullage 30.9506, and the dry ullage 16.942 imperial gallons. MISCELLANEOUS QUESTIONS CONCERNING G A U G I N G. 1. The perpendicular depth of a vessel in the form of a parallelopipedon is 52, its breadth 75, and the diagonal of its bottom 125 inches; what is its content in imperial gallons? Ans. 1406.419 gallons. 2. A vessel in the form of a parallelopipedon contains 6'75 imperialgallons ; its length is 85, and its breadth 64 inches; What is its depth ? Ans. 34.4044 inches. 3. The diagonal of a cylindrical vessel is 45, and its diameter 27 inches ; What is its content in imperial gallons ? ~’ Ans. 74.3456 gallons. 4. The greatest diameter of a vessel in the form of the frustum of a cone, is 96, the least diameter 48, and its slant height 51 inches; how many imperial gallons will it contain ? Ans. 685.3257 gallons. 5. A reservoir measures 144 inches in length, 122 in breadth, and 85 in depth; how long will a person be in filling it with water, by means of a pump ; supposing he makes 30 strokes in a minute, and lifts 3 pints of water at each stroke ? Ans. 7 hours, 58.717 minutes. 6. The slant height of a cistern in the form of the ram vm. IN GAUGING. 317 frustum of a square pyramid is 153, the perpendicular height 135, and the side of the least end 92 inches ; What is its content in imperial gallons? Ans. 13.935.2326 gallons. 7. The altitude of a vessel in the form of a hexagonal prism is 60, and the side of its base 30 inches; what is its content in imperial bushels? Ans. 63.2477 bushels. 8. The greatest depth of, the liquor in a cylindrical vessel, placed upon an inclined plane, is 38 inches, the least depth 32 inches, and the diameter of the vessel 36 inches; how many imperial gallons does it contain? Ans. 128.4985 gallons. 9. The transverse diameter of an elliptical bath mea- sures 144, the conjugate 112, and the depth 60 inches; how many imperial gallons will it contain? Ans. 2740.7712 gallons. 10. If the internal diameter of a hollow sphere be 100 inches; how many bushels of corn, imperial measure, will it hold ? Ans. 236.0472 bushels. 11. The top diameter of a conical vessel measures 32, and its slant height 34 inches; how many imperial gallons of wine will it contain? Ans. 29.002 gallons. 12. The top diameter and depth of a vessel, in the form of the greater segment of a globe, are 32 inches each ; how many imperial gallons will it hold? Ans. 108.2773 gallons. 13. If the linear side of each Platonic body be 30 inches ; required their respective contents in imperial gallons. Ans. The content of the tetraedron is 114748, the hexaedron 97 '3674, the octaedron 45.8994, dodeeaedron 746.1385, and the icosaedron 212.4261 imperial gallons. 14. Two porters agreed to drink of a pot of strong beer at two pulls, or a draught each; now, the first having given it a black eye, as it is called, or drunk till the surface of the liquor just touched the opposite edge of the bottom, gave the remaining part to the other; what was the difference of their shares, supposing the pot was the frustum of a cone, whose top diameter was 3.7, bottom diameter 4.23, and perpendicular depth 5.7 inches? Ans. 7.06511 cubic inches. 15. At Konigstein, near Dresden, in Germany, is a cash whose head diameter is 25, bung diameter 26, and perpendicular altitude 28 feet; how many gallons of wine, imperial measure, Will it contain ? P 3 318 , COMPUTING DISTANCES BY PART VIII. Am. The content found by Rule II., Problem X, is 89674.38?) gallons, which exceeds the content of the cash at Heidelberg, by 31387.4255 imperial gallons. (See Example 4, Problem III.) . The Konigstein cask was begun in the year 1722, and finished in 1725, under the direction of General Kyaw ; and is considered to be the largest cask in the world. It consists of 157 staves, each 8 inches in thickness ; and one of its heads is com- posed of 26, and the other of 28 boards. The top or upper head of this enormous cask is railed round, and affords suffi- cient room for twenty persons to regale themselves ; and there are several sorts of large goblets, called “ Welcome Cups,” offered to strangers, who are invited to _ drink by a Latin inscription, which in English is as follows :-—“ Welcome, Tra- veller, and admire this Monument, dedicated to Festivity, in order to exhilarate the Mind with a cheerful Glass, in the year 1725, by FREDERIC AnGUs'rus, King of Poland, and Elector of Saxony, the Father of his Country, the TITUS of his Age, the Delight of Mankind: Drink, therefore, to the Health of the Sovereign, the Country, the Electoral Family, and Baron KYAW, Governor of Kom‘gstez‘n ; and if thou be able, according to the Dignity of this Cask, the most 'capacious of all Casks, drink to the Prosperity of the whole Universe; so farewell." Note. Those who wish to see this subject more fully treated, are referred to Nesbit’s and Little’s Treatise on Practical Gauging. THE METHOD OF COMPUTING DISTANCES BY THE VELOCITY OF SOUND. THE velocity of sound, or the space through which it is propagated in a given time, has been very differently estimated by authors who have written upon this subject. Roberval states it at the rate of 560 feet in a second of time; Gassendus at 1473; Musenne at 1474:; Duhamel, in the History of the Academy of Sciences at Paris, at 1338 ; Newton at 968; Derham, in whose measure Flam- stead and Halley acquiesce, at 1142. The reason of this variety is ascribed, by Derham, partly to some of those gentlemen using strings and plummets instead of regular pendulums; partly to the too small distance between the sonorous body and the place of observation ; and partly to no regard being paid to the winds. ' By the account since published by M. Cassini de Thury, in the Memoirs of the Royal Academy of Science at Paris, 1738, where cannon were fired at various, as well as great distances, under many varieties of wind, weather, and other circumstances, and where the measures of the different places had been settled with the utmost exact-. ness, it was found that sound was propagated, on a PART VIII. ' THE VELOCITY or SOUND. 319_ medium, at the rate of 1038 French feet, in a second of time; and as the French foot is-to the English, in the proportion of 15 to 16, it follows that 1038 French feet are equal to 1107 English feet. Therefore, the difference of the measures of Derham and Cassini is 35 English feet, or 33 French feet in a second. The medium velocity of sound, therefore, is nearly at the rate of 1 mile, or 5280 feet, in 4% seconds, or 1 league in 14: seconds, or 13 miles in 1 minute. But sea miles are to land miles nearly as 7 to 6; and, therefore, soured moves over 1 sea mile in nearly 5% seconds, or 1 sea league in 16 seconds. It is also commonly observed, that persons in health, have about 75 pulsations, or beats of the artery at the wrist, in 1 minute; consequently, in 75 pulsations, sound flies about 13 land miles, or 1117 sea miles, which is nearly 1 land mile in 6 pulsations, and 1 sea mile in 7 pulsations, or 1 league in about 20 pulsations ; and hence the distance of objects may be found by knowing the time occupied by sound, in passing from those objects to an observer. ffaving given the time, in seconds 07' pulsations, that sound is in passing from an object to an observer, tofind flee distance qf the object. RULE S. 1. As 1 second is to 1107 feet, so is the time in seconds, to the distance of the object in feet, which being divided by 3, will give the distance in yards; and this again divided by 1760, will give the distance in miles. 2. As 75 pulsations is to 22140 yards, so is the time, in pulsations, to the distance in yards. 3. When accuracy is not required, say, as 6 pulsations is to 1 mile, or 1760 yards; so is the time in pulsations, to the distance in miles, or yards. Or take T3; of the time in seconds, for the distance in miles. N015. 1n 1% pulsations, sound flies a quarter of a mile; in 3 pulsations,im1f a mile ; in 4% pulsations, three quarters ofa mile ; in G pulsations, one mile, 850. &c.: hence the distance of an object from which sound proceeds, may be easily ascer- tained without the trouble of calculation. EXAMPLES. 1. After observing a flash of lightning, I found by my watch that it was 16 seconds before I heard the thunder ; how far was I from the cloud whence it came? P 4 320 TONNAGE Os snirs. PAR-T VIII. ‘ I, feet. I, W‘- AS 1 : 1107 I: 16 _}§ , 664.2 $.12”: 1107 3)17712 yards. 1760)§99fi(3 miles, 624 yards. A728. 2. If the report of the Tower-guns be heard at Shooter’s Hill, 38 seconds after they are fired 5 what dis- tance are the two places from each other? Here (38 X3) -—2- 14: = 114 —:— 14 = 8.142 miles. 3. After seeing the flash of a gun at sea, I counted 58; pulsations, at the wrist; what was the distance bet seen the ship from which the gun was fired, and the place- where I stood P Ans. 9.7 28 miles; 4. Perceiving a man at a distance felling a tree, I remarked that 5 of my pulsations passed between seeing; him strike with the axe, and hearing the report of the blow ; what was the distance between us ? A728. 1476 yards, 5. The report of a clap of thunder was heard 4 pulsa— tions after seeing the flash of lightning; what was the distance of the cloud from which the thunder issued? Ans. 1180.8 yards. Note 1. Here it may not be improper to remark, that when the report of thunder is heard nearly at the same instant the lightningis seen, the observer is in great dan- ger ; as it is evident that the thunder cloud is very near him. 2. It may also be remarked that persons should never take shelter, in a thunder- storm, under a tree; as the branches will attract the lightning, if it comes within the sphere of their attraction. Many have met premature deaths by taking shelter under trees, during thunder-storms. METHODS OF ASCERTAINING THE TONNAGE OF SHIPS. PRELIMINARY OBSERVATIONS. IF the number of cubic feet of water which a ship dis- places, in sinking from the light water-maria to the load water-maria, be divided by 85, the number of cubic feet n mar VIII. TONNAGE or SHIPS. 021 of sea-water in 1 ton, the quotient will be the number of tons, which the vessel is capable of carrying, or the vessel’s true burthen. Now the number of cubic feet of water so displaced, is exactly equal to the solid content of so much of the body of the ship, as is contained between the two water—lines or marks; but in consequence of the great variety of forms given to this part of the vessel, no exact practical Rule can be given that will apply in all cases. If, indeed, the areas of three or five horizontal sections of that part of the hull contained between the said two water—lines, could be obtained by means of equi-distant ordinates, then the content might be correctly found by Problem VI, Part VEL; but as taking of the dimensions of these sections would be attended with great difiiculty, we shall give the following Parliamentary Rules for ascertaining the ton- nage of Merchants’ and King’s ships. CASE I. ”/7167; the vessel is laid dry. RULE. Measure the length on a straight line along the rabbit of the keel of the ship, from the back of the main stern- post, to a perpendicular line let fall from the fore-part of the main-stem, under the bow-sprit; from this length subtract 35‘— 0f the extreme breadth; and the remainder will be the length of the keel, for tonnage. The breadth must be taken from outside to outside of the plank, in the broadest part of the ship, whether above or below the main-wales, exclusive of all manner of doubling planks, or sheathing, that may be wrought upon the sides of the vessel ; then multiply the length of the keel, in feet, by the breadth ; divide the last product by 94; and the quo- tient will be the tonnage required. EXAMPLES. l. The length from the back of the stern-post to a line let fall from the fore-part of the main-stem, is 88 feet 6 inches ; the extreme breadth from outside to outside of the plank, 26 feet 6 inches 3 required the tonnage of the ship. P 5 322 TONNAGE or SHIPS. PART VIII. feet. Gross length 88.5 of the heel. 26.5 X 2 = 15.9 the deduction. True length E6 dg'fl‘erence. 26.5 breadth of the beam. 376370 4356 1452 :1923.90first product. Then 1923.90 X 13.25, half the breadth : 25491.6750 second product. And 25491'675 -:—- 94 : 271.188 tons, the tonnage re~ quired. 2. The length from the back of the stern-post to a line let fall from the fore-part of the main-stem, is 108 feet 9 inches; the extreme breadth from outside to outside of the plank, 29 feet 6 inches ; required the tonnage of the vessel. ‘ Ans. 421.469 tons. CASE II. When the vessel is afloat. RULE. Let fall a plumb—line over the stern of the ship, and measure the distance between this line and the aft-part of the stern-post, at the land water-mark; then measure from the top of the said plumb—line, in a parallel direction with the water, to a perpendicular point immediately over the load water-mark, at the fore-part of the main-stem ; from the last measured distance subtract the former ; and the remainder Will be the ship’s extreme length. From this length deduct 3 inches for every foot of load-draught of water, for the rake abaft ; and also ~2— of the ship’s ex- treme breadth, for the rake of the stem ; and the remainder will be the true length of the keel for tonnage. The extreme breadth must be measured, and the tonnage found as directed in the first ”Case. EXAMPLE. The true length of an eighty—gun ship, after all deduc- tions are made, in taking the dimensions, is 150 feet 9 inches; and the extreme breadth 50 feet 6 inches; re— quired the tonnage of the vessel. Ans. 20449478 tons. Note 1. It is found, by experience, that ships of war carry less ; and most mer- chant ships carry considerably more tonnage than they are rated at, by the pre- ceding Rules. 2. Some writers on this subject, divide by 100, instead of 94, for King’s ships. 0n the same principles, the divisors for Merchants’ ships should be decreased, perhaps to 90 or 92. 3. Solutions to the last two Questions may be found in the Key to Nesbit’s and Little’s Practical Gauging. PARTIX. fHflGONOMETBY. 323 PART IX. TRIG-ONOMETRY. TRIGONOMETRY is an important branch of the mathe- matical sciences; by it we determine the magnitudes of the earth and planets, their mutual distances and motions: it is that part of science which teaches how to measure the sides and angles of triangles, plane and spherical. Plane trigonometry is the art of measuring plane tri- angles: and of determining the sides and angles which principally depend on the properties of the circle, and circular arcs. Every circle is divided into 360 equal parts called de- grees, each degree into 60 equal parts called minutes, each minute into 60 equal parts called seconds, and so on by a sexigesimal division into thirds, fourths, fifths, &c. An angle is spoken of as containing as many degrees, minutes, seconds, 850. as are contained in the are, or part of the circumference by which it is measured. Degrees, minutes, seconds, 8m. are marked at the top of the figures of their quantity in the order °, ’, ”, ”’, W, &C. as 20° 34’ 40” 37’ ’ ’ 48iv are 20 degrees, 84 minutes, 40 seconds, 37 thirds, 48 fourths; but it is not usual in modern practice to extend the division by 60 beyond seconds, lower denominations, if there be any, being eX pressed in decimal parts of a second; thus, in the above number, 37’ ’ ’ 481" = T5536 of a second; hence it would be written 20° 84’ 40”.63. A right angle is measured by the fourth part of the circumference, or 90° ; an obtuse angle is greater than 90°, and an acute angle is less than 90°. The complement of an angle is What it wants of a right angle or 90°, and the supplement is what it wants of two right angles, or 180°. The complement of an arc is what it wants of a quad- rant, or 90°, and the supplement is what it wants of a semicircle, or 180°. The chord of an arc is a straight line drawn from one extremity of the arc to the other, as AB is a chord to the P 6 324 TRIGONOMETRY. PART 1);. are ADB; it is also the chord of the arc AEB, the re- mainder of the circle. The sine of an arc, is a straight I: line drawn from one extremity of H L the are, perpendicular to, and ter— {:74 minated by, the diameter drawn \ from the other extremity of the arc. / I As AF perpendicular to, and ter- ll:——T--—)D minated by the diameter DE, is the \ L/ sine of the are AD; it is also the ' . L sine of the are AE, the supplement \ of the are AD. I The tangent of an arc, is a straight line drawn per- endicularly to the extremity of the diameter, just touch- ing the are, in a point at one extremity of that are, and continued from thence to meet a line drawn from the centre of the circle through the other extremity, which line is called The secant of , the arc : thus DGr is the tangent of the are DA, and CG is the secant of the same are; DC is also the tangent of the are AE, and CG the secant of the are AE, the supplement of AD. The versed sine of an arc is that part of the diameter contained between the sine and the are ; thus FD is the versed sine of the arc AD, and FE is the versed sine of the supplemental arc AE. If the arc DH be taken equal to a quadrant, and the diameter HI be drawn perpendicular to the former, DE, then the are AH is the complement of the arc AD, and AK, HL, CL are the sine, tangent, and secant respect- ively of the arc AH; we have, in other words, AK the sine of the complement of the are AD, HL the tangent of the complement of the same are ; or, more briefly, AK is the co-sine of the are AD, HL is the co-tangent, CL the co-secant, and HK is the co-versed sine, of the are AD. In place of the are we may put the angle at the centre, which the arc subtends and measures: thus, AF, DG, CG is the sine, tangent, and secant respectively of the angle ACD, or angle ACE, and AK, HL, CL is the sine, tangent, and secant respectively of the angle ACH, or the co-sine, co-tangent, and co-seeant of the angle ACD : ACH being the complement of ACD. Conversely, ACD is the complement of ACH, and AF, the sine of ACD or ACE, is the co-sine of ACH, and similarly of the tangent PART 1x. TRIGONOMETRY. . 325 and secant; hence, when the sine, cosine, tangent, co- tangent, secant, cosecant of any obtuse angle is required, we may take the sine, cosine, &c. of the supplemental acute angle : or, for the sine of an obtuse angle take the cosine of its excess above a right angle; for the cosine of an obtuse angle take the sine of its excess above a right angle: in place of the tangent, cotangent, secant, cosecant of an obtuse angle, we may take the cotangent, tangent, cosecant, secant of the excess of the given angle above a right angle. The. are ADB is double of the arc AD, and its chord AB is double of the sine AF ; hence the sine of any are is half the chord of twice the are. The versed sine of an arc and the versed sine of its supplement are together equal to the diameter; thus FD, F E are together equal to DE. “ The figure KACF is a parallelogram, and KA and CF are equal; taking CF, therefore, in place of AK for the cosine of the angle ACD, or of the arc AD, it is manifest when the arc is less than a quadrant that its versed sine and cosine are together equal to the radius, and that when the arc is greater than a quadrant the excess of its versed sine above its cosine is equal to the radius : thus FD and CF are together equal to the radius CD, and the excess of EF above CF is equal to the radius CE. In the right-angled triangle ACF, whereof the radius CA is the hypothenuse, and AF, CF the sine and cosine of the same angle, viz. of the angle ACF, the squares of the sine and cosine of any angle are together equal to the square of the radius. In the right-angled triangle CDG, we find the squares of the radius and tangent are together equal to the square of the secant. From the right-angled triangle LCH, the squares of the radius and cotangent are together equal to the square of the cosecant. From the similar triangles CFA, CDG, we have CF : FA: 2 CD : DG ; or the rectangle under CF and DG equal to the rectangle under FA and CD, or the rectangle under the cosine and tangent equal to the rectangle under the sine and the radius. Again, CF : CA1: CD : CG; or the rectangle under CF and CG is equal to the rectangle under CA and CD, £326 «TRIGONOMETRY. PART IX. or the square of CA or of CD ; that is, the rectangle under the cosine and secant is equal to the square of the radius. In like manner from the similar triangles ACK, LCH, we find the rectangle under the sine and cotangent equal to the rectangle under the cosine and radius; and the rectangle under the sine and cosecant equal to the square of the radius. Comparing the triangles CDG, CHL together, they are similar, and DG 1 CD: : CH 1 HL; whence the rectangle under the tangent and cotangent (i. 8. under DG and HL) is equal to the square of the radius (2'. e. to the rect- angle under CD and CH). By the known property of the circle, the rectangle under CF and FD is equal to the square of AF; or the square of the sine is equal to the rectangle under the versed sine, and the versed sine of the supplement (which Mendoza in his Tables calls the suversed sine, as HK is the coversed sine). These are the most important and useful properties of the lines drawn in and about the circle. It is plain, from an inspection of the diagram, that the sine and cosine can never exceed the radius, the secant. and cosecant can never be less than the radius, and the tangent and cotangent may have any degree of magnitude whatever. And in the same proportion that the tangent is less or greater than the radius, so is the cotangent greater or less. When the angle is 90°, its sine becomes the radius The side of a hexagon which is the chord of 600 is equal to the radius, whence the sine of 30° is half the radius. When the angle ACD is 45°, the triangle CDG has the angle CGD also 45°, and therefore DG equal to CD; whence the tangent of 45° is equal to the radius. In every plane triangle the greater side is opposite the greater angle, and conversely the greater angle is opposite the greater side. And when two sides are equal, the angles opposite are equal; and when two angles are equal, the sides opposite to them are equal. In every plane triangle the three angles are together equal to two right angles, or 180°; hence if one of the angles be a right or obtuse angle, the rest are acute. PART IX. TRIGONOMETRY. 