% I mm- &®i ir*^ IN MEMORIAM FLOR1AN CAJORI A. StnUtr' HAWNEY'S COMPLETE MEASURER: OR, THE WHOLE ART OF MEASURING A PLAIN AND COMPREHENSIVE TREATISE ON PRACTICAL GEOMETRY AND MENSURATION. Corrected and improved by T. Keith. THIRD EDITION. WITH AX APPENDIX CONTAINING RULES AND EXAMPLES FOR tflND- LSTG THE WEIGHT AND DIMENSIONS OF BALLS AND SHELLS, WITH THEIR APPROPRIATE QUANTITIES OF POWDER. REVISED AND CORRECTED BY JOHN D. CRAIGj PROFESSOR OF MATHEMATICS, BALTIMORE : PUBLISHED BY F. LUCAS, JR. AND NEAL, WILLS & COLE, S. P. CHILD & CO. PRINTERS. 1813. \w*\ v^ \S'I3 <: .' pjsti-ici ?f >W>lanclj to wit : *••'••••* BE II' REMEMBERED, that on this first day of * seal. * May, in the thirty-seventh year of the Independence ******** of the United States of America, FjgJ i djngLiicas, junr. of the said district, liath deposited in this office, tli'e title of a book the right whereof he claims as proprietor, in the words and figures following-, to wit : " Hawncy's Complete Measurer : or, the whole art of meas- uring. Being a plain and comprehensive treatise on "practical geometry and mensuration. Corrected and "improved by T. Keith. Third edition, with an appen- dix containing- rules and examples for finding- the weight "and dimensions of balls and shells, with their appropri- ate qviantities of powder. Revised and corrected by " John D. Craig-, professor of mathematics." In conformity to the act of the congress of the United States, intitled, " An act for the encouragement of learning by securing the copies of maps, charts, and books, to the authors and pro- prietors of such copies during the times therein mentioned." And also to the act, intitled " An act supplementary to the act, intitled, 'An act for the encouragement of learning, by secur- ing the copies of maps, charts, and books, to the authors and proprietors of such copies during the times therein mentioned,' and extending the benefits thereof to the arts of designing, engraving, and etching historical and other prints." PHILIP MOORE, t'krk of the District of Maryland. cajor: PREFACE BY THE EDITOR. THE many editions which Mr. Hawney's Treatise on Mensuration has gone through, are evident proofs of its general utility. But at the time when this work was first pub- lished, it was not the practice in schools for each scholar to have a printed book by him, while engaged in the study of arithmetic and mensuration; consequently Mr. Hawney gave no particular examples as exercises for a learner, without working them at full length, and explaining every step. Though this was perhaps originally an advantage to the book, it precluded the use of it in our modern schools; for with such assistance, a boy of good abilities would naturally become indo- lent, for want of something to exert Ins ge- nius ; and a boy of a heavy disposition would be induced to copy all his work from the book. To remedy these inconveniences, the pro- prietors of the work engaged the present edi- tor to make such alterations and additions as IV PREFACE. would render it useful in schools, without diminishing its former plainness and perspi- cuity. On examination of the sixteenth edition, the editor found it to he replete with errors, owing to the inattention, or incapacity of those who had the care of the press, since the death of the author. Hence arose the necessity of working every example anew, and examining every rule with the greatest care. This was attended with considerable labour, and at the same time was not a very agreeable task ; for as the nature and plan of the work has undergone little or no change, so the merit is still due to the author, while the errors, if any of consequence are left, will be attached to the present editor. The first nine chapters are in substance the same as in former editions. The Xth, Xlth, and XII th are added by the editor, besides various rules and observations throughout the work, the principal of which are distinguish- ed in the table of contents by prefixing an asterisk. - The mensuration of the five regular bodies is placed immediately after the mensuration of solids, and these are succeeded by board and timber measure. PREFACE. V In the former editions, timber measuring was divided into several sections ; as, squar- ed timber, or pieces of timber in the form of a parallelopipedon ; unequal squared timber, or pieces of timber in the form of the frustum of a pyramid ; round timber with equal bases, or pieces of timber in the form of a cylinder ; round timber tviih unequal bases, or pieces of timber in the form of the frustum of a cone : and the contents were found, both by the customary method of measuring timber, and likewise by the rule for each respective solid. In this edition these examples are placed under the several rules to which they belong, and are distinguished by asterisks* ; and in timber measuring they are brought forward, and solved by the customary method, with reference to the pages where they are truly solved. By this means several pages have been saved for the introduction of more useful matter. The gauging and surveying, which were given in the former editions by way of ap- pendix, are introduced in this edition with- out that distinction. The former has under- gone such alterations as will render it very useful for those who are candidates for the excise, or who want to find the contents of different kinds of vessels. And the latter - % M OlIEFACE. will be found to contain sufficient information for such as want to find the true content of any single field, or parcel of land, by the chain and cross-staff only. To form the complete gauge* and surveyer, recourse must be had to treatises written expressly for the purpose. The practical questions, which proceeded the appendix, are now placed at the end of the book and in the same order in which they were formerly arranged. The rest of the practical examples through- out the work, which amount to about four hundred, are given by the editor ; several of which were copied from a manuscript, con- taining notes and observations on Hawney, by Mr. Rawes, master of the academy at Bromley, in Kent. In this edition, eighty new geometrical fig- ures, illustrating the different problems, are likewise given ; so that neither pains nor ex- pence have been spared, to render this work of equal utility with any other of a similar nature. Heddon- Court, Swallow -street, September, 1798. CONTENTS. PART I. ARITHMETIC AND GEOMETRY. Chap. Page I. Notation of Decimals, 1 II. Reduction of Decimals, .... 3 III. Addition of Decimals, 10 IV. Subtraction of Decimals, . . . 11 V. Multiplication of Decimals, . . .12 VI. Division of Decimals, . . .17 VII. Extraction of the Square Root, . . .25 VIII. Extraction of the Cube Root, . . . 31 IX. Duodecimals, . . . . .40 X. *Of Gunter's Scale, ... . .48 *0* the Diagonal Scale, ... 51 XI. *Of the Cahfenteh's Rule, . . .52 XII. *Piiactical Geometry, . . . .57 PART II. CIIAFTEH I, ^Mensuration of Superficies. 1. To find the area of a square, . . 79 2. To find the area of a rectangled parallelogram, 80 3. To find the area of a rhombus, ... 82 4. To find the area of a rhomboides, . . .83 5. To find the area of a triangle, . . . . 84 *T\vo sides of a right-angled triangle given to find the third, .... 89 6. To find the area of a trapezium, . .91 VJii CONTEXTS rvr.r Skit. * To find the area of* a trapezoid, 93 7- To find tin- ana of an irregular figure, . . vi 8. To find the area of a regular Polygon, . 96 y. Of a Circle, . 101 10. To find the area of a semicircle, . . 115 11» To find the area of a quadrant, . . . 116 To find the length of any arch of a circle, 117 To find the diameter of a circle, having- the chord and versed sine given, . . 119 \2- To find the area of the sector of a circle, . 120 13. To find the area of the segment of a circle, . 122 14. To find the area of the compound figures, . 125 15. To find the area of an ellipsis, . . ' 126 * To find the area of an elliptical segment, . 128 * To find the circumference of an ellipsis, . 129 * To find the area of an elliptical ring-, . . 130 16. To find the area of a parabola, . . 132 cnAr-TEE ii. Mensuration of Solids. Sect. 1. To find the solidity of a cube, . . . 135 2. To find the solidity of a parallelopipedon, . 137 3. To find the solidity of a prism, . . . 139 4. To find the solidity of a pyramid, . . . 142 To find the superficial content of a pyramid . 143 * The perpendicular height of a pyramid given to find the slant height, and the contrary, . 144 5. To find the solidity of a cylinder, . . . 147 To find the superficies of a cylinder, . . 149 6. To find the solidity of a cone, . . . 150 To find the superficies of a cone, . . . 151 7. To find the solidity of a frustum of a pyramid, 153 To find the superficies of the frustum of a py- ramid, 155 * To find the slant height, or perpendicular height of the frustum of a pyramid, . . . 155 CONTENTS. IX PAGE Sect. * To find the perpendicular height of that pyramid of which any given frustum is the part, . 155 * To find the area of the front of a circular arch 161 8. To find the solidity of the frustum of a cone, 163 9. * To find the solidity of a wedge, . , 165 10. To find the solidity of a prismoid, . . 167 To find the solidity of a cylindroid, . . 171 11. To find the solidity of a sphere or globe, . 174 To find the solidity of the segment of a sphere, 179 * To find the solidity of the frustum or zone of a sphere, 180 * To find the convex surface of any segment or zone of a sphere, . . . 181 12. Tq find the solidity of a spheroid, . . 185 * To find the solidity of the segment of a sphe- roid, . . ... 187 * To find the solidity of the middle zone of a spheroid, ..... 190 13. To find the solidity of a parabolic conoid, . 191 * To find the solidity of the frustum of a parabo- lic conoid, 193 14. To find the solidity of a parabolic spindle, . 194 * To find the solidity of the middle frustum of a parabolic spindle, . . . 197 * To find the solidity of the middle frustum of any spindle, 198 15. To find the solidity of the five regular bodies, 199 16. To measure any irregular solid, . . . 203 CHAPTEIt III. The Mensuration of Board and Timber. Sect. 1. To find the superficial content of a board or plank, 205 2. The customary method of measuring timber . 208 * A table for measuring timber, . . . 209 * A general scholium, or remarks, on timber mea- suring, 215 • "ON'TRNTS. nnrrr.n iv. 77a' Mensuration of Artificer*' Work. Sbct. 1. Carpenters' and .Toiners' work, . . '221 2. Bricklayers' work, 229 3. Plasterers' work, 238 4. Painters' work, .... 240 5. Glaziers' work, 242 6. Mason's work, .... 244 7. * Paviors' work, .... 245 CHAPTER V. Of Gauging. Pkob. * Of the sliding rule, 247 1. To find multipliers, divisors, and gauge-points, 250 2. To find the area in gallons, of any rectilineal plane figure, .... 252 3. To find the area of a circle in ale gallons, &c. 254 4. To find the area of an ellipsis in ale gallons, &c, 255 5. To find the content of any prism in ale gallons, 256 6. To find the content of any vessel whose ends are squares, or rectangles of any dimensions, 257 7. To find the content in ale gallons, 8cc. of the frustum of a cone, 259 8.*To gauge and inch a tun in the form of the frus- tum of a cone, ..... 261 9. To gauge a copper, 263 10. To compute the content of any close cask, . 265 * To find the content of am cask, . . . 269 * A general table for finding the content of any cask by the sliding rule, .... 270 11. Of the ullage of casks, .... 272 * A table of the areas of the segments of a circle, 278 CONTENTS. XI CHAPTER VI. Of Surveying. Prob. * Of the surveying-cross, or cross-staff, . 283 1. * To measure off-setts with a chain -and cross- staff, 284 2. To measure a field in the form of a trapezium, 287 3. * To measure a four-sided field with crooked hedges, 290 4. To measure an irregular field, . . . 291 5. * To cut off from a plan a given number of a- cres, 295 CHAPTER VII. Practical questions in measuring, .... 295 Explanation of characters made use of in the work, . 308 APPENDIX. Of the weight and dimensions of Balls and Shells. THE COMPLETE MEASURES. PAIIT I. CHAPTER I. Notation of Decimals, A. DECIMAL fraction is an artificial way of setting down and expressing natural, or vulgar fractions, as whole numbers. A decimal fraction has always for its denominator an unit, with a cypher or cyphers an- nexed to it, and must therefore be either 10, 100, 1000, 10000, &c. and consequently in writing down a decimal fraction there is no necessity for writing down the denominator : as by bare inspection, it is certainly known, consisting of an unit with as many cyphers annexed to it as there are places (or figures) in the numerator. Examples. The decimal fraction T **y may be writ- ten thus, .25, its denominator being known to be an unit with two cyphers ; because there are two figures in the numerator. In like manner, ^** v may be thus written, .125: ^gfo thus, .3575 ; -Jf^ thus, 1078} and iHtns thus > - 00 * 5 ' B lotion of Decimals. \.i -aIwIi lei i rease in a ten-fold proportion, tOwardl the li -I't hand, si) on the contrary, decimals de- crease towards the right hand in the same proportion. as in the following table. A c Xi .3 H a _g li — ! 5 -3 ,: - 9 ~ 2 «4 K f ^ Oj „ 3 PlbTl Z eg I a fi ^I^I 76543210, 123456 Hence it appears, that cyphers put on the right hand of whole numbers, increase the value of those numbers in a ten-fold proportion : But being annexed to the right hand of a decimal fraction, neither in- crease nor decrease the value of it : So T 2 ^°^y is equi- valent to T 2 ^ or .25. And, on the contrary, though in whole numbers, cyphers before them, neither increase nor diminish the value $ yet cyphers before a decimal fraction diminish its value in a ten-fold proportion : For .25, if you put a cypher before it, becomes -ft 2 ^ or .025: And .123 is ■fgjgfo by prefixing two cyphers thus, .00125. And therefore when you are to write a decimal fraction, whose denominator has more cyphers than there are figures in the numerator, the places of such figures must be supplied by placing cyphers be- fore the figures of your numerator ; as, suppose -j^^ were to be written down, without its denominator; here, because there are three cyphers in the denomi- nator, and but two figures in the numerator, therefore put a cypher before 19, and set it down thus, .010. Reduction of Decimals. CHAPTER II. Reduction of Decimals. IN Reduction of Decimals, there are three eases : 1st, To reduce a vulgar fraction to a decimal. 2dly, To find the value of a decimal in the known parts of coin, weights, measures, &c. and 3dly, To reduce coin, weights, measures, &c. to a decimal. I. To reduce a Vulgar Fraction to a Decimal. THE RULE. As the denominator of the given fraction is to its numerator, so is an unit (with a competent number of cyphers annexed) to the decimal required. Therefore, if to the numerator given, you annex a competent number of cyphers, and divide the result by the denominator, the quotient is the decimal equivalent to the vulgar fraction given. Example 1. Let | be given, to be reduced to a decimal of two places, or having 100 for its denomi- nator. To 3 (the numerator given) annex two cyphers, and it makes 300, which divide by the denominator 1, and the quotient is .75, the decimal required, and is equi- valent to 1 given. Note. That so many eyphers as you annex to the given numerator, so many places must be pointed oft* in the decimal found ; and if it should happen, that there are not so many places of figures in the quoti- ent, the deficiency must be supplied, by prefixing cy- phers to the quotient figures, as in the next example. .! Reduction of Decimal*- Example 2. Let -^ be reduced to a decimal hav- ing six places. To tlie numerator annex six cyphers, and divide by the denominator, and the quotient is 5235, hut it was required to have six places, therefore you must put two cyphers before it, and then it will be .005235, which is the decimal required, and is equivalent ro^. See the work of these two examples. 4)3.G0(.75 573)3.000000(.005.233 28 20 ' 1350 20 2040 3210 345 in the second example there remains 345, which re- mainder is very insignificant, it being less than totto? part of an unit, and therefore is rejected. PRACTICAL EXAMPLES. 3. Reduce -J to a decimal. Jlns. .57142 rem. 4, 4. Reduce 2 fj to a decimal. Jns. .0041152263 rem. 91. 5. Reduce £ of § of A to a decimal. Jlns. .20833, &c'. 6. Reduce 15 fa to a mixed decimal. Jins. 15.38461 rem. 7. Jlns. .17241379 rem. 9. 8. Reduce T ^ T to a decimal. Jlns. ,026178010471 rem. 39. Reduce ^ to a decimal. Reduction of Decimals. & Note. A finite decimal is that which ends at a certain number of places, such for instance as ex- ample 1. But an infinite decimal is that which no where ends, but is understood to be indefinitely con- tinued, such as example 3. — In short, all fractions whatever, whose denominators are not composed of 2" or 5, or both, will have their correspondent decimal infinite. The method of managing circulating deci- mals may be met with in KeiWs Arithmetic, and se- veral others ; but for common use, all the decimals, beyond three or four places, may be safely rejected, without affecting the truth of the conclusion. II. To find the value of a Decimal in the known part? of money, weight, measure, <$*c. THE RULE. Multiply the given decimal by the number of parts in the next inferior denomination, and from the pro- duct point oft* so many figures to the right hand as there were figures in the decimal given ; and multiply those figures pointed oft" by the number of parts in the next inferior denomination, and point off so many places as before, and thus continue to do till you have brought it to the lowest denomination required. Example. 1. Let .7565 of a pound sterling be given to be reduced to shillings, pence and farthings. Multiply by 20, by 12, and by 4, as the rule directs, and always point oft' four figures to the right hand, and you will find it make 15s. id. 2q. See the work* .7563 ? 20 s. . 15.1300 12 d. 1.5600 4 t ?• 2.2400 tf Reduction of Decimals. Example 2. Let .50755 of a pound troy be reduced to ounces, penny- we it^hts :ind grain* Multiply by 12, by 20, and by 21, and always point oft* five figures towards the right band, and you will find the answer to be 7 oz. Sdwts. i0.gr. fere. Bee the Vorfc. .59755" 12 ,17000 20 oz. pivts. pr. FacifT 3 9.8SS 3.41200 24 164800 82400 9.88800 Example 3. Let .43569 of a ton be reduced fo hundreds, quarters, and pounds. Multiply by 20, by 4, and by 28, and the answer will be 8 C. 2 qrs. 24 lb. fere. .43569 20 8.71380 4 C. qrs. lb. Facit 8 2 23.9456 2.85520 23.94560 Reduction of Decimals* ? Example. 4. Let .9595 of a foot be reduced info inches and quarters. .9595 12 11.5140 4 Facit 11 inches 2 quarters. 3.0560 PRACTICAL EXAMPLES. 5. What is the value of .7575 of a pound sterling ? J&ns. 15s. lfd. .2, 6. Required the value of .75435 of a shilling ? Jins. 9.0522 pence, 7. What is the value of .375 of a guinea ? Jlns. 7s. 10^,d. 8. What is the value of .4575 of a hundred weight? Jlns. 1 qr. 23 lb. 3 oz. 13.44 drams. 9. What is the value of .175 of a ton avoirdupois t 5 Jlns. 3 cwt. 2 qrs. 10. Required the value of .02575 of a pound troy ? Jins. 6 dwt. 4.32 grs. 11. What is the value of .04535 of a mile ? Jlns. 14 p. 2yds. 2ft. 5 in. 1.128 barley-corn, 12. What is the value of .6375 of an acre ? Jlns. 2 roods 22 perches. 13. What is the value of .574 of. a hogshead of beer ? Jlns. 30 gal. 3 qt. 1.968 pt. 14. What is the value of .4285 of a year ? Jlns. 156 days, 12hrs. 13 m. 51 sec. 36 thirds. III. To reduce the known parts of money, weights, mea* sure, Sfc. to a decimal. THE RULE. To the number of parts of the less denomination given, annex a competent number of cyphers, and 8 Reduction of Decimals. divide by the number of Mich parts that are contain- ed in the greater denomination, to which the deei- m;il is to be brought; and the quotient is the decimal sought. Example 1. Let Qd. be reduced to the decimal of a pound. To G annex a competent number of cyphers (sup- pose 3,) and divide the result by 210 (the pence in a pound,) and the quotient is the decimal required. 240)6.000(.023 1200 Facit .023 Example 2. Let 3c?. | be reduced to the decimal of a pound, having six places. In 3d. % there are fifteen farthings, therefore to 13 annex six cyphers (because there are to be six places in the decimal required,) and divide by 960 (the far- things in a pound,) and the quotient is .013625. 9610)13.00000l0(.013625 540 600 240 480 Example. 3. Let 3 \ inches he redueed to the deci- mal of a foot, consisting of four places. Reduction of Decimals. 9 In 3£ inches, there are 13 quarters ; therefore fo 13 annex four cyphers, and divide by 48 (the quarters in a foot) and the quotient is .2708. 48)13.0000(.2708 340 400 16 Example 4. Let 9 C. ±qr. 16/6. be reduced to the decimal of a ton, having 6 places. C. qr. Ik 9 1 16 2240)1052.000000(.469642 ** 37 qrs. 28 15600 302 75 21600 14400 1052 Pounds 9600 6400 1920 Facit .469642. PRACTICAL EXAMPLES. 5. Reduce 7s. 5|d. to the decimal of a pound r Ans. 1. 3729166, &c. 6. What decimal part of a pound sterling is three- halfpence ? Jins. 1 .00625. 7. Reduce 4s. 7 1 9 T d. to the decimal of a pound ster- ling i Ms, 1 .2325757, &c 10 Addition of Decimals. 8. Reduce 10 o%, Ji dwt. 3 gr. to the decimal of a pound Ti Ans. ,1296870 ft. 9. Reduce Scui. 1 qr. 1 1 lb. to the decimal of a Ton? Ans. .16870 Ton. 10. Reduce 22 feet 7 inches to the decimal of a Foot ? Ans. 22.08*8 Feet 11. Reduce 2qrs. 15 lb. to the decimal of a hundred weight? Ans. .6336285711, &c. cwt. 12. What decimal part of a year is 3w. 4?d. 55 hours, reckoning 363 days 6 hours a Year ? Ans. .074720511065 yr. 13. Reduce 2.15 shillings to the decimal of a pound ? Ans. 1 .1225. 14. Reduce 1.074 roods to the decimal of an Acre ? Ms, .2685 Acre. 13. Reduce 17.69 yards to the decimal of a mile ? Ms. .010031136 m. CHAPTER m. Addition of Decimals. ADDITION of decimals is performed the same way as Addition of whole numbers, only yon must observe to place your numbers right, that is all the decimal points under each other, units under units, tenths un- der tenths, &c. Example. Let 317.25; 17.125; 275.5; 47.3579: and 12.75 ; be added together into one sum. 317.25 17.125 275.5 47.3079 12.75 Sum 669.9825 Subtraction of Decimals. 11 PRACTICAL EXAMPLES. 2. Add 5.714; 3.456; .543; 17.4957 together. Sum 27.2087. 3. Add 3.754; 47.5; .00857; and 37.5 together. Sum S8.7G257. 1. Add 54.34; .375; 14.795; and 1.5 together. Sum 71.01, CHAPTER IV. SuB'TRACflON Of DECIMALS. SUBTRACTION of decimals is likewise perform- ed the same way as in whole numbers, respect being had (as in addition) to the right placing of the num- bers, as in the following examples. (1) (2) From 212.0137 From 201.125 Subtr. 31.1275 Subtr. 5.57846 Rests 180.8862 Rests 195.54634 Note. If the number of places in the decimals be more in that which is to be subtracted, than in that which you subtract from, you must suppose cyphers to make up the number of places as in the second Ex- ample. PRACTICAL EXAMPLES. 3. Required the difference between 57.49 and 5.768 ? Jins. 51.722, 12 Multiplication of Decimals. 4. "What is the difference between .3054 and 3.051 ? Jlns. i.risG. 5. Required (lie difference between 1745.3 and 173.45? 4n$. 1.371.85. 6. What is (he difference between seven tenths of an unit, and fifty-four ten thousand parts of an unit? Jlns. .6940. 7. What is the difference between .105 and 1.00075? Jlns. .89575. 8. What is the difference between 150.43 and 755.355? Jlns. 604.925. 9. From 1754.754 take 375,494478 ? Jlns. 1379.259522. 10. Required the difference between 17.541 and 35.49 ? Jlns. 17.949. CHAPTER V. Multiplication of Decimals. MULTIPLICATION of decimals is also performed the same way as Multiplication of whole numbers ; but to know the value of the product, observe this Rule : Cut off, or separate by a comma or point, so many decimal places in the product, as there are places of decimals in both factors, viz. both in the multiplicand and multiplier. Example 1. Let 3.125 be multiplied by 2.75. Multiply the numbers together, as if they were whole numbers, and the product is 8.59375 : And because there are three places of decimals pointed off in the multiplicand, and two places in the multiplier, there- fore you must point off five places of decimals in the product, as you may see by the work. Multijilication of Decimals * Multiplicand 3.125 Multiplier 2.75 15625 21875 6250 Product 8.59375 Example 2. Let 79.25 be multiplied by .439. In this example, because two places of decimals are pointed oft' in the multiplicand, and three in the mul- tiplier, therefore there must be five pointed off in the- product. Multiplicand 79.25 Multiplier .459 71325 39625 31700 Product 36.37375 Example 3. Let 1.35272 be multiplied by .00423. In this example, because in the multiplicand arc six decimal places and in the multiplier live places ; therefore in the product there must be eleven places of decimals $ but when the multiplication is finished, the product is but 57490600 viz. only eight places ; therefore, in this case, you must put three cyphers be- fore the product figures, to make up the number of eleven places: So the true product will be .00057490600. Multiplicand .135272 Multiplier .00425 676360 270544 541088 Product .00057490600 U tracted Multiplicati PRACTICAL K: i. Multiply .001 172 by .1013. Product .0001538240, -». Multiply 4)17533 by 3*7. Product 6.083604. 6. Multiply 2T9.23 by .445. Product I2t.j> Multiply 32.07.32 by .032 0. Product 1.0.1211 MM). 8. Multiply 1. 1 13 bv 15.98. Product 70.99914. Multiply 20.0291 by 35.45. Product 710.031595. 10. Multiply 7.3564 by .0126. Product .09269064. 11. Multiply .75432 by .0356. Product .026853792. 12. Multiply .004735 by .0374. Product .0001770890. Contracted Multip lic j r jos of Decimals. Because in multiplication of decimal parts, and mix- ed numbers, there is no need to express all the figures of the product, but in most cases two, three or four places of decimals will be sufficient; therefore, to con- tract the work, observe the following Write the unit's place of the multiplier under that place of the multiplicand, which you intend to keep in the product: then invert the order of all the other figures; that is write them all the contrary way; and in multiplying, begin always at that figure in the mul- tiplicand which stands over the iigure you are then multiplying withal, and set down the first figure of each particular product directly one under the other: But take care to increase the first figure of every Hue Contracted Multiplication. 15 of the product, with what Mould arise by carrying 1 from 5 to 15 ; 2 from 15 to 25 ; 3 from 25 to 35, &c. from the product of the two figures (in the multipli- cand) on the right-hand of the multiplying figure. Example 1. Let 2.38645 be multiplied by S.2175, and let there be only four places retained in the de- cimals of the product. First, according to the directions, write down the multiplicand, and under it write the multiplier, thus : place the 8 (being the unit's place of the multiplier) under 4, the fourth place of decimals in the multipli- cand, and write the rest of the figures quite contrary to the usual way, as in the following work : Then begin to multiply, first the 5 which is left out, only with regard to the increase, which must be carried from it i saying, 8 times 5 is 40 ; carry four in your mind, and say, 8 times 4 is 32, and 4 I carry, is 36 ; set down 6, and carry 3, and proceed through the rest of the figures as in common multiplication. Then begin to multiply with 2; saying, 2 times 5 is 10, nought and carry 1; 2 times 4 is 8, and 1 is 9, for which I carry 1, because it is above 5; then 2 times 6 is 12, and 1 that I carry is 13; set down 3 and carry 1 ; and proceed through the rest of the figures as in common multiplication. Then multiply with 1 : saying, once 6 is 6, for which I carry 1, and say, onee 8 is 8, and 1 is 9 ; set down 9, and proceed as in com- mon multiplication. Then multiply with 7: saying, 7 times 6 is 42, 2 and carry 4; 7 times S is 56, and 4 is 60, nought and carry 6 ; 7 times 3 is 21, and 6 is 27 ; set down 7 and carry 2, and proceed. Lastly, multi- ply with 5 : saying, 5 times 8 is 40, nought and carry 4; 5 times 3 is 15, and 4 is 19 ; for which carry 2, and say, 5 times 2 is 10, and 2 that I carry is 12: which set down, and add all the products together, and the total will be 19.6107. — See the work. 16 Contracted Multiplication. Contracted. Common. 9«#8 8. 2 173 100916 11 167 0513 615 167 90 12 190916 1 9.6 107 19.6106 | 52873 I kave here set down the work of the last example, w rooght by the common way, by which you may see the reason of the contracted way, all the figures on •he right-hand of the line being wholly omitted. Example 3. Let 375.13758 be multiplied by 16.7324, so that the product may have but four places of de vinials. 375.13753 the -Multiplicand. 4237.61 the Multiplier reversed. 513758 the product with 1 22508255 the prod, with 6 increased with 6X8 5963 the prod, with 7 increased with 7x^X3 112511 the prod, with 8 increased with 3X5X7 7503 the prod, with 2 increased with 2X7X3 1500 the prod, with 4 increased with o 6276.9520 the product required. PRACTICAL EXAMPLES. Example 3. Multiply 395.3756 by .75642. Ans. 299.0699. Division of Decimals. 17 4. Let 54.7494367 be multiplied by 4.724733 reserv- ing only five places of decimals in the product. Jlns. 258.67758. 5. Multiply 475.710564 by .3416494 and retain three decimals in the product. Jlns. 162.525. 6. Let 4745.679 be multiplied by 751.4519, and re- serve only the integers, or whole numbers, in the pro- duct. Ms. 3566163; CHAPTER VI. Division of Decimals. DIVISION of decimals is performed in the same manner as division of whole numbers: to know the value or denomination of the quotient, is the only dif- ficulty ; for the resolving of which, observe either of the following RULES. I. The first figure in the quotient must be of the same denomination with that figure in the dividend which stands (or is supposed to stand) over the unit's place of the product of the first quotient figure by the divisor. II. When the work of division is ended, count how' many places of decimal parts there are in the divi- dend more than in the divisor; for that excess is the number of places which must be separated in the quo- tient for decimals. But if there be not so many figures in the quotient as there are in the said excess, that deficiency must be supplied, by placing cyphers before the significant figures, towards the left-hand, with a point before them; and thus you will plainly discover the value of the quotient. c 2 th Division of Decimals. 'fliese following directions ought also to be carefulTtf observed. If the divisor consist of more places than the divi- dend, there must be a competent number of cyphers annexed to the dividend, to make it consist of as many (at least) or more places of decimals than the divisor 3 for the cyphers added must he reckoned as decimals. Consider whether there be as many decimal parts in the dividend as there are in the divisor; if there be not, make them so many, or more, by annexing. cyphers. In dividing whole or mixed numbers, if there he a remainder, you may bring down more cyphers; and, by continuing your division, carry the quotient to as many places of decimals as you please. Example 1. Let 48 be divided by 144. Tn this example the divisor 144- is greater than the dividend 48; therefore, according to the directions above, I annex a competent number of cyphers (viz. (bur,) with a point before them, and divide in lire i*ual way, i44)4S.0000(.333:'i 432 430 lint, first, in seeking hoir often 144 in 48.d (t]\Q first three figures of the dividend,) I find the unit's place of the product of the first quotient figure by ther Division of Decimate. $9 divisor to fall under the first place of decimals ; there- fore the first figure in the quotient is in the first place of decimals : Or, by the second rule, there being four places of decimals in the dividend, and none in the divisor ; so the excess of decimal places in the divi- dend, above that in the divisor, is four ; so that when the division is ended, there must be four places of de- cimals in the quotient. Example 2. Let 217.75 be divided by 65. First, in seeking how often 65 in 217 (the first three figures of the dividend) I find the unit's place of the product of the first quotient figure by the divisor to fall under the unit's place of the dividend ; therefore the first figure in the quotient will be units, and all the rest decimals: Or, by the second rule,, there being two places of decimals in the dividend, and no deci- mals in the divisor, therefore the excess of decimal places in the dividend, above the divisor, is two ; so when the division is ended, separate two places in the quotient, towards the right-hand by a point, 65)217.75(3.35 195. . 227 225 Example 3. Let 2F6.15975 be divided by 13..2& 13.25)267.15975(20.163 2650 2159 8347 3975 Igj Div'mon of Dcc.lv. In this third example the unit's place of the prd- duet of the first quotient figure by the divisor falls Ullder 6, the ten's place of the dividend; therefore, (by the first rule) the first figure in the quotient is tcus: Or, by the second rule, the excess of decimal places in the dividend, above the divisor, is three ; there being five places of decimals in the dividend, and but two in the divisor, so there must be three pla- ces of decimals in the quotient. Example 4. Let 13.673139 be divided by sY8t89\ ;;;.i.s9)i. , 5.r,r.jt59(.04ir "130355 03935 263669 346 In this fourth example, the unit's place of the pro- duct of the first quotient figure by the divisor, falls under 7, the second place of decimals in the dividend; therefore (by the first rule) the first figure in the quo- tient is in the second place of decimals ; so that you must put a cypher before the first figure in the quo- tient ; and by the second rule, the excess of decimal places in the divisor is 4; for the decimal places in the dividend are 6, and the number of places in the divisor but two ; therefore there must be four places of decimals in the quotient : But the division being finished after the common way, the figures in the quo- tient are but three, therefore you must put the cypher hefore the significant figures. Division of Decimals. 21 Example 5. Let 72.1564 be divided by .134?, .134:7)72.1564(535.68 6735 , . 4806 '654 9190 410S0 304. In this example, the divisor being a decimal, the last figure of the product of the first quotient figure by the divisor falls under the ten's place in the divi- dend, therefore the units (if there had been any) would fall under the hundreds place in the dividend, and so the first figure in the quotient is hundreds. And by the second rule, there being four places of decimals in the dividend, as many in the divisor, so the excess is nothing; but in dividing I put two cy- phers to the remainders, and continue the division to two places further ; so I have two places of deci^ mals. Example 6. het .125 be divided by .0457:. ,0457).1250000(2.735 0914. . . 3560 1610 2390 Ifl Division of Decimals. In this example, the unit's place of the product of the tirst <| not ii* nt figure by the divisor (if there had any) Would fall under the unit's place of the dividend : therefore the first figure of the quotient IB units. And, by the second rule, their being seven pla- ces of decimals in the dividend, and but lour places in the divisor, so the excess is three ; therefore there must be three places of decimals in the quotient. PRACTICAL EXAMPLES. 7. Divide .0000059791 by .00436. Quotient .00131 8. Divide an unit by 282, or, in other words, find the reciprocal of 282. Quotient .0033-101 9. Divide A by .323. Quotient 1.2307 10. Divide 493 by .012. Quotient 11783.71 11. Divide .475321 by 97.453. Quotient 0048774 12. Divide 17.543273 by 123.7. Quotient .13930 13. Divide 143734.35 by .7493. Quotient 191851.528 14. Divide 16 by 960. Quotient .01666, &c. 13. Divide 12 by 1728 Quotient .006944, Sec. 16. Divide 47.5493 by 34.75 Quotient 1.36832517 17. Divide 70.3571 by .00573. Quotient 12976.3062 18. Divide .3754 by 73.714. Quotient .004958131 Division of Decimals contracted. In division of decimals the common way, when the divisor has many figures, and it is required to con- tinue the division till the value of the remainder be but small, the operation will sometimes he long and tedious, but may be contracted by the following method. Division of Decimals. 23 THE RULE. "By the first rule of this chapter (page 17,) find v hat is the value of the first figure in the quotient : then by knowing the first figure's denomination, you may have as many or as few places of decimals as you please, by taking as many of the left hand figures of the divisor as you think convenient for the first di- visor; and then take as many figures of the dividend as will answer them ; and, in dividing, omit one figure of the divisor at each following operation ; observing to carry for the increase of the figures omitted, as in contracted multiplication. Note. When there are not so many figures in the divisor as are required to be in the quotient, begin the division with all the figures, as usual, and continue dividing till the number of figures in the divisor is equal to the number of figures remaining to be found in the quotient ; after which use the contraction. Example 1. Let 721.17562 be divided by 2.257432; Sclet there be three places of decimals in the quotient. Contracted. Common. 25743)721.175—62(319.467 677229 43946 22574 21372 20317 1055 903 152 135 17 10 2.25743)721.17562(319.467 077229 •* 43946 6 22574 3 21372 3 3 20316 87 1055 450 902 972 152 4780 135 4458 17 03220 15 80201 1 23019 |fl Contracted Divisit Tn this example, the unit's place of the product of the first quotient figure by the divisor falls under the hundred's place in the dividend, and it is required, that three places of decimals he in the quotient, so there must he six places in all ; that is, three places of whole numbers, and three places of decimals. Then, because 1 can have the divisor in the first six figures of the dividend, I cut oft* the 62 with a dash of the pen, as useless; then I seek how often the divi- sor is in the dividend, and the answer is three times ; put three in the quotient, and multiply and subtract as in common division, and the remainder is 43940. Then point oft' three in the divisor, and seek how oft- en the remaining figures may be had in 43946, the re- mainder, which can be but once ; put 1 in the quotient, and multiply and subtract, and the next remainder is 21373. Then point o IT' the 4 in the divisor, and seek how often the remaining figures may be had in 21372, which will be 9 times ; put 9 in the quotient ; multi- ply as in contracted multiplication, and thus proceed till the division is finished. I have set down the w ork of this example at large, according to the common way, that thereby the learn- er may see the reason of the rule ; all the figures on the right -hand side of the perpendicular line being whol- ly omitted. PRACTICAL EXAMPLES. 2. Let 5171.59165 be divided hy 8.758615, and let it he required, that four places of decimals he pointed oft' in the quotient. Jlns. 590.4577 3. Let 25.1367 be divided by 217.3543, and let there be five places of decimals in the quotient. Jins. .11564. 1. Divide 7414.76717 by 2.756756, and let there be five places of decimals in the quotient. Ms. 2689.6flJS Extraction of the Square Root. 25 5. Divide 514.75498 by 12.34254, and let there be six places of decimals in the quotient. Jliis. 41.705757. 6. Divide 47194.379457 by 14.73495. and let the quotient contain as many decimal places as there will be integers, or whole numbers, in it. JillS. 3202.88G9. CAA CERVIX Extraction of the Square Root. If a square number be given ; J O find the Root thereof, that is, to find out such a number, as being multiplied into itself, the product shall be equal to the number given ; such operation is called, The Extraction of the Square Root ; which to do, observe the following directions. 1st, You must point your given number; that is, make a point over the unit's place, another over the hundred's, and so over every second figure throughout. 2dly. Then seek tlie greatest square number in the first period towards the left hand, placing the square number under that point, ami the root thereof in the quotient, and subtract the said square number from the first point, and to the remainder bring down the next point, and call that the resolvend. 3dly, Then double the quotient, and place it for a di- visor on the left hand of the resolvend ; and seek how often the divisor is contained in the resolvend (re- serving always the unit's place) and put the answer in the quotient, and also on the right hand side of the divisor ; then multiply by the figure last put in the quotient, and subtract the product from the resolvend (as in common division) and bring down the next point D Extraction of the Square Hoot. to the remainder (if there he any more) and proceed as before. Ji Table of Squares and their Hoots. .t 1 1 3 8 1 * ■ 1 • j 8 | U | Square | 1 * 1 » 16 23 | 36 | ♦9 | fit | SI Example 1. Let 4489 be a number given) and let the square root thereof be required. 4489(67 36 127)889 Resolveml. S89 Product. First, point the given number, as before directed^ then by the little table foregoing, seek the greatest square number in 44 (the first point to the left-hand) which you will find to be 36, and 6 the root ; put 36 under 44, and 6 in the quotient, and subtract 36 from 44, and there remains 8. Then to that S bring down the other point 89, placing it on the right-hand, so it makes 889 for a resolvond ; then double the quotient 6, and it makes 12 ; which place on the left-hand for a divisor, and seek how often 12 in 88 (reserving the unit's plaee) the answer is 7 times ; which put in the quotient, and also on the right-hand side of the divisor, and multiply 127 by 7, as in common division, and the product is S89, which subtracted from the resolvend, there remains nothing ; so is your work finished ; and the square root of 4489 is 67 ; which root if you multi- ply by itself, that is 67 by 67, the product will be 4489, equal to the given square number, and proves the Extraction of the Square Root. 27 work to be right. Had there been any remainder it must have been added to the square of the root found. Example 2. Let 106929 be a number given, and M the square root thereof be required. 108929(327 62)169 Resolvend, . 124 Product. 647)4529 Resolvend. 4529 ProdueU First, point your given number, as before directed, putting a point over the units, hundreds, and tens of thousands ; then seek what is the greatest square num- ber in 10 (the first point) which by the little iable you will find to be 9, and 3 the root thereof; put 9 under 10, and 3 in the quotient ; then subtract 9 out of 10, and there remains 1 ; to which bring down 69, the next point, and it makes 169 for the resolvend; then double the quotient 3, and it makes 6, which place on the left-hand of the resolvend for a divisor, and seek how often 6 in 16 ; the answer is twice : put 2 in the quotient, and also on the right-hand of the divisor making it 62. Then multiply 62 by the 2 you put in the quotient, and the product is 124 ; which subtract from the resolvend, and there remains 45 ; to which bring down 29, the next point, and it makes 4529 for a new resolvend. Then double the quotient 32, and it makes 64, which place on the left side of the resolv- end for the divisor, and seek how often 64 in 452, which you will find 7 times : put 7 in the quotient, and also on the right-hand of the divisor, making it 647, which multiplied by the 7 in the quotient, it makes 28 Extraction of the Square Boot. 4529, which subtracted from the resolrend, there re- mains nothing. 80 327 is the square root of the giv-. en number. Note. The root will always contain just so many figures, as there are points over the given number to J>e extract! (1 : And these figures will he whole num- bers or decimals respectively? according as the points stand over whole numbers or decimals. — The method of extracting the square root of a decimal is exactly the same as in the foregoing examples, only if the number of decimals be odd, annex a cipher to the right hand to make them even, before you begin to point. The root may be continued to any number of figures you please, by annexing two cyphers at a time to each remainder, for a new resolvend. PRACTICAL EXAMPLES. 3. It is required to extract the square root of 2268741. Jns. 1506.23. Hem. 121871. t. What is the square root of 7596796 ? J$tl8. 2756.228. Rem. 3212016.. 5. What is the square root of 751427.5745 ? Jlns. 866.84. Rem. 59889. 6. Extract the square root of 656714.37512. Jlns. 810.379. Rem. 251479. 7. What is the square root of 15241578750190521 ? Jlns. 123456789. 8. What is the square root of 75.347 ? Jlns. 8.6802649729. Rem. 24536226559. 9. What is the square root of .4325 ? Jlns. .65764. Rem. 96304. To extract the Square Root of a Vulgar Fraction. RULE. 1. Reduce the given fraction to its lowest terms, if Extraction of the Square Root. 29 it be not in its lowest terms already ; then extract the square root of the numerator for a new numerator, and the square root of the denominator for a new de- nominator. 