ORTON'S 1 LIGHTNING CALCULATOR, ACGOMTANT'S ASSISTANT, iViLiiJliiiiiiiiiiiiiiiiiiil of u 1 1 within bne having the Fig 1 " BY HOYm ORTON, 705 JAYNE STREET, PHILADELPHIA.. ENERGY IS THE PRICE Or ^ IT, B, Any infringement upon the copyright of this book will be prosecuted to the fullest extent of the law. Address all Orders for this Book to T. K. COLLINS, 705 Jayne St., Philadelphia. FOJR SALE HY ALL WHOLESALE BOOKSELLERS. Single Copies, by Mail, One Dollar, IN MEMORIAM FLOR1AN CAJORI LIGHTNING CALCULATOR, ACCOUNTANT'S ASSISTANT. THE SHORTEST, SIMPLEST, AND MOST RAPID METHOD OP COMPUTING NUMBERS, ADAPTED TO EVERT KIND OF BUSINESS, AND WITHIN THE COMPREHENSION OF EVERY ONE HAVING THE SLIGHTEST KNOWLEDGE OF FIGURES. BY HOY D. ORTOX, 705 JAYNE STREET, PHILADELPHIA. ENERGY IS THE PRICE OF SUCCESS. N. B. Any infringement upon the copyright of this book will be prosecuted to the fullest extent of the law. Address all Orders for this book to T, K, COLLINS, 705 Jayne Street, Philadelphia, For Sale by all Wholesale Booksellers, Single Copies, by Mail, One Dollar. 1869. Entered according to Act of Congress, in the year 1866, by HOY D. ORTON, fcl the Clerk's Office of the District Court of the United State* for the Southern District of Ohir. COiLINS, PRINTER. INTRODUCTION. QUANTITY is that which can be increased or diminished by augments or abatements of homo* geneous parts. Quantities are of two essential kinds, Geometrical and Physical. 1. Geometrical quantities are those which occupy space ; as lines, surfaces, solids, liquids, gases, etc. 2. Physical quantities are those which exist in the time, but occupy no space ; they are known by their character and action upon geometrical quan- tities, as attraction, light, heat, electricity and mag- netism, colors, force, power, etc. To obtain the magnitude of a quantity we com- pare it with a part of the same ; this part is im- printed in our mind as a unit, by which the whole is measured and conceived. No quantity can be measured by a quantity of another kind, but any quantity can be compared with any other quantity, and by such comparison arises what we call caZcu- lation or Mathematics IV INTRODUCTION. MATHEMATICS. MATHEMATICS is a science by which the com- parative value of quantities are investigated ; it ia divided into : 1. ARITHMETIC, that branch of Mathematics which treats of the nature and property of num- bers ; it is subdivided into Addition, Subtraction, Multiplication, Division, Involution, Evolution and Logarithms. 2. ALGEBRA, that branch of Mathematics which employs letters to represent quantities, and by that means performs solutions without knowing or noticing the value of the quantities. The subdi- visions of Algebra are the same as in Arithmetic. 3. GEOMETRY, that branch of Mathematics which investigates the relative property of quantities that occupies space ; its subdivisions are Longemetry^ Planemetry, Stereometry, Trigonometry and Conic Sections. 4. DIFFERENTIAL-CALCULS, that branch of Math- ematics which ascertains the mean effect produced by group of continued variable causes. 5. INTEGRAL-CALCULS, the contrary of Differen- tial, or that branch of Mathematics which investi- gates the nature of a continued variable cause that has produced a known effect. PREFACE. xj . MATEJMATICAL LAWS are the acknowledged of all science. Ever since the streets of Athens i ^sounded with that historical cry of "Eu- reka," emanating from one of antiquity's greatest mathematicians, the science has been steadily pro- gressing. It is not our purpose, in this small work, to in- troduce any of the higher branches of mathematics, viz.: Algebra, Conic Sections, Calculus, etc. Our object is merely to present to the public a system of calculation that is practical to every business man. It consists of the addition of numbers on a principle entirely different from the one ordinarily used. In the practical application of this new prin- ciple of addition, scarcely any mental labor is re- quired, compared with the principle of addition set forth in standard works. The superiority we claim for this principle above all others, is this, that it requires no great mental exertion, affording the VI PREFACE. greatest facilities to the calculator in the additioi of D umbers, enabling him to add a whole day with- out any mental fatigue ; whereas, by the ordinar} way, it is very laborious and fatiguing. Our system of calculation also embraces a concise rapid, and at the same time practical method of Multiplication, by which one is enabled to arrive at the product of any number of figures, multiplied by any number, immediately, without the use of partial products. This small work also embraces the shortest and most concise method for the computation of Interest ever introduced to the public. Our system for com- puting interest is entirely different from any rule ever introduced, for the computation of either Sim- ple or Compound Interest. A student having gone no further than Long Division in Arithmetic, can, by our rule, calculate Simple or Compound Interest at any given rate per cent., for any given timo, in one-tenth of the time that the best calculators will compute it by the rules laid down in other books. By using our rules, you can entirely avoid the use of fractions, and save the calculation of 75 to 100 figures, where years, months and days are giv^Q OD a note. ADDITION. To BE able to add two, three or four columns of figures at once, is deemed by many to be a Her- culean task, and only to be accomplished by the gifted few, or, in other words, by mathematical prodigies. If we can succeed in dispelling this illusion, it will more than repay us ; and we feel very confident that we can, if the student will lay aside all prejudice, bearing steadily in mind that to become proficient in any new branch or principle a little wholesome application is necessary. On the contrary, we can not teach a student who takes no interest in the matter, one who will always be a drone in society. Such men have no need of this principle. If two, three, or more, columns can be carried up at a time, there must be some law or rule by which it is done. We have two principles of Addi- tion; one for adding short columns, and one for adding very long columns. They are much alike, differing only in detail. When one is thoroughly learned, it is very easy to learn the second. By a little attention to the following example, much time in future will be saved. 10 OKTON'S LIGHTNING CALCULATOR. ADDITION OF SHORT COLUMNS OF FIGURES. ADDITION is the basis of all numerical opera- tions, and is used in all departments of business, To aid the business man in acquiring facility and accuracy in adding short columns of figures, the following method is presented as the best: PROCESS. Commence at the bottom of 274 g^g the right-hand column, add thus: 16, 22, 134 32 ; then carry the 3 tens to the second 342 column ; then add thus : 7, 14, 25 ; carry the 2 hundreds to the third column, and add the same way: 12, 16, 21. In this 2152 wa y y OU name tbs sum of two figures at once, which is quite as easy as it is to add one figure at a time. Never permit yourself for once to add up a column in this manner : 9 and 7 are 16, and 2 are 18 and 4 are 22, and 6 are 28, and 4 are 32. It is just as easy to name the result of two figures at once and four times as rapid. The following method is recommended for the ADDITION OF LONG COLUMNS OF FIGURES. In the addition of long columns of figures which frequently occur in books of accounts, in order to add them with certainty, and, at the same time, with ease and expedition, study well the following method, which practice will render familiar, easy, rapid, and certain. ADDITION. 11 THE EASY WAY TO ADD. EXAMPLE 2 EXPLANATION. Commence at 9 to add, and add as near 20 as pos- sible, thus: 9+2+4+3=18, place the 8 to tho right of the 3, as in example; commence at 6 to 7 7 add 6+4+8=18 ; place the 8 to the right of 4 the 8. as in example ; commence at 6 to add 6 6+4+7=17 ; place the 7 to the right of the 3 6 7, as in example ; commence at 4 to add 4+ 9 9+3=16 ; place the 6 to the right of the 3, 4 as in example; commence at 6 to add 6+4 7 T +7=17; place the 7 to the right of the 7, 4 as in example; now, having arrived at the 6 top of the column, we add the figures in the 8* new column, thus: 7+6+7+8+8=36; place 4 the right hand figure of 36, which is a 6, 6 under the original column, as in example, and 3 8 add the left hand figure, which is a 3, to the 4 number of figures in the new column; there 2 are 5 figures in the new column, therefore 9 3+5=8; prefix the 8 with the 6, under the original column, as in example ; this makes 86 86, which is the sum of the column. Remark 1. If, upon arriving at the top of th< column, there should be one, two or three figures whose sum will not equal 10, add them on to thfr Bum of the figures of the new column, never placing 12 ORTON'S LIGHTNING CALCULATOR. an extra figure in the new column, unless it be an excess of units over ten. Remark 2. By this system of addition you can stop any place in the column, where the sum of the figures will equal 10 or the excess of 10 ; but the addition will be more rapid by your adding as near 20 as possible, because you will save the form- ing of extra figures in your new column. EXAMPLE EXPLANATION. 2+6+7=15, drop 10, place the 5 to the right of the 7; 6+5+4=15, drop 10, place the 5 to the right of the 4, as in example; 8+3+7=18, drop 10. place the 8 to the right of the 7, 4 as in example; now we have an extra figure, 7 8 which is 4 j add this 4 to the top figure of the 3 new column, and this sum on the balance of 8 the figures in the new column, thus: 4+8+ 4* 5+5=22; place the right hand figure of 22 5 under the original column, as in example, and 6 add the left hand figure of 22 to the num- 7* ber of figures in the new column, which are 6 three, thus: 2+3=5; prefix this 5 to the 2 figure 2, under the original column ; this makes 52, which is the sum of the cokmn. 52 ADDITION. 13 RULE. For adding two or more columns, com- mence at the right hand, or units' column; proceed in the same manner as in adding one column ; after the sum of the first column is obtained, add all except the right hand figure of this sum to the second column, adding the second column the same way you added the first ; proceed in like manner with all the columns, always adding to each successive column the sum of the column in the next lower order, minus the right hand figure. N. B. The small figures which, we place to the right of the column when adding are called integers. The addition by integers or by forming a new column, as explained in the preceding examples should be used only in adding very long columns of figures, say a long ledger column, where the foot- ings of each column would be two or three hundred, in which case it is superior and much more easy than any other mode of addition ; but in adding short columns it would be useless to form an extra column, where there is only, say, six or eight fig- ures to be added. In making short additions, the following suggestions will, we trust, be of use to the accountant who seeks for information on this subject. In the addition of several columns of figures, where they are only four or five deep, or when their respective sums will range from twenty-five 14 ORTON'S LIGHTNING CALCULATOR. to forty, the accountant should commence with the unit column, adding the sum of the first two figures to the sum of the next two, and so on, naming only the results that is the sum of every two figures. In the present example in adding the unit 346 column instead of saying 8 and 4 are 12 and 235 5 are 17 and 6 are 23, it is better to let the 724 eye glide up the column -reading only, 8, 12, 598 17, 23 ; and still better, instead of making a separate addition for each figure, group the figures thus : 12 and 11 are 23, and proceed in like man- ner with each column. For short columns this is a very expeditious way, and indeed to be preferred; but for long columns, the addition by integers is the most useful, as the mind is relieved at intervals and the mental labor of retaining the whole amount, as you add, is avoided, which is very important to any person whose mind is constantly employed in various commercial calculations. In adding a long column, where the figures are of a medium size, that is, as many 8s and 9s as there are 2s and 3s, it is better to add about three figures at a time, because the eye will distinctly see that many at once, and the ingenious student will in a short time, if he adds by integers, be able to read the amount of three figures at a glance or as quick, we might say, as he would read a single figure. ADDITION. 15 Here we begin to add at the bottom of the 6 26 8 anit column and add successively three fig- " ares at a time, and place their respective 004 sums, minus 10, to the right of the last fig- 554. are added; if the three figures do not make 62 10, add on more figures; if the three figures 87j make 20 or more, only add two of the fig- ^ tires. The little figures that are placed to ^. 4 the right and left of the column are called 877 integers. The integers in the present ex- 33 ample, belonging to the units column, are 4, 4, 5, 4, 6, which we add together, making ^ 23; place down 3 and add 2 to the number of integers, which gives 7, which we add to 803 the tens and proceed as before. REASON. In the above example, every time wo placed down an integer we discarded a ten, and when we set down the 3 in the answer we dis- carded two tens; hence, we add 2 on to the num- ber of integers to ascertain how many tens were discarded; there being 5 integers it made 7 tens, which we now add to the column of tens; on the same principle we might add between 20 and 30, always setting down a figure before we got to 30; then every integer set down would count for 2 tens, being discarded in the same way, it does in the present instance for one ten. When we add be- tween 10 and 20, and in very long columns, it 16 OETON'S LIGHTNING CALCULATOR. would be much better to go as near 30 as possible, and count 2 tens for every integer set down, in which case we would set down about one -half ai many integers as when we write an integer foi every ten we discard. When adding long columns in a ledger or day * book, and where the accountant wishes to avoid the writing of extra figures in the book, he can place a strip of paper alongside of the column he wishes to add, and write the integers on the paper, and i this way the column can be added as convenient almost as if the integers were written in the book. Perhaps, too, this would be as proper a time as any other to urge the importance of another good habit; I mean that of making plain figures. Some persons accustom themselves to making mere scrawls, and important blunders are often the result. If letters be badly made you may judge from such as are known; but if one figure be illegible, its value can not be inferred from the others. The vexation of the man who wrote for 2 or 3 monkeys, and had 203 sent him, was of far less importance than errors and disappointments sometimes result- ing from this inexcusable practice. We will now proceed to give some methods of proof. Many persons are fond of proving the cor- rectness of work, and pupils are often instructed to do so, for the double purpose of giving them ADDITION. 17 x exercise in calculation and saving their teacher the trouble of reviewing their work. There are special modes of proof of elementary operations, as by casting out threes or nines, or by changing the order of the operation, as in add- ing upward and then downward. In Addition, some prefer reviewing the work by performing the Addition downward, rather than repeating the ordinary operation. This is better, for if a mis- take be inadvertently made in any calculation, and the same routine be again followed, we are very liable to fall again into the same error. If, for instance, in running up a column of Addition you should say 84 and 8 are 93, you would be liable, in going over the same again, in the same way to slide insensibly into a similar error ; but by begin- ning at a different point this is avoided. This fact is one of the strongest objections to the plan of cutting off the upper line and adding it to the sum of the rest, and hence some cut off the lower line by which the spell RULE. Multiply the whole number by the next higher whole number, and annex ^ to the product. Ex. 1. What is the square of 7? Ans. 56. We simply say, 7 times 8 are 56, to which we add I. 2. What will 91 Ibs. beef cost at 9^- cts. a lb.? 3. What will 12^ yds. tape cost at 12^ cts. a yd.? 4. What will 54- Ibs. nails cost at 5^ cts. a lb. ? 5. What will ll| yds. tape cost at 11-j- cts. a yd. ? 