327 In every plane triangle, if one angle be known the sum of the other two is found by subtracting the known one from 180°; and if two of the angles are known, the third is found by subtracting the sum of the given angles from 180°. Hence whenever two angles of a plane triangle are known, the whole three may be considered as known. And in a right-angled plane triangle, if one of the acute angles be known the other is found by subtracting the known one from 90°, the acute angles being complements one of the other. Every plane triangle consists of six parts, viz. three sides and three angles; and it is necessary that three of these be known, in order to determine the other three, and consequently the triangle itself. The combinations of six things taken three together are twenty, whence it would appear that there are twenty cases in general in plane triangles; but‘several of these cases fall under one general case, so that there are in reality but three, the case of the three angles being given being eX- cluded, for the triangle cannot be determined from these data alone: since an infinite number of triangles have their three angles respectively equal, it is therefore neces- sary that a side be included among the given quantities. The theee general cases are: 1. Given -——A side and angle, opposite each other, to- gether with another side or another angle. 2. Given— Two sides and the angle included. 3. Given— The three sides. These include right—angled triangles of course; but these latter are of sufficient importance to form a distinct class, and be treated of separately : and by sides we shall understand not any indifferently, but only those about the right angle, that opposite being always called the hypo- thenuse ; and here are three cases, viz. 1. Given——-An acute angle, and with the hypothenuse or side. 2. Given —-— The hypothenuse and a side. 3. Given —-— The two sides. Now the T rigonometrical Tables of Sines, Tangents, 8w. are a register of 5400 right- -angled triangles, in which the hypothenuse or a side is made radius, and expressed by a power of 10; and 1n this table a right- angled triangle may always be found similar to any proposed right- 328 TRIGONOMETRY. PART IX. angled triangle, whence the latter can be determined by proportion from the former : the actual operations being performed by logarithms, Whose uses are copiously ex- plained in directions which accompany the tables. And, note—in the following pages no subtraction by loga- rithms is admitted in the operations, except a preliminary subtraction for obtaining a certain result called arithme- tical complement, and the subtraction of 10 or its multi- ples, 100 and its multiples, &c. when necessary. RIGHT-ANGLED PLANE TRIANGLES. PROBLEM I. Given an acute angle, wit/2 the fiypotfienuse 07' a side, to determine the triangle. RULE. As sine of any angle : side opposite :1 sine of any other angle : its opposite side. DEMONSTRATION. Let ABC be the proposed triangle right-angled at C, and ADE the tabular triangle similar to it; let AD be made radius: then DE will be the sine of the opposite angle A, and AE the cosine of the same angle, or sine of the complement of B ; that is, the sine of the Opposite angle D, and AD the radius, is equal to the sine of 900, the opposite angle E. Thus each side of ADE will be the sine of its opposite angle, and the sides of ABC being as the similar sides of ADE, are also as the Sines of the opposite and equal angles A, B, and C; whence in the triangle ABC any one side : sine of angle opposite :: any other side 2 sine of its opposite angle. And conversely, Sine of any one angle: side opposite :: sine of any other angle : its opposite side. In using logarithms, the logarithm of the first term is subtracted from the sum of the logarithms of the second and third; and when the first term is radius, its logarithm, which is 10, will be written after the logarithm of the PART IX. TRIGONOMETRY. 329 second term with the sign of subtraction, thus taking up only one line instead of two. And when radius occurs in the second or third terms, its logarithm 10 will be added at once to the logarithm of the third or second term, thus saving a line. And when the subtractive term is other than radius, we take its arithmetical complement in the following manner : — When the number is less than 1, its arithmetical com— plement is its difference to 1; when the number is greater than 1 and less than 10, the arithmetical complement of it is its difference to 10; if it be greater than 10 and less than 100, then it is taken from 100, and so 011. New supposing the number decimally expressed, the arithmetical complement is found at once by subtracting each digit, beginning at the highest or left hand, from 9, and the last significant figure on the right from 10. This avoids carriage in the subtraction, and may be effect- ed at sight; so that while any number is being read its arithmetical complement may at the same time be written. Thus, of .375 the arithmetical complement is .625, of .3750 it is .6250, of .03074990 it is .96925010, of 3.75 it is 6.25, of 37.5 it is 62.5, and so on; and the arithme— tical complement of the arithmetical complement of any number being taken, reproduces the original number. Thus the arithmetical Complement of 375 being 625, the arith- metical complement of this reproduces the original num- ber 375. ' This is likewise true in angles and arcs. The comple- ment of the complement of an arc reproduces the original angle or are; and the supplement of the supplement reproduces the original angle or arc; and thus the cosine of the complement of an angle or are is the sine of the angle or are itself : similarly the cotangent of the complement is the tangent, the cosecant of the complement is the secant. That number from which a lesser number is taken to obtain its arithmetical complement, may be called the con- stant, being the sum of the number and its arithmetical complement. The constant of two acute angles or arcs less than 90°, which are complements of each other, is 90°. The constant of two arcs or angles, which are sup- plements of each other, is 180°. When instead of subtracting a number its arithmetical 330 TRIGONOMETRY. PART IX.‘ complement is added, the constant must be subtracted in place of the original number. Thus, 1463 —— 358 = 1463 +642 —— 1000 ; 642 being the arithmetical complement of 358, and 1000 the constant. When several numbers are to be subtracted, their arithmetical complements may be added instead, and the sum of the constants subtracted from the sum of the added quantities. ' This method is made use of in the modern astrono- mical tables, wherein it frequently happens that some quantities are to be added and some subtracted; but by increasing each quantity by a constant, these all become additive, and the sum of the constants which is known is subtracted once for all from the sum, the computer having nothing to do but to add previous to this. Single sub- traction avoids the mistake of adding instead of subtract- ing, and subtracting instead of adding, —-—mistakes which might be said to be unavoidable in the former case, when the number of quantities are very numerous (as they are in Damoiseau’s Tables of the Moon, wherein 46 cor- rections are required to obtain the moon’s true place in the heavens). Thus, for‘instance, instead of performing the diversified operations 90° —43° + 76° ——32° — 15°, let the constant 100° be applied to each subtractive quantity, which are three, and we have the uniform operations 90° + 57° + 76° + 68° +85° —- 300°. EXAMPLES. l. The hypothenuse is 420, and one of the acute angles is 41° 30’: to find the sides, and the other acute angle. The other acute angle is 90°——41° 30’ :48° 30’. As sine 90° 1 hypothenuse 420 . . . 262325—10 :2 sine 41° 30’ . . . 9.82126 : Opp. side 278.3 . . . 2.44.451. As sine 9O ; hypothenuse 420 . . . 2.62325— 10 :2 sine 48° 30' . . . 9.87446 : opp. side 314.6 . . . 2.49771‘ By the Gunter’s Scale. 1st. Extend the compasses from 90° to 41° 30’ on the line of sines ; that extent will reach from 420 to 278.3 on the line of numbers. 2d. Extend the compasses from 90° to 48° 30’ on the PART IX. TRIGONOMETRY. 331 line of sines; that extent will reach from 420 to 314.6 on the line of numbers. 2. Let the greater side be 276, and one of the acute angles 40° 10’: to find the hypothenuse, the remaining side, and the other acute angle. Since 400 10’ is less than 45°, it is the lesser of the two acute angles; the greater is 49° 50', and therefore opposite the greater side 276. As sine 49° 50’ . . . . . ar. co. 0.11681—10 2 276 I: Sine 90° . . . . . . 12.44091 : hypothenuse 361.2 . ' . . . 2.55772 As sine 49°50 . . . . . ar. co. 0.11681—10 I 276 . . . . . . . . . . 2.44091 I: sine 40° 10' . . . . .‘ . . 9.80957 opposite side 233 . . . . . . 2.36729 By tlze Ganter’s Scale. 1st. Extend the compasses from 49° 50’ to 90° on the line of sines; that extent will reach from 276 to 361.2 on the line of numbers. 2d. Extend the compasses from 49° 50’ to 40° 10’ on the line of sines ; that extent will reach from 276 to 233 on the line of numbers. 3. The hypothenuse is 744, one of the acute angles 30° 44’ ; required the other acute angle, and the sides. Ans. Angle 59° 16’, tlae sides 380.2 and 639.5. 4. One of the sides is 470 links, the angle adjacent to it 40°; required the opposite angle, the other side, and the hypothenuSe, and also the area of the triangle. Ans. Angle 50°, side 394.4 links, figpotfienase 613.5 links ; area 3 mods, 28.3 perches. 5. The hypothenuse of a right-angled triangle is 924 yards, one of its angles 50° 10’; required the other acute angle, and the sides. Ans. Angle 39° 50’, sides 709.5 and 591.9 yards. 6. Standing on the bank of a river, and wanting to know the distance of an object on the opposite side, I measured 250 yards in a straight line at right angles to the direction of the object, and then found the angle be— tween this line and the direction of the object at the second station to be 66° 30’ ; required the breadth of the 332 TRIGONOMETRY. PART IX. river, and the distance of the object from the second Station. Ans. Breadth of the river 575 yards, the distance of the objectfro-m the first station ; and 627 yards the distance from the second station. PROBLEM II. Given the hypothenase and a side, to find the other Side and the angles. RULE I. , By the proportions in page 328., we have hypothenuse : sine 90" :1 given side : sine of the opposite angle. Then find the other acute angle, which is opposite the required side; and sine 90° : hypothenuse :: sine of angle opposite the required side : the required side. When the angles are not required, the side may thus be found :— The square of the hypothenuse is equal to the squares of the sides, whence the square of one side is equal to the difference of the squares of the hypothenuse and of the other side, which is equal to the rectangle of the sum and the difference of the hypothenuse and side. “Thence, in logarithms, The logarithm of the square of a side, equal to twice the logarithm of the side, is equal to the logarithm of the sum of the hypothenuse and other side, added to the loga- rithm of the difi‘erence between the hypothenuse and this other side. Consequently, RULE II. Add to the logarithm of the sum of the hypothenuse and given side the logarithm of the difference between the hypothenuse and given side, and half the sum of the two logarithms will be the logarithm of the side required. EXAMPLE S. 1. The hypothenuse is 420, one of the sides is 310; required the angles, and the other side. PART 12:. TRIGONOMETRY. 333 As hy pothenuse 420.. . . ar. co. 7.37675— 10 'sine 90O ‘ ' Side 310 . . . . . 12.49136 sine 470 34’. . . . . . . 9.86811 The other angle 42° 26’. As sine 90O ' hypothenuse . . . . 2.62325-—- 10 :: sine 420 26’. . . . . . . 9.82913 , other side 283.4 . . . . . . 2.45238 By the Ganter’s Scale. 1st. Extend the compasses from 420 to 310 on the line of numbers; that extent will reach from 900 to 470 34’ on the line of sines. 2d. Extend the compasses from 90’ to 42° 26’ on the line of sines; that extent Will reach from 420 to 283.4 on the line of numbers. 2. The hypothenuse is 420, one of the sides 283; re- quired the other side. Hypothenuse . . . . . . 420 Side . . . . . . . . 283 Sum . . . . . . . . 703 . . 2.84696 Difilerence . . . . . . 137 . . 2.13672 Sum . . . . . . . . .\ . .498368 Other Side . . . . . .310 3 .2 49184 3. The hypothenuse is 564, one of the sides 372; re- quired the angles, and the other side. Ans. Angles 41° 16’ and 48° 44’, side 423. 9. 4. Given the hypothenuse 196, one side 142, to find the other side. Ans. 135.1. 5. The hypothenuse of a right-angled triangle is 670 yards, one of its sides 526.2 yards 5 required the angles, the other side, and the area of the triangle. Ans. Angles 380 15’ and 51°45, side 414.8, area 109133.88 yards. PROBLEM III. ‘ Given the two sides, to determine the angles and hypotltenuse. RULE. As one of the sides: other side :: 1adius: tangent of the angle adjacent to the first side, or opposite the second. 334 V TRIGONOMETRY. PART IX; And,—As sine of this angle : second side :: sine 900 : hypothenuse. DEMONSTRATION. By the proportion, page 328., transposed, As one of the sides : other side :: sine of angle opposite the first side 1 sine of angle opposite second side. ::cosine of angle adjacent to first side: sine of same angle. . Multiply the last two terms of this proportion by the tangent of the angle; then, . As one of the sides : other side:: cosine of angle ad- jacent to the first side x tangent :2 sine >< tangent. But cosine >< tangent : sine >< radius; by page 326. Therefore, as one of the sides : other side :: sine of angle adjacent to first side >< radius 2 sine >< tangent. Dividing out the sine from the last two terms, As one of the sides 1 the other or second side :2 radius : tangent of the angle adjacent to the first side, or oppo- site the second. ’ ' And as sine of this angle : second side i: sine 90o : hy- pothenuse. ' ' ‘ EXAMPLE S. , , 1. The sides are 560 and 444 links; required the angles and hypothenuse. ' AS560 . . . . . . .ar. co. 7.25181—10 I 444 I: rad. . . . . . . . 12.64738 2 tan. 380 25 . . . . . . . 9.89919 Other angle 51 035’. As sine 38° 25 . . . . . ar. co. 020665—10 I 444 :: sine 90° . . . . . . 12.64738 ' :hyp. 714.6 . . . . . . . 2.85403 By the Gunter’s Scale. “E 1st. Extend from 444 to 560 on the line of numbers; this extent on \the line of tangents will reach from 450 to 380 25’ in one direction, and to 51° 35’ in the opposite direction. 2d. Extend the compasses from 38° 25’ to 90° on the line of Sines; that extent will reach from 444 to 714.6 on the line of numbers. PART 1x. TRIGONOMETRY. 335 2. The sides of a right-angled triangle are 375 and 428 yards ; required the angles and hypothenuse. Ans. Angles 41° 13’ and 48° 47’, figpotlzenuse 569. 3. The sides of a triangle are 300 and 400; required the hypothenuse. Ans. 500. OTHER EXAMPLES. 1. The side of a triangle is 125, the angle adjacent to it 26° 35’ ; required the other angle, the other side, and the hypothenuse. Ans. Angle 63° 25’, side 62.55, ligpot/ienuse 139.8. 2. The hypothenusc is 637, and one of the angles 47° 37’; required the other angle and the side. Ans. Angle 42° 23’, sides 470.5 and 429.4. 3. The sides of a right-angled triangle are 25 and 125; required the angles and hypothenuse. Ans. Angles 11° 19’ and 78° 41’, hgpotlzenuse 127.5. 4. The hypothenuse is 1023, and one of the sides 473 ; required the angles, and the other side. Ans. Angles 27° 33’ and 62° 27', side 907. 5. The hypothenuse is 813, and one of the sides 312; required the other side only. ' Ans. 750.8. OBLIQUE-ANGLED PLANE TRIANGLES. PROBLEM I. ' Given a side and its opposite angle, together wit/z another side or angle, to determine the triangle. RULE. As the sine of any one angle : side opposite :: sine of any other angle : its opposite side. - And, Any one side 2 sine of angle opposite :1 any other side I sine of its opposite angle. DEMONSTRATION. Let ABC be the proposed triangle, and DEF a circle described with the radius of the tables, in which is in- scribed the triangle DEF, similar to the triangle ABC; so that DE : DF : EF::AB : AC : BC. Now the side DE is the chord of the arc DE, or of the angle at the centre which this are subtends, and which is double the 336 TRIGONOMETRY. PART IX. angle DFE at the circumference; hence we may say DE is the chord of twice the opposite 0 angle F, DF the chord of twice ’\ the opposite angle E, and EF the chord of twice the opposite angle D. But the chord of an arc is I‘- B twice the sine of half the arc (page 325.) ; consequently DE is equal to 2 sine F, DF to 2 sine E, and EF to 2 sine D 3 whence we have 2 sin. F : 2 sin. E : 2 sin. thAB : AC 3 BC. Trans- posing, dividing out by 2, and putting the equal angles, Viz. C for F, B for E, and A for D, there results AB : AC : BC :1 sin. C : sin. B : sin. A; that is, the sides of plane triangles are as the sines of their opposite angles, and conversely; hence, in the same triangle, the sine of any given angle : its opposite side : 1 sine of any other given angle : its opposite side. Note: If the sum of two angles, being known, be less than 90°, its sine may be used for that of the third angle; and if greater, the cosine of the excess may he used for the sine of the third angle, these quantities being equal. See page 325. . EXAMPLES. l. The two angles of a triangle being 50° 30’ and 68° 20’, and the side opposite this second angle 500, it is required to determine the triangle. First angle . . . . . . . . . 50° 30’ Second angle . . . . . . . . . 68° 20 Sum . . . . . . . . . . . 1180 5C" Excess . . . . . . . . . . . 28° 50’ As sine 68° 20’ . . . . ar. (:0. 0.03182—10 : opposite side 500 . . . . 2.69897 :: sine 50° 30’ . . . . . . 9.88741 : opposite side 415.1 . . . . 2.61820 AS sine 680 20’ . . . . . . 0.03183—10 : opposite side 500 . . . . 2.69897 :: cosine excess 280 50’ . . . 9.94252 :third side 471.3 . . . . . 2.67332 By the Gunter’s Scale. Knowing the three angles 50“ 30’, 68° 20’, and 180°— (500 3OI+68° 20’) 2:1800—1180 50’2600 10’. PART Ix. ' TRIGONOMETRY. 337 1st. Extend the compasses from 68° 20’ to 50° 30’ on the line of sines; this extent will reach from 500 to 415.1 on the line of numbers. 2d. Extend the compasses from 68° 20’ to 61° 10’ on the line of sines; this extent will reach from 500 to 471.3 on the line of numbers. When one angle and two sides are given, the second angle found by the proportion of p. 335 is sometimes doubt- ful; that is, it may be either acute or obtuse, since the same sine answers to two angles supplements of each other. This happens when the given angle is opposite the lesser of the two given sides ; for if opposite to the greater. it must be acute. See the following example. 2. Two sides of a triangle are 710 and 406, and the angle opposite the less is 34° ; required the other side and angles. As the side 406 . . . ar. co. 7.39147—10 : sine of its opposite angle 34° . 9.74756 22 Side 710 . . . . . . . . 2.85126 : sine of its opposite angle . . 9.99029 Second angle is 77° 56’, or 102° 4’; and the third is 68° 4', or 4.30 56’. As sine 34° ar. co. 0.25244— 10 ar. co. 0.25244—10 : opposite side 406 2.60853 . . . . 2.60853 :: sine 68° 4’ . . 9.96737 or 43° 56 . . 9.84125 2 third side 673.5 . 