2. If the fraction will not extract even, reduce it to a decimal, and then extract the square root. 3. When the number to be extracted is a mixed fraction, reduce the fractional part to a decimal, and annex it to the whole number, then extract the square root. Example 1. Extract the square root of £££ First, |$£ is equal to •§§ in its lowest terms, the square root of 25 is 5, and the square root of 36 is 6 5 therefore £ is the root required. Example 2. Let seven-eighths be a vulgar fraction . wliose square root is required. 8)7.000 61 (.87500000(.935-l 81 183)650 519 1S65)10100 9325 1ST 01)77500 71S16 >6S1 Reduce this | to a decimal, it makes .875 ; (0 which annex cyphers, and extract the square root, as if it was a whole number. So the root is .9351. d2 30 Extraction of ike Square Root. Example 3. Let -^ be a vulgar fraction, whose square root is required. )3.000000 288 (.00312500(.0559 Root. 120 90 103)625 525 240 192 ' 1109)10000 9981 480 480 19 Example 4. What is the square root of 15 j ? Here | reduced to a decimal is .625, which annex ed to the 15 makes 15.625, the square root of which is 3.95284. Hem. 559344. PRACTICAL EXAMPLES. FKACT1CAJL tXAMrLfcb i. What is the square root of |££ ? i. What is the square root of ||£? '. What is the square root of -p f ? Ms. .691S984, &c. ;. What is the square root of 29^\ ? Ms. 5.4. I. What is the square root of |? Ms. 3333. Rem. 8W* Extraction of the Cube Root, CHAPTER VIII. Extraction of the Cube Root. TO extract the cube root, is nothing else but to find such a number, as being first multiplied into itself, and then into that product, produceth the given num- ber; which to perform, observe the following direc- tions. is*, You must point your given number, beginning with the unit's place, and make a point, or dot, over every third figure towards the left-hand. 2dbj, Seek the greatest cube number in the first point, towards the left-hand, putting the root thereof in the quotient, and the said cube number under the first peiut, and subtract it therefrom, and to the re- mainder bring down the next point, and call that the resolvend. 3f%, Triple the quotient, and place it under the re- solvend; the unit's place of this under the ten's place of the resolvend ; and call this the triple quotient. tehly. Square the quotient, and triple the square, and place it under the triple quotient; the units of this under the ten's place of the triple quotient, and call this the triple square. Bthly, Add these two together, in the same order as they stand, and the sum shall be the divisor. Gthly, Seek how often the divisor is contained in the resolvend, rejecting the unit's place of the resol- vend (as in the square root,) and put the answer in the quotient Extraction of the Cube Rout. 7'thly, Cube the figure lust put in the quotient, and put the unit's place thereof under tht unit's plaee of the resolvend. Sthly, Multiply the square of the figure last put in the quotient into the triple quotient, and plaee the product under the last, one place more to the left- hand. Qthly, Multiply the triple square by the figure last put in the quotient, and place it under the last, one place more to the left-hand. \Othly, Add the three last numbers together, in the same order as they stand, and call that the sub- trahend. Lastly, Subtract the subtrahend from the resolvend, and if there be another point, bring it down to the re- mainder, and call that a new resolvend, and proceed in all respects as before. Note. To square a number is to multiply that num- ber by itself. And, To cube a number is to multiply the square of the number by the number itself. » Jl Table of Cubes and their Roots. Roots 1 2 3 | 4 5 6 7 8 | 9 Cubes 1 8 27 | 64 125 216 343 512 j 729 Extraction of the Cube Root. 3$ Example 1. Let 314132 be a cubic number, whose loot is required. 314132(68 Root. 216 98132 Resolvent! . 18 Triple quotient of 6. 108 Triple square of the quotient 0. 1098 Divisor. 512 Cube of 8, the last figure of the root. 1152 The square of 8, by the triple quotient. 864 The triple square of the quotient 6 by 8. 98432 The subtrahend. After you have pointed the given number, seek what is the greatest cube number in 314, the first point, which, by the little table annexed to the rule you will find to be 216, which is the nearest that is less than 314, and its root is 6 ; which put in the quo- tient, and 216 under 314, and subtract it therefrom, and there remains 98 ; to which bring down the next point, 432, and annex it to 98 ; so will it make 98432 for the resolvend. Then triple the quotient 6, it makes 18, which write down the unit's place, 8, under 3, the ten's place of the resolvend. Then square the quo- tient 6, aud triple the square, and it makes 108, which write under the triple quotient, one place towards the left-hand ; then add those two numbers together, and they make 1098 for the divisor. Then seek how often the divisor is contained in the resolvend, (rejecting the unit's place thereof) that is, how often 1098 iu 9843. 8* d ruction of the Cube Hoot. which is S times; put 8 in the quotient, and the cube (hereof below the divisor, the unit's place under the unit's place of the rcsolvend. Then square the 8 last put in the quotient, and multiply 64, the square there- of*, by the triple quotient IS: the product is 1102) set this under the cube of 8, the units of this under the tens of that. Then multiply the triple square of the quotient by 8, the figure last put up in the quotient, the product is 864; set this down under the last pro- duct, a place more to the left-hand. Then draw a line under these three, and add them together, and the sum is 08432, which is called the subtrahend; and be- ing subtracted from the rcsolvend, the remainder is nothing; which shews the number to be a true cubic number, w hose root is 6S ; that is, if 68 be cubed, it will make 314132. For if 68 be multiplied by 68, the product will be 4624; and this product, multiplied again by 68, the last product is 314432, which shews the work to be right. Example 2. Let the cube root of 5735339 be re- quired. After you have pointed the given number, seek what is the greatest cube number in 5, the first point, which, by the little table, you w ill find to be 1 ; which place under 5, and 1, the root thereof, in the quotient; and subtract 1 from 5, and there remains 4; to which bring down the next point, it makes 4735 for the rcsolvend. Then triple the 1, and it makes 3; and the square of 1 is 1, and the triple thereof is 3; which set one under another, in their or.der, and added, makes 33 for the divisor. Seek how often the divisor goes in the resol- vend, and proceed as in the last example. Extraction of the Cube Root, 35 5735339(179 Root. 1 4735 3 The triple of the quotient 1, the first figure. 3 The triple square of the quotient 1. 33 The divisor. 343 The cube of 7, the second figure of the root. 147 The square of 7, multipl. in the triple quot. 3. 21 The triple square of the quot. multiplied by 7. 3913 The subtrahend. 822339 The new resolvend. 51 The triple of the quot. 17, the two first fig. 867 The triple square of the quotient 17. 8721 Divisor. 729 The cube of 9, the last figure of the root. 4131 The squ. of 9, multipl. by the triple quo. 51. 7S03 The triple square of the quotient 867 by 9. 822339 The subtrahend. In this example, 33, the first divisor, seems to be contained more than seven times in 473, the resolvend, after the unit's place has been rejected ; but if you work with 9, or 8, you will find that the subtrahend will be greater than the resolvend. 30 ruction of the t Example 3. Required the cube ro 22069310125(2805 8 1 fc069 Resolveud. G Triple of 3. \z Triple square <>i' 2. 126 Divisor. 512 Cube of 8. Square of s by f>. 96 Triple square by S. 13952 Subtrahend. 117810125 New resolvcnd. 8 l Triple of Triple square of :s. 23604 Divisor. 8-10 Triple of 280. m) Triple square of 2S0. :<> New divisor. In this example 2, being tracted from (lie re- solvend 14009. the remainder is 1 1 T ; to which bring down 810, the 3d. point, and it makes 117810 for a new resolvend; and the next divi- sor is 23604, which you cannot have in the said resolvend (the unit's place being rejected:) so you must put in the quotient, and seek a new divi- sor (after you have brought down your last point to the re- solvend;) which new divisor is 2852640; and you will find it to be contained ."> times. So proceed to finish the rest of the work. l M Tube of five. 21000 Square of 5 by 840. l IT nooo Triple square by 5. £17810125 Subtrahend. Extraction of the Cube Root. Note. The root will always contain just so many figures, as there are points over the given number to be extracted; and these figures will be whole numbers or decimals respectiv.lv, according as the points stand over whole numbers or decimals. The method of extracting the cube root of a mixed number, or deci- mal, is the same as in the above examples; only the number of decimals must be made to consist of three, six or nine, &c. figures, by annexing cyphers. PRACTICAL EXAMPLES. Example 4. What is the cube root of 32461759 ? Ms. 31 9 5. What is the cube root of 84604519 ? Ms. 439 6. What is the cube root of 259697989 ? Ms. 638 7. What is the Cube root of '33917056 ? Ms. 295.9. Rem. 8995931 8. What is the cube root of 93759.57507 ? Ms. 45.42. Rem. 59186982 9. Required the cube root of .401719179. Ms. .737. Rem. 1403626 10. Required the cube root of .0001416 ? Ms. .052. Rem. 992 11. Required the cube root of 122615327232. Ms. 496 S 12. What is the cube root of 705.919947284? Ms. 8.904. Rem. 20. 13. What is the cube root of } «„ Ai ~~<.^r- 1531328.215978518623 ? \ MS ' li5 " « 25 14. The cube root of .57345 is required. Ms. .S308. Rem. 8045888 To extract the Cube Root of a Vulgar Fraction. rule. 1. Reduce the given fraction to its lowest terms, if it be u.o% in it* lowest terms alreadv ; then extract E 3$ Extraction of the Cuhe Hmt. the cube root of the numerator tor a new numerator, ami the cube root of the denominator for a new deno- minator. 2. If the fraetion will not extract even, reduce it to a decimal, and then extract the cube root. 3. When the number to be extracted is a mixed fraction, reduce the fractional part to a decimal, and annex this decimal to the whole number, then extract the cube root. Example 1. What is the cube root of |^| ? First. ^*| is equal to f £ in its lowest terms, the cube root of 27 is 3, and the cube root of 64< is 4; therefore the cube root of -|| is J, the answer. Example 2. Let ? |y be a vulgar fraction, whose cube root is required. By the first rule of Chapter II. reduce the vulgar fraction to a decimal. 276)5.000000000(.018U5912 276 22 tO 1640 2600 1160 060 Extraction of the Cube Moot. .39 .018115942(.262 Root. 8 10 115 Resolvend. 6 Triple of 2. 12 Triple square of 2. 126 Divisor. 216 Cube of 6. 216 Square of 6 by the triple of ,2. 72 Triple square by 6. 9576 Subtrahend. 539942 Resolvend. 78 Triple of 26. 2028 Triple square of 26, 20358 Divisor. S Cube of 2. 312 Square of 2 by 78. j.056 Triple square 2028 by 2. 408728 Subtrahend. 131211 Remainder. You may prove the truth of the work, by cubing the root found, as was shewn in the first example: and if any thing remains, add it to the said cube, and the sum will be the given number, if the work is rightly performed. Multiplication of Feet, %'c. Example 3. What is the cube root of 56623 T V T - ; Here ^ reduced to a decimal is .104, which annex- ed to 36623 makes 36623.104, the cube root of which 8.4. PRACTICAL EXAMPLES. 4. What is the cube root of i£ 5 6 ? Ms. 7163S Bf. Required the cube root of j£te. Ms. 6. What is the cube root of -fifor Ms. .}. 7. What is the cube root of | ? Ms. .82207 5. What is the cube root of |J ? Ms. .98591 9. What is the cube root of 5333|? Jins. .17.471 10. What is the cube root of 108220l|? Ms. loi.or CHAPTER IX. Multiplication of Fee*, Inches, and Part's; ov Duodecimals, THE multiplication of feet and inches is generally called duodecimals, because ever} 7 superior place is i2 limes its liext inferior in this scale of notation. This way of conceiving an unit to be divided, is chiefly in use among artificers, who generally take the linear di- mensions of their work in feet and inches : It is like- wise called cross midtiplication, because the factors are sometimes multiplied crosswise. RULE I. 1. Under the multiplicand write the corresponding denominations of the multiplier; that is, feet under feet, inches under inches, parts under parts, &c. Multiplication of Feet f $c. 41 2. Multiply each term in the multiplicand, begin- ning at the lowest, by the feet in the multiplier ; write each result under its respective term, observing to carry an unit for every 12, from each lower denomi- nation to its next superior. 3. In the same manner multiply every term in the multiplicand by the inches in the multiplier, and set the result of each term one place removed to the right- hand of those in the multiplicand. 4. AVork in a similar manner with the parts in the multiplier, setting the result of each term removed two places to the right-hand of those in the multipli- cand. Proceed in like manner with the rest of the denominations, and their sum will be the answer re- quired. Examfle 1. Let 7 feet 9 inches be multiplied by 3 feet 6 inches. F. I. Multiplicand 7 •• 9 ' Multiplier 3 -6 23-3 Fts. 3 •• 10 •• 6 Product 27 -. 1 .. 6 First, Multiply 9 inehes by 3, saying, 3 times 9 is 27 inches, which make 2 feet 3 inches ; set down 3 under inches, and carry 2 to the feet, saying, 3 times 7 is 21, and 2 that I carry make 23; set down 23 un- der the feet. Then begin with 6 inches, saying, 6 times 9 is 54 parts, which is 4 inches and 6 parts ; set down 6 parts, and carry 4, saying, 6 times 7 is 42, and 4 that I carry- is 46 inches, which is 3 feet 10 inches; which set down, and add all up together, and the product is 27 feet 1 inch 6 parts. E 2 Multiplication of Feet. • Example 2. Let r feet inches 9 parts be multi- plied b) 3 leet 5 inches 3 pari-. F. Multiplicand 7 .. Multiplier 3 .. I. 5 5 P. .. 9 > 3 .. 5 1 1 ..3 V .. 9 T. .. 10 .. 5 .. .: Product 26 .. 8 .. .. 2 .. 3 In this example, I first begin with 3 feet, and there- by multiply 7 feet 5 inches and 9 parts : First, I say, 3 times 9 is 27 parts, that is 2 inches and 3 parts 5 set down 3 under the parts, and carry 2, saying, 3 times 5 is 15, and 2 I carry is 17, that is, 1 foot 5 inches ; set down 5 inches, and carry 1, and say, 3 times 7 is 21, and 1, 1 carry is 22; set down 22 feet: Then begin with 5 inches, saying, 5 times 9 is 45, which is 45 seconds, which makes 3 parts and 9 se- conds ; set down 9 seconds a place towards the right- hand, and carry 3 parts, saying, 5 times 5 is 25, and 3 I carry is 28, which is 2 inches and 4 parts ; set •down 4 parts and carry 2, saymg 5 times 7 is 35, and 2, I carry is 37, whieh h 3 feet 1 inch; set down 3 feet 1 inch ; and begin to multiply by 3 parts, say- ing, 3 times 9 is 27 thirds, that is, 2 seconds and 3 thirds ; set down 3 thirds, and carry 2, saying 3 times 5 is 15, and 2, 1 carry is 17, that is, 1 part and 5 se- conds -, set down 5 seconds, and carry 1, saying 3 times 7 is 21, and 1, 1 carry is 22, whieh is 1 inch and 40 parts, which set down, and add all up, and the prx>~ duct is 25 feet 8 jnehes 6 parts 2 seconds 3 thirds. When h •Multiplication of Feet, $c. 43 RULE II. 'ten the Feet in the Multiplicand are expressed by a large number. Multiply first by the feet in the multiplier, as be- fore. Then, instead of multiplying by the inches and parts, &c. proceed as in the Rule of Practice, by tak- ing such aliquot parts of the multiplicand as corres- pond with the inches and parts, &e. of the multiplier. Then the sum of them all will be the product required. Example 3. Let 75 feet 7 inches be multiplied by 9 feet 8 inches. In. F. I. $ 75 .. 7 Multiplicand J 9 .. 8 Multiplier. 680 .. 3 25 .. 2. .. 4 25 .. 2 .. 4 I 'roduet 730 .. 7 . . S Multiply by 9 feet, first, as above directed ; ihea instead of multiplying by the 8 inches, let them be di- vided into aliquot parts of a foot, as 4 and 4, because 4 is the third part of 12. So, if you take the third part of 75 feet 7 inches, and set it down twice, and add all together, the sum will be 730 feet 7 inches 8 parts. To take the third part, say, how often 3 in 7, which is twice ; set down 2 ; then because twice 3 is 6, say, 6 out of 7, and there remains 1, for which you laiust add 10 to the 5, and it makes 13 ; then the threes in 15 are 5 times: set down 5: and, because three times 5 is 15, there is remains. Then go to the 7 inches, saying, the three in 7 are twice : set down '2 41 Multiplication of Fee in the inches ; and because <\\ lake fi out of 7, arid there remains l inch, which is 12 parte; then threes in 1 Jure Mimes, and remains. So the third part of 75 feet 7 inches is $0 feel 2 incites 4 parts; which set twice over, and add them together as in the example. Example l. Let 37 feet 7 inches 5 parts he multi- plied by 4 feet S inches 6 parts. I. P. I .. 0) 1 •• 05 I. p. .. 6 F. I. P. 37 .. 7 - 13 Multiplicand 4-8 - 6 Multiplier. 150 - 5 - 8 S. 12 •• 6 - 5 - 8 12 » 6 - 5 - 8 T. 1-6 - 9 ..8-6 Product 177 In this example I first multiply by 4 feet as usual.— Then for the 8 inches I say 4 inches is the third of a foot, therefore I take the third part of 37 feet 7 inches 5 parts, which is 12 feet 6 inches 5 parts 8 seconds, and set it down twice. Then for 6 parts, I say, 6 parts are the eighth of 4 inches, because 12 parts make 1 inch ; hence it follows, that whatever be the value, or product, by 4 inches, the value of 6 parts will be one- eighth thereof; therefore I take one-eighth of 12 feet 6 inches 5 parts 8 seconds, and find it to be 1 foot 6 inches 9 parts 8 seconds 6 thirds ; so that the sum of the whole is 177 feet 1 inch 5 parts 6 thirds. Multiplication of Feet, $c* 46* RULE III. When the Feet both in the Multiplicand and Multiplier are large numbers. Multiply the feet only into each other: Then, for the inches and parts in the multiplier, take parts of the feet, inches, &c. of the multiplicand : And, for the inches and parts of the multiplicand, take parts of the feet o?Uy in the multiplier. The sum of all will be the product. Example 5. Let 75 feet 9 inches be multiplied by 17 feet 7 inches. I. p. I. i T 75 17 .» 9 •• 7 525 75. P. i 1 37 •• 10 •• 6 6 -. 3 •• 9 12 .. 9 •• •od net 1331 .. 11 .. 3 I. F. 6 l 17 • T 3 l 8 • • 6 4 . • 3 12 In this example, because there are more than 12 feet in the multiplier, I first multiply the 75 feet by 17 feet. Then, I say, 6 inches are the half of a foot, and take the half of 75 feet 9 inches, which is 37 feet 10 inches 6 parts ; but I ought to take the parts for 7 inches, therefore I say 1 inch is the sixth of 6 inches, and take the sixth part of 37 feet 10 inches 6 parts, which I find to be 6 feet 3 inches 9 parts. Then, be- cause there are 9 inches in the multiplicand, I take parts with them out of the 17 feet in the multipli- er, saying, 6 inches are the half of a foot, I there- Multiplication of 7< ke the half of 17 feet* which is 8 fee$ 6 inches; inches left, and whatever the : by 6 inches may be, that by 3 inches must half thereof \ I inches are the lialt; and the half of 8 feet 6 inches, which is 4 feci 3 in- li<^. the sum of these is 12 feet 9 inches, which 1 place under the former parts, and the sum of the whoii feet 11 inches 3 parts. . 6. Let 3JI feet l inches 7 parts be mul tiplied by 36 feet 7 inches 5 parts. F. 36 1..G 0..3 13-. 9 In. P. 6 •• i I 1-. 0.-4, CI ~5 l T i T F. I. P. 311 ..4-7 36-" 1866 933 . I. P. 8. 150-. 8 •• \ •• 6 25 .11 •• 4 •• 7 T 8- 7 - 9 •• 6 .4 2» 1..11 .. 4 • .7 13- 9 I. P. 4..0 T 0-6 0-1 i Product 11402 •• 2 « 4 •• 11 •• 11 In this example I first multiply the feet as in • - pic 5th. Then I say 6 inches are the half of a foot, and take the half of 311 feet 4 inches 7 parts, which 1 find to be 155 feet 8 inches 3 parts 6 seconds; then, as 1 inch is one-sixth of 6 inches. I therefore take pne sixth of 155 Feet 8 inches 3 parts 6 seconds, which is 25 i'eet 1 J. inches 4 pai ads : Then, because 4 parts are one third of an inch, I take one third of 25 I inches 4 parts 7 seconds, and find it to be 8 feet 7 inches 9 parts 6 seconds 4 thirds ; and as I have one ft, I say 1 is the fourth of 4, and take the fourth of s feet 7 inches 9 parts 6 Seconds i thirds, which is 2 feet 1 inch 11 parts 4 seconds 7 thirds. — Then for Multiplication of Feat, $c. 47 the four inches in the multiplicand, I take a third part of 36, the feet in the multiplier ; because 4 inches are one-third of a foot ; and 6 parts are the eighth of 4 inehes, 1 therefore take the eighth of 12 feet which is I foot 6 inches ; then I have 1 part left, which is the sixth of 6 parts, so I take the sixth of 1 foot 6 inches, which is 3 inches. The sum of these parts is 13 feet 9 inches, which I place under the former parts, and add them together, so that the whole is 11402 feet 2 inches 4 parts 11 seconds 11 thirds. PRACTICAL EXAMPLES. 7. Let 97 feet 8 inches he multiplied hy 8 feet 9 inches. Jlns. 854 feet 7 inches. 8. Let 87 feet 5 inehes he multiplied by 35 feet 8 inches. Jlns. 3117 feet 10 inches 4 parts. 9. Let 259 feet 2 inches be multiplied by 48 feet II inches. rfns. 12677 feet 6 inches 10 parts, 10 Multiply 179 feet 3 inches by 38 feet 10 inehes. •ins. 6960 feet 10 inches 6 parts. 11. Multiply 246 feet 7 inches by 46 feet 4 inches. Jlns. 11425 feet inches 4 parts. 12. Multiply 246 feet 7 inches by 36 feet 9 inches. Jlns. 9061 feet 11 inches 3 parts. 13. Multiply 257 feet 9 inches by 3W feet 11 inch- es. Jlns. 10288 feet 6 inches 3 parts. 14. Let 8 feet 4 inches 3 parts 5 seconds 6 thirds he multiplied by 3 feet 3 inches 7 parts 8 seconds 2 thirds. Jins. 27 feet 7 inches 3 parts 5 seconds 1 third 8 fourths 8 fifths 11 sixths. 15. Multiply 321 feet 7 inehes 3 parts by 9 feet 3 inches 6 parts. Jlns. 2&SS feet 2 inches 10 pads 4 seconds 6 thirds, 16 Multiply 42 feet 7 inches 8 parts by 7 feet 3 inches 6 parts. Jlns. 310 feet 10 inches 10 parts 10 second?* Of C Scale, $c. Mult iply l„M fool 7 inches 9 parts by 14 feet fi iiielit Ji 3 pa I feet l inch 1 part seconds 6 thirds. Muhiph to inches 8 parts by 18 feet • indie* i pert*. 99 i\ el inches parts 10 seconds 8 thirds. 19. Multiply 207 feet 7 inches 10 parts by 23 feet 9 inches 7 parts. - Jus. 60(1.3 fefet inches 5 parts seconds 10 thirds. 20 Multiply 317 feet inches 7 parts by 37 I inches | parts. Jn<. liuio feet 9 inches 11 parts 1 second 3 thirds. CHAPTER X. nation of the line of numbers on Gusrsn^s Scale, and tlte construction and use of the Common Diagonal Scale. THE line of numbers on the two feet Gunter's Scale, marked Number, is numbered from th» left-hand of the Scale towards the right, with the figures 1, 2, 3, 4, 5, 6, 7, 8, 9 ; 1, which stands exactly in the middle of the Scale : the numbers then go on 2, 3, 4, 5, 6, 7, 8, 9 ; 10, which stands at the right-hand end of the Scale. These two equal parts of the Scale are also equally divided, the distance between the first, or left-hand 1, and the first 2, 3, 4. &,e. is exactly equal to the dis- tance between the middle I, and the numbers 2, 3, 4, &.C which follow it Of Gunter' > s Scale, $c. 4S The subdivisons of these two equal parts of the scale are likewise similar, viz. they are each one tenth of the primary divisions, and are distinguished by lines of about half the length of the primary di- visions. These subdivisions are again divided into ten parts, where room will admit, and where that is not the case, the units must be estimated, or guessed at by the eye, which is easily done by a little practice. The primary divisions on the second part of the scale are estimated according to the value set upon the unit on the left-hand of the scale : Thus, if you call the unit on the left-hand of the scale 1, then the first 1, 2, 3, 4, &c. stand for 1, 2, 3, 4, &c. the middle 1 is 10; and the 2, 3, 4, &c. following, stand for 20, 30, 40, &c. and the 10 at the right-hand is 100. — If you call the unit on the left-hand of the scale 10, then the first 1,2, 3, 4, &c. stand for 10, 20, 30, 40, &c„ the middle 1 will be 100 ; and the 2, 3, 4, &e. follow- ing will be 200, 300, 400, &c. and the 10 at the right- hand will be 1000 — If you call the unit on the left- hand of the scale, 100, then the first 1, 2, 3, 4, &e. will stand for 100, 200, 300, 400, &c. the middle 1 will be 1000 ; and the 2, 3, 4, &e. following, 2000, 3000, 4000, &c. and the 10 at the right-hand end will be 10000. — Lastly, if you consider the unit on the left- hand of the scale as one-tenth of an unit, then the first 1, 2, 3, 4, &c. will be ^, ^ T %, T 4 ^, &e. the mid- dle 1 will stand for an unit, and the 2, 3, 4, &c. fol- lowing it will be 2, 3, 4, &c. and the 10 at the right- hand end of the scale will stand for 10. From the above description it will be easy to find the divisions representing any given number. Sup- pose 12 was required ; take the division at the figure 1 , in the middle of the scale, for the first figure of 12 ; then for the second figure, count two of the longer strokes to the right-hand, and this last is the point representing 12, where there is a brass pin. — If 34 were required : Call the figure 3 on the right -hand / of a . and count forward four of the rUions towards the right-hand: if 340 were it must be found in the same manner, if the point i ng 345 were required, lind 3 to as then the middle distance between the point of «h>. and the point representing 330, will be the point 1>\ the line of numbers and a pair of compasses al- 'I the problems in mensuration may be readily for they in general depend upon proportion . Ami, as in natural numbers, the quotient of the first term, of any abstract proportion, by the second, is equal to the quotient of the third term by the fourth ; so in logarithms (for the line of numbers is a loga- rithmical line) the difference between the first and se- cond term, is equal to the difference between the third and fourth : consequently, on the line of numbers, the distance between the first and second term, will be c- qual to the distance between the third and fourth. — And for a similar reason, because four proportional quantities are alternately proportional, the distance between the first and third term will be equal to the distance between the second and fourth. Hence the following GENERAL RULE. The extent of the compasses from the first term to the second, will reach, in the same direction, from the third to the fourth : Or, the extent of the compasses from the first term to the third, will reach in the same direction, from the second to the fourth. By the same direction must be understood, that if the second term lie on the right-hand of the first, the fourth term will lie on the right-hand of the third, and the contrary. Hence, 1. To find the product of two Niiinbers. As an unit is to the multiplier, so is the multipli- cand to the product. Of the Diagonal Scale. 51 II. To divide one Number by another. As the divisor is to the dividend, so is an unit to the quotient. III. To find a mean proportional between two Numbers. Because the distance between the first and second term, is equal to the distance between the third and fourth ; therefore, if you divide the space between the point representing the first term, and that represent- ing the fourth, into two equal parts : the middle point must necessarily give the mean proportional sought. IV. To extract the Square Root. The square root of a quantity is nothing more than a mean proportional between an unit aud the given number to be extracted ; the unit being the first term, and the number to be extracted the fourth ; therefore it may be done by the preceding direction. Note. These rules are all applied in the succeed- ing parts of the book. Of the Diagonal Scale. The diagonal scale, usually placed on the two feet Gunter's scale, is thus contracted. A. B Draw eleven lines of equal length and at equal dis- tances from each other, as in the above figure; divide the two outer lines, as AD, CE, into any conveni- ent number of equal parts, according to the largeness you intend your scale. Join these parts by straight lines, as AC, Bo, &c. first taking care that the cor- ners C, A. D, E, are ail square. Again divide the Of the Carpenter's Rule. lengths AB, and Co, into to equal parts, join these riz. from the point A to the first division in Co, ami from the first division in AH, fo the second division in Co, &c. as in the figure, and number the several divisions. The chief use of sue!: a scale as this, is to lay down any line from a given measure; or to measure any and thereby to compare it with others. If the divisions in oE he ealled units, the small divi- sions in v_'o will be LOths, and the divisions in the al- titude oB will be 100th parts of an unit. If the large us be tens, the others will be units, and tenth parts. If the large divisions be hundreds, the others will be tens and units, &.c. each set of divisions be- ing tenth parts of the former ones. For example, suppose it were required to take off 244 from the scale : fix one foot of the compasses at .2 of the larger divisions in oE, and extend the other to the number 4 in Co 3 then move both points of the compasses by a parallel motion, till you come at the fourth long line, taking care to keep the right-hand point in the line marked 2 : then open the compasses a small matter, till the left-hand foot reaches to the intersection of the two lines marked 1, 4, and you have the extent of the number required. In a similar man- ner any other number may be taken oft*. CHAPTER XI. Description and use of the common Carpenter's Rule. TlllS rule is generally used in measuring of timber, and artificers works : and is not only useful in taking dimensions, but in casting up the contents of such work. It consists of two equal pieces of box, each one foot in length, connected together by a folding joint; in one of the^e equal pieces there is a slider, and four lines Of the Carpenters Rule, 5% marked at (he right-hand A, B, C, D ; two of these lines, B, C, are upon the slider, and the other two, A, J}, upon the rule. Three of these lines, viz. A, B, C, are called double lines, because they proceed from 1 to 10 twice over; these three lines are all exactly alike, both in numbers and division. They are numbered from the left-hand towards the right 1, 2, 3, 4, 5, 6, 7, 8, 9, 1 which stands in the middle ; the numbers then go on, 2, 3, 4, 5, 6, 7, 8, 9, 10 which stands at the right-hand end of the rule. These numbers have no determinate value of their own, but depend upon the value you set on the unit at the left-hand of this part of the rule ; thus if you call it 1, the 1 in the middle will be 10, the other figures which follow will be 20, 30, &e. and the 10 at the right-hand end will be 100. If you call the first, or left-hand unit 10, the middle 1 will be 100, and the following figures will be 200, 300, 400, &c. and the 10 at the right-hand end will be 1000. Or, if you call the first, or left-hand unit, 100, the middle 1 will be 1000, and the following figures 2000, 3000, 4000, &c. and the 10 at the right-hand 10,000. Lastly, according as you alter, or number, the large di- visions, so you must alter the small divisions propor- tionality. The fourth lineD, is a single line, proceeding from 4 to 40 : it is also called the girt-li?ie, from its use in easting up the contents of trees and timber. Upon it are marked WG at 17.15, and AG at 18.95, the wine and ale guage points, to make it serve the purpose of a guaging-rule. The use of the double lines A and B, is for work- ing the rule of proportion, and finding the areas of plane figures. And the use of the girt-line D, and the other double line C, is for measuring of timber. — On the other part of this side of the rule, there is a table of the value of a load, or 50 cubic feet, of tim- ber, at all prices, from 6 pence to 24 pence, or two shillings, per foot. f 2 Of the Carpenters Ride. On tike other side of the rule arc several plane scales divided into 12th parts, marked inch, |, £, \, gmtying, that the inch, J inch, &c. are each di- vided into 13 parts. These scales are useful for plan- ning dimensions that are taken in feet and inches. The edge of the ruler is divided into inches, and each of these inches into eight parts, representing half in- ches, quarter inches, and half quarters. In this description we have supposed the rule to he folded ; let it now he opened, and slide out the slider, you will find the hack part of it divided like the edge of the rule, so that altogether will measure 1 yard or 3 feet in length. Some rules have other scales and tahles upon them ; as a table of hoard measure, one of timber measure; a line for shewing what length for any breadth will make a foot square ; also a line shewing what length for any thickness will make a solid foot ; the. former Jine serves to complete the table of hoard measure, and the latter the table of timber measure. The thickness of a rule is generally about a quar- ter of an inch; this face is divided into inches, and tenths, and numbered, when the rule is opened, from the right-hand towards the left, 10, 20, 30, 40, &c. towards 100, which falls upon the joint ; the other half is numbered in the same manner, and the same way. This scale serves for taking dimensions, in feet, tenths, and hundredths of a foot, which is the most commodious way of taking dimensions, when the con- tents are cast up decimally. The use of the Sliding Rule. PROBLEM I. To multiply Numbers together, as 12 and 16. Set one on B. to the multiplier (12) on A; then against the multiplicand (16) on B, stands the product (192) on A. Of the Carpenter's Rule. 55 II. Find the product of 35 and 19. Set 1 on B, to the multiplicand (35) on A; then be- cause 19 on B runs beyond the rule, I look for 1.9 on B, and against it on A, I find 66.5; but the real mul- tiplier was divided by 10, therefore the product 68.5 must be multiplied by 10, which is done by taking away the decimal point, so the product is 665. PROBLEM II. To divide one Number by another, as 360 by 12. Set the divisor (12) on A, to 1 on B ; then against the dividend (360) on A, stands the quotient* (30) on B. II. Divide 7680 by 24. Set the divisor (24) on A, to 1 on B; then because 7680 is not contained on A, I look for 768 on A, and against it I find 32 on B, the quotient; but because one-tenth of the dividend was taken to make it fall within the compass of the scale A, the quotient must be multiplied by 10, which gives 320. PROBLEM III. To Square dny Number, as 25. Set 1 upon C to 10 upon D — Then observe, that if you call the 10 upon 1), 1, the 1 on C will be 1 ; if you call the 10 on D, 10, then the 1 on C will be 100; if you call the 10 on D, 100, then the 1 on C will be 1000, &c. This being well understood, you will ob- serve that against every number on D, stands its square on C. Thus against 25- stands 625 against 30 stands 900 against 35 stands 1225 against 40 stands 1600 Reckoning the 10 on D, to be 10. 56 Of the Carpenter's Rule. PROBLEM IV. To extract the Squart Root of a Number. Fix (!) • slider exactly as in the preceding problem, and estimate ! Ik- value of the lines D and C in the same manner; then against every number found on C stands its square root on D. problem v. To find a mean Proportional between two given Num- bers, as 9 and 25. Set the one number (9) on C, to the same (9) on D; then against 25 on C, stands 15 on D, the mean proportional sought. For, 9 : 15 :: 15 : 25. 2. What is the mean proportional between 29 and 430 ? Set 29 on C, to 29 on D $ this being done, you will find that 430 on C will either Fall beyond the scaJe D> or it will not be contained on C. Therefore take the 100th part of it, and look for 4.3 on C, and against it on 1) stands 11.2, which multiply by 10, and 112 is the mean proportional required. PROBLEM VI. To find a fourth Proportional to three Numbers; or, to perform the Pule of Three. Suppose it were required to find a fourth propor- tional to 12, 28, and 57. — Set the first term (12) on B to the second (28) on A; then against the third term (57) on B stands the fourth (133) on A. If either of the middle numbers fall beyond the line, take one-tenth part of that number, and increase the answer 10 times. Note. The finding a third proportional between two numbers is exactly the same, for in such cases the second number is repeated to form the third: Thus the third proportion to 12 and 28 is 65.33, and is thus found 12 : 28 :: 28 : G5.33. viz. set 12 on B, to 29 on A 5 then against 28 on B stands €5.3 on A. Practical Geometry* Bt CHAPTER XII. Practical GeomeTrt. definitions. 1. GEOMETRY is a science which teaches and cfe° monstrates the properties, affections, and measures of all kinds of magnitude, or extension 5 as solids, sur- faces and lines. 2. Geometry is divided into two parts, theoretical and practical. Theoretical geometry treats of, and con- siders, the various properties of extension abstract- edly ; and practical geometry applies these considera- tions to the various purposes of life. 3. A solid is a figure, or body, of three dimensions, viz. length, breadth and thickness. And its boundaries are su- perficies, or surfaces. Thus A sents a solid. 4. A superficies, or surface, is an extension of two dimensions, viz. length and breadth, with- out thickness : And its boundaries are B lines. Thus B represents a surface. 5. A line is length without breadth, and is form- ed by the motion of a point. Thus ~- CD represents a line. Hence the ex- C D tremities of lines are points, as are likewise their in- tersections. 6. Straight lines are such as cannot coincide in two points without coinciding altogether, and a straight line is the shortest distance between two points. 7. A point is position, without magnitude. 8. A plain rectilineal angle is the inclination, or opening, of two right lines meeting in a point, as E. 58 Prt nj. 9. One angle is said to be less than another, when the lines which form it are nearer to . other. Takfc two lines AH and BC lunching each other in the point B: Conceive these two lines to open, like the legs of a pair of compasses, so as always to remain fixed to each other in B. — While the extremity A moves from the extremity C the greater is the opening or angle ABC; ami on the contrary, the nearer yon bring them toge- ther, the lt^s the opening or angle will.be. to. A circle is a plane figure eontain- G>— -B fed by one line called the circumference, J* y. which is every where equally distant f C J from a point within it, called the centre, V J as C : And an arch of a circle is any ^ » » S part of its circumference, as Gil. 11. The magnitude of an angle does not consist in the length of the lines which form it, but in their opening or inclination to each other. Thus the angle ABC is less than the angle DBE, though the lines AB and CB, which form the former angle, are longer than the lines DB and BE, which form the latter. 12. When an angle is expressed by three let! vBC, the middle letter always stands at the angular point, and the other two letters at the extremities of the lines which B form the angle. Thus the angle ABC is formed by the lines AB and BC, that of DBE is formed by the lines DB and BE. 13. Every angle is measured by an arch of a cir- cle, described about the angular point as a centre ; thus the arch 1)E is the measure of the angle DBE, and the arch GH is the measure of the angle ABC. Practical Geometry. 5& 14. 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. Angles are measured by the number of degrees cut off from the circle, by the lines which form the angles. Thus, if the arch GH contain 20 degrees, or the eighteenth part of the circumference of the circle, the measure of the angle ABC is said to be 20 de- grees. 15. When a right line EC stand- ing upon a right line AB, makes the adjacent angles ACE and BCE equal to each other; each of these angles is said to be a right angle, and the line EC is perpendicular to AB. The measure of a right angle is therefore 90 degrees, or the quar- ter of a circle. 16. Jin acute angle is less than a right angle, as DCB, or ECD. 17. Jin obtuse angle is greater than a right angle as ACD. 18. A plane triangle is a space included by three straight lines, and contains three angles. 19. A right angled triangle is that which lias one right angle in it, as ABC. The side AC, opposite the right angle, is ealled the hypothVnuse, the side BC is called the perpendicular, and the line AB, on which the triangle stands, is called the base. 20. An obtuse angled triangle has one obtuse angle in it, as B. 21. An acute angled triangle has all its three angles acute, as C. M Practical Geometry. is lliat which !ia^ three equal sides, ami three equal an- gles, as I). An isosceles triangle lias two equal sideband the third side either greater or less, than each of the equal sides, as E. 24-. A scalene triangle has all its three sides uneqal, as F. 2/5. A quadrilateral figure is a space included by four straight lines, and contains four angles. 26, A parallelogram is a plane figure hounded by four right lines, whereof those which are opposite are parallel one to the other ; that is, if produced ever so far would never meet. 27. A square is an equilateral parallel- ogram, viz. bavins all its sides equal, and all its angles riglft angles, as G. 28. An oblong is a rectangled parallel- ^____ ogram, whose length exceeds its breadth, [ H asH. ' ' 29. A rhombus is a parallelogram having all its sides equal, but its angles are not right angles, as I. 30. A rhomboides is a parallelogram having its opposite sides equal, but its /"" length exceeds its breadth, and its an- £•» gles are not right angles, as K. y 31. A trapezium is a plane figure contained under four right lines, no two of which are parallel to each oth- er, as L. A line connecting any two opposite angles of the trapezium, as AB. is called a diagonal. \ u Practical Geometry, 61 32. .# trapezoid is a plane quadrilaler- y t al figure, having two of its opposite sides y^ %f \ parallel, and the remaining two not par- Z. \ allel ; as M. 83. Multilateral figures, or polygons, have more than four sides, and receive particular names, accord- ing to the number of sides. Thus, a Pentagon, is a Polygon of five sides ; a H&cagon } has six sides ; a Heptagon, seven ; an Octagon, eight ; a Nonagon, nine 5 a Decagon, ten ; an Undecagon, eleven ; and a Duodecagon has twelve sides. If all the sides and an- gles are equal, they are called regular polygons 5 if unequal, they are called irregular Polygons, or Fi- gures. 34. The diameter of a circle is a right line drawn through the centre, and terminated by the circumfer- ence both ways 5 thus AB is a diameter of the circle. The diameter divides the surface, and circumference, into two equal parts, each of which is called a semicircle. If a line CD be drawn from the centre C perpendicular to AB, to cut the circumference in D, it will divide the semicircle into two equal parts, ACD and DCB, each of which will be a quadrant, or one-fourth of the cir- cle. The line CD, drawn from the centre to the cir- cumference, is call d the radius. 35. A sector of the circle is comprehended under two radii, or semidiameters, which are supposed not to make one continued line, and a part of the circum- ference. Hence a sector may be either less or greater, than a semi- circle 5 thus ACB is a sector less than a semicircle, and the re- maining part of the circle is a sec- tor greater than a semicircle. 36. The chord of an arch is a right line less than the diame- G &4 Practical GLeometry. ter, joining flu- extremities of* the arch ; thus DB is the chord of the arch DGE, or of the arch EBAD. pneni i* any part of a circle bounded by an arch and its chord, and may be either greater or less than a semicircle. . Concentric circles are those which have the same centre, and the space included between their circumferences lied a ring, as D. 39. If two pins he fixed at the points F, /, and a thread YPf he put round them and knotted at J* : then if the point P and the thread he moved about the fixed centres F, /, so as ^ ^ to keep the thread always stretched, the point P will de- scribe the curve PBDACP A( called an ellipsis. 40. The points or centres F, jf. are called the foci, and their distance from C, or D ? is equal to the half of AB. 4t. The line AB, drawn through the foci to the curve, is called the transverse axis, or diameter. The point O, in the middle of the axis AB, is the centre ot the ellipsis. The line CD, drawn through the centre O, per- pendicular to the transverse diameter AB, is called the conjugate axis, or diameter. 43. The line LR, drawn through the focus F, per- pendicular to the transverse axis, is called the para- meter^ or latus rectum. ll. A line drawn from any point of the curve, per- pendicular to the transverse axis, is called an ordinate to the transverse, as EG. If it go quite through the figure, as EH, it is called a double ordinate. Practical Geometry* m 45. The extremity of any diameter is called the vertex; thus the vertices of the transverse diameter AB, are the points A and B. 46. That part of the diameter between the vertex and the ordinate is called an abscissa; thus GB and AG, are abscisses to the ordinate GE. 47. If one end of a thread equal in length to CD, be fixed at the point F, and the other end fixed at D, the end of the square BCD ; and the side CB of the square be moved along the right line AB, so as always to coincide therewith, the string being kept stretched and close to the side of the square PD, the point P will describe a curve HROLPG, called a parabola. 4S. The fixed point F rs called th£ focus. 49. The right line AB is called the directrix. 50. The line ON is the axis of the parabola, and O is the vertex. 51. A line LR, drawn through the focus F, perpen- dicular to the axis, is called the parameter, or latus rectum. 