6. What will 19 bu. bran cost at 19| cts. a bu.? REASON. We multiply the whole number by the next higher whole number, because half of any number taken twice and added to its square is the same as to multiply the given number by ONE more than itself. The same principle will multiply any two like numbers together, when the sum of the fractions is ONE, as 8 by 8|, or llf by llf, etc It is obvious that to multiply any number by any two fractions whose sum is ONE, that the sum of the products must be the original number, and adding the number to its square is simply to multiply it by ONE more than itself; for instance, to multiply 7-J by 7|, we simply say, 7 times 8 are 56, and then, to complete the multiplication, we add, of course, the product of the fractions (| times are A)> making 56 T 3 g the answer. MULTIPLICATION. 29 Where the sum of the Fractions is ONE. To multiply any two like numbers together when the sum of the fractions is ONE. RULE. Multiply the whole number by the next higher whole number; after which, add the product of the fractions. N. B. In the following examples, the product of the fractions are obtained first for convenience. PRACTICAL EXAMPLES FOR BUSINESS MEN. Multiply 3| by 3^ in a single line. Here we multiply ^X|j which gives T 3 5 , 3| and set down the result ; then we multiply 3^ the 3 in the multiplicand, increased by - unity, by the 3 in the multiplier, 3X^, 12 T V which gives 12 and completes the product. Multiply 7f by 7f in a single line. Here we multiply f Xf> which gives 5 \, 7f *nd set down the result; then we multiply 7^ the 7 in the multiplicand, increased by unity, - oy the 7 in the multiplier, 7x$, which gives 66, and completes the product. Multiply 11-| by llf in a single line. Here we multiply -| X, which gives f, and 11 J- set down the result; then we multiply the 11 llf in the multiplicand, increased by unity, by -- the 11 in the multiplier, 11X.12, which gives 132| 132, and completes the product. 3) ORTON'S LIGHTNING CALCULATOR. EXAMPLE THIRTY-THIRD. Multiply 16f by 16 J in a single line. Here we multiply Jxf which gives 1, and 16$ set down the result, then we multiply the 16 J 16 in the multiplicand, increased by unity by the 16 in the multiplier, 16X17, 272 which gives 272 and completes the product. EXAMPLE THIRTY-FOURTH. Multiply 29J by 29J in a single line. Here we multiply JX^ which gives J, 29 J and set down the result, then we multiply the 29 in the multiplicand, increased by unity by the 29 in the multiplier, 29 X 870J 30, which gives 870 and completes the pro- duct. EXAMPLE THIRTY-FIFTH. Multiply 999f by 999f in a single line. Here we multiply fXf 3 which gives 999f J|, and set down the result, then we 999|- multiply the 999 in the multiplicand, increased by unity by the 999 in the 999000f multiplier, 999X1000, which gives 999000 and completes the product. NOTE. The system of multiplication introduced in the preceding examples, applies to all numbers, \Vhcre the sum of the fractions is onej and the whole numbers are alike, or differ by one, the learner is requested to study well these useful properties of numbers. ORTON S LIGHTNING CALCULATOR. 81 WJiere the sum of the Fractions is ONE. To multiply any two numbers whose difference ip one, and the sum of the fractions is one, RULE. Multiply the larger number, increased by ONE, by the smaller number; then square the frac- tion of the larger number , and subtract its square from ONE. PRACTICAL EXAMPLES FOR BUSINESS MEN. 1. What will 9 Ibs. sugar cost at 8| cts. a lb.? Here we multiply 9, increased by 1, by 8, 91 thus, 8x10 are 80, and set down the result; g 3 then from 1 we subtract the square of -J, -- thus, squared is T ^-, and 1 less -^ is |-|. &OJ-J 2. What wih 8| bu. coal cost at 7| cts. a bu.? Here we multiply 8, increased by 1, by S-f 7, thus, 7 times 9 are 63, and set down the 7^ result; then from 1 we subtract the square - of -f , thus, -| squared is f , and 1, less , is f . "S 3. What will 11^ bu. seed cost at $10f| a bu.? Here we multiply 11, increased by 1, by 10, thus, 10 times 12 are 120, and set 11 i\ down the result; then from 1 we subtract "^T5 the square of ^ thus, ^ squared is T fo, 120 i5 and 1 less is 4. How many square inches in a floor 99|^ in- wide and 98| in. long? Ans. 9800&J. 32 ORTON'S LIGHTNING OALCILATOE. METHOD OF OPERATION. EXAMPLE FIRST. Multiply 6J by 6J- in a single line. Here we add 6J-|- J, which gives 6J ; this 6] multiplied by the 6 in the multiplier, 6J 6X6*2, gives 39, to which we add the pro- duct of the fractions, thus JXi gives y'g, added 39_i g to 39 completes the product. EXAMPLE SECOND. Multiply 11 J by llf in a single line. . Here we would add llj-j-f, which gives 11 J 12; this multiplied by the 11 in the multi- 31f plier gives 132, to which we add the product of the fractions, thus f Xj gives T 3 5 , which 132 7 8 g added to 132 completes the product. EXAMPLE THIRD. Multiply 12J by 12f in a single line. Here we add 12J+f, which gives 13J; 12J this multiplied by the 12 in the multiplier, 12| 12X13^, gives 159, to which add the pro duct of the fractions, thus |Xi gives f, 159f which added to 159 completes the product. ORTON'S LIGHTNINGI CALCULATOR. 33 Where the Fractions have a Like Denominator. To multiply any two like numbers together, each of which has a fraction with a like denominator, as 4f by 4|, or 11^ by 11 1, or lOf by lOi, etc. RULE. Add to the multiplicand the fraction of the multiplier, and multiply this sum by the whole number; after which, add the product of the fractions PRACTICAL EXAMPLES FOR BUSINESS MEN. N. B. In the following example, the sum of the frac- tions is ONE. 1. What will 9| Ibs. beef cost at 9^ cts. a lb.? The sum of 9|- and ^ is ten, so we simply 9-| say, 9 times 10 are 90; then we add the * product of the fractions, | times | are T 3 ^-. 90^ N. B. In the following example, the sum of the frao tions \s less than ONE. 2 What will 8J yds. tape cost at 8| cts. a yd. ? The sum of 8| and f is 8|, so we simply 8 say, 8 times 8f are 70; then we add the product of the fractions, -f times ^ are ^ or -J-. 70| N. B. In the following example, the sum of the frac- tions is greater than ONE. 3. What will 4| yds. cloth cost at $4| a yd.? The sum of 4| and |- is 5, so we simply 4| say, 4 times 5^ are 21; then we add the product of the fractions, J times | are |J. 21-| N. B. "Where the fractions have different denominatoru, reduce them to a common denominator. 3 34 ORTON'S LIGHTNING CALCULATOR. Rapid Process of Multiplying Mixed Numbers. A valuable and useful rule for the accountant in the practical calculations of the counting-room. To multiply any two numbers together, each of which involves the fraction -i, as 7 by 9, etc., RTJLE. To the product of the whole numbers add "kalf their sum plus . EXAMPLES FOR MENTAL OPERATIONS. 1. What will 3doz. eggs cost at 7| cts. a doz.? Here the sum of 7 and 3 is 10, and half this 31 sum is 5, so we simply say, 7 times 3 are 21 7-J and 5 are 26, to which we add 4. 264 N. B. If the sum be an odd number, call it one less tc make it even, and in such cases the fraction must be }. 2. What will 11| Ibs. cheese cost at 9 cts. a lb.? 3. What will 8 yds. tape cost at 15 cts. a yd.? 4. What will 7 Ibs. rice cost at 131 c ts. a lb.? 5. What will lOj- bu. coal cost at 12 cts. a bu.? REASON. In explaining the above rule, we add half their sum because half of either number added to half the other would be half their sum, and we add -^ because % by % is \. The same principle will multiply any two numbers together, each of which has the same fraction; for instance, if the fraction was -J-, we would add one-third their sum ; if f, we would add three-fourths their sum, etc.; and then, to complete the multiplication, we would add, of course, the product of the fractions. MULTIPLICATION. 35 GENERAL RULE For multiplying any two numbers together, each of which involves the same fraction. To the product of the whole numbers, add tht product of their sum by either fraction; after which add the product of their fractions. EXAMPLES FOR MENTAL OPERATIONS. 1. What will llf Ibs. rice cost at 9f cts. a lb.? Here the sum of 9 and 11 is 20. and three- ]\z fourths of this sum is 15, so we simply say, 9| 9 times 11 are 99 and 15 are 114, to which we add the product of the fractions (3^-). ra 2. What will 7f doz. eggs cost at 8| cts. a doz. ? 3. What will 6f bu. coal cost at 6| cts. a bu. ? 4. What will 45f bu. seed cost at 3f dol. a bu.? 5. What will 3| yds. cloth cost at 5f dol. a yd. ? 6. What will 17f ft. boards cost at 13f cts a ft.? 7. What will 18| Ibs. butter cost at 18| cts. a lb. ? N. B. If the product of the sum by either frac tion is a whole number with a fraction, it is bettei to reserve the fraction until we are through with the whole numbers, and then add it to the product of the fractions; for instance, to multiply 3^ by 7J, we find the sum of 7 and 3, which is 10, and one fourth of this sum is 2.1 ; setting the | down in some waste spot, we simply say, 7 times 3 are 21 and 2 are 23 ; then, adding the \ to the product of the fractions (y 6 ), gives T \, making 23^, Ans 36 ORTON'S LIGHTNING CALCULATOR. Rapid Process of Multiplying all Mixed Numberi. N. B. Let the student ren) ember thai this is a goneral and universal rule. GENELAL RULE. To multiply any two mixed numbers together, 1st. Multiply the whole numbers together. 2d. Multiply the upper digit by the lower fraction. 3d. Multiply the lower digit by the upper fraction. 4th. Multiply the fractions together. 5th. Add these FOUR products together. N. B. This rule is so simple, so useful, and so true that every banker, broker, merchant, and clerk should post it ap for reference and use. PRACTICAL EXAMPLES FOR BUSINESS ME3 N. B. The following method is recommended to begin- ners : EXAMPLE. Multiply 12f by 9|. 12| 1st. We multiply the whole numbers. 9f 2d. Multiply 12 by -| and write it down. jQg 3d. Multiply 9 by f and write it down. 9 4th. Multiply f by -| and write it down. 5th. Add these four products together, _ T2 and we have the complete result. 123 T 6 Tj- 1ST. B. When the student has become familiar with the above process, it is better to do the inter- mediate WDrk in the head, and, instead of setting down the partial products, add them in the mind as you pass along, and thus proceed very rapidly. MULTIPLICATION. 37 Multiply 8^- by 10. Here we simply say 10 times 8 are 80 8| and | of 8 is 2, making 82, and | of 10 is 10j 2 3 which makes 84; then times is -fa, flaking 84^ the answer. 81^ PRACTICAL BUSINESS METHOD For Multiplying all Mixed Numbers. Merchants, grocers, and business men generally, in multiplying the mixed numbers that arise in the practical calculations of their business, only care about having the answer correct tG the near- est cent; that is, they disregard the fraction. When it is a half cent or more, they call it an- other cent ; if less than half a cent, they drop it. And the object of the following rule is to show the business man the easiest and most rapid process of finding the product to the nearest unit of any two numbers, one or both of which involves a fraction. GENERAL RULE. To multiply any two numbers to the nearest unit, 1st. Multiply the whole number in the multiplicand by the fraction in the multiplier to the nearest unit. 2d. Multiply the whole number in the multiplier by the fraction in the multiplicand to the nearest unit 3d. Multiply the whole numbers together and add the three products in your mind as you proceed. N. B. In actual business the work can generally be done mentally for on?y easy fractions occur in business. 38 ORTON'S LIGHTNING CALCULATOR. N B. This rule is so simple and so true, according to all business usage, that every accountant should make himself perfectly familiar with its application. There being no such thing as a fractian to add in, there is scarcely any liability to error or mistake. By no other arithmetical process can the result be obtained by so few rfgures. EXAMPLES FOR MENTAL OPERATION. EXAMPLE FIRST. Multiply 11^- by 8| by business method. 11 1 Here ^ of 1 1 to the nearest unit is 3, and ^ of 8 J 8 to the nearest unit is 3, making 6, so we sim- ply say, 8 times 11 are 88 and 6 are 94, Ans. 94 REASON. J of 11 is nearer 3 than 2, and J of 8 is nearer 3 than 2. Make the nearest whole number the quotient. EXAMPLE SECOND. Multiply 7f by 9| by business method. Here | of 7 to the nearest unit is 3, and J 7-f of 9 to the nearest unit is 7 ; then 3 plus 7 9$ is 10, so we simply say, 9 times 7 are 63 and 10 are 73, Ans. 73 EXAMPLE THIRD. Multiply 23i by 19 by business method. Here ^ of 23 to the nearest unit is 6, and 23 of 19 to the nearest unit is 6 ; then 6 plus 19J 6 is 12, so we simply say, 19 times 23 are 437 and 12 are 449, Ans. N. B. In multiplying the whole numbers together al ways use the single-line method. MULTIPILCATION. 39 EXAMPLE FOURTH. Multiply 128f by 25 by business method. Here f of 25 to the nearest unit is 17, so 128 we simply say, 25 times 128 are 3200 and ^ 17 are 3217, the answer. 3217 PRACTICAL EXAMPLES FOR BUSINESS MEN. 1. What is the cost of 17 Ibs. sugar at 18| cts, per lb.? Here | of 17 to the nearest unit is 13, 17^ and | of 18; is 9 13 plus 9 is 22, so we 18| simply say. 18 times 17 are 306 and 22 are Qi'l 9Q 328, the answer. 2. What is the cost of 11 Ibs. 5 oz. of butter at 33 J cts. per lb.? Here -J of 11 to the nearest unit is 4, HT$ and ^ of 33 to the nearest unit is 10 ; 33} then 4 plus 10 is 14, so we simply say, 33 times 11 are 363, and 14 are 377, Ans. 3. What is the cost of 17 doz. and 9 eggs at 12^ cts. per doz.? Here % of 17 to the nearest unit is 9, l?^ and ^ of 12 is 9 ; then nine plus 9 is 18, lf> BO we simply say, 12 times 17 are 204 and 18 are 222, the answer. 4. What will be the cost of 15| yds. calico at 12J cts. per yd.? Ans. $1.97. N. B. To mul tiply by aliquot parts of 100, see page 44 D 40 ORTON'S LIGHTNING CALCULATOR. RAPID PKOCESS OF MARKING GOODS. A VALUABLE HINT TO MERCHANTS ANP ALL RETAIL DEALERS IN FOREIGN AND DOMESTIC DRY GOODS. RETAIL merchants, in buying goods by whole- sale, buy a great many articles by the dozen, such as boots and shoes, hats and caps, and notions of various kinds. Now, the merchant, in buying, for instance, a dozen hats, knows exactly what one of those hats will retail for in the market where he deals ; and, unless he is a good accountant, it will often take him some time to determine whether he can afford to purchase the dozen hats and make a living profit in selling them by the single hat ; and in buying his goods by auction, as the merchant often does, he has not time to make the calculation before the goods are cried off. He therefore losei the chance of making good bargains by being 1 afraid to bid at random, or if he bids, and the goods are cried off, he may have made a poor bar- gain, by bidding thus at a venture. It then be- comes a useful and practical problem to determine instantly what per cent, he would gain if he re- tailed the hats at a certain price. To tell what an article should retail for t* make a profit of 20 per cent., RULE. Divide what the articles cost per dozen by 10, which is done by removing the decimal point onf place to the left. M CTLTIPLICTATION. 41 For instance, if hats cost $17.50 per dozen, re- move the decimal point one place to the left, mak- ing $1.75, what they should be sold for apiece to gain 20 per cent, on the cost. If they cost $31.00 per dozen, they should be sold at $3.10 apiece, etc. We take 20 per cent, as the basis for the following reasons, viz. : because we can determine instantly, by simply removing the decimal point, without changing a figure; and, if the goods would not bring at least 20 per cent, profit in the home mar- ket, the merchant could not afford to purchase and would look for goods at lower figures. REASON. The reason for the above rule is ob- vious : For if we divide the cost of a dozen by 12, we have the cost of a single article ; then if we wish to make 20 per cent, on the cost, (cost being 1 or iO> we a( ^ ^ e ^O per cent., which is , to the ^, making -| or i-J- ; then as we multiply the cost, divided by 12, by the J-f to find at what price one must be sold to gain 20 per cent., it is evident that the 12s will cancel, and leave the cost of a dozen to be divided by 10, which is done by re- moving the decimal point one place to the left. 1. If I buy 2 doz. caps at $7.50 per doz., wliat shall I retail them at to make 20%? Ans. 75 sts. 2. When a merchant retails a vest at $4.50 and makes 20%, what did he pay per doz.? Ans. $45. 3. At what price should I retail a pair cf boots that cost $85 per doz., to make 20%? Ans. $8.50. 