2.82834 or 503.8 . . 2.70221 By the Gunner’s Scale. lst. Extend the compasses from 406 to 710 on the line of numbers; this extent will reach from 34° to 77° 56’ on the line of Sines. Then 180° —(34° + 77° 56’) :180°~ 111° 56’:68° 4’, the third angle. 2d. Extend the compasses from 34° to 68° 4’ on the line of sines ; this extent will reach from 406 to 673.5 on the line of numbers. And this is one solution. ‘ Taking new 180° -— 77° 56’2102° 4’ in place of 77° 56’, the third angle is 180°—— (102° 4’+34°) =180° — 136° 4’ = 43° 56’: 77° 56’— 34°. And extending the compasses from 34° to 43° 56’ on Q 338 TRIGONOMETRY. PART IX. the line of Sines, the same extent will reach from 406 to 503.8 on the line of numbers; which is the other solu- tion. Note. When the question admits of two solutions, the second angle in the second solution is the supplement of the second angle in the first solution, and the third angle of the second solution is the difference between the second angle in the first solution and the given angle. Thus, in the above example, first solution, given angle is 340, second angle 77056’, third angle 680 4’ ; in the second solution, second angle is 1800--77O 56’=1020 4’, and the third angle is 77° 5G’—-34°=4 0 56’. 3. Two angles of a triangle are 42° and 98°, the side opposite the lesser is 124; required the third angle, and the remaining sides. Ans. An 36 40° sides 183.5 and 119.1. I n o a 4. Two Sides of a triangle are 300 and 480, the angle opposite the lesser is 38° 40’ ; required the other angles, and remaining side. Ans. 880 30’ and 52° 50’, side 382.6. Or, 910 30’ and 49° 50’, side 367. 5. Two sides of a triangle are 300 and 480 the angle I I OL/ / I ’ c oppos1te the greater 1s 38 4:0 ; required the other angles, and the remaining side. Ans. 22° 59’ and 118°2l’,side676.l PROBLEM 11. Given two sides and the included angle, tofind the rest. RULE. As the sum of the sides : their difference: 1 cotangent of half the given angle : tangent of an angle E. Then the lesser of the unknown angles of the triangle is the complement of the sum of the angle E and half the given angle. And the greater of the unknown angles is the sum of the complement of half the given angle, and the angle E. The other side is found by Problem I. DEMONSTRATION. Let ABC be the proposed triangle; let DE be the radius of the tables and FG passing through E at right angles to DE ; let DF, DG meet FGr in the points F and Gr, making equal angles EDF, EDG, each equal to half the supplement of the angle C, i. e. half the exterior angle 0 PART IX. TRIGONOMETRY. ‘ 039 at C ; then will the angle FDG equal the supplement of C, and if FD be produced, the C angle GDH will equal the angle C. Let GrH be drawn meeting PE in H, and making the angle DGH B A equal to one of the angles A, B, equal to A the greater for in- stance; then angle DHG equal to the angle B, whence the triangles GHD, ABC are similar; bisect EH in I, and join El; and since FG, FH are bisected in E and I, EI Will be parallel to GH, and angle EID equal to the angle GED; and drawing from D, DK parallel to IE the angle KDF, equal to the angle GHD, equal to the angle B; the angle EDF, equal to half the supplement of C, is equal to half the sum of the angles A and B, and the angle EDK is equal to this half sum diminished by B, and therefore equal to half the differ-- ence of the angles A and B ; and DE being radius, EF is the tangent of EDF the half sum, and EK is the tangent of EDK the half difference. New, FH being the sum of HD, DF, or HD, DGr, and F1 half that sum, and DI being the difference between Fl, FD, or their equals HD, DGr, we have by similar triangles FI : DI: :EF : EK, or twice PI 1 twice DI: :EF : EK; that, is the sum of the sides DH, DG : their difference :2 tangent of half the sum of their opposite angles : tangent of half their difference; and since the sides BC, AC are proportional to DH and DG, the sums and differences will be proportional. ‘Whence by substitution, Sum of the given sides (AC, BC) 1 difference of these sides:: tangent of half the sum of their opposite angles (A, B) : tangent of half their difference. Adding the half difference to the half sum gives the greater angle, and subtracting the half difference from the half sum gives the lesser angle. Note. Instead of taking the tangent of half the sum of the unknown angles, we may take the cotangent of half the given angle ; and adding the half difference of the unknown angles to half the known angle gives the complement of the lesser angle, and adding the half difference to the complement of half the given angle gives the greater angle. All the angles being known, the third side may be found by the lst Problem. Q2 h 340 TRIGONOMETRY. PART IX. EXAMPLE S. 1. Two sides of a triangle are 90 and 120, the included angle 104°; required the other two angles, and the re- maining side. Here 210 is the sum of the sides, and 30 the difference, half angle 52°. Sum 210 . . . . . . ar. co. 7.67778—10 Difference 30 . . . . . . . 1.47712 Cotangent 52° . . . . . . 9.89281 Tangent % difi'erence 6° 22’ . . 9.04771 52° . . . . . . . . 38° 6° 22/ . . . . . . . 6° 22’ Sum 58° 22’ EEO—22’ 31° 88’ lesser angle. 44° 22’ greater angle. Since sine 104° equal cosine 14°, _ As sine 31° 38’ . . . . ar. co. 028027—10 : opposite side 90 . . . . . 1.95425 :2 cosine 14° . . . . . . . 9.98690 I third side 166.5 . . . . . 2.22142 By the Gunter’s Scale. lst. Extend the compasses from 210 (the sum) to 30 {the diff.) on the line of numbers; the same extent will reach from 38° (half sum of the unknown angles) to 6° 22’ (half their diff.) on the line of tangents. Then 88° +6° 22’ 244° 22’ greater angle, 38°—6° 22’:31° 38’ lesser angle. 2d. Extend the compasses from 31° 38’ to 76° (the supplement of 104°) on the line of sines, and this extent will reach from 90 to 1665 on the line of numbers. ’2. Two sides of a triangle being 180 and 212, and the included angle 72° 30’; required the other two angles, and the remaining side. Ans. 47° 24’ and 60° 6’, side 233.2. 3. The distance of the earth from the sun 100 millions of miles, the distance of Jupiter from the sun 520 millions of miles, and the angle of these distances 20° ; required the distance of Jupiter from the earth in millions of miles. ‘ Am. 428 millions of miles PART IX. TRIGONOMETRY. 341 PROBLEM III. Gieen the three sides, tofind the angles RULE As the base (for which the longest side is taken): sum of the other two sides :: difference of these two sides : difference of the segments of the base made by a per- pendicular from its opposite angle. Add this number to, and subtract it from, the base 5 then Half sum : greater segment. Half difference : lesser segment. Then as the greater of the other two sides: greater segment :: sine 90° : sine of angle opposite the greater segmentzcosine of angle adjacent to the greater seg- ment. And as lesser side : lesser segment :: sine 90° : sine of angle opposite the lesser segment 2 cosine of the angle adjacent to the lesser segment. Hence, the angles at the base being known, the angle opposite is known ; for angle opposite =1800—(angles at the base) = sum of the complements of the angles at the base. DEMONSTRATION. Let ABC be the proposed triangle, with either angular point, say C, as a centre, and with either of the two sides A C, BC, ———the larger, AC 101- instance, ~—as a radius, describe the circle AFD- , produce AB to meet the circle in D, and BC both ways to meet the circle in E and F; join CD, and let fall A ‘ ——~ the perpendicular CG which bisects AD, and the rectangle under FB“, BE . will be equal to the rectangle under AB 'BD; or AB : FB :ZBE : BD. Now CA and CF being equal, BB is the sum of the sides AC, BC; and CA, CF, CE being equal, BE is the difference of the sides AC, BC. The side AB is divided in G into the segments AG, GB; and since AG is equal to GD, therefore BD is equal to the difference of the seg- ments AG, GB, the side AB being the sum. Hence, call« ingtl e greatest side the base (for then the perpendicular Q 3 342 ZIRIGONOMETRY.‘ PART IX. drawn from the opposite angle will fall within the triangle), there results base : sum of the other two sides:: difference of these two sides 1 difference of the segments of the base. Adding this difference to the whole base, and halving the sum, gives the greater segment, which subtracted from the whole base leaves the lesser segment ; and the triangle ABC is thus divided into two right-angled triangles, Whose hypothenuses AC, BC, and bases AG, GB, are known; thence determining the angles ACG, BCG, their complements give the angles CAB, CBA, and their sum the angle ACB. Thus AC : radius :: AG : sine ACG; or side of the given triangle ABC 3 adjacent segment of the base :: radius : sine of angle opposite the segment. Note. The greater segment is adjacent to the greater side. EXAMPLE S. 1. The sides of a triangle are 184, 140, and 92; re- quired the angles. Here 184: is made base. As base 184 . . . . . a1“. co. 7.73518—10 : sum of sides 232 . . . . . 2.36549 :: difference of sides 48 . . . 1.68124 1 difference of segments 60 . . 1.78191 Base . . . . . . . . 184 Diff. seg. . . . . . . . 60 Sum . . . . . . . . 24'4 Greater secr. . . . . . . 122 Lesser seg. . . . . . . 62 As side 140 . . . . . ar. c0. 7.85387—10 Adj. segment 122 . . . . . 2.08636 +10 Cosine 29° 22’ . . . . . . 9.94023 As side 92 . . . . . . . . 8.03621 —10 Adj. segment 62 . . . . . . 1.79239+10 Cosine 47° 38’ . . . . . . 9.82860 One angle 29° 22’ . . . . . . . 60° 38’ Another angle 47° 38 . . . . . 42° 22’ Third angle . . . ' . . . . . . 103° 0’ PART IX. TRIGONOMETRY. 343 By the szter’s Scale. 1st. Extend the compasses from 184 the base, to 232 the sum of the sides on the line of numbers; this extent will reach from 48 the difference of the sides, to 60 the difference of the segments of the base. Then add this to, and subtract it from, the base; half the sum and difference gives 122 the greater, and 62 the lesser segment. 2d. Extend the compasses from 140 the greater of the two sides, to 122 the greater of the segments on the line of numbers; this extent will reach from 90° to 60° 38’ on the line of sines; then 90°~60° 38’: 29° 22’ is one of the angles. 3d. Extend the compasses from 92 the lesser side, to 62 the lesser segment on the line of numbers; this extent will reach from 90° to 42° 22’, the complement of which, 47° 38’, is another angle; and 60° 38’+42° 29’2103O, is the third angle. 2. The sides of a triangle are 76, 148, and 110; re- quired the angles. Ans. 29° 54’, 1030 56’, 46° 19’ PART X. THE APPLICATION OF TRIGONOMETRY TO HEIGHTS AND DISTANCES. THE mensuration of heights and distances, and of inac- cessible objects, depends on plane and oblique trigo- nometry, as also on oblique sailing. The most proper instruments for taking the horizontal and vertical angles are the theodolite, the sextant, the quadrant, and the compass; the former being furnished with one or two telescopes and a vertical arc. The base lines are generally measured by a tape—line of 50 or 100 feet, or by the Gunter’s chain. Q 4 344 ‘ TRIG-ONOMETRY. PART X. , THE COMPASS. EXAMPLES. 1. Wanting to know the height of a tower which is inaccessible, or intercepted by a C moat, I found by the quadrant the angle at the side of the meat and the top of the tower to be 38° 30’ ; I then measured 200 yards in a straight line from the tower, and found the angle to be 27°_40’ : required the height of the tower, and the width of the moat. By Theorem 6, Part 1., ang. CDB —ang. CAB:ang. ACD=38° 30’—-27° 20/2110 10’. Then, as sine ang. ACD ll0 10’ ar. co. 0.71295 — 10 2 side AD 200 . . . . . 2.30103 2 2 sine ang. CAB 27° 20’ . . 9.66197 : side DC 474.2 . . . . . 2.67595 T? {j , w .____.__.._. PART X. TRIGONOMETRY. 345 And, as sine 90° 1 hyp. DC 474.2 . 2.67595— 10 ‘2 sine ang. CDB 38° 30’ . . 9.79415 ': side 130 295.2 . . . . . 2.47010 And, as sine 90° 1 hyp. DC 474.2 . 2.67 595—10 :: sine DCBzcos. 38° 30’ . . 9.89354 2 Side DB 371.1 . . . . . 2.56949 Hence 371.1 yards is the width of the moat, and 295.2 yards the height of the tower. 2. Wanting to know the height of a cloud, I found it formed an angle of 23° 30’ ; I then moved 300 yards in a straight line towards the cloud, and found the angle to be 40° 30’ : required the height of the cloud, and its distance from my first and latter stations. See Fig. of EX. 1. Here, ang. CDB—ang. CAB = ang. ACD =40° 30’“ 230 302170. As sine ACD 17° 2 side AD :: sine CAB 23° 30’ : side DC. As sine ACD : side AD :: sine ADC=sine CDB (its supplt.) 1 AC. As sine 90° : hyp. DC :: sine CDB 40° 30’ : BC. Ans. BC is 265.7 yards, DC 409.1, and AC 6664. 3. Two observers, at the distance of 880 yards from each other, took at the same instant the angle of elevation of a balloon; the angle at the nearest station was 34° 30’, and at the other 26° : required the height of the balloon. See Fig. of EX. 1. - Here, ang. CDB ——ang. CAB = ang. ACD =34° 30’ ~26° 2 8° 30’. As sine ACD 8° 30’ 1 side AD 880 :2 sine CAB 26° 2 side DC. As sine 90° 2 hyp. DC :3, sine CDB 34° 30’ : perp. BC. Ans. BC 1478 yards, the height of the balloon. 4. Being on the bank of a river, and observing a tower standing upon a hill on the opposite side of the river at station A, I took E my observation, or angles C and D, at the top and bottom of the tower, and found the angle CAB to be 56°, and DAB 28° 30’. Q 5 346 TRIGONOMETRY. PART X. There being a moat on my other side, I measured 240 yards on the bank of the river to station E; I then found the angle CAE to be 78° and CEA 76°: required the height of the tower and hill, and the distance from station A, to the top and bottom of the tower. Here, ang. ACE=180° —— (ang. CAE + ang. CEA): 180° -— 154° :260. ang. ADB=complt. of DAB :2 90° -— 28°80’=Sl°30’. =supplt. of ang. ADC. . ang. CAD = ang. CAB —ang. DAB = 56° —28°30’ =27O 30’. ang. ACD=complt. of CAB: 90° — 56°:34O, and AE=940. As sine ang. ACE ,' side AE :: sine ang. CEA .' side AC. As sine ang. ADC=sine ang. ADI) : side AC :: sine ang. ACD : side AD. As sine ang. ADC=sine ang. ADB 3 side AC 2: sine ang. CAD 3 side CD. And, As sine 90° : hyp. AD :: sine ang. DAB : side BD. Ans. CD the height of the tower 279.1 yards, BD height of the hill 161.3. The two dis- tances AC, AD, 531.2 and 338 yards. ,5. Sailing from Nelson Haven to New Plymouth, New Zealand, on the Taranaki coast, I observed an object on the summit of Mount Egmont, the angle of elevation being 9° 50’ ; I then sailed six miles upon the same point of the compass, and found the angle of the same object to be 18° 34’ : required the height of the mountain above the level 01" the sea. See Fig. EX. 1. Ans. 2.15 miles, or 3784 yards, or 11352feet. 6. Sailing along the shore of Port Nicholson, New Zealand, I saw a cape of sand which bore from me N.N.E. ; I then sailed NTW. by W. 24: miles, when the same cape bore E.N.E.: required the distance of the cape at both stations. Let A represent the place of observation, and AB the N.N.E. direction, which is 2 points, and AC the NWV. by W. course, being 5 points; then the angle BAC is 7 points, or 78° 45’. Draw the dotted line a I) parallel to the N. S. line, and take angle aCB:67° 30’26 points for the E.N.E. course, then angle hCA=angle CAN :5 pointS, PART X. TRIGONOMETRY. 347 and aCB = 6 points; whence angle BCA: (16 - 11) points = 5 points, or 56° 15’, and angle B = (16— 5—7) points: 4 points, or 450. Then sine ang. B . 450 ; side AC 24 miles A :2 sine ang. BAG 78° 45’ : ide BC; and sine ang. B 1 side AC: 1 sine ang. BOA : side AB. Ans. BC is 33.29 miles, and AB 28.22 miles. 7. From a station at A an object at B, the distance being 8 miles, bore N.E. byN. ; another , N B object at C, distance 53:, w A A E miles, bore SE. : re- quired the distance of the two objects. C Ans. 10.56 miles. s 8. Suppose two ships sail from the same port at the same time; one sails NE. by E. 28 miles, the other sails SE. %— S. 34: miles: I demand the distance of the two ships. Ans. 41.86 miles. 9. Saint Paul’s Cathedral at London bore from me N.N.E., and travelling N.W. by W. 12 miles it then bore N.E.; required the distance of Saint Paul’s from both stations. Ans. The distances are the two equal sides of cm isoceles triangle, being 30% miles. A.) 10. I observed a tower, C CD, on the top of an inaccessible hill BD ; the angle of elevation of the top of the hill is BAD D 360, and the top of the tower BAC 49° 15’. I then measured in a di- rect line AE 120 yards, / i and found the angle of E A 3 Q6 348' TRIGONOMETRY. PART x. the top of the tower to be BEG 30° 30’ ': required the height of tower and the hill, and its distance from my first station, allowing 5 feet for the height of the instru- ment. Ans. Height of tower 53.7 yards, heightof hill 94.87 yards, distance from station A 123.7 yards. 11. From a window near London Bridge, which ap- peared to be on a level with the 0 bottom of the Monument on the opposite side of the river, the D angle of elevation of the top is 10° 02’, and from another . / window in the same building 24 A B feet above the former the angle is 8056; required the height and distance of the Monu- ment. Ans. Ileight 0f Monument215.l feet, distance 1216 feet. Note. The erection of the Monument was commenced in the year 1671, and finished in the year 1677, in remembrance of the dreadful fire of London in l666, and the rebuilding of the City in the reign of Charles 11., under the inspection of Sir Christopher Wren. Leigh, in his “ Picture of London,” says this noble pile is undoubtedly the first modern column in the world : it is of the Doric order, fluted, and exceeds in height the pillars of Trajan and Antoninus. 12. A celebrated navigator sailing on the Caspian Sea, observed through his glass an object on the summit of Mount Ararat, which bore SW. by W. ;. he then sailed 36 miles N.W., when the same object bore SEW. by 8.; required the distance of the object at each station. Ans. Distance 92.26 miles each station. Note. Mount Ararat is a celebrated mountain of Armenia, on which Noah’s ark rested at the time of the flood, where it is supposed still to remain on the summit. The capacity of the ark was 2.730782 cubic feet, which is about twenty times larger than a first-rate man of war, and may be considered the object ob— served. A Russian traveller says this mountain is 17260 feet above the level of the sea, and is doubtless volcanic. Several attempts have been made to reach the top of the mountain, but no one has got much beyond the snow limit. , Sir R. K. Porter visited this region, and is of opinion the inaccessible summits have never been trodden by the foot of man since the days of Noah. 13. Being at sea I observed two headlands, whose bear- ings from each other I found by the chart to be W.N.W. and E. SE, and distance 20 miles; the first bore from me 83V. by S., and the second S.E. : required my distance from each, headland. Ans. The SW. by S. headland 7.8, and SE. 20 miles. 14. Two ships sail from the same port or station A; one sails N .‘W. by N. 28 miles to station B, and the other PART X. TRIGONOMETRY. 349 sails N.E. by N. 38 miles to station C; required the dis— tance and bearings from each other. Ans. The hearing of B from C, is W. by S., and C from B, E. by N. ; distance 37.6 miles. 15. “Wanting to know the height of a tower CD, on the opposite side of a river, which appeared to be on a level where I stood at A, I there fixed a station, and B then ascended the acclivityr of a hill AB, in a straight line from the tower 250 yards, when I found the angles of depression to be as follows, viz. that of my 1 first station EBA 44°, of the bottom of tower EBC 280, and the angle EBD of depression with the top of tower was 18°. Required the height of the tower, distance from my first station to its bottom, as also the horizontal distance; and my height above the top of the tower. Ans. Horizontal distance 326.6; distance from sta- tion to tower 146.8; height of the tower 67.53; and my height above the top 106.16 yards. 16. The distance between a tower T, and a church C, was known to be 20 miles; '1‘ from a ship at anchor A, I r saw a windmill at M, in a W right line with the tower; the bearing of both was NNV. by VV._; the church at the same time bore N.N.