52. A right line IK, drawn from the curve perpen- dicular to the axis, or parallel to the directrix, is cal- led an ordinate ; if it go quite through the figure, as IM it is called a double ordinate. 53. The part of the axis, KO, between the vertex, O, and ordinate, IK, is called the abscissa. Note. The definitions of the solids are given in Part II. Chap. II. in the respective sections where each solid is considered. 64 Practical Geometry. To bisect, or divide a given right line AB, into two From the points A and B with any ^.O* radius, or opening of the compasses, . ?\ greater than half AB, describe two arches cutting each other in C and I) ; draw CD. and it will cut AB in the point E. making AE equal to EB. 7 *-'£fr+ PROBLEM II. At a given distance E, to draw a right line CD parallel to a given ", , _ fl right line AB. ■ ' ^ >. From any two points m n in the *» /»"■"** line AB with the extent of E in R your compasses describe two arches o s; — draw CD to touch these arches, without cutting them, and it will be parallel as required. PROBLEM III. Through a given point o, to draw a right line CD, par- allel to a given straight line AB. Take any point m at pleasure in the line AB 5 with m as a centre and distance in 0, describe the arch o n ; with n as a centre and the same distance describe the arch sm. Take the arch on in your compasses, and apply it from m tost through o,s, draw CD, and it will be par- allel as required. ££ Practical Geometry, eg PROBLEM IV. To divide a given right line AB into any number of equal parts. From the ends A and B of the given line, draw two lines AP and BK of any length, parallel to each other. Then set any num- ber of equal parts from A towards P, and likewise from B towards K 5 draw lines be- tween the corresponding points HM, GL, FT^ &e. and they will divide AB into the equal parts AC, CD, DE EB. PROBLEM V. To divide a given rjght line. AB into two such parts, as shall be to each other, as mr, torn. From A draw any line AC, equal to mn, and upon it transfer the divisions of the line mn, viz. mv and m. Join BC, and parallel to it draw DE; then will AE : EB :: mr: rn. fly- PROBLEM VI. To find a third Proportional to two given lines AB, AD. Place the two lines B ~D given and AD so as to make an angle with each other at A; in AB the greater, cut off' apart AC, equal to AD the less given line ; join BD, and draw CE parallel to it, then will AEbe the third proportional required, m. AB : AD :: AD : AE. G 2 ■ TROBLI.M VII. To find a fourth Proportional to three AB, AC, AD. given lines. Place two of the given lines AB A., and AC so as to make an angle with A- each oilier at A, and join BC. On ^" AB, set oft" the distance AD, and draw DE parallel to BC ; then AE is the fourth proportional required, viz. AB : AD :: AC : AE. A*r*T%n PROBLEM VIII. From a given point V in a right line AB to erect a Per- pendicular. 1. When the point is in, or near, the middle of the line. On opposite sides of the point P take two equal distances P?n, Pn from the points m and nas centres, with any opening of the compass- es greater than Pw, describe two arches cutting each other in r; through r, draw CP, and it will be the perpendicular required. ~?k 2. When the point P is at tlie end of the line. With the centre P and any ra- dius describe the arch m n o ; set off Pm from m to w, and from n with the same radius describe an archr: through m and n draw the line m n r to cut the arch in r; then through r and P draw CP and it will be the perpendicu- lar required. f A Practical Geometry. or thus: Set one foot of the compasses in P and extend the other to a- ny point n, out of the Hue AB ; from n as a centre and dis- tance nP, describe a circle cut- ting AB in m 5 through m and \ n 9 draw m n o to cut the circle ^ :i \^ in o, then through o draw CP, »v* which will be perpendicular to AB. PROBLEM IX. From a given point C to let fall a Perpendicular up- on a given line AB. 1. When the point is nearly opposite the middle of the line. From the centre C describe an {J arch to cut AB in m and n ; with the centres m and n, and the same ra- dius, describe arches intersecting in £\fq, p n/ -& o ; through C and o draw CP, the ***— perpendicular required. 2. When the point is nearly opposite the end of the line, Frodpe given point C draw any line Cm * meeting AB in the point m; besectraC in n, and with n as a centre and radius n m, or wC de- » scribe an arch cutting AB in o, draw Co, and it will be the perpendicular required. Oh Practical Geometry. PROitLKM x. In any Triangle ABG to draw a Perpendicular from any tingle h> t<» it* opposite side. Bisect either of the sides con- taining the ankle from which the perpendicular is to be drawn, as BC in m ; with the radius m IJ and centre m describe an areh cutting AB (produced if necessary) in I), draw CI), and it will be the per- pendicular required. PROBLEM XI. Upon a given right Une AB to describe an Equilatera Triangle. With B as a centre and radius equal to AB describe an arch 5 with A as a centre and distance, AB, cross it in C ; draw AC and BC, then will ABC be the equilateral triangle required. PROBLEM XII. To make a Triangle with three given lines AB, A and BC, of which any two taken together are gret than the third. With A as a centre and radius AC, describe an arch, with the centre B and distance BC cross it in C, draw AC, and BC, then ABC is the triangle required. A Practical Geometry. $9 PROBLEM XIII. Two sides AB and BC of a right angled Triangle are given, to find the Hypotlienuse. From the point B in AB draw BC perpendicular and equal to BC : join AC, and it will be the A~" -B hypotlienuse required. „ n PROBLEM XIV. The Hypotlienuse AC, and one side AB, of a right an- gled Triangle are given, to find the other side. Bisect AC in m, with m as a centre and distance m A describe an arch, with A as a centre and distance AB cross it in B: join BC : then ABC is a right angled triangle, and BC the required side. "7 PROBLEM xv. To find a mean Proportional between two given lines AB and BC. Join AB and BC in one straight A «* une viz. make AC rqual to the sum _ ~" yf them, and bisect it in the point o. " !£— ^.n With the centre o and radius o A /^y^k descrij^J^ semicircle ; at B erect Ax^^ j >* the pe^(Micular BD, and it will be Js *? k d the mean proportional required, viz. AB : BD :: BD : BC. Note. If AD and DC be joined, AD will he a mean proportional between AB and AC, also CD will be a mean proportional between BC and AC, viz. AB : AD :: AD : AC and BC : CD :; CD : AC. Practical Geometry. PROBLEM XVI. To bisect, or divide t into two equal parjs. From the centre C With any radius describe an arch win from m and u as centres with the game radius de scribe two arches crossing other in o, draw Co, and it will vide the given angle ACH into the /w Ivvo equal angles ACo and BCo. problem xvli. At a given point A, in a given line AB to make an Jingle equal to a given Angle C. With the centre C and any radius describe an arch m n; with the centre A A and the same radius describe the arch os: Take the distance mn in your compasses, ana apply it from s to o ; then a line drawn from A through o will make the angle A equal to the angle C. PROBLEM XVIII. To make an Angle of any proposed number of degrees upon a given right line, by the Scale of Chords, Upon the line AB to make an angle of 30 deg. (Fig. 1.) Take the extent of (50 degrees from the line of chords with which and the centre A describe the arch am. Take 30 degrees from the same scale of chords, and set them off from o to n ; through n draw An, then 7?AB is the an gle required. Practical Geometry. iTi To make an angle of 150 degrees, produce the line BA to m, with the centre A and the chord, of Go de- grees ; describe a semicircle ; take the given obtuse angle from 180 degrees, and set off the remainder, viz. 30 degrees from m to n ; through n draw An, then n AB is the angle required. PROBLEM XIX. *Sn Angle being given to find how many degrees it con- tains, by a Scale of Chords. With the chord of 60 degrees in your compasses and centre A (Fig. 1.) describe the arch on, cutting the two lines which contain the angle in o and n; take the distance on in your compasses, and setting one foot at the beginning of the chords on your scale, ob- serve how many degrees the other foot reaches to, and that will be the number of degrees contained in the arch o n, or angle ftAB, If the extent o n reach beyond the scale, which will always be the case when the angle is obtuse, produce the line BA from A towards m, "and measure the arch m n in the same manner, the degrees it contains, de- ducted from ISO degrees, will give the measure of the angle nAB. PROBLEM XX. Upon a given right line AB, to describe a Square. With A as a centre and distance & £ AB describe the arch EB, with B as a centre and the same extent describe the arch AC cutting the former in o ; make o E equal to B o, and draw BE, make oC and ol) each equal to AF, or Fo, and A " H join the points AD, DC and CB, then will ABCD be the square required. Practical Geometry. or thus: Draw BC perpendicular, and equal to AB : with the extent AB and one foot in A describe an arch; with the same extent and one foot in C cross it in ]>. join AD and DC, then ABCD is the square re- quired. PROBLEM XXI. To make an Oblong, or Rectangled Parallelogram, of a given length BA, and breadth BC. Place BC perpendicular to AB ; with the centre A and distance BC describe an arch, with the centre C and distance AB cross it in D; join AD and DC, then ABCD is the ob- long required. it PROBLEM XXII. Ujmn a given right line AB to des- cribe a Rhombus, having an angle to the given angle A. Upon AB make the angle DAB equal to the given angle A; also make AD equal to AB. Then with D and B as centres and radius AB, describe arches crossing each other in C : join DC and BC, then ABCD is the rhombus required. Practical Geometry. £3 PROBLEM XXIII. To find -the centre of a given Circle Take any three points. A, C, B, in the circumference of the circle, and join AC and BC Bisect AC and BC with the lines no, and mo, meeting each other in the point o. Then o is the centre required. Note. By this problem it will\e easy to describe a circle through any three given points that are not in the same straight line : Or, to describe a circle about. a given triangle. x\lso, if any areh of a circle be given, the whole circle may be readily described. PROBLEM XXIV. To draw a right line equal to any given arch of a Circle, A B. Divide the chord AB into four equal parts ; set one part AC on the arch from B to D: draw CD, and it will be nearly equal to half the length arch ADB. OR THUS : Through the point A, and o the centre of the circle draw A m ; divide o n into four equal parts, and set off three of them from m to n. Draw AC perpen- dicular to A m, then through m and B draw m C ; then will ilC be equal to the length of the arch AB very nearly. H Practical Geometry. PROBLEM XXV. To make a square equal in Jlrea to a given circU, Divide the diameter AB into fourteen equal parts, and make. AX! equal to eleven of these parts; erect the perpendicular GC, and join AC ; then the square AEDC, formed upon AC, is equal to the whole cir- cle whose diameter is AB, ex- ceedingly near the truth. PROBLEM XXV] In a given circle to describe a Sana re. Draw any two diameters AB and CD perpendicular to each other, then connect their extremities, and that will give the inscribed square A DBC. Note. If a side of the square, as DB, be bisected in m, and aline om n be drawn from the centre of the circle to cut the circumference: then if the line Bn be drawn, it will be the side of an octagon inscribed in the circle. PROBLEM XXVII. To make a regular Folygon on a given line AB. Divide 360 degrees by the num- p her of sides contained in your poly- gon, subtract the quotient from ^ ISO degrees, and the remainder will be the number of degrees in each angle of the polygon. From each end of AB draw lines AG, BO. makini angles with the gitenline equal to half the angle o£ the polygon: then with the centre and radius Practical Geometry, iu ©A describe a circle to the circumference of which ap- ply continually the given side AB. or thus: Take the given line AB from any scale of equal parts : multiply the side of your polygon by the number in the third column of the following table, an- swering to the given number of sides, and the product will give you the length of AO or OB, with which pro- ceed as above. M. of JVame of the Had. of the circum- An. OAI3, or Sides. Polygon. scribing- c- OBA. 3 Trigon, .5773503 30 4 Tetragon, .7071068 45 5 Pentagon, .8506508 54 b 6 Hexagon, 1, Side=Radius. 60 | 7 Heptagon, 1.1523825 64? 8 Octagon, 1.3065630 67]; s 9 Nocagon, 1.4619022 70 10 Decagon, 1.6186340 72 i 11 Undecagon, 1.7747329 73t\ 12 Duodecagon, 1.9318516 75 PROBLEM XXVIII* In a given circle to inscribe any regular Polygon $ or to divide the circumference of a given circle into any number of equal parts. Divide the diameter AB into as many equal parts as the figure has . sides ; from the centre o draw the per- pendicular om divide the ra- dius on into four equal parts and set off three of these parts from n to m, from m, through the second division s 9 of the diameter AB draw mC : join AC, and it will be the side of the polygon re- quired. Practical Geometry. PROBLEM XXIX. the two diameters of an Oval, a figure re~ semblins the conic section called an Ellipsis, to describe it. Bisect the longer diameter AB in the point C by the line 6\ Yards 5 40 Poles 8 Furlongs- Y Foot Yard Fathom Pole or Rod Furlong Mile ^ II. Square Measure. 144 Inches 9 Feet 36 Feet 27 2\ Feet 30-1- Yards 1600 Poles 64 Furlongs 1 Foot 1 Yard 1 Fathom 1 Pole or Rod 1 Furlong 1 Mile •Land is measured by a chain, called Gunter's chain, «f 4 poles, or 22 yards in length, and consists. of 100 equal links, each link being T %% of a yard in length, <#r 7.92 inches. Ten square drains, or ten chains in Mensuration of Superficies. ?9 length and one in breadth, make an acre; or 4840 square yards, 160 square poles, or 100,000 square links, each being the same in quantity. Forty perches, or square poles, make a rood, and 4 roods make an acre. The length of lines measured with a chain, are ge- nerally set down in links, as whole numbers: every chain being 100 links in length. Therefore after the dimensions are squared, or the superficies is found, it will be in square links ; when this is the ease, it will be necessary to cut off five of the figures on the right hand for decimals, and the rest will be acres. These decimals must be then multiplied by 4 for roods, and (lie decimals of these again, after five figures are cut. •iff, by 40 for perches. § I. To find the JIjrea of a Square. Example 1. Let ABCD be a square given, each sida feeing 14. Required the area, or superficial content, Is£ 14X14=196, the area of the square ABCD, Sfi Mensuration of Superficies, By Scale and Conqmsses. Extend the compasses from 1, in the line of num- bers, to 14; the same extent will reach from the same point, turned forward, to 106. Or, extend from 10 to 14, that extent will reach to 19.6, which multiply by io. Demonstration. Let each side of the given square he divided into 14 equal parts, and lines drawn from one another crossing each other within the square ; so shall the whole great square lie divided into 196 little squares, as you may see in the figure, equal to the number of square feet, yards, poles, or other measure, by which the side was measured. 2. What is the area of a square whose side is 35.25 chains ? Ans. 124 acres 1 rood 1 perch. 3. Required the area of a square whose side is 5 feet 9 inches. Jliis. 33 feet inches 9 parts. 4. What is the area of a square whose side is 3723 links ? Jlns. 138 acres 3 roods 1 perch. § II. Examples of an Obloxg, or rectangled PARAL- LELOGRAM. To find the area of a parallelogram; whether it he a square, a rectangle, a rhombus, or a rhomboid. Multiply the length by the breadth, or perpendicu- lar height, and the product will be the area. Mensuration of Superficies* Si l£ B j 1 1 ~T~ T 9 - ] \ 1 1> . J 1 c Example 1. Let ABCD be an oblong, the length of it 18 feet, and the breadth 9 feet; these multiply to- gether, the product is 162, the superficial content. By Scale and Compasses. Extend the compasses in the line of numbers from 18 to 162, the square feet. Or, call the middle unit on the scale 10, and extend from it to 9, that extent will reach from 18 to 16.2, which multiply by 10. Demonstration. If the sides AB and CD he eaeU divided into 18 equal parts, representing 18 feet; and the lines AD and BC each divided into 9 equal parts, and lines drawn from point to point, crossing each other within the figure ; those lines will make thereby so many little squares as there are square feet, viz. ±Q2. 2. What is the superficial content of a rectangular hoard, whose length is 14 feet 6 inches, and breadth 4 feet 9 inches. Jns. 68 feet 10 inches 6 parts. 3. Required the area of a rectangular piece of ground, whose length is 1375 links, and breadth 950. Jfris, 13 acres roods 10 perches. 4. What is the superficial eontent of an oblong which is 2 feet 10 inches 6 parts long, and 9 inches broad ? tins. 2 feet 1 inch 10 parts 6 second's. Mensuration of Superfic. § HI. Examples of a RhombI's. Example i. Let ABCD be a rhombus given, whose sides are each 13.5 feet, and the perpendicular BA is 13.42 5 these multiplied together, the product is 208.010 ; which is the superficial content of the rhom- bus, that is, 208 feet and one hundredth part of a foot. By Scale and Compasses. Extend the compasses from 1 to 13.42, that extent will reach from 15.3 the same way to 20Sfeet, the con- tent. Or, call the middle 1 upon the scale 10 and ex- tend from it to 13.42, that extent will reach from 15.5 to 20.8, which multiply by 10. Demonstration. Let CD be extended out to F, making DF equal to CE, and draw the line BF; so shall the triangle DBF be equal to the triangle ACE: For DF and GE are equal, and BF is equal to AE, because AB and CF are parallel. Therefore the par- allelogram ABEF is equal to the rhombus ABCD. 2. "VVhat is the area of a rhombus, whose length is 6.2 chains and perpendicular height 5.45 chains? Jlns. 3 acres 1 rood 20.64 perches. The length of a rhombus is 12 feet 6 inches, and perpendicular height 9 feet 7 inches, what is the area? Jlns. 119 feet 9 inehes 6 parts. 4. The length of a rhombus is 725 links, and per- pendicular height 635 links, what is the area ? Jlns. 4 acres 2 roods 16.6 perches. Mensuration of Superficies. § IV. Examfles of a Rhomboides. irv- Ex ample. 1. Let ABCD be a rhomboides given*, whose longest side AB or CD, is 19.5 feet, and the perpendicular AE is 10.2; these multiplied together, the product is 198.9, that is, 198 superficial feet and 9 tenth parts, the content. Demonstration. If DC be extended to F, making CF equal to DE, and a line drawn from B to F ; so will the triangle CBF be equal to the triangle ADE, and the parallelogram AEFB be equal to the rhom- hoides ABCD ; which was to be proved. 2. A piece of ground, in th^ibrm of a rhomboides, measures 4784- links in length,^and its perpendicular breadth is 1908 links, what is its area ? •Sns. 91 acres 1 rood 4.5952 perches. 3. Required the area of a rhomboides, whose length is 12 feet 6 inches, and perpendicular breadth 5 feet inches. Jlns. 68 feet 9 inches. 4. Plow many square yards of painting are in a rhomboides, whose length is 37 feet and breadth 5 feet 3 inches? Jlns. 21 f 2 yards. 84 Mensuration of Superficies. § V. To find the JIrea of a Triangle when the base and perpendicular are given. RULE. Multiply the base by the perpendicular, and half he product will be the area. EXAMPLES. a 1. Let ABC he right angled triangle, whose base is 11.1 and the perpend ieular 12 feet. Multiply 1 1.1 by 6, half the perpendi- cular, and the product is 84.6 feet, (he area. Or multiply 14.1 by 12, the product is 169.2; the half of which is S4.6, the same as before. By Scale and Compasses. Extend the compasses from 1 to 14.1; that extent will reach the same way from 6 to 84.6 feet, tfee con- tent. Or, call the middle 1 on the scale 10, extend from it to 1 4.1, that extent will reach from 6 to 8.46, which multiply by 10. -$. S * Note. The per- pendicular in the figure ought to be 7.8. 2. Let ABC (Fig. 2.) be an oblique-angled trian- gle given, whose base is 15.4, and the perpendi- cular 7.8 ; if 15.4 be multiplied by 3.9 (half the perpendicular,) the product will he 60.06 for the Mensuration of Superficies, 8o . area, or superficial content : Or, if the perpendicular 7.8 be multiplied into half the base 7.7, the product will be 90.06 as before: Or, if 15.4, the -base, be mul- tiplied by the whole perpendicular 7.8, the product will be 120.12, which is the double area; the half whereof is 60.06 feet as before. By Scale and Compasses. Extend the compasses from 2 to 15.4, that extent will reaeh from 7.8 to 60 feet, the content. Or, extend from 20 to 15.4, that extent will reach from 7.8 to 6, which multiply by 10. — Here the middle 1 on the scale is considered as 10. Demonstration. If AD (Fig. 1.) be drawn parallel to BC, and DC parallel to AB ; the triangle ADC shall be equal to the given triangle ABC. Hence the parallelogram ABCD is double of the given triangle ; therefor- half the area of the parallelogram is the area of the triangle. In Fig. 2. the parallelogram ABEF is also double of the triangle ABC; for the triangle ACF is equal to the triangle ACD, and the triangle BCE is equal to the triangle BCD ; therefore the area of the parallelogram is double the area of the given triangle : Which was to be proved. 3. The base of a triangle is 28.2 yards, and the per- pendicular height 18.4 yards, what is the area? Jlns. 259.44< square yards. 4. There is a triangular field whose base measures 1236 links, and perpendicular 731 links; how many acres does it contain ? Ans. 4 acres 2 roods 2.8128 perches. 5. What is the area of a triangle whose base is 18 feet 4 inches, and perpendicular height 11 feet 10 inches ? Jlns* 10S feet 5 inches 8 parts. I &6- Mi a 8 u rat ion of Superfi c ie& , To find the area of any plane Triangle by having the three sides given, without the help of a Perpendi- cular. ruli:. Add the three sides together, and take half that sum; from which stibstract each side severally; then multiply the half sum and the three differences con- tinually, and out of the last product extract the square root 5 which will be the area of the triangle sought. EXAMPLES. !• Let ABC be a triangle, whose three sides are as follows; viz. AB 43.3. AC 20.5, and BC 31.2, the area is required. ("43.3 | 4.2") 1 31.2 16.3 I Di [20.5 J 27.0 J Sides < 31.2 ir>.3 >. Differences. Sum 95.0 Half 47.5 Mensuration of Superficies, *7 Area 296.31 47.5 The half sum. 27 Difference* 3325 950 1282.5 Product. 16.3 Difference. 3S4Jb 76950* 12825 20904.75 Product. 4.2 Difference. 4180950 S361900 87799.950 Last product the square root of which is 296.31 (remainder 3339) the area required. 2. What is the area of a triangle whose sides are 50, 40 and 30 ? Jins. 600. ' 3. The sides of a triangular garden are 41, 29 and 56 yards, what is the area thereof? Jins, 574.34 yards. 4. How many acres are contained in a triangle whose three sides are 4900, 5025 and 2569 links. Jins. 61 acres 1 rood 39.68 perches. 5. A field of a triangular form, whose sides are 3S0, 420 and 765 yards, rents for 55 shillings per acre j what is the annual rent ? Jins. The area is 44699.0347 yards, or 9,235337 acres, which at 55 shillings per acre amount* to 1.25 : 7 : 11^28, the annual rent. $s iitioU of >*- Demonstration of the r.ule.— -In the triangle BDC, I say, if from the half sum of Ihc sides, you subtract each particular .side, and multiply the half sum and the three differences together con- tinually, the square root of the product shall be the area or the triangle. First, by the lines Bl, CI, and l)i, bisect the, thro. which lines will all meet in the point I: by which lines the given triangle is divided into three new triangles CBI, DCI, and BDI; the perpendiculars of which new triangles are the lines AI, EI, and OT, being all equal to one another, because the point 1 is the centre of the inscribed circle (Eutlid, Lib. IV. Vrob.. 4.) : Wherefore to the side BC join CF equal to DE, or DO; so shall BF be equal to half the sum of the sitles : vix. = 4 BC + * BD 4. -V CD. AndBA—BF— CD; for CA=CO and OD—CF ; therefore CD—AF; and AC— BF—BD, for BE= BA, and ED—CF; therefore BD^BA+CF, and CF=BF— BC. Make FH perpendicular to FB, and produce BI to meet it in II. Draw C1I and HK. perpendicular to CD. Because the angles FCK-f-EHK are equal to two right angles (fbr the .angles F and K are right angles) equal also to FCK+ACO (by Euclid, I. 13.) and the angles ACOxAIO are equal to two right angles; therefore the quadrangles FCKH and AIOC are simi- lar; and the triangles Oil and AIC are also similar. And the triangles BAI and BFH are likewise similar. From this explanation, I say, the square of the area of the given triangle: that is, BF 2 "xlA 3 ~BF XBAxCAxCF. In words: Mensuration of Superficies. 89 The square of BF the half sum of the sides mul- tiplied into the square of IA (—1 E— I O) will be equal to the said half sum multiplied into all the three differences. ForIA: BA :: FH:BF;andIA: CF : : AC : FH 5 because the triangles are similar. By Euclid ', Lib. VI. Prop. 4. Wherefore multiplying the extremes and means In both, it vvillbe IJ^+BFxFH=BAxCAxCF-{- FH ; but FH being on both sides of the equation, it may be rejected ; and then multiply each part by BF, it will be "BT^xiA^zrBFxBAxCAxCF. Which was to be demonstrated. Jf any two sides of a right angled Triangle be gi~ :en, the third side may be found by the following Pi RULE. 1. To the square of the base add the square of the perpendicular, the square root of the sum will give the hypothenuse, or longest side ; or from the square of the hypothenuse take the square of either side, and the square root of the remainder, will be the other side. Example 1. Given the base AB (see Fig. 1. § V.) 14.1, the perpendicular BC 12, what is the length of the hypothenuse AC ? 14.1 the base AB. 12 the perpend. BC* 14.1 12 ill 144 square of BC. 564 ill 198.81 square of AB 144. square of BC. 3*2.81 Sum of the squares, the square root of which extracted gives AC 18.51513. 90 Mensuration of Superjl Gtiven the base *'0, and the hypothenuse 2j ; to find the perpendicular. From the square of 25— Take the square of 15z_; - -And the square root of 400 will be the pei pendieular—^O. i'iie base of a right angled triangle is 24, and the perpendicular 18. what is the hypothenuse ? Ms. 30. 4. The wall of a fort standing on the brink of a ri- ver is 42.426 feet high, the breadth of the river is 23 yards : what length must a eord be to reach from the top of the fort across the river? Ms. 27 yards. 5. The hypothenuse of a right angled triangle is 30, and the perpendicular IS, what is the base ? Ms. 21. 6. A ladder, 50 feet long, will reach to a window .JO feet from the ground on one side of a street ; and without moving the foot will reach a window 40 feet high on the other side. The breadth of the street is re- quired. Ms. 2Z\ yards. 7. A line of 380 feet will reach from the top of a precipice, which stands close by the side of a brook, to the opposite bank, and the precipice is 128 feet high, how broad is the brook ? Ans. 357.29 feet. 8. If a ladder, 50 feet in length, exactly reach the co- ping of a house, when the foot is 10 feet from the up- right of the building ; how long must a ladder be to reach the bottom of the second floor window, which is 17.9897 feet from the coping, the foot of this ladder Mensuration of Superficies. 01 and standing 6 feet from the upright of the building ; what is the he£$ht of the wall of the house ? JLns. The height of the wall is 48.9897 feet, and the length of a ladder to reach the second floor win- dow must be 31.5753 feet. § VI. To find the Area of a Trapeziu. rule. Add the two perpendiculars together, and take half the sum, which multiply by the diagonal : the product is the area. Or, find the areas of the two triangles, ABC and ACD (by section V.) and add them toge- ther, the sum shall be the area of the trapezium. Mensuration of Superjl EXAMPLES. i. Let ABCD be a trapezium given, the. diagonal of which is 80.0, and the perpendicular 13F 30.6j and ilie perpendicular DBJ, 34.0 : these two added together, the sum is 54.6, the half of which i- and this being multiplied by the diagonal, 80.3, the product is 2197.63, which is the area of the trape- zium. By Scale and Compasses. Extend the compasses from 2 to 51.6 ; that extent will reach from 80.5 to 2197.65, the area. Or call the unit at the beginning of the scale 100, and extend from 200 to 316, that extent will reach from S05 to 2197.65. Demonstration. This figure ABCD is composed of two triangles, the triangle ABC is half the paral- lelogram AGHC: also the triangle ACD is equal to half the parallelogram ACIK, as was proved, sect. V. Wherefore the trapezium ABCD is equal to half the parallelogram GH1K. To find the area, HI=BF-f- DE; therefore! HI x AC (=KI=GII) area of the trapezium, which was to be proved. 2. There is afield in the form of a trapezium, whose diagonal is 1660 links, the perpendiculars 702 and 713 links, what is the area? Ms. 11 acres 2 roods 37.792 perches. 3. What is the area of a trapezium whose diago- nal is 34 feet 9 inches, and the two perpendiculars 19 feet 9 inches and 8 feet 9 inches ? Jins. 493 feet 2 inches 3 parts. 4. Required the area of a four-sided field, whose south side is 2740 links, east side 3575 links, north Mensuration of Superficies* A3 side 3755 links, west side 4105 links, and the* diagonal from south-west to north-east 4835 links. Jhis. 123 acres rood 11.8673 perches. 5. Suppose in the trapezium ABCD (see the fore- going figure) the side AB to be 15, BC 13, CD 14, and AD 12 ; also the diagonal AC 16 ; what is the area thereof? Ms. 172.5247. 6. Suppose in the trapezium ABCD, on account of obstacles, I could only measure as follows, viz. the diagonal AC 878 yards, the side AD 220 yards, and the side BC 265 yards : But it is known that the per- pendicular DE will fall 100 yards from A ; and the perpendicular BF will fall 70 yards from C ; required the area in acres. Jlns. The perpendicular DE will be 195.959 yards, BF 255.5875 yards ; and the area of the trapezium 85342.2885 yards, or 17.6327 acres ; or 17 acres, 2 roods 21 perches. When any two sides of a Trapezium are parallel to each other, it is then generally called a Trapezoid; the area may be found by the following RULE. Multiply half the sum of the two parallel sides by the perpendicular distance between them, and the pro- duct will be the area. EXAMPLES. 1. Let x\3CD be a trape- zoid, the side AB 23, DC 9.5, and CI 13, what is the area First, the sum of AB 23, and DC 9.5, is 32.5 ; the half whereof is 16.25, which mul- tipled by CI 13, gives the product 211.25 for the area required. Mensuration of Superficies, Demonstration. Bisect AI) in II, through H draw FE parallel to BC, ami then produce CI) to E; also draw Gil parallel to AB or CD. Now AH is equal to HO. and the angle AUF is equal to the ingle EIIl) {Euclid. I. 15.) also the angle HED is equal to the angle UFA {Euclid,!. 29.) Therefore the triangles being equiangular, and having one side equal, are equal in all respects {Euclid,!. 26.) consequently ED is equal to AF. Again, IIG is half the sum of EC and FB, for it is equal to each of them, consequently it is half the sum of AB and DC, but FB multiplied by CI is the area of the figure ECBF, or of its equal ADCB, consequently TIG multiplied into CI is the area thereof 5 which was to be proved. 2. Required the area of a trapezoid whose parallel sides are 750 and 1225 links, and the perpendicular distance between them 1540 links ? Jims, 15 acres roods 33.2 perches. 3. What is the Area of a trapezoid whose parallel sides are 12 feet 6 inches, and 18 feet 4 inches ; and the perpendicular distance between them 7 feet 9 inches ? Jins. 119 feet 5 inches 9 parts. 4. A field in the form of a trapezoid, whose parallel sides are 6340 and 4380 yards, and the perpendicular distance between them 121 yards, lets for Z.S3 : 1 : 7^ fer annum, what is that per acre ? Ms. 1.0 : 12 : 4f. § VII. Of Irregular Figures. Irregular figures are all such as have more sides than four, and the sides and angles unequal. All such figures may be divided into as many triangles as there are sides, wanting two. To find the area of such figures, they must be divided into trapeziums and tri- angles, by lines drawn from one angle to another ; and Mensuration of Superficies. S3 so'find the areas of the trapeziums and triangles se- parately, and then add all the areas together; so will you have the area of the whole figure. Let ABCDEFG be an irregular figure given to be measured ; first, draw the lines AC and GD, and there- by divide the given figure into two trapeziums, A CDG and GDEF, and the triangle ABC; Of all which, I find the area separately. First, I multiply the base AC by half the perpen- dicular, and the product is 49.6, the area of the tri- angle ABC. Then for the trapezium ACDG, the two perpen- diculars, 11 and 6.6, added together, make 17.6 ; the half of which is 8.8, multiplied by 29, the dia- gonal ; the product is 255.2, the area of that trape- zium. And for the trapezium GDEF, the two perpendicu- lars, 11.2 and 6, added together, make 17.2; the half is 8.6; which multiplied by 30.5, the diagonal, the product is 262.3, the area. All these areas added to- gether, make 567.1, and so much is the area of the whole irregular figure. See the work. 06 Mensuration of Superficies. *>aso AT. 11 perpendicular. -' half perpendicular. o.r> 49. G area of ABC. 17.6 sum. 8.8 half. 29 diagonal CG. 792 176 £ perpendiculars. 255.2 area of ACGD. 11.2? v„.i„_ 30.5 8.6 17.2 sum. 1830 2440 8.6 half sum. 262.30 area of GDEF. 233.2 area of ACIMI. 49.6 area of ABC. 567.10 sum of the areas. This figure being composed of triangles and trape- ziums, and those figures being sufficiently demonstrated in the Vth and Villi sections aforegoing, it will be needless to mention any thing of the demonstration in this place. § VIII. Of Regular Poltgqss. To find the area, or superficial content, of any regu- lar polygon. RULE i. Multiply the whole perimeter, or sum of the sides by half the perpendicular let fall from the centre •Mensuration of Superficies, {,7 to the middle of one of the sides ; and the product is the area. EXAMPLE.S 1. Let HIKLMN be a regular hexagon, each side being 14.6, the sum of all the sides is 87.6, the half sum is 43.8, which mmltiplied by the perpendicular GS 12.64397, the product is 553.805886. Or if 87.6, the whole sum of the sides, be multiplied by half the perpendicular 6.321985, the product is 553.805886, which is the area of the given hexagon. By Scale and Compasses. Extend the compasses from 1 to 12.64, that extent will reach from 43.8, the same way to 553.8. Or, ex- tend from % to 12.64, that extent will reach from K titration of Superficies. S7.r> to 003.8. Or, as before, call (he first number 100 times as much as it is, and each of the two follow- ing 10 times as much as they are. Demonstration. Every regular polygon is equal to the parallelogram, or Jong square, whose length is equal to half the sum of the sides, and breadth equal to the perpendicular of the polygon, as appears by the foregoing tigure ; for the hexagon H1KLMN is made up of six equilateral triangles : and the parallelogram OPQU is also composed of six equilateral triangles, that is, five whole ones, and two halves ; therefore the parallelogram is equal to the hexagon. 2. The side of a regular pentagon is 25 yards, and the perpendicular from the centre to the middle of one of the sides is 17.204775 ; required the area. Jlns. 1075.2984. 3. Required the area of a heptagon, whose side is 19.3S, and perpendicular 20.1215. ' Jlns. 136 l.S 11345. 4. Required the area of an octagon, whose side is 9.941, and perpendicular 12. Jlns. 477.168. 5. Required the area of a decagon, whose side is 20, and perpendicular 30.776836. Jlns. 3077.6836. 6. Required the area of a duodecagon. whose side is 102, and perpendicular 190.3345908. ' Jlns. 116484.7695696. 7. Required the area of a nonagon, whose side is 40, and perpendicular 54.949548. Jlns, 9890.91864. Mensuration of Superficies. 99 8. Required the area of an undecagon, whose side is 20, and perpendicular 34.056874. Ms. 3746.25614. 9. Required the area of a trigon, viz. an equilate- ral triangle, whose side is 20, and perpendicular 5.773502. Ms. 173.20506, 10. Required the area of a tetragon, viz. a square, whose side is 20, and perpendicular from the centre to the middle of one of the sides 10* Ms. 400. A Table for the more ready finding the Area of a Po- lygon, and also the Perpendicular. No. of Name of the I. Areas. II. Perpend. Sides. Polygon. The side I. The side T. 3 Trigon, .433013 .28867-51 ! 4 Tetragon, 1.000000 .5000000 5 Pentagon, 1.720477 .6881910 6 Hexagon, 2.59S076 .8660254 i 7 Heptagon, 3.633912 1.0382617 ! 8 Octagon, 4.828427 1.2071068 9 Nonagon, 6.181824 1.3737387 10 Decagon, 7.694209 1.5388418 11 Undecagon, 9.365640 1.7028437 12 Duodecagon, 11.196152 1.8660254 J RULE II. Multiply the square of the side by the tabular area, and the product is the area of the polygon. Or, mul- tiply the side of the polygon by the tabular perpendic- ular, and the product will give the perpendicular of the polygon j then proceed by the first rule. 1U0 Mensuration of Superficies. How to find the Tabular Numbers. These numbers are found by trigonometry, thus : find the ancle at the centre of the polygon by dividing 360 degrees Dy the number of sides of the polygon. Example. Suppose each 9ide of the duodecagon an- nexed be l, and the area be required. Divide 360 by 12 (the number of sides) and the quo- tient is 30 degrees for the angle ACB ; the half of which is 15, the angle DCB, whose complement to 90 degrees is 75 degrees, the angle CBD : then say As s, DCB 15 degrees, is to .5 the half-side DB log. so is s, CBD 75 degrees, Co-ar. 0.587004 9.698970 9.984944 to the perpendicular CD 1.866025 0.270918 Then 1.866025 multiplied by 6, (the half perimeter) the product is 11.196152 the area of the duodecagon required. If the side of a hexagon be 14.6. what Example. is the area ? 14.6 14.6 876 584 146 213.16 squ. 2.598076 213.16 15588456 2598076 7794228 2398076 5196152 tab. area. 553.80588016 area, the same as at page 97. Mensuration of Superficies. 101 Note. Should more examples he thought necessary, take any of those under rule I. j > •*, § IX. Of a Circle. There is no figure that affords a greater variety of useful properties than the circle. It is the most ca- pacious of all plane figures, or contains the greatest area within the same perimeter, or has the least peri- meter about the same area. The area of a circle is always less than the area of any regular polygon circumscribed about it, and its cir- cumference always less than the circumference of the polygon. But on the other hand, its area is always greater than that of its inscribed polygon, and its cir- cumference greater than the circumference of its in- scribed polygon ; hence the circle is always limited between these polygons. The area of a circle is equal to that of a triangle, whose base is equal to the cir- cumference, and perpendicular equal to the radius. Circles, like other similar plane figures, are in pro- portion to one another, as the squares of their diame- ters ; and the circumferences of circles, are to one another as their diameters, or radii. The proportion of the diameter of a circle to its circumference has never yet been exactly determined. This problem has engaged the attention, and exercised the abilities of the greatest mathematicians for ages ; no square, or any other right-lined figure, has yet been found, that shall he perfectly equal to a given cir- cle. But though the relation between the diameter k2 io J Mensuration of Superficies, and circumference has not been accurately expressed itt. number?, it may be approximated to any assigned .degree of exactilesV. Jlrchimedes, about two thousand the proportion to be nearly as 7 i nearer, ratios have since been suc- cessively assigned, viz. As 106 to 333, As 113 to 355, &C. This last proportion is very useful, for being turned into a decimal, it agrees with the truth to the sixth fi- gure inclusively. Victa, in his universaliwn inspectio- ncm ad canonem mathematicum, published in 1579, by means of the inscribed and circumscribed polygons of 363216 sides, carried the ratio to ten places of figures, shewing that if the diameter of a circle be 1 the cir- cumference will be greater than 3.1415926535, but less than 3.1415926537. And Ltidolf Van Ceulen, in his book de circulo et ad- 'scriptis, by the same means carried the ratio to 36 pla- ces of figures ; this was thought so extraordinary a C performance, that the numbers were cut on his tomb- alone in St. Peter's church-yard, at Leyden. These numbers were afterwards confirmed by Willebrord Snell. Mr. Abraham Sharp, of Little Horton, near Bradford, Yorkshire, extended the ratio to 72 places of figures, by means of Dr. Halley's series, as may be seen in Sherwin's logarithms. Mr. Machin, professor of astronomy in Gresham college, carried the ratio to 100 places of figures ; his method may be seen in Dr. Jlutton's large treatise on mensuration. Lastly, M. de Lagny, in the memoirs de VAcad. 1719, by means of the tangent of an arch of 30 degrees, has carried the Mensuration of Superficies* 103 ratio to the amazing extent of 128 figures ; finding that if the diameter be 1, the circumference will be 3.1415, 92653, 58979, 32384, 62643, 38327, 95028, 84197, 16939, 93751, 05820, 97494, 45923, 07816, 49628, 62089, 98628, 03482, 53421, 17067, 98214, 80865, 13272, 30664, 70938, 446-f or 447 But on ordinary occasions 3.1416 is generally used, as being sufficiently exact. PROBLEM I. Having the Diameter to find the Circumference, or cir- cumference to find the diameter. RULES. 1. As 7 is to 22, so is the diameter to the circum- ference. Or, as 113 is to 355 so is the diameter to the cir- cumference. Or, as 1 is to 3.1416, so is the diameter to the cir- cumference. 2. As 22 is to 7, so is the circumference to the di- ameter. Or, as 355 is to 113, so is the circumference to the diameter. Or, as 3.1416 is to 1, so is the circumference to the diameter ; or, which is the same thing, as 1 is to .318309, so i& the circumference to the diameter. EXAMPLES. l. The diameter of a circle is 22.6 ? what is the cir- cumference ? Mensuration of Superf. Multiply the the product is : which divided by 7 gives 71.02s for the cii Or, (by the second proportion) if 22.fi he multiplied by 835, the product will be 8023; this di- vided by 113, the quotient is 71, the circumference. Or (by the third proportion) if 22.6 be multiplied into 3.1416, the product is 71.00016, the circumference : the two last proportions are the most exact. By Scale and Compasses. Extend the compasses from 7 to 22, or from 113 to 355, or from 1 to 3.1416 ; that extent will reach from 22.6 to 71. 2. The circumference of a circle is 71, what is the diameter ? If .318309 he multiplied by 71, the product will be 22.5999239 for the diameter. Or, 113 multiplied by 71, the product is 8023; which divided by 355, the quotient will be 22.6 the diameter : Or 7l multiplied by 7, the product is 497 ; this divided by 22 f the quo- tient as 22,5999, the diameter. Mensuration of Superficies. 10S By Scale and Compasses, Extend the compasses from 3.1416 to 1, that extent will reach from 71 to 22.6, which is the diameter sought. Or, you may extend from 1 to .31S309. Or from 22 to 7. Or from 355 to 113 5 the same will reach from 71 to 22.% as before. Note. That if the circumference he 1, the diameter will be .318309 ; this number being found by dividing an unit by 3.1416. 3. If the diameter of the earth be 7970 miles, what is its circumference, supposing it a perfect sphere ? Ms. 25038.552 miles. 4. Required the circumference of a circle whose diameter is 50 feet. Ms, 157.08 feet. 5. If the circumference of a circle be 12, what is tho diameter ? Ms. 3.819708. PROBLEM II. To find the Area of a Circle, RULES. 1. Multiply half the circumference by half the dia- meter: Or, take £ of the product of the whole cir- cumference and diameter, and it will give the area. 2. Multiply the square of the diameter by '.7854, and the product will be the area. 3. As 452 is to 355, so is the square of the diameter to the area. 4. As 14 is to 11, so is the square ^>f the diameter to the area. 106 Mensuration of Superf 5. Multiply the square of the circumference by 58, and the product will be the area. 6. As 88 is to 7, so is the square of the circumfer- ence to the area. 7. As 1420 is to 113, so is the square of the circum- - Terence to the area. EXAMPLES. 1. The diameter of a circle is 22.6, and its circum- } ference 71, what is the area by the first rule ? 35.5 half the 11.3 half diai circumference, meter. 22.6 diameter. 71 circumference. 1065 355 355 401.15 area. 226 1582 4)1604.6 401.15 area. Demonstration of rule 1. Every circle may be conceived to be a polygon of an infinite number of sides : Now the semidiameter, must be equal to the perpendicular of such a polygon ; and the circum- ference of the circle equal to the periphery of the polygon; therefore half the circumference, multiplied by half the diameter, gives the area. The other part is self-evident. 2. If the diaireter of a circle be 1, and its circum- ference 3.1416, what is the area ? Jins, .7854. Mensuration of Superficies. 