42 ORTON'S LIGHTNING CALCULATOR. RAPID PROCESS OF MARKING GOODS AT DIFFERENT PER CENTS. Now, as removing the decimal point one place to the left, on the cost of a dozen articles, gives the selling price of a single one with 20 per cent, added to the cost, and, as the cost of any article is 100 per cent., it is obvious that the selling price would be 20 per cent, more, or 120 per cent.; hence, to find 50 per cent, profit, which would make the selling price 150 per cent., we would first find 120 per cent., then add 30 per cent., by increasing it one-fourth itself; to make 40 per cent., add 20 per cent., by increasing it one-sixth Hself ; for 35 per cent., increase it one-eighth itself, etc. Hence, to mark an article at any per cent, profit, we have the following GENERAL RULE First find 20 per cent, profit, by removing the decimal point one place to the left on the price the articles cost a dozen; then, as 20 per cent, profit is 120 per cent., add to or subtract from this amount the fractional part that the required per cent, added to 100 is more or less than 120. Merchants, in marking goods, generally take a per cent, that is an aliqot part of 100, as 25%. 33^%, 50%, etc. The reason they do this is be- cause it makes it much easier to add such a per cent, to the cost; for instance, a merchant could MULTIPLICATION. 43 mark almost a dozen articles at 50 per cent, profit in the time it would take him to mark a single one at 49 per cent. For the benefit of the student, and for the convenience of business men in mark- ing goods, we have arranged the following table : TABLE For Marking all Articles bought ~by the Dozen. N. B. Most of these are used in business. To make 20$ remove the point one place to the left, " " 80$ " " and add J itself. " " 60$ " " " " " 4 " " " 50% " " " " " 4- " " " 44:% " " " " U -i- U u n 40% a " u " (( -i u u ci 32* ' A " " " " A " " " " " T V " a a a u i u " " " " X " " " subtract^ " U U U U 1 a w If I buy 1 doz. shirts for $28.00, what shall I retail them for to make 50%? Ans. $3.50. EXPLANATION. Remove the point one place to the left, and add on | itself. 44 ORTON'S LIGHTNING CALCULATOR. Where the Multiplier is an Aliquot Part of 100. Merchants in selling goods generally make the price of an article some aliquot part of 100, as in Belling sugar at 12^ cents a pound or 8 pounds for 1 dollar, or in selling calico for 16| cents a yard or 6 yards for 1 dollar, etc. And to be- come familiar with all the aliquot parts of 100, so that you can apply them readily when occasion requires, is perhaps the most useful, and, at the same time, one of the easiest arrived at of all the computations the accountant must perform in the practical calculations of the counting-room. TABLE OF THE ALIQUOT PARTS OF 100 AND 1000 N. B. Most of these are used in business. Ill is i- part of 100. 8J is ^ part of 100 25 is f or J of 100. 16| is T 2 ^ or J- of 100 371 i g I part of 100. 33* is ^ or \ of 100 50 is f or J of 100. 66| is & or f of 100 62J is f part of 100. 83J is ^ or f of 100. 75 is f or | of 100. 125 is part of 1000, 87 is | part of 100. 250 is f or J of 1000. 6 i is Tir P art of 10 - 375 is I P arfc of !000. 18| is T 3 F part of 100. 625 is f part of 1000. 31i i s ^ part O f 100. 875 is f part of 1000. To multiply by an aliquot part of 100, RULE.- Add two ciphers to the multiplicand, Jie* Mite such part of it as the multiplier is part of 100, N. B. If the multiplicand is a mixed number reduce the fraction to a decimal of two places before diTiding. MULTIPLICATION. 45 3. To multiply any number by 125 ad if three ciphers, and divide by 8. Multiply 3467 by 125. Product, 433375. 8)3467000 433375 NOTE. By annexing three ciphers the number is increased one thousand times ; and by dividing by 8, the quotient will be only one-eighth of 1000, that is 125 times. 4. To multiply any number by 16f add two ciphers, and divide by 6. Multiply 3768 by 16f . Product, 62800. 6)376800 62800 5. To multiply any number by 166$ add three ciphers, and divide by 6. Multiply 7875 by 166f . Product, 1312500. 6)7875000 1312500 6. To multiply any number by 33J add tw* ciphers, and divide by 3. Multiply 9879 by 33 J. Product, 329300 )987900 329300 16 ORTON'S LIGHTNING CALCULATOR. RATIONALE. As in the last case, by annexing two ciphers, we increase the multiplicand one hun- dred times ; and by dividing the number by 3, we only increase the multiplicand thirty-three and one-third times, because 33 J is one-third of 100. 7. To multiply any number by 333J add three ciphers, and divide by 3. Multiply 4797 by 333J. Product, 1599000. 3)4797000 1599000 8. To multiply any number by 6f add two ci- phers, and divide by 15 ; or add one cipher and multiply by f . Multiply 1566 by 6f . 15)156600 10440 First method. 15660 2 3)31320 10440 Second method. 9. Tc multiply any number by 66 add three ciphers, and divide by 15 ; or add two ciphers and multiply by f. MULTIPLICATION. 47 Multiply 3663 by 66f . 15)3663000 244200 First method. 366300 2 3)732600 244200 Second method. 10. To multiply any number by 8J add two ci- phers, and divide by 12. Multiply 2889 by 8J. Product, 24075. 12)288900 24075 11. To multiply any number by 83 J add thref ciphers, and divide by 12. Multiply 7695 by 83. Product, 641250, 12)7695000 641250 12. To multiply any number by 6J add two A jhers, and divide by 16 or its factors 4X4. Multiply 7696 by 6J. Product, 4810C 4)769600 4)192400 48100 48 ORTON'S LIGHTNING CALCULATOR. 13. To multiply any number by 62J add three ciphers, and divide by 16 or its factors 4x4. Multiply 3264 by 62J. Product, 204000. 16)3264000 204000 14. To multiply any number by 18f, add two ciphers, and multiply by 3, and divide by 16 or *ts factors 4X4. Multiply 768400 by 18f . Product, 144075. 768400 3 16)2305200 144075 15. To multiply any number by 31J add three ciphers, and divide by 32 or its factors 4x8. Multiply 7847 by 31J. Product, 245218g. 4)7847000 8)1961750 245218| 16. To multiply any number by 37 J, aid two ciphers and multiply by 3, and divide by 8. MULTIPLICATION. 49 Multiply 2976 by 37J. Product, 111600. 297600 3 8^892800 111600 17. To multiply any number by 87J add two ciphers, divide by 8, and subtract the quotient. Multiply 6768 by 87J. 8)676800 84600 592200 Ans. 18. To multiply any number by 75 add two ci- phers, divide by 4, and subtract the quotient. Multiply 4968 by 75. 4)496800 124200 372600 Ans. 19. To multiply by 99 add two ciphers, and subtract the given number from the result. Multiply 31416 by 99. 3141600 31416 Product, 3110184 As 99 times is 1 time less than 100 times, and 60 ORTON'S LIGHTNING CALCULATOR. adding 00 is in effect multiplying by 100, one time the multiplicand is deducted, which leaves 99 times, as required. This principle is applicable where any number of 9's is the multiplier ; as many ciphers being added of course as there are 9's. If the multi- plier were 98, we would subtract twice the multi- plicand ; if 97, three times, and so on. 20. To multiply by any number of 9's. E.ULE. Annex as many ciphers to the multipli- cand as there are 9's in the multiplier, and from this number subtract the number to be multiplied, and th* remainder is the product required. NOTE. To multiply by any number of 3's, pro- ceed as above, and divide the product by 3 ; but if it be required to multiply by 6's, proceed as above, and then multiply the product by two, and divide the result by 3, and the quotient is the product. 21. To multiply by any number between 10 and 20, as 16 or 18, multiply by the units' figure, and set the product under the multiplicand ; but put it one place to the right ; then add the lines together. The reason is evident on looking at the calculation. Multiply 3854 by 16. 3854 23124 61664 An*. MULTIPLICATION. 51 On the same principle, if any number of ct- ^flers intervene, as 106, 1006, 10006, etc., so* the product so many places farther to the right. Multiply 3854 by 1006. 3854 23124 3877124 An*. Cross Multiplication is a mode of multiplying by large multipliers in a single line ; and by prac- tice the operation may be performed with great expedition. It is necessary to begin with small numbers, say of two places, and carry the calf dili- gently, if you would carry the ox successfully, Here we multiply 5x2 and set down and 32 carry as usual ; then to what you carry UCkl 45 5X3 and 4X2, which gives 24; set down 4 and carry 2 to 4x3, which gives 14. It 1440 is obvious that this is just the usual mode, with the intermediate work done in the head. Here the first and second places are 123 found as before; for the third, add 6X1, 456 4X3, 5X2, with the 2 you had to carry, making 30 ; set down and carry 3 ; then 56088 drop the units' place and multiply the hundreds and tens crosswise, as you did the tens and units, and you find the thousands' figure } then, 4 E 62 ORTON'S LIGHTNING CALCULATOR. dropping both units and tens, multiply the adding the 1 you carried, and you have 5, which completes the product. The same principle may be extended to any number of places ; but let each fttep be made perfectly familiar before advancing to another. Begin with two places, then take three, then four, but always practicing some time on each Bumber, for any hesitation as you progress will con- fuse you. To Multiply by 21, 31, etc., to 91 in a single line, multiply by the tens' figure and set the pro- duct one place to the left underneath the multipli- cand ; then add. Multiply 3854 by 21. 3854 7708 80934 Ans. il ciphers intervene, as 201, 3001, etc., multiply us before, but set the product as many additional places to the left as there are ciphers. Multiply 3854 by 6001. * 3854 23134 2316254 Ans. The following is a convenient mode of multi- MULTIPLICATION-. 53 plying by any two figures, and is not difficult to apply: Multiply 3754 By 27 Product, 101358 I here multiply 27 by 4, setting down the first product figure and carrying the others ; I then multiply by 5 and set down and carry in the same way so proceeding to the highest place of the multiplicand. Where the multiplier is not too large, and can be divided into two or more factors, there is a sav- ing in adopting the following mode, and it may be used as a proof of the common mode. If I seek to multiply, say 7864 by 24, it requires in the usual way two product lines, and finding their sum by addition, making a third operation; but if I multiply by 6, and that product by 4, or by 8 and 3, or 12 and 2, the business is dispatched in two lines. But being able to multiply in a single line is still better. The same remarks apply in Division, and hence there is often economy of figures in dividing by factors of your divisor ; and if a remainder occurs only in your first division it is the true one ; but if in the second only, then multiply such remain- der by the first divisor, or all if more than one ; fft ORTON'S LIGHTNING CALCULATOR. '*nd if there was a remainder on the first division also, it must be added in, and the sum will be the true remainder. Divide 73640 by 24. 6 ) 73640 4)12273 and 2 over. 3068 and 1 over. 1x6+2=8, the true rem'der. Or 4)73640 6)18410 3068 and 2 over, and 2 x4=8, as before. Another mode of proving Division is to divide the dividend by the quotient, and the result will be the divisor the same remainder occurring in one case as in the other, but not of the same fractional value ; for as the dividend exceeds some multiple of the divisor and quotient, just the amount of the remainder, that remainder will be the same, with- out regard to which of the factors occupies the di- visor's place ; and as the divisors would differ, th0 value of the fraction formed by the remainder and divisor must differ. To make the dividend an ex tct multiple, subtract the remainder from it. MULTIPLICATION. 55 Multiply 3756 By 182 7512 30048 3756 683592 In tlie above example the second product Lne may be conveniently found by multiplying the first by 4 ; but if there is an error in the first it will run into the second. A shorter mode would be to mul- tiply the first product line by 9, which would give the product by 18 in one line, and save work. Trj it. As a large number of the problems solved ii, this book are solved by different modes, they fur- rdsh a great variety of modes of proof. To multiply a decimal or a mixed number by 10, 100, 1000, etc., remove the decimal point one, two, or three places to the right ; and to divide by such numbers remove the point as many places to the left ; and if there be not a sufficient number of places on the left, ciphers must be prefixed. . To multiply or divide by any composite number ; multiply, or divide, as the case may be, by the fac- tors of the numbers successively. This was sufficiently illustrated under the head of Proofs. 56 ORION'S LIGHTNING CALCULATOR. Tlie French mode of operating in Long Dh ision has some advantage over ours. They place the divisor on the right of the dividend, as we do the quotient, and place the quotient underneath the divisor, by which the figures to be multiplied together are brought near each other. Thus : Divid. Divis. 3936)96 384 41 Quotient. 96 96 For sake of brevity they fre- qtiently omit the product figures, setting down only the remain- 3936)96 ders, which they find as they 96 41, Ans. pass along. Thus : This, however, applies to our mode as well as theirs. The following mode of multiplying"^ large multipliers in a single line, is both curious and useful : Multiply 7865 by 432 in a single line. On a slip of paper, separate from that on which the multiplicand is written, place the multiplier in Inverted order : thus, 234 and close to the upper edge of the paper. Then bring the multiplier so that the 2 shall be directly under the 5, or units' MULTIPLICATION. 57 place of the multiplicand: multiply those figures, set down and carry 1. Slide the paper to the left one place, that 2 may be under G, and 3 under 5 ; and to the 1 you carried add the products of the 2 by 6, and 3 by 5, making 28 set down 8 and carry 2. Again move your paper one place to the left, and to the 2 you carried, add the several products of the multiplicand figures with the figures of the multiplier that are under them, viz: 8X2, 6X3, 5X4, and the result will be 56 ; set down 6 and carry 5. Slide again and you have 5 (that you carried) +14+24+24=67. Thus pro- ceed toward the left until the multiplier passes from under the multiplicand, each time adding what you carry, to the several products of the figures that stand one over the other, the result will be 3397680. These additions will soon be performed at a glance, as the products are obvious without the formality of naming the factors. To understand these directions clearly, factors must be placed on slips of paper, and the directions strictly complied with ; by which the mode of op- eration and the reason will be better understood in ten minutes, than in three hours without them. When familiar with the slide, the operator may proceed without it, and perform operations aston- ishing to tke uninitiated; the largest numbers being multiplied together readily in a single liae. 68 ORION'S LIGHTNING CALCULATOR. 01 THE PROPERTIES OF NUMBERS. PROPOSITION 1. Digits in our system of notation increase in oaluefrom right to left in a tenfold ratio. PROPOSITION 2. r n any series of digits expressing a number, the value of any digit is greater than the value of all the digits on its right. This property results also from value according lo place ; and that the proposition is true is obvi- mis, for if we take the smallest digit (1) and place it on the left of the largest (9) we form 19; the 1 expresses 10 units, while the 9 expresses but 9 units; and let us add what numbers of nine we may, the unit will constantly retain its greater value: e.g. 19, 199, 1999, etc. Not only is the left hand digit higher in value than all upon its right, but the same remark applies to each digit, in reference to those on its right. PROPOSITION 8. If the sum of the digits in any number be a multi- ple of 9, the whole number is a multiple of 9. This is one of several peculiar properties of the number 9, all arising from its being just one less than the radix of our system of notation, and MULTIPLICATION. 59 hence the highest number expressed by a single ' character; and these properties will belong to the highest number so expressed in any system. We might go a step further in reference to this prop- erty, and say that it belongs to any number that will divide the radix of the system, and leave one as a remainder. If we carefully examine the genesis of numbers, we must see that so far as the number 9 is con- cerned, this is an accidental property, resulting from our scale of notation. We constantly express each successive number from unity to 9, inclusive, by a digit of greater value than any preceeding it ; but when we pass 9 we express the next number, 10, by a unit and a cipher. The number is one greater than 9, and the sum of its digits is 1. Eleven is 2 greater, and the sum of its digits is 1-j-l 2. Thus we proceed, the sum of the dig- its constantly expressing the excess over 9, until we reach 18, or twice 9. Nineteen is 1 and 9, and it is one over twice 9. 