‘W. %W. After which we weigh» ed anchor, and sailed upon a W. by N. course AH: 25 miles, and the mill and the church were in a straight line HC, and bore N.E. by N. Required the distance between the church, tower, and windmill, at each station, or place of observation, A and H? lst. In the triangle ACH the angles are known, and the side AH is known. For AC bears N.N.VV. %W., or 2% points, AH —- W. by N. or 7 points, whence angle CAH is 4-},- points, or 500 37’. The bearing of HC is N.E. by N., or 3 points; the com— E . / 350 TRIGONOMETRY. PART X. plement of this added to the angle HAW, which 1s one point, the sum is 6 points, or 670 30’, the angle AHC, whence angle ACH IS 5 i points, 01 610 52’. As sine ACH: AH: :sine AHC: AC. sine ACH: AHZ' sine CAH: HC. 2d. In the triangle ACT, the angle TAC and the oppo- site side TC are known, together with AC, whence AT may be found. For AC bears N. N. W. ~VV., or 21 points, AT -—~ NW. by W. or5 points. Hence angle TAC IS 2% points, or 28° 7’. As TC: sine TAC: ::AC sine ATC. Angle ACT: BOO—(angle TAC+angle ATC). As sine TAC: TC:' sine ACT: AT. 3d. In the triangle AMH the angles are known, and the side AH is known, For AM bears N.W. by VV., or 5 points, AH -- . W. by N. or 7 points; whence angle MAH is 2 points, or 22° 30’. AHM or AHC is 6 points, or 67° 30’, whence angle AMH 1s 8 points, or a right angle. As sine 90° _ AH: , sine AHM : AM, sine 90° ' AH: ' sine MAH ‘ HM. 4th. In the triangle TMH, WhiCh 1s right- -angled at M (for since AMH 1s a right angle, TMH 1s likewise one), TM:AT—— AM 1s known, and HM 1s known. Then, as HM: radius:' , TM' . tangent THM, sine THM : TM: : sine 90O : TH. Ans. AM:23.09, Ans. HM: 9.57, AC:26.19, 110221.91, AT:38.84, HT: 18.413. 17. Coasting along the sea shore, N c I observed at A, two headlands, B B and C, the fi1 st b01e N. N. W., the second N. N. E. 3E. Then stee1ing E..NE.1E. 164 miles to D, the first headland bore W...NVV, the second N. W. by N. l—VV. Required the bearing and distance of the two headlands from each other? /D V A "E ’1 PART X. TRIGONOMETRY. 351 1st. In the triangle BAD the angles, and the side AD are known, For AB bears N.NJV., or 2 points, And AD -— E.N.E.%E., or 6%, points, whence angle BAD is 8?3 points, or 95° 37’: And DB bears W’.N.VV., 0r 6 points, AD —- ENE. %E. or 6% points; whence angle ADB is (16—12%) points=3% points, or 390 22’, and the angle ABD 4 points, or 45°. As sine ABD : AD: 2 sine BAD : BD, sine ABD : AD: : sine ADB : AB. 2d. In the triangle ADC the angles, and the side AD are known, For DB bears VV.N.W., or 6 points, DC -— NJV. by NiW. or 3%; points; whence angle BDC is 2% points, and angle ADB is 3% points, whence angle ADC is 6% points, or 70° 19’. AC bears N.N.E. 43E, or 2% points, AD —— E.N.E.%E., or 6% points ; whence angle DAC is 3% points, or 42° 11’, and angle ACD is 6 points, or 670 30. As sine ACD 2 AD: : sine ADC 1 AC, sine ACD : AD: : sine DAC 1 CD, 3d. In the triangle BCD the sides CD, BD are known, and also the included angle BDC, which is 2_E—; points, or 300 56’. Then BD+CD : BD—CDRcotand. fin BDC : tang. of half the difference of the angles DBC, DCB : And sine DBC : CD: 2 sine BDC : BC, the distance of the two headlands B and C. The bearing of ED is WVNWV” and of BD, it is E.S.E., which is 6 points, or 67° 30’; to this add the angle DBC, which is 25° 29’, the supplement of the sum is 87° 1’, the bearing of BC, that is, C bears N. 87'0 1’ E. from B, and B bears S. 87° 1’ WV. from C, and their distance is 13.81 miles. NOTE. — The determination of the distances and the bearings of any number of ints is effected in a similar manner by taking observations of each in succession; If this be applied to buildings, we may, in a similar manner, determine the ground plan of the outline, and the elevation of the building, whether it be perpendicular, 352 TRIGONOMETRY. PART X. or inclining as the leaning tower of Pisa,ox combine vertical and inclining iines as the Pyramids of Egypt. 18. Wishing to know the length of the object AB, which is oblique to the 3 ground plane, and its per- pendicular altitude BC, I took a first station at D, from which the object appeared upright, and ob- A 0 served the angle ADB, D 630 20’; fixing a station staff at D, I measured DE, 150 feet, and at the second station E, I took the angles DEB, 55°, DEA, 33° ; leaving a station stafi“ at E, I returned to my former station D, and took the angles: BDE 72° 10’; ADE, 102° 50’. Required the length AB, the vertical altitude BC, and the inclination BAG. 1st. In the triangle BDE the angles, and the side DE are known, for angle DBE: 180° ~{BDE +DEB). Then, as sine DBE : DE: 1 sine DEB : BD. 2d. In the triangle ADE the angles, and the side DE are known, for angle DAE:180° —(ADE+DEA). Then, as sine DAE : DE: : sine DEA : AD. 3d. In the triangle ABD the sides BD, AD are known, and also the included angle ADB, whence AB and angles BAD, ABD are determined by Prob. II. of Oblique-angled Triangles. ' The angle BAD is the angle of inclination of the object 7w IL 4th. In the right-angled triangle BCD, As sine 90° : BD: : sine ADB : BC, the altitude of AB. Ans. Length AB:l45.7 feet, inclin. BAD, 71° 4’, perp. BC: 137.8 feet. 19. I wanted to know the distance between two places A and B, but could not meet with any station :, ~ /: from whence I could see ' / ’I‘ both objects. I measured a line CD:200 yards; from C the object A was visible, and from D the object B was visible, at each of which places I set C23 PART X. TRIGONOMETRY. 53 up a pole. I also measured FC=200 yards, and DE=200 yards, and at F and E set up poles. I then took the angles : AFC, 83° ; ACE, 54° 31’; ACD, 53° 30’; BDC, 1560 25’ ; BDE, 54° 30’; BED, 880 30’. Required the distance AB. 1st. In the triangle AFC angle FAC=180° —(AEC + ACE). Then, as sine FAC : EC: : sine AFC : AC. 2d. In the triangle ACD there are known AC, CD, and the included angle ACD, whence the angle CDA and the side AD may be found. 3d. In the triangle BDE angle DBE=18OO—(BDE +BED). Then, as sine DBE : DE: : sine BED 3 BD. 4th. In the triangle ABD there are known AD, BD, and the angle BDA: angle BDC ——angle ADC; whence AB is determined =845A yards. 20. From a station at D, I perceived three objects A, B, C, whose distances from each other I knew to be as follow, Viz. AB=12 miles, BC=9 miles, and AC:6 miles. At D I took the angles : ADC=92O 30’, and BDC:33° 45'. Hence it is required to find my distance from the objects. Nora—In the triangles ADC, BDC, one side and an angle opposite only is given in each, whence it would not seem possible to determine the other sides : but be- cause AB is given, we may. from a property of the circle, determine what is re. quired. . lst. Let the triangle ADB be circumscribed by the circle ADBE, and CD joined, cutting the circle in E, and AE, BE be joined. The angles ADE, ABE, which stand on the same seg- ment AE are equal, as also the angles BDE, BAE ; therefore in the triangle ABE all the angles, and the side AB are known ; whence AE, BE may be found by Prob. I. Oblique—angled Triangles. 2d. In the triangle ABC, whereof the sides are given, the angles may be found by Prob. III. Triangles. Hence the angles CAE, CBE become known. [If the angles ADC, BDC taken together are less than the angles CAB, CBA taken together, then the angle 354 TRIGONOMETRY. PART X. AEB, the supplement of ADB, is greater than ACB, the supplement of the angles CAB, CBA, and then C lies beyond E, out of the triangle AEB : and the angles CAE, CBE are the differences of the angles at the base in the triangles ABC, ABE. If the point C lie within the triangle ABE, in this case the angles at the station, viz. ADC, BDC taken together I are greater than the angles CAB, CBA taken together; and the angles CAE, CBE are the differences of the angles CAB, EAB; CBA, EBA as before. If the point C be within the triangle ABD, the angles CAE, CBE will be the sum of the angles at the base in the two triangles ABC, ABE, instead of the difference, as in the former case] In the triangle ACE, having AC, AE, and the angle CAE, find the angle ACE; whence the angle BCE is known, since BCA is known. 3d. Now, in the triangles ADC, BDC, all the angles are known in each, and likewise a side; whence the sides AD, CD, BD can be found. Ans. AD=14, CD:15.6, BD=10.7. 21. Three objects A, B, C, forming a triangle, were visible from one sta— tion, at D within the triangle ; the angle ADB was observed to be 123° A 45’, CDB: 132° 22’, and consequently ADC: 1030 53’; the distance AB was known to be 12 miles, from B to C 9 miles, and from A to C 6 miles. Be- quired the distance of each object from the station at D. lst. Here angle ABEzangle ADEzsuppl. of ADC angle BAEzzangle BDEzsuppl. of CD13. Hence, in the triangle BAE, all the angles are known, and also the side AB, whence AE, BE are found by Prob. I. of Oblique-angled Triangles. 2d. Finding the angles of the triangle ABC from the sides being given, we have, by adding, the angles GAE, CBE; and the sides about the angle CAE being known, viz. AC, AE, all the angles of the triangle ACE may be found: and angle BCE becomes known, as also angle CBE. 3d. In the triangles ADC, BDC all the angles are PART X. TRIGONOMETRY. 3 5 5 known, and a side AC, BC in each is known; whence the other sides are found, viz. AD 5.5, DC 1.4, DB 8. 22. From a station at D, I perceived three objects A, B, C, whose distances were as follow: AB 12 miles, AC 13 miles, BC 7 miles ; at D the angle ADB‘ was 20°. D, was in CA produced. Required the distances DA, DB, and DC. In the triangle CAB find the angle CAB, from the sides given ; the supplement of this angle is angle BAD; hence, B in the triangle ABD, all the angles are known, and the side AB is known; whence AD, D BD may be found, and CD=AD+AC. Ans. AD 7.4: miles, BD 18.6, CD 20.4. In a similar manner the distances may be found when D is on one of the sides of the triangle ABC. 23. Three objects A, B, C, E in a straight line whose dis- tances were, AC 36 yards, A B BC 84 yards, AB 120 yards, were Visible from one station at D, from whence the angle ADC was 18° 56’, and BBC D 25° 21. Required the dis— tances DA, DC, DB. let. The triangle ABD being circumscribed by the circle, and DC produced to E, angle ADC =angle ABE, and angle BDCzangle BAE. Hence, in the triangle ABE, all the angles, and the side AB are known; whence the sides AE, BE are determined. 2d. In the triangle CAE, the sides AE, AC, and the included angle CAE, being known, the angles AEC, BEC are known, for BEA is known. 3d. In the triangles AED, BED all the angles are known, and a side AE, BE in each; whence AD is found =94.8, CD=108.5, and BD:168. 24. Wishing to know the dimensions of the pyramid ABCDE, which was inaccessible on all sides, I chose a station G, such that the edge AE appeared perpendicular to the ground plane; I measured the angles: AGB, 20° 4’; AG'D, 43° 52’; DGE, 58° 29’; AGE, 43° 49’; C 356 TRIGONOMETRY. PART X. BGE, 47° 22’. I then took a second station at II, measuring GH 36 feet, AE appearing vertical from H, as at Gr; I measured the angles: AHB, 17° 21’; AHD, 37° 12’; AHE, 39° 18’. I then took a third station at I, from which the edge BE appeared vertical, and measured the angles: BIA, 49° 37’; BIC, 40° 32’; BIE, 58° 58’; EIA, '70° 32’; E10, “66° 55 ’. I then moved from I to K, measuring IK 57 feet, where, as at I, BE appeared ver- tical : the angles observed were, BKA, 39°; BKC, 29° 43’ ; BKE, 48° 27’. And at station L, from whence the edge CE appeared vertical, I observed the angles: OLE, 57° 31’; CLD, 31° 48’; at M, in the direction of CL produced, I measured LM 60 feet, and the angle CMD 25° 35’. From these data it is required to find the dimensions of' the pyramid, its vertical height, the area of the ground plan, and of each of its faces, and its solid con- tent, supposing it to be solid. 1st. In the triangle BGII the side GE is known, and all the angles ; then, As sine GBH : GH::sine AHB : BG. In a similar manner DG may be found in the triangle DGH; and EG in the triangle EGH. 2d. In the triangle BGE, the sides BG, EG are known, and the included angle BGE; whence the edge EB is determined by'Prob. II. Oblique-angled Triangles. In a similar manner the edge ED is determined from the triangle EGD. 3d. In the triangles AIK, CIK we may find IA, IC, Just as 13G, DGr were found in the triangles BGH, DGH. PART X. TREGONOMETRY. 357 And from the triangle IEK, EI may be found in a similar manner as EG was found in the triangle EGH. And in the triangles EIA, EIC, the edges EA, EC are determined in the same manner as EB, ED, in the triangles EGB, EGD, were determined. 4th. In the triangle EAG the sides EG, EA are known, and the opposite angle AGE; then As EA : sine AGE: :EG : sine EAG; and the supplement 'of this is the inclination of EA to the ground plan, which is known, as also the angle AEG; the vertical altitude EF is thus found: As sine 90° .2 EA: 2 sine of the inclination of EA : elevation EF. And sine AGE : AE: ; sine AEG : AG. In a similar manner may BI, CL be found in the tri- angles BEI, CEL. 5th. In the triangle DAG the sides AG, DG are known, with the included angle AGD; whence the base AD is known, and the angles GDA, DAG. Similarly, in the triangle BGA, the base line AB may be found, and the angles BAG, GBA; and in the triangle CBI the base BC may be found, and in the triangle CLD the base CD. Hence all the base lines of the pyramid are known, like- wise all its edges, and its altitude. 6th. In the ground plan ABCD let the diagonal BD be drawn ; since the angles BAG, DAG are known, the angle BAD is known; whence the diagonal BD may be found by the second Problem of Oblique—angled Triangles: thus the triangles ABD, CBD have a common base BD, which is known, and likewise their sides known; whence the ground plan may be drawn, and its area found, as in p. 130. 7th. In each of the faces ABE, ADE, CDE, BCE the sides are known; whence their areas are determined as in p. 130. 8th. The area of the ground plan being known, and the altitude EF, the solid content is found, as in p. 132. Am. BG: 226.5, DG=187.5, EG:289.6, BE: 215.2, DE=249.4, IA=194:.7, IC=150.6, EI: 233.7, EA:249.4—, EC:222.9, any. EAF: 53029’, EF:200.5, AG: 60.5, any. BAE: 27°3’, AB:171, (mg. DAF:60° 7’, any. BAD 358 TRIGONOMETRY. PART X. 287° 10’, AD:150, B12416, BC: 122, any. ECF264O 3’, CL=30.1, LD=939.3, CD: 213.8, BD=221.8. Area of ground plan: 25670 square feet; area of face ABE218094 square feet; area of face BCE: 12810 square feet; area of face CDE:22344 square feet; area of face ABEL-17839 square feet; solid content: 1715612 culzic feet. MENSURATION OF ALTITUDES BY THE BAROMETER AND THERMOMETER. The following method for determining difl'erence of levels by the barometer and thermometer is the reduction of a formula marked 7: in p. 64. of the translation of Fran- coeur’s Hydrostatics. 1. At each of the stations Whose difference of levelis required an observer is placed, provided with a barometer and two thermometers, one thermometer being attached to the barometer to ascertain the temperature of the mer- cury of the barometer, the other being detached and placed in the open air to ascertain the temperature of the air; each observer at the same time, previously agreed upon, notes the states of his barometer and thermometers. Both barometers, it is almost needless to say, must be similar in their construction, viz. if one be graduated by inches and parts, such as is generally used in England, the other must also be of the same construction, or if one be a metrical barometeri‘é, such as is generally used in France, the other must also be of the metrical construc— tion: a similar remark applies to the four thermometers, viz. if one be of Fahrenheit’s graduation, all must be so; if one be of the centigrade graduation, all four must be so. 2. 'Whatever kind of barometers be used, the difference of the logarithms of their indications, omitting the index, 9' The barometer used in England is graduated by inches and decimal parts ; the metrical barometer is graduated by millemetres or thousandths of a. metre, and 1 inch English=25.39 millcmetres. In Fahrenheit’s thermometer, the freezing point is marked 320, the boiling point by 2120, the interval being 180°: in the cen- tigrade thermometer, the freezing point is marked 00, and the boiling point is marked 1000; hence 1800 of Fahrenheit’s thermometer=100° of the centigrade thermometer. And note, 1 metre=3.280899 English feet. PART x. TRIGONOMETRY. 359 being taken, is to be increased or diminished according to the states of the attached thermometers; thus, take the difference of the heights of the attached thermometers, and multiply it by .00008, if they are of the centigrade construction, or divide four times the difference by 90000, if they be of F ahrenheit’s construction ; then, if the degrees ' of the lower thermometer are less than those of the upper one, this product or quotient is to be added to the differ— ence of the barometer logarithms, but added in the con- trary case. The result is the first factor. 3. Add together the degrees of the detached thermo- meters, and multiply the sum by .002, if they are centi- grade, but for Fahrenheit’s, diminish the sum by 64°, and divide the remainder by 900 ; the result added to 1 gives the second factor. 4. If the first factor be less than .07, multiply together the number 18393 and the two factors; the product is the difference of the heights of the upper and lower station, nearl , in French metres ; or multiply the number 60345.6 and the factors together, the result is the height, nearly, in English feet. When the difference of the barometer logarithms is 0.7 or above, 18336 is to be used instead of 18393 for the metres, and 60158.6, instead of 60345.6, for the feet. 5. A correction for the rotation of the earth may be found as follows, —- Take the differencebetween 90° and twice the latitude, and then, As radius : sine of the difference :: .002837 x approximate height: correction for latitude. This cor- rection is to be added to the approximate height when the latitude is less than 45°, but subtracted when the latitude is greater than 45°. 6. A second correction, when the difference of the baro- meter logarithms is not less than .07, may be found as follows, —— Add to the approximate height, if in metres, the num- ber 15926.6, if in feet, the number 52253.8; divide for metres by 6366198, for feet by 20898000, and multiply by the approximate height; the result is the second cor- rection, which is always to be added. A part of this correction, viz. 33335‘5’5 of it is due to the diminution of the weight of the mercury at the upper station. The whole of this second correction may be omitted *360 VTRIGONOMETBY. PART X. when the difference of the barometer logarithms is less than 0.7. The approximate altitude corrected gives the altitude very near to' that determined by geometrical measure- ,ment. It is hardly necessary to say that the instruments should be very carefully constructed, similar in their indications and furnished with verniers, that we may be enabled to estimate the smallest fractions of the scale. The time se— lected should be such as to afford the circumstances most favourable for observations, viz. calm weather, the middle of the day, 8m. 5 NOTE. —— The reader may consult on this subject a very elegant paper by M. Ra- gnond, from whom the above numbers are taken. See also Mécam‘gue Celeste, tom. 'v. p. 289. Log. of 18393=4.26465 Log. of 18336=4L26324 Log. of .002837=7.45287—10 Log. of m-%T©~g=3.19613 — 10 Log. of 60345.6:478065' Log. of 60158.6:4.77923 Log. of 1.8686=0.27151 Log. Office—9%nnn=2-67990 — 10 EXAMPLES. 1. We shall take as an example, the measurement of Mont Blanc, the highest mountain in Europe. The first who ascended to the summit of it was Saussure ; he made there observations stated in his Voyage aux Alpes, No. 2003.; and at the same time other observers were stationed at Geneva in the valley of Chamouny. Let us confine ourselves to the observations made at mid-day, as they are attended with more certainty. At the distance of one metre below the summit of Mont Blanc, the barometer indicated 4342’” and the centigrade thermometer 2.87 below the freezing point; at an elevation of 35.55” above the Lake of Geneva, thealti- tude of the barometer was 738.5%, and that of the ther- mometer 28.25. ‘ (No mention is made here of thermometers attached and detached; we may suppose them to agree at each sta- tion.) PART X. TRIGONOMETRY. 361 Barometers. Lower station . . 738.5 . Log. .86835 Upper station '4 . 434.2 . Log. .63769 Difference. . . . . . =.23066 Subt1 act .2200249 1 1st factor . . . . . . . . =.22817 Thermometers. _ Above freezing . . 28.25 . . . 28.25 above freezing. Below freezing . . 2.87 . . . 2.87 below freezing. Difference . . . 31.12 X Sum 25.38 X .00008 .002 Subtract .00249 2.05076 2d factor . . . . . . . =1.05076 Log. 18336 . =4.26324 Log. 1st factor 29.35826— 10 Log. 2d factor =0.02l50 Sum . . . . =3.64300=L0g. 4395.6 metres, approxi- mate height. lst Correction. Latitude of Chamouny 45° 58’. Log. . . . 23.64300 Log. .002837 . :7.45286—10 Log. sine 1O 56’ 28.52810—10 ‘ Log. sum . . :9.62396——10=log. .4207 , hence, .4 is the first correction, and is subtractive. 2d Correction. 