107 3. If the diameter of a circle be 22.6, what is the area by the second rule ? 22.6 diameter. 510.76 22.6 .7854 1356 204304 452 255380 452 40860S 357532 510.76 squ. of the diameter. 401.150904 area. \ • Demonstration of rule 2. All circles are to each other as the squares of the diameters, and the area of a circle whose diameter is 1, is .7854 (by the second example); therefore as the square of 1, which is 1, is to .7854, so is the square of the diameter of any circle to its area. 4. If the diameter of a circle be 22.6, what is the ,area by the third rule ? As 452 : 355 :: 510.76 the square of the diameter: 401.15, the same as before. Demonstration of rule 3. By problem I. we have, as 113 : 355 :: diameter 1 : circumference; but the area of a circle whose diameter is 1, is * of this cir- cumference by rule 1. problem II. viz. it is ||| ; but the areas of circles are to each other as the squares of their diameters, therefore square 1: |f|: : square of any diameter : the area ; viz. 452 : 355 : : square of any diameter : the area. 5. If the diameter of a circle be 22.6, what is the area by the fourth rule ? 108 Mensuration of Superjl As 14 : 11 :: 610.7ft the square of the diameter : 401.31 the area which is larger than by the other me- thods. Demonstration of rule 4. By problem I. we have as 7 : 23 :: diameter 1 : circumference ; and, by the same argument as above, the area of a circle whose diameter is 1, is || or \l» hence, 14: 11 :: square of the diameter : the area. 6. If the circumference of a circle be 71, what is the area by rule 5 ? 71 circumference. .07958 71 0041 71 7958 497 31833 5041 squ.of the circumference. 39790 401.16278 area. The reason of the above operation is easy ; for as the diameter of one circle is to its circumference, so is the diameter of any other circle to its circumfer- ence : therefore the areas of circles are to each other as the squares of their circumferences ; and .07958 is the area of a circle whose circumference is 1 ; for when the circumference is 1, the diameter is .318309, as has been observed before. 7. If the circumference of a circle be 71, what is the area by the sixth rule ? As ss : 7 :: 5041 the square of the circumference : 400.98 area. Mensuration of Superficies. 109 Demonstration of rule 6. By problem I. we have as 22 : 7 :: circumference 1 : diameter ; but by rule I. problem II. the area of a circle whose circum- ference is 1, is i of this diameter, viz. ¥ \. Now the areas of circles are to each other as the squares of their circumferences; therefore square 1 : -^ :: square of any circumference: area; viz. 88: 7 :: square of any circumference : area. vSk,* 8. If the circumference of a circle be 71, wh^t is the area by the 7th rule ? ** As 1420 : 113 :: 5041 the square of the circumfer- ence : 401.15 area, Demonstration of rule 7. By problem I. we have as 355 : 113 :: circumference 1 : diameter; and, by the same argument as above, the area of a circle whose circumference is 1, is t Vt 3 o ? therefore it follows, that 1420 : 113 :: square of any circumference : area. 9. What is the area of a circle whose diameter is 12? Ms. 113.0976. 10. The diameter of a circle is 12, and its circum- ference 37.6992, what is the area ? Jlns. 113.0976. 11. The circumference of a circle is 37.6992, what is the area? Jlns. 113.104579, &c. 12. The surveying-wheel is so contrived as to turn just twice in length of a pole, or 16 £ feet; in going rou^d a circular bowling-green, I observed it to turn exactly 200 times; what is the area of the bowling- green ? Ans. 4 acres 3 roods 35.8 perches. 13. The length of a line, with which my gardener formed a circular fish pond, was exactly 27| yards : pray what quantity of ground did the fish-pond take ,iin ? Jlns. 2419.22895 yards, or half an acre nearly. no Mensuration of Superficies. PROBLEM III. Having the Diameter, Circumference, or Jirea of a Cir- cle given : to find the side of a Square equal in Jlrea to the Circle, and the side of a Square inscribed in the Circle ; Or, having the side of a Square given, to find the Diameter of its Circumscribing Circle, and also of a Circle equal in Area, <$*c. RULES. 1. The diameter of any circle, multiplied by .8862269, will give the side of a square equal in area. 2. The circumference of any circle, multiplied by .2820948, will give the side of a square equal in area. 3. The diameter of any circle, multiplied by .707106S, will give the side of a square inscribed in that circle. 4. The circumference of any circle, multiplied by .2250791, will give the side of a square inscribed in that circle. 5. The area of any circle, multiplied by .6366197, and the square root of the product extracted, will give the side of a square inscribed in that circle. 6. The side of any square, multiplied by 1.414236, will give the diameter of its circumscribing circle. 7. The side of any square multiplied by 4.4428829, will give the circumference of its circumscribing circle. 8. The side of any square, multiplied by 1.1283791, will give the diameter of a circle equal in area to the square. 9. The side of any square, multiplied by 3.5449076, will give the circumference of a circle equal in areato the square. Mensuration of Superficies, 111 Explanation of these Rules. Let ABBC represent a square equal in area to the circle; and EH.FG a square inscribed in the circle ; GH and EF diameters of the circle, perpendicular to eaeh other. 1. The area of a circle, whose diameter is 1, is ►7854, or .78539816, &c. the square root of which gives the number in rule 1 And as the diameter of one circle is to the diameter of another, so is the side of a square equal in area to the former, to the side of a square equal in area to the latter, &c. 2. The area of a circle, whose circumference is 1 is .07 95775, &c. the square root of which gives the number in the 2d rule. 3. To the square of EO=|, add the square of GO = |, the square root of the sum will give the third number. 4. The diameter of any circle, when the circum- ference is 1, is .318309, &e. hence EO or GO is 1.159154, &c. with which proceed as above, and you will iind the fourth number. •J 13 Mensuration of Superficies. Hence it appears, that if the diameter of the cir- cle is given, or can be found, the side of the inscri- bed square may be found ; and if (he side of a square equal in area to the circle is given, the diameter and circumference of the circle may be readily found, the former by dividing the area by .78539, &c.and extract- ing the square root of the quotient (see rule 2. prob. II.) and the latter by dividing the area by .0795775, &c. and extracting the square root of the quotient (see rule 5. prob. II.) — The rest of the numbers are left for the learner to examine at his leisure. EXAMPLES. 4. If the diameter of a circle be 22.6, what is the side of a square equal in area to the circle ? The side of a square equal in area to a circle, whose diameter is 1, by rule 1, is .8862269; — hence, 22.6X 8862269=20.02872794, the side >f the square. 2. If the diameter ef a circle be 50, what is the side of a square equal in area ? Ms. 44.311345. 3. If the circumference of a circle be 40, what is the side of a square equal in area to the circle ? The side of a square equal in area to a circle whose circumference is 1, by rule 2, is .2820948; and .2820948X40=11.283792, the side of the square. 4. If the circumference of a circle be 71, what is the side of a square equal in area to the circle ? Ms. 20.0287308. 5. If the diameter of a circle be 50, what is the side of a square inscribed in that circle ? The side of a square inscribed in a circle whose diameter is l, by rule 3, is .7071068, therefore, 50 x .7071068=35.35534, the side of the square. 6. If the diameter of a circle be 22.6, what is the side of a square inscribed in that circle ? Ms. 15.98061368. Mensuration of Superficies. 118 7. If the circumference of a circle be 40, what is the side of a square inscribed in that circle ? The side of a square inscribed in a circle whose circumference is 1, by rule 4, is .2250791 ; and 40X .22507011=9,003164, the side required. 8. If the circumference of a circle be 71, what is the side of a square inscribed in that circle ? Ms. 15.9806161. 9. If the area of a circle be 1963.5, what is the side of a square inscribed in that circle ? The area of a square inscribed in a circle whose area is 1, by rule 5, is .6366197; therefore, 1963. 5X .6366197=1250.00278095, whose square root is 35.3553, the side of the inscribed square. 10. If the area of a circle be 401.15, what is the side of a square inscribed in that circle. Jns. 15.9806. 11. If the side of a square be 35.36, what is fhe diameter of a circle which will circumscribe that square ? The diameter of a circle that will circumscribe a square whose side is 1, by rule 6, is 1.4142136 ; and .35.36X1-4142136=50.006592896, the diameter. 12. If the side of a square be 15.98, what is the diameter of a circle which will circumscribe that square. Ans. 22.599133328. 13. If the side of a square be 35.36, what is the circumference of a circle which will circumscribe that square ? The circumference of a circle that will circum- scribe a square whose side is 1, by rule 7, is 4.4428829 and 35.36X4-4428829x^157.1000339344, the circum- ference. 14. If the side of a square be 15.98, what is the circumference of a circle which will circumscribe that square ? Ans. 70.997268742. 15. If the side of a square be 35.36, what will be the diameter of a circle having the same area as the square ? l 2 114 Mensuration of Superficies. The diameter of a circle having the same area as a square whose side is 1, by rule 8, is 1.1283791; hence 35.36X1.1 -83791=39.899484976, the diameter. IK. It' the side of a square be 20.029, what will be the diameter of a circle having the same area as the square ? Ms. 22.6003049939. 17. If the side of a square he 33.36, what is the circumference of a circle, having the same area as the square ? The circumference of a circle that will be equal to the area of a square whose side is 1, by rule 9, is 3.5449076 ; therefore — 35.36X3.5449076= 125.347932736, the side of the square. 18. If the side of a square be 20.029, what is tli3 circumference of a circle, having the same area as the square ? Jins. 71.000954. Note. All the foregoing examples are easily work- ed by scale and compasses, for they consist only of mul- tiplication. But in the 9th example a square root is to be extracted ; and in order to do this right, it will he necessary to call the middle unit upon the scale a square number, as 1, 100, 10000, &c. or 1, T ^ T , &c — Thus, if the number to be extracted consist of one whole number, call the 1 in the middle of the scale an unit; if of two or three whole numbers, call it 100; if four or five whole numbers, call it 10000, &c. for whole or mixed numbers. For pure decimals, if the number to be extracted consist of one or two places, call the unit in the middle of the scale 1, then the 1 at the left-hand will be T \j, the figure 2 will be T 9 , &c. or the l at the left-hand may be called T ^, then the figure 2 will be T 2 ^. &c. so that you will more readily obtain the intermediate places. — This being under* stood, extend the compasses from the middle unit thus estimated, to the number to be extracted ; divide the space of this extent into two equal parts, and the m&V Mensuration of Superficies. 115 die point will be the root required : but here you must observe that the middle unit will change its val- ue (except you call it an unit), that is, if in the first extent you estimated it at 100, it must now be consid- ered as 10; if at 10000, it must now be called 100, &c. Thus in the 9th example the square root of 1250 is to he extracted ; here I consider the middle unit as 10000, and extend from it towards the left-hand to 1250, the middle point between these is 35.3, estimating the mid- dle unit now at 100. A little practice will render these observations familiar. § X. To find the Area of a Semicircle. Multiply the fourth part of the circumference of the whole (that is, half the arch line) by the seinidi- ameter, the product is the area. EXAMPLES. 1. Let ACB be a semi- circle, whose diameter is ^ *L 22.6, and the half circum- ference, or arch line, ACB, is 35.5, the half of it is 17.75, which multiply by / iM the semidiameter ll.a, and y^amo^J^. the product is 200.575, the area of the semicircle. By Scale and Compasses. N Extend the compasses from 1 to 11.3; that extent will reach from 17.75 to 200.575, the area. Or, as before, call the first number 100 times as much as it is> 116 Mensuration of Superficies. and each of the two following 10 times as much as they are. If only the diameter of the semicircle he given you may say, by the Rule of Three, As 1 is to .39 27, so is the square of the diameter to the area ; or multiply the square of the diameter by .7854, and take half the product. 2. The diameter of a circle is 30, and the semicir- cumference 47.124 ; what is the area of the semicir- cle ? Ms. 353.43. 3. The diameter of a circle is 200, what is the area of the semicircle ? Jlns. 15708. 4. If the circumference of a circle he 157.08, and the diameter 50 ; what is the area of the semicircle ? dns. 981.75. § XI. To find the Area of a Quadrant. rule. Multiply half the arch line of the quadrant, (that is, the eighth part of the circumference of the whole circle,) by the semidiameter, and the product is the area of the quadrant. EXAMPLES. 1. Let ABC be a quadrant, or _ fourth part of a cirele, whose radi- us, or semidiameter is 11.3, and the half arch line 8.875; these multipli- ed together, the product is 100.2875 \ for the area. \ These are the rules commonly given for finding the Mensuration of Superficies. 117 area of a semicircle and quadrant; but it is the best way to find the area of the whole circle, and then take half that area for the semicircle, and a fourth part for the quadrant. 2. Required the area of a quadrant, the radius being 15, and the length of the arch line 23.562. Ms. 176.710. 3. The diameter of a circle is 200, what is the area of the quadrant ? Ms. 7854. 4. If the circumference of a circle be 157.08, and the diameter 50, what is the area of the quadrant ? Ms. 490.875. Before I proceed to shew how to find the area of the sector, and segment of a circle, I shall shew how to find the length of any arch of a circle. To find the. length of any Mch of a Cirde. RULE I. Multiply, continually, the radius, the number of de- grees in the given arch, and the number *.0l745329 ; the product will be the length of the arch. Example. 1. If the arch ACB contain 118 degrees 46 minutes 40 seconds, and the radius of the circle, of which ACB is an arch, be 19.9855: what is the length of the arch ? First, 118 degrees 46 minutes 40 seconds, are equal to 118| degrees. 19.9855X118|X-0l745329=r41.431199, &C. the length of the arch. * When the radms is 1, the semi-circumference of the cir- cle is 3.14159265, &c. and this number divided by 180, the degrees in a semicircle, gives .01745329 for the length of one degree when the radius is unity. 113 Mensuration of Superficies, RULE II. From eii^lit times the chord of half the arch, sub- tract t he chord of the whole arch, ami one third of the remainder will be the length of the arch nearly. Example 2. If the chord, AC or BO, of half the arch be 1 .8, and the chord AB of the whole arch 34,4 j what is the length of the arch ACB ? 8X19.8 — 34.4 — 124 j which divided by 5, gives 41 j- for the length of the arch. 3. If the chord AB of the whole arch be 6, and the chord AC of half the arch 3.0*3836, what is the length of the arch ? Ms. 6.116896. 4. If the radius of the circle be 9, and the arch contain 38.9424412 degrees, what is its length ? Ms. 6.117063. 5. There is an arch of a circle whose chord AB is 50.8 inches, and the chord AC 30.6, what is the length of the arch ? Ms. 64| inches. 6. The chord AB of the whole arch is 40, and the versed sine DC 15, what is the length of the arch ? •ins. 5B±. 7. If the radius of the circle be 11.3, and the arch contain 52 degrees 15 minutes, what is its length ? Ms. 10.304858, &C. Mensuration of Superficies. 119 To find the diameter of a Circle, by having the chord and versed sine of the Segment given. RULE. Square half the chord ; divide it by the versed sine, and to the quotient add the versed sine ; the' sum will be the diameter. Demonstration. The half chord AD, is a mean propor- tional between the segments, CD,DE, of the diameter CE; that is AD*=CD XDE ( by 13.6. ) AD* therefore, = CD DE ; to these equal AD add CD, and CD xCD=DExDC :CE : which, in words, gives the above rule. ' EXAMPLES. 1. Let ACB be a segment given, whose chord AB is 36, and the versed sine CD 6; half 36 is 18, which squared, makes 324; this divided by 6, the quotient is 04 : to which add 6, the sum is 60, the diameter of the circle CE. 2. If the chord AB be 18, and versed sine DC 4 ; what is the diameter EC ? dns. 2±\, 120 Mensuration of Stiperjl § XII. To find the Area of the Sector of a Circle. rule. Multiply lialf the length of the arch by the radius of the circle, and the product is the area of the sector. EXAMPLL3. I. Let ADBE he the sector of a circle given, whose' radius AE or BE is 24.5, and the chord AB 39.2 ; what is the area ? First, find the chord of half the arch, and then the length of the arch by section XI. From the square of AE 600.2.5, take the square of AC 38*. 16, the square root of the remainder, 216.09. gives EC, 14.7 ; from ED, or its equal AE, take EC, the remainder 9.8 is CD — then to the square of AC 384.16, add the square of CD 96.04, the square root of the sum 480.2 gives AD 21.9135, the chord of half the arch, hence the arch AD is found to be 22.68466 ; this multi- plied by the radius, 24.5, gives 555.77417 for the area of the sec- tor. , 2. LetLMNO be a sector greater than a semicircle,^ whose radius LO or NO, is 20.6, and the chord Nc of half the arch McN 20.35962; also the chord MN of the whole arch McN, that is, of half the arch of the sector, is 33.4; what is the area? Mensuration of Superficies. 12i The arch McL will be found, by the following, to be 42.49232, this multiplied by the radius, 20.6, gives 875.341792, the area. 3. Required the area of a sector, less than a se- micircle, whose radius is 9, and the chord of the arch 6. Ms. The length of the arch is 6.116896, and area 27.526032. 4. Required the area of a sector, whose arch con- tains 18 degrees, the radius being 3 feet. Jlns. The length of the arch is .£4247766, and the area 1.41371649. 5. Required the area of a sector, greater than a semicircle, whose radius LO or NO is 10, and the chord MN of half the arch LM c N, viz. the chord of the whole arch M c N, is 16. Jlns. As in example 1, you will find Nc= 8.9442719, and the length of the arch McN 18.518050384; hence the whole area will be 185.18050584. 6. What is the area of a sector, less than a semi- circle, the chord of the whole arch AB being 50.8, and the chord of half the arch AD 30.6. Jlns. The length of the arch is 64§ \ and by taking the square root of the difference of the squares of AD M I 1^2 Jlensurution of Supcrjl and AC, DC will be found to be t7.or>i5; then by the method of finding a diameter, Section XI. as 17.0940: 35.4 (=AC) :: 20.4 : 37.8071, the remaining part of the diameter; hence the diameter is 5 4.8716, radius 27A.i3S, and area of the sector 887.0908. §XIII. To find the Area of the Segment of a Circle. rule I. Find the area of the whole sector CADBC by Section XII. and then (by Section V.) find the area of the triangle ABC, and subtract the area of the trian- gle from the area of the sector, the remainder will be the area of the segment. — If the segment be greater than a semicircle, add the area of the triangle to the area of the sector, and the sum will be the area of the segment. EXAMPLES. 1. Required the area of the segment AD BE whose chord AB is 35, and versed sine DE, 9.6. First. By the method of ^ finding a diameter, sect. XI. as 9.6: 17.5 (=AE) :: 17.5 : 31.9, the re- maining part of the dia- meter, which, added to DE, gives 41.5 for the diameter, the half of which, 20.75, is the radius CD ; from which take DE, the remainder 11.15, is the perpendicular CE of the tri- angle ACB. Mensuration of Superficies. 123 Second. To the square of AE 306.23 add the square of ED 92.16, the square root of the sum 398.41 is 19.96, the chord AD of half the arch : then, the area of the triangle ABC will be = 195.125 the arch AD=20.7S; the area of the sector CADB=43 1.1850$ and that of the segment ABD,= 236.06. 2. Let MACBM be a segment greater than a semicircle ; there are given the base of the segment or chord AB 20.5, the height MC 17.17, the radius of the circle 1 1 .64, the chord of half the arch ACB, viz. AC 20, and the chord of one -fourth of the areh 11.5, to find the area of the segment. Half the arch line will be found, as before, to be =24 ; which mul- tiplied by the radius 1 1.64, gives 279.36 for the area of the sector LACB; and half the base AM, multiplied by the perpendicular LM ? = 5.53 gives 56.6825 for triangle the area of the ABL. Hence 336.0425 is the ABC. area of the segment 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, nearly. Note. When the segment is greater than a semicir- cle, find the area of the remaining segment, and sub tract it from the area of the whole circle. fgft Mensuration of Sii])erf< EXAMPLES. 1. Required the area of a segment ADBE, less than a semicircle, whose chord AB is 3D, and versed sine, or height of the segment, DE 9.6. | of 35 X 9.6=r224, and 9.6X 9.6 X 9.6—2 X 35 = 12.639 Area of the segment=236.639 2. The area of a segment greater than a semi- circle is required, the chord being 20.5, and height 17.17. As 17.17: 10.25 :: 10.25 : 6.119, the height of the remaining segment by sect. XI ; hence the diameter is 23.289, and the area of the whole circle 425.983305. By rule 2, the area of the remaining segment is 89.211347, which subtracted from the area of the whole circle, leaves 336.768958 for the area required- See example 2. 3. Required the area of a segment, less than a semi- circle, whose chord is 18.9, and height 2.4. „ 530.601875 by rule 1. **" S ' I 30.6057 by rule 2. 4. Required the area of a segment, less than a se- micircle, whose chord is 23, and height 3.5. a 5 54.5844 by rule 1. " n I 54.598732 by rule 2. 5. Required the area of a segment, greater than a semicircle, whose chord is 12, and height 18. Jlns. 297.826 by rule 2. This may also be done by rule 1 ; for, by what is given, every given line marked in the figure to example 2, may be found. Mensuration of Superficies. 123 §XIV. To find the Area of Compound ^Figures. Mixed or compound figures are such as are compos- ed of rectilineal and curvilineal figures together. To find the area of such mixed figures, you must find the area of the several figures of which the whole compound figure, is composed, and add all the areas together, and the sum will be the area of the whole compound figure. Let AaBCcDA he a compound figure, AaB being the arch of a circle whose chord AB is 18.9, and height 2.4 ; CcD is likewise the arch of a circle whose chord CD is 23, and height 3.5 : And ABCD is a tra- pezium, whose diagonal BD is 34, and the two per- pendiculars 16 and 16.2 3 required the area of this compound figure. By example 4. section XIII. the area of the segment A«B is 30.60187.5 By example 5. section XIII. the area of the segment CcD is 54.5844 Area of the trapezium ABCD - - 547.4 Area of the compound figure - - - 632.586275 M 2 1 26 Mensuration of Superficies. > § XV. To find the Area of an Ellipsis. rule. Multiply the transverse, or longer diameter, hy the conjugate or shorter diameter, and that product by 7851 5 the last product is the area of the ellipsis. &0 5 EXAMPLES. 1. Required the area of an ellipsis whose longer diameter AB is 61.6, and shorter diameter CD 44.4. 61.6 X44.4X .7854= 2148.100416, the area of the ellipsis. 2. The longer diameter of an ellipsis is 50, and the shorter 40, what is the area? Am. 1570.8. 3. The longer diameter of an ellipsis is 70, and the shorter 50, what is the area ? Am. 2748.9. 4. Required the area of an ellipsis, whose transverse diameter is 24, and conjugate 18. Ans. 339.2928, Mensuration of Superficies. Demonstration of the Rule. IZT Put the transverse axis ST=a; the conjugate Nw=c % the height Ta. or Ncr, of any segment =0-, and its cor- responding ordinate ab—y; then by the property c of the curve, y—^^ ax — x 2 au d the fluxion of the a 2c . iegment bTb, or bNb= — Xxy/ax — x 2 . Now, ,2x a ^/ax — x* is known to be the fluxion of the correspond- ing circular segment, BTB or bNb. Therefore, if the circular area be denoted be A, the elliptical area will c be expressed by — xA. Hence it appears that the a area of the segment of the ellipsis it to that of the corresponding circle as a to c : and consequently the whole ellipsis, to the whole circle, in the same pro- portion. Fig. 1. Fig. 2. Prom a due consideration of the two figures, and the foregoing demonstrations, the following rule, for finding the area of an elliptical segment^ is easily de- duced. 12S Jle nsuration of Superfi c its. Given the height of an elliptical segment, and the two diameters, to find the area. RULE. 1. Subtract the height of the segment, from that diameter of which it is a part? multiply the remain- der by the height of the segment, and twice the-square root of the product will be the chord of a circular segment, having the same height. 2. Find the area of this gegment by rule 2, for cir- cular segments. 3. Then, as that diameter, of which the height of the segment is a part, is to the other diameter; so is the circular area, to the elliptical area. Note. If the chord of the elliptical segment also be given ; then, instead of the first part of the above rule, say, as that diameter to which the chord of the segmant is parallel, is to the other diameter, so is the chord of the elliptical segment, to the chord of the circular segment, and proceed as above. EXAMPLES. 1. In an ellipsis, whose longer axis TS is 120, and shorter Nn 40, what is the area of a segment thereof, 6T5, cut parallel to the shorter axis N«, the height of aT being 24 ? TS— T«=Sa=:96 ; this multiplied by Ta, 24, gives 2304; twice the square root of which, is=96, the chord BaB, of the circular segment BTB, whose area by the second rule for that purpose, will be found to be 160S. Then, as 120 : 40 ; or, as 3 to 1 :: 1608 : 536, the area of the elliptic segment bTb. Note. If the chord bab had been given 32 ; BaB mighUiave been found thus, as 40 : 120 :: 32 : 96. 2. In an ellipsis whose longer axis TS is 120, and shorter Nw40, what is the area of a segment thereof, Mensuration of Superficies. 129 BNB, cut parallel to the longer axis, TS, the height dN being 4 ? Proceeding as before, the chord, bob, of the circular segment bNb {Fig. 2.) will be found=24, and its area = 65^-; therefore, as 40 : 120; or as 1:3 :: 65£: 196, the area of the elliptic segment BNB. v 3. Required the area of an elliptical segment cut offby a chord or double ordinate, parallel to the short- er diameter, the height of the segment being 10, and the two diameters 35 and 25? Ms. 161.878. 4. What is the area of an elliptical segment cut off by a chord, parallel to the longer diameter, the height of the segment being 5, and the two diameters 35 and 25. Ms. 97.7083. 5. What is the area of an elliptical segment cut off by a chord, parallel to the shorter diameter, the height being 20, and the two diameters 70 and 50 ? Ms. 647.513997. Given the two diameters of an Ellipsis to find the cir- cumference. * RULE I. Multiply the sum of the two diameters by 1.5708, and the product will be the circumference; near enough for most practical purposes. RULE 11. To half the sum of the two diameters add the square root of half the sum of their squares 5 multiply this last sum by 1.5708, and the product will be the cir- cumference, extremely near. EXAMPLES. 1. The two diameters of an ellipsis are 24 and 18, what is the circumference ? By rule 1. n -j-is /21+is |.^ + is«\ By rule 2. f L.^/ Jx 1.3roS=66.30S5, the circumference of the elUf -. W hat i» the eireumference of an ellipsis whose trau? »o, and conjugate l s - The 1 ,d the short v, by rule 1. <>6 by rui w of an Elliptical Ring, or the space ided between the Circumferences of two concen- RULE. Find the area of each ellipsis, and subtract the less from the greater, the remainder will be the area of the ring. Or, from the produet of the two diameters of the greater ellipsis, subtract the produet of the two diameters of the less : the remainder multiplied bv .t^34, will be the area of the ring. Xote. This rule will also serve for a circular ring; for when the diameters of each ellipsis become equal, the square of the diameter of the greater circle, di- minished by the square of the diameter of the less, and the remainder multiplied by he area of the circular ring: Or, multiply the sum of the diame- * It is here supposed, that the difference between the conji gate diame ers is equal to the difference between the tr. diameters ; archt, tween the semi-transverse, or semi-conjugate diametei theler. i between the arc! Mensuration cf Superficies. 131 ters by their difference, and that product into .7854 for the area of a eireular ring. MPLES. 1. The transverse diame- ter AB of an ellipsis is 60, the conjugate CD 47; and the transverse diameter EF of another ellipsis^having the same centre Of is 43, and the conjugate GH 32: required the area of the space contained between their circumferences. AB*CD=2820. EFXGH=i440. Their difference=l3S0. Multiplied by .7854 Gives the area of the ring=10S3.S32 2. The ellipsis in Grosvenor-square "measures 840 links the longest way, and 612 the shortest, within the wall ; the wall is 14 inches thick, what quantity of ground does it inclose, and how much does it stand upon ? a CThe V ' a ^ encloses 4 acres roods 60.13 p. I And stands on 193.6103 yards. ^3. A gentleman has an elliptical fountain in his gar- den, whose gr atest diameter is 30, and less 24 feet; and has ordered a walk to be paved round it of 3 feet jit inches in width, with Purbtck stone, at 4 shillings per square yard, what will the expence be ? Ms. The area of the walk is 361.561 feet Expence 1.12 : 9s. : 6d|. .968. LJ2 Mensuration of Superficies. 4. The diameters of two concentric circles are 24 and 18, what is the area of the space included be- tween their circumferences ? Jins. 197.9208. § XVI. To find the JIrea of a Parabola. RULE. Multiply the base or greatest double ordinate, by the perpendicular bright, and two thirds of the pro- duct will be the area. EXAMPLES. 1. If the greatest double ordinate GH be 53.75, and ■the abscissa, or height of the parabola EF, be 39.25, what is the area ? ■f of 53.75X39.25=1406.4583, the area of the pa- rabola. Mensuration of Superficies. 13S Let the altitude EF be denoted by x, and the semi- ordinate FH, by y, then if a be the latus rectum of the parabola, we have from the nature of the curve, a x y* • 2 yy • 2 y*y —y* and x= — ; hence x= , and y x= -the a a a y 3 fluxion of the areaHEF; and its fluent=fx — =|X a a?y=fxEFxFH$ therefore the area of the whole parabola GEH, is=f xGHxEF, or two thirds of the circumscribing parallelogram. Note. If the ordinate GH was not at right angles to the axis EF, still the included area would be ex- pressed by two thirds of the base, multiplied into the perpendicular height of the curve above said base or ordinate. Demonstration of the rule. Let FH, the ordi- nate, he divided into an infinite number of equal parts n, and suppose lines ef, parallel and equal to EF, to be drawn through each of these equal divisions. Then, by the nature of similar triangles, assuming ±23 n EF=1, ge will be equal to o, — j J — ,&e. — to — ; n n n n and by a property of the parabola. (Emerson's Conies. B. III. Prop. T.) EF=e/ : ge :: ge : pe continually, viz. 1 1 1 1 : — :: — : — n n rc* . 2 2 4 1 : n n w 2 3 3 9 n* 1 : — :: — . : — &c — — to — n . n n % /i 3 N t34 'Hon of Solids. Hence the area of the space BKHAE will b< 14 n % pressed by o-| 1 J &c. to — , but the sum ., n* n* n* of this series (Emerson's Arithmetic of Difnites, Prop. 3)=}w. Now the area of the parallelogram EKHF ■=FHx'EF=»Xl = n, and n — \n~jn. Hence the I i' Ihe semi-parabola is | of the area of its cir- eumscribing parallelogram. 2. AVhat is the area of a parabola, whose, height or abscissa is 10 ; and the base, or double ordinate, 16 ? Ms. 106f. 3. Required the area of a parabola whose base, or double ordinate, is 38, and the height, or abscissa, 12 ? Ms. 301. CHAPTER II. Ilie MExsvRA; that extent turned over twice from l7.j will reach to 5850, the solid content in inches. Or, extend the compasses from 100 to 17.>, that extent, turned twice over will reach G, which multiplied by 10, gives 53G0. Then ex^ tend the compasses from 1728 to 1 ; that extent, turn- ed the same way, from 5339, will reach to 3.1 feet. Demonstration. If the square ABCD be conceived to be moved down ihe plane ADEF, always remaining pa- rallel to itself, there will be generated, by such a motion, a solid, having six planes, the two opposite ones of which will be equal and parallel to each other; whence it is called a parallelo- pipedon, or square prism. And if the plane ADEF be a square equal to the generating plane ABCD, then will the generated solid be a cube. From hence such solids may be conceived to be constituted of an infinite series of equal squares, each equal to the square ABCD; and AE or DF will be the number of terms. Therefore, if the area of ABCD be multipli- ed into the number of terms AE, the product is the sum of all the terms, and consequently, the solidity of the parallelopipedon or cube. Or, if the base ABCD, be divided into little square areas, and the height AE be divided in a similar manner, you may conceive as many little cubes to be generated in the whole solid, as are equal to the number of the little areas of the base multiplied by the number of divisions which the side .\E contains. From this demonstration it is very plain, that, if von multiply the area of the base of any prism into its length or height, the product will be its solidity. — Any solid figure, whose ends are parallel, equal, and similar, and its sides are parallelograms, is called a Prtsm Mensuration of Solids. 137 2. What is the solidity of a cube, whose side is 19 feet 4 inches? Jlns. 7226 feet 4 inches 5 parts 4 seconds. 3. A cellar is to be dug, whose length, breadth, and depth, are each 10 feet 4 inches 5 how many solid feet does it contain, and what will it cost digging, at 1 shil- ling per solid yard ? I 'the content is 1103^ feet. sins. £ Expence l2 . . 10 ^. 4. How many three inch cubes may be cut out of a 12 inch cube? Jlns. 64. § II. Of a Parailelopipedon. A parallelopipedon is a solid having six rectangu- lar sides, every opposite pair of which are equal and parallel. To find the Solidity. RULE. Multiply the breadth by the depth, and that pro- duct by the length. EXAMPLES. 1. Let ABCDEFG be a parallelopipedon, or square prism, representing a square piece of timber or stone, each side of its square base ABCD being 21 inches : and its length AE 15 feet, required its solidity. *l „___., ™e 3? N 2 138 Mensuration of Solids. First, then, multiply 21 by 21, the product i* hi, the area of the base in inches; which multiply by ISO, the length in inches, and the product i^ 79380, the so- lid content in inches. Divide the lust product by and ihe quotient is 13.9, that is, 45 solid feet and 9 tenths of a foot. Or thus, by multiplying feet and inches. Multiply l foot. 9 inches by l foot 9 inches, and the product is S feet inches 9 parts ; this multiplied a- gain by 15 feet, gives 45 feet 11 inches 3 parts. By Scale and Compasses. Extend the compasses from 12 to 21 (the side of the square) and that extent will reach to near 46 feet, be- ing twice turned over from 15 feet; so the solid con- tent i* almost 46 feet. Note. When the breadth and depth of the solid are unequal, a mean proportional between them must be found, by dividing the space between them into two equal parts, for the side of a mean square in inches ; then proceed as above. For, as 12 is to the side of a mean square in inches, so is the length in feet to a fourth number; and so is this fourth number to the content in foot measure, of any parallelopipedon. 2. If a piece of timber be 25 inches broad, 9 inches deep, and 25 feet long, how many solid feet are con- tained therein ? Jlns. 39 feet inches 9 parts. 3. If a piece of timber be 15 inches square at the end, and 18 feet long, how many solid feet are con- tained in it ; and suppose I want to cut off a solid foot from one end of it, at what distance from the end mustl cut it? q C The solidity is 28.125 feet.. I And 7.68 inches in length will make 1 foot. 4. If a piece of squared timber be 2 feet 9 inches long, 1 foot 7 inches broad, and 16 feet 9 inches long, Mensuration of Solids. 139 how many feet of timber are in that piece ; and how much in length will make a solid foot ? The solid content is 72.93 feet, or 72 feet 11 Ans. <{ inches 2 parts 3 seconds; and 2.75% inches '•{! in length make 1 solid foot. § III. Of a Prism. A prism is a solid contained under several planes, and having its bases similar, equal, and parallel. To find the Solidity. RULE. Multiply the area of its base or end, by the height or length, and the product will be the solidity. EXAMPLES. 1. Let ABCDEF be a triangu- l :ism, each side of the base be ing 10.6 inehes, the pcrpeadicular Ca 13.51 inches, and the length of the solid 19.5 feet. How many solid feet are contained therein ? Multiply the perpendicular of the triangle 13.51 by half the side 7.8, and the product is 105.378, the area of the base ; which mul- tiply by the length 19.5, and the product is 2054.871 ; which divide by 144, and the quotient is 14,27 feet fere, the solid content. Mensuration of Solids. By Scale and Compasses. First, find a mean proportional between the perpen- dicular and half side, by dividing the space upon the line, between 13.51 and 7. 8 into two equal parts; so shall you find the middle point between them to be at 10.26, which is the mean proportional sought : by this means the triangular solid is brought to a square one, each side being 10.26 inches. Then extend the compasses from ]2 to 0.26 ; that extent, turned twice downwards from 19.5 feet, the length, will at last fall upon 14.27, which is 14 feet and a little above a quarter. 2. Let ABCDEFGHTK represent a prism, whose base is a hexagon, each side be- ing 16 inches, the perpendic- ular from the centre of the base to the middle of one of the sides (ab) 13.84 inches, and the length of the prism 15 feet : the solid content is required. Multiply half the sum of the sides 48 by 13.84, and the product is 864.32, the area of the hexagonal base (by § VIII. p. 96.) which multiplied by 15 feet, the length, the product is 9964.8; which divided by 14*, the quotient will be 6 9.2 feet, the solid content required. Mensuration of Solids. 141 By Scale and Compasses. First, find a mean proportional between the perpen- dicular, and half the sura of the sides ; that is divide the space between 13.84 and 48, and the middle point will be 25.77, the side of a square equal in area to the base of the solid. Then extend the compasses from 12 to 25.77 ; that extent will reach (being twice turned over) from 15 feet, the length, to 69.2 feet, the content. To find the Superficial Content of any Prism. RULE, Multiply the circumference of the base by the length of the prism ; the product will be the upright surface* 5 to which add the area of the bases 3 the sum will be the whole superficial content. EXAMPLES. In the hexagonal prism last mentioned, the sum of the sides, or circumference of the end, being 96, and the length 15 feet, that is, 180 inches : which multi- plied by 96, the product is 17280 square inches ; to which add twice 664.32, the areas of the two bases, the sum is 18608.64, the area of the whole, which is 129.226 feet. 1. Each side of the base of a triangular prism is ft inches, and the length 12 feet 5 inches 5 what is the solidity and surface ? a J The solidity 10.79 feet ' \ Surface 54.50S9 feet 142 Mensuration of Solids. \ hexagonal prism measures 28 inches across (ho '(ntre of thf end, from corner to corner, and ■ a in length; required the solidity and surface. a J The solidity is 39.48835 feet. £ Surface 85.2893 feet. A decagonal prism, or pillar, measures 50 inches rcumferenee, and is 30 feet high; required the solidity and surface. q ^ The solidity 40.074 feet. "*• £ Surface 137.6718 feet. 4. The gallery of a church is supported by 10 oeta* gonal prisms of wood, which measure each 48 inches in circumference, and are each 12 feet highj what will he the expence of painting them at 10 pence per square yard ? dns. 1.2 : 7s : 9i. 5. A trapezoidal prism of earth, or part of a canal, is to he dug, whose perpendicular depth is 10 yards, the width at the top 20 yards, at the bottom 16, and the length 50 yards, the two ends being cut perpendi- cularly down ; how many solid yards of earth are con- tained in this part ? %ins. 9000 yards. £ §IV. OfaPrkAMiD. A pyramid is a solid figure, the base of which is a polygon, and its sides plain triangles, their several vertical angles meeting together in one point. Mensuration of Solids. 14ci To find the Solidity. RULE. Multiply the area of the base by a third part of the altitude, or length 5 and the product is the solid con- tent of the pyramid. Example 1. Let ABD be a square pyramid, having each side of its base 18 5 inches, and the perpendicular height CD 15 feet, what is the soli- dity? Multiply 13.5 by 18.5, and the product is 342,25, the area of the base, in inches, which multiply by 5, a third part of the height, and the product is 1711.25 ; this divided by 144 gives 11.83 feet, the solid con- tent. To find the Superficial Content rule. Multiply the slant height by half the circumference of tike base, and the product will be the upright sur- face. — To which the area of the base may be added, for the whole surface .Mensuration of Solids. Note. This rule will serve for the surface of all pyramids. Perhaps it may he proper here to acquaint the learner, that the slant height of any pyramid is not the height from one of the corners of the hose to the vertex or top, hut from the middle of one side of the base. And the perpendicular height of a pyramid, is a line drawn from the vertex, to the middle or centre of the base; henee it will he necessary to find the dis- tance between the centre of the hase of a pyramid, and the middle of one of the sides. — This distance may always be found by multiplying the tabular per- pendicular in section V. p. 84, by one of the sides of the base; then, if to the square of this number, you add the square of the perpendicular height of the pyramid, the square root of the sum will give the slant height. EXAMPLES. 1. Required the surface of the foregoing pyramid. To the square of the perpendicular height De 15 feet or 180 inches, add the square of de 9.25, the dis- tauce from the centre e, of the base, to the middle d of one of the sides ; which, in a square base, is always equal to half the side. The square root of the sum, viz. 32485.5625 is 180.24 nearly, the slant height Drf. IV ow half the circumference of the base is 37, which multiplied by 180.24, gives 666S.88 inches for the up- right surface ; to which add 342.25 the area of the base, the sum is 7011.13 inches, the whole surface equal to 48.69 feet nearly. Demonstration of the rule. Every pyramid is a third part of a prism, that has the same base and height (by Euclid, XII. 7.) That is, the solid content of the pyramid ABD (in the last figure) is one third part of its circumscribing prism ABEF. Now the solidity of a prism is found Mensuration of Solids. 145 by multiplying the area of the base into the height; therefore the solidity of a pyramid will be found by multiplying the area of the base by the height, and taking one third of the product. You may very easily prove a triangular pyramid to be a third part of a prism of equal base and altitude, mechanically, by making a triangular prism of cork, and then cutting that prism into three Pyramids, in a diagonal direc- tion. 2. Let ABCD be lar pyramid, each side of the base being 21.-5 inches, and its perpendicular height 16 feet 5 required the solidity and sur- face. First, the area of the base, by sect. V. is 200.19896 inches, which multiplied by 64?, the third part of the height in inches, gives 12812.73344 inch- es, the solidity, :qual to 7.41477 feet. Again, the distance from the centre of the triangle ABC to the middle of one of the sides, as d in the side AC, is 6.20652 ; to the square of this number, which is 38.5208905104, add the square of the height 36864 inches, the square root of the sum 36902.5208905104, is 192.10028 inches, the slant AJ height dD. Hence the upright surface is 6195.23403 inches, and the whole surface 6395.43299 inches, or 44.41272 feet. O H6 JMenmration of Solids. 8. Let 1BCDEFGH be s pyramid, whose base is a hep- ■ach side of it being 1/5 inches, ami the perpendicular heigh*! of the pyramid. Ill, ee< ; required the solidity and superficies. First, 1.3 multiplied by 1.0382017, the tabular perpen- dicular for a heptagon (sect. VIII.) gives 15.573<)255 inches, the distance from the centre 1 to the middle of the side AB; this multiplied by 52.5, half the sum of the sides of the base, gives 817.63108873 inch- es, the area of the base. This last number multiplied by 4.3, one third of the height, and divided by 144, will give feet, the solidity. — Again, if to the square of 15.5739255,' vou add the square of the height in inches, the square root of the sum 26486.5471." 