20 is 2 over twice 9, and the sum of its digits is 2. The same course con tinued to millions, would but produce the same recurring result. Nine is 1 less than 10; twice 9 are 2 less than 20; 3X9 are 3 less than 30 , and BO on; and hence the 1 of 10, 2 of 20, 3 of 30 etc., come just in the proper place to keep up the excess above 9 and its multiples. If tr e multiples 60 ORION'S LIGHTNING CALCULATOR. of 9 did not constantly fall at each product, one further behind the corresponding multiples of 10, th* two of 20, 3 of 30, etc., would not fall in the right place, to' keep up the regular order of the aeries. PROPOSITION 4. If tlie sum of the digits in any number be a multi- ple of 3, the number is a multiple of 3. The same reasoning applied to the number 9 to *how the correctness of the preceding proposition, will show the correctness of this. Ten, the sum of whose digits is 1, is 1 over 3 times 3; 11, the sum of whose digit is 2, is 2 more than 3 times 3 ; 12, the sum of whose digit is 3, is a multiple, etc., etc. PROPOSITION 5. Dividing any number ty 9 or 3, will leave the same remainder as dividing the sum of its digits by 9 or 3. This proposition follows as a matter of course from the two next preceding it ; and we shall ad- duce .no other proof of its correctness. Like the former, it is an accidental property of the highest number expressed by a single digit in any system, and of all its factors. If 9 were the basis of our system, these properties would belong to 8, 4, and 2; if 8, then 7 only, since 7 has no factors; and MULTIPLICATION. 6 1 if 7 were the basis, then 6, 3, and 2 would possess these properties; and if 12 were the basis, then 11 only would possess such properties; for it would in that case be expressed by a single digit, and would be the highest number so expressed. Twelve would be written with a unit and a cipher as 10 now is ; and 11 being prime, it would be the only number that would divide the radix of the system and leave 1 as a remainder. As early as 1657, Dr. Wallis, of England, applied this principle to prove the correctness of operations in the elementary rules of Arithmetic, and the practice has been continued to the present time. The operation is performed thus : We cast the nines out of each Add 79864=7 number separately, and set the ex- 32075=8 cess on the right. We then cast 83214=0 the nines out of the sum total 61840=1 305160, and also out of the sum 48167=8 of the excesses 8+1+8+7, and they are equal : both being 6, and 305160=6 we hence " infer that the work is right. To cast out the nines, the number may be divided by 9 ; but a better way is to add the digits together in each number, rejecting 9 whenever it occurs, and carrying forward only the excess. Thus 7 and 8 are 15; 9 being rejected, we carry 6 to d is 12; rejecting 9, we carry 3 to 4=7; the num- 62 ORTON'S LIGHTNING CALCULATOR. ber carried in each place is the excess over 9 ; ana where 9 occurs it is passed over. In Subtraction cast out the nines from the minu- end and subtrahend, and also from the remainder If the excess in the remainder is equal to the dif- ference of excesses in the minuend and subtrahend, the work is right. Here, as we can not take 8 From 6894321=6 from 6, we take from 9 and Take 2960864=8 add 6; the result, 7 agrees with the excess above 9 in Leaves 3933457=7 the difference of the numbers. In Multiplication, find the excess in the factors, and if the excess in the product of these two ex- cesses equals the excess in the product of the fac- tors the operation is correct. Multiply 48756=3 By 245=2 243780 3 195024 - 97512 6 11945220=6 This is often called proving by the cross ; and instead of placing the excessss after marks of equality, they are placed in the angles of a cross as on the right hand of the above operation. MULTIPLICATION. 63 In Division cast the nines out of the divisor, lividend, quotient, and remainder; then to the product of the excesses in the divisor and quo- tient, add the excess in the remainder, and cast the nines out of the sum, and if the excess equal that in the dividend the work is right. Excess in Divisor Excess in Quotient 8 27)465 Product of Excess 17+6 Add Excess of Remainder 6 Excess in Dividend 6=6 Hence the work is right, the excesses being equal. It is proper to remark that this mode of proof is liable to much objection. If the figures become transposed, or if mistakes are made that balance each other, the work will prove right when it is wrong. The work will, however, never prove wrong when it is right. In the product above, you may transpose the digits as you please, the work will prove, since the excess is the same what- ever is the order of the digits ; and ciphers may al- ways be omitted. Or if mistakes balance each other, as if instead of 91145 it be 83216. The excess here will be the same and the work will prove, though not a figure is right. 64 ORION'S LIGHTNING CALCULATOR. PROPOSITION 6. If from any number the sum of its digits le sub* tractedj the remainder is a multiple of 9. Fm 31416 For, Take 15=3+1+4+1+6 31416^-9=3490+6 15+9= 1+6 9)31401 31401-7-9=3489 3489 times 9, diff. As the remainder on dividing the given number by 9 will be just the same as on dividing the sum of its digits, (Prop. 5,) it is obvious that the differ- ence must be an even multiple. The given num- ber is 6 more than 3490 times 9 ; the sum of digits is 6 more than 1 time, hence their difference is 3489 times. This principle is sometimes used as a puzzle; you may let a person write down any num- 6745 ber for a minuend, then have the party 22 add the figures and place their sum for a subtrahend; then, after subtracting, let the 6723 person rub out any one figure in the re- mainder and give you the figures left in the re- mainder. You, by adding the figures given and subtracting their sum from the next multiple of 9, may tell the figure rubbed out, although you did not see a figure that was written. Try it. MULTIPLICATION. 65 PROPOSITION 7. The difference between a given number and the dig- its composing such number reversed or any how . arranged, is always a multiple of 9. The difference, for instance, between 7425, and any arrangemsut you can make of the same figures is a multiple of 9. From 7425 7425 7425 Take ' 5247 5724 2457 9)2178 9)1701 9)4968 242 189 552 This is based on the same reason as the preced- ing; for whether you take the sum of the digits or transpose the digits, it is the same in effect. The excess over an even multiple being the same as in the given number, the difference must nec- essarily be an even multiple. A practical application is sometimes made of this principle by a person setting down two rows of Ggures for subtraction, but being careful to have the figures of the subatrahend and minuend the same, though differently arranged. One figure of the remainder is then stricken out, and the puzzlo is to restore it without seeing the minuend and sub- trahend It is done by taking such number as will make the remainder a multiple of 9. 66 ORTON^S LIGHTNING CALCULATOR. Here if 5, 4, or 1, be From 7351681 erased, any one may restore Take 1864537 it ; but if the 9 or be re- moved he can not know 5490144 whether a 9 or a cipher should be supplied, as either will make the num> ber a multiple of 9. The mode we have adopted in explaining the four last preceding propositions appears to us plain and sufficiently satisfactory. In addition to the use of these properties as modes of proof, they are the key to many numeral puzzles and amusing questions ; and hence the care we have bestowed in explaining the principle. What has been said may be a sufficient explanation of the following article on the " Wonderful Proper- ties of the Number Nine : " " Multiply 9 by itself or any other digit, and the figures of the product added will be 9. Take the sum of our numerals 1+2+3+4+5+ 6+7+8+9=45, the digits of which, 4+5=9. Multiply each of these digits by 9, and their sum will be 405 ; which added 4+0+5=9 ; and 405-7- 9=45, also a multiple of 9. Multiply any number, large or small, by 9, or 9 times any digit, and the sum of the digits i)f the product will be a multiple of 9. Multiply the 9 digits in their order, 123456 MULTIPLICATION. 67 7 6 9, by 9, or any multiple of 9 not exceeding 9 times 9, and the product, except the tens' place, will be all the same figures, while the tens' place will be filled with 0. Th;3 significant figure will always be the number of times 9 is contained in the multiplier. 27, or 3 times 9, will 123456789 produce all 3s; 4 times 9 18=9X2 all 4s. Omit 8 in the multi- 987554312 plicand and the product 123456789 will be all the same dig- its, the having disap- 2222222202 peared." To a superficial observer the above results may Beem accidental, but investigation will show that they all flow from the laws and principles we have laid down ; and that a much longer list might be made of apparently simple and detached facts, but really of results flowing from well established laws. There are no unaccountable properties in numbers. While the number 9 has some peculiar proper- ties from being the next below the radix of the system, the number 11 has some peculiarities from being next above the radix. Among these are the following : " If from any number the sum of tho digits standing in the odd places be subtracted, and to 5 68 ORTON'S LIGHTNING CALCULATOR. the remainder the sum of the digits standing in the even places be added, then the result is a mul- tiple of 11." Again, "If the sum of the digits standing in the even places be equal to the sum of the digits standing in the odd places, or differ by 11 or any of its multiples, the number is a multiple of 11." As these however are of no practical utility, we shall not discuss them. The number 7 has also some peculiarities, but we shall name only one, as they are useless. If a number be divided into periods of three figures each, beginning at the units 1 place, when the difference of the sums of the alternate periods is a multiple of 7, the whole number is a multiple of 7. Here 862428 is a mul- 7)382,907,428,862 tiple of 7 ; and so is 907 382; therefore the whole 54,701,061,266 number is a multiple of 7. The division of numbers into Even and Odd seems to arise from considering them in pairs. The following facts growing out of this division will be readily understood : The sum of two even numbers is even, and so ia their difference : 8+4=12; 8 4=4. The sum of an odd number of odd numbers is odd; but the sum of an even number of odd num- bers is even : 3+5+7=15; 3+5=8; and 5+7=1 2. MULTIPLICATION. 69 An even and an odd number being added to- gether, or one subtracted from the other, the result will be odd: 8+311; 83=5. If a number has 0, 2, 4, 6 or 8 in the units' place, it is divisible by 2, and is consequently even. No odd number can be divided by an even num- ber without a remainder. If an odd number measure an even one, it will also measure the half of it. 7 measures 42, and therefore measures 21, the half of it. If the sum of the digits standing in the EVEN places, be equal to the sum of the digits standing in the ODD places, then the number is divisible by II. Thus the numbers 121, 363, 12133, 48422, etc., are all divisible by 11. Multiplication and Division. . To multiply one -half, is to take the multiplicand one-half of one time; that is, take one-half of it, or divide it by 2. To multiply by J, take a third of the multipli- cand, that is, divide it by 3, To multiply by f , take J, first, and multiply that by 2; or, multiply by 2 first, and divide the pro- duct by 3.* *Sometimes one operation is preferable, and sometimes the other; good judgment alone can decide when thi case is bef>re us. 70 ORTON'S LIGHTING CALCULATOR. EXAMPLES. I. What will 360 barrels of flour come to at 5J dollars a barrel. At 1 dollar a barrel it would be 360 dollars ; at 5 J dollars, it would be 5J times as much. 360 5 times, 1800 J of a time, 90 Am. $1890 Before we attempt to divide by a mixed number, such as 2J, 3J, 5f , etc., we must explain, or rather observe the principle of division, namely: That the quotient will be the same if we multiply the divi- dend and divisor by the same number. Thus 24 divided by 8, gives three for a quotient. Now, if we double 24 and 8, or multiply them by any num- ber wnatever, and then divide, we shall still have 3 for a quotient. 16)48(3; 32)96(3, etc. Now, suppose we have 22 to be divided by 5J; we may double both these numbers, and thus be clear of the fraction, and have the same quotient. 5|)22(4 is the same as 11)44(4. How many times is 1J contained in 12? Ans. Just as many times as 5 is contained in 48. The 5 is 4 times 1 J, and 48 is 4 times 12. From these observations, we draw the following rule for divid- ing by a mixed number. MULTIPLICATION AND DIVISION. 71 RULE. Multiply the whole number "by the lower term of the fraction ; add the upper term to the pro- duct for a divisor ; then multiply the dividend by the lower term of tJie fraction, and then divide. How many times is 1^ contained in 36? An*. 30 times. N. B. If we multiply both these numbers by 5, they will have the same relation as before, and a quotient is nothing but a relation between two numbers. After multiplication, the numbers may be considered as having the denomination of fifths. How many times is J contained in 12 ? Ans. 48 times. One-fourth multiplied by 4, gives 1; 12, multi- plied by 4, gives 8. Now, 1 in 48 is contained 48 times. Divide 132 by 2|. Ans. 48. Divide 121 by 15 J. Ans. 8 How many times is f contained in 3 ? Ans. 4 times. By a little attention to the relation of numbers, we may often contract operations in multiplication. A dead uniformity of operation in all cases indi- cates a mechanical and not a scientific knowledge of numbers. As a uniform principle, it is much easier to multiply by the small numbers, 2, 3, 4, 5, than by 7, 8, 9. 72 ORTON'S LIGHTNING CALCULATOR. Multiply 4532 Multiply 4532 by 639 by 963 (63=9X7.) 40788 40788 285516 285516 Product, 2895948 Product, 4364316 In both the foregoing examples we multiply the product of 9 by 7, because 7 times 9 are equal to 63. Because 9 is in the place of hundreds in exam- ple 2, the product for the other two figures is set two places toward the right. In this last example we may commence with the 3 units in the usual way ; then that product by 2, because 2 times 3 are 6 ; then the product of 3 by 3, which, will give the same as the multiplicand by 9. The appearance of the work would then be the same as by the usual method, but would be easier, as we actually multiply by smaller numbers. Multiply 40788 by 497 285516 1998612 20271636 Product of the 7 units. As 7X7=49, multiply the product of 7 by 7. Every fact of this kind, though extremely sim- ple, should be known bj all who seek for knowl- edge in figures. MULTIPLICATION AND DIVISION. 73 First multiply by 12, then that product by 12. Multiply 785460 by 14412 9425520 113106240 11320049520 Multiply 576 Multiply this last by 186 number, 3456, (which is 6 times 576,) by 3> (6X3 18.) 3456 and place the product 10368 in the place of tens, and we have 180 times 107136 576. Observe the same principle in the follow- ing examples : Multiply 576 Multiply 4078 by 618 by Commence with 6. (6X3=18.) 497 3456 (7X7=49.) 285516 10368 1998612 355968 20271636 Multiply 61524 by 7209 553716 4429728 Product, 443646516 Multiply this pro- duct of 9 by 8, because 9 times 8 are 72, and place the product in the place of 100, be* cause it is 7200. 74 ORTON'S LIGHTNING CALCULATOR. Multiply 1243 by 636 7458 First by 600. 44748 Multiply 7458 by 6. Product, 790548 Multiply 7 8 6 '4 by 246 This may be done by commencing with the 2; tnen that product by 2 and 3 ; or we may com- mence with the 6 units, and then that product by 4 ; because 4 times 6 are 24. Multiply 3764 by 199. Take 3764 200 times, and from that product sub- tract 3764. Multiply 764 by 498J. Take 764 500 times, and from that product sub- tract 1| times 764. Multiply 396 by 21|, or, (which is the same,) 99X87=8700 86 8613. N. B. Ninety-nine is J of 396, and 87 is 4 times 2 If. How many times is 125 contained in 2125 ? Same as 250 in 4250 ; Same as 25 in 425 . Same as 50 in 850 , Same as 5 in 85 ; Same as 10 m 170 j that is, 17 timeu. MULTIPLICATION AND DIVISION. 75 The object of these changes is to give the learner an accurate and complete knowledge of numbers and of division; and the result is not the only object sought for, as many young learners suppose. How many times is 75 contained in 575 ? or di- dde 575 by 75. An*. 