15926.6 4395.6 Sum 2031222 _ . . Log. 2 63661753 . . Log. = 3.19613‘——10 3.95 e . . Log. = 3.64300 Sum = 1.14710=L0g. 14.1. And 4395.6 + 13.7 = 4409.3 for the difference of levels, to which, adding 1 metre, wewhave 4410.3 metres. And again, adding 35.55, we have 4445.83, the elevation of the summit of Mont Blanc above the Lake of Geneva. Francceur makes it 4436.2 above the level of the Lake of Geneva. Now the difference of levels is determined in the for- mulae by the multiplication of five factors, three of which are nearly equal to 1, and their product is nearly equal to 1; the other two factors are 18336 and the barometer factor, 2'. e. the 1st factm ; the height of Mont Blanc is R . 362 TRIGONOMETRY. PART X. three miles nearly, or about T3155 of the earth’s radius; the barometer factor being increased by this quantity, and multiplied into 18336, gives 14 metres for the correction, Which Francoeur makes only 3.55 metres. M. Coraboeuf has, by Geodesic operations,_ made by means of triangles of the first order, and the accuracy of which may be depended on, found for the height of Mont Blanc, 4435.92 metres. The height of the level of the Lake of Geneva above the level of the sea is 376.165 metres, whence the summit of Mont Blanc is elevated 4812.085 metres, or 15788 feet above the level of the sea; the number of feet in the Cabinet Atlas is stated at 15735 feet, which is not very different in such a great height. 2. At Leith quay, latitude 56°, the barometer stood at 29.567 inches, attached thermometer 55%, detached ther- mo‘meter 54°. At Arthur’s seat, barometer 28.704 inches, attached thermometer 51%, detached thermometer 50%,. Required the elevation of Arthur’s seat above Leith quay. Barometers. Attached therm. Detached therm. Lat. Lower 29.567 _ 55.25 54.0 560 Upper 28.704 51.75 50.5 X 2— 90 gives 220 Diff. 3.5 Sum —64=40.5 x4 +900 90000)] 4' .045 Subtract .00016 2d factor: 1.045 Log. lower = .47080 Log. upper = .45794 Difference = .01286 Subtract == 16 _ lst factor := .01270 Log. . =8.10880-—10 ‘Which is less than .07 Log. 1.045 =0.01912 ‘ Log. 60345.6:4.78065 2. 90357 = Log. 800. 9 Add Log. .002837 =7.45287 -— 10 { Log. sin. 220:9.57358 — 10 9. 93002 — 10:. Log. '85 = cor» rection for latitude which is subtractive. Ans. Height of Arthur’s seat above Leith quay = 800.05 feet The height by geometrical measurement is 802.66 feet. , 3. Find the height of the summit of Mont Blanc abovs the level of the sea in English feet from the following data. Height of barometer at an elevation of 116.6 feet PART X. TRIGONOMETRY. . . 360 above the level of the Lake of Geneva 29.086 inches, thermometers 82°.85 Fahrenheit: at 3.3 feet below the summit, barometer 17.101 inches, thermometer 26°.834 Fahrenheit; the altitude of the Lake of Geneva above the level of the sea being 1234.2 feet. ‘ Ans. 15819.7 feet. 4. Find the height of Guanaxuato in Mexico, from the following observations of H. Von Humboldt :— Lower station—~Height of barometer 763.15, attached thermometer 25.3 centigrade, detached thermometer the same. Upper station—height of barometer 600.95, ther- mometers 213. Am. 2083.6 metres. 5. Find the elevation of Puy-de-Dome on the Alps above Clerrnont, latitude 45° 46’, from the following ob- servations of NL Ramond : —— Lower station—Barometer 728.52, attached thermo- meter 24°.7 centigrade, detached thermometer 28°.3. Upper station—barometer 705.65, attached thermometer 27.8, detached thermometer 25.5. Ans. 287.2 metres. 6. Suppose the barometer at the lower station to be 29.4 inches, attached thermometer 50° of Fahrenheit, de- tached thermometer 45°; and at the upper station baro» meter 25.19 inches, attached thermometer 46°, detached thermometer 39°; the latitude 45°: required the elevation of the upper station above the lower in feet. Ans. 4129 feet. ON THE FIGURE AND MAGNITUDE OF THE EARTH. In the foregoing examples of heights and distances, we supposed the measurements of the base lines to be taken on a plane surface ; but the curvature of the earth’s sur- face, has a considerable influence, when the distances are , considerable. This influence is so much the greater, as the magnitude of the earth is smaller, other things remaining the same. We shall take for granted that the earth is somewhat of a spherical form, and revolves uniformly on its axis in about 24 hours, and proceed to investigate its magnitude. Supposing the earth, therefore, to be a sphere, if any are of the meridian be measured which bears a known proportion to the Whole meridian circle, we shall obtain the magnitude of the meridian, and therefore the length of the radius of the sphere. Norwood, by measuring from R 2 364 TRIGONOMETRY. . PART X. York to London, found a degree of the meridian to answer to 69—;— geographical miles: the ‘circle of the meridian is therefore 360x69; or 25020 miles: dividing this by 3.1416, we have 7964 miles for the diameter of the sphere}i If the degrees of the meridian at different latitudes are unequal, it is manifest that the figure of the earth is not spherical; and, by trigonometrical measure taken in India and Sweden, such is found to be the case. . i The figure of the earth has been investigated on th principles of hydrostatics, considering the earth to have been originally a fluid mass. Assuming this hypothesis, and considering the only force concerned to be gravity, the figure takes a spherical form to insure equilibrium: but, in taking into account the diurnal rotation, a part of the gravitating force is destroyed in consequence. If in addition to the assumption of fluidity, the specific gravity of it is considered to be the same in every part, Newton , has demonstrated, that the figure of equilibrium, is that of an ellipse revolving on its shorter diameter, and that the ratio of the diameters, is that of 230 to 231. Whatever be the figure of equilibrium, and in Whatever manner the specific gravity of the fluid mass may vary, all the forces concerned at any point may be reduced to one, which one must be perpendicular to the surface at that point. The force produced by the rotation is called the centri- fugal force; and, When a body revolves in a circle, this force is proportional to the square of the linear velocity divided by the revolving radius. This force acts in the direction of the prolongation of the radius, but the square of the linear velocity, is as the square of the revolving radius multiplied into the square of its angular velocity. The centrifugal force is therefore as the revolving radius multiplied by the square of the angular velocity, or as the revolving radius divided by the square of the time of rotation: this is the centrifugal force at any latitude in the direction of the prolongation of the radius of the it He took the sun’s meridian altitude at London near the Tower, 11 June, 1633, and deduced the latitude 510 30’; and also at York, near the middle of the city, and deduced the latitude of York 530 58’, having 2° 28' for the difference of latitudes ; and be measured from York to London 9149 chains, each chain being 6 poles, €301! pole 16% feet, that is, every chain 99 feet; then as 148‘, the difference of latitudes : 9149 chains 1 60’ : 2 3709 chains, or 367196 feet, or 367200 feet in a degree; whence he deduces 25036 miles for the circumference of the meridian, and 7966 miles for the diameter. — See Norwaod’s Seaman’s Practice, pp. 6, 7, 8. London, 1637. ‘PART X. TRIGONOMETRY. 365 parallel of that latitude. To find its efficacy in the direction of the perpendicular to the surface of equilibrium, it must be diminished in the ratio of the cosine of the latitude to radius. This is nearly the ratio of the revolving radius to the distance from the surface to the centre. The centri fugal force opposed to gravity is therefore very nearly as the square of the revolving radius, divided by the product of the square of the period of rotation by the distance from the surface to the centre. . Newton has shown that the effective force of gravity is at any point on the surface inversely as the distance of that point from the centre ; and the distance of this point from the plane of the equator reckoned by the perpen- dicular to the surface, is inversely as the distance of this point from the centre. It follows that the effective power of gravity on any point on the surface is proportional to the distance of that point from the plane of the equator, reckoned by the perpendicular to the surface. This agrees with Stirling’s construction (P/zz'l. T Tans. No. 438. for part of the year 1735). If, therefore, we can obtain the ratio of the effective gravities at two points of the earth’s surface, we can also obtain the ratio of the distances of these points from the centre, and, consequently, the ratio of the diameters of the elliptic meridian; this if the fluid mass be homogeneous. Now we may estimate gravity by the effect it produces; as, for example, by the space through which a heavy body falls in a small portion of time; and this space by writers on mechanics is demonstrated to be proportional to the length of a pendulum performing its vibrations in the same time: the force of gravity is then proportional to the length of the pendulum. The length of a pendulum vibrating seconds: space through which a heavy body falls in a second :1 twice the square of the radius of a circle : square of the semi- circumference; or, as 2 3 9.916, or 1 1 4.953. The length of the pendulum vibrating seconds in the latitude of Lon- don is 39.]393 inches, whence in one second a heavy body falls 16.155 feet. The length of a pendulum vi- brating seconds in Melville Island, in lat. 74° 47’, is 39.21 inches, and at the North Pole it may be reckoned at 39.22; and the descent of a heavy body at 16.6 feet... and 16 feet at the equator. In comparing the force of gravity with other forces, R 3 36-6 TRIGONOMETRY. PART 2:; the double of the fall is taken as its measure; if, there- fore, we compare it with the centrifugal force at the equator, we must take it in the ratio of 32 feet to the square of the velocity divided by the revolving radius. Then, as 24 hours : 25020 miles :1 1 second ‘ 25020 X 5280 __ 1251 x 11 __ m feet—T“ feet—1529 feet. 1529 x 1529 Then, 32 feet . 3982X—52‘8—6 Thus at the equator the centrifugal is 5-2;? of gravity. Whence the gravity at the pole : gravity at the equa- tor :: 283 : 282, if the earth were a solid sphere of rota- tion. And Newton’s investigation leads to the conclusion, that for the fluid spheroid, equatorial diameter : polar diameter :1 4 times the above polar gravity: 5 times the above equatorial gravity—e- polar gravity :1 1132 ; 1127 :: 226.4 : 225.4 » In the fluid spheroid of equilibrium he deduces the conclusion that the effective force of gravity on any point of the surface is inversely as the distance of that point from the centre, whence the polar gravity: equatorial gravity :: 226.4 2 225.4. He has also shown that the effective force of gravity is proportional to the length of a pendulum vibrating in small circular arcs in a very small portion of time; thus. the length of a pendulum vibrating seconds at the pole : length of a seconds pendulum at the equator :: 226 : 225.* . In an ellipse, when the difference of the diameters is very small in comparison to the greater diameter, the excess of the equatorial radius above the distance of any point on the surface to the centre : equatorial radius :: differ- ence of the diameters of the spheroid divided by the equatorial diameter and multiplied by the square of the sine of the latitude of the given point : square of radius. feet I: l I 283. EXAMPLE. Let a point on the surface be in latitude 30° ; then sine 30°=~§~ radius and (sine 30°)2 :44 (radius)2. Hence the 9* The above is only to be considered an approximation. Newton, by using more accurate numbers, finds the above ratios to be 289 : 288, and 231 2 230. PART X. TRIGONOMETRY. 367 above ratio becomes gal xi— (radius)? : (radius 2 :: 5331 I 1. Hence equatorial radius : distance from the centre :1 1 Z —8%% I: 924 Z 923. Similarly for a point at latitude 45° equatorial radius .‘ distance :2 1 : l—fiTx—lg- :: 462 : 461 z: 924 z 922. Hence distance of the former point : distance of the latter :: 923 : 922 :: gravity of the latter point 3 gravity of the former :1 length of the seconds pendulum in latitude 45° : length of the seconds pendulum in latitude 30°. N ow in fact the lengths of the pendulums are as 39.12650 inches: 39.06393 inches, or as 923 : 921.5, which does not agree with the former ratio; neither do the trigonome- trical measurements of arcs of the meridian : the hypothesis of a. uniform specific gravity is therefore untenable. ' Abandoning the assumption of homogenity, but retain- ing the hypothesis of fluidity, the conditions of equili- brium are still fulfilled if we suppose a spheroid consisting of spheroidal strata, in each of which every particle is of the same specific gravity, the density varying from one spheroidal surface to another according to any assigned law. And Clairaut has deduced this remarkable conclusion : Whatever be the law of the earth’s specific gravity, if its ellipticity be added to the ratio which the excess of the polar above the equatorial gravity has to the equatorial gravity, the sum will be equal 2—;— times the ratio which the centrifugal force at the equator bears to the equatorial gravity. Or, as he has stated it in other terms : If from twice the ratio which the difference of the polar and equatorial diameters bears to the equatorial diameter, on the suppo- sition of homogenity, there be taken the ratio which in the actual case the difference of the polar and equatorial gravities bears to the equatorial gravity, the difference will be equal to the ratio, which in the actual case the difference of the polar and equatorial diameters bears to the equatorial diameter. Now Airy has shown (Mathematical Tracts, p. 106. No. 63. edit. of 1826) that the increase of gravity above the equatorial gravity is as the square of the sine of the latitude ; and the increase of the pendulum’s length is in the same proportion. We may therefore, from known observations of the R 4 568 TRIGONOMETRY. PART X. seconds pendulums at any two latitudes, find the ratio of the polar and equatorial gravities thus. RULE. Multiply the length of each of the pendulums, by the square of the sine of the latitude, corresponding to the other pendulum; take the difference of these products, divided by the square of radius for a dividend, which di-v vide by the difference of the lengths of the pendulums; the quotient gives the ratio of' the equatorial gravity, to the increase of the polar gravity, above the equatorial. ' EXAMPLE. At Madras, lat. 18° 4’, the length of the seconds pen-~ dulum was observed to be 390237 inches. At Melville: Island, lat. 74° 47’, length of seconds pendulum was observed 39-207 inches. Sine 18° 4’ . . Log: 9.491535 Log. sine square . =18.983070 Log. 39.207 : 1.593364 Sine Log. . . . . 220.576434 1020 x 3.37708. Sine 74° 47’ . Log: 9.984500 Log. sine square . :19.969000 Log. 39.0237 = 1.591328 Sine . 221.560328 1.020} 36.33530. Difi‘. : 1020 X 32.95822 :Divisor 1020::(radius) 2. 39.2070—— 39.0237 . . = .1833zdividend‘ Log. 32.95822 . .. . . =1.517964 Log. .1833 . ar. co. 20.736838 Difi‘. Log. 180 . . . :2.254802 Hence, difference of polar and equatorial gravities: equat. grav. :: 1 : 180. By Clairaut’s rule, we have for the ellipticity 2 X gfi - 1__—w—_Lai_—Tin T80—231X180_41580—‘0I7‘ Therefore, polar diameter : equatorial :: 316 : 317. This is conformable to experience. 2. The length of a pendulum vibrating seconds at Lon- don, 1at. 51° 31’, is 39.13929 inches ; and at Unst, lat. 60° 45%], it is 39.17146 inches. Required the ratio of the PART X. TRIGONOMET'RY. 369 polar, and equatorial gravities, and the ratio of the polar and equatorial diameters. Ans. 179 to 180, 321 to 322. PROBLEM I. To find the length of a pendulum vibrating seconds in any latitude. RULE . Multiplythe square of the sine of the latitude by the number .20862; divide the product by the square of the radius; add the result to 39.01146 3 the sum is the length of the seconds pendulum in inches. EXAMPLE S. 1. Required the length of the pendulum vibrating sea conds in lat 30°. Here (sine 30°)2 —:— (radius)2 2 71,-, and i— of .20862 = .05215. VVhence 39.01146 + .05215 = 39.06361 inches, the length required. . 2. Required the length of the seconds pendulum in the following latitudes, viz. 0°, or at the equator, at lat. 45°, at lat. of London 51° 31’, and at the north and south poles. Ans. 39.01146, 39.11577, 39.13929, and 39.22008 inches. Note. It has been found by measurement of meridian arcs, that the ratio of the earth’s diameter is more nearly as 299 to 300 ; from this the polar is to the equa- torial gravity as 188 to 187: this ratio, combined with the measure of the pendulum by Captain Kater, at-London, has furnished the above rule. To determine the magnitude as well as the ratio of the diameters, recourse must be had to trigonometrical surveys. PROBLEM II. To determine the figure and magnitude of the earth’s diameters, having given the lengths of the arcs of meridian corresponding to two given latitudes. RULE. Divide each measured are by the number of degrees and parts contained in it; call the greatest quotient the R 5 370 TRIGONOMETRY. PART X. first term, the other the second term, and the difi’erence is the third term. To double the third term, add three times the second term multiplied by the square of the sine of the latitude nearest the equator, corresponding to the first term, and divided by the square of the radius, and from the sum subtract three times the first term multi- plied by the square of the sine of the less latitude corre- sponding to the second term divided by the square of the radius; call this result the fourth term; then, as fourth term : third term :: equatorial diameter : difference of the equatorial and polar diameters. ' Divide the third term by the fourth, and call the quo- tient the fifth term. Multiply three times the fifth term by the square of the least latitude, corresponding to the first or second term, and add 1 to it, and from the sum subtract twice the fifth term; call this last result the sixth term; then as sixth term .‘ 2 :2 1st or 2d term (accord- ing to which the latitude has been used): 7th term 9% which, multiplied by 57°29578, gives the equatorial diameter. The equatorial diameter diminished by the product of itself and the fifth term gives the polar diameter. EXAMPLES. 1. Let the arc of the meridian measured from lat. 30° to lat. 320 be 137.756 miles, and the arc of the meridian from lat. 450 to lat 46° be 69.047 miles. Required the equa- torial and polar diameters. Here are 2 degrees of arc in the first, and 1 degree in the latter arc. 137.756 2 69.047 the first, and .169 the third. Lat. for 1st term 115"; its square divided by radius squared :7}; lat. for second term 30°, its square divided by radius squared ——1 '_ gL' =68.878; hence this is the second term, and 2 x .169+3 x 68.878 x %=103.655 3 x 69.047 x 71.: 51.785 Diff. = 51.870:4th term. As 51.870 : .169 I I 306 : 1 : equatorial diameter: difference of equatorial and polar diameters. Hence 31133 is the fifth term. * This seventh term is the length of two degrees on the equator. PART X. TRIGONOMETRY.' 371 Taking lat. 45° square of its sine, divided by square of radius—.2355, which belongs to the 1st term. 1 —-4 1 +§%g x%j—§%€:6»——2611—: — = 81712—226511 term. Then, as 24.17 : 2 :2 69.047 : 138.320; and 138.320X 57.29578 =7925.l53 miles, the equatorial diameter; and 7925.153 —:- 306 : 25.899 ; whence 7925.153 -—25.899 :2 7799.254 miles, the polar diameter. diameter x 180 Note. 57.29578: . . - Circumference 2. Colonel Lambton, in India, measured an arc of the meridian extending from latitude 8° 9’ 38.’ ’4; to latitude 10° 59’ 48” .9 its length =1029100.5 feet 2194.905 miles. Svanberg, in Sweden, measured an arc of the meridian extending from latitude 65° 31’ 32” .2 to latitude 67° ,8’ 49”.8; its length was found =593277.5 feet 2 112.363 miles. Find from these data the ratio and magnitudes of . the equatorial and polar diameters. Ans. Ratio 302 to 303; equatorial diameter 7924 miles ; polar diameter 7897.85 miles. The lines drawn perpendicular to the surface of a sphe- roid do not, as in the sphere, meet at the centre; the latitudes assumed above are the angles which their per— pendiculars make with the plane of the equator, and are immediately determinable from astronomical observations. If we call the true latitudes those angles which lines drawn from the surface to the centre make with the plane of the equator, the above may be called the elliptic latitudes, which are always greater than the true latitudes. The difference of these two latitudes is equal to the angle' which the two straight lines, one perpendicular to the sur- face the other drawn from the same point of the surface to the centre, make with each other at the surface —- this is called the angle of the vertical. This difference is greatest at latitude 45°, and bears the same proportion to O iii-8357926 : 57° 17’ 44’ ’ .8, that the difference of the equatorial and polar diameters bears to the equatorial diameter. If we take this ratio as 1 to 800, this greatest difference is 11’ 27’ ’ .5: at equal angular distance from the equator and pole the difference is equal: thus, in latitude 15° the difference is 5’ 43” .75, the same at latitude 75°. , R 6 372 TRIG ONOMETRY. PART X. PROBLEM III. Tog/incl the true latitude, having the elliptic latitude given, and conversely. RULE. Take the difference between 450 and the given latitude; then, ‘ ' As radius : cosine of twice this difference I I 11’ 27” .5 : angle of the vertical. Then, true lat. =elliptic lat. -— angle of the vertical; elliptic lat. :true lat. +angle of the vertical. EXAMPLE. Find the true latitude answering to 65° elliptic latitude, and the elliptic latitude answering to 32° true latitude. Ans. First angle qf t/ie vertical 8’ 43” .6 ; true lati- tude 64° 51’ 16”.4 ; second angle qf the vertical 10’ 18”; elliptic latitude 320 10’ 18”. Note. A more recent discussion of the figure and magnitude of the earth gives the ellipticity =§gg and 69.156 miles, the length of an equatorial degree ; whence 24896.16 miles, the circumference of the equator, and 7924.7 miles, the equatorial 7924.7 300 diameter, which values we shall follow. diameter =26.4 miles difference of diameters, and 7898.3 miles, the polar PROBLEM IV. Tofind the length of an arc of the meridian from the equator to any given elliptic latitude. - RULE . Multiply 69.156 by the degrees and parts 01" the latitude; from the product subtract its 600th part; subtract besides 9.9 multiplied by the sine of twice the latitude, divided by radius; the result is the answer in miles.* EXAMPLE S. 1. Required the arc of the meridian from the equator to 15° of latitude. 9* The whole subtractive quantity shows by how much the proposed are of the meridian is less than a similar arc of the equator, the quantity 69.156—its 600th part=69.04l. See Prob. IX. p. 374. PART X. TRIGONOMETRY. 3 4 3 Here, sine 2 x 150—2—radius:%. .‘.69.156 x15=1037.340 . ., the 600th=1.729 Subtract 6.679 . . 9.9 X % :495 V Sum=6.679 Ans. 1030.661~miles. Or thus :— 69.041 x 15 2 1035.615 and 1035.615 —- 4.95.: 1030.665, as before, nearly. 2. Required the arc of the meridian from the equator to the parallel of 54.0 13:54.30. Ans. 3739.5 miles. 3. Required the are of the meridian from the equator to the pole. Ans. 6213.67 miles. PROBLEM V. To find the length of an arc of the meridian from the equator to any given true latitude. RULE. Multiply 69.041 by the degrees and parts in the lati- tude, and add 3.302, multiplied by the sine of twice the latitude divided by the radius. EXAMPLES. 1. Required the arc of the meridian from the equator to 150 of lat. Here, sine 2 X l5°~:—radius :%. 69.041><15=1035.615 3.302 x %:1.651 Add . . . 1.651 Ans. 1037.266 miles. 2. Required the are of the meridian from the equator to the parallel of London 510 32’. Ans. 3561.13 miles. 3. Find the length of the elliptic quadrant, and the Whole meridian, and show the difference between the meridional circumference and the equatorial. Ans. Quadrant =6213.67 miles ,- whole meridian : 24854.68, which is 41.493 miles shorter than the equator. Note. According to the estimate of the French mathematicians, the length of the quadrant of the meridian, reduced to English measure, is 6213.822 &c. mites, or 32808992 feet, the ten-millionth part of which is 3.2808992 feet, the length of the French metre, which is the basis of their measures. 374 TRIGONOMETRY. PART X. PROBLEM VI. To find the are qf the meridian between two given latitudes. RULE. Find the length of distance of each lat. from the equator by 4 or 5, and take their difference. PROBLEM VII. To find the length of a degiee of the meridian to any elliptic latitude, in the middle q)" the degree. RULE. Subtract from 69.041 the quantity .346 multiplied by the cosine of twice the latitude divided by radius, if the latitude be less than 45°; otherwise, add. The difi"er- ence or sum is the length of the degree in miles required. EXAMPLES, 1. What is the length of the degree from lat. 290 30’ to lat. 30° 30’. And from lat. 590 30’ to lat. 600 30’. Ans. Ist. 68.868; 2d. 69.214 miles. 2. What is the length of the mean degree bisected by the parallel of London, whose trite lat. is 51° 30’, elliptic Lat. 510 41’. Ans. 69.121 PROBLEM VIII. To find the length of a meridional degree to any true latitude, in the middle of the degree. RULE. Fro-m 69.156 subtract .231 multiplied by the square of the sine of the latitude divided by the square of the ra— dius; the difference is the answer in miles. EXAMPLES. 1'. Required the length of the degree of the meridian from lat. 290 30’ to lat. 30° 30’. ‘ Ans. 69.098 miles. 2. W hat is the length of the mean degree bisected by the parallel of London ? Ans. 69.014. PART X. TRIGONOMETRY. 375 PROBLEM IX. To find the length of the mean degree by elliptic latitude, and. what latitude it refers to. This degree is 69.156—6-396:()1—:§=69'041 miles, and ex- tends from lat. 44° 30’ to 45° 30’. ' The same results from using the true latitude. Note. This mean degree is the sum of all the degrees divided by their number. Similarly, the mean length of the seconds pendulum is the sum of all the se- conds pendulums divided by their number, and is equal to the mean of the equa- torial and polar pendulums, it is 39.11577 inches, and answers to lat. 45°. PROBLEM X. Tofind the distance of any point on the earth’s surface to the centre, corresponding to any given latitude. RULE . From 3962.35 subtract 13.21 multiplied by the square of the sine of the latitude divided by the square of radius. The result is the answer in miles. EXAMPLE. In lat. 30° 3962.35———3.30:: 3959.05 miles. Ans. PROBLEM XI. To find the length of a degree of longitude on any parallel of latitude. RULE. 1st. For elliptic latitude- Multiply .115 by the square of the sine of the latitude divided by the square of radius. Add the result to 69.156, and multiply the sum by the cosine of the latitude divided by radius. The result is the answer in miles. 2d. For true latitude. Multiply .115 by the square of the sine of the latitude divided by the square of radius. Subtract the result from 69.156, and multiply the difference by the cosine of the latitude divided by radius, the result is the answer in miles. 376 TRIGONOMETRY. PART X. EXAMPLE . Let the elliptic latitude be 60°, then square of sine of 60° .5. square of radius:-}, and cosine of 60° -:-radius: 3. And (69.156+% of .115) >< %: 39.621 miles. Ans. ~ ' If the true latitude be 60°, then (69156—72 of .115) x —}Z~39.535 miles. Ans. Note. If the number of miles in the degree of the parallel be multiplied by 57.29578, the result is the radius of the parallel in miles; and multiplying by 360 gives the circumference of the parallel in miles. Log. 59.29578=1.758123. The ratio of the miles of any parallels of latitude is the ratio of their velocities in the diurnal rotation. In the same manner as the mean meridional degree of the true latitude was found, may be found the mean dis— tance from the surface to the centre, this is the mean of the equatorial and polar semidiameters, it is 3955.75 miles and answers to the latitude of 45°. The mean parallel of latitude is the sum of all the parallels of latitude divided by their number, and similarly of the mean degree of longitude. Its length is 44.075 miles, and answers to the elliptic latitude 50° 25.’ 8 By using the true latitude, it comes out 43.977 miles, and answers to 50° 34' true latitude, the mean is 44.026 miles. , By finding the surface of a sphere, having its diameter equal to the equatorial diameter, and subtracting the . 450th part, we obtain the surface of the earth as a spheroid, this is equal to the surface of a sphere having a diameter equal to the equatorial diameter diminished by a 900th part. This answers to 196856266488 miles. By finding the solidity of a sphere having its diameter equal to the equatorial diameter, and diminishing it by its 300th part we obtain the solidity of the earth, this is equal to the solidity of a sphere having a diameter equal to 3%3— of the equatorial diameter, the same will result by multiplying the surface by one-third of the equatorial diameter, and subtracting a three hundredth part in this manner, we find the solid content 518282128509519 cubic miles. That sphere which has its surface and solidity equal to the terrestrial spheroid, has its radius equal to the distance from the surface to the centre of the terrestrial Spheroid, at a latitude the square of whose sine is T} the square of radius, this latitude is 35° 16’, true latitude this distance is 3957.95 miles.‘ This is the radius used in the discussion of the tides. PART X. TRIGONOMETRY. 377 ON THE MEASUREMENT OF AN ARC OF THE MERIDIAN. The following extract from Keith’s Trigonometry will fully explain the method of conducting a trigono- metrical survey for the purpose of being furnished with such data as those given in EX. 2. p. 371. A country may be surveyed, and the distances of re- mote objects may be obtained by means of a Series of triangles. Thus, measure a base AB, and let G and D be two objects which can be seen from A and B. Measure the horizontal angles BAC, \T BAD, ABC, ABD, and cal- *- culate the distance DC, as in Example 12, and likewise the distances DB, DA, BC, AC. In like manner, if E and K be two objects visible from C and D, the distance EK may be found, and its position with respect to CD ; and thus the measurement may be continued to any dis- tance. But, in order that the conclusion may be more accurate, the mensuration from one base to another may be carried on by dif- ferent sets of triangles; for instance, two objects V and W might have been chosen instead of A and B, by means of which the distance EK might have been ascertained ; and, it is plain, that by taking several series of triangles which lead to the same two objects, the mean of the results will be more accurate than the measurement obtained from one triangle. When a series of triangles has been carried on for a considerable distance, the interval of two objects, whose distance has been determined by calculation, should be actually measured, in order to detect any error which may have been made in the calculated distance. This is called a Base of verification. The trigonometrical sur- vey of England and Wales had its first commencement in. 1784; and from a base of 27404.2 feet measured on Hmmslow Heath, and continued by a series of triangles to Salisbury Plain, the medium distance between Beacon 378 TnIGONOMETRY. PART x. Hill and Old Sarum was 36574.3 feet, and from actual admeasurement the distance was 36574.4 feet, a difference of little more than one inch.* By the assistance of a survey, executed with such ac- curacy, the length of an arc of a meridian may be mea- sured. Let NS be the meridian of anyplace A, and let 6 represent the point of the horizon where the sun sets ; measure the angle BA 6); then, by having the latitude of the place A given, the amplitude NA (9 is easily ascer- tained, and consequently their difference BAN is given; also if from BAC there be taken BAN, the remainder CAN is given. Let BF, CI, be drawn perpendicular to the meridian, then because AB and the angle BAN are given, AF and FB are easily found: in the manner, be- cause AC and the angle CAN are given, the lines AI and IC may be found. ’ If from the angle ABC there be taken the angle ABF, the remainder FBC, added to DBC, gives DBF; by means of which, and the given side DB, DC, or FL will be found, and likewise BG. By adding AF to FL we get AL, and by deducting BG from BF we shall have GrF=DL. In a similar manner the several distances from the point A to the perpendiculars, let fall from each of the angular points of the figure upon the meridian NS, may be de- termined, as also the perpendiculars themselves. By this method the extensive survey of the kingdom of France was carried on from the measure of nineteen bases. The measures being all reduced to the meridian N S, the difference of latitude between A and D may be de- termined. The latitudes of A and D determined direct from astro- nomical observations are elliptic latitudes. OBSERVATIONS ON THE ADEIEASUREMENT OF A BASE LINE. Where the ground is perfectly level, the manner of - measuring a straight line from one object to another ap- pears to be simple and easy; yet, on account of the curvature of the earth, no two points on its surface can be exactly situated in the same horizontal line ; the chord of the arc, and not the arc itself, being the horizontal distance. N ow *9 Trigonometrical Survey of England and Wales, vol. i. pp. 279, 280., published by Mr. Faden, Charing Cross, London. PART X. ,TRIGONOMETRY. 379 the radius of one circle is to the radius of any other circle, as any are of the former is to a similar arc of the latter. If we take, for instance, the base line mea— sured on rounslow Heat/z 27404.2 feet, the radius of the earth 3958 miles {the radius of the equal sphere at lat. 85° 16’ of the spheroid) or 20898240 feet, we shall .— . . 27404.2 have 20898240 feet . 21404.2 feet . 1 . 20898240 the length of the measured arc in parts of the radius. But the difference between any are and its chord, the radius being 1, is 51—4 of the cube of the length of the arc; “274042 )3 1 — 000000000094 Vill x r sth 20898240 X3?“ ‘ e p 68 6' difference in parts of the radius, which multiplied by 20898240 feet, the radius of the earth, produces .001965 feet, the extent by which a terrestrial arc of upwards of 5 miles, exceeds the chord of the same arc, a difi'erence scarcely worth notice, even where the greatest accuracy is required. ‘When the ground on which a base line is to be mea— sured is sloping, it will be necessary, in some cases, to reduce it to a horizontal level. Thus, after having deter- mined the direction of the base AI”, by poles LA, 1712, H72, feet, hence < G0, pointed at one end, and fixed perpendicularly in the ground by means of a plumb-line; the sum of the hori- zontal distances Lm, In, H0, GrF will evidently be equal to the whole horizontal distance AE; and if the heights- AL, mI, 72H, Go, be successively measured, their sum will give the whole height EF. If the ground be irregular, or if it ascend and descend alternately, it is evident that the difference between the heights of the poles must be added when ascending, and subtracted when descending, in order to determine the different elevations and depressions of the ground. Surveyors generally ascertain the altitudes of irregular 9* The point A, and the summits of the hills, m, n, 0, F, should be connected, so as to form a regular slope AF. 380 TRIGONOMETRY. PART X. hills by the assistance of a spirit-level, and perpendicular poles placed at convenient distances from each other. This practice is called levelling. A base line, on a sloping ground, may likewise be mea- sured by taking angles at its extremities With a theodolite. Thus, let Im represent a theodolite, AL a pole fixed per- pendicular to the horizon and equal in length to the height- of the instrument; also, let KI be a horizontal line , (which may be ascertained by the bubble of air in the spirit-level 0f the telescope resting in the middle) and KIL the angle of depression between the top of the pole AL and the horizontal line KI. Then, because KI is parallel to AB, the angle KIL is equal to the angle mAB ; then, rad. : Am :: cos.4 KIL : AB, or AB:Am 0:214 KIL. If Am2400 yards, and /_ KIL=4°, AB will be 399.026 yards, hence the difference between Am and AB is less then 1 yard. It appears from this example, that when the measured base is inclined to the horizon in a small angle, a reduction of this kind will be unnecessary, except in cases where great accuracy is required. OF THE ERRORS WHICH OCCUR IN TAKING ANGLES OF ELE- VATION AND DEPRESSION WITH A THEODOLITE. W’hen the observer is at a considerable distance from the ob- ject, the altitude taken with a theodolite will require correction. In _ the first place, the ho- rizon of the observer and that .of the object observed are not the same. Let C be the centre of the earth, D the summit of a moun— tain, and IIPOB the horizon of the ob- server. Through D draw EDF perpendi- / cular to DC, and it fl PART X. TRIGONOMETRY. 381 will be the horizon of the point D. Now SD will be the true height of the mountain above the horizontal line PS; DPS the true angle of elevation, and DPB the observed angle, from which the height of the mountain is deter— mined to be BD, instead of SD. Again, the LPBD is supposed to be a right angle, in ordinary calculations, but in reality it is equal to the sum of the two interior angles BPC and PCB (32 Euclid, I.) = 90° -l- L C. The French academicians measured an inclined base at Peru, the length PD of which was found to be 6274.057 toises‘ié, the angle of elevation DPB was 1° 5’ 43” (the effect of refraction being deducted). Now rad. : PD 2: sin. 'DPD ; BD, hence BD=PD sm' DPB rad“: 119.93 toises. In order to calculate BS, PD may be used instead of PB, and CS and CB may be considered as equal to each other without sensible error. And as {2CS+BS) > For it is equal to the angle in the alternate segment (33 Euclid, 111.), which is an angle at the circumference, which is half the angle at the centre (20 Euclid, III.) 1- ’I‘ngonometrical Survey of England and Wales, vol.ii. part 2. p. 113. PART X. TRIGONOMETRY. 383 THE NATURE OF TERRESTRIAL REFRACTION, AND ITS EFFECTS ON ANGLES OF ELEVATION.T As terrestrial refraction arises from the gross vapours, and exhalations of various kinds, which are suspended in the air near the surface of the earth, and which are per- petually changing, it is very difficult to ascertain the exact quantity of it at any particular time. The course of a ray of light in its passage through the atmosphere is, in general, that of a curve which is con- cave towards the earth, and the observer views the object in the direction of a tangent to this curve; hence the apparent, or observed angle of elevation is always greater than the true angle. The altitudes of the heavenly bodies when Within 5° or 6° of the horizon, should never be used where a very accurate result is required. The figures of the sun and moon, when near the horizon, are sometimes elliptical, having the minor axis perpendicular, and the major aXis parallel to the horizon. This change of figure arises from the refraction of the under limb being greater than that of the upper. But a perpendicular object, situated on the surface of the earth, will not have its length altered by refraction, the refraction of the bottom being the same as that of the top'. , The allowances usually made for refraction are too uncertain for any reliance to be placed on them, as scarcely two writers agree on this subject. Dr. Maskelynev makes it T16 of the intermediate arc PS between the observer and the object; Bouguer lg; Legendre T11: General Roy from % to @113 and in the second volume of the Trigo- nometrical Survey, the variation is found to be from —% to 31? of the intermediate arc.T This difference does 'not arise from inaccuracy of observation, but from circum- stances which cannot be avoided, as the evaporation of rains, dews, &c. which produce variable and partial re- fractions. The following method is used in the Trigonometrical Surveyi for ascertaining the quantity of refraction : it See a paper by Mr. Huddart, in the Philosophical Transactions for 1797, p. 29.; and another by the Rev. S. Vince, 1799, p. 13. Also the Trigonometrical Survey of England and Wales, v01. i. p. 175. 1 Pages 177,178. part I. :1: V01. i. p.175. 384 TRIGONOMETRY PART x. Let C be the centre of the earth, P and S two stations on its surface; PB, SA the horizontal lines at right an- gles to CA and CB ; also, suppose A and B to be the true places of the objects observed, and a and 6 their apparent places. Then the 4 bPS will be the refraction at P, and the 4 aSP that at S. * In the quadrilateral figure CSOP, the angles at P and S are right angles, therefore the [_ SOP+ /_ C: two right angles, but the three angles of the triangle SOP: two right angles, hence 4 OS]? + A SOP+ [_ OPS: A SOP-{— A C, consequently 4 OSP+ A OPS: 4 C, which is mea- sured by the intermediate arc PS. Now the sum of both refractions A aSP+ /_ bPS=( 4 ASP+ ABPS)-—(4 ASa + LBPb) :(4 OSP + A OPS) ~(L ASa + A BPb): A C—(L ASa + /_ BPb). Hence the following ’ RULE. Subtract the sum of the two depressions from the con- tained arc, and half the remainder is the mean refraction. If one of the objects (A), instead of being depressed, be elevated, suppose to the point e, the A of elevation being eSA ; then the sum of the angles mSP, mPS will be greater than /_ OPS+ /_ OSP (the A C, or contained arc PS) by the /_ of elevation eSA. Hence /_ mSP+ / mPS = [_ C + A eSA, from each of these equals take the /_ BPb, then the sum of the two refractions A mSP+ A bPS: /_ C+ /_ eSA— ABPb; that is, subtract the depression from the sum of the contained arc and elevation, and half the remainder is the mean refraction. Perhaps it may be necessary to remark, that previous to the observations the error of the instrument must be accounted for. (See p. 381.) EXAMPLE. , The refraction between Dover Castle and Calais church was thus determinedxl‘ , Let C be the centre of the earth, PS the surface ; D the station on Dover Castle ; A the top of the great balustrade of Calais steeple; EDF the horizontal line ; also let PG: SD ; then the /_ FDG=Tj [_ C, or half the arc PS. * In observing these angles two instruments are used, one at P and another at S.; and the reciprocal observations are made at the same instant of time by means of signals, or by watches previously regulated for that purpose. The observer at P takes the depression of S, at the same moment which the observer at S takes the depression of P. f Trigonometrical Survey, vol.i. p. 178. PART X. . TRIGONOMETRY. 