31 8 will be 162.74688, the slant height of the pyramid 5 this multiplied by 52.5 gives 8544.21 ti inches, the upright surface, to which add the area oi the base, and the whole surface will be 9361.8422887* inches, or 65.0128 feet. By Scale and Compasses, First, find a mean proportional between 15.57 an< 52.5, by dividing the space between them into twc equal parts, and you will find the middle point to be 28.6, the side of a square equal in area to the base oi the pyramid; then extend the compasses from 12 to 28.6, that extent will reach from 4.5 (twice turned over) to 25.55 feet, the solid content. Or, extend the compasses from 1 to the area of the base, that extent Mensuration of Solids. 147 will reach from one-third of the height to the solidity ; the dimensions being all of the same name. 4. There is a triangular pyramid, the sides of its base are 13, 14, and 15 feet, and its altitude 63 feet, what is the solidity ? Ms. 1764 feet. . 5. Eacli side of the base of a hexagonal pyramid is 10, and the perpendicular height 43, what is the soli- dity and superficies ? q C Solidity 3897.1143 feet. . ' \ Superficies 1634.580328 feet. 6. Required the solidity and surface of a square py- ramid, each side of the base being 3 feet, and the per- pendicular height 24 feet. - C Solidity 72 feet. * aw{ " I Surface 153.2808 feet. 7. A square pyramidal stone, whose slant height is 21 feet, and each side of its square base 30 inches, is to be sold at 7s. per solid foot ; and the polishing of the upright surface will cost 8d. per foot ; what will be the expence of the stone when finished ? Ms. Us : 15s. : 8^d. 8. The spire of a church is an octagonal pyramid (built of stone) each side of the base being 5 feet 10 inehes, and its perpendicular height 45 feet ; also each side of the cavity, or hollow part, at the base is 4 feet 11 inches, and its perpendicular height 41 feet ; I de- sire to know how many solid yards of stone the spire contains ? ("Solidity of the whole 2464.509614 feet. Ms. 1 The cavity 1595.180393 feet. |_And the stone-work 32.19738 yards. § V. Of a Ctlisder. A cylinder is a solid, having its bases circular, equal, and parallel, in form of a rolling stone. Ml Mensuration of Solid*. To find the Solidity. RULE. Multiply the area of the base by the length, and the product is the solid content. EXAMPLES. l. Let ABC be a cylinder, whose diameter AB is 21.5 inch- es, and the length CD is 16 feet, the solid content is required. First, sqnare the diameter .21.3, and it makes 462.2") ; which multiply by .7854, and the product is 363.03113: then multiply, this by 16, and the product is 5808.8184. Divide this last product by 144, and the quotient is 40.339 feet, the solid content By Scale and Compasses. Extend the compasses from 13.54 to 21.5, the dia- meter, that extent (turned twiee over from 16, the length) will at last fall upon 40.34, the solid content. Note. 13.54 is the diameter of a circle whose area is 144 inches and > to the diameter of any eylinders base in inches, so is the length of that cy- linder, in feet, to a fourth number; and so is this fourth number, to the solid content of the cylinder in feet. When the circumference is given in inches, and the length in feet, extend from 42.54 to the circumference of the cylinder, that extent will reaehfrom the length, Mensuration of Solids. 149 turned twice over, to the content in feet. For. as 42.54 is to the circumference of any cylinder, in inches, so is the length of that cylinder in feet to a fourth num- ber; and so is this fourth number to the solidity of the cylinder in feet. 42.54 is the circumference of a circle whose area is 144. To find the Superficial Content. First, (by chap. I. sect. IX. prob. 1,) find the circum- ference of the base 67.54, which multiplied by 16, the product is 1080.64; and this divided by 12, the quo- tient is 90.05 feet, the upright surface : to which add 5.04 feet, the sum of the areas of the two bases, aud the sum is 95.09 feet, the whole superficial content. 2. If a piece of timber be 96 inches in circumfe- rence, and 18 feet long, how many feet of timber are contained in it, supposing it to be perfectly cylindri- cal ? Jns. 91.676 feet. 3. If a piece of timber, perfectly cylindrical, be 86 inches in circumference, and 20 feet long: how many solid feet are contained therein? Jlnj± 81.74634 feet. 4. The diameter of the base of a cylinder is 20.75 inches, and its length 4 feet 7 inches ; what is its so- lidity and surface r C Solidity 10.7633 feet. I Surface 29.595 feet. 5. I have a rolling stone, 44 inches in circumfe- rence, and am to cut off 3 cubic feet from one end; whereabouts must the section be made ? . At 33.64 inches, 6. A person bespoke an iron roller for a garden, the outside- diameter to be 20 inches, the thickness of the metal i\ inch, and length of the roller 50 inches; now supposing every cubic inch to weigh 41- ounces, what will the whole come to at3id. per pound ? f Solidity 4358.97 inches. Ans. < Weight 1157.8514 pounds. (Co8t7. 15:13s. : 7d. o 2 150 Mensuration of Solids, § VI. Of a Cone, A cone is a solid, having a circular base, and grow- ing proportionally smaller, till it ends in a point, called the vertex, and may be nearly represented by a sugar loaf. 2b find the Solidity. RULE. Multiply the area of the base by a third part of the perpendicular height, and the product is the solid con- tent. EXAMPLES. 1. Let ABC be a cone, the di- ameter of whose base AB is 26.5 inches, and the height DC 16.5 feet : required the solidity. First, square the diameter 26.5, and it is 702.25 ; this multiplied by .7854 gives 551.54715 inches, the area of the base ; which mul- tiplied by 5.5, one third of the height, gives 3033. 509325; this divided by 144 gives 21.066 feet, the solid content. By Scale and Compasses. Extend the compasses from 13.54 to 26.5, the diame- ter, that e ent turned twice over from 5.5 (a third part of the height,) it will at last fall upon 21.06 feet, the content. — See the remark on the number 13.54, in section V. Mensuration of Solids, 151 To find th$ Superficial Content The rule given in sect. IV. for finding the superficies of a pyramid, will serve to find the superficies of a cone. First, find the slant height thus : To the square of the radius, 13. 5, which is ±75.5625, add the square of the height in inches, 39204, the square root of the sum, 39379.5625, is 198.4428, the slant height. The radius of the base, 13.25, multiplied by 3.1416 gives half the circumference of the base 41.6262, and this pro- duct multiplied by the slant height, gives 8260.4196S136 for the upright, or convex surface ; to which add the area of the base 551.54715,and the sum is 8811.96683136 inches, the whole surface, equal to 61.194 feet. Demomstration. Every cone is the third part of a cylinder of equal base and altitude (by Euclid, XII. 10.) The truth of this may easily be conceived, by only considering, that a cone is but a round pyramid ; and therefore it must needs have the same ratio to its circumscribing cylinder, as a multangular pyramid hath to its circumscribing prism, viz. as 1 to 3. For, the base of the multangular pyramid may consist of such a number of sides, that the difference between its circumference, and that of the circle, will be less than any assignable magnitude. The curve superficies of every right cone, is equal to half the rectangle of the circumference of its base into the length of its side. For the curve surface of every right cone is e- qual to the sec- tor of a circle, whose arch BC is equal to the peri- phery of the base \ / of the cone, and radius AB equal to the slant side of the cone: which / / Mensuration of Solids. will appear very evident, if you cut a piece of paper in the form of a sector of a circle, as ABC, and bend the aides AB and AC together, till they meet, and you will find it to form a right cone. I have omitted the demonstrations of tin 1 superfi- cies of all the foregoing solids, because I thought it needless, they being all composed of squares, paral- lelograms, triangles. &e. which figures arc all demon- strated before. And if the area of all such figures as compose the surface of the solid, be found separately, and added together, the sum Mill be the superficial content of the solid. 2. The diameter of the hase of a cone is 10, and its perpendicular height 6S.1 ; required the solidity and superficies. a 5 Solidity 1782.858. I Superficies 1151.134076. 3. What is the solidity of an elliptical cone, the greater diameter of its base being 15.2, the less 10, and the perpendicular height 22 ? dns. 875.4592. 4. What is the solidity and superficies of a cone, whose perpendicular height is 10.5 feet, and the cir- cumference of its base 9 feet ? a S Solidity 22.56093 feet. X Superficies 54.1336 feet. 5. What is the solidity and superficies of a cone, whose slant height is 32 feet, and the circumference of its hase 24 feet ? a J Solidity 4S5.4433316 feet. ' i Surface 429.83804 feet. 6. What will the painting of a conical church spire come to at 8d. per yard; supposing the circumference of the base to be 61 feet, and the perpendicular height 118 feet rw - • $ Superficies 3789.76 feet. %ms - ^Expence J.14 : Os. : bU, Mensuration of Solids* 153 § VII. Of the Frustum of a Ptramid. A frustum of a pyramid is the remaining part, when the top is cut off by a plane parallel to the base. To find the Solidity. GENERAL RULE. Multiply the areas of the two bases together, and to the square root of the product add the two areas; that sum, multiplied by one third of the height, gives the solidity of any frustum. rule 11. If the Bases be Squares. To the rectangle (or product) of the sides of the two bases add the sum of their squares : that sum, being multiplied into one-third part the frustum's height, will give its solidity. RULE III. If the Bases be Circles. To the product of the diameters of the two bases add the sum of their squares 5 this sum, multiplied by the height, and then by .2618, or one-third of .7854^ the last product will be the solidity. RULE IV. If the Bases be regxdur Polygons. Add the square of a side of each end of the frus- tum, and the product of those sides into one sum; multiply this sum by one-third of the tabular area be- 154 Mensuration of Solids. longing to the polygon (sect. VIIT. p. 99.) and this product by the height, for the solidity. EXAMPLES. 1. Let ABCD he the frus- tum of a square pyramid, each side of the greater base 18 inches, each side of the Jo** 12 inches, and tbe height 18 feet; the solidity is re- quired. BY THE GENERAL RULE. The square of 18 is 324, and the square of 42 is 141; also the product of these two squares is 46656, the square root of which is 216, a mean area. Area of the greater base - - - - 324 Area of the less base 144 Mean area - - - 216 ,'XI » Sum - 6S4 which multiplied by 6, one third of 4he height, gives 4104; this divided by 144 gives 28.5 feet the solidity. BY RULE 2. Product of the sides of the bases 18 and 12 is Square of the greater side 18 is - - - - Square of the less side 12 is Then proceed as above. Sum 216 324 144 684 Mensuration of Solids. 4 >;> To find the Superficial Content. Multiply half the sum of the perimeters of the two bases by the slaut height, and to that product add the areas of the two bases for the whole superficies. Note. The slant height of any frustum, whose ends are regular polygons, is a line drawn from the middle of one side of the less end, to the middle of its paral- lel side at the greater end. And, the perpendicular height, is a line drawn from the centre of the less end, to the centre of the greater; or it is a perpendicular let fall from the middle of one of the sides of the less end, upon the surface of the greater end. Hence the slant height, and perpendicular height, will be two sides of a right angled triangle ; the base of which will be equal to the difference between the radii of the inscribed cireles of the two ends of the frustum. And this base may always be found, by multiplying the dif- ference between a side at each end of the frustum, by the tabular perpendicular in section VIII. p. 100. The perpendicular height of the pyramid of which any frustum is a part, may readily be found, by saying, as the base found above, is. to the perpendicular height of the frustum 5 so is the radius of the inscribed cir- cle of the frustum's base, to the perpendicular height of the pyramid. The radius of the inscribed circle is found by multiplying a side of the base, by the tab- ular perpendicular, section VIII. Example. Required the superficies of the foregoing frustum. The perimeter of the greater base 72 inches. The perimeter of the less base is 48 inches. Sum - - 120 Half sum - - 00 i56 Mensuration of Solids. From 19, a side of the greater end, take 12, a side of the less; the difference 6, multiplied by the tabular perpendicular .5, gives 3. The perpendicular height of the frustum in inches is 216 : to the square of 216, which is 46656, add the square of 3, which is 9, the square root of the sum 46656 is 216.02083 inches, the giant height, which multiplied by 60 gives 12^61.2498 inches. To this product add the areas of the two ends 324, and 144, and the whole surface will be 13429.2498 inches, or 93.2586 feet. Example 2. Let ABCD be the frustum of a triangular py- ramid, each side of the greater base 25 inches, each side of the less base 9 inches, and the length 15 feet ; the solid content of it is required. BY RULE 4. The square of 25 is The square of 9 is Product of 25 and 9 is 625 81 225 Sum 931 The tabular area is .433013, one-third of which is .1443377, CI this multiplied by 931, and then by 15, produces 2015.6759805, which divided by 144 3 the quotient is 13.9977 feet, the solidity. OR THUS, BY THE GENERAL RULE. The square of 25, multiplied by the tabular area, (section VIII.) gives the area of the greater base 270.63312.^: in a similar manner the area of the less base wiirbe found to be 35.07405 ; the square root of the product of these two areas 9492.200569805625 is 97.427U25. Mensuration of Solids. 157 Area of the greater base - - - 270.633125 Area of the less base - 35.074053 Mean proportional between them - 27.427925 Sum - - 403.135103 which multiplied by 5, one third of the height, and di- vided by 144, gives 13.9977 feet, the solidity as be- fore. To find the Superficial Content. The perimeter of the greater base is 75 inches. The perimeter of the less base is - 27 inches. Sum - 102 Half sum - 51 From 25, a side of the greater end, take 9, a side of the less, the difference 16, multiplied by the tabular perpendicular .2886751, gives 4.6 1 880 i 6. — The per- pendicular height of the frustum in inches, is 180; to the square of 180, which is 32400, add the square of 4.6188016, which is 21.33332822, is 180.0592 inches, the slant h. ight ; which multiplied by 51 gives 9183.0192 inches. To this product add the areas of the two ends 270.63312 -, and 35.074053, and the whole surface will be 9488.726378 inches, or 65.893y33 feet. Mem \\\( \) to be id" frustum of a pyramid, having an octagonal base. i-a< \\ suit* of tt being 9 inches, each side of tlie less 1) 5 inches, ami the length, or height, 10.8 feet ; the solidity is required. by rule 4. The square of 9 is 81 The square of 5 is 25 Product of 5 and A 9 is - - 45 >um - 151 Tabular area is 4.828427, one third of which is 1.6094756 ; this multiplied by 151, and then by 10.5, produces 2551.8235638, which divided by 1.44, the quo- tient, is 17.721 feet, the solidity. To find the Superficial Content. 72 inches. 40 112 56 The perimeter of the greater base is The perimeter of the less base is Sum - Half sum - From 9, a side of the greater base or end, take 5, a side of the less, the remainder 4 multiplied by the tabu- lar perpendicular 1.2071068, gives 4.8284272. The perpendicular height of the frustum in inches is 126 ; to the square of 126, which is 15876, add the square of / i \ Mensuration of Solids. 159 4.8284272, which is 23.3137092257, the square root of the sum 15899.3137092237, is 126.09248, the slant height ; which multiplied by 56 gives 7061.17888 inch- es. To this product add the areas of the two ends, 391.102587, and 120.710675 (found by rule 2, section VIII.) and the whole surface will be 7572.992142 in- ches, or 52.590223 feet. Demonstration. From the rules delivered in the IVth and Vlth sections, the preceding rules may easi- ly be demonstrated. Suppose a pyramid ABY, to he cut by a plane at ab 9 parallel to M its base AB, and it were required / j \ to find the solidity of the frustum, or part AaoB. i i Let D=AB, a line, or diameter , of the greater base. h d=.ab, a similar line of the l\ less base. /*=PC, the height of thefrus turn. A=area of the greater base, ffczrarea of the less. AP— oC : PC : : aC : CV, by similar trian-les ; D d d hd a viz. — - — : h i : — : =CV. 2 2 D— d hd AD but PC+CV—Zi-f- -r =PV D— rf D— d hd a X — = solidity of the pyramid abV l)—d 3 /*— D A '■11 X — zzsolidity of the pyramid ABV. D— d 3 JiD A hd a AD— ad. h Hence X X--= X— the solidi- D— d 3 D— d 3 D— d 3 i«)i> Mensuration of Solids. (y of the frustum ABab. Now all similar figures are h other as the squares of their homologous sides. (Euclid VI, ami 19th ami 20th ; also XII. and 2d.) A a Therefore A t a : : D* :d* or — =r — put each of these equal to ?7i ; then A=7JiD 2 and « — mcZ% and multiply- ing these qu anti ties together, we get aA = m 2 D 2 de, con- sequently \/ aA=mY)d. For A and a substitute their values in the expression for the frustum's solidity, and it 7*il) 3 — mdth m~- d* k becomes X — = X — Xm=D 2 -f-Dc/-f d x -, D — d 3 D— • d 3 h h X — x m = m D 2 4- m *^+ w ^* X — ; restore the values, 3 3 of A and a, &c. and the rule becomes A-j-v'aA-j-aj h X — which is the general rule. 3 Corollary 1. If the bases be squares, of which D and h d are each a side, then D*-fD, the circumference of the less base of the frustum ; AD the slant height=S, now, because simi- lar arches of circles are as their radii. DE : BC : : AD : AB S/> viz. P : p : : S : — P Sjp P— .p Hence B D=A D— A B=S = xS P P PS The area of the sector A D E= — , and 2 P Sp Sp* The area of sector ABC= — x — = — 2 P 2P Also ADE — ABC = curve surface of the frustum, PS 8p* PS T — ®*S P*_ p* V+p BDEC= = = X &= — : 2 2Y 2P 2V % V-p X- xS> which is the rule. P v2 ifvj Memura\ - UiL<. Note. From the foregoing demonstration we have the following rule for finding the area of a segment of a sector B 1) E C, or the front of a circular arch, built with stones of equal length. KUI.E. Multiply half the sum of the bounding arches (RE and BC) by their distance (BD), and the product will £?ive the area. Example 4. If each side of the greater end of a piece of squared timber be 25 inches, each side of the less end 9 inches, and the length 20 feet ; how many solid feet are contained in it ? Jins. By rule 2, 43.1018 feet. 5. If a piece of timber be 32 inches broad, and 20 inches deep, at the greater end; and 10 inches broad, and G inches deep, at the less end ; how many solid Feet are contained therein, the length being 18 feet? Jins. By the general rule, 37.3316 feet. 6. A portico is supported by four pillars of marble, each haying eight equal sides, whose breadth at the greater end is 7.5 inches, and at the less end 1* inches, and their length 12 feet 9 inches; I desire to know how many solid feet they contain, and what they will come to at 12s. lOd. per solid foot ? j ns C Solidity 60.52927823 feet. I Expence I. 3S : 16s. : 9}d. 7. In a frustum of a square pyramid, each side of the greater end is 5 feet, each side of the less end 3 feet, and the height 8 feet; required the height and solidity of that pyramid of which this frustum is a part ? n 5 TJie height is 20 feet, see p. 143. JlnS ' I The solidity 166$ feet. 8. If each side of the greater base of the frustum of a hexagonal pyramid be 13, each side of the less base S, and the length 24, what is the solidity and superfi- cies ? '^ C Solidity Vo04.412S96. *\ Superficies 2141.7G42, Mensuration of Solids. 103 § VIII. Of the Frustum of a Cone. A frustum of a cone, is that part which remains when the top end is cut oft* by a plane parallel to the base. The solidity may be found either by the general rule, or rule 8, of the preceding section. The superficies may also be found by the rule given in that section. EXAMPLES. 1. Let ABCD be the frus- tum of a cone, whose greater diffmeter CD is IS inches, the less diameter AB 9 in- ches, and the length 14.25 feei ; the solid content is re- quired. BY THE GENERAL RULE. The square of flp is 324, which multiplied hy .7854 gives 254.4696 inches, area of the greater base. The square of 9 is 81, which multiplied by .7854 gives 63.6174, area of the less hasc. The square root of the 16188.69433104 is 127.23AS. Area of the greater base Area of the less base, Mean proportional product of these areas 254.4696 - ' -• - 63.6174 - 127.2348 Sum 445.3218 This multiplied by 4.75, one third of the height, gives 2115.27855, whieh^divide by 144, and the quo tient is 14.68943 feet, the solidity. 16* Jhnsuraiion of Solids. Note. The diameter of the less base being exact- ly half that (if the greater, in this example', the ope- ration might have been performed shorter: for the area of the h'^s base in sueh eases is one fourth of that of the greater, and the mean area double that of the less base, or half that of the greater. OR THUS, BY RULE 3. The square of 18 (the greater diameter) is 324, and the square of 9 (the less diameter) is 81, and the rect- angle, or the product of £8 by 9, is 162, the sum of these three is 567, which multiplied by the height 11.25 feet, gives 8079075 ; which multiplied by .2618, and divided by 144, gives 14.68943 feet as before. To find the Superficial Content. The circumference of the greater base, by chap. I. section IX. is 56.5488 And the circumference of the les^s base is 28.2744 Sum 84.8232 Half sum 42.4110 To the square of 4.5, the difference between the ra- dii of the two bases, add the square of 171, the per- pendicular height in inches ; the square root of the sum 2 ) 261.27, is 171.0592 inches, the slant height; which multiplied by 42.4116 produces 7254.894J6672, to which add the sum of tm\,areas of the two ends, and the whole superficies will be 7572.98136672 inches, or 52.r> feet. 2. If a piece of timber be 9 inches in diameter at the less end, 36 inches, at the greater end, and 24 feet long ; how many solid feet are contained therein ? Jns. 74.2203 feet. 3. If a piece of timber be 136 inches in circumfer- Mensuration of Solids. 165 ence at one end, 32 inches at the other, and 21 feet long 5 how many solid feet are contained therein ? Jns. 95.34816 feet. § IX. Of a Wedge. A wedge is a solid, having a right angled parallel- ogram for its base, and two of its sides meeting in an ed^e. To find the Solidity. RULE. To twice the length of the base add the length of the edge, multiply the sum by the breadth of the base, and that produet by the perpendicular height of the wedge; and one -sixth of the last product will be the solidity. EXAMPLES. 1. The perpendicular height 01, of a wedge is 24.S inches ; the length CK of the edge, 110 inches; the length AE of the base 70, and its breadth AB 30 inch- es; what is the solidity ? 70 length of the base AE. ■3' 7500 140 24.8 perp. 01. 110 length of the edge CK. 60000 250 sum. . 30000 30 breadth of the base. 15000 7500 product. 6)186000.0 31000 inches; which divided by 1728 gives 17.9398 feet, the solidify. '-to Mensuration of Solnh 2. Required the solidity of a wedge, whose altitude ^s 11 inehes, in inches, and the length and breadth of its base 32 and 4.8 inches ? Jlns. 892J inches. 3. Required the solidity of a wedge, whose altitude is J feet -i inches, length of the edge 3 feet 6 inch- es, length of the base 5 feet 4 inches, and breadth 9 inches ? Jns. 1, 13 tut J, &c. feet. DEMONSTRATION OF THE RULE. Let ABCDEF represent a wedge : now when the length of the edge CD is equal to the length AE of the base ABF E, the wedge is evid ntly equal to half a parallelopipe- dt» ABGCHDEF, having the same base and altitude as the wedge ; and this will al- ways be true, whether the end ABC of the wedge be perpendicular to its base ABFE, or inclined as ABI, since parallelopipedons standing on the same base and between the same parallels are equal to each other {Euclid, XT. and 29.) But when the edge CD of the wedge is longer or shorter than the base, by any quanti- ty DK or CI, it is evident that the wedge will be great- er or less than the half parallelopipedon aforesaid, by a pyramid whose base is EFD, or ABC, and perpendic- ular altitude DK or CI. Let the length AE of the base of the wedge be represented by L, and the breadth AB by B : call the length of the edge I, and the per- pendicular height AC, «C, or 01, h. Then, the so- BL lidity of the wedge will be expressed by v^ or B/i Bhh L 2 *— XL= , when the edge is of the same length Mensuration of Solids. 1G7 as the base : The solidity of the pyramid ABCI will Bh L—l be expressed by X -when the length ID of the 2 3 edge is less than that of the base ; and the solidity of Bh I— L the pyramid EFDK will be expressed by x ? 2 3 when the length of the edge GK is greater than that Bhh Bh l—L 2L+1 of the base. Hence, •-£ X -= X Bh, 2 2 3 6 the rule, when the edge is longer than the base, and Bhh Bh L—l 2L+1 5 X =- XBh, the rule when the 2 2 3 6 edge is shorter than the base. § X. Of a Vrismqid. A prismoid is a solid somewhat resembling a prism ; its bases are right angled parallelograms, and parallel to each other, though not similar, and its sides four plain trapezoidal surfaces. To find the Solidity. RULE. To the greater length add half the less length, mul- tiply the sum by the breadth of the greater base, and reserve the product. Then, to the less length, add half the greater length, multiply the sum by the breadth of the less base, and add this product to the other product reserved j multi- 16S ration of Solids. ply the sum by a third part of the height, and the product is the solid content. EXAMPLES. 1. Let ABCDEFGft be a prismoid given, the length of the greater base AB 38 inches, and its breadth AC 16 inches; aud the length of the less base EF 30 inches, and its breadth 12 inches, and the height 6 feet; the solid content is required. To the greater length AB .?8, add half EF the less length 15, the sum is 53 ; which multiplied by in, the greater breadth, the product is 848 : which reserve. Again, to EF 30, add half AB 10, and the sum is 49; which multiplied by 12 (the less breadth EG) the product is 588; to which add 848 (the reserved pro- duct,) and the sum is 1436 ; which multiply by 2, (a third part of the height,) and the product is 2872 y divide this product by 144, and the quotient is 19.94 feet, the solid content. 2. One end of a prismoid is a square, each side of which is 13 ; the other a right angled parallelogram, Mensuration of Solids. 169 whose length is 12 and breadth 5; what is the solidity, the perpendicular height being 20 ? Jlns. 22&3^. 3. In the neighbourhood of Newcastle, and in the county of Durham, &c. the coals are carried from the mines in a kind of waggon, in the form of a prismoid. The length at the topis generally about 6 feet 9^ inch- es, and breadth 4 feet 7 inches ; at the bottom the length is 3 feet 5 inches, and breadth 2 feet 5] inches; and the perpendicular depth 3 feetfli inches. Requir- ed the solidity. Jlns. 126340.59375 cubic inches, or 73.11376 feet. 4. How many gallons of water, reckoning 282 cubic inches to a gallon, are contained in a canal 304 feet by 20 at top, 300 feet by 16 at bottom, and 5 feet deep ? Ans. 166590.6383 gallons. DEMONSTRATION OF THE RULE. The prismoid is evidently compassed of two wedges whose bases are equal to the bases of the prismoid, and their height equal to the height of the prismoid. Let L equal the length of the greater base AB, which of course will be the length of the edge of the less wedge ; I equal the length of the less base EF, or the edge of the greater wedge : B the breadth of the great- er base AC, b the breadth of the less base EG, and h 1~0 . 'ids. th<> common height Then the solidity of the wedge whose base is ABDC will be expressed by 2L+1 l y XB//, and the solidity of the wedge whose base is 6 2L+1 EFIIG by xbh ; henec the solidity of the 6 2Z4-L 2/+L whole prismoid will be XBA, X X 6 6 L+l L h bh— XB; -f/-| Xb; X— , which is the rule 2 2 3 exactly. This rule is demonstrated section II. Prob. XIV. of Emerson's Fluxions, and in a similar manner at page 179, second edition, of Simpson's Fluxions ; also in Holiday's Fluxions, page 302. L+/ B+& Note. If =M, and =?n, the above rule 2 2 h becomes BL-J-W-f4rM?n ; X — , that is, To the sum of 6 the areas of the two ends add four times the area of a section parallel to, and equally distant from, both ends; this last sum multiplied into one-sixth of the height, will give the solidity. It is shewn in proposition III. page 456, part IV. of Dr. Button's mensuration, quarto edition, that this last rule is true, for all frustum's whatever, and for all sol- ds whose parallel sections are similar figures. And, Mr. Moss, in his Guaging, page 175, third edition, Mensuration of Solids. 171 says, that it is nearly true, let the form of the solid be what it will. The following rule for measuring a cylindroid given by Mr. Hawney, from Mr. Everard's Gu aging, has been by some teachers considered as erroneous, others again have imagined it to be true. The general rule, which has just been described, will most certainly give the content either exactly, or as near as possible, pro- vided the figure of the middle section between CD and AB (see the following figure) can be accurately deter- mined. Half the sum of CD and AB, in the follow- ing example, is 85, and half the sum of AB and EF is 20, the two diameters of the middle section ; therefore it must either be an ellipsis, or a curve of the oval kind, whose area, perhaps, cannot be easily determin- ed. If the section be an ellipsis (and it cannot mate- rially differ from one) the rule given by Mr. Hawney exactly agrees with the above general rule. But Mr. Moss, in prob. VI. sect. VIII. of his Guaging, has shewn that the figure of the middle section between CD and AB, can never be an ellipsis, unless the par- allel ends AB and CD are similar ellipsis, and simi- larly situated ; viz. the transverse and conjugate diam- eters of each end, respectively parallel to each other, and this circumstance can never happen but when the solid is the frustum of an elliptical cone. The rule therefore cannot be strictly true, though sufficient- ly near for any practical purpose. To measure a Ctlindroid ; or, the Frustum of an elliptical Cone. RULE. To the longer diameter of the greater base, add half the longer diameter of the legs base, multiply the I7J Mensuration of Solids. sum by (he shorter diameter of the greater base, and e the product Then, to the longer diameter of the less base, add half the longer diameter of the greater base; and multiply the sum by the shorter diameter of the less add tliis product to the former reserved pro- duet, and multiply the sum by .2018 (on -third of ) and then by the height, the last product v ill be the solidity. EXAMPLES. 1 . Let ABCD be a eylindroid, whose bottom base is an ellipsis, the transverse diameter being 41 inches ; and the conjugate diameter 14 inches; and the upper base is a circle, of which the diameter is 26 inches ; Mensuration of Solids. 173 and the height of the frustum is 9 feet ; the solidity is required. 44= CD 26=AB I3=half AB 22=half CD — — 57 sum 48 sum 14=EF 26=AB 228 288 57 96 798 product reserved. 1248 product. * 798 add reserved product. 2046 sum. 2046 .2618 16368 2046 12276 4092 533.6428 9= height. 4820.7852 which, divided by 144, gives 33.4776 feet the solidity. 2. The transverse diameter of the greater base of a cylindroid is 13, conjugate 8; the transverse diameter of the less base 10, and its conjugate 5.2. If the length of the cylindroid be 20, what is the solidity ? Ms. 1203.2328. 3. I desire to know what quantity of water an ellip- tical bath will hold, the longer diameter .at the top be- Q 2 Mmsuration of Sulich. 2 feet, ami shorter?; (he longer diameter at the boll 0111 10 feet, and the shorter 6 feet ; and the depth i feet : reckoning 2%2 eubic inches to a gallon ? Jlns. 1379.6303 gallons. § XI. Of a Sphere or Globe. A sphere, or globe, is a round solid body, every part of its surface being equally distant from a point within, called its eentre; and it may be conceived to be form- ed by the revolution of a semicircle round ils diameter. To find the Solidity. RULE I. Multiply the cube of the diameter by .5236, and the product will be the solidity. RULE II. Multiply the diameter of the sphere into its circum- ference, and the product will be the superficies; which multiplied by one-third of the radius, or a sixth part of the diameter, will give the solidity. EXAMPLES. 1. Required the soli- dity of a globe, ABCD, / whose diameter AB or $.„ CD is 20 inches. \ Mensuration of Solids. 175 BY RULE 1. 20X20X 20 X .5236=4188.8, the solidity. BY RULE 2. 20X3.1416=62.832, the circumference. 20X62.832=1256.64. the superficies. 20xl256.64~6=41SS.S, the solidity, in inches, = 2.424 feet. 2. The diameter of the earth is 7970 miles, what is its surface in square miles, and solidity in cubic miles ? a ( Surface 199537259.44 miles. * i)lS ' I Solidity 265078559622.8 miles. 3. The circumference of a sphere is 1, what is its solidity and superficies ? a S Soli(lit y -0168868. " * \ Superficies .3183099. 4. What is the solid and superficial content of a globe, whose diameter is 30 ? - J Solidity 14137.2. J1VS ' I Superficies 2827.44. 5. The circumference of a globe is 50.3, what is its solidity and superficies ? ^ S Solidity 2149.073728. £ Superficies 805.3526, 6. A globe, ti cube, a cylinder, All three in surface equal are,* viz. 3.1416 In solidity what do they ditt'er .? each. Jlns. The solidity of the globe is .5237, the cube .378877, and cylinder .4275176 ; so that of all solids, under the same superficies, the globe is the greatest. * First, for the globe. If the superficies of a globe be divided by 3.1416, the quotient will be the square of the diameter, the square root of whieh will be the diameter. — Hence the solidity is easily found, 170 Mensuration. of Solids. Second, for the cube. The cube lias 6 equal surfaces.* therefore the whole surface divided by 6 ? will give the square of the side of the cube ; the square ruol will be the hide. — Hence the solidity is found. Third, for the cylinder. The diameter and depth of the cylinder are here supposed equal. — The superficies of a cylinder, whose diameter and depth are each an unit, will be found to be 4.7124; and the superficies of similar cylinders are to each other as the squares of their diameters, therefore 4.7124 is to the square of 1, as the superficies of the cylinder is to the square of its diameter; the square root of which will be the diameter, and likewise the depth. — Hence the solidity of the cylinder is found. DEMONSTRATION OF THE RULE, &C. Tliat every Sphere is two-thirds of its Circumscribing Cylinder, may be thus proved. Let the square ABCD, the quardrant CBD, and the right angled triangle ABD, be supposed all three to revolve round the line BI) as an axis : Then will the square generate a cylinder, the quadrant a hemisphere, and the triangle a cone, all of the same base and altitude. By Euclid, I. and 47, FD*--FH*-fDH% but FIX EH, and GHrrDH because EH is parallel to CD, therefore EH*=FH s 4-GH a ; and as circles are to each other as the squares of their diameters, or radii (Euclid, XII. and 2,) it follows that the circle describ- ed by EH, is equal to the two circles described by FH Mensuration of Solids. 177 and GH ; take away the circle described by FH, and there remains the circle described by EH— -circle de- scribed by FH (viz. the annul us, or ring, described by EF between the sphere and cylinder)— circle describ- ed by GH. And it is evident, that this property will hold in every section similar to EH ; but the cone, hemisphere, and cylinder, may be conceived to be made up of an infinite number of such sections, therefore when the hemisphere is taken out of the cylinder, the remaining part is equal to the solidity of the cone ; but the cone is one-third of the cylinder, therefore the hemisphere must be two-thirds, and what is here proved with respect to the halves of the proposed so- lids, holds equally true with the wholes. Therefore every sphere is two-thirds of its circumscribing cylin- der. Now if D be the diameter and height of a cylin- der, its solidity will be D*x.~S54xDzrD 3 X.78.34, hence the solidity of its inscribed sphere will be D 3 X .7S3-1X|=D 3 X^236, which is the first rule. 'The Surface of a Sphere is equal to the Curve Surface of its Circumscribing Cylinder, Take FK, an extremely small part of the quadran- tal arch CB, and suppose lines LM and EH to be drawn through K and F parallel to each other. Then, because FK is extremely small, it may be considered as a straight line, and the angle FKD as a right an- gle. The figure FKMH, by revolving round MH, will form the frustum of a cone, whose slant side is FK ; and the figure ELMH, by revolving in the same manner, will form a cylinder, whose perpendicular height is LE. But the surface of the frustum is equal to FKXhalf the sum of the circumferences of two circles whose radii are KM and FH ; or these be- ing extremely near to each other, the surfaee of the frustum will be equal to FKX circumference of a cir- ele, whose radius is KM, or FH. Let C— the circum- 178 Mensuration of Solids. ferenee of a circle whose radius is ( D er EH, ami c— the circumference of a circle whose radius is FH or KM: then the surface formed bj the revolution of FkMH about Mil, is equal to cXFKj ami the cylin- drical surface formed by the revolution of ELMH, is equal to CxLE. But. the triangles FeK and DMK are equiangular and similar: for FKD and eKM are right angles; take away the common angle eKG, and there remains FKc— DKM; and the angles FeK and KMD are right angles : therefore DK or CD : KM : : KF : Ke or LB. But the circumferences of circles are to each other as their radii, therefore C -. c : : CD or LM : KM, consequently. C : c : : KF : Ke or LE. Hence CxLE=cXKF. That is the cylindrical sur- face formed by the revolution of ELMH, is equal to the spherical surface formed by the revolution of FKMH. And, if more parallel planes be drawn, ex- tremely near to each other, the small parts of the cylindrical surface will be equal to the correspondent parts of the spherical surface : and therefore the sum of all the parts of the cylindrical surface, will be equal to the sum of all the parts of the spherical sur- face; that is, the surface of the half cylinder will be equal to the surface of its inscribed hemisphere, and the surface of the whole cylinder equal to the surface of its inscribed sphere. Hence if D=^the diameter as before, then D DX3.1416XH; X— is the second rule,= 6 D 3 X5236, the same as the first rule. Corollary 1. The surface of the sphere is | of the whole surface of its circumscribing cylinder. For D*X3. 1416 is the surface of the sphere=rcMrre surface of the cylinder ; and the area of the two ends -Mensuration of Solids. 179 ofthecyliiKler=D 2 X.7S54;+D 2 X.7854:=l,570SD% therefore the surface of the whole cylinder is=3.1416 D 2 -|- 1.5708 D 2 =4.7124 D 2 j two-thirds of which = 3.1416 D 2 , the surface of the sphere. Cor. 2. Hence the surface of the whole sphere is equal to the area of four great circles of the same sphere; or to the rectangle of the circumference and diameter. Cor. 3. The surface of any segment or zone of a sphere, is equal to the curve surface of a cylinder, of the same height; and whose diameter is equal to that of the sphere. For the zone formed by the revolution of FKMH about MH, is equal to the surface of the cylinder form- ed by the revolution of ELMH. Investigation of General Rules for finding the Solidity of any Segment, or Zone of a Sphere. Let ACy represent a tri- angular pyramid, whose side Aw is infinitely small. The sphere may be considered to he constituted of an infinite P, number of such pyramids, Jj yCe, eCrt, &c. whos^ bases *1 compose the spherical sur- face, and altitudes are equal « to the radius of the sphere, their common vertex being the centre C. And, conse- quently, the sphere, or any sector thereof, is equal to a pyramid, the area of whose base is equal to the spher- ical surface, and height equal to the radius of the sphere. Let 6T the height of the segment aTm be called h, the diameter of the sphere D, then will its circumference be 3.1416 D; hence by Cor. 3, prece- ding, the surface of the spherical segment aTm will be expressed by 3.1116 Bh. And the solidity of the spher- I so Mensuration of Solids. ical sector CaTm (supposing CM to be drawn) by what has just been observed, will be 3.1410 D/*X — = 6 ,5230 D 2 //. By the property of the circle D — hXh—ab 2 ; but bU ' D—2h «6'X3.1416X— =D— /iX/iX3.1416X =D-WtX/t 3 6 X.5236X13— 2/l = D T /i— 3D/l 8 4-2/i 3 ;X.3236, the so- lidity of the cone Cam. Now, the solidity of the cone taken from the solidity of the sector, leaves the solidity of the segment, viz. D " h X .523 6 5— -D * /t-f3D/i ' —2h 3 ; X .5236 = 3D/t T - -2/i 3 ; X .5236 = 3D— 2hXh'*X .5236, which is one rule for the solidity of a segment ; and, it is evi- dent from the property of the circle, that this rule will be true, whether the segment be greater or less than the hemisphere, or whatever be the magnitude of h, provided it be not greater than D. Or, if r—ab the ra- dius of the segment's base, then D — hxh=r*, there* r € xh J fore D= , for D in the above rule, substitute h 8r*X3&* its value, and the rule becomes 2h; xh* X — h .5236 = 3r 9 -j-/£ 2 x/iX.5230,an useful rule, when the dia- meter of the sphere is not given. To find the Solidity of a Zone of a Sphere. Call the height of the greater segment H, and the height of the less/* ; also 11 the radius of the greater base, and r that of the less : then it is evident the dif- ference between the solidity of these two segments will be the solidity of the zone. Hence 3R*-fH T xH Mensuration of Solids. jsi X.5236;-3r 2 4-/i 2 X/iX.^36 = (3R 2 H4.H3;-3^Vt+/^) X.5236. Put a—H—h the breadth of the zone, and D the diameter of the sphere : then, by the prop- erty of the circle, D—H xH=R 2 ,and D— hxh=r* ItXH 2 from the former of these D= , and from thelat- H r 2 +/i 2 R 8 -fH* r 2 -f/* 2 ter D= ~; therefore = , and from h H h the above a=-H — h. Exterminate the values of H and h, and the above Theorem (3R 2 H+H T ;— -3r 2 /i-f-/i 3 ) : « 2 X.5236, will become R 2 -fr 2 -j ;XaX1.5708. 3 If one of the radii pass through the centre, as in the I) 2 — — zone A G m a, then R 2 =— =r«& 2 -H>C*=r 2 -f a* ; 4 henee the last Iheorem becomes r 2 -j-|a 2 ; X«X 3.1416 = -lD* — !-«*; XaX3.1416. Hence r T -f fa 2 : X«X6.2832=:iD 2 — 1« 2 X«X6.2S32 will express the solidity of the middle zone amKN, being double of the former, where a is half the alti- tude, and r=half the diameter of each end. Put A for the whole altitude, and d=2r the diameter of each end, and the theorems become rf'+fA* ; XAX .7854 — D ? — |A 2 ~; X AX- 7854. To find the Solidity of the Segment of a Sphere. RULE I. From three times the diameter of the sphere sub- tract twice the height of tfie segment ; multiply the remainder by the square of the height, and that pro- duct by .5236, and the last product will be the solidity. . R is j Mm isurat km of Solids. Rt'LE I! To three times the square of the radius of tli- incut's base, add (lit- square of its height ; multiply thii sum by the height, and the product by .5236, and (he last product will be the solidity. EXAMPLES. 1. Let ABCD be the frustum a ^^ T^Vti of a sphere ; suppose AB, the / ti " \ diameter of the frustum's base / \ to be 16 inches, and CI), the I height, 4 inches ; the solidity is I required ? V BY RILE Al)*-j-lX\.f!)C=CE; that is, 8X8—4,4-4=20 the diameter ; therefore, (20X3 — 4X2)X4X^ * .5236 = 433.6352, the solidity of the frustum. by rule 2. ( 3AD 2 +CD2)XCDX.3236 = :435.6352, the solidi^ ty, as before. 2. What is the solidity of a segment of a sphere,- whose height is 9, and the diameter of its base 20? Ms. 1795.4214. 3. What is the solidity of the segment of a sphere, whose diameter is 20 feet, and the height of the seg- ment 5 feet ? Ms. 654.5 feet. 4. The diameter of a sphere is 21, what is the so- lidity of a segment thereof, whose height is 4.5 ? Ms. 572.5366^i 5. The diameter of the base of a segment of *y sphere is 2$, and the height of the segment 6.5 ; re- quired the solidity ? Jins. 2144.99283. Mensuration of Solids. 183 To find the Solidity of the Frustum, or Zone, of a Sphere, GENERAL RULE. Add together the square of the radii of the ends, and one-third of the square of their distance, viz. the lieight; multiply the sum by the height, and that pro- duct by 1.5708, the last product will be the solidity. Or, for the middle Zone of a Sphere. i To the square of the diameter of the end, add two- thirds of the square of the height; multiply this sum by the height, and then by .7854, for the solidity. f 2? Or, from the square of the diameter of the sphere, subtract one-third of the square of the height of the middle zone; multiply the remainder by the height, and then by .7854, for the solidity. EXAMPLES. 