7f . Divide 800 by 12J. Quotient, 64. Divide 27 by 16f . Quo. 3 T %, or If J. A person spent 6 dollars for oranges, at 6J cents a-piece; how many did he purchase? Ans. 96. "When two or more numbers are to be multiplied together, and one or more of them having a cipher on the right, as 24 by 20, we may take the cipher from one number and annex it to the other with- out affecting the product ; thus, 24x20 is the same as 240X2; 286X130028600X13; and 350X 70x40=35x7x4X1000, etc. Every fact of this kind, though extremely simple, will be very useful to those who wish to be skillful in operation. NOTE. If there are ciphers at the right hand either of the multiplier or multiplicand, or of both, they may be neglected to the close of the opera- tion, when they must be annexed to the product. REMARKS. We now give a few examples, for the pur- pose of teaching the pupil how to use his judgment; he will then have learned a rule more valuable than aU others. G 76 ORTON'S LIGHTNING CALCULATOR. Multiplication and Division Combined. WHEN it becomes necessary to multiply two or more numbers together, and divide by a third, or by a product of a third and fourth, it must be lit* erally done if the numbers are prime. For example : Multiply 19 by 13 and divide that product by 7. This must be done at full length, because the numbers are prime ; and in all such cases there will result a fraction. But in actual business the problems are almost all reduceable by short operations ; as the prices of articles, or amount called for, always corresponds with some aliquot part of our scale of computation. And when two or more of the numbers are composite numbers, the work can always be contracted. Example : Multiply 375 by 7, and divide that product by 21. To obtain the answer, it is suffi- cient to divide 375 by 3, which gives 125. The 7 divides the 21, and the factor 3 remains for a divisor. Here it becomes necessary to lay down & plan of operation. Draw a perpendicular Hue and place all numbers that are to be multiplied together under each other, on the right hand side, and all numbers that are Uvisors under each other, on the left hand side. MULTIPLICATION AND DIVISION. 77 EXAMPLES. Multiply 140 by 36, and divide that product by 84. We place the numbers thus : 84 I 14 4 I 36 We may east out equal factors from each side of the line without affecting the result. In this case 12 will divide 84 and 36 j then the numbers will Btand thus : 140 3 But 7 divides 140, and gives 20, which, multi- plied by 3, gives 60 for the result. Multiply 4783 by 39, and divide that product by 13. r* 4783 *0 3 Three times 4783 must be the result. Multiply 80 by 9, that product by 21, and di- vide the whole by the product of 60 X 6X14. In the above divide 60 and 80 by 20, and 14 and 21 by 7, and those numbers will stand canceled as above, with 3 and 4, 2 and 3, at their sides. Now, the product 3X@X2, on the divisor side, is equal to 4 times 9 on the other, and the remain' ing 3 is the result. 78 ORTON'S LIGHTNING CALCULATOR. General Rules for Cancellation. RULE IST. Draw a perpendicular line; obsetre this line represents the sign of equality. On the right hand side of this line place dividends only; on the left hand side place divisors only ; having placed dividends on the right and divisors on the left, as above directed. 2d. Notice whether there are ciphers both on the right and left of the line ; if so, erase an equal number from each side. 3d. Notice whether the same number stands both on the right and left of the line ; if so, erase them both. 4th. Notice again if any number on either side of the line will divide any number on the opposite side without a remainder ; if so, divide and erase the two numbers, retaining the quotient figure on the side of the larger number. 5th. See if any two numbers, one on each side, can be divided by any assumed number without a remainder; if so, divide them by that number, and retain only their quotients. Proceed in the same manner, as far as practicable, then, 6th. Multiply all the numbers remaining on the rignt hand side of the line for a dividend, am* those remaining on the left for a divisor. 7th Divide, and the quotient is the answer. INTEREST, DISCOUNT AND AVERAGE. 79 NOTE. If only one number remain on either iide of the line, that number is the dividend or divisor, according as it stands on the right or left of the line The figure 1 is net regarded in the operation, because it avails nothing, either to mul- tiply or divide by. REMARKS. This method may not work a great many problems, as they are found in some books, but it will work 90 out of every 100 that ought to be found in books. In a book we might find a problem like this : What is the cost of 21b. 7oz. 13pwt. of tea, at 7s. 5d. per pound. But the person who should go to a store and call for 31b. 7oz. and 13pwt. of tea would be a fit subject for a mad-house. The above problem requires downright drudgery, which every one ought to be able to perform ; but such drudgery never occurs in business. INTEREST, DISCOUNT, AND AVERAGE. BEFORE entering upon an investigation of the different modes of calculating interest, it may be interesting to bestow some attention upon the his- tory of the subject, that we may be better prepared to understand it. 80 ORTON'S LIJHTNING CALCULATOR. Among the Jews a law existed that they should not lake interest of their brethren, though they wero permitted to take it of foreigner*. " Thou sh?Jt not lend upon usury to thy brother: usury of money, usury of victuals, usury of any thing that is lent upon usury ; unto a stranger thou may- cst ]end upon usury ; but unto thy brother thou ehalt not ^nd upon usury." (Deuteronomy xxiii, 19, 20.) Alter the dispersion of the Jews they wandered through the earth, but they yet remain a distinct people, mixing, but not becoming assim- ilated with the people among whom they reside. Still looking to the period when they shall return to the promised land, they seldom engage in per- manent business, but pursue traffic, and especially dealing in money ; and if their national policy for- bids their taking interest of each other, they show no backwardness in taking it unsparingly of the rest of mankind. For ages they have been the money lenders of Europe, and we may safely at- tribute to this circumstance the prejudice, in some measure, that still exists even in our own country against such as pursue this business as a profession. The prejudice of the Christian against the Jew has been transferred to his occupation, and from the days of Shakspeare, who painted the inexorable Shylock contending for his pound of flesh-, down Jx> the present time, the grasping money lender, no INTEREST, DISCOUNT, AND AVERAGE. 81 less than the grinding dealer in other matters, has been sneeringly called a^ew. For ages the taking of any corapensatioii wnLt- ever for the use of money was called usury, and was denounced as unchristian ; and we find Aris- totle, the heathen philosopher, gravely contending that as money could not beget money, it was bar- ren, and usury should not be charged for its use. The philosopher forgot that with money the bor- rower could add to his flocks and his fields, and profit by the produce of both. Definition of Terms. Interest is premium paid for the use of money, {roods, or property. It is computed by per centage a certain per cent. OB the money being paid for its use for a stated tirce. The money on which interest is paid is called the PRINCIPAL. The per cent, paid is called the RATE ; the prin- cipal and interest aaded together is called the AMOUNT. When a rate per cent, is stated, without tho mention of any term of time, tn^ time is under- v stood to be 1 year. ' The first important step in the calculation of simple interest is the arranging of tne tm:e for which it is computed. The student must study the 82 ORTON^S LIGHTNING CALCULATOR. following Propositions carefully, if lie would be expert in this important and useful branch of bus- Sness calculations : PROPOSITION 1. / the time consists of years, multiply the principal ly the rate per cent., and that product by the number of years. EXAMPLE 1. Find the interest of $75 for 4 tears at 6 per cent. Operation. $75 The decimal for 6 per cent, is .06 06. There being two places of decimals in the multiplier, we 4.50 point off two in the product. 4 818.00 4ns. PROPOSITION 2. If the time consists of years and months, reduce the time to months, and multiply the principal by the rate per cent, and number of months together, and divide the result by 12. NOTE. The work can always be abbreviated at I, 6, 8, 9, 12, and 15 per cent., by canceling the per cent., or time, or principal, with the common divisor 12. INTEREST, DISCOUNT, ^ND S3 EXAMPLE 2. Find the interest of $240 for 2 years and 7 months at 8 per cent. First method. Second method : Principal, $240 Per cent., .08 In. forlyr., 19.20- 2yrs.-j-7nios., Slmos. by cancellation. ^020 It 8 rate. 31 time. 49.60 Ans. 12)595.20 $49.60 Ans. The operation by canceling is much more brief. We simply place the principal, rate, and time, on the right of the line, and 12 on the left; then we cancel 12 in 240, and the quotient 20 multiplied with 8 and 31 gives the interest at once. NOTE. After 12 is canceled the product of the remaining numbers is always the interest. PROPOSITION 3. If the time consists of years, months, and days, re- duce the years to months, add in the given months, and place one-third of the days to the right of this number, which we multiply by the principal and rate per cent., and divide by 12, as before; or cancel and divide by 12 before multiplying. EXAMPLE 3. Find the interest of $231 for 1 year, 1 month, and 6 days, at 5 per cent. 84 ORTON'S LIGHTNING CALCULATOR, First method. Principal, Per cent., In. for lyr., $231 .05 11.55 , 13.2mo, 12)152.460 Second method : by cancellation. 231 prin. 5 rate. 11 $12.705 Ans. $12.705 Am. By the second method we cancel 12 in 132, and multiply the quotient 11 by 5 and 231. NOTE. When the principal is $, and the time is in years or months, the interest is in cents ; if the time is in years, months, and days, the interest is in mills, unless the days are less than 3, in which case it would be in cents, as before. NOTE. The reason we divide the days by 3 is because we calculate 30 days for a month, and di- viding by 3 reduces the days to the tenth of months, NOTE. The three preceding propositions will work any note in interest for any time and at any given rate per cent. How to Avoid Fractions in Interest. PROPOSITION 4. If, when the time consists of years, months, and days, are not divisible by 3, you can divide the days ly 3, and annex the mixed number as in Proposition INTEREST, DISCOUNT, AND AVERAGE. 3, or if you wish to avoid fractions, you can reduce the time to interest days, and**tnultiply the principal^ rate and days together, and divide the result by 36 or its factors, 4x9. NOTE. In this case as in the preceding, the work can almost always be contracted by dividing the rate or time or principal with the divisor 36. NOTE. We use the divisor 36, because we cal- culate 360 interest days to the year. We discard the 0, because it avails nothing to multiply or di- vide by. EXAMPLE 4. Find the interest of $210 for 1 year, 4 months, and 8 days, at 9 per cent. Year. Months. Days. 1 4 8=16.2f months or 488 days. Operation Operation By Prop. 3. By Prop. 4. $210 $210 .9 18.90 16.2J 12)307440 $25.620 Ans. 18.90 488 36)922320 $25.620 Ans. We will now work the example by cancellation to show its brevity. 66 ORTON'S LIGHTING CALCULATOR. Operation ty Cancellation. Time 488 days. 210 m 122 122 210 $25.620 *W cancel 9 in 36 goes 4 times, then 4 into 485 122. Now multiply remaining numbers to- gether, thus, 210X122 and we have the interest at once. When the days are not divisible by 3 we reduce t\i3 whole time to days ; then we place the princi- pal rate and time on the right of the line. Now, because the time is in days, we place 36, on the left of the line for a divisor. (If the time was months we would place 12 on the left.) NOTE. A very short method of reducing time to interest days is to multiply the years by 36 ; add in 3 times the number of months and the tens' figure of the days, and annex the unit figure; but if the days are less than 10 simply annex them. EXAMPLE 1. Reduce 1 year, 2 months, and 6 days, to days. Years. Months. Days. 36X1+3X2=:42 annex 6=426 Ant. SIMPLE INTEREST BY CANCELLATION. 87 EXAMPLE 2. Reduce 2^ears, 3 months and 17 days, to interest days. Years. M'ths. Days. . Days. 36x2+3x3+1=82. annex 7=827 days An*. NOTE. The student should commit to memory the multiplication of the number 36 up as far as 9 times 36, and then he can reduce almost in- stantly years, months, and days, to days. SIMPLE INTEREST BY CANCELLATION, RULE. Place the principal, time, and rate per cent, on the right hand side of the line. If the time consists of years and months, reduce them to months, and place 12 (the number of months in a year) on the left hand side of the line. Should the time con- sist of months and days, reduce them to days or deci- mal parts of a month. If reduced to days, place 36 on the left. If to decimals parts of a month, place 12 only as before. Point off two decimal places when the time is in months, and three decimal places when the time u in days. NOTE. If the principal contains cents, point off four decimal places when the time is in months, and five decimal places when the time is in days. H 88 ORTON'S LIGHTNING CALCULATOR. NOTE. We place 36 on the left because there u 360 interest days in a year. (Custom, has made thu lawful.} EXAMPLE 1. What is the interest on $60 for 117 days at 6 per cent? Operation. Here 117X0 must be the $ answer. Both sixes on the right cancels 36 on 117 the left, and we have nothing left $1.170 Ans. to divide by. In this case we point off three decimal places be- cause the time is in days. If the time had been 117 months, we would have pointed off but two deci- mal places. EXAMPLE 2. What is the interest of $96.50 for 90 days at 6 per cent? Operation. 96.50 9650 00 15 15 A ___ 1.44.750 Ans. Now cancel 6 in 36 and the quotient 6 into 90, and we have no divisor left. Hence 15X96.50 must be the answer. NOTE As there are cents in the principal, we point off five decimals ; three for days and two for cents. Pay no attention to the decimal poot Until the clo of the operation. SIMPLE INTEREST BY CANCELLATION. S9 EXAMPLE 3. What is the interest of $480 ftr 361 days at 6 per cent? **0 80 361 361 80 $28.880 Ans. Now cancel 6 in 36 and the quotient 6 into 480, and we have no divisor left. Hence 80X361 must be the answer. EXAMPLE 4.- What is the interest of $720 for 9 months at 7 per cent? W0 60 60 9 9 7 _ 540 7 $37.80 Ans. Now cancel 12 in 720 there is nothing left to divide by. Hence 60X9X7 must be the answer. N. B. When interest is required on any sum for days only, it is a universal custom to consider 30 days a month, and 12 months a year ; and, as the unit of time is a year, the interest of any sum for one day is 3 ^ , what it would be for a year. For 2 days, 3 1 , etc.; hence if we multiply by the days, we must divide by 360, or divide by 36 and save labor. The old form of this method was to place 360, or 12 and 30, on the left of the line, but using 36 is much shorter. 90 ORTONS LIGHTNING CALCULATOR. WHEN THE DAYS ARE NOT DIVISIBLE BY THREE NOTE. When the time consists of months and days, and the days are not divisible by three, re- d\ce the time to days. EXAMPLE 5. What is the interest of $960 foi 11 months and 20 days at 6 per cent? Months. Days. Operation. 11 20=350 days. 000160 350 36 350 160 $56.000 Now cancel 6 in 36 and the quotient 6 into 960, and we have no divisor left. Hence 160X 350 must be the answer. EXAMPLE 6. What is the interest of $173 for 8 months and 16 days at 9 per cent? Months. Days. Operation. 8 16=256 days. 173 173 64 $11.072 Ans. Now cancel 9 in 36 and the quotient 4 into 256, and we have no divisor left. Hence 64X173 must "be the answer. N. B. Let the pujil remember that this is a gen- eral and universal method, equally applicable to any per cent, or any required time, and all ri 'ier rules must be reconcilable to it; and, in ft"* *1I other rules are but modifications of this. SIMPLE INTEREST BY CANCELLATION. 91 EXAMPLE 7. What is Jhe interest on $1080 A)r 7 months and 11 days at 7 per cent? Months. Days. 7 11=221 days. Operation. 0#0 30 221 221 30 7 _, 6630 7 $46,410 Ans. Now cancel 36 in 1080 and we have no divisor left, hence 30X221X7 must be the answer. WITH MORE DIFFICULT TIME AND RATE PER CENT. EXAMPLE 8. What is the interest of $160 for 19 months and 23 days at4J per cent? Months. Days. 19 23=593 days. Operation. 16020 593 593 20 $11.860 Ans. Now cancel 4J in 36 and the quotient 8 into 160 we have no divisor left, hence 1 20x5 93 must be the interest. WHEN THE DAYS ARE DIVISIBLE BY THREE. EULE. Place one-third of the days to the right v/ the months, and place 12 on the left of the line. 92 ORTON'S LIGHTNINQ CALCULATOR. NOTE. Dividing the days by 3 reduces them to tenths of months, as 3 days 1 tenth of a month, 6 days=2 tenths, etc. EXAMPLE 9. What is the interest of $240 for 1 year 11 months and 12 days at 5 per cent? Years. Months. Days. 1 11 12=23.4 months. Operation. &I020 20 234 5 5 100 234 $23.400 Am. Now Cancel 12 in 240 and we have no divisor left, hence 20x5X234 must be the interest. EXAMPLE 10. What is the interest of $500 for 2 years 5 months and 24 days at 6 per cent ? Yeare. Months. Days. 2 5 24=z29.8 months. Operation. 500 149 20$ 149 500 $74.500 Ans. Now cancel 6 in 12 and the quotient 2 into 298 and we have no divisor left, hence 500x1 *9 equals the interest. SIMPLE INTEREST Bl CANCELLATION. 93 EXAMPLE 11. What is the interest of $350 for 3 years 7 months and 6 days at 10 per cent? Years. Months. Days. 3 7 6=43.2 months. Operation. 