380 The distance from Dover to Calais is 137455 feet, hence 364950 feet : 60’ :1 137455 feet : 22’ 35” the A C; hence. =FDG=11’ 17%”. ~ The height of D above low-water spring tides =479 feet. The height of A (communicated from, France) 2 140;, feet. AG=32893 The triangle DGA may be considered as isosceles, and DG or DA=137455 feet, the distance between Dover and Calais. Hence 7} AG : rad. I: DA : sec. ADAG: 890 55’ 53’ ’ , the double of which deducted from 180°, leaves 8’ 141” for the /_ GDA, to which add the A FDG: 1 1’ 17—5”, and the whole angle FDA:19’ 31:1,” supposing there was no refraction ; but the 4 FDA was determined from ob- servation to be 17’ 59”, hence the refraction was (19’ 31%” — 17’ 59’ ’ z) 1’ 3271—”, being about T13 of the contained arc. Mr. Huddart is of opinion, that a true correction for the effect of terrestrial refraction cannot be obtained by taking any part of the contained arc *; for different points, though nearly at the same distance from the observer, will have various refractions. OF THE REDUCTION OF ANGLES TO THE CENTRE OF THE STATION. In surveys of kingdoms and counties, where signals on the steeples of churches, vanes of spires, 8:0. are used for points of observation, the instrument cannot be placed exactly at the centre of the signal, and consequently the angle observed will be different from that which would have been found at the centre. The correction is generally very small, and is only necessary where great accuracy is required. The observer may be considered in three different po- sitions with respect to the centre, viz. he is either in a line with the centre and one of the objects; or a line drawn from the centre through his situation would, if produced, pass between the objects ; or a line drawn from the centre to the place of the observer, when produced, would pass without the objects. * Philosophical Transactions for 1797:, p. 29., ct seq. S 386 TRIGONOMETRY. PART X. First. Let the ob- server be at D, in a line C between the obj eetsB and C, viz. on one side of the triangle ABC; B being the proper centre of the station and the LABC that required. It is plain E ‘ '-, that the 4 CDA, being """"""" ‘_ ___::=-.==:::ZZZ 1111 \\ the exterior A of the ' H ‘1) triangle ADB, is too A b large by the interior ADAB. Therefore 4 ABC: 4 CDA— A DAB. Secondly. Let the observer be at 0, within the triangle ABC, and let B be the centre of the station, and the [_ ABC that required. N ow 4 AOC+ A OAC + A OCA : A ABC+ A OAB+ A OAC + /_ OCA+ [_ OCB,eachof the sums being equal to two right angles; therefore the 4 A00: 4 ABC+ 4 CAR + 4 00B, that is, the 4 ADC is greater than the ,4 ABC by the sum of the angles CAB and OCB. Therefore 4 ABC* = A AOC—( A CAB ~5— 1. OCB). Thirdly. Let the observer be at E, without the tri- angle ABC, and let A be the centre of the station, and the /_ CAB that required. Now /_ CAB+ LACE—t— [_ABE+ AEBC: /_ CEB + /~ ECA+ Z. ACB + 4 EEC, each of the sums being equal to two right angles ; therefore 4 CAB + 4 ABE: /_ CEB + 4 BOA, and consequently 4 CAB: A CEB + /_ ECA—A ABE. Assume 4 CAB: A CEB, from which it does not much differ, and 4 ABC: /_ AOC, to which it is nearly equal, and find the angle ACB then in the triangle ABC, a side being given and all the angles, the remaining sides may be found by Prob. I., Oblique Angled Triangles. In the triangle ACE there are known AC, AE and the angle CEA, whence the angle ECA may be found. Similarly in the triangle BEA, the sides AB, AE and the angle BEA are known, whence the angle ABE may be found. [If E is upon the side AC, angle ECA=O, this becomes the .first case, and angle CAB: angle CEB—angle ABE] *5 If the proper centre of the station were at O, and the observer at B, it is plain that the angles OAB and OQB must be added to the; ABC to obtain the AAOC. PART X. TRIGONOMETRY. 887 In a similar manner in the triangles BCO, ABO the angles OCB, OAB may be found. Thus the angles CAB, ABC may be corrected with these corrected angles, and a side given, the other sides of the triangle ABC may be corrected. EXAMPLE . Let A and B represent the vanes on two steeples, E the situation of the theodolite upon the steeple A, and O its ituation upon the steeple B. Then, suppose AE=12 feet BO:10.5 feet A CEB=74° 32’ /_ AOC:49° 27’ A CEA2139° 39’ L COB:137° 55' It is required to find the angles CAB and ABC, the distance AB being 5000 feet. Solution. ’The A CEB = [_ CAB nearly, and 4 A00 = /_ ABC nearly, with these angles and AB, find AC, BC:4~581.7 and 5811.5. Then, ‘ AC : sin./_CEA :2 AE : sin.[_ECA=5’ 50” AB : sin.LBEA :: AE : sin./_ABE::7’ 29” Hence 4 CAB274° 30’ 21”. BC : sin.[_ COB :: BO : sin.[_OCB=4l’ 10” AB : sin.4AOB :: BO : sin.LOAB=O’ 56” HenceLABszlQ" 21’ 54”. With the corrected angles CAB, ABC, and the distance AB, the sides AC and BC may be determined, viz. 4570 and 5803. ‘ OF THE REDUCTION OF ANGLES FROM ONE PLANE TO ANOTHER. Angles which are inclined to the horizon, may be re- duced to the corresrionding horizontal angles, in cases Where very great accuracy is required. Let the lines PS, PB, BS, be three chords of terrestrial arcs, that is, let the points 13’, B, and S, be all equally distinct from the centre of the earth, and let the point D be elevated so as to be farther from the centre of the earth than any of the points P, B, S, itjsreguired to reduce the triangle BDP to the triangle BSD. ' The line SD may be supposed to be perpendicular to s 2 388 TRIGONOMETRY. PART x. each of the chords SB and SP without sensible error, though strictly speaking the angles DSB and DSP are each equalto 90°+§ the are which the chords SB and SP subtend, by art. Likewise the chords SB and SP may be used instead of their corresponding arcs. By inspection of the figure it is plain that ED is greater than BS, and PD greater than PS; but the base PB is common to the two triangles BDP and BSP, therefore the /_ BSP is greater than the /_ BDP. (21 Euclid, I.) , The horizontal angle SBP is obtained from the angle of elevation DES, and the oblique angle DBP by the fol- lowing proportion : As cosine of the angle of elevation DBS : cosine of the oblique angle DBP :: radius : cosine of the horizontal angle SBP. , Exactly in the same manner the angle SPB may be found. , Then angle BSP:180°—— (ang. SBP+ ang. SPB). But the angle BDB taken on the elevated station may be reduced-to the corresponding angle BSP by using only the. observed angles BDP, and the angles of depression mDB and ”DP. The angle mDB = ang. DB8, and the ang. nDPzang. DPS. Then by De Lambre’s formula {sine % (BDP+ DES—DPS) x Sine half ang. BSP=rad. x 4/ sine % (BDP + DPS— DBS) } cos. DBS x cos. DPS Which is in a form adapted for logarithmic computation. This formula of De Lambre’s may be applied to any two triangles, Whether their bases be in the same hori- zontal plane, and their vertices elevated, or their vertices be in the same horizontal plane, and the extremities of the base of the one triangle be elevated above the ex- tremities of the base of the other. In the annexed figure, if the angle PDB be measured with a sextant, and the vertical angles BDO, PDS be measured with a theodolite, PART x. TRIGONOMIZTRY. 389 by 1%. de Lambre’s formula, Viz. sin. % (PDB+PDS—BDO) . sin. % (PDB+BDO—P1>S) Sin. 54 SDO=rad. cos. PDS . cos. BDO EX AMPLE. From a station at D in the horizontal plane DSO, I took the angle PDB, subtended by the tops of two towers =37° 523’ 02”, and also the angles of elevation BDO: 4° 23’ 55”, and PDS— *- 4° 17’ 21”. The height of the tower B0 is known to be 40 yards, and that of PS 30 yards; from which it is required to find the horizontal distance of my station from each of the towers, and their horizontal distance from each other. Ans. /_ SDO :: 38° 0’; D8: 400 yards, DO— _ 50,2 and 802320 yards. SOME TRIGONOMETRICAL QUESTIONS WHICH REGARD THE EARTH’S CURVATURE. At the distance of 25 miles from a tower its top just appeared in the horizon; required its height. The dia- meter of the earth being 7916 miles, and its circumference 24869 miles. F Solution. 24869 miles: 360° :1 25 121’ 43”“ —_ A DAB. In the right angled triangle ABD, AB and the angle DAB 'are given, by which DB may be found: 25 miles. Then CE: 7916 miles: DB 1: DB: DC very nearly 419 feet. s3 390 TRIGONOMETRY. PART x. EXAMPLE . Supposing it were possible to see a light-house or other object D, in the horizon, at the distance of 200 miles, it is required to find its height CD, the diameter EC of the earth being 7916 miles, and its circumference 25019.7024 miles. Ans. The 4 DAB=2° 52’ 40”, AD=3987.028 miles, andDC=5.028 miles :: 26547.84feet, being 5914.84 feet higher than Chimborago, the highest of the Andes ,- and 285.84 feet higher than Dhawaligiri in Bootan, the highest hnown mountain in the world. Similarly, when DC and therdjt‘meter CE are given DB : 47 CE >< DC. EXAMPLE. Let DC:26.262 feet, the height of Dhawaligiri, and CE=7916 miles to find the extent of the visible distance DB. ' Ans. 197.6 miles. OF THE DIP, OR DEPRESSION OF THE HORIZON AT SEA. The dip or depression of the horizon at sea is the angle con- tained between the horizon of the observer, and the farthest visible point on the surface of the sea. For, if an observer whose eye is situated at D, takes the alti— tude of a celestial object by a sextant, or Hadley’s quadrant, and brings that object to the surface , of the water at B, instead of the E horizon DF, he evidently makes the altitude too great by the A FDB = A BAC. Log. tangent dip FDB:6.49047 + % log. of DC in feet. Let the height DC be 40 ; required the dip. Ans. 6’ 44”. THE METHOD 0F LEVELLING, AS GENERALLY PRACTISED BY SURVEYORS, ENGINEERS, ETC., FOR THE RAILVVAYS. LEVELLING is the art of determining the heights or de- pressions of points on the earth’s surface, or, 2then the extent of ground to be levelled is inconsiderable, with respect to an horizontal plane passing through some given point on the ground. in order to obtain the difference of level between any number of places, 1eference must be had to some fixed mark, from which an imaginary line, called the datum line, is drawn, upon which line all the levels are based; and all variations on the surface of the ground are reck- oned from this assumed line, some fixed mark being generally chosen at either end of the line of levels, to which they are all referred. It is sometimes deemed advisable to fix the datum line lower than either end, whichemay be done by assuming a line of 100 or any number of feet lower than the fixed mark, on which assumed datum line the levels will be based. In setting out lines of railways it is very obvious that the centre line of the road should be very accurately de- termined, and for this purpose a stake should be firmly driven into the ground at every chain’s length, or at every 100 feet, as may be deemed most advisable : this being done, the centre line should be carefully levelled, and sights taken by the spirit-level at each stake respectively, which are termed bench marks; these stakes should be numbered in regular succession ; each section must be plotted and the gradient marked thereon, showing the declivity or acclivity of the line, as also the height of the embank- ment or depth of the cutting at each stake, the respective widths should also be noted, and the names of the pro- prietors of the grounds through which the line passes, the s A: 392 LEVELLING. quality and quantity of land that may be required, and the estimated damage they may probably sustain either by severance or otherwise. It is not a difficult matter to set out a straight line of railway, but to do it correctly a good theodolite is indispensable: should any impediments, such as woods, buildings, 850., obstruct your line, recourse must be had to tie lines, see Prob. 8, Land Surveying; or two sides of an equilateral triangle may be measured, by first setting off the supplementary angle of 120° between the first side of the triangle and the line run ; let the station forming the triangle be placed in such a position as to form an angle of 60° with another station forming the second side of the triangle, which two sides being mea- sured equal, the third side and angles will also be determined and equal. When a line is curved the straight line becomes a tangent to it, which admits of a variety of trigonometri- cal calculations; but this is disregarded by surveyors generally, it being evident that the circle cannot be struck from a centre, but that the curve may be easily formed by a combination of straight lines. When the gradient of a line is known, you may easily determine the respective widths that will be required on every part of the line: knowing the embankment or cutting where stakes are driven, multiply the height or depth of the section by 4, to which add the width of the top of the railway and the ditches, if any, and you will have the width of the bottom; this rule is founded (though subj ect to variations) on the principle that every foot of perpendicular height requires two feet of base. The gradients or rates of clivity may be ascertained by dividing the difference of level by the distance in chains; the quotient will be the rise or fall at each chain’s length. The method of calculating the contents of embank— ments or earth-work must be performed by the rules for the prismoid, Prob. 10, Part 4, as also Probs. 3 and 4, Part 6. When the sections are numerous the calculations by these rules become very tedious, whence the reader is referred to Bidder’s tables, which will be found very useful to railway surveyors in performing their various calcula4 tions. The figure of the earth is said to be known by a sur- LEVELLING. 393 face perpendicular to the direction of gravity, or the direction which heavy bodies fall or gravitate towards the _ centre, the visible horizon being a tangent, or at right angles to the line of direction. T wo or more places are on a true level when they are , equally distant from the centre of the earth; also, one place is higher than another, or out of level with it, when it is farther from the centre of the earth, and a line equally distant from that centre in all its points is called the line of true level. But the earth being round, that line must be a curve, which forms a part of the earth’s circumference, or at least parallel to’ it, or concentrical with it, as the line BCFG, which has all its points equally distant from A, the centre of the earth, considering it as a perfect globe. But the line of sight BDEH, given by the operations of levels, is a tangent, or a right line per- pendicular to the semi-diameter of the earth at the point of contact B, rising always high- er above the true . line of level, which '3" is also called the apparent line of level. Thus CD is the height of the apparent level above the true level, at the distance BC or BD; also EF is the excess of height at F, and GH at G, 85c. The difi‘erence, it is evident, is always equal to the excess of the secant of the arc of distance above the radius of the earth. Now the difference CD, between the true and apparent level, at any distanCe BC or BD may be found thus. By a well-known property of the circle, 2A0 + CD: BD :1 BDzCD; or, because the diameter of the earth is great with respect to the line CD, at all distances to which an operation of levelling commonly extends; sup— pose ZAC be taken for 2A0 +CD in that proportion, without any sensible error, it will be 2A0 : BD :1 BD : CD, which is :: BD2 —:- 2A0, or BC2 —:— 2A0, that is, s 5 ‘ 39a LEVELLING. the difference between the true and apparent level is equal to the square of the distance between the places, divided by the diameter of the earth; consequently it is always proportional to the square of the distance. . Suppose the diameter of the earth to be 7958 miles; if we first take BC = 1 mile, then the excess B02 ~3— 2AC becomes WIS—g of a mile, which is 7.962 inches, or nearly 8 inches, for the height of the apparent above the true level at the distance of one mile. In order to make prOper correction for the curvature of the earth, observe the following rules. When the distance is in yards. RULE I. Multiply the square of the distance in yards by 2.57, mi divide by 1000000, or cut off six places on the right- hand for decimals ; the rest is the curvature in inches. When the distance is in chains. RULE 1L Divide the square of the distance in chains by 800, the quotient will be the depression in inches. When the distance is in miles. RULE III. Multiply the square of the distance in miles by 66 feet 4 inches, and divide by 100, or prick off two decimals, and you will have the depression in feet. Or, two-thirds of the square of the distance in miles will be the amount of curvature in feet, nearly. Note. It must be observed that an allowance of one-seventh is deducted for refraction, it being opposed to the curvature. The terrestrial refraction is found to vary with the state of the atmosphere in regard to heat, cold, and humidity, so that what may be deemed correct in one state of the atmosphere will not answer correctly for another. EXAMPLE I. Suppose the distance of the line of level to be 900 yards, what is the curvature ? LEVELLING. . 395 Here 9002 x 2.57 = 810000 x 2.57 = 2081700, and sir decimals being cut of leaves 2.0817 inches ,- then 2.0817 g ——%=1.7833 inches, the'true curvature. EXAMPLE II. Suppose the distance of the line of level is 40 chains, Here 402+800—-7—:1600+800—+:2~%«=1.7l in. EXAMPLE III. . Suppose the distance of the line of level to be 2% miles, required the curvature. . Here 2.52 x 66.33 —:— 100 = 6.25 X 66.83 —:- 100 :2 414.5625 —:— 100 = 4.145625; then deduct % leaves 3.55 . 2 2 1 _. 2 1 .— .feet for cnmatnre. 0r, 2.5 x g—7 __ 6.25 X? —- 7 .— 423 azae feet. Note. The correctiou for refraction of one-seventh will be found sufficiently correct for every state of the atmosphere, except in very extensive trigonometrical operations. It will be necessary to observe that in the commonpractice of levelling, the surveyors and engineers for the railways seldom make any allowance for either curvature or refraction. A DESCRIPTION OF THE INSTRUMENTS , USED IN SU’RVEYING AND LEVELLING. SPIRIT-LEVEL is a tube of glass nearly filled with spirits of Wine or distilled water, and hermetically sealed at both ends, so that when held with its axes in an horizontal posi- tion, the air which occupies the part not filled with the spirit or water places itself contiguously to the upper surface. The tube being supposed to be perfectly cylin- drical, the exact horizontality of its axes is ascertained by the extremities of the air—bubble being at equal distances from the middle point in the lengthof the glass. The spirit-tube is used in determining the relative heights of ground at two or more stations, in order to render it available : for this purpose it is placed Within a brass case having a long opening on the side Which is s 6 1 396 LEVELLING. ' the uppermost, and is attached to a telescope; the teles- cope and tube are then fitted to a frame or cradle of brass, which is supported on three legs; in the interior of the telescope, at the common focus of the object-glass and eye—glasses, are fixed two wires, at right angles to each other, their intersection being in the line of colli- mation, or that which joins the centres of all the lenses. —\ =- =- = . § «mm: mm mm Im- fi . The case containing the spirit—tube is made to turn on a joint at one extremity, as a, by the revolutions of a screw I), at the opposite extremity; and the telescope rests, near each end, within two agns at the top of a small pillar, A or B, the pillar and its arms resembling the letter Y, and the interior sides of the arms being tangents to the tube of the telescope. One of these pillars is made capable of a small movement in a vertical direction by turning a screw, C, at its base, for the pur- pose of elevating or depressing one end of the telescope and spirit-tube, and, in the more perfectly constructed instruments, both the pillars may be so moved. The pillars are at the extremities of a strong brass plate, EF, the under side of which is connected with the tripod- stand, which supports the whole instrument; and a compass-box, G, is attached immediately to the plate, as in the cut, or is raised above the telescope by means of four small pillars. A hollow conical socket, H, of brass, is screwed to the under side of the plate, and is intended LEVELLING. 397 to receive a piece of bell-metal of a corresponding form, which constitutes the upper part of the stand. This piece serves as a vertical axis, upon which the telescope, the spirit—level, and the compass are turned round hori— zontally; sometimes, however, the conical pivot projects from the under part of the plate, EF, and the socket is ‘ on the stand. ' The three legs which are to support the instrument are firmly fixed to a circular plate, K, perforated at its centre, and having about the perforation a hollow spherical zone, resembling a small inverted cup. In the simpler kinds of spirit-levels a circular plate L, of the same dimensions as the last, carries above it the pivot before mentioned ; and from below it projects a stem terminating in a ball, which fits the inverted cup or socket. By means of four screws which pass through one of those two plates (the upper plate in , the cut) nearly at the extremities of two diameters at right angles to one another, the upper plate is made parallel to the horizon, and consequently the conical pivot which it carries is brought to a vertical position. The telescope should, by a proper opening of the legs of the stand, be first reduced as nearly level as can be estimated by the eye: then, being turned so as to lie ver- tically above the line joining two opposite screws in what are called parallel plates, K and L, the spirit tube is brought to an horizontal position by relaxing the screw nearest to its higher end, and tightening that which is opposite to it. The like operation is to be performed with the other pair of screws, after placing the telescope vertically above them. In order to render the spirit-tube parallel to the axes of the telescope, after the bubble of air has been made to occupy the middle place by the pro— cess just mentioned, let the telescope be reversed in the arms, the Y’s as they are called; then if the bubble does not still occupy the middle, it must be made to do so by successive trials, endeavouring to correct half the error by means of the screw 2'), and the other half by the screw G. The eye-piece of the telescope must be moved inwards or outwards till the wires in the field of view are dis- tinctly seen; and the obj ect-glass must also be moved by means of the pinion M, till the station-staff; placed at any 398 LEVELLING. convenient distance, suppose 100 yards, is also distinctly seen. By a few trials, the distance between the eye and the ob} ect—glass may be made such that the intersection of the wires will appear to remain constantly at one point on the staff, while the observer on looking through the telescope varies the position of his eye. It is necessary besides that the intersection of the wires should be pre— cisely in the line of collimation, or the optical axes of the telescope; for this purpose the point of intersection should be directed to some well-defined mark at a con- siderable distance. The telescope must then be turned on its axis; and if the intersection remains constantly on the mark, the adjustment is complete, otherwise it must be rendered so by means of the screws, 0, d, &c., on the telescope; those screws being placed at the extremities of two diameters at right angles to one another, on being turned they move the plate carrying the wires in the direction of those diameters. In order that the correction may be made, the apparent displacement of one of the wires, in consequence of the telescope being turned half round on its axis, should be observed, and the screws turned till half the displacement is corrected; the like observation and correction may then be made for the other wire; a few repetitions of each adjustment will probably be necessary before the error is wholly removed. The level constructed by the late Mr. Troughton difi‘ers from that which has been above described in having the spirit-tube sunk partly in the telescope, and the latter, being incapable of a movement about its axis, does not admit of a separate adjustment for the intersection of the wires. Mr. Gravatt, who has within a few years made con- siderable improvements in the mechanism of these instru- ments, recommends the following method by which the error in the positions of the cross-wires and spirit-tube may be ascertained and corrected. ' Let three pickets be driven into the ground in a line and at equal distances from one another, and let the spirit- level be set up successively in the middle between the first and second, and between the second and third pickets; then having by the screws of the instrument adjusted the spirit-tube so that the bubble of air may retain the same place while the telescope is turned round LEVELLING. 399 on the vertical axis, direct the object-end of the telescope successively to the station-staves held up on the different pickets, read the several heights, and take the differences between those on the first and second, and on the second and third staves. Now the staves being at equal dis- tances from the instrument, it is obvious that any error which may have existed in the line of collimation, or from the spirit-tube not being parallel to that line, will be destroyed, and the differences between the readings on the staves are the differences in the levels of the heads of the pickets; but unless the adjustments are perfect, this will not be the case if the instrument be set up at any point which is unequally distant from all the pickets ; therefore from such point direct the telescope to the staves, and take the differences of the readings as before; on comparing these differences with the former, a want of agreement will prove that the intersection of the wires is not in the optical axis, and the error may be corrected by means of screws belonging to the wire plate. After the agreement has been obtained, should the bubble of air not stand in the middle of the tube, it may be brought to that position by the screw 6, at one extremity of the case, and the instrument is then completely adjusted. The spirit-level is usually provided with a clamp, N, and a screw, P, by which, when the axis of the telescope has by hand been brought near the object, the coincidence may be accurately made by a slow and steady motion about the vertical axis. The spirit-tube or level which is employed for the adjustment of transit telescopes or astronomical circles is contained in a case with feet or with loops at its ex- tremities, in 01 der that it may rest above or be suspended below the horizontal axis of the instrument to be levelled: also the upper part of the case is furnished with a gradu‘ ated scale, the divisions of which are numbered on each side of a zero point, this point being usually placed near each of the two extremities of the air-bubble when the tube is in an horizontal position. Having set up or suspended the spirit-tube, the two particular graduations at which the extremities of the air-bubble rest are marked; and half the sum, or half the difference of these numbers, according as the extremities of the bubble are 400 LEVELLING. in the same or in opposite directions from the two zero- ‘points, being taken, gives the distance of the centre. of the bubble from the middle between those points. The level being then reversed, the graduations at which the air-bubble rests are again marked, and half the sum or half the difference is taken as before. A mean of the two distances thus found is the true distance of the centre of the bubble from the middle point on the scale; and the screw which elevates or depresses one end of theaxis of the telescope being then turned, till either extremity of the bubble has moved, in a direction contrary to that in which the centre of the bubble had moved from the middle of the scale, through a number of divisions equal to that mean distance, that axis will be brought to an horizontal position. This method is used in preference to that of successive trials, in order to avoid the trouble of making several reversions of the whole instrument. ‘ The levelling staff is a rod consisting of two parts, each six feet long, which, by being made to slide on one another, will indicate differences of levels nearly as great as twelve feet. The face of the rod is divided into feet, inches and tenths, and a vane or cross-piece of wood, perforated through the middle, is moved up or down upon the rod by an assistant till a chamfered edge at the perforation is seen by the observer at the spirit- level to coincide with the horizontal wire in the telescope. The height from the ground to the vane must be read by the assistant. The staves now in general use, invented by Mr. Gravatt, are without any vane, and are divided into feet, tenths, and hundredth parts of a foot, the lines and numbers being sufficiently distinct to enable the observer to note the reading of the instrument, with the telescope now applied to spirit—levels, at the distance of ten chains: the arrange- ment of the staves is simple; they are in three pieces, with joints similar to a fishing-rod, and when put together for use are 17 feet in length. Mr. Sopwith and Mr. W. P. Barlow have improved on Mr. Gravatt’s invention; the divisions are nearly the same, but the subdivisions are more minute; when closed it is 5 feet in length, but draws out to 15 feet, and a strong catch retains each joint in its place. LEVELLlNG. 401 Another staff has been invented by Mr. Peter Bruft; author of a treatise on Surveying and Levelling; this stain is also frOm 5 to 15 feet in length, but the figures are inverted, whereby, when viewed through an inverting telescope, they appear in their natural order; this staff has a shoe attached to it, which the author considers of considerable importance in practice. THE THEODOLITE. The Theodolite is an instrument used for measuring horizontal angles; it consists of a circular plate, which is to be set parallel with the horizon, which is divided quite round the circle into 360°. A semicircular are for taking vertical angles is also fixed to the Y’s, the whole mo- ving in a vertical plane; from the centre of the circle describing this are are projecting arms resting on stan- dards, which are fixed to the upper plate of the instrument on which the verniers are marked, which plate is called the limb. Brufi, in his treatise on Surveying and Levelling, says he used a five-inch theodolite, which he recommends as the most useful size for general purposes, made by Troughton and Simms. Captain Everett’s improved theodolite is said to have some advantages over those in common use; it has three verniers for taking horizontal angles, the mean angle being taken, as also the two vertical angles. Ramsden’s great theodolite, the circle of which is three feet in diameter, was used for a triangulation to connect the observatories of Greenwich and Paris ; the principal triangles of the English, Irish and Indian surveys have been observed with this instrument, or with those nearly identical in size and construction. The course of observation, after the instrument is pro- perly adjusted, is very simple. The question is to measure the horizontal angles between two objects. Turn the telescope two or three times round in the direction in which you intend to observe, bisect one of the objects, read ofi’ the verniers, take a mean, and the difference of the two means is the angle required. By means of a repeating table the operation is continued thus: bring the telescope back on the first object by the motion of the repeating table, using its clamp and tangent screw, and by 402 LEVELLING. the motion of the instrument, bring the telescope on the second object; it is evident the motion of the repeating- table has nearly restored the telescope to its original direction, without altering the readings of the circle; and that if the telescope be turned on the second object by its motion alone, without distributing the circle, the difference between the mean of these new readings and the preced- ing mean will also be the angle required : and by continu— ing the process, the angle may be measured as often as the observer pleases, or, in order to bisect objects correctly, a slow-motion screw is attached to the upper plate, the clamp-screw securing it when the object is nearly bisected, and the slow-motion screw, moving the vernier through the least possible space, will complete the bisection. A vernier is fixed to the plate through which the are for taking vertical angles passes; two spirit-bubbles are attached to this plate at right angles to each other, for the purpose of setting the instrument in an horizontal posi— tion. A small compass is also attached to this plate, the line NS ranging with the line of sight; the bearings at the different stations being noted will serve as a check on the angles, the parallel plates being mounted similar to those of the spirit-level. The method of performing adjustments in the theodo- lite is similar to that of the level; the axis of the hori- zontal plate must be vertical, that is, set up the instrument as level as you can, the telescope lying over the plate- screws ; then, by means of the clamp and slow—motion screws attached to the vertical arc, bring the bubble into the centre of the tube, move the instrument half round, and the telescope will be over the other platevscrews ; and if the bubble remain in the centre of the tube it is right, if not, it may be corrected thus, one-half by the clamp and slow-motion screw attached to the vertical arc, and the other half by the parallel plate-screws ; this pro- cess may be repeated, if necessary, until the result is satisfactory, and the bubble stands in the centre of the tube. The adjustment of the vertical are :-—when the vernier stands at zero, and the former adjustments are perfect, it is correct; if not, alter the vernier by means of the screws, attaching it to the plate until it does, and note the error, if any; then allow for it in each vertical angle; and for LEVELLING. 403 the purpose of determining this error accurately, it will be necessary to take the vertical angle of some conspicuous object with the telescope, reversed in the Y’s; this must be done by turning the instrument half round; the tele- scope will then occupy the. same position as at first, but the vertical arc will be reversed ; and the mean of these. two readings will be the error required. s TO TAKE ANGLES WITH THE THEODOLITE. Fix the theodolite exactly over some station, which can be easily done with the assistance of a plummet and line being suspended under the centre from a hook at- tached to the stand; fix it level with the parallel plate- screws, clamp the plates together, turn the instrument towards some station or B mark ; clamp the lower plate, and the bisection must be made with the slow-motion screw, always observing to bisect the object as near the ground as possible; observe the number of degrees, minutes, and seconds, taking the mean of two or three verniers, (two, however, should always be used); the upper plate may now be loosed, but the lower one must remain clamped and kept steady; turn towards another station, which clamp as before, then bisect with its own slow- motion screw, observe the number of degrees, minutes, and seconds, as before, and their difference will be the angle required. But in general practice the vernier is set at 360° and clamped; the instrument is then to be turned quite round, in the direction of the first object; this being done, clamp the lower plate, and bisect with its slow—motion screw, release the upper plate, the lower remaining clamped; then turn the instrument in the direction of the second station, which clamp and bisect, and you will have the angle as before. The latter method of taking angles is generally adopted by surveyors, although not so correct as the former, it being a difficult point to set the instrument exactly to 360°, and in exten- sive operations, where many angles are required to be taken from one station, the former method is decidedly preferable. At each principal station the bearing is the angle pointed out by the compass needle, which will serve as a check on the accuracy of the angle when 404 LEVELLING. the bearing can be read correctly; set the plates and needle at zero, and the angle on the limb, when the object is bisected, will be the bearing. Vertical angles are taken for the purpose of reducing hypothenusal measure to horizontal; on one side of the vertical are are the degrees and minutes for determining the angle, and on the other the number of links to be deducted from each chain’s length, to reduce it to hori- zontal measure: to do this, set up a mark at the exact height of “the optical axis of the telescope, .also at the extreme point to be measured to the angle on which this deduction is to be made, which can be allowed for in the field; thus suppose the angle of elevation to be 18° 15’, on the other side of the arc is five links to be deducted from the hypothenusal line ; or the chain may be drawn forward on the ground five links to bring it to horizontal measurement. Surveyors generally find it the most ad- vantageous to make the allowances in the field, especially where many cross fences occur and you take a good quantity of offsets ; it would be found very troublesome in plotting, and to make the necessary allowance for each distance ; but if great accuracy is required, the angle should be noted, and the reduction made in plotting the work, as it will be seen you can very well allow for even links on each chain in the field, but not for the decimal parts, which might be considerable on a line of great extent ; hence it depends on the description of the work whether the allowance is made in the field or office. It is required to find the difference of level between the two points A and L. Set up the level as at B, also the staff A, the spirit-level must be brought to an horizontal position, and on looking through the telescope attached to it, you have the dotted line of sight towards the staff at A, LEVELLING. 405 which is graduated or divided into feet, inches, 860., which can be read by the observer with the aid of the telescope. Suppose the line of sight to cut the staff A at 3 feet 6 inches, the spirit-level being turned horizontally in the direction of the staff at C, which is bisected at 7 feet; then the difference between the two points A and C will be 3 feet 6 inches. The spirit-level is then moved to D, the staff at C still remaining; the spirit-level being in an horizontal position towards the staff at C, which it bisects at 1 foot 6 inches, it is then turned towards the staff at E, which it bisects at 7 feet 9 inches; the difference of level- between the points 0 and E will be 6 feet 3 inches, and from A to E 9 feet 9 inches. Remove the spirit-level to F, the staff being placed alternately at E and G; at E it is 3 feet, and at Gr 7 feet, which is a fall of 4 feet from E to G, making altogether the fall from A to Gr 13 feet 9 inches. Remove the level to H, and on the staff at G is 5 feet 3 inches, then on the staff at I is 2 feet 3 inches, which is a rise of 3 feet from G to I, making a fall from A to I 10 feet 9 inches; place the level at K, and on the staff at I is 4 feet, and L is 2 feet 9 inches, making a rise of 1 foot 3 inches; the two rises together make 4 feet 6 inches, which being deducted from 13 feet 9 inches will be equal to 9 feet 6 inches, the difference of the level from A to L; and in this manner the process may be continued to any length, adding or subtracting as the ground rises or falls. The difference of the back and fore sights, between the two extreme points A and L, is the difference of level. Ft. In. Ft. In A 3 6 7 0 C l 6 7 9 E 3 0 7 0 Cr 5 3 2 3 I 4 O 2 9 Back sights l7 3 26 9 fore sights. l7 3 Diff. of level between A and L 9 6 But in order to draw the various sections so as to form a plan and show the undulation of the ground between the 406 LEVELLING. points A and L, you must meaSure the respective distances between A and C, C and E, 8w. The staff being set up at each change of inclination of the ground, and knowing the distance from one staff to another, you may proceed to plot the section, and the form of the field-book would be similar to the followin