1. Required the solidity of the zone of a sphere AGKN (see the figure, p. 179,) whose greater diame- ter AG is 20 inches, less diameter NK 15 inches, and distance between the ends, or height CB 10 inches. (AC2+NB 2 +1CB 2 )XCBX1.5708=2977.9741764, the solidity. 2. Required the solidify of the middle zone am KN of a sphere, whose diameter AG is 22 inches ; the top and bottom diameters a m and NK of the zone being each 16.971 iuches, and the height B6 14 inches ? Jlns. 4603.4912 inches, 3. Required the solidity of a zone whose greater diameter is 12, less diameter 10, and height 2 ? Alls. 195.8264. 4. Required the solidity of the middle zone of a sphere, whose top and bottom diameters are each 3 feet, and the breadth of the zone 4 feet ? Ms. 61.7848 feet. is* Mensural ulids. 5. Required the solidity of the middle zone of" a sphere, whose diameter is 5 feet, and the height of the zone 4 feet ? Jfiw. 61.7848 feet. To find the Convex Surface nf any Segment, or Zone, of a Sphere. RULE. Multiply the circumference of the whole sphere by the height of the segment, or zone, and the product will be the convex surface. EXAMTLES. 1. If the diameter of the earth be 7970 miles, the height of the frigid zone will be 552.361283 miles ; required its surface ? Ms. 6318761.10718.2210 miles. 2. If the diameter of the earth be 7970 miles, the height of the temperate zone will be 2143.6235335 miles ; required its surface ? Ms, 53673229.812734532 miles. 3. If the diameter of the earth be 7970 miles, the height of the torrid zone will be 3178.030327 miles ; required its surface ? Ms. 79573277.600166501. Note. The surfaces of the . P OT zones ' b - y } The surfaces of the twolemO B367322 c,. 81273i332 perate zones, by example 2, I 03a73229m273is3i are J * The torrid zone, example 3,5 7957 3277.600166504 is Surface of the whole globe,! agreeing exactly with ex- I 199557259.440000000 ample 2, p. 175. - - - J Mensuration of Solids. 185 § XII. Of a Spheroid. A spheroid is a solid formed by the revolution of an ellipsis about its axis. If the revolution be made about the longer axis, the solid is called an oblong, or prolate spheroid, and resembles an egg : but if the revolution be made round the shorter axis, the solid is called an oblate spheroid, and resembles a turnip, or an orange. The earth aud all the planets are consi- dered as oblate spheroids. In an oblong or prolate spheroid, the shorter axis is called the revolving axis, and the longer axis the fixed axis : but in an oblate spheroid, the longer axis is called the revolving axis, and the shorter axis is called the fixed axis. To find the Solidity. Multiply the square of the revolving axis by the fixed axis, and that product by .5236 for the solidity. EXAMPLES. 1. Let ADBC be a prolate sphe- roid, whose longer axis CD is 55 .inches, and shorter axis AB 33 ; re- quired the solidity ? AB*XCDx .5236=31361.022, the solid content, required. 2. What is the solidity of an oblate spheroid, whose longer axis is .55 inches, and shorter axis 33 inches ? Ans. 52263.37 inches. r 2 ISC, •WaisaraiioH of Solids. 3. Th<> axes of an oblong or prolate spheroid are SO ami 30; required the solidity? Jlns. 3S06& 4. The axes of an oblate spheroid are 50 and 30 ; required the solidity? Ms. 392,0 re Demonstration. Suppose the figure NTwSN to represent a spheroid, formed by the rotation of the semi-ellipsis TNS, about its transverse axis TS. Let D=TS, the length of the spheroid, and the axis of the circumscribing sphere ; and rf=Nn, the diameter of the greatest circle of the spheroid : Then TS* : Nw» :: A6* : ab\ by section XV. p. 127, that is, D 2 : d* :: Aft 2 : a& 2 ; but all circles are to each other as the squares of their radii. Now, the solidity of the sphere may he imagined to be constitut- ed of an infinite numb r of circles, whose radii are parallel to Aft ,• and the spheroid of an infinite number of circles whose radii are parallel to ab. Therefore D 2 : d- :: solidity of the sphere whose diameter is D : Solidity of the spheroid whose revolving axis is d; 3230 D 3 d 2 viz. D 2 : &* :: D 3 X.5236 : =.5236 Bd z . D 2 Again for the oblate spheroid, by Emerson's Conies, Book I. prob. 7, NC2 : TC 2 :: naXaS : Ba° ; but by Mensuration of Solids. 1S7 the property of the circle naXfflN=6d% therefore NO : TC 2 : : 6a 2 : Ba 2 , viz. d 2 : D 2 :: ba* : Ba 2 ; hence from the same manner of reasoning as above, d 2 : D 2 : : solid- ity of *he sphere, whose diameter is d : the solidity of a spheroid, whose revolving axis is D. .523Qd 3 & 2 Hence d 2 : D 2 :: d 3 x-5236 : =.52SQdD 7 -. d 2 Now from these proportions, between the sphere and its inscribed or circumscribed spheroid, it will be very easy to deduce theorems for finding the solid content, either of the segment or middle zone of any spheroid, having the same height with that of the sphere ; for, As the solidity of the whole sphere is to the solidity of the whole spheroid, so is any part of the sphere to* the like part of the spheroid. 1st. To find the solidity of a segment of a spheroid. TS a :Nrc 2 :: Ab* : ab 2 in the prolate spheroid, Nw 2 :TS 2 :: ab 2 : aB 2 in the oblate spheroid, Hence TS 2 : Nn 2 : : solidity ATM : solidity aYm, Nw 3 : TS 2 :: solidity bNb : solidity BNB. But the solidity of ATM or 6N6 may be found by rule the first, section XL p. 181. 2d. To find the solidity of the middle zone of a spheroid, Let/=TS the fixed axis. } for the oblong r=Nn the revolving axis $ spheroid. h — Bb the height of the middle zone or frustum. D=AM the diameter of one end of the spherical zone. d=am the corresponding diameter of the spheroid- al zone. 184 Mensuration of Solids. By section XI. the solidity of the middle zone of the sphere is D 8 -j X//X .7854=3l)*-f 2/t« ; X^X D 2 .2618 ; hut A6* = AC 2 — Co* vfcfc. -- s& ; 4 4 4 therefore J) 2 =/ a — // 2 , hence the solidity of the spher- ical zonc=3/ 2 — h 2 : xhX-^iH. JB.ut.5236/ 3 : ..1236 fr r :: Zf 2 —h; Xhx. 2(518 : the solidity of the spher- oidal frustum, vis. /■ : ?•:: 3]f *•— A* $ X/*X .2618 : the solidity of the spheroidal frustum. By tlie property of the ellipsis I\S B : N?* 2 : : AM 2 : am 2 , that is, r 2 h 2 f 2 : r' ::f 2 —h 2 ~ : rf 2 ? hence / 2 = $ r 2 — . The axes of an oblate spheroid are 50 and 3o/j~S what is the solidity of a segment whose height is 6, J and its base parallel to the transverse or revolving ax- is ? Jlns. 4084.08. 5. If the axes of a prolate spheroid be 10 and 6, re- quired the area of a segment whose height is 1, and its base parallel to the conjugate or revolving axis ? Jlns. 5.27788S. 6. The axes of a prolate spheroid are 36* and^.8; what is the solidity of a segment whose height and its base parallel to the conjugate or revolving axis ? Ms. 517.1306. id 18; #4* X 6 - oTvins 190 Mensuration of Solids. To find the Solidity of the middle Zone of a Spheroid, having circular ends ; the middle Diameter, and that of either of the eitds being given. RULE, To twice the square of the middle diameter, add the square of the diameter of the end; multiply the sum by the length of the zone, and the product again by .2618 for the solidity. Note. This rule is useful in guaging. A cask in the form of a middle zone of a prolate spheroid, is by guagers called a cask of the first variety. EXAMPLES. 1. What is the solidity of the middle zone of a pro- late spheroid, the diameter am of the end being 36, the middle diameter Nw 60, and the length B6 80 ? (2N?i 2 -j-am 2 )xB&X.26ls= 177940.224, the solid content. 2. Required the solidity of the middle zone of an oblate spheroid, the middle diameter being 100, the diameter of the end 80, and the length 36 ? Ans. 248814.72. 3. Required the content in ale gallons of spheroidal cask, whose length is 40 inches ; the bung diameter being 32, and head diameter 24 inches. A gallon of ale being 282 cubic inches ? Jlns. 97.44. 4. Required the content in wine gallons of a spher- oidal cask, whose length is 20 inches ; the bung diam- eter being 16, and head diameter 12 inches. A gallon of wine being 231 cubic inches ? MS. 14.869, § XIII Of a Parabolic Conoid. A parabolic conoid, or paraboloid, is a solid, gene- rated by supposing a semiparabola turned about its axis. Mensuration of Solids. To find the Solidity. 191 RULE. Multiply the square of the diameter of its base by the height, and that product by .3927 ; the last pro- duet shall be the solid content. EXAMPLES. 1. Let ABCD be a par- abolic conoid, the diameter of its base, AB, 36 inches, and its height, CD, 33 inches ; the solidity is re- quired ? AB 2 X CD X. 3927= 16794.9936, the solid content which divided by 172S gives 9.719 feet the solidity. Demonstration of the rule. The parabolic co- noid is constituted of an infinite number of circles, whose diameters are the ordinates of the parabola. Now according to the property of every parabola, it will be, AB» SA : AB : : AB" : =L, the Lotus Rectum, SA rSaxL ==ba* Then-] SeXL=/f? a ISyxL-^S $c. Here SftxL, HexL, »%XL, &c. are a series of terms in arith- metical progression. Therefore ba 2 , /e 2 , gy z , &c. are also, a series of terms in the same pro- gression, beginning at the point B, wherein AB 2 is the greatest .:•<£ i9~ JWetituration uf Solid*. term, and S A. the number of all the terms*. There- fore, AB 1 Xl*A=rthe sum of all the terms : and cir- cles are to each other as the squares of their radii. Therefore, if we put D=AB, and H=8A. Thru .78M DDXiH=.3927 1)1)11 will he the so- lid content of the conoid: which is just half the cy- linder, whose has? i&=D,and heights H. This being rightly understood, it will he easy to raise a theorem for finding the lower frustum of any parabolic conoid. * For supposing /i=«A, the height of the frustum, and o=S«, the height of the part b*b cut off, then h+p=SA, the height of the whole conoid; lctD = BB and d=bb, then the solidity of BSB will be D 2 /t+ l) 2 ;;; X-3927, and the solidity of bSb will he d^px .3927, hence the solidity of BMB must be D 2 /i-|-l) 2 /; — d 2 p ;X-3'.>27, & hut, by the property of the para- /i N « hola SA : Sa : : AB* : ab 2 , viz. I 1)2 cF hd* 7 i t>tl h+p: p:: — : — hence p = / v \ 4 4 1)*— d* / \ for p substitute its value in the ex- qj k" pression for the solidity of the v *- " hd 2 frustum BMB, and it becomes (I) 2 /*;-f-D a — d*X ) D2— d* X .3927=D 2 -f r^ 2 -f /iX.3927. 2. What is the solidity of a parabolic conoid whose height is 84, and the diameter of its base 48 ? Jns. 76001.5872. 3. Required the solidity of a paraboloid whose height is 30, and the diameter of its base 40 ? Ms. 18S49.6. 4. Required the solidity of a paraboloid whose height is 22.S5, and the diameter of its base 32 ? Ms. 9188.55168. Mensuration of Solids. 193 To find the Solidity of the Frustum of a Paraboloid, or Parabolic Conoid. RULE. Multiply the sum of the squares of the diameters of the two ends by the height, and that product by .3927 for the solidity. Note. This rule is useful in cask-gauging. For if two equal frustums of a parabolic conoid B&6B, be joined together at their greater bases, they form a cask of ^p. JBPf the third variety. The less dia- B ~ meter of the frustum bb will be the head diameter of the cask, and the greater diameter BAB will represent the bung diameter. EXAMPLES. 1. The greater diameter BAB, of the frustum B6&B of a paraboloid is 32, the less diameter bb 26, and the height Aa, 8; required the solidity ? (BB 2 4-^2) xAax .3927—5340.72, the solidity. 2. Required the solidify of the frustum of a para- bolic conoid, whose greater diameter is 30, less dia- meter 24-, and the height 9 ? Jins. 5216. 626S. 3. Required the solidity of the frustum of a para- boloid, the diameter of the greater end beins: 60, that of the less end 4*3, and the height 13 ? Jlns. 41733.0144. 4. There is a cask in the form of two equal frustums of a parabolic conoid ; the length is 40 inches, the bung diameter 32, and head diameter 24. Required its content in ale gallons ? 282 cubic inches being one gallon. Jlns. 89.1234 gallons. 5. There is a cask in the form of two equal frus- tums of a paraboloid : the length is 20 inches, the bung S 194 Mensuration of Sol'nls. diameter Ifi, and head diameter 1 >. Required the eon tent in wine gallons i zm cubic inches being one gal Ion. Ans. u.fi gallons. § XIV. Of olVarabolic Spinels. A parabolic spindle is formed by the revolution of parabola about its base, or greatest double ordinate. To find the Solidify. RULE. Multiply the square of the middle diameter by .11888 (being j T of .7854) and that product by its length; the last product is the solid content EXAMPLES. 1. Let ABCI) be a parabolic spin- dle, whos^ middle diameter CD is 36 inches, and its length AB 99 inches; fhe solidity is required ? ^ CD a XABx.4-lSS=53743.979;52, inches, or 31.10184 feet, the solidity. Mensuration of Solids. 195 2. The length of a parabolic spindle is 60, and the middle diameter 34 ; what is the solidity ? JhlS. 29053.516S. 3. The length of a parabolic spindle is 9 feet, and the middle diameter 3 feet; what is the solidity ? Jim. 33.92928. 4. What is the solidity of a parabolic spindle, whose length is 40, and middle diameter 16 ? Ms. 4289.3312. Demonstration of the rule. A parabolic spin- dle is constituted of an infinite number of circles, whose diameters are all parallel to the axis SA of the parabola, as m a, n e,py, $*c. Let us suppose the line Sa parallel to AB, &c. — Then it has already been proved, (section XIV.) that the lines/ m, gii, h p, Sfc. are a series of squares, whose roots are in a- rithmetical pro- r ^ gression, consequently their squares, viz.fm*,gn 2 ,hp' i i ifr&. will be the series of biquadrates, whose roots will be in arithmetical progression : which being premised, we may proceed thus : 1. &A—fm=.mu 2. SA — gn—ne 3. SA — hpzzpyi 4" c - and squaring each equation, we set 1. &A i --2$AXfm+fni z =ma 2 2, SA 2 — 2HAXgn+gn*=ne i 3. S\*—2SAxhp+hp*=py? $c. The sum of these equations will evidently give the sum of the squares of the radii, m a,ne,n y, f equal square*, where the number of terms is AB, the sum of the terms will be SpXAB. The sum of the second terms of th< qua- (jftfit {Emu-son's arithmetic of infinites, PropJ II!.) is 8A4.AB L*xAB — 2SAx = wherein SA is the 3 8 greatest term, and AB the number of terms. The sum of the third terms of the above equations (Emerson's arithmetic of infinites, Prop. V.) is l SA'XAB XSVxAB= jforfni'', gn*, hp* 9 $C. 4-j-l 5 is a scries of biquadrates, whose greatest term is and number of terms AB. Hence, the sum of all the terms of .the above equations will be 2SA*xAB SA«XAB 88 A « SA 2 x AB ; — + = X AB, 8 5 15 the sum of all the series of squares, nut 2 , ne* 9 pjf*i # c - But as circles are to each other as the squares of their radii or diameters ; it evidently follows that , 8 T SA* X-7SJ4XAB will he the solidity of one-fourth of the spindle ASB, that is T 8 T of the cylinder SaBA circum- scribing one-fourth of the spindle; therefore the whole spindle ty °* lis circumscribing cylinder : hence if D rzithe middle jdiameter, and L the length of the spin- dle, its solidity will be ^Xl) 2 XLX^834zzD a X .41888X1'. Which is the rule. To find the Solidity of one-fourth of the middle Fnis* turn of the Spindle, SpyA. Exactly on the same principles as above SA 2 2$Axhp W^ X Ay ;- — |- Ay ;-j . X Ay=(sum of all the Mensuration of Solids. 19^ series of squares 8A a , ma' 1 , -we 5 , and ^#%)= 2SAX/?,w tip* \ SA* 1+ ) 3 5 /XAt/=S. Hence 3SA 8 —- 3hp 3S 2SAX/^H = — • But SA a — 2SAx/^+/ip 2 ~- g Ay ^y*, and subtracting this equation from the former ify? 3S we get 2$ A* -J % 2 =z i>#% hence by re- duction, &c. 5 Ay 2SA*-f;»y 2 — f/*/> 2 ; XjApS. And circles are to each other as the squares of their radii, hence 1.5708SA 2 4-.7S54py 2 — 3.i4>16hp z X i Ay=S, or 2SA 4 -fjoy 2 — lfyo 2 X.261S Ay=8, the solidity of one- fourth of the middle frustum of the spindle. Hence To, find the Solidity of the middle Frustum of a Para- bolic Spindle. To twice the square of the middle diameter, add the square of the diameter of the end; and from the sum subtract four-tenths of the square of the differ- ence between these diameters; the remainder multi- plied by the length, and that product by .2618 will give the solidity. Note. This rule is useful in cask-gauging. A cask in the form of the middle frustum of a parabolic spin- dle, is called by gaugers a cask of the second variety ; and is the most common of any of the varieties. EXAMPLES. 1. Required the solidity of the middle frustum of a parabolic spindle EFGH, the length AB being 20, the greatest diameter CD 16, and the least diameter EForGHl2? s 2 Mensuration of Sol D -f-GH* — .4X(CI) — GH)2)xAlix.-'« (5134-144 — .4X4X4) >C«0 X .S618 = ($iW — 6.4) X .■>.2i6 = 640.6X3.236 =3401.3056, the solid cont; 2. The bung diameter CD of a cask is 32 inches, head diameter EF, 21 inches, and length AB, 40 inches ; required its content in ale gallons ? 282 cubic inches being t gallon. Ans, 96.4909 gallons. 3. Tl:e bung diameter ( 1), of a cask is JO inches, head diameter EF 20 inches, and length AB 36 inches; required its content in nine gallons? 231 cubic inches being 1 gallon. Ans. 117.89568 gallons. To find the Solidify of the middle, Frustum of any Spin- die, formed by the revolution of a Conic Section about the Diameter of that Section. To the square of the greatest diameter add the square of the least, and four times the square of a rliameter taken exactly in the middle between the two'; multiply the sum by the length, and that product by ,1309 for the solidity. Note. See the latter part of the demonstration of the rule, iu section X. EXAMPLES. 1 . Required the solidity of the middle frustum EFGH of any spindle ; the length All, being 40, the greatest or middle diameter CD, 32, the least diameter EF or Gil, 24, and the diameter IK in the middle between GH and CD, mt37568 ? Ans. 2742.3.72624. 2. The bung diameter of a cask being 36 inches ; head 20; length 36, and a diameter exactly in the middle, 31.93 inches; what is the content in wine gallons ? Ans. 117 gallons 3* quarts'. Mensuration of Regular Bodies. 190 § XV. Of the five Regular Bodies. A regular, or pi atonic body, is a solid contained under a certain number of similar and equal plane figures. Only three sorts of regular plane figures joined together can make a solid angle ; for three plane angles, at least, are required to make a solid angle, and all the plane angles which constitute the solid angle, must be less, when added together, than four right angles, {Euclid, XL and 21.) Now each angle of an equilateral triangle is 60 degrees ; each angle of a square 90 degrees; and each angle of a pentagon 180 degrees. Therefore there can be only five regular bodies, for the solid angles of each must consist either of three, four, or jive triangles, three squares, or three pentagons. 1. The tetraedron, or equilateral pyramid, which has four triangular faces ; hence all the plane angles about one of its solid angles make 180 degrees. 2. The octaedron, which has eight equilateral trian- gular faces; hence all the plane angles about any one ol its solid angles make 240 degrees. 3. The dodecaedron, which has twelve equilateral pentagonal faces ; hence all the angles about any one of its solid angles, make 324 degrees. 4. The icosaedron, which has twenty equilateral tri- angular faces ; hence all the angles about any one of its solid angles, make 300 degrees. 5. The he.vacdron, or cube, which has six equal square faces ; hence all the angles about any one of its solid angles make 270 degrees. Note. If the following figures be exactly drawn on pasteboard, and the lines cut half through, so that the parts may be turned up and glued together, they will represent the five regular bodies above described. .200 The Mensuration df or i /* Ifie-lcojaedron, the Five Regular Bodies. 201 A table, shewing the solidity and superficies' of the five regular bodies, the length of a side in each be- ing 1, or unity. JSTames of the bodies. Solidity. Superficies. Tetraedron .11785113 205081 Oetaedron .47110-152 8.46410162 Hexaedron 1.00000000 6.00000000 Icosaedron 2.18 169499 8.6602540-1 Dodeeaedron 7.6631189 20.6457288 To find the Solidity or Superficies, of any regular Body, by the Table. RULE, 1. Multiply the cube of the length of a side of the body, by the tabular solidity, and the product will give the solidity of the body. 2. Multiply the square of the length of a side of the body, by the tabular superficies, and the product will give the siperficies of the body. EXAMPLES. 1. Let ABCD be a tetraedron, / j \ whose side AB is 12 inches; requir- / ( \ ed the solidity and superficies ? / g \ 12=AB 12 Tab. sol. .11785113 1728 144= square of AB 12 1728= cube of AB 94280904 23570220 82195791 11785113 203.64675264 solidity. The Mensuration of And. ill- multiplied by t. 78205 081, the tabular su perlicies, gives 249.41531664 inches, the superfieio oj the tetraedron. 2. Each side of a tetraedron is 3, required its sur face and solidity ? Atis, Superficies 15.58843729; solid. 3.18198051 a.LetABCDE be an ottae- dron, each side being L2 inch- es; required the solidity and superficies ? * / Answer. 814.58701056 inch, solid. 49S.S306332S inch, super. 4. Let B be a h&vaedron, or cube, whose side is 12 inches; required the solidity and superficies? a C1728 inches, solidity. ' \ 864 inches, superiieies. [ 5. Let ABCDEPGH1 fr's he an icosaedron, each side thereof being 12 inches ; j \ required the solidity and superficies ? ■ Hj & a 5 3769.97470272 inches, solidity. * 1 1247.07638176 inches, superficies. VI \ \ V -V the Five Regular Bodies. 203 6. Let ABCDEFGHIK be a dodecaedron, each side thereof being 12 inches; required the solidity and superficies ? a C 13241.8694592 inches, solidity. 1 ' £ 2972.9849472 inches, superficies. § XVI. To measure any Irhegulam Solid. If you have any piece of wood or stone that is craggy p uneven, and you desire to find the solidity, put the solid into any regular vessel, as a tub, a cistern, or ;he like, and pour in as much water as will just, cover t; then take out the solid, and measure how much the pall of tho water is, and so find the solidity of that part of the vessel. EXAMPLES. 1. Suppose a piece of wood or stone to he measured, md suppose a cylindrical tub 32 inches diameter, into which let the stone or wood be put, and covered with .vater: then, when the solid is taken out, suppose the all of the water 14 inches ; square 32, and multiply the square by .7854, the product will be 804.2496, the :oi The Mensuration of §c. area of the base ; which multiplied by 14, the depth or fall of the water, the product is 11359.40, &c. which divided by t: 38, the quotient is 0.31 feet; and so much is tin* solid content required. 2. It is reported that Hiero, king of Sicily, furnish- ed a workman with a quantity of gold to make a crown. When it came home, lie suspected that the workman Itad used a greater alloy of silver than was necessary in its composition, and applied to Archi- medes* a celebrated mathematician of Syracuse, to dis- cover the fraud, without defacing the crown. He pro- cured a mass of gold and another of silver, exactly of the same weight with the crown ; considering that if the crown were of pure gold, it would be of equal hulk and displace an equal quantity of water with the golden ball: and if of silver, it would be of equal bulk, and displace an equal quantity of water with the silver ball; but if a mixture of the two, it would dis- place an intermediate quantity of water. Now, for example, suppose that each of the three masses weighed loo ounces; and that on immersing them severally in water, there were displaced 5 ounces of water by the golden, 9 ounces by the silver, and G ounces by the crown ; then their comparative bulks are 5, 9, and 0. From 9, silver. From 6, crown. Take 6, crown. Take 5, gold. 3 remainder. 1 remainder. The sum of these remainders is 4; Then 4 : 100 :: 3 : 73 ounces of gold, and 4 : 100 :: 1 : 25 ounces of silver. That is, the crown consisted of 75 ounces of gold; and 25 ounces of silver. See Cotes' Hydros. Led. p. SI, The Measuring of Boards and Planks. 205 CHAPTER III. The Measuring of Board and Timber. § I. Of Board Measure. To find the Superficial Content of a Board or Plank, RULE. MULTIPLY the length by the mean breadth. When the board is broader at one end than the other, add the breadths of the two ends together and take half the sum of the mean breadth. By the Carpenter's Rule. Set 12 on B to the breadth in inches on A ; then against the length, in feet, on B, you will find the su- perficies on A, in feet. By Scale and Compasses. Extend the compasses from 12 to the length in feet, that extent will reach from the breadth, in inches, to the superficies in feet. EXAMPLES. 1. If a board be 16 inches broad, and 13 feet longj how many feet are contained in it ? Multiply 16 by 13, and the product is 208 ; which divided by 12, gives 17 feet, and 4 remains, which is a third part of a foot. T ■.'06 The M of Boards and 1'lr . fly the Carj'Otfn's Jlitlc. As 12 on B : 16 on A : : 13 on B : 17£ on A. fly Scale and Compasses. Extend tlie compasses from 12 to 18, the length in feet, that extent will reach from 16, the hreadlh in inches, to IT \- t lie superficies in feet. 2. What is the value of a plank, whose length i-> s feet inches, and breadth throughout 1 foot 3 inches; at 2U\. per foot ? Ms. 2 shill. 2 hi 3. Required the superficies of a board, whose mean breadth is 1 foot finches, and length 12 feet 6 inches ? Jins. 14 feet 7 inches. 4. Having occasion to measure an irregular mahog- any plank of 14 feet in length, I found it necessary to measure several breadths at equal distances from each other, biz, at every two feet. The breadth of the less end was 6 inches, and that of the greater end 1 foot ; the intermediate breadths were 1 foot, 1 foot 6 inches, 2 feet, 2 feet, 1 foot, and 2 feet ; how many square feet were contained in the plank ? Ans. t&j feet; — meau breadth 1| feet, found by divi- ding the sum of the breadths by their number. 5. Required the value of 5 oaken planks at 3d. per foot, each of them being 17j feet long; and their se- veral breadths as follows; viz. two of i8£ inches in the middle, one of 14V inches in the middle, and the two remaining ones, each IS inches at the broader end, and 11 \ inches at the narrower ? +1ns. Li : 5 : H\. The Measuring of Boards and Planks. 207 Having the breadth aj a rectangular Plank given in inches, to find how much in length will make afoot, or any other assigned quantity. Divide 144, or the area to he cut off, by the breadth in inches, and the quotient will be the length in inches. Note. To answer the purpose of the above rule, some carpenters' rules have a little table upon them, in the following form, called A Table of Board Measure. 5 n 6 12 6 4 3 * 2 i 1 1 2 3 4 5 6 7 8 By this table you are to understand, that if the breadth be 1 inch, the length must be 12 feet ; if 2 inches, the length is 6 feet ; if 5 inches broad, the length is 2 feet 3 inches, &e. If the breadth be not contained in the table on the rule, shut the rule, and look for the breadth in the line of board measure, which runs along the rule from the table of board measure, and over against it on the opposite side, in the scale of inches, is the length re- quired. Thus, if the breadth be 9 inches, you will find the length against 15 inches ; if the breadth be 41 inches, you will find the length a little above 13 inches, &c. EXAMPLES. i. If a board be 19 inches broad; how many inch- in length will make a foot ? Am. 7.58 inches 203 Tlie Mensuration of Timber. 2. From a mahogany plank 26 inches broad, a yard and a half is to be cutott'; what distance from the end must the line be struck ? .flws. 74.7692 inches, or 6.23 icet § II. The Customary Method of Measuring Timber. I. The customary method of measuring round tim- ber, is to gird the piece round the middle with a string ; one fourth part of this girt squared and multiplied by the length gives the solidity. Or, if the piece of timber be squared, half the sum of the breadth and depth in the middle is considered as the quarter-girt, and used as above. II. If the piece of timber be very irregular, gird it in several places equally distant from each other, and divide the sum of their circumferences by their number, for a mean circumference 5 the square of one quarter of the mean circumference, multiplied by the length, will give the solidity. Note. If the circumference be taken in inches, and the length it feet, divide the last product by 144. "The Mensuration of Timber. 209 III. Otherwise by the following Table for measuring Timber. Quart. Quart. Quart. girt. Inches. Area. girt. Area. girt. Area. Feet. Inches. Feet. Inches. Feet. 6 .230 12 1.000 18 2.250 H .272 i2\ 1.042 18 J- 2.376 6£ .294 13$ i.085 19 2.506 6| .317 i$t 1.129 m 2.640 7 .340 13 1.174 20 2.777 a .364 13* 1.219 20 2 2.917 * i .390 1*1 1.265 21 3.062 rt .417 ia* 1.313 211 3.209 8 .444 14 1.361 22 3.362 8^ .472 14 \ 1.410 2*1 3.516 »2 .501 141 1.460 23 3.673 s* .531 l4| 1.511 2;>1 3.835 9 .562 15 1.562 24 4.000 K .594 IS* 1.615 24' 4.168 9* .626 ±5\ 1.668 25 4340 ft .659 15| 1.722 "°2 4.5 1 6 10 .694 16 1.777 26 4.694 10| .730 16} 1.833 26j 4.876 10] .766 iftj 1.890 27 5.062 iOf .803 16| 1.948 sp 2 5.252 n .840 17 2.006 28 5.444 Hi ■ .878 17 V 2.066 28 £ 5.640 Ui .918 171 2.126 29 5.840 fit .939 17| 2.187 29i 30 6.044 6.230 T 2 tittt The Mensuration of Timber. Tlie use of the Table. Multiply the area corresponding to the quarter-girt . in inches, by the length of the piece of timber in feet. and the product will be the solidity. IV. By tlie Carpenters Rule. Measure the circumference in the middle of the piece of timber, and take a quarter of it in inches, call this the girt. Then set 12 on D to the length in feet on C, and against the girt in inches on D, you will find the con- lent in feet on C V. By Scale arid Comjwsses. Measure the circumference in the middle of the piece of timber, and take a quarter of it in inches. Then extend from 12 to this quarter-girt, that ex-' rent will reach from the length in feet, being turned twice over, to the solidity. Note. The buyer is allowed to take the girt any where between the greater end, and the middle of the tree, if it taper. All branches, or boughs, whose quarter-girt is not less than six inches, are reckoned as timber ; and any \ part of the trunk less than 2 feet in compass is not considered as timber. An allowance is generally made to the buyer on account of bark ; thus for oak, one-tenth or one-twelfth part of the circumference is deducted; but the allow- ance for the bark of ash, beech, elm, &c. is a small matter less. EXAMPLES. 1. If a piece of square timber be 2 feet 9 inches deep, and 1 foot 7 inches broad 5 and the length 16 ■ The Mensuration of Timber. 211 feet 9 indies, (or, which is the same thing,* if the I quarter-girt be 26 inches, and length 16 feet 9 inches) how many solid feet are contained therein ? 26 inches, 26 quarter ile. feet. -girt. 16.75=16 feet 9 inch 676 156 52 676 square. % the Ted 4 694 10050 11725 10050 144)11323.00(7S.63 feet. 1243 16.75 910 23470 32858 28164 4694* 460 28 remainder. 78.62450 Note. The true content, measured as a parallel- opipedon, § II. of chap. II. is 72.93 feet. See example 4, page 137. * That is, according- to the customary method of measuring", mentioned at the beginning of this section. And when the breadth and depth of a piece of timber are nearly equal, this method is nearly true. — Otherwise, the breadth in the middle, is generally multiplied by the depth in the middle, and that product by the length. When the timber tapers regularly, half the sum of the breadths of the two ends is the breadth in the middle, and half the sum of the depths of the two ends is the depth in the middle. :i- The Mensuration of Tin liy the Carpenter's Rule. As 12 on D : 16£ on C : : 26 on D : 78£ an C. /?.4J feet, in the second T.92, in the third, tf.i5 9 in the fourth 4.74, and in the fifth Feet ? Jns. 42.52 feet. To find how much in length ivill make a foot of any squared Timber, of equal thickness from end to end. Divide 1728, the solid inches in a foot, by the area of the end in inches, and the quotient will be the length of a solid foot, in inches. Note. To answer the purpose of the above rule, the carpenters' rules sometimes have a little table up- on them in the following form, called A Table of Timber Measure. O | 0| o » ■■• [•; 11 | 3 | 9 j Inches. 14* | 36 15 • » * 2\2 1 | Feet. * 1 ».l »| * ? • 1 r|sj 9 j hide of the sq. By this table you are to understand, that if the side of the square be one inch, the length must be 14 f feet; if 2 inches be the side of the square, the length must be 36 feet, to make a solid foot. If the side of the square be not in the little table, look for it in the line of timber measure, running along the rule from the table, and against it, in the line of inches is the length required. Thus, if the side of the square be 16 inches, you will find the length to be 6 inches and 7 tenths, &c. EXAMPLES. 1. If a piece of timber be 18 inches square, how much in length will make a solid foot ? .fyis. ."U inches. The Mensuration of Timber, 215 2. Ti" apiece of timber be 22 inches deep, and 15 Riches broad : how much in length will make a solid toot ? Sins. 5.23 inches. SCHOLIUM. The foregoing section contains the whole substance of timber measuring, as practised in all timber yards ; and though the method has been noticed as erroneous by various different writers for near a century past, it still stands its ground ; nor does it appear probable that it will soon be abolished. In measuring round timber by the foregoing method, of taking a quarter of the circumference in the mid- dle, for the side of a mean square ; it is objected that it makes the content too little, and that such timber ought to be measured as a cylinder, or the false con- tent should be increased in the ratio of .0625 to .0795S, or as 11 to 14. But this objection is answered by saying, that be- fore the wood can be squared, and made fit for use, a great part of it goes to waste ia chips, and therefore the quantity of round timber ought to be reckoned no more than what the inscribed square will amount to. Xow, if the circumference be 1, the area of the sec- tion will be .0795S, the square of the quarter-girt .0625, and the side of the inscribed square .22">1, the square of which is .05067. Therefore to make the content by the inscribed square correspond with the cylindrical content, it ought to be increased in the ra- tio of .03067 to .07958, or 7 to 11 which is a greater, increase than 1 1 to 11, and therefore instead of the content by the quarter-girt being too little, it is on this consideration too much. Again, when you take the mean girt of a tree, it is very probable that the tree is not perfectly circular in that place, and the more it differs from a circle the greater the quarter-girt will be ; so that if you 216 Tlie Mensuration of Timber. were to measure such a tree as a cylinder, by the pro- per rule for that purpose, the content Mould in reality be too much on account of the error in taking the girt. It is likewise objected that in round tapering tim- ber, taking the side of a square in the middle of the piece makes the content too little, and that even in a greater proportion than above. The answer given to this objection is, that in almost in all cases the great- ' er must be cut away till it be of the same dimensions as the less end, otherwise the timber cannot be sawn into useful materials ; and therefore there is no just reason for any objection to the rule on this ground ; and particularly, as no general rule has hitherto been« published, of sufficient merit, to supersede the use of the quarter-girt rule. Dr. Ifutton, in the quarto edition of his Mensura- tion (published in 17/0) p. 607, has given an easy rule for measuring round timber : thus, " Multiply the square of one-fifth of the girt or ctr- cumference by twice the length, and the product will be the content {extremely near the truth.)" Dr. Hutton says that many reasons may be alledg- ed for changing the customary method of measuring timber, and introducing this rule instead thereof, and the principal reason is, (see p. 6 14 of the quarto edi- tion of his Mensuration) " the preventing of the sel- lers from playing any tricks with their timber by cut- ting trees into different lengths, so as to make them measure to more than the whole did ; for, by the false method, this may be done in many respects." Mr. Mpnnycastk gives the same problems, " to shew the artifices that may be used in measuring timber ac- cording to the false method now practised, and the absolute necessity there is for abolishing it." It is rather singular that neither of those gentle- men should perceive, that the very same tricks and artifices may be practised, with equal success, if Dr. The Mensuration of Timber. 217 Hutiotfs rule be used. The truth of these remarks will easily appear to those who are qualified to read the Doctor's demonstrations; and those who are not, may consult the following EXAMPLES. 1. Dr. Hutton, p. 613, Prob. IV. Supposing a tree to girt 14 feet at the greater end, 2 feet at the less, and 8 feet in the middle; and that the length be 32 feet. Rale. If this tree be cut through, exactly in the middle, the two parts will measure to the most possi- ble, by the common method, and to more than the whole. By the common Method. The whole tree measures 1.28 feet. The sreater end measures 121 feet. The tess end measures 25 feet. ing the whole by 18 feet. Sum 146 exceed By Dr. Hutton' 's Rule. The whole tree measures 163.84 feet. The greater end measures 154.88 feet. The less end measures 32. feet. Sum 186.88 exceeding the whole by 23.04 feet. Measured as the Frustums of Cones; The whole tree measures 193.53856 feet. The great end measures 157.88672 feet. The less end measures 35.65184 feet. Sum 193.53856 equal to the whole, as it ought. U 218 Tne Mensuration of Timber. 2. Dr. Button, p. 610, Prob. V. Supposing a tn>e to girt 11 Reel at the greater end, 2 feet at the lest, and its length to be 83 Feel ; where must it be cut that the part next the greater end may measure to more than the whole, by the customary measure ? RULE. Cut it through (if possible) where the girt is 4 of the greatest. This, according to Dr. Ilutton's rules, will be 7-£ feet from the less end, and 24-| feet from the greater end ; and the girt at the section will be \ 4 or ;et : By the common Method. The greater end measures - 135£i feet. The whole tree measures - -128 feet. Diff. 7*\ so that the jpart exceeds the whole by 7 feit. By Br. Hutton's Rule. The greater end measures - 173.44 feet. The whole tree measures - 163.84 feet. Diff*. 9.60 so that the part exceeds the whole by 9.6 feet. 3. Dr. Ilutton, p. 617, Prob. VI. Supposing a tree to girt 14 feet at the greater end, 2 feet at the less, and its length to be 32 feet : where must it be cut that the part next the greater end may measure exactly to the same as the whole, by the cus- tomary method ? By Dr. Huttonh rules the lengths of the two parts must be 13.599118 feet, and 18.100882 feet ; also the girt at the section must be 7.099669. The Mensuration of Timber. 219 By the common Method. The greater end measures - - 128 feet. The whole tree measures - - 128 feet. Notwithstanding above ^ part is cut off the length. By Dr. Huttotfs Rule. The greater end measures - 163.8399 feet, The whole tree measures - - 163.84 feet, notwithstanding above -} part is cut off the length. The following rules for measuring timber are very accurate, provided the dimensions can be truly taken, and that the timber be in the form of a parailelopi- pedon, cylinder, frustum of a rectangular pyramid, or the frustum of a cone ; but for the reasons already given, it is not probable that they will ever be brought into general use. PROBLEM I. To find the Solidity of Timber Scantling, or Squared Timber, being of equal breadth aud thickness through- mit. RULE. Multiply the breadth by the thickness, and that pro- duct by the length. Thus, you will find the answer to the first exam- ple, p. 211, to be 72.93 feet, or 72 feet 11 inches. PROBLEM II. To find the Solidity of a piece of Timber, supposing it to be PERFEcfLr cylindrical. RULE. Multiply the square of one-fourth of the circumfe- rence by the length : then say, as 11 is to 14, so is this content, to the true content. — Or, use Dr. Hutton's rule, see page 26 1 . C20 Tlie Mensuration of Timber. Thus, yon wiil find the answer to the 5th example, rage Jl), to bey 1.(53 feet; and to the 6th example •1. IS feet PROBLEM III. To find the Solidity of Squared Timber, tapering regularly, RULE. Multiply the breadth at each end by the depth, and also the sum of the breadths by the sum of the depth* These three products added together* and the sum mul- tiplied by one-sixth of the length, will give the so- lidity. Thus, you will find the answer to the 3d example, page 212, to be 43.101 feet, and to the 4th example 37.33. To find the Solidity of round tapering Timber, hav- ing the Girt or Circumference of the two ends given in inches, and the length in feet. RULE. To the squares of the two circumferences add the square of their sum; multiply this sum by, the length, cut off four figures from the right hand for decimals, or move the decimal point four places to the left hand, and J | of the product will be the content. Thus, you will find the answer to the 7th example, page 213, to be 74.38 feet, and the 8th example 92.54 feet. Note. By taking the first example in the scholium, page 217, you will find The content of the whole tree 193.931 feet. The content of the greater end 158.231 feet. The content of the less end 35.729 feet. Sum 193.960 equal to the whole very nearly. Tiie Mensuration of Carpenters' Work, #c. 221 CHAPTER IV. Of Measuring the Works of the several Artificers relating to Building ; and what Methods and Cus- toms are observed in doing it § I. Of Carpenters' and JotitsRs 9 Work. THE Carpenters' and Joiners' works, which are mea* surable, are flooring, partitioning, roofing, wainscot- ing, &c. I. Of Flooring. Joists are measured by multiplying their breadth by their depth, and that product by their length. They receive various names according to the position in which they are laid to form a floor ; such as trimming joists, common joists, girders, binding joists, bridging joists, and cieling joists. In boarded flooring, the dimensions must be taken to the extreme parts, and the number of squares of 100 feet must be calculated from these dimensions. De- ductions must be made for stair-cases, chimneys, &c. EXAMPLES. 1. If a floor be 57 feet 3 inches long, and 28 feet 6 inches broad ; how many squares of flooring are there in that room ? 77/e Mensuration of Carpenters' Work, #c. ft y Decimals. By Duodecimals, tf.25 F. 1. 57 : 6 28 : -; 28625 40800 11450 450 114 28 7 : 6 7 : 100 I 1631.625 feet. Squares 16.31625 16 I 31 : 7 : 6 , Facit 16 squares and 31 feet. 2. Let a floor be 53 feet 6 inches long, and 47 feet 9 inches broad; how many squares are contained in that floor ? Jlns. 25 squares 54 feet. 3. A floor being 36 feet 3 inches long, and 16 feet 6 inches broad; flow many squares are contained in it ? Jlns. 5 squares 98 feet. 4. In a naked floor the girder is 1 foot 2 inches deep, A foot broad, and 20 feet long; there are 8 bridging joists, whose scantlings (viz. breadths and depths) are & inches by 6| inches, and 20 feet long; 8 binding joists, whose lengths are 9 feet, and scantlings 8| inches by 4 inches : the cieling joists are 24 in number, each 6 feet long, and their scantlings 4 inches by 2\ inches. Required the solidity of the whole? Jlns. 72 feet. 5. Suppose a house of three stories, besides the ground floor, was to be floored at £.6 10s. per square ; ♦he house measures 20 feet 8 inches, by 16 feet 9 inches ; there are 7 tire-places, the measures whereof are, two, each of 6 feet by 4 feet 6 inches ; two other, each of 6 feet by 5 feet 4 inches ; and two, each of 5 feet 8 inches by 4 feet 8 inches ; and the seventh, 5 feet 2 inches by 4 feet. The well-hole for the Stairs is 10 feet 6 inches by 8 feet 9 inches. What will the whole tome to ? Ms. 1.53. 13s. 3£& Tfie Mensuration of Carpenters 9 Work, <$*c. 25& II. Of Partitioning. Boarded partitions are measured by the square in the same manner as flooring, and deductions must be made for doors and windows, except they are included by agreement. The strongest partitions are those made with framed timber, all the parts of whieh are measured as in flooring, except the king-posts, and these are of the same kind as in roofing, an example of which will be given in the next article. EXAMPLES. 1. If a partition between rooms be in length 82 feet 6 inches, and in height 12 feet 3 inches ; how many squares are contained therein ? Jlns. 10 squares 10 feefc 2. If a partition between rooms be in length 91 feet 9 inches, and its breadth 11 feet 3 inches; how many squares are contained in it ? Jlns. lo squares 32 feefc III. Of Roofing. In roofing take the whole length of the timber for the length of the framing, and for the breadth gird over the ridge from wall to wall with a string. This length and breadth must be multiplied together for the content. It is a rule also among workmen, that the flat of any house, and half the flat thereof, taken within the walls, is equal to the measure of the roof of the same house; but this is when the roof is of a true pitch. The pitch of every roof ought to be made according to its covering, which in England is lead, pantiles, plaJHtiles, or slates. The usual pitches, are the pedf- Tlie, Menswution of Carpenters' Work, $c. ment pitch, used when the covering; is lead ; the per- pendicular height is -| of the breadth of the building. The common, or true pitch, where the Length of the ratters are } of the bread." of the building; this is used when the covering is plainliles. The Gothic pitch, is when the length of the principal rafters is equal to the breadth of the building, forming an equi- lateral triangle : This pitch is used when the covering is of pantiles. In the measuring of roofing for workmanship alone, all holes for chimney-shafts and sky-lights are gene- rally deducted. But in measuring for work and mate- I rials, they commonly measure in all sky-lights, luthern- lights, and holes for the ehimney -shafts, on account of • their trouble, and waste of materials. EXAMPLES. 