350 350 It 4#2 36 36 10 12600 10 $126.000 Ans. Now cancel 12 in 432 and we have no divisor left. Hence 350x36X10 equals the interest. EXAMPLE 12. What is the interest of $241 for 13 months and 9 days at 8 per cent? Months. Days. 13 9=13. 3 months. Operation. 241 241 13.3 133 32053 2 3)64106 $21.368f Ans. In this example I canceled 8 and 12 by 4, and then multiplied all on the right of the line and di P4 ORTON'S LIGHTNING CALCULATOR. vided by 3. If I could have divided by 3 before multiplying I would have saved labor, but when the numbers are prime the whole work must be liter* ally done. CLOSING REMARKS. We have now fully ex- plained the canceling system of computing inter- est. Any and every problem can be stated by this method, and the beauty and simplicity of the system ranks it high among the most important abbreviations ever discovered by man. As we have before remarked, at 6, 4, 8, 9, 12, 15, and 4J per cents., every problem in interest can be canceled, besides a great many can be abbreviated at 5, 7, and other per cents.; and after the problem has beec stated and we find that we can not cancel, what have we done? We have simply stated the prob- lem in its simplest and easiest form for working it by any other method. Hence we have a decided advantage of all notes that will cancel, and if we can not cancel we have stated the problem in its correct and proper form for going through tie whole work ; but it is only when the principal^ time, and rate per cent, are all prime, that the WHOLE work must be LITERALLY done. At 6 per sent, we can cancel through, and 6 is the rate most commonly used 6HORT PRACTICAL RULES. 95 SHORV PRACTICAL RULES, DEDUCED FROM THE CANCELING SYSTEM, For calculating interest at 6 per cent., either for months, or months and days. To find the interest for months at 6 per cent. RULE. Multiply the principal by half the num- ber of months, expressed decimally as a per cent.; that is, for 12 months, multiply by .06 ; for 8 months, multiply by .04. NOTE 1. It is obvious that if the rate per cent, were 12, it would be 1 per cent, a month ; if, there- fore, it be 6 per cent., it will be a half per cent, a month ] that is, half the months will be the per cent. NOTE 2. If any other per cent, is wanted you can proceed as above, and then multiply by the given rate per cent, and divide by 6, and the quo- tient is the interest. 1. Whtt is the interest of $363 for 8 inoflths? $368 .04=half the months. NOTE 3. When the months are not even ; that **< will not divide by 2, multiply oiw Ix^f tkc f rv*^ 2 96 ORTON'S LIGHTNING CALCULATOR. by tl.e whole number of months, expressed deci- mally. To find the interest of any sum at 6 per cent, per annum for any number of months and days. RULE. Divide the days by 3 and place the quo- tient to the right of the months; one-half of the num- ber thus formed multiplied by the principal, or one- half of the principal multiplied by this number, will give the interest pointing off three decimal places when the principal is $. 2. What is the interest of $76 for 1 year, 5 months, and 12 days, at 6 per cent? Years. Months. Days. 1 6 12=18.4 months half 9,2. $76 Or, 184 9.2 38=half prin. $6.992 An*. $6.992 Ans. NOTE. Dividing the days by 3 reduces them to the tenth of months. To find the the interest of any sum at 6 per cent, per annum for any number of days. RULE. Divide the principal by 6 and multiply the quotient by the number of days; or divide the days by 6 and multiply the quotient by the principal, pointing off three decimal places when the principal **. NOTE. Always divide 6 into the number that SHORT PRACTICAL RULES. 97 will divide without a remainder; if neither one will divide, multiply the principal and days to- gether and divide the result by 6. 3. What is the interest of $240 for 18 days at 6 per cent ? 18-.-63 240-^-6=40 $240 Or, $40=4 of P rin - 3z=J of the days. 18 $0.720 Ans. $0.720 Ans. 4. What is the interest of $1800 for 72 days at 6 per cent. $1800 Or, $300= of prin. 12=% of the days. 72 $21.600 Ans. $21.600 Ans. Useful Suggestions to the Accountant in Computing Interest at 6 per cent. If the principal is divisible by 6, always reduce the time to days; then multiply the number of days by one-sixth of the principal. EXAMPLE. 5. Find the interest of $240 for 1 year, 5 months, and 17 days, at 6 per cent. 6)240 lyr, 5mos., 17da.=527 days. Multiplied by 40 $ of prin.~40 $21.080 An*, 90 ORTON'S LIGHTNING CALCULATOR. If the days are only divisible by 3, multiply one -third of the principal by one-half of the days. 6. What is the interest of $210 for 80 days at 6 per cent. ? $70=J of the principal. 40=| of the days. 2.800 An*. When the Rate of Interest is 4 per cent RULE. Multiply the principal by one-third the number of months, or by one-ninth the number of days, and the product is the interest. NOTE. This principle is also deduced from the canceling method of computing interest ; the stu- dent can readily see that 4 is J of 12 and ^ of 36, When the Rate of Interest is 9 per cent. RULE. Multiply the principal by three-fourth* (he number of months, or one-fourth the number of days, or vice versa. BANKERS METHOD CF COMPUTING INTEREST. 99 BANKEBS' METHOD OF COMPUTING INTEREST, AT 6 PER CENT. FOR ANY NUMBER OF DAYS. RULE. Draw a perpendicular line, cutting off the two right hand figures of the $, and you have tht interest of the sum for 60 days at 6 per cent. NOTE. The figures on the left of the line are $, and those on the right are decimals of $. EXAMPLE 1. What is the interest of $423 60 days at 6 per cent. ? $423=the principal. $4 | 23 cts.=interest for 60 days. NOTE. When the time is more or less than 60 days, first get the interest for 60 days, and from that to the time required. EXAMPLE 2. What is the interest of $124 for 1 > -days at 6 per cent. ? Days. Day. 15= J of 60 $124 principal. 4)1 | 24 cts.=interest for 60 days. | 31 cts. ^interest for 15 days. 7 I 100 ORTON'S LIGHTNING CALCULATOR. EXAMPLE 3. What is the interest of $123.40 fo* 90 days at 6 per cent.? Days. Days. Days, 90=60+30 $123.40=principal. 2)1 I 2340=interest for 60 days. | 6170=interest for 30 days. Ans. $1 | 851=iinterest for 90 days. EXAMPLE 4. What is the interest of $324 for 75 days at 6 per cent.? Days. Days. Days. $324=rprincipal. 75=60+ 15 4)3 24 cts. interest for 60 days. 81 cts. interest for 15 days. 8. $4 | 05 cts. interest for 75 days. REMARKS. This system of Computing Interest is very easy and simple, especially when the days are aliquot parts of 60, and one simple division will suffice. It is used extensively by a large ma- jority of our most prominent bankers ; and, indeed, is taught by most all Commercial Colleges as the snortest system of computing interest. Method of Calculating at Different Per Cents. This principle is not confined alone to 6 per cent, as many suppose who teach and use i\* It is their custom first to find the interest at 6 per cent., and from that to other per cents. But it is equally ap- plicable for all per cents., from 1 to 15 inclusive. BANKERS 1 METHOD OF COMPUTING INTEREST. 101 The following table shows the different per cents., with the time that a given number of $ will amount to the same number of cents when placed at interest. RULE. Draw a perpendicular line, cutting off the two right hand figures of $, and you have th interest at the following percents. Interest at 4 per cent, for 90 days. Interest at 5 per cent, for 72 days. Interest at 6 per cent, for 60 days. Interest at 7 per cent, for 52 days. Interest at 8 per cent, for 45 days. Interest at 9 per cent, for 40 days. Interest at 10 per cent, for 36 days. Interest at 12 per cent, for 30 days. Interest at 7-30 per cent, for 50 days. Interest at 5-20 per cent, for 70 days. Interest at 10-40 per cent, for 35 days. Interest at 7J per cent, for 48 days. Interest at 4J per cent, for 80 days. NOTE. The figures on the left of the perpen- dicular line are dollars, and on the right decimals of $. If the $ are less than 10 prefix a 0. EXAMPLE 1. What is the interest of $120 for 1 5 days at 4 per cent? Days, Bays, $120 principal. 15=r of 90. 6)1 20 cts.~ int for 90 days. 20 cts. int. for 15 days. 1 D2 ORTON'S LIGHTNING CALCULATOR. EXAMPLE 2. What is the interest of $132 for 13 days at 7 per cent. ? Days. Days. $132=prmcipal. 13=J of 52. 4)1 I 32 cts.=int. for 52 days. I 33 cts.=int. for 13 days. 9 EXAMPLE 3. What is the interest of $520 for 9 days at 8 per cent. ? Days. Days. $520=principal. 9=r of 45. 20 cts.r=int. for 45 days. 04 cts.r=int. for 9 days. EXAMPLE 4. What is the interest of $462 for for 64 days at 7J per cent. ? Days. Days. Days. $462=principal. 64=48+16. 3)4 62 cts. =int. for 48 days. 54 cts. =iut. for 16 days. $6 | 16 cts.=int. for 64 days. REMARK. We have now illustrated several ex- amples by the different per cents. ; and if the stu- dent will study carefully the solution to the above examples, he will in a short time be very rapid in this mode of computing interest. NOTE. The preceding mode of computing in- terest is derived and deduced from the canceling system ; as the ingenious student will readily see. It is a short and easy way of finding interest for days when the days are even or aliquot parts ; but when they are not multiples, and three or four di BANKERS' METHOD OP COMPUTING INTEREST. 103 visions are ncessary, the canceling system is much more simple and easy. We will here illustrate an example to show the difference : Eequired the in- terest of $420 for 49 days at 6 per cent. Bankers' method. Canceling meth. 2)4 2)2 5)1 3) 070 20 cts.i=int. for 60 days. 10 cts.=int. for 30 days. 05 cts. int. for 15 days. 21 cts. int. for 3 days. 7 cts.nrint. for 1 day. $3.430 Ans. 49 70 $3 | 43 cte.r^int. for 49 days. The canceling' method is much more brief; we simply cancel 6 in 36, and the quotient 6 into 420 ; there is no divisor left; hence 70x49 gives the in- terest at once. If the time had been 15 or 20 days, the Bankers' Method would have been equally as short, because 15 and 20 are aliquot parts of 60. The superiority the canceling system has above all others is this : it takes advantage of ths principal as well as the time. For the benefit of the student, and for the con- venience of business men, we will investigate this system to its full extent and explain how to take advantage of the principal when no advantage can be taken of the days. This is one of the most im- portant characteristics of interest, and very often eaves much labor. It should be used when the day* are not even or aliquot parts. 104 ORTON'S LIGHTNING CALCULATOR. The following table shows the different sums of money (at the different per cents.) that bear 1 cent interest a day ; hence the time in days is always the interest in cents ; therefore, to find the interest * on any of the following notes at the per cent, at- tached to it in the table, we have the following rule: RULE. Draw a perpendicular line, cutting off the two right hand figures of the days for cents y and you have the interest for the given time. Interest of 90 at 4 per cent, for 1 day is 1 cent Interest of 72 at 5 per cent, for 1 day is 1 cent. Interest of 60 at 6 per cent, for 1 day is 1 cent. Interest of $52 at 7 per cent, for 1 day is 1 cent. Interest of $45 at 8 per cent, for 1 day is 1 cent. Interest of $40 at 9 per cent, for 1 day is 1 cent. Interest of $36 at 10 per cent, for 1 day is 1 cent Interest of $30 at 12 per cent, for 1 day is 1 cent. Interest of $50 at 7.30 per ct. for 1 day is 1 ct. Interest of $70 at 5.20 per ct. for 1 day is 1 ct. Interest of $35 at 10.40 per ct. for T day is 1 ct. Interest of $48 at 7-J per. cent, for 1 day is 1 cent. Interest of $80 at 4J per cent, for 1 day is 1 cent. Interest of $24 at 15 per ct. for 1 day is 1 cent. NOTE. The 7.30 Government Bonds are calcu- lated on the base of 365 days to the year, and the 5.20's and 10.40's on the base of 364 days to the year BANKERS METHOD OF COMPUTING INTEREST. 105 NOTE. This table should be committed to mem- ory, as it is very useful when the days are not even or aliquot parts. If the days are less than 10 pre- fix a before drawing the line. EXAMPLE 1. Required the interest of $60 for 117 days at 6 per cent. 11 7= the days. Here we cut off the two $1 | 17 cts. Ans. right hand figures for cents. The student should bear in mind that the inter- est on $60 for 117 days is just the same as the interest on $117 for 60 days. By looking at the table we see that the interest for $60 at 6 per cent, is 1 cent a day ; hence the time in days is the answer in cents. If this note was $120, instead of $60, we would first find the interest for $60, and then double it; if it was $180, we would multiply by 3, etc. EXAMPLE 2. Required the interest of $45 for 219 days at 8 per cent. 219=the days. Here we cut off the two $2 | 19 cts. Ans. right hand figures for cents. The student should bear in mind that the inter- est on $45 for 219 days is just the same as the interest on $219 for 45 days. By looking at the table we see that the interest on $45 at 8 per cent, is 1 cent a day ; hence the time in days is the answer in cents. If this Dote 106 ORTON'S LIGHTNING CALCULATOR. was $22.50, instead of $45, we would first get the interest for $45, and then divide by 2 ; if it was $75^ we would add on f ; if $60, add on J, etc. EXAMPLE 3. Required the interest of $48 for 115 days at 9 per cent. 115the days. $48=$40+$8. 5)$1 I 15 cts.=the int. of $40 for 115 days. I 23 cts.=the int. of $8 for 115 days. Ans. $1 | 38 cts.^the int. of $48 for 115 days. Here we first find the interest of $40, because the days is the interest in cents ; then we divide by 6 to find the interest for $8 ; then by adding both we find the interest for $48, as required. EXAMPLE. 4 Required the interest of $260 for 104 days at 7 per cent. Aiis. 104=the days. 04 cts=the int. of $52 for 104 days. 20 cts. Multiply by 5. Here we first find the interest of $52, because the days is the interest in cents ; then we multiply by 5 to get it for $260. We could have worked this note by the Bankers' Method, just as well, by cutting off two figures in the principal, making $2.60 cts. the interest for 52 days, and then multi- ply by 2 to get it for 104 days. GShe student must remember that the interest of $260 for 104 days is just the same as the interest of $104 for 260 days. BANKERS' METHOD OF COMPUTING INTEREST. 107 Problems Solved by Both Methods. We will now solve some examples by both meth- ods, to further illustrate this system, and for the purpose of teaching the pupil how to use his judg- ment. He will then have learned a rule more val- uable than all others. EXAMPLE 5. What is the interest $180 for 75 days at 6 per cent.? Operation by taking advantage of the $. 75=the days. $60X3 $130. 75 cts.=the int. of $60 for 75 days. 3 Multiply by 3. Ans. $2 | 25 cts.=the int. of $180 for 75 days. Operation by the Bankers' Method. $180 the principal. 60da.-f 15da.=75da. 4)$1 80 cts.^the int. for 60 days. 45 cts.~the int. for 15 days. Ans. $2 | 25 cts. the int. for 75 days. By the first method we multiplied by 3, because 3X$60=$180; by the second method we added on J, because 60da.-|- 6 4 da.:=75da. N. B. When advantage can be taken of both time and principal, if the student wishes to prove his work, he can first work it by the Bankers' Method, and then by taking advantage of the prin- cipal, or vice versa. And as the two operations are entirely different, if the same result is obtained by each, he may fairly conclude that the work is correct 108 ORTON'S LIGHTNING CALCULATOR. LIGHTNING METHOD OP COMPUTING INTEREST On all notes that bear $12 per annum, or any all* quot part or multiple of $12. IF a note bears $12 per annum, it will certainly bear $1 per month; hence the time in montha would be the interest in $ ; and the decimal parts of a month would be the interest in decimal parts of a $; therefore when the note bears $12 per annum we have the following rule : RULE. Reduce the years to months, add in the given months, and place one-third of the days to the right of this number, and you have the interest in dimes. EXAMPLE 1. Required the interest of $200 for 3 years, 7 months, and 12 days, at 6 per cent. 200 J of 12 days=4. 6 - Yh. Mo. Da. $12.00z=int. for I yr. 37 12=:43.4mo. Hence 43.4 dimes, or $43.40cts., Ans. We see by inspection that this note bears $12 in**rest a year ; hence the time reduced to months, LIGHTNING METHOD OF COMPUTING INT. 105 with one-third of the days to the right, is the in- terest in dimes. If this note bore $6 a year, in- stead of $12, we would take one-half of the above interest ; if it bore $18, instead of $12, we would add one-half; if it bore $24, instead of $12, we would multiply by 2, etc. EXAMPLE 2. Required the interest of $150 for 2 years, 5 months, and 13 days, at 8 per cent. 150 J of 13 days=4J. 8 Tr. Mo. Da. $12.00=int. for 1 yr. 25 13=r29.4Jmos Hence $29.4J dimes, or $29.43J cts., Ans. We see by inspection that this note bears $12 interest a year ; hence the time reduced to months, with one -third of the days placed to the right, gives the interest at once. EXAMPLE 3. Required the interest of $160 for 11 years, 11 months, and 11 days, at 7J per cent. 160 J of 11 days=3f. 1% $12.00=int. for 1 yr. 11 11 1 i=143.3|mo8. Hence $143.3f dimes, or $143.36f cts., Ans. When the Interest is more or less than $12 a Tear. RULE. First find the interest for the given time on the base of $12 interest a year ; then, if the in- terest on the note is only $6 a year } divide by 2 j if 110 ORTON^S LIGHTNING CALCULATOR $24 a year, multiply 1y 2 ; if $18 a year, ado* on one-half, etc. EXAMPLE 1. What is the interest of $300 for 4 years, 7 months, and 18 days, at 6 per cent. J of 18 days=6. 300 4yr. 7mo. 18da.=55.6mo. 6 $i8.00:=int. for 1 year. 2)55.6, int. at $12 a year. 18=li times $12. 278 $83.4 An*. If tne interest was $12 a year, $55.60 would be the answer; because 55.6 is the time reduced to months ; but it bears $18 a year, or 1 \ times 12 ; hence 1J times 55.6 gives the interest at once. EXAMPLE 2. .Required the interest of $150 for 3 years, 9 months, and 27 days, at 4 per cent. of 27 days=9. 150 Syr. 9mo. 27da.r=45.9mo. 4 2)45.9, int. at $12 a year. $6.00z=int. for 1 year. $22.95 Ans. $6:= times $12. If the interest was $12 a year, $45.90 would be the answer ; because 245.9 is the time reduced to months; out it bears $6 a year, or \ times 12 j hence \ nines 45.9 g'ves the interest at once. MERCHANTS METHOD OF COMPUTING xNT. Ill MERCHANTS' METHOD OF COMPUTING INTEREST. FOR YEARS, MONTHS, AND DAYS. THE computation of simple interest, where the lime consists of years, months, and days, is quite difficult. Taking the aliquot parts for the differ- ent portions of time almost invariably involves the calculator in fractions, and, unless he is well versed in vulgar fract' )ns he will not be able to arrive at the correct result. We have three bases by which sve compute interest at different rates per cent, and by which we are enabled to entirely avoid the use of fractions. These three bases are each obtained different from the other, and consequently we have three rules for computing interest : one at a base of sne per cent., a second at a base of twelve per \ent., and a third at a base of thirty-six per cent RULE for computing interest at 1 per cent. : Take one-third of the number of days and annex o the number of months ; divide the number thu$ formed by 12 ; annex the quotient thus obtained to ihe number of years, and multiply the principal by (his number ; if the principal contains cents, point iff five decimal places ; if not, point off three deci- K 112 ORTONS' LIGHTNING CALCULATOR. mal places; this will give the interest at one per cent. For any other rate per cent., multiply the in- terest at one per cent, by the required rate per cent. Remark. This rule applies to all problems in interest where the days are divisible by 3, and this number, annexed to the number of months, divisi- ble by 12. EXAMPLE. Required the interest on $112, at 1 per cent., for 3 years, 3 months and 18 days. SOLUTION. Take one-third of the number of days, J of 18 :nr6, annex this number to the months given, 36, divide this number by 12, 36-^-12=3, annex this number to the year gives, 33, multiply the princi- pal by 33, $112X33=3.69 6, point off three deci- mal places, and we have the required interest, $3.696. EXAMPLE. Required the interest on $125 12, at 7 per cent., for 2 years, 8 months and 12 days. SOLUTION. Take one-third of the number of days, $ of 12rr:4, annex this number to the number of months we have 84, divide this number by 12, MERCHANTS' METHOD OF COMPUTING INT. 113 34-7-12=7, annex this number to the number of years' we have 27, multiply the principal by this number, and point off five decimal places, and you have the interest at one per cent.; mul- tiply this interest by 7, and you have -- the interest at 7 per cent., the required $23 .64768 rate. EXAMPLE. Kequired the interest on $1,023, at 8 per cent., for 1 year, 9 months and 18 days. SOLUTION. Take one-third the number of days and annex to the number of months, \ of 18=6, we have 96-r-12=8, annex this number to the years $1023 we have 18, multiply the principal by 18 this number, and point of three decimal - places, which gives the interest at 1 per $18 .414 cent.; multiply the interest at one per 8 cent, by 8, and you have the required in -- terest. $147 .312 . This rule will apply to all problems in interest if one-third of the number of the days be taken decimally and annexed to the number of months, and this number, divided by 12, carried out decimally. But this makes the multiplier very large ; hence, to avoid this large number in 114 ORTON'S LIGHTNING CALCULATOR. the multiplier, where the days are divisible by $ y and this number, annexed to the months, is uot divisible by 12, we use the following rule, called our base at 12 per cent. : RULE. Reduce the years to months^ add in the months, take one-third of the number of days and annex to this number, multiply the principal by the number thus formed; if there are cents in the prin- cipal, point off five decimal places ; if there are no cents in the principal, point off three decimal places ; this gives the interest at 12 per cent. For any other rate per cent., take such part of the base before mul- tiplying as the required rate is a part of 12. EXAMPLE. Required the interest on $123, at 12 per cent., for 2 years, 2 months and six days. SOLUTION. Reduce the 2 years to months gives us 24 months, add on the 2 months gives us 26 months, take one-third of the days, J of $123 6=2, annexed to the 26 months gives 262 262, which constitutes the base ; multiply the principal by this base, and you have $32 .226 the interest at 12 per cent. EXAMPLE. Required the interest on $144, at 6 per cent., for 4 years, 5 months and 12 days. MERCHANTS' METHOD OF COMPUTING INT. 115 SOLUTION. Reduce the 4 years to months gives 48 mcntha, add in the 5 months gives 53 months, take one- third of the days and annex to the number of months, J of 12=4. annex to the 53 months, 534 ; this number multiplied into the principal would give the interest at 12 per cent. But we want it at 6 per cent. We will now take such part of either principal or base as 6 is a part of 12 ; 6 is | of 12, therefore we will take J of 14472 one-half of the principal, and mul- 534 tiply it by the base, which will give the interest at 6 per cent. $38.448 EXAMPLE. Required the interest on $347 25, at 8 per cent., for 2 years, 3 months and 9 days. SOLUTION. Reduce the 2 years to months, 24 months, add the 3 months, 27 months, take one-third of the days, J of 9=3, annex to the months, 273, the base; this, multiplied into the principal, would give the interest at 12 per cent. But we want the interest at 8 per cent ; we will take two-thirds of the base before multiply- $347 25 ing: f of 273=182; the principal 182 multiplied by this number gives the interest at 8 per cent. $63.19950 Remark. This base is used where the days are divisible by 3, and the number formed by annex- 8 1 1 ORTON'S LIGHTNING CALCULATOR. ing one- third of the days to the months not divis) ble by 12. We now come to time in which neithei days nor months are divisible. .Where such time as this occurs, we use a base at 36 per cent. RULE. Reduce the time to days, by multiplying the years by 12, adding in the months, if any, and multiplying this number by 30, adding in the days, if any; multiply the principal by this number , pointing off 5 decimal places, where cents are given in the principal, and 3 places where no cents arc given. This will give the interest at 36 per cent. EXAMPLE. Required the interest on $144, at 36 per cent., for 3 years, 2 months and 2 days. SOLUTION. Reduce the time to days gives 1142 $144 days; multiply the principal by this base, 1142 and you have the interest at 36 per : - cent $164.448 EXAMPLE. Required the interest on $144, at 9 per cent., ror 5 years, 7 months and 5 days. SOLUTION. Reduce the time to days gives 2,015 days ; if we multiply the principal by this base, we would get the interest at 36 per cent.; but we want it at 9 per cent. We can take such part of either MERCHANTS' METHOD OF COMPUTING INT. 117 principal or base as 9 is a part of 36 before multi- plying ; 9 is J of 36 ; we will take J- of the prin- cipal, it being divisible by 4 ; J of 144=36, 2915 which, multiplied into the base, will give 35 the interest at 9 per cent., by pointing off 3 decimal places. $72.540 EXAMPLE. Required the interest on $875 15, at 6 per cent., for 5 years, 7 months and 12 days. SOLUTION. Reduce the time to days gives 2022 days ; 6 is J of 36 ; take one sixth of the base, of 2022=337; multiply the prin- $875 15 cipal by this number, point off 5 dec- 337 imal places, and you have the interest at 6 per cent., the required rate. $294.92555 Remark. We have now fully explained our method of computing interest at the three different bases. Any and every problem in interest can be solved by one of these three bases. Some prob- lems can be solved easier by one base than another. Where the days are divisible by 3, and their num- ber, annexed to the months, divisible by 12, it is the shortest and best method to use the base at 1 per cent. By using one or the other of these three bases, the student can avoid the use of vulgar fractions. The student must study these thiee principles carefully, and learn to adopt readily the base best suited to the problem to be solved. 118 ORTON'S LIGHTING CALCULATOR. PARTIAL PAYMENTS ON NOTES, BONDS, AND MORTGAGES. Ta compute interest on notes, bonds, and mort- gages, on which partial payments have been made, two or three rules are given. The following is called the common rule, and applies to cases where the time is short, and payments made within a year of each other. This rule is sanctioned by custom and common law ; it is true to the principles of simple interest, and requires no special enactment. The other rules are rules of law, made to suit such cases as require (either expressed or implied) an- nual interest to be paid, and of course apply to no business transactions closed within a year. RULE. Compute the interest of the principal sum for the whole time to the day of settlement, and find the amount. Compute the interest on the several pay- ments, from the time each was paid to the day of settlement; add the several payments and the inter- est on each together, and call the sum the amount of the payments. Subtract the amount of the payment! from the amount of the principal, will leave the sum due. PARTIAL PAYMENTS. 119 EXAMPLES. 1. A gave his note to B for $10,000 ; at the end of 4 months, A paid $6,000 ; and at the expiration of another 4 months, he paid an additional sum of $3,000 ; how much did he owe B at the close of the year? By the Common Rule. Principal $10,000 Interest for the whole time 600 Amount $10,600 1st payment $6,000 Interest, 8 months 240 2d payment 3,000 Interest, 4 months 60 Amount $9,300 9,300 Due $1300 PROBLEMS IN INTEREST. There are four parts or quantities connected with each operation in interest: these are, the Principal, Hate per cent., Time, Interest or Amount. If any three of them ar3 given the other may be found. Principal, interest, and time given, to find the rate per cent. 1. At what rate per cent, must $500 be Dut 0* interest to gain $120 in 4 years? [20 ORTON'S LIGHTNING CALCULATOR. Operation. By analysis. $500 The interest of .01 $1 for the given time at 1 per cent. 5.00 is 4 cents. $500 4 will be 500 times asmuch 500X-04 2ft. 00)120.00(6 per cent., Ans. =$20.00. Then if 120.00 $20 give 1 per cent., $120 will give \$> =6 per cent. RFLE. Divide the given interest "by the interest of the given sum at 1 per cent, for the given time y and the quotient 'will be the rate per cent, required Principal, interest, and rate per cent, given, to find the time. 2. How long must $500 be on interest at 6 per cent, to gain $120 ? Operation By analysis. $500 We find the in- .06 terest of $1.00 at the given rate for 30.00)120.00(4 years, Ans. 1 year is 6 cents 120.00 $500, will therefore be 500 times as much=500X .06=$30.00. Now, if it take 1 year to gain $30, it will require yy> to gain f 120=4 years, Ans. PARTIAL PAYMENTS. 121 RULE. Divide the given interest by the interest of the principal for 1 year, and the quotient is the time. Given the amount, time, and rate per cent., to find the principal. RULE. Divide the given amount by the amount 0/*$l, at the given rate per cent., for the given time. REMARK. This rule is deduced from the fact that the amount of different principals for the same time and at the same rate per cent., are to each other as those principals. BANK DISCOUNT. Bank Discount is the sum paid to a hank for the payment of a note before it becomes due. The amount named in a note is called the face of the note. The discount is the interest on the face of the note for 3 days more than the time specified, and is paid in advance. These 3 days are called days of grace, as the borrower is not obliged to make payment until their expiration. Hence, to compute bank discount, we have the fol- lowing RULE. Find the interest on the face of the note for 3 days more than the TIME specified ; this will be the discount. From the face of the note deduct the discount, and the remainder will be the PRESENT VALUE of the note. 122 ORTON'S LIGHTNING CALCULATOR. DISCOUNT, OR COUNTING BACK. The object of discount is to show us what al- lowance should be made when any sum of money is paid before it becomes due. The present worth of any sum is the principal that must be put at interest to amount to that sum ^ in the given time. That is, $100 is the present worth of $106 due one year hence; because $100 at 6 per cent, will amount to $106 ; and $6 is the discount. 1. What is the present worth of $12.72 due one year hence ? First method. Second method. $12.72 $ 100 1.06)12.72($12 Ans. 10.6 106)1272.00($12 Am. 106 2.12 2.12 212 212 As $100 will amount to $106 in one year at 6 per cent., it is evident that if {{Jg of any sum be taken, it will be its present worth for one year, and that T $g will be the discount. And as $1 is the present worth of $1.06 due one year hence, it is evident that the present worth of $12.72 must be equal to the number of times $12.72 will contain 81.06. EQUATION OF PAYMENTS. 123 RULE. Divide the given sum by the amount of 81 for the given rate and time, and the quotient will be the present worth. If the present worth be sub- tracted from the given sum, the remainder will be the discount. EQUATION OF PAYMENTS. EQUATION OF PAYMENTS is the process of find- ing the equalized or average time for the payment of several sums due at different times, without loss to either party. To find the average or mean time of payment, when the several sums have the same date. RULE. Multiply each payment by the time that must elapse before it becomes due; then divide the sum of these products by the sum of the payments, and the quotient will be the averaged time required. NOTE. When a payment is to be made down, it has no product, but it must be added with the other payments in finding the average time. EXAMPLE 1. I purchased goods to the amount of $1200; $300 of which I am to pay in 4 months $100 in 5 months, and $500 in 8 months. How long a credit ought I to receive, if I pay tho wholo sum at once? Ans. 6 months. L 124 ORION'S LIGHTNING * CALCULATOR. Mo. Mo. f A credit on $300 for 4 months if >< and links, or acres. EXAMPLES. 1. How many square feet of boards are contain- ed in the floor of a room \vhich is 20 feet square ? 20X20=400 feet, the answer. 2. Suppose a square lot of land measures 36 rods on each side, how many acres does it contain ? 36X36=1296 square rods. And 1296-^-160=8 acres, 16 rods, Ans. As 160 square rods make an acre, therefore we di- \ide 1296 by 160 to reduce rods to acres. N. B. The shortest way to work this example is, to cancel 36X36 with the divisor 160. Arrange the example as below ; (divide both terms by 4X^ :) 36X36 9X9 same as =8.1 acres, or Sac. 16 160 10 144 ORTON'S LIGHTNING CALCULATOR. To measure a parallelogram or long square. RULE. Multiply the length by the breadth, and the product will be the area, or superficial content, in the same name as that in which the dimension teas taken, whether inches,, feet, or rods, etc. EXAMPLES 1. A certain garden, in form of a long square, is 96 feet long, and 54 feet wide; how many square feet of ground are contained in it ? Ans. 96X54=5184 square feet. 2. A lot of land, in form of a long square, is 120 rods in length, and 60 rods wide ; how many acres are in it? 120X60=7200 sq. rods. And 7200^-160=45 acres, Ans. NOTE. The learner must recollect that feet in length, multipled by feet in breadth, produce square feet ; and the same of the other denominations of lineal measure. NOTE. Both the length and breadth, if not in units of the same denomination, must be made so before multiplying. 