1. If a house within the walls be 44 feet 6 inches' long, and 18 feet 3 inches broad; how many square* «f roofing will cover that house ? By Decimals. ±8.25 44.5 ft The flat JThe half Sum 1 Facit 12 squan 1 Duodecimals. F. I. 44 : 6 18 * 3 9125 t 7300 7300 352 44 11:1*6 Flat 812.125 Half 406. 9:0:0 812 : 1 : 6 100 | 12 18 406 12.18 L2|18 es 18 fee*. The Mensuration of Carpenters' Work, Sfc. 22& 2. What < i»< the roofing of a house at los. 6d. per square; the length, within the walls, being 52 feet 8 inches, and the breadth 30 feet 6 inches ; the roof being of a true pitch? Ms. 1.12 12s. llfd. 3. The roof of a house is of a true pitch ; and the house measures 40 feet 6 inches in length, within the walls, and 20 feet 6 inches in breadth; how many squares of roofing are contained therein ? * Ms. 12.45375. Note. All timbers in a roof are measured in the same manner as in floors, except king-posts, &c. such as A in the annexed figure, where there is a necessity for cutting out parallel pieces of wood from the sides, in order that the ends of the braces B, that come against them, may have what the workmen call a square hutment. To measure king-posts for workman- ship, take their breadth and depth at the widesf place, and multiply these together, and the product by the length. To find the quantity of materials, if the pieces sawn out are 2- inches broad, or upwards, and more than 2 feet S>* long, they are con- ^^\ sidered as pieees of f ~^ timber fit for use. In measuring these pieces, the shortest length must always be taken, because the sawing of them from the king-post, renders a part of them useless. The soli- dity of these must be deducted from the solidity of the king-post. "When the pieces sawn out are of smaller dimensions than above described, the whole post is measured as solid without any deduction, the pieces cut out being esteemed of little or no value. 4. Let the tie-beam D, in the above figure, be 36 feet long. 9 inches broad, and 1 foot 2 inches deep; the king-post All feet 6 inches high, 1 foot broad Vie Mensuration of Carpenters" ffbrh, feet long, S inches broad, and 10 inches deep: the struts, C, (', are 3 feet 6 inches long, 4 inches broad, and 5 inches deep. Required #4 he measurement for workmanship, and likewise for materials ? F. I. P. 31 : 6:0 solidity of the tie-beam D. 4 : 9:6 solidity of the king-post A. 2 : 7:3 solidity of the bracks B, B. 13 : 2:4 solidity of the rafters E, E. — : 11 : 8 solidity of the struts C ? C. I 53 : 0:9 solidity for workmanship. 1 : 5:6 solidity cut from the king-post. 51 : 7:3 solidity for materials. IV. Of Wainscoting, Sfc. Wainscoting is measured by the yard square, con- sisting of 9 square feet. The dimensions are taken in feet and inches ; thus, in taking the height of a room they girt over the cornice, swelling pannels, and mouldings, with a string; and for the compass of the room, they measure round the floor. Doors, window- shutters, and such like, where both sides are planed, are considered as work and half; therefore in mea- suring a room they need not be deducted, but the room may be measured as if there were none; then the contents of the doors and shutters must be found, and the half thereof added to the content of the whole room. Tlie Mensuration of Carpenters' Work, $c. %$2 Windows, where there are no shutters, must be de- ducted; also chimneys, window-seats, cheek-boards, sopheta-boards, linings, &c. must be measured by them- selves. Weather-boarding is measured by the yard square^ and sometimes by the square. Windows are generally made and valued by the foot superficial measure, and sometimes at so much per window. When they are measured, the dimen- sions must be taken in feet and inches, from the under side of the sill to the upper side of the top-rail, for the height ; and for the breadth, from outside to out- *side of the jambs. This length multiplied by the breadth will be the superficies. Stair-cases are measured by the foot superficial, and the dimensions are taken with a string, girt over the riser, and tread ; and that length, or girt, multiplied by the length of the step, will give the su- perficies. The rail is taken at so much per foot in length, according to the diameter of the well-hole; archi- trave string-boards by the foot superficial ; brackets and strings at so much per piece, according to the workmanship. Door-cases are measured by the foot superficial, ami the dimensions must be taken with a string girt round the architrave and inside of the jambs, for the breadth; and for the length, add the length of the two jambs to the length of the eap-pieee (taking the breadth of the opening for the length) the product of these two will be the superficies. Frame doors are measured by the foot, or sometimes by the yard square. Modillion cornices, coves, &e. are generally measur- ed by the foot superficial. Frontispieces are measured and valued by the foot superficial, and every part is measured separately, viz\ architrave, frieze, and cornice. I 229 The Mensuration of Carpenters* Work, cjv. EXAMPLES. 1. If a room, or wainscot", being girt downwards over the mouldings, be. u feet 9 inches high, and 126 inches in compass; how many yards does that room contain ? By Duodecimals. F. I. 126 : 3 15 : 9 By Decimals. 126.25 15.75 630 126 63 : 1 : 6 31 : 6 : 9 3:9:0 63125 88375 63125 12625 9)1988.4375 9)1988 : 5 : 3 220.8 Jlns. 220 yards S feel. Ms. 220.8 2. If a room of wainscot he 16 feet 3 inches high, and the compass of the room 137 feet 6 inches; how many yards are contained in it ? Jlns. 2-18 yards 2 feet. 3. If the window-shutters about a room he 69 feet 9 inches broad, and 6 feet 3 inches high, how many yawls are contained therein at work and half ? Jlns. 72.656 yards. 4. What will the wainscoting of a room come to at 6 shillings per square yard, supposing the height of the room, including the cornice and moulding, be 12 feet 6 inches, and the compass 83 feet 8 inches ; three window-shutters, each 7 feet 8 inches by 2 feet 6 inch- es, and the door 7 feet by 3 feet 6 inches ; the shutters and door being worked on both sides, are reckoned work and half ? dns. 1.36 4s. 6|d. Jim The Mensuration of Bricklayers' Work, d. per square yard. After dua allowance for girt of cornice, &c. it is 16 feet 3 inches high; the door is 7 feet by 3 feet 9 inches: the window shutters, two pair ai - 7 feet 3 inches by 4 fret 6 inches ; the cheek-boards round ihem come 13 inches below the shutters, and are 14 inches breadth; the fining-boards round the door-way are 16 inches broad; the door and window-shutters being worked on both sides, are reckoned as work and half, and paid for accordingly : the chimney 3 feet 9 inches by 3 feet, not being enclosed, is to be deducted from the superficial content of the room. The estimate of the charge is required ? Jlns. Z.43 :4s. :6fd. § II. Of BmcKLArERS' Work. The principal is tiling, walling, and chimney -work, I. Of Tiling. Tiling is measured by the square of 100 feet, as .flooring, partitioning, and roofing were in the carpen- ter's work ; so that between the roofing and tiling, the difference will not be much ; yet the tiling will be the most; for the bricklayers sometimes will require to have double measure for hips and vallies. When gut- ters are allowed double measure, the way is to measure the length along the ridge-tile, and by that means the measure of the gutters becomes double : it is usual al- so to allow double measure at the eaves, so much as the projection is over the plate, which is commonly about IS or 20 inches. \v iration of Bricklayers' Work, $c. mi-i.i.s. I ii< iv i> ;i roof covered willi tiles, whose depth on both sides (v itli (lie usual allowance at the eaves) is .;: feet -J inches, and the length 43 feet ; 1 demand how many squares of til jug are contained therein? By Duodecimals. By Decimals. \\ I. 87,28 ;}7 : 3 45 13 : 18623 1S3 11900 ^L4S 11. : 3 16 | 76.23 16176 : 3 Jns. 16 squares, 76 feet. -. There is a roof covered with tiles, whose depth yii both sides (with the allowance at the eaves) is 35 feet" 9 inches, and the length 43 feet 6 inches; I de- mand haw many squares of tiling are in the roof? Jln§. 13 squares 55 feet. 3. What will the tiling a barn cost at lA 5s. 6d. per square, the length being 43 feet 10 inches, and breadth 27 feet 5 inches, on the Hat, the eaves projecting 16 inches on eacli side ? Jlns. 1,24 9s. 5^d. 2' IT. Of Walling. Bricklayers commonly measure their work by the rod square of 16 feet and an half; so thai one rod in length, and one in breadth, contain 272.25 square feel ; for f1i.5, multiplied into itself, produces 272.23 square feet. But in some places the custom is to allow 18 The Mensuration of Bricklayers' Work, Sfc. 231 feet to the rod ; that is, 321 square feet. And in some places the Usual way is, to measui-6 by the rod of 2t feet long and 3 feet high, that is, 63 square feet ; and here they never regard the thickness of the wall in measuring, but regulate the price according to the thickness. When you measure a piece of brick-work, the first thing is to enquire by which of these ways it must be measured, then, having multiplied the length and breadth in feet together, divide the product by the pro- per divisor, viz. 272.23 ; 324 ; or 63 ; according to the measure of the rod> and the quotient will be the answer in square rods of that measure. But commonly brick-walls, that are measured by the rod, are to be reduced to a standard thickness ; viz. of a brick and a half thick (if it be not agreed on the contrary;) and to reduce a wall to standard thick- ness, this is THE RULE. Multiply the number of superficial feet that are found to be contained in any wall by the number of half-bricks which that wall is in thickness ; one-third part of that product shall be the content in feet, re- duced to the standard thickness of one brick and a half. EXAMPLES. 1. If a wall be 72 feet 6 inches long, and 19 feet 3 inches high, and 5 bricks and a half thick ; how ma- ny rods of brick-work are contained therein, when re- duced to the standard ? The Mensuration of Bricklayers- Work, tj*c. By Decimals. I 9625 3850 1305.625 11 :) 153 5 1.875 .2:2.25)5117.291(18 rods. 239479 63.06)216.79(3 quar. 12.61 feet. Bit Duodecimals. V. J. 72 : 6 19 : 3 648 IS : 1 : 6 9 : 6 : - 1395 : 7 : 6 11 3)15351 10 : 6 272)5117 : 2397 (18 quar rods 68)221(3 ters. feel. Note. That 68.06 is one -fourth part of 272.25, aud 68 is one-fourth of 272. In reducing feet into rods, it is usual to reject the odd parts, and divide only by 272, as is done in the second way of the last example ; here the auswer is 18 rods 3 quarters and 17 feet ; about 4£ feet more than by the first way, where it is done decimally; a difference too trilling to be considered in practice. To find proper Divisors for bringing the Answer in feet, or rods, of the Standard thickness; without multiplying the Superficies, by the number of half- bricks, <$*c. Divide 3, the number of half-bricks in l l 2 , by the number of half-bricks in the thickness, the quotient will be a divisor, which will give the answer in feet. The Mensuration of Bricklayers' Work, Sfc. 23 3 But if you would have a divisor to bring the rnswer in rods at once, multiply 272 by the divisor found for feet, am? the product will be a divisor for rods ; as in the following table. Divisors Divisors for The thickness for the the answer of the wall. answer in feet. in rods. 1 Brick 1.5 408. 1§ Brick 1. 272. 2 Bricks .75 204. 2± Bricks .6 163.2 3 Bricks .5 136. 3f Bricks .4285 116.6 4 Bricks .373 102. 5\ Bricks .2727 74.17 By Scale and Compasses. Extend the compasses from the tabular divisor against the given thickness, to the length of the wall, that extent will reach from the breadth to the con- tent. Or by the Carpenter' 's Rule, As the tabular divisor, against the thiekness of the wall, is to the length of the wall ; so is the breadth to the content. Taking the preceding example, extend the eompas ses from 74.17 to 19.23 tol8|rods. '2.5. that extent will reach from By the Carpenter's Mule. B A 74.17 : 72.5 B : 19.23 A 181 The Mensuration of Bricklayers* Work, $ei The dimensions of b building arc generally taken by measuring half round the outside, and half round the inside, for the whole length of the wall; this length being multiplied by the height gives the super- ficies. And to reduce it to the standard thickness, 8a proceed ns above. All the vacuities, such as doors, windows, window-backs, &c. must be deducted. To measure an arched-way, arched-window, or doors, &c. take me height of the window, or door, from the crown or middle of the arch, to the bottom or sill ; and likewise from the bottom or sill to the spring of the arch, that is, where the arch begins to turn. Then to the latter height add twice tlie for- mer, and multiply the sum by the width of the win- dow, door, &c. and one-third of the product will be the area, sufficiently near for practice. 2. If a wall be 245 feet 9 inches long, 16 ftet 6 inches high, and two bricks and a half thick ; how many rods of brick-work are contained therein, when reduced to the standard thickness. dns. 24 rods 3 quarters 24 feet. 3. How many rods are contained in a wall 63| feet long, 14 feet 11 inches high, and 2£_ bricks in thick- ness, when reduced to the standard ? Jlns. 5 rods 218 feet. 4. A triangle gable-end is raised to the height of 15 feet abo\e the end-wall of a house, whose width is 45 feet, and the thickness of the wail is 2\ bricks; required the content in rods at standard thickness ? Jlns. 2 rods 18 feet. 5. Admit the end-wall of a house to be 28 feet 10 inches in breadth, and the height of the roof from the The Mensuration of Bricklayers' Work, $c. 233 ground 55 feet 8 inches, the gable (or triangular part above the side walls) to rise 42 courses of bricks, reck- oning 4 courses to a foot ; and that 20 feet high be 2\ bricks thick, 20 feet more 2 bricks thick, and the re- maining 15 feet 8 inches t£ brick thick ; what will the work come to at 1.5 10s. per rod, the gable being pne brick in thickness ? Jins. £.48 13s. 5£d, III. Of Chimneys. If you are to measure a chimney standing by itself,. without any party-wall being adjoining, then girt it about for the length, and the height of the story is the breadth ; the thickness must be the same as the jambs are of, provided that the chimney be wrought upright from the mantle-tree to the cieling, not deduct- ing any thing for the vacancy between the floor (or hearth) and the mantle-tree, because of the gatherings of the breast and wings, to make room for the hearth in the next story. \i the chimney-back be a party-wall, and the wall be measured by itself, then you must measure the depth of the two jambs, and the length of the breast for a length, and the height of the story for the breadth, at the same thickness your jambs were of. . When you measure chimney-shafts, viz. that part which appears above the roof, girt them with a line round about the least place of them, for the length, and take the height for the breadth : and if they be four inches thick, then you must set down their thick* ness at one brick-work ; but if they be wrought 9 inches thick (as sometimes they are, when they stand high and alone above the roof,) then you must account your thickness i\ brick, in consideration of plastering? and the trouble in scaffolding. novation of Bricklayers' Work, any of (lie floors, is drawn in the above figure. Also (he jambs HIKw, Jik(«/>. are supposed to he perpen- dicular to the breast 1L, viz. IUL and ILK are right angle* ; the tame must be observed in the other two floors. 2. Suppose the breadth TL of the breast of a chim- ney, not standing in an angle, to be 7 feet :; inches, the depth III or LIv of the jambs 3 bricks thick; the height of the room Irt 14 feet 6 incites ; the vacuity for the tire-place 4 feet high, 3 feet 6 inches wide, and 3 bricks deep. The chimney and fire-place in each of the other two rooms are of the same dimen- sions, but the height of the room aD is 12 feet; and the jamb cB, which is the same height as the story, 10 feet 6 inches. The shaft BA stands 6 feet above the roof, and its compass is 10 feet; also its thickness is estimated at !*■ brick; how many rods of brick- work standard thickness, allowing double measure, are contained in the chimney, deducting the three fire- places. *2ns. 3 rods 3 quarters 5 feet. §111 Of Plasterers' Work. The plasterers' works are principally of two kinds \ namely, 1. plastering upon laths, called cieling, and 2. plastering upon walls, or partitions made of framed timber, called rendering. In plastering upon walls, no deductions are made except for doors, windows, &c. but in plastering timber partitions, in large ware- houses, &e. where several of the braces and larger timbers project from the plastering, a fifth part is commonly deducted. Plastering between these timbers is generally called rendering between quarters. Whitening and colouring are measured in the same manner as plastering : as in rendering between these projecting timbers one fifth-part is deducted, so in gu- The Mensuration of Plasterers 9 Work, Sfc. 239 louring, it will be necessary to add one-fourth or one-fifth of the whole, for the trouble of colouring the projections. Plasterers' work is measured by the yard square, consisting of 9 square feet. In arches, the girt round them, multiplied by the length, will give the superficies. EXAMPLES. 1. If a cieling be 59 feet 9 inches long, and 24 feet <5 inches broad; how many^ards does that cieling contain ? By Decimals. 59.75 21.5 By Duodecimals. F. I. 59 : 9 24 : 6 / 236 118 29 : 10 : 6 18 : : 9)1-163 : 10 : 6 29873 23900 11950 9)1463.873 feet. :s. 162.65 yards. 162 yards 5 feet. 2. If the partitions between rooms be 141 feet 6 inches about, and 11 feet 3 inches high; how many yards are in those partitions ? Jins. 176.87 yards. 3. AVhat will be the expence of plastering a ciel- ing, at llj-d. per yard, supposing the length 22 feet 7 inches, and breadth 13 feet 11 inches ? Ms. I A 13s. 0fd. 4. There is a partition which measures 234 feet 8 inches round, and 14 feet 6 inches high, this partition is rendered between quarters, that is, it is made of framed timber, and the interstices are filled up with 960 The Mensuration of Painters' Work. hit h and plastering , The lathing and plastering wiH be Bd. per yard, and the whitening 2d. per yard; what will the whole eome to ? Ans. j.18 IV*. J'd. 5. The length of a room is 14 feet .1 inches, breadth IS feet C inches, and height 9 feet •> inches, to the un- der side of the cornice, which projects 5 inches from the wall, on the upper part next the cieling ; required the quantity of rendering and plastering, there being no deductions but for one door, the size whereof is 7 feet by i . : Ans. 5o ya#*ls 3 feet of rendering, and 18 yard* 3 feet of cie'ing. 6. The circular vaulted iyf of a church measures 103 feet 6 inches in the arch, "and 273 feet 5 inches in length 5 what will the plastef ing come to at is. per yard? Ms. U5Q 17s. B\d. U/ § IV. Of Paint-ers^ Work. The taking the dimensions of painters' work is the same as that of joiners, by girting over the mouldings and swelling pannels in taking the height; and it is but reasonable that they should be paid for that on which their time and colour are both expended. The dimensions thus taken, the casting up, and reducing feet into yards, is altogether the same as the joiners' work, but the painter never requires work and half, but reckons his work once, twice, or thrice coloured over. Only take notice, that window-lights, window- bars, casements, and such-like things, they do at so much a piece. EXAMPLES. 1. If a room be painted, whose height (being girt over the mouldings) is 10 feet 6 inches, and the com- Tlie Mensuration of Painters' Work. Jll pass of the room 97 feet 9 inches ; how many yards are in that room ? By Duodec F. I. 97 : 9 16 : 6 imals. : 6 By Decimals. 97.5 16.5 584 98 48 : 10 4S875 58630 ' 9775 1612 : 10 6- 9)1612.875 179 : 1 179.2. Facit 179 yards 2 feet, nearly. 2. A gentleman had a room painted at S^d. per yard, the measure whereof is as follows*; the height 11 feet 7 inches, the compass 74 feet 10 inches, the door 7 feet 6 inches by 3 feet 9 inches ; five window-shutters, each 6 feet 8 inches by 3 feet 4 inches, the breaks in the windows 14 inches deep and 8 feet high ; the open- ing for the chimney 6 feet 9 inches by 5 feet, to be de- ducted; the shutters and doors are coloured on both sides ; what will the whole come to ? Ans. lA 6s. lid. 3. Suppose a room were to be painted, and that its length is 24 feet 6 inches, breadth 16 feet 3 inches, and height 12 feet 9 inches ; also the size of the door 7 feet by 3 feet 6 inches, and the size of the window- shutters to each of the windows, there being two, is 7 feet 9 inches by 3 feet 6 inches ; but the breaks of the windows themselves are 8 feet 6 inches high, and 1 foot 3 inches deep ; what will be the expence of giving it 3 coats, at 2d. per yard each; the size of the fire- place to be deducted, being 5 feet by 5 feet 6 inches ? Jins. 1,3 3s. 104-. X 143 The Mensuration of Glaziers' Work. § V. Of Glaziers' Work: Glaziers take their dimensions in feet, inches, and eighths or tenths, or else in feet and hundredth parts of a foot, and estimate their work by the square-foot Windows are sometimes m asored by taking the di- mensions of one pane, and multiplying its superficies hv the nnniher of panes. But more generally, they measure the length and breadth of the window over all the pan s and their frames, for the length and breadth of the glazing. Circular or oval windows, as fan lights, See. are measured as if they were square, taking for their di- mensions the greatest length and breadth ; as a com- pensation for the waste of glass, and labour in cutting it to the nec/ssary forms. Plumbers' work is rated at so much per pound, or by the hundred-weight of 112 pounds. Sheet-lead, used in rooting, guttering, &c. weighs from 7 to 12 pound per square foot, according to the thickuess. And a pipe of an inch hore weighs com- monly 13 or 14 pounds per yard in length. EXAMPLES. 1. If a pane of glass he 4 feet 8 inches and 3 quar- ters long, and 1 foot 4 inches and 1 quarter broad : how many feet of glass are in that pane ? By Duodecimals. By Decimals. ft I. P. 4.729 4:8:9 1.334 1:4:3 18916 4:8:9 21645 1 : v 6 :11 : 14187 1:2:2:3 4729 6 : 4 :10 : 2 : 3 6.403066 Ms. 6 feet 4 inches. The Mensuration of Glaziers 9 Work. 243 2. If there be 8 panes of glass eaeh 4 feet 7 inches 3 quartos long, and 1 foots inches 1 quarter broad; how many feet of glass are contained in the said 8 panes ? Ms. 53 feet 5 inches. 3. If there be 16 panes of glass, each 4 feet 5 inch- es and a half long, and 1 foot 4 inches 3 quarters broad; how many feet of glass are contained in them? Jlns.. 99 feet 6 inches. 4. If a window be 7 feet 3 inches high, and 3 feet B inches broad ; how many square feet of glazing are contained therein ? Jlns. 24 feet 9 inches. 5. There is a house with three tiers of windows, 7 in a tier ; the height of the first tier is 6 feet 11 inches, of the second 5 feet 4 inches, and of the third 4 feet 3 inches ; the breadth of each window is 3 feet 6 inches ; what will the glazing come to at 14*d. per foot ? Ms. l.24< 8s. 5 Id. 6. What will the glazing a triangular sky -light come to at lOd. per foot : the base being ±2 feet 6 inches long, and the perpendicular height 16 feet 9 inches ? Ms. Ia. 7s. 2^d. 7. What is the area of an elliptical fan-light, of 14 feet 6 inches in length, and 4 feet 9 inches in breadth? Ms. 68 feet 10 inches. 8. W T hat cost the covering and guttering a roof with lead, at 18s. per hundred-weight ; the length of the roof being 43 feet, and the girt over it 32 feet ; thfe guttering being 57 feet in length and 2 feet in breadth ; admitting a square foot of lead to weigh 8f pounds. Ms. Z.104 15s. 3 |cL The Mensuration of Masons- Work; §c. §VI. Of Ma soy s' Work. • Masons measure their work sometime* by the foot solid, sometimes by the foot superficial, and sometimes by ihe foot in length. In taking dimensions they girt all their mouldings as joiners do. The solids consist of blocks of marble, stone, pil- lars, columns, &c. The superficies are pavements, slabs, chimney-pieces, &c. Masons reckon all such stones as are above two inches thick, at so much per foot, solid measure. And for the workmanship, they measure the superficies of that part of the stone which appears without the wall. EXAMPLES. 1. If a wall be 97 feet 5 inches long, 18 feet 3 inches high, and 2 feet 3 inches thick; how many solid feet are contained in that wall ? By Duodecimals. F. I. 97 : 5 length. 18 : 3 breadth or height 776 97 24 : 4 : 3 6 : : o 1 : 6 : 1777 : 10 : 3 superficies. 2 : 3 thickness. 3555 : 8 : 6 444 : 5:6:9 4000 : 2:0:9 solidity. By Decimals. 97.417 18.25 4S7085 194834 779336 97417 1777.86025 2.25 888930125 355572050 255572050 4000.1855625 The Mensuration of Paviors' Work. 245 2. If a wall be 107 feet 9 inches long, and 20 feet 6 inches high ; how many superficial feet are contain- ed therein ? Jlns. 220$ feet 10 inches. 3. If a wall he 112 feet 3 inches long, and 16 feet 6 inches high ; how many superficial rods of 63 square feet each, are contained therein ? Jlns. 29 rods 25 feet. 4). What is a marble slab worth, whose length is 5 feet 7 inches, and breadth 1 foot 10 inches, at 6s. per foot superficial ? Jlns. 1.3 Is. 5d. § VII. Of Pa viors* Work. Paviors' work is measured by the square yard, con- sisting of 9 square feet. The superficies is found by multiplying the length by the breadth. EXAMPLES. 1. What cost the paving a foot path at 2s. 4d. per yard ; the length being 35 feet 4 inches, and breadth S feet 3 inches ? x 2 246 Tfie Mensuration of Paviors' Work. By Decimals. Jhj Duodecimals. F. I. 35.3383 &C 33 : 1 8± 8:3 382.6066 282 : 8 8.8333 8 : 10 9)291.4099 &C. 9)291 3s. id.is 1)32.3888 &C\ 3s. 4d. is £)32.3 : 6 J.5.3981 Feet - - - 5 6 8 20 3—-L of a yard 1 1|33 6=z^ of 3 feet 2.88 S. 7.9620 12 1.5 7 11' 21 d. 11.5140 4 2.1760 Answer 1.5 7s. ll^d. 2. What will the paving a court-yard come io at 3s. id. per yard, the length being 24 feet 5 inches, and breadth 12 feet 7 inches? dns. 1.5 13s. 9 t ]d. 3. What will be the expence of paving a rectangu- lar court-yard, its length being 62 feet 7 inches, and breadth 44 feet 5 inches ; and in which there is laid a foot-path the whole length of it. 54 feet broad, with flat stones, at 3s. per yard, the rest being paved with pebbles at 2s.. 6d. per yard? dns. I.B9 lis. S^d 1 . Of the Sliding Rule, 2if CHAPTER V. Of Gauging*. GAUGING is the art of measuring and finding tlie contents of all sorts of vessels, in gallons, or cubi£ inches ; such as casks, brewers' vessels, &c. &c» Of the Sliding Rule. The sliding rule is an instrument, particularly use- ful in gauging, made generally of box, in the form of a parallelopipedon. There are various kinds, but the the most convenient, or at least that which is most used in the excise, was invented by Mr. Verie, collec- tor of the excise. 1st, The line, marked A, on the face of this rule, is called Gunter's Ride, and is numbered 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. At 2150.42 is a brass pin, marked MB, signifying the cubic inches in a bushel of malt ; at 282 is another brass pin, marked A, signifying the number of cubie inches in a gallon of ale. 2dly, The line marked B is on the slide, and is di* vided in exactly the same manner as that marked A ; there is another slide B, which is used along with this ; the two brass ends are then placed together, and so make a double radius numbered from the left-hand towards the right. At 231, on the second radius, is & '„>is Of the Sliding Rule. brass pin, marked AV, signifying the cubic inches in a gallon of wine; at 314- is another brass pin, marked C, signifying the circumference of a circle whose diameter is 1. The manner of reading, and using these lines, is exactly the same as the lines A and 13, described in chap. XI. page 52, &c. 3dly, The back of the first radius, or slide, marked B, contains the divisors for ale, wine, mash-tun gallons, malt, green starch, dry starch, hard soap hot, hard soap cold, green soft soap, white soft soap, flint glass, Sfc. Sfc. as in the table, prob. 1, following. — Tlie back of the second radius, or slide, marked B, contains the gauge- points correspondent to these divisors, where S stands for squares, and C for circles. ^thly, The line M. D. on the rule, signifying malt depth, is a line of numbers beginning at 2150.42, and is numbered from the left to the right-hand 2. 10. 9. 8. 7. 6. 5. 4. 3. This line is used in malt gauging. 5thly, The two slides, B, just describ d, are always used together either with the line A; Ml); or the line D, which is on the opposite face of the rule to that already described. This line is numbered from the left-hand towards the right 1. 2. 3. 31. 32. which is at the right-hand end: it is then continued from the left-hand end of the other edge of the rule 32. 4. B. 6. 7. 8. 9. 10. At 17.15 is a brass pin marked WG. signifying the circular gauge-point for wine gallons. At 18.95 is a brass pin marked AG for ale-gallons. At 46.37 MS signifies the square gauge point for malt bushels. At 52.32 MR signifies the round, or circular gauge-point for malt bushels. The line D on this rule is of the same nature as the line marked D on the Carpenter's Bule, described in chap. XL page 52. The line A and the two slides B are used toge- ther, for performing multiplication, division, propor- tion. &c. and the line D, and the same slides B, are used together for extracting the square root, &c. Of the Sliding Rule. 249 Uhly, The other two slides belonging to this rule are marked C, and are divided in the same manner, and used together, like the slides B. The back of the first radius or slide, marked C, is divided, next the edge, into inches, and numbered from the left-hand towards the right 1. 2. 3. 4. &c. and these inches are again divided into ten equal parts. The second line is marked spheroid, and is numbered from the left-hand towards the right 1. 2. 3. 4. 5. 6. 7. The third line is marked second variety, and is numbered 1. 2. 3. 4. 5. 6. These lines are used, with the scale of inches, for finding a mean diameter, see prob. X. following. The third and fourth variety are omitted on this slide, and with good reason, for it is very probable that there never was a cask made resembling either of these forms. The back of the second radius, or slide, marked C, contains several factors for reducing goods of one denomination to their equivalent value in those of another. Thus | X to VI 6. | signifies that to re- duce strong beer at 8s. per barrel to small beer at Is. 4d. you must multiply by 6. | VI. to X. 17. | signi- fies that to reduce small beer at Is. 4d. per barrel to strong beer at 8s. per barrel, you must multiply by .17. | C 4s. to X. .27. | signifies that 27 is the mul- tiplier for reducing cider at 4s. per barrel to strong beer at 8s. &c. 7thly, The two slides C, just described, are always used together, with the lines on the rule marked seg. st. or 88, segments standing ; and seg. ly or SL, seg- ments, lying, for ullaging casks. The former of these lines is numbered 1. 2. 3. 4. 5. 6. 7. 8, which stands at the right-hand end ; it then goes on from the left- hand on the other edge S. 9. 10, &c. to 100 ; the latter is numbered in a similar manner, 1. 2. 3. 4, which stands at the right-hand end ; it then goes on from the left-hand on the other edge 4. 5. 6. to 100, &c 23% Of Gauging. PROBLEM I. To find the several Multipliers, Divisors, and Gauge- points, belonging to the several Measures now us England. \-v For Square figures, the following multipliers and divisors are to be used: 1.0000(.0Q3546 multiplier for ale gallons. 231)1.0000(.00l: .29 multiplier for wine gallons. 268. S)i.OOO(. 0037 202 multiplier for malt gallons. 2150.42)1.00()(.0004f)502 multiplier for malt bushels. 227)1. 000(. 00405 multiplier for mash-tun gallons. So, if the solid inches in any vessel be multiplied by the said multipliers, the product will be the gallons in the respective measures ; or dividing by the divisors 2S2, 231, or 268.3 ; the quotient will likewise be gal- lons. Note. That 282 solid inches is a gallon of ale or beer measure : 231 solid inches is a gallon of wine measure ; 268.8 solid inches is a gallon, and 2150.42 solid inches is a bushel of malt, or corn measure. For circular areas, the following multipliers and di- visors are to be used : 282).785398(. 002785 multiplier for ale gallons. + 231).785398(. 003399 multiplier for wine gallons, -f- .785398)282.(359.05 divisor for ale gallons. .785398)281.(294.12 divisor for Mine gallons. .785398)2150.42(2738 divisor for corn bushels. The square root of the divisor is the gauge-point. r^ x . . fAle measure, is 16.79 The gauge-point J Wine mea8u ; is ±5±Q for squares in j Ma , t ^^ .' ^7 The gauge-point fAle measure, is 18.95 for circular fi- -< Wine measure, is 17.15 gures in ^Malt bushel, is 52.32 And thus the numfljBin the following table were cal- culated. IT** Of Gauging >31 * if a c 5*5 1 «3 Cj ■r, h & • 2 "W 5 H 3 a .V 5y er. -- fc. « =u St ex .&•<« | *S ^ ^ ^ a Jl <>, It so CO ^ h v) io ^ i-~ oo h. o» ctcho dTcT^ - ' £> -2f 0^ >o o> a h co *n o » oi k N cj (o h iq c« c^ eo £ 2 COOo6K^c6lsl*rjiO*OAO^£J«3islcO'^cO S 6 rt rf H rt >0 r( rl ~5" KOiOiKOi HOi(0>0 »0 Vj tJ< 00 a , v) ts h c; r ; rt q cn q o o a. co n k t}i c* el W^^^u5OVJiOioio ^ i-< *C GS> ■*? o <£> O CO •S S. ^KWO)*OWCNWC^(NC»COCO-*r-lrtH R 5 Si to 00 ^ ^ V) Oi *0 O CT> O rtHV)rt<0 <£> t> »3$ o >9>ocoa<0(W(Oco»o co o o a> o ^»0<* CO OOOOOCOONtNCOOWtNHK'OC 5*< 05 K oooqoqppqqqooqoop ^MOOiOO^iO^OCO^OCOKCOM 8 2 •^K^CT^NO^Hto^rfco^CTiNO O--W5»OC0^K'<} | CCKOlH00KC0«OH'H *fi tOOCO^OcO^^iOCOOlHW^^HIM *» Si >3 O O O O O O O CO CO CO CO CO !M « O) N CO c^ c^ c^ <^ <=> c^ a> '* rH ■2 ? e^ ^ w CO *> B $ - ' *• •• f ,^ e ?£* s 1 n 3 ■ g a i « § "S 3 1 1 '^3 u o . -j 201.55 (_ 21.63 dns. -( 201.55 wine gallons, malt bushels. PROBLEM VIII. To Gauge and Inch a Tun in the form of the Frustum of a Cone, and to make an allowance for the drip or fall. The inching of a tun, or vessel, is finding how much liquor it will hold at every inch of its depth. When a vessel does not stand even, or with its base parallel to the horizon, the quantity of liquor mDw, which will just cover the bottom, is called the drip or fall of the tun. Let water be poured into a tun till the bottom is just covered, which will be when the water touches the point m, and suppose the measure of the water in this case to be 30.92 gallons. Find the horizontal line FB, which will be parallel to m n 9 the surface of the liquor ; the line FB will represent the surface of the liquor when the vessel is full. Find the perpendicular depth on to the surface of the liquor, which suppose to be 26 inches. Take mean diameters pa- .A^— " — ' — rallel to FB and m n at every Ffc^ " % "" fv^N r 6 or 10 inches depth. Sup- tf ,fi§^§:- mSSff pose the first mean diame- filliP SMlf * ter a a at 5 inches from FB ifillL-- ^8l~h to be 83.6 inches; the second lERi "jjl|l>f mean diameter b b at 13 inch- _g 6l| g^^;ejfe^ es from FB—7S.7 inches and *" eg ^Sii||||^ the third mean diameter c c at 23 inches from FB=74.3 inches. 8M Of Gait" Find the areas correspondent to each of those di- ameters, as in the third column of the? following ta- ble; multiply these areas by their respective depths and you will have the fourth column. Lastly, these contents being brought into barrels, &e. allowing 24 gallons to a barrel, and 8*- gallons to a firkin, will give the remaining columns. Depths. Diam. Area. Content. Content. in gallons. B. F. G. oP=lo PS=lo Sw= 6 S3.6 78.7 74.5 19.465 17.250 15.458 194.65 172.50 92.75 30.92 5 2 2 3 7.65 2.50 7.75 5.42 whole 26 Cont.of the drip. Whole content. 490.82 : i 1 ft. 32 Now to find the content at every inch of the depth, reduce the first area 19.465 into barrels, &c. and it will be 2 F. 2.465 G. which subtracted from the whole content 14 B. 1 F. 6.32 G. leaves 13 B. 3 F. 3.8555 G. the content when 1 inch is dry; and thus continue subtracting the first area from the several remainders until 10 inches are dry, when there will remain 8 B. 2 F. 7.17 G. — Then take the second area 17.25, having first brought it into firkins, &c. and subtract it from the last remainder, and thus continue to do till you have 20 inches dry, when there will remain 3 B. 2 F. 4.67 G — Lastly, take the third area 15.458, having first brought it into firkins, &c. and subtract it from the area when 20 inches are dry ; and thus proceed till you have 26 inches dry, and there will remain, if you have made no mistake, 3 F. 5.42 G. the quantity of liquor contained in the drip or fall. Of Gauging. 263 The point F may be marked as a constant dipping place, and the several contents entered into a dimen- sion book. Then to find how much liquor there is in this fixed vessel, at any future time, take the depth of the liquor, and against that depth in your table you will find the content. PROBLEM IX. To gauge a Copper. Let ABCD be a small copper to be gauged. Take a small cord of packthread, make one end fast at A, and extend the other to the opposite side of the copper at B, where make it fast. Then with some convenient instrument take the nearest distance from the deepest place in;the copper, to the thread, as «C, which suppose to be 47 inches. In like manner, set the end of the instrument or rule upon the top of the crown at d. and take the nearest distance to the thread, as dg, which suppose 4.2 inches : this subtracted from aC, 47, the remainder 5 is the altitude of tUf crown. j it l Of Gaugi To find CD, the diameter of the hoi loin of the crown. Measure AB, the diameter ef the top, which admit to hi- y ( .> Inches; then hold a thread so as a plummet at the end thereof may hang just over C, by which means you will find the distance Aa. Do the like on the other side: so will yOu find also the distance, which suppose 17.5 inches each; add these two togeth- er, and subtract their sum (viz. 35) from 99, and the remainder is 04 inches, equal to CD, the diameter at the bottom 'of the crown. The diameter hk 9 which touches the top of the crown, may be found, by mea- suring, to be 65 inches. Now to find the content of the copper from tlie crown upwards, that is the part ATM//, the depth i ' being 42 inches, you may take the diameter in the mid die of every 6 inches of the depth, which suppose to he as in the second column of the following table, the numbers in the third column are the respective areas in ale gallons, found by prob. ID. the fourth column shews the content of every 6 inches ; all which being added together, the sum will be the content of that part, AB/i7i ; that is, so much as it will hold after the crown is covered. Now, if the crown be taken for the segment of a sphere, the content (by the latter part of sect. XI. p. 74.) will be found to be 28.75 gallons. But may be more readily found, very near the truth, thus: The diameter CD was found to be 64, and the area to this diameter is 11.408; this multiplied by half the crown's altitude, viz. by 2.5, gives 28.52 gallons, the content of the crown. The content of the part hkiDC is 57.935 gallons ; from which subtract the content of the crown, 28.52, and the remainder is 29.415 gallons, and so much li- quor will just cover the crown. Of Gauging. 265 Parts of the depth. Diam. Areas. Content of every six inches. 6 6 6 6 6 6 6 i>5.3 90.1 85.0 80. 75.2 70.5 66. 25.2943 22.6095 20.1223 17.8246 15.7499 13.8426 12.1310 151.767 135.657 120.734 106.947 94.499 8 .056 72.791 Thesu To jus 765 4 1 t cover the crown - - hole content - - - - 29.415 Thew - 794.866 The contents in the last column may be brought in- to firkins and barrels, and then the content at every inch in depth, as in the Vlllth problem, may be found and entered into the dimension book. PROBLEM X. To compute the Content of any close Cask, In order to perform this difficult part of gauging, te three fol truly taken : the three following dimensions of the cask must be ! } cask, J within the cask. f The bung-diameter, Viz. < The head-diameter (The length of the In taking these dimensions, it must be carefully'ob- served, 1st, That the bung-hole be in the middle of the cask ; also, that the bung-stave, and the stave op- Of Gauging, posite to the bung-hole, arc both regular and within. 2dljfj That the heads of the en-k'- arc equal, and truly circular: if so. the distance between the inside of the chhnb to the outside of the opposite stave will be the head-diameter within the cask, ven near. The diameters and length of one cask may be equal to those of another, and yet one of those casks may contain several gallons more than the other. \> for instance, the figure MH'DF is supposed to rep- resent a cask : then it is plain, that if the outward curve lines, ABC, and FGD, are the hounds or staves of the cask, it will hold more than if the inner dotted lines """ '&' were the hounds, or staves ; and yet the bung-diame- ter BG, and head-diameters CD and AF, and the length LH, are the same in both those casks. Whence it appears, that no one general rule can be given, by which all sorts of casks may be gauged ; and therefore gaug«>rs usually suppose every cask to be in some of these forms : ls£, The middle zone or frustum of a spheroid ; see page 190. 2dly, The middle frustum of a parabolic spindle ; see page 18. Sdly, The lower frustums of two equal parabolic conoids ; see page 193. ( 4umg- diameter 31.5, and length iz inches; required its content in ale and wine gallons ? Jlns J 95 - 7 ' 66 a * e gallons. \ 117.047 wine gallons. The content by the sliding rule is exactly the same. 3. Required the content of a cask in ale and wine gallons, whose head-diameter is 24 inches, bung-diam- eter 32 inches, and length 40 inches ? q C 112-28 wine gallons. ** ns ' I 91.86 ale gallons. The content by the sliding rule is the same. 4. The bung-diameter of a cask is 48 inches, head- diameter 35.8 inches, and the length 55 inches 5 re- quired the content in ale and wine gallons ? - C 2S3.178 ale gallons. 1 346.106 wine s-allons. The content is the same by the sliding rule. 5. The head-diameter of a cask is 28.2 inches, bung- diameter 33.8 inches, and length 48 inches ; required the content in ale and wine gallons ? 132.367 ale gallons. 61.782 wine gallons. Ms.\l r The content by the sliding rule is the same. PROBLEM XI. Of the Ullage of Casks. The ullage of a cask is what it contains when only partly filled, and is considered in two positions, viz, standing on its end, or lying on one side. Of Gauging. 273 To Ullage a lying Cask. RULE. Divide the wet inches by the bung-diameter : find the quotient in the column height, in the table at the end of this chapter; take out the corresponding area seg. multiply this area by the content of the cask, and that product by 1.27324; or divide it by .7854, the last product, or quotient, will give the ullage nearly. By the Sliding Rule. Set the bung-diameter on C to 100 on the line mark- ed seg. ly. or SL, viz. segments lying : then look for the wet inches on C, and observe what number stands against it on the segments, which call a fourth num- ber. Then set 100 on A to the content of the cask upon B, and against the fourth number, before found, on A is the quantity of liquor in the cask on B. EXAMPLES. 1. Supposing the bung-diameter of a lying cask to be 32 inches, its content 97.6 ale gallons ; required the ullage for 19 wet inches ? 1.000 whole diameter. 32)19.000(.594 height. ,7854 whole area* 0.406 height. Area seg. .299255 .486145 rem. The content 97.6 2916870 3403015 4375305 .7854)47.4477520(64.4 gal. Note. Because, in this example, the quotient of the wet inches, divided by the bung-diameter, ex T on C. nnSl, onC. : 100 : : 19 on A. onB. on A. As loo : 97.6 : : 8&0 Of Hanging. eeedtthe heights in the (able, it is necessary to sub- trait it from the whole diameter, &e. Thj the Sliding Hulr. on & 09.5 fourth number, on STi. : 61 gallons, .tins. 2. The bung-diameter of a lying cask is 20 inches, the content 18.3 ale gallons and the wet inches 11 ; re- quired the quantity of liquor in the cask? a $20.46 by calculation. Mns ' 1 19.9 by the sliding rule. 3. The bung-diameter of a lying cask is 32 inches, its content 91.6 gallons, and the wet inches 8 ; required the quantity of liquor in the cask? a C 1?.9 gallons, by calculation. * US ' £ 16. gallons, by the sliding rule. To Ullage a standing Cask. Multiply the difference between the squares of the bung and head diameters, by the square of the dis- tance of the liquor's surface from the middle of the cask; and divide the product by the square of half the length of the eask, subtract one-third of the quotient from the square of the bung-diameter, and multiply the remainder by the distance of the liquor's surface from the middle of the cask. The last product divided by 359.05 for ale, or 294.12 for wine, will give the quantity of liquor above, or un- der half the content of the cask, according as the wet inches exceed, or fall short of half the length of the eask. Of Ganging. By the Sliding Rule. Set the length of the cask on C to 100 on the line marked seg. st. or SS, viz. segments standing ; then look for the wet inches on C, and observe what num- ber stands against it on the segments, which call a fourth number. Then, set 100 on A to the content of the cask upon B, and against the fourth number, before found, on A is the quantity of liquor in the cask on B. EXAMPLES. 1. Suppose in the annexed cask, whose content is 97.6 ale gallons, that the bung diameter EF be 32 inches, head-diameter AB 24 inch- es, length 40 inches, and the wet inches SH 26 ; what quantity of ale is contained in the caskP * o s — *\ H 3 2~ =EF 24—AB 2fc=SH 32 24 20^1H 64 96 6 diff.=SI 96 48 6 1024 diff. 076 20=IH 36 square of SI 448 36 20 26S8 400=IIP 1344 quot. 1024 13. square of EF 44 0) 16128 3)40.32 1010 56 6= SI 13.44 359.03)6063.36(16.837 gallons, 89* Of Banging, above half the content of the. cask, viz. in ad EF. To which add 48.8 gallons, half the content of the cask, and the gum is 65.687, the quantity of the liquor in the cask. By the Sliding Rule* on C. on SS. on C. on SS. As 40 : 100 : : 26 : «6.1 fourth numher. on A. on B on A. on B. As 100 : 71.6 : : 66.1 : 64.5 gallons, Jlns. 2. The bung-diameter of a standing cask is 35 inch- es, head-diameter 28.7, length 40, wet inches 30 ; con- tent in ale gallons 121.5, in wine gallons 148.5; re- quired the content of the ullage iu ale and wine»gal- lons? a J 93 - 9 -» ale gallons. * £ 114.76 wine gallons. The answer by the sliding rule is the same. 3. The bung-diameter of a standing cask is 26.5 inches, head-diameter 23 inches, length 28.3, wet inch- es 11 ; and the content in ale gallons 48.3 ; required the ullage? a J I 801 gallons, by calculation. Jln ' 1 18.3 by the sliding rule. The use of the fallowing table in gauging has been shewn '.at the beginning of the 11th problem ; but its chief use is for finding the area of a segment of a cir- cle. Thus, Divide the height of the segment, by the diameter of that circle of .which it is the segment, to three pla- ces of decimals ; find the quotient in the column height, take, out the corresponding Area Seg. which multiply by the square of the aforesaid diameter, and the pro- duct will be the area of the segment required. If the quotient of the height by the diameter be greater than 15, subtract it from an unit, and find the pf Gauging, 277 area seg. corresponding to the remainder; which sub- tract from .7854 for the area seg. If the quotient of the height by the diameter do not terminate in three figures, find the area seg. answering to the first three decimals of the quotient; subtract it from the next greater area seg. multiply thv* remainder by the fractional part of the quotient, and add the pro- duct to the first area seg. taken out. —— EXAMPLES. 1. Required the area of the segment of a circle, whose height is 3.25 ; the diameter of the circle be- ing uO ? 50)3.25(.065 quotient, or tabular height. The tabular segment is .021659, which multiplied by 2500, the square of the diameter, gives 54.1475, the area/required. ™ Required the area of the segment of a circle, whose height is 46.75, and diameter of the circle 50? \1.000 50)46.75/ .93.1 quotient. Whole area .7854 .065 tab. height. Area seg. .021659 Remains, tabular area seg. .763741 which multiplied by 2500 gives 190j.3525, the area required. 3. Required the area of the segment of a circle whose height is 2, and diamet r of the circle 52? 52) 2.000(.0 8||=. 03 8 T 6 T = quotient. The area seg. answering to .038 is .009763 ; the next greater area seg. is .010148; the difference is .000^85, T 6 T of which is 000177, which added to .009763 gives .009940, the area seg. corresponding to .038 T \ : heuce .009940 X square of 52=26.87776 answer. A a Areas of the Segments of a Circle. A TABLE OF THE AREAS SEGMENTS OF A CIRCLE, Hlwse Diameter is Unity, and supposed to be divided into 1000 equal jiarts. Beig. .bf a Se$, f J1 <'i$- . trea Meg. "V/^-. . iff a Seg. .001 .000042 .036 .009008 .071 .024680 .002 .000119 .037 .009385 .072 .025195 .003 .000219 .038 .009763 .073 .025714 .004 .000337 .039 .010148 1 .074 .026236 .005 .000470 .040 .01^537 | .075 .026761 . 06 .000618 .041 .010931 .076 .027289 .007 .000770 .042 .011330 .077 .027821 .008 .000950 .043 .011730 .078 .028356 .009 .001135 .044 .012142 .079 .028894 .02940k .010 .001320 .045 .012554 .080 .011 .001533 .046 .012971 .081 .029979 .012 .001746 .047 .013392 .082 .030526 .013 .001968 j .048 .013818 .083 .0310, 6 .014 .002199 1 .049 .014247 .084 .031629 .015 .002438 j .050 .014681 .085 .032186 .016 .002685 .051 .014119 ,086 .032745 .017 .002940 .052 .015561 .087 .033307 .018 .003202 .053 .016007 .088 .033872 .019 .003471 | .054 .016457 .089 .034441 .020 .003748 .055 .016921 .090 .035011 .021 .004031 .056 .017369 .091 .035585 .022 .004322 .057 .017831 .092 .036162 .023 .004618 .058 .018296 .093 .036741 .024 .004921 .059 .018766 .094 .037323 .025 .005230 .060 .019239 .095 .037909 .026 .005546 .061 .019716 .096 •038496 .027 .005867 .062 .020196 .097 .039087 .028 ' .006194 .063 .020691 .098 .039680 .029 .006527 .064 .021168 .099 .040276 .030 .006865 .065 .021650 .100 .040875 .031 .006909 .066 .022154 .101 .041476 .032 .007558 .067 .022652 .102 .042080 .033 .007913 .068 .023154 .103 .042687 .034 .008275 .069 .023659 .104 .043296 .035 .008638 .070 .024168 .105 .043908 Areas of the Segments of a Circle. 279 Heig. Area Seg. Heig. Area Seg. Heig. Area Seg. ■ — — _____ __ — ■ — _— . .106 .044522 .152 .075306 .198 .110226 .107 .045139 .153 .076026 .199 .111024 .108 . .045759 .154 .076747 .200 .111823 .109 .046381 .155 .077469 .201 .112624 .110 .047005 .156 .078194 .202 .113426 .111 .047632 .157 .078921 .203 .114230 .112 .048262 .158 .079649 .204 .115035 .113 .048894 .159 .080380 .205 .115842 .114 .049528 .160 .081112 .206 .116650 .115 .050165 .161 .081846 .207 .117460 .116 .050804 .162 .082582 .208 .118271 •117 .051446 .163 .083320 .209 .119083 .118 .052090 .164 .084059 .210 .119897 .119 .052736 .165 .084801 .211 .120712 .120 .053385 .166 .085544 .212 .121529 .121 .054036 .167 .086289 .213 .122347 .122 .054689 .168 .087036 .214 .123167 .123 .055345 .169 .087785 .215 .123988 .124 .056003 .170 .088535 .216 .124810 .125 .056663 .171 .089287 .217 .125634 **26 .057226 .172 .090041 .218 .126459 .127 .057991 .173 .090797 .219 .127285 .128 .058658 .174 .091554 .220 .128113 .129 .059327 .175 .092313 .221 .128942 .130 .059999 .176 .093074 | .222 .129773 .131 .060672 .177 .093836 1 .223 .130605 .132 .061348 .178 .094601 I .224 .131438 .133 .062026 .179 .095366 j .225 .132272 .134 .062707 .180 .096134 , .226 .133108 .135 .063389 .181 .096903 .227 .133945 .136 .064074 .182 .097674 .228 .134784 .137 .064760 .183 .098447 .229 .135624 .138 .065449 .184 .099221 .230 .136465 .139 .065140 .185 .099997 .231 .137307 .140 .066833 .186 .100774 .232 .138150 .141 .067528 .187 .101553 .233 .138995 .142 .068225 .188 .102334 .234 .139841 .143 .068924 .189 .103116 .235 .140688 .144 .069625 .190 .103900 .236 .141537 .145 .070328 .191 .104685 .237 .142387 .146 .071033 .192 .105472 .238 .143238 .147 .071741 .193 .106261 .239 .144091 .148 .072450 .194 .107051 .240 .144944 .149 .073161 .195 .107842 .241 .145799 .150 .073874 .196 .108636 .242 .146655 .151 .074589 .197 .109430 .243 .147512 eso Areas of the Segments of a Circle. Hdir. . //•<•(.' 1 Bag. ! 590 Area 8eg. Heisr. . Irea Seg. '1 1 1HJ71 .189047 .336 .231689 l 19330 I ::\n .189055 ' .337 634 .246 1 1 091 1 .292 .190804 .338 , m .150953 .293 .191775 .339 .234546 .248 .151816 .294 .192684 .340 .473 .249 .295 .193596 .341 .236421 .153546 .296 .194509 i .342 .237369 .154412 ' .297 .195422 .343 .23 83 18 .252 .155280 .298 .196337 .344 .239268 .253 .156149 .299 .197252 .345 .240218 .254 .157019 .300 .198168 .346 .241169 .255 .157890 .301 .199085 .347 .242121 .256 .158762 .302 .20000 > .348 .243074 .257 .159636 .303 .200922 .349 .244026 .258 .160510 .304 .201841 .350 .244980 .259 .161386 .305 .202761 .351 .245934 .260 .162263 .306 .203683 .352 .246889 .261 .163140 .307 .204605 .353 .247845 .262 .164019 .308 .205527 .354 .248801 .263 .164899 .309 .206451 .355 .249757 .264 .165780 .310 .207376 .356 .250715 .265 .166663 .311 .208301 .357 .251673 .266 .167546 .312 .209227 ! .358 .252631 .267 .168430 " .313 .210154 .359 .253590 .268 .169315 .314 •211082 .360 .254550 .269 .170202 .315 •212011 | .361 .255510 .270 .171089 .316 .212940 j .362 .256471 .271 .171978 .317 .213871 .363 .257433 .272 .172867 .318 .214802 .364 .258395 .273 .173758 .319 .215733 , .365 .259357 .274 .174649 1 .320 .216666 I .366 .260320 .275 .175542 1 .321 .217599 .367 .261284 .276 .176435 \ .322 .218533 1 .368 •262248 .277 .177330 .323 .219468 .369 .263213 .278 .178225 .324 .220404 J .370 .264178 .279 .179122 .325 .221340 .371 .265144 .280 .180019 1 .326 .222277 1 .372 .266111 .281 .180918 1 .327 .223215 .373 .267078 .282 .181817 j .328 .224154 | .374 .268045 .283 .182718 .329 .225093 .375 .269013 .284 .183619 1 .330 .226033 .376 •269982 .285 .184521 .331 .226974 1 .377 .270951 .286 .185425 .332 .227915 .378 •271920 .287 .186329 .333 .228858 ! .379 .272890 .288 .187234 ! .334 .229801 ! .380 .273861 .289 .188140 | .335 .230745 | .381 .274832 ~~ Areas of the Segments of a Circle. 281 Heig. \ Area Seg. \\ Heig. \ Area Seg. | Heig. | Area Seg. .382 .275803 1 .422 .315016 .462 .344736 .383 .276775 I .423 .316004 .463 .355732 .384 .277748 j .424 .316992 .464 .356730 .385 .278721 1 .425 .317981 .465 .357727 .386 .279694 1 .426 .318970 .466 .358725 .387 .280668 | .427 .319959 .467 .359723 .388 .281642 1 .428 .320948 .468 .360721 .389 .282617 | .429 .321938 .469 .361719 .390 .283592 .430 .322928 .470 .362717 .391 .284568 J .431 .323918 .471 .363715 .392 .285544 .432 .324909 .472 .364713 .393 .286521 1 .433 .325900 .473 .365712 .394 .287498 | .434 .326892 .474 .366710 .395 .288476 | .435 .327882 .475 .367709 .396 .289453 .436 .328874 .476 .368708 .397 .290432 .437 .329866 .477 .369707 .398 .291411 .438 .330858 .478 .370706 .399 .292390 .439 .331850 .479 .371705 .400 .293369 .440 .332843 .480 .372704 .401 .294349 .441 .333836 .481 .37*3703 .302 .295330 .442 .334829 .482 .374702 .403 .296311 .443 .335822 .483 .375702 .404 .297292 .444 .336816 .484 .376702 .405 .298273 .445 .337810 .485 .377701 .406 .299255 .446 .338824 .486 .378701 I .407 .300238 .447 .339798 .487 .379700 .408 .301220 .448 .340793 .488 .380700 .409 .302203 .449 .341787 .489 .381699 .410 .303187 .450 .342782 .490 .382699 .411 .304171 .351 .343777 .491 .383699 .412 .305155 .452 .344772 .492 .384699 .413 .306140 .453 .345768 .493 .494 .385699 .414 .307125 .454 .346764 .386699 .415 .308110 .455 .347759 .495 .387699 .416 .309095 .456 .348755 .496 .388699 .417 .310081 .457 .349752 .497 .389699 .418 .311068 j .458 .350748 .498 .390699 .419 .312054 .459 .351745 .499 .391699 .420 .313041 1 .460 .352742 .500 .392699 .421 1 .314029 .461 .353739 a: a 2 Ml Surveying. CHAPTER VI. SuRVEriSG. SURVEYING is the art of measuring, planning, and finding the superficial content, of any field, or parcel of land. In this kind of measuring, th - area or su- perficial content id always expressed in acres ; or acres, . and perches: and the lengths of all lines, in the field, or parcel of land, are measured with a chain, such as is described at page 78. A line, or distance on the ground, is thus measured. Having procured to small arrows or iron rods, to stick in the ground at the end of each chain ; also some sta- tion-staves, or long poles with coloured flags, to set up at the end of a station-line, or in the angles of a field \ two persons take hold of the chain, one at each end ; the foremost, for the sake of distinction, is called the leader, the hindermost the follower. A station-staff is set up in the direction of the line to he measured, if there be not some ohject, as a tree, a house, &c. in that direction. The leader takes the 10 arrows in the left hand, and one end of the chain, by the ring, in his right hand, and proceeds towards the station -staff, or other object. The follower stands at the beginning of the line, hold- ing the other end of the chain, by the ring, till it is stretched straight, and laid, or held level, by the lead- er, whom he directs, by waving his hand to the right or left, till he see him exactly in a line with the object towards which they are measuring. The leader then sticks an arrow upright in the ground, as a mark for the follower to come to, and proceeds forward another Surveying;. 288 chain, at the end of which he is directed, as before, by the follower ; or he may now, and at the end of every other chain, direct himself, by moving to the right or left, till the follower and the object measured from, be in one straight line. Having stuck down an arrow, as before, the follower takes up the arrow which the leader first stuck down. And thus they proceed till all the 10 arrows are employed, or in the hands of the follower, and the leader, without an arrow, is arrived at the end of the eleventh chain length. The follow- er then sends or carries the 10 arrows to the leader, who puts one of them down at his end of the chain, and proceeds with the other nine and the chain as be- fore. The arrows are thus changed from the one to the other, till the whole line is finished, if it exceed 10 chains ; and the number of changes shews how ma- ny times 10 chains the line contains. Thus, if the whole line measures 36 chains 45 links, or 3645 links, the arrows have been changed three times, the follower will have 5 arrows in his hand, the leader 4, and it will he 45 links from the last arrow, to be taken up by the follower, to the end of the line. Of the Surveying-Cross, or Cross-Staff. The surveying cross consists of two pair of sights at right an- gles to each other : these sights are sometimes pierced out in the circumference of a thick tube of brass; and sometimes the cross- staff consists of four sights strongly fixed upon a brass cross, and when used is screwed on a staff, having a sharp point to stick in the ground. The accu- racy of the cross-staff depends on the sights being exactly at right 2S* Surveying. angles to each oilier. A cross-slaff may be easily made by any carpenter. Thus, take a piece of beech or box, ADBC, of tour or live inches in breadth, and three or four indies in depth, and upon ADBC draw two lines, AB ai\d CD, crossing each other at right angles. Then wilh a fine saw make two slits, ABG and CDH, of about two inches in depth ; fix this piece of wood upon a staff 8, of about 4 and a half or 5 feet in length, pointed at one end, so that it may easily stick into the ground. PROBLEM I. To Measure Off-Sets with a Chain and Cross-Staff. Let Abcdefg be a crooked hedge, river, or brook, &c. and AG a base line. First, begin at the point A, and measure towards G : when you come to B, where you judge a perpendicular must be erected, place the cross-staff in the line AG in such a position that both G and A may be seen through two of the sights, look- ing forward towards G, and backward towards A. — Then look along one of the cross-sights, and if it point directly to the corner, or bend at b, the cross-staff is placed right; otherwise move backward or forward a- long AG till the cross-sights do point to 6, and measure B6, which set down in links: proceed thus till you have taken all the off-sets, as in the following Surveying. 285 FIELD-BOOK. Off-sets left. Base line AG, or, O, stations. Off-sets right. 62 84 70 98 57 91 OA. 43 220 340 510 634 785 dS To lay down the Plan. Draw the line AG of an indefinite length. Then, by a diagonal scale, such as described at page 48, set oft'AB equal to 45 links, draw B6 perpendicular to AG, and equal to 62 links. Next set off AC equal to 220 links, or 2 chains, 20 links ; draw Cc perpendicular to AG, and equal to 34 links : then set off AD equal to 340 links; or 3 chains 40 links, and make Dd equal to 70 links : proceed thus till you have completed the figure. To cast up the Content. AB6 must be measured as a triangle ; BCcfr, CDc?c, DEec?, &c. must be measured as trapezoids; see page 93. Some authors direct you to add all the perpen- diculars B6, Cc, &c. together, and divide their sum by the number of them, then multiply the quotient by the length AG; but this method is always erroneous, ex- cept the off-sets B6, Cc, &c. be equally distant from each other. 286 Surveying bs> 3 1 ■ -> 1 1 1 II II a* «£ © CO 1 05 *> 1 O 1 O 1 to cm c:x oca do >»>• to cm n = II 3 H II n 11 1 6SO II II CM ►*• H* ►*• to Cx ** 4* 00 3s *' I 4- to © CM Ci | 4- to e» 1 CM O nap CO O >> £« o OS > 00 II II II 1 II II r 4> — — >*• 10 oo ^ CO to cm -« 00 ' ' 1 «^ .— "< o © 4- 1 O 4* o 1 o o f gsp £e o i>> > K| » a. H ow o to 00 'I 1 II II I' 1 II II J CM ►*■ H» ~ W CM a •<» <3J eS •> > ©* (J<3 ^s $ ^o to II 3 1 II II 11 1 II II So h*. ^ CO ^ <3» -1 m 3i 4> k_ Ox 9 CO 00 00 M" 00 I ^ H- 1 08 4- 2790 = do i blc area of A 3ft. 25550 sa doi bl e area of B Ccb. 48480 = doi ihle area of C ndc. 28560 qp doi ihlearea of D Red. 19220 = doi bio area of E Ffi. 22348 = doi lble area of F< %/■ 2)116948 = double area of the whole in sq. links. 58474 = area in square links. .58474 = area in acres =0 A. 2 R. 13.5584 P. 2. Required the plan and content of part of a field^ from the following field-book. Surveying. 2S7 Off-sets left. Base line AG. Off-sets right. 60 150 0A. 100 250 325 450 550 700 100 60 150 46 Cross Hedge. The figure must be laid down, and the content cal- culated as in the first example. Thus you will find the area of the part E d c b a A to be 1 R. 13. 4. P. and of the part E/g G to be 30 perches ; so that the whole is 2 roods 3.4 perches. PROBLEM II. To measure a Field in the form of a Trapezium. Set up station-staves, or long poles, at the corners A, B, C. Then begin at D, and measure along the diagonal DB, in a right line, till you come to the place of the perpendicular AF, which will be known by looking backward and forward through the sights of the cross-staff,* as before directed. Make a mark at F, and measure the perpendicular AF ; then proceed from F towards B, and when you come to E, the place where the second perpendicular will fall, make a mark, and measure the perpendicular EC : lastly, continue your measure from E to B. You may either draw a rough plan of the field by the eye, and write the length of the diagonal and perpendiculars on it, which some 233 Survcj/iui writers recommend as the best method; or set them down in a field-hook, thus : Offsets 'left. S Id lion, or base line. Off-sets right. 849 on 600 1100 1360 620 To lay down the Plan. Draw the station-line DB equal to 1360 links, or 13.60 chains; from 1) set oft' I)F equal to 600 links, or 8 chains ; draw AF perpendicular to DB, and equal to o42 links, or 3 chains 42 links: make DE equal to 1190 links, and at E erect the perpendicular EC equal to 620 links. Join DA, Aii, BC and DC, and the field is constructed. CALCULATION. This field being a trapezium, its content must be found as directed in section VI. chapter 1. part II. Surveying 289 AF=r342 links. EC=625 Sum 967 680 half the diagonal DB. 77380 5S02 6.57560 area in acres:=:6 acres 2 rods 12.096 perches. 2. Required the plan and content of a field from the following; field-book. Off-sets left. Stations, or base lines. Off-sets right. 306 0D. 214 362 592 210 Jlnswer. The content is 1 acre 2 roods, 4.3776 per. Note. In either of the above figures, if the sides DC, CB, BA, AD, and the diagonal DB had b en measured with the chain, the figures might have been planned, and their contents found by the rule, page 84, without using a cross-staff, or measuring the^M^ pendiculars AF and CE. Some, who pretend to measure land, always measure round a four-sided field, and cast up the content by adding every two opposite sides together, and taking half their sum ; and then multiply these half sums into each other. It may be necessary to inform the learner, that this method is very erroneous, and ought never to be practised. Bb Z90 Surveying. PROBLEM III. To measure a four-sided Field with crooked Hedges. Set ii|) staves, or poles, at the corners D, C, B, as before directed. Then begin at A, and measure, from AB, noting all the necessary oft-sets, and in this man- ner go round the field, then measure the diagonal AC. FIELD-BOOK. V Stations, or base lines. Off-sets right. OA. 300 400 550 750=AB 100 130 80 OB. 200 700 1500=BC 200 150 oc. • 300 500 700 900 iooo=CD 200 100 150 50 150 200 100 OD. 200 400 700 soozrDA Diagonal AC =1294. Surveying. 29.1 CALCULATION. Acres. The sides of the triangle ABC are 750, "J 1500, and 1291 links, and its content, by I 4.85239 the rule, page 84 J The sides of the triangle ACD are 1000,1 800, and 1294 links, and its content, by I 3.99907 the rule, page 84 - ' J Content of the oft-sets along AB= 0.50250 Ditto : along BC= 1.67500 Ditto along CD= 1.07500 Sum= 12.10396 Content of the off-sets along DA, deduct 1.00000 Remains the content of the field=li.l0396 PROBLEM IV. How to Measure an Irregular Field. a' The way to measure irregular land, is to divide it into trapeziums and triangles, thus : First, look over the field, and set up marks at every angle, and by those marks you may see where to have a trapezium, as -ABCI in the following figure. Muvveyiug> H<< _ b* "^ \r WAG % .S/ /v- I !5» hi begin and measure in a direct line from A to- warils C 5 but when you come to a> set up your cross, and try whether you be in a square to I (as is before directed) : and then measure the perpendicular a I, which is 482 links; then measure forward again to- ward C, but when you come to b, set up yoar cross, and try whether you be in the place where the perpen- dicular will fall 5 then measure the perpendicular 6B, 4 Surveying. 293 which is 206 links ; then continue your measure to C, and you will find the whole diagonal 942 links. Then proceed to measure the trapezium CDHI, he- ginning at C, and measuring along the diagonal line towards H : but when you come to rf, set up your cross, and try if you be in the place where the perpendicu- lar will fall : measure the perpendicular rfD, which is 146 links ; and then measure forward till you come to c, and there, with your cross, try if you be right in the place where the perpendicular will fall, and measure the perpendicular el, which is 3 chains; and from c continue your measure to H, and you will find the whole diagonal 1236 links. • Then proceed to measure the trapezium HGED, beginning at H, and measuring along the diagonal line towards E; but when you come to /, try with your cross if you be in the place where the perpendicular will fall ; and measure the perpendicular /G, which is 448 links; then continue on your measure from/ till you come to «•, and there try if you be in a square with the perpendicular gD ; and measure the said perpendicular, which is 294 links ; then measure on from g to E, and you will find the whole diagonal to be 1144 links* Then measure the triangle EFG, beginning at E, and measuring along the base EG, till you come to h, and there with your cross try if you be in the place where the perpendicular will fall ; and measure the perpendicular /iF, which is 314 links, continue your measure to G, and you will find the whole base to be 912 links; so you have finished your whole field. But to draw a planiof the field, it would be neces- sary to measure a few more lines, or mark the points $, 6, &c. b b2 l\)h Of Sunrffh}^ l ION. The area of the trapezium ABCI =324043 The area of the trapezium CDH1 Thfl area of the trapezium HGED=420714 The area of the triangle EFG =14.; 1 8 1 The area of the whole - 1163574 Cut off five figures from the right-hand, and the re- sult will he 11.63574 acres= 11 acres 2 roods 21.71S4 p. 2. Required the plan and content of an irregular field, from the following FIELD-ROOK. Off-sets Station, or Off-sets left. Base Line. right. OK. 380 240 , 380 480 553 / 35 265 . 940 600 140 1100 1410 435 — — — 1620 This field, when constructed, is exactly similar to the former one. The content of the off-sets must be found as in problem 1. Thus, Square Links. The content of ABCDEF will he found 601000 The content of AIHGF 439390 The area of the whole - 1040390 Cut off five figures, and the result will be 10.40390 acres = 10 A. 1 11. 24.624 P. Practical Questions. 295 problem v. To cut off from a Plan a given number of Acres, Sft\ by a line drawn from any point in the side of it. Let A be the given point in the annexed plan, from which a line is to be drawn towards B, so as to cut off 5 acres 2 roods 14 perches. Draw AB, so as to cut off a quantity ABC, as near the quantity proposed as you can judge ; and suppose the true quantity of ABC, when calculated to be only 4 A. 3 R. 20 P. which is 3 R. 34 P. =114 perches =71250 square links too little. Then measure AB, which suppose equal to 1234 links, by the half of which, viz. 617 links, let 71250 links be divided ; the quotient, 115 links, will be the alti- tude of the triangle to be added, whose base is AB, therefore make BD=115 links, and draw AD, which will cut off the quantity required. CHAPTER VIL Practical (Questions in Measuring. Question 1. IF a pavement be 47 feet 9 inches long, and 18 feet 6 inches broad, I demand how many yards are contained in it? Jins. 98 yards 1 foot. Quest. 2. There is a room, whose length is 21.5 feet, and the breadth 17.5 feet, which is to be paved with 196 Practical Questions. stoii< s, each 10 inches square ; 1 demand how many such stones will pave it? Jlns. 167| stones. Quest. 3. There is a room 109 feet 9 inches about, and 9 feet 3 inches high, which is all (except two win- dons, each 6 feet 6 inches high, and 5 feet i) inches broad) to be hung with tapestry that is ell-hroad ; I desire to know how many yards will hang the said room ? From the content of the room, subtract the content of the windows, and divide the remainder by the square feet in a yard of tapestry, viz. i4\ feet, and you will find the answer to be 83.59 yards. Jlns. Quest. 4. If the axis of a globe be 27.5 inches ; I demand the content solid and superficial r 10889.24375 inches=6.3 feet solid. 2373.835 inches— 1 6.49 feet superficies. Quest. 5. There is a segment of a globe, the diam- eter of whose base is 24 inches, and. its altitude 19 inches ; what is the content solid and superficial including the area of its base ? a J .'785.552 inches the solidity. ' ^ 1218.9408 inches the superficies. The diameter of the whole sphere may be found as- in page 118, to be 24.4 inches. Quest 6. If a tree girt 18 feet 6 inches, and be 24 feet long, how many tons of timber are contained in that tree, using the customary method of measuring, and allowing 40 feet of timber to a ton ? Jlns. 12 tons 33 feet 4 inches 6 parts. #. Practical Questions. 297 Quest. 7. There is a cellar to be dug by the floor, the length of which is S3 feet 7 inches, and the breadth 18 feet 9 inches, and its depth is to be 5 feet 9 inches ; I demand how many floors of earth are in that cellar ? Ms. 11 floors 56 feet 8 inches 5 parts. Note. That 18 feet square and a foot deep is a floor of earth, that is, 324 solid feet. Quest. S. There is a roof covered with tiles, whose depth on both sides (with the usual allowance at the eaves) is 35 feet 6 inches, and the length 48 feet 9 inches ; how many squares of tiling are contained in it? Jins. 17 squares 30 feet 7\ inches. Quest 9. There is a cone, the diameter at the base being 42 inches, and the perpendicular height 94 inches ; and it is required to cut off two solid feet from the top end of it 5 I demand what length upon the perpendicular must be cut oft'? First, find the solidity of the whole cone, 43410.628S cubic inches, or 25.1219 feet, and the cube of the alti- tude 830584 inches. Then, since all similar solids are in proportion to each other as the cubes of their like parts, Feet. Inches. Feet. Inches. As 25.1219 : 830584 : : 2 : 66124.297, the cube of the altitude to be cut oft", the cube-root of which is 40.436 isches, Answer. * Quest. 10. If a square piece of timber be 12 feet long, and if the side of the square of the greater base be 21 inches, and the side of the square of the lesser base be 3 inches 5 how far must I measure from the greater end, to cut off" five solid feet ? You will find the length of the whole pyramid by the rule, page 118, to be 14 feet. Then find the solid Practical Questions. content of the whole pyramid 14*294 feet, from which deduct r> feet, the remainder is 9.291; feet. Solidity. Cube length. Solidity. Cube length. As 14.29 if. : .2744 :: 9.29 1^ : y 1784 Or 42.875 : 2744 : : 27.873 : 1784 the cube-root of which is 12.128 feet, which subtracted from the whole length, 14 feet, leaves 1.872 i'eet^ the length of 5 solid feet at the greater enjj. Quest. 11. Three men bought a grindstone of 40 inches diameter, which cost 20 shillings ; of which sum the first man paid 9 shillings, the second 6 shil- lings, and the third 5 shillings : I demand how much of the stone each man must grind down, proportion- able to the money he paid ? AH circles are to each other as the squares of their diameters ; and each man must grind away a surface proportionable to the money he paid. 20 : 20X20 :: 5 : 100, the square-root of 10, 20 : 20X20 : : 6 : 120 Sum 220, the square-root is 14.832397, the radius which two men must grind down from the centre, from which take 10, the radius which one man must grind down, there remains 4.S32397,the breadth of the ring which the second man must grind away : from the whole radius 20 subtract 14.832397, the radius which two men must grind away; the re- mainder is 5.167603, the breadth of the ring the first man must grind away. Practical Questions. 299 OR GEOMETRICALLY THUS : First, upon the centre O describe the circle ABCD, and cross it at right angles with the two diameters AB and CD : then divide the semidiameter A © in proportion to 9s. 6s. and 5s. the several sums paid by the three men ; viz. make AE 9, EF 6, and F O ft:: then divide EB into two equal parts in d, and upon d 9 as a centre, describe the semicircle EoB; divide FB into two equal parts in c, and upon c, as a centre, with the radius e F, describe the semicircle F b B ; which will divide the semidiameter © C into three such parts as the stone ought to be divided; and circles described through these points, will shew how much each man must grind for his share. This construction is derived from the above me- thod of calculation for B©=20, QF—5, and ©E=ll, by the pro perty of the circle B©X©F=©& a , and B*©X©E=©« 2 ; also ©«— ©&=«& and ©C— Qa=aQ : hence ©&=lo, a&=4.83, and aC=5.l7 nearly. .loo "Practical ({usstiotu. ({ifst. i j, A gard'ner had an upright cone, Out of which should he cut him a rolling-stone, The biggest thai e'er it could make: The mason he said, that there was a rule For such sort of work, but he had a thick skull: Now help him for pity'* sal Jlns. It must he cut at one-third part of the altitude. Note. This is property a question in fluxions, and not dependant upon any rule in this book. S e Simp- son's Geometry, on the Maccima and Minima of geo- metrical quantities, theorem XXX. Qu<>r,t. 13. There is a cistern, whose depth is seven tenths of the width, and the length is 6 times the depth, and the solid capacity is 367.3 feet; I demand the depth, width, and length, and how many bushels of corn it will hold ? First, you must find three numbers in proportion to the depth, width, and length thus : suppose the width 1, the depth will be .7, and the length 4.2; hence the solidity will be 2.94 feet. But solids arc to each other as the cubes of their like parts, consequently 2.94 feet : 1 cube width :: 367.5 feet : 125 the cube of the real width, the cube-root of which is (' et the width; hence the depth is 3.5 feet, and length 21 feet. The content is 295 bushels 1 peck 4 pints. Quest. 14. Suppose, sir, a bushel be exactly round, Whose depth being measured, 8 inches is found ; If the breadth IS inches and half you discover, This bushel is legal all England over. But a workman would make one of another frame ; Sev'n inch and a half must be the depth of the same; Now, sir, of what length must the diameter be, That it may with the former in measure agree ? Ans. 19.107 inches must be the diameter when the depth is 7' £ inches. Practical Questions. 301 Quest. 15. In the midst of a meadow, well stored with grass, I took just an acre, to tether my ass : How long must the cord be, that, feeding all round, He may'nt graze less nor more than his acre of ground? Jlns. 117 feet 9 inches, Quest. 16. A malster has a kiln, that is 16 feet 6 inches square; but he is minded to pull it down, and build a new one, that may be big enough to dry three times as much at a time as the old one will do ; I de- mand how much square the new one must be ? Jlns. The side of the new one must be 28 feet and near 7 inches, or 28.578 feet. Quest. 17. If a rfcund cistern be 26.3 inches diame- ter, and 52.5 inches deep; how many inches diameter must a cistern be to hold twice the quantity, the depth being the same ? and how many ale gallons will each cistern hold ? Jlns. The diameter of the greater cistern is 37.19 inches, and its content 202.275 gallons ; hence the con- tent of the less cistern must be 101.137 gallons. Quest. 18. If the diameter of a cask at the bung be 32 inches, and at the head 25 inches, and the length 40 inches; how many ale gallons are contained therein ? Jlns. 94.41 gallons, by the general rule, page 269. Quest. 19. There is a stone, 20 inches long, 15 inches broad, and 8 inches thick, which weighs 217 pounds ; I demand the length, breadth, and thickness of another of the same kind and shape, which weighs 1000 pounds ? The cube of 20, the length, is 8000. Then 217 : 8000 : : 1000 : 36866.3594, whose cube root is 33.282 inches, the length of the stone weighing 1000 pounds. Then say, Ce inches. 302 Practical questions. 20 : .;.;.JsJ :: 10 : 21.901 20 : J.J.2S2 : : 8 : 18, ("The length sjl Ms. < The breadth jj.'.kh li I^The thickness 13.31 2 J Quest. 20. If an iron bullet, whose diameter is 4 inches, weighs 9 pounds; what will he the weight of another bullet (of the same metal) whose diameter i- 9 inches ? Jins. 102.515 pounds. Quest, 21. There is a square pyramid of marble, each side of its base* is 5 inches, and the height 15 inches, and its weight is 12 pounds and a quarter; I demand the weight of another like square pyramid, each side of who se Jjase is 30 inches? The cube *of U is 125, and the cube of 30 is 27000. Then (by End. XII. 12.) • lb. lb. 125 : 12.25 : : 27000 : 2646 Ms. The weight is 2646 pounds. Quest. 22. There is a ball or globe of marble, whose diameter is 6 inches, and its weight 11 pounds; what will be the diameter of another globe of the same marble, that weighs 500 pounds ? Ms. 21.4 inches. Quest. 23. There is a frustum of a pyramid, whose bases are regular octagons ; each side of the greater base is 21 inches, and each side of the less base is 9 inches, and its perpendicular length is 15 feet; I de- mand how many solid feet are contained in it ? Ms. 119.2 feet* Quest. 24. There is a frustum of a cone, the diame- ter of the greater base is 36 inches, and the diameter of the less base is 20 inches, and the length or height is 215 inches ; I demand the length and solid content of the whole cone, and also the solid content of the ^iven frustum ? Practical Questions* 303 First, find the length of the whole cone, thus : As 8, the difference between the radii of the two ends : 215, the length of the frustum : : 18, the radius of the greater end : 483.75 inches, the whole length of the cone. The solidity of the whole cone is 94.98 feet. The solidity of the frustum is 78.7 feet. quest. 25. If the top part of a cone contains 26171 solid inches, and 200 inches in length, and the lower frustum thereof contains 159610 solid inches ; I de- mand the length of the whole cone, and the diameter of each base ? Inches. ("The length of the whole cone 384.3766 Arts. \ The diameter of the greater base 42.9671 I The diameter of the less base 22.3568 Quest. 26. There is a frustum of a cone, whose solid content is 20 feet, and its length 12 feet ; and the greater diameter bears such proportion to the less as 5 to 2 ; I demand the diameters ? First, I find the content of a conical frustum, whose diameters are 5 and 2, and depth 12 feet, to be 122.5224 feet. Then, as 122.522 : 20 feet : : 25, square of the greater diameter : 4.080886, the square root of which is 2.02012 feet, the true greater diameter; and 5:2'.'. 2.02012 : .S0804 feet, the less diameter required. Or, the two required diameters are 24.2414 inches and 9.6964 inches. Quest. 27. There is a room of wainscot 129 feet 6 inches in circumference, and 16 feet 9 inches high (be- ing girt over the mouldings ;) there are two windows, each 7 feet 3 inches high, and the breadth of each, from cheek to cheek, 5 feet 6 inches ; the breadth of the shutters of each is 4 feet o inches.; the cheek-boards «M-4 Practical ({urstions. and top and bottom-boards of each window, taken to- gether, are 24 feet 6 inches, and their breadth 1 foot inches; the door-case 7 feet high, and > feet B inches uidc: the door 8 feet 3 inches vide : I demand how many yards of wainscot are contained in that room ? The content of the room 2169 1 6 The shutters, at work and half 97 10 6 The door, at work and half 84 1 6 The cheek-boards 83 o The sum 2J86 10 6 The window-lights and > door-case deduct $ 9)2282 7 6 253 Ans. 253 yards 5 feet. Quest. 2S. There is a wall which contains 18225 cu- bic feet, and the height is 5 times the breadth, and the length 8 times the height ; what is the length, breadth, and height ? Assume the breadth 1, then the height must be 5, and length 40 ; hence the solidity will be 200. Then say 200 : 1, cube of the breadth : : 18223 : 91.125, the cube of the real breadth ; the cube-root of which is 4.5 the breadth; hence the height is 22.5, and the length 480 feet Practical Questions. Quest. 29. There is a may-pole, whose top-end was broken oft' by a blast of wind, and, in falling, struck the ground at 15 feet distance from the bottom of the may -pole : the broken pieee was 39 feet ; now I de- mand the length of the may -pole ? From the square of 39, the length of the broken pieee, subtract the square of 13 ; the square-root of the remainder is the piece standing; to which add the piece broken oil', and you have the whole length. Thus you will find The pieee standing to be 36 feet. The piece broken off is 39 feet. The whole length is 73 &05 j I \ Quest. 30. A may -pole there was, whose height I would know, The sun shining clear, strait to work I did go. The length of the shadow, upon level ground, Just sixty-five feet, when measurd, I found ; AstaffI had there.just five feet in length — The length of its shadow was four feet one-tenth : How high was the may -pole, I gladly would know? And it is the thing jou're desired to show. Here A B represents the length of the shadow of the may -pole, andBC its height : «A the sha- dow of the stall', and A6 its height. By similar triangles. ah : Ab : : AB : BC. The height BC will be found to be T9.26 feet. / C / A ' / / J together. f \f 1 * ne s *S n °^ subtraction, as 8 — 3 sig- — i Minus > Unifies that 3 is to be subtracted 1 orless J from 8. f multipli- 1 the sign of multiplication, as 7X5 X i edinto or y signifies that 7 is to be multiplied (. D J> J * nto or D y 5 * f r 'A A 1 ine sl S n °^ division, as 9~3 signi- ^ i uiviaea ( fiesl h a t9 isto be divided by 3:and (. ^' J I or 3 " 9 5 signifies the same. _ a the sign of equality, as 9—9 signi- J equal / fies that 9 is equal to 9, or 5-f-4 — to f 2=7 signifies that 5, increased by *- } 4- and diminished by 2 is equal to 7. C Propor- ) as 2 : 4 : : 8 : 1 6 signifies that 2 is : : : ; l tion, $ *o 4 as S is to 16. w 2 , m% signifies the square or cube of the letter m. TS 2 signifies the square of the line TS. y/a\ signifies the square root of a A. APPENDIX. WEIGHTS AND DIMENSIONS BALLS AND SHELLS. THE weights of bodies, composed of the same sub- stance, are proportional to their magnitudes; and the magnitudes of similar bodies are proportional to the cubes of their corresponding lineal dimensions 5 there- fore, sinee all globes are similar bodies, the weights of those, which are formed of the same substance, must be proportional to the cubes of their diameters. It is well known, however, that different portions of metal, even from the same casting, will differ a small matter, in density ; and also that a still greater difference is produced by different degrees of temperature : the theory here laid down, is therefore not to be understood as absolutely true, but sufficiently so for the purposes to which it is applied. FEOBLEM I. Given the diameter of an Iron Ball, to find its weight. An iron ball of 4 inches diameter, is known, from experiment, to weigh 9 lb. avoirdupoise ; therefore, since the weights are as the cubes of the diamaters, it will be, as the cube of 4, (or 64) is to 9, so is the cube of Dd APPENDIX. the given diameter, to the weight required. Hem* ^®j of the cube of the diameter will be the weight in But ^ =J-|-.i of £; which gives the following RULE. To ^ of the cube of the diameter, add £ of that | £ the sum will be the weight in lbs. EXAMPLES. 1. Required the weight of an iron ball whose diam- eter is 6.4 inches. The cube of 6.1 is 262.14* i of which is - - 32.768 | of | is - - - - 4.096 Weiffht of the ball - 36.864 lb. 2. The diameter of an iron ball is 8 inches, what is its weight. Ans. 72 lb. 3. What is the weight of an iron ball, its diameter being 2.4 inches. Jins. 1 lb. 15.1 oz. PROBLEM II. To find the weight of a Leaden Ball. A. leaden ball of 3 inches diameter weighs 6 lb. Therefore as the cube of 3 is to 6 ; that is, as 9 to 2 so is the cube of the diameter to the weight of a leaden ball. Hence the following practical Take | of the cube of the diameter of a leaden ball, for its weight in pounds. APPENDIX. EXAMPLES. 1. Required the weight of a leaden ball, whose di- amter is 6.3 inches. | of 6.3X6.3X6.3=5.3.366 lb. the weight required. 2. What is the weight of a leaden ball, 8.1 inches diameter? Jlns. 118 lb. 3. The diameter of a leaden ball is 1.8 inches, what is its weight ? Jlns. 1 lb. 4.736 oz. PROBLEM III. Given the weight of an Iron Ball, to find its diameter. The converse of prob. 1, will give this RULE. Multiply the given weight by 7-|, and the cube root of the product will be the diameter: or, to the con- stant logarithm 0.85194 add the logarithm of the weight -ifl pounds, and | of the sum will be the loga- rithm of the diameter, in inches. EXAMPLES. Required the diameter of a 42 lb. iron ball. Constant log. 0.8/5194 Los. of 42 1.62325 3)2.47519 Diam. 6.683, log. 0.82506 2. What is the diameter of an iron ball, weighing 18 lb. Ans. 5.04 inches. 3. An iron ball weighs 6 lb. what is its diameter. Jlns, 3.494 inches. APPENDIX. PROBLEM IV. The weight of a Leaden Ball being given, to find it* diameter. The converse of prob. 2, will give this RULE. Multiply the weight by 4*, and the cube root of the product will be the diameter : or, to the constant log- arithm 0.65321, add the logarithm of the weight, and | of the sum will be the logarithm of the diameter. EXAMPLES. 1. Required the diameter of a 64 lb. leaden ball. Constant log. 0.65321 Log. of 64 1.80618 Diam. 6.605, log. 3)2.45939 f» % 0.81979 2. What is the diameter of a leaden ball weighing 18 lbs ? Ans. 4.327 inches. 3. A leaden ball weighs 9 lbs. what is its diameter ? Jins. 3.434 inches. PROBLEM V. To find the weight of an Iron Shell. To ^ of the difference of the cubes of the external and internal diameters, add its j, the sum will be the weight of the shell. APPENDIX. EXAMPLES. 1. Required the weight of an iron shell, the exter- # ^j nal diameter being It, and the internal, 9 inches. Tire cube of 11 is 1331 The cube of 9 729 Difference 602 |of| Hi Weight of the shell **§1 2. To find the weight of a shell whose external diameter is 9.6 inches, and internal 7.2 inches. Jins. 71.928 lb. 3. What is the weight of an iron shell, the outer diameter being 18 inches, and the inner 16. 4g£, Jins. 244 J lb. PROBLEM VI. To find what weight of Powder will fill a given ShelL > ^»- ~V*I%* : "RULE. Divide the cube of the internal diameter, in inches, by 58, the quotient will be the lbs. of powder. EXAMPLES. 1. AVhat weight of powder will fill a shell, whose internal diameter is 9 inches. 9X9X9—58— 12|4 lb. the quantity required. 2. What quantity of powder will fill a shell 7.2 inches in diameter. Jins. 6.435 lb. 3. Required the weight of powder to fill a sh-'ll whose inner diameter is 16 inches. Jins. 70 £ | lb. L U'PEXDIX. rnoHi.v.N: vn. To find the diameter of a Shell to contain a given weight of Fowder. Multiply the pounds of powder by 58; the eube root of the product will be the internal diameter, in inches. EXAMPLES. 1. To find the diameter of a shell to contain 9 lb. of powder. 58X9=522; the cube-root of which is 8.05 inches nearly. 2. Required the diameter of a shell, to contain 47| lb. of powder. Jlns. 14 inches, nearly. 8. To find the diameter of a shell, to contain 27 lb. of powder. «te.«flMP ncnes * N. D. The two last rules were deduced from experiments on the powder manufactured for the government of the United States. cSm+Mr/iL^- ^ THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL FINE OF 25 CENTS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO SO CENTS ON THE FOURTH DAY AND TO $1.00 ON THE SEVENTH DAY OVERDUE. AUG 19 mi JAN 23 1947 — —— ;>f >$s& *_ ^ • '■ ° |ftAR-&$-^ PHOTOCOPY APR '87 LD 21-100m-7,*40 (6936s) -% 91833** &4f£i /w THE UNIVERSITY OF CALIFORNIA LIBR