3. How many acres are in a field of oblong form, whose length is 14,5 chains, and breath 9,75 chains? Ans. 14ac. Orood, 22rods. NOTE. The Gunter's chain is 66 feet, or 4 rods, iOng, and contains 100 links. Therefore if dimen- sions be given in chains and decimals, point off from the product one more decimal place than are MENSURATION OE PRACTICAL GlftMETflY. 145 contained in both factors, and it will be acres and decimals of an acre ; if in chains and links, do the Bame, because links are hundredths of chains, and therefore the same as decimals of them. Or, as 1 chain wide, and 10 chains long, or 10 square chains, or 100000 square links, make an acre, it is the same as if you divide the links in the area by 100000. 4. If a board be 21 feet long and 18 inches broad, how many square feet are contained in it? 18 inches 1,5 foot; and 21x1,5=31,5 ft., Ans. Or, in measuring boards, you may multiply the length in feet by the breadth in inches, and divide the product by 12 ; the quotient will give the an- swer in square feet, etc. 21x18 Thus, in the last example, =31Jsq. ft., as before. 12 5. If a board be 8 inches wide, how much :a length will make a foot square ? BULE. Divide 144 by the width; thus, 8)144 Ans. 18 in. 6. If a piece of land be 5 rods wide, how many rods in length will make an acre ? RULE. Divide 160 by the width, and the quo- tient will be the length required; thus, 5)160 Ans. 32 rods in length. 146 ORTON'S LIGHTNING CALCULATOR. NOTE. When a board, or any other surface, is wider at one end than the other, but yet is of a true taper, you inay take the breadth in the middle.. or add the widths of both ends together, and halve the sum for the mean width ; then multiply the said mean breadth in either case by the length ; the product is the answer or area sought. 7. How many square feet in a board, 10 feet long and 13 inches wide at one end, and 9 inches wide at the other? 13+9 =11 in., mean width. 2 ft. in. 10x11 =9Jft., Ans. 12 8. How many acres are in a lot of land which is 40 rods long, and 30 rods wide at one end, and 20 rods wide at the other ? 30+20 =25 rods, mean width. 2 Then, 25x40 =6^ acres, Ans. 160 9 If a farm lie 250 rods on the road, and at one end be 75 rods wide, and at the other 55 rods wide, bow many acres does it contain ? Ans. 101 acres, 2 roods, 10 rods. N. B. Always arrange your example as above and cancel the factors common to both terms before multiplying >iJtf PS RATION OR PRACTICAL GEOMETRY. 147 3. To measure the surface of a triangle. DEFINITION. A triangle is any three-cornered p:ire which is bounded by three right lines.* RULE- Multiply the base of the given triangle mto half its perpendicular hight, or half the base mto the whole perpendicular , and the product will kc the area. EXAMPLES. 1. Required the area of a triangle whose hase or longest side is 32 inches, and the perpendicular bight 14 inches. 14-j-2=7=half the perpendicular. And 32X7 224sq. in., Ans. 2. There is a triangular or three-cornered lot of land whose base or longest side is 51 J rods; the perpendicular, from the corner opposite to the hase, measures 44 rods ; how many acres does it contain ? 44-j-2=22=half the perpendicular. And 51,5X22 --- =7 acres, 13 rods, Ans. 160 Joists and planks are measured l>y the following : RULE. Find the area of one side of the joist or plank by one of the preceding rules ; then multiply it by the thickness in inches, end the last product will be the superficial content. * A triangle may be either right-angled or oblique. 10 N 148 ORTON'S LIGHTNING CALCULATOR. EXAMPLES. 1. What is the area, or superficial content, ,r board measure, of a joist, 20 feet long, 4 inches wide, and 3 incLes thick? 20X4 X3=20ft., Ans. 12 2. If a plank be 32 feet long, 17 inches wide, and 3 inches thick, what is the board measure of it ? Ans. 136 feet NOTE. There are some numbers, the sum of whose squares makes a perfect square ; such are 3 and 4, the sum of whose squares is 25, the square root of which is 5 ; consequently, when one leg of a right-angled triangle is 3, and the other 4, the hypotenuse must be 5. And if 3, 4, and 5, be multiplied by any other numbers, each by the same, the products will be sides of true right-angled tri- angles. Multiplying them by 2, gives 6, 8, and 10, by 3, gives 9, 12, and 15 ; by 4, gives 12, 16, and 20, etc.; all which are sides of right-angled tri- angles. Hence architects, in setting off the corners of buildings, commonly measure 6 feet on one side, and 8 feet on the other ; then, laying a 10-foot pole across from those two points, it makes the corner a true right-angle. N. B. The solutions of the foregoing problems are all very brief by canceling. MENSURATION OR PRACTICAL GEOMETRY. 149 To find tlie area, of any triangle when the three sides only are given. RULE. From half the sum of the three sides sub- tract each side severally ; multiply these three re- mainders and the said half sum continually together ; then the square root of the last product will be the area of the triangle. EXAMPLE. Suppose I have a triangular fish-pond, whose three sides measure 400, 348, and 312yds j what quantity of ground does it cover ? Ans. 10 acres, 3 roods, 8-f rods. NOTE. If a stick of timber be hewn three square, and be equal from end to end, you find the area of the base, as in the last question, in inches ; multiply that area by the whole length, and divide the product by 144, to obtain the solid content. If a stick of timber be hewn three square, be 12 feet long, and each side of the base 10 inches, the perpendicular of the base being 8f inches, what is *4s solidity? Ans. 3,6+feet. PROBLEM 1. The diameter given, to find the circumference. RULE. As 1 are to 22, so is the given diameter to the circumference; or, more exactly, as 113 are to 355, 50 is the diameter to the circumference, etc 150 ORTON^S LIGHTNING CALCULATOR. EXAMPLES. 1. 'What is the circumference of a wheel, whose diameter is 4 feet ? As 7 : 22 : : 4 : 12,57+ft, the cirram., An*. 2. What is the circumference of a circle, whose diameter is 35 rods ? As 7 : 22 : : 35 : 110 rods, Am. NOTE. To find the diameter when the circum- ference is given, reverse the foregoing rule, and say, as 22 are to 7, so is the given circumference to the required diameter; or, as 355 are to 113, so is the circumference to the diameter. 3. What is the diameter of a circle, whose cir- cumference is 110 rods? A.S 22 : 7 : : 110 : 35 rods, the diam., Am. CA.SE 5. To find how many solid feet a round stick of timber, equally thick from end to end y will contain , when hewn square. RULE. Multiply twice the square of its semi-di- ameter, in inches, by the length in the feet; then divide the product by 144, and the quotient will be the an- swer. N. B. When multiplication and division aro combined, always cancel like factors. When the numbers are properly arranged, a lew clips with the pencil; and, perhaps, a irijting multiplie^on will suffice. MENSURATION OE PRACTICAL GEtfME^HY. 151 EXAMPLES. 1. If the diameter of a round stick of timber be 22 inches, and its length 20 feet, how many solid feet will* it contain when hewn square ? 11XHX2X20 Half dismeter=ll, and -- =33, 6+ft. 144 the solidity when hewn square, the answer. CASE 6. To find how many feet of square edged boards, of a given thickness, can be sawn from a log of a given diameter. RULE. Find the solid content of the log, when made square, by the last case; then say, as the thickness of the board, including the saw calf, is to the solid feet, so is 12 inches to the number of feet of boards. EXAMPLES. 1. How many feet of square edged boards, 1J inch thick, including the saw calf, can be sawn from a log 20 feet long, and 24 inches diameter? 12X12X2X20 -- =40ft. solid content when hewn sq. 144 As 1J : 40 : : 12 : 384 feet, Ans. 2. How many feet of square edged boards, 1J inch thick, including the saw gap, can be sawn from a log 12 feet long, and 18 inches diameter? Ans. 108 feet 152 ORTON-'S LIGHTNING CALCULATOR. NOTE. A short rule for finding the number of feet of one inch boards that a log will make, is to deduct J of its diameter in inches, and J of its length in feet ; then for each inch of diameter that remains, reckon 1 board of the same width as this reduced diameter, and of the same length as this reduced length of the log : thus a log 12 feet long, and 12 inches through, gives 9 boards, 9 feet long, 9 inches wide, or 60f feet a log 16 feet long, and 16 inches through, gives 12 boards, 12 inches wide, 12 feet long, or 144 feet. In measuring timber, however, you may multiply the breadth in inches by the ^epth in inches, and that product by the length in feet; divide this last product by 144 and the quotient will be the solid content in feet, etc. How many solid feet does a piece of square tim- ber, or a block of marble contain, if it be 16 inches broad, 11 inches thick, and 20 feet long? 16X 11 X 20=3520, and 35 2 0--1 44=24,4+ sol. ft. CASE 8. To Jlnd the solidity of a cone or pyramid whether round, square, or triangular. DEFINITION. Solids which decrease gradually from the base till they come to a point, are generally called cones or pyramids, and are of various kinds, according to the figure of their bases ; round, square, oblong, triangular, etc.; the point at the top is called the vertex, and a line drawn frcrn the MENSURATION OR PRACTICAL GEOMETIJJ. 153 vertex, perpendicular to the base, is called the flight of the pyramid. RULE. Find the area of the base, whether round t square, oblong, or triangularly some one oj the fort going rules, as the case may be; then multiply this area by one-third of the hight, and the -product will be the solid content of the pyramid. EXAMPLES. 1. What is the content of a true-tapered round stick of timber, 24 feet perpendicular length, 15 inches diameter at one end, and a point at the other ? 15xl5x,7854x8 ' =9,8175 solid feet, Ans. 144 To find the solid content of a frustum af a cone. What is the solid content of a tapering round stick of timber, whose greatest diameter is 13 inches, the least 6J inches, and whose length is 24 feet, calculating it by both rules ? RULE. 2 Multiply each diameter into itself; mul- tiply one diameter by the other; multiply the sum of these products by the lengths ; annex two ciphers to the product, and divide itr by 382 ; the quotient will be the content , which divide by 144 for feet a* in other cases. ;13X13)+(6,5X6,5)+(13X6,5)X2400 =1858,115+ 382 And 1858,115-j-144=12,903-i-ft, Am. 154 ORTON'S LIGHTNING* CALCULATOR. To find the content of timber in a tree, multiply the square of i of the circumference at the middle of the tree, in inches, by twice the length in feet, and the product divided by 144 will be the content, extremely near the truth. In oak, an allowance of ye or T \j must be made for the bark, if on the tree; in other wood, less trees of irregular growth, must be measured in parts. To find the solid content of a frustum or segment of a globe. DEFINITION. The frustum of a globe is any part cut off by a plane. RULE. To three times the square of the semi-di- ameter of the base, add the square of the hight; multiply this sum by the hight, and the product again by .5236 ; the last product will be the solid content. EXAMPLE. If the hight of a coal-pit, at the chimney, be 9 feet, and the diameter at the bottom be 24 feet, how many cords of wood does it contain, allowing nothing for the chimney? 24-=-2=rl2=:h'fdiain. 12X12X3=432. 9X9=81 And 432+ 81 X9X, 5236 18,886+ cords, Am. 128=rsolid feet in a cord. MENSURATION OR PRACTICAL aEOJgSTRjjT. 155 NOTE. A pile of wood that is 8 feet long, 4 feet high, and 4 feet wide, contains 128 cubic feet, or a cord, and every cord contains 8 cord-feet; and as 8 is y 1 ^ -of 128, every cord-foot contains 16 cubic feet; therefore, dividing the cubic feet in a pile of wood by 16, the quotient is the cord-feet; and if cord-feet be divided by 8, the quotient is cords. NOTE. If we wish to find the circumference of a tree, which will hew any given number of inches square, we divide the given side of the square by .225, and the quotient is the circumference re- quired. What must be the circumference of a tree that will make a beam 10 inches square? NOTE. When wood is " corded" in a pile 4 feet wide, by multiplying its length by its hight, and dividing the product by 4, the quotient is the cord- feet ; and if a load of wood be 8 feet long, and its hight be multiplied by its width, and the product; divided by 2, the quotient is the cord-feet. How many cords of wood in a pile 4 feet wide, 70 feet 6 inches long, and 5 feet 3 inches high? NOTE. Small fractions rejected. To find how large a cube may be cut from any given sphere, or be inscribed in it. RULE. Square the diameter of the sphere, divide that product ly 3, and extract the square root of the quotient for the answer. 116 ORTON'S LIGHTNING CALCULATOR. To find the contents of a round vessel, widei at one end than the other. RULE. Multiply the great diameter by the less, to this product add \ of the square of their differ* ence, then multiply by the hight, and divide as in the last rule. Having the diameter of a circle given, to find the area. RULE. Multiply half the diameter by half the circumference, and the product is the area; or, which is the same thing, multiply the square of the diameter by .7854, and the product is the area. To find the solidity of a sphere or globe. RULE. Multiply the cube of the diameter by .5236. To find the convex surface of a sphere or globe. RULE. Multiply its diameter by its circumfer- ence. To find the solidity of a prism. RULE Multiply the area of the base, or end, by the hight. How many wine gallons will a cubical box con- tain, that is 10 long, 5 feet wide, and 4 feet high? RULE. Take the dimensions in inches; then mul- tiply the length, breadth, and hight together; di- vide the product by 282 for ale gallons, 231 for icine and 2150 for bushels. MENSURATION OR PRACTICAL GEOMETRY. 157 I have a piece of timber, 30 inches in diameter ; how large a square stick can be hewn from it ? RULE. Multiply the diameter by .7071, and the produot is the side of a square inscribed. I have a circular field, 360 rods in circumference ; what must be the side of a square field that shall contain the same quantity? RULE. Multiply the circumference by .282, and the product is the side of an equal square. I have a round field, 50 rods in diameter; what is the side of a square field that shall contain the same area? Ans. 44.31135-|-rods. RULE. Multiply the diameter by .886, and the product is the side of an equal square. There is a certain piece of round timber, 30 inches in diameter ; required the side of an equi- . lateral triangular beam that may be hewn from it. RULE. Multiply the diameter by .866, and the product is the side of an inscribed equilateral tri- angle. To find the area of a globe or sphere. DEFINITION. A sphere or globe is a round solid body, in the middle or center of which is an imag- inary point, from which every part of the surface is equally distant. An apple, or a ball used by children in some of their pastimes > may be called a sphere or globe. 158 ORTON'S LIGHTNING CALCULATOR. RULE. Multiply the circumference by the diam- eter, and the product will be tlie area or surface. To find the force of the wedge. RULE. As half the breadth or thicJtness of the head of the wedge is to one of its slanting sides, so is the power which acts against its head to the force produced at its side. Suppose 100 pounds to be applied to the head of a wedge that is 2 inches broad, and whose slant is 20 inches long, what force would be affected on each side? * To find the solidity of a cone or pyramid. RULE. Multiply the area of the base by one-third of its height, or vice versa. II. TIMBER MEASUBE PROBLEM 1. To find the superficial contents of a board or plank > RULE. Multiply the length by the breadth. NOTE. When the board is broader at one end than the other, add the breadth of the two ends together, and take half the sum for a mean breadth. N. B. When the breadth of the board is in inches, or feet and inches : RULE. Multiply the length of the board, taken in feet, by its breadth taken in inches, and divide this produci by 1 2 ; the quotient is the contents in square TIMBER MEASURE. 159 PROBLEM III. To find tlit solid contents of squared or fow+iided Timber. By the Carpenters 1 Rule. As 12 on D : length on c : Quarter girt on D : solidity on c. RULE I. Multiply the breadth in the middle by the depth in the middle, and that product by the length for the solidity. NOTE. If the tree taper regularly from one end to the other, half the sum of the breadths of the two ends will be the breadth in the middle, and half the sum of the depths of the two ends will bo the depth in the middle. RULE II. Multiply the sum of the breadths of the two ends by the sum of the depths, to which add the product of the breadth and depth of each end ; one-sixth of this sum multiplied by the length, wilt give the correct solidity of any piece of squared tim- ber tapering regularly. PROBLEM IV. Ft? find how much in length will make a solid foot, or any other asssigned quantity, of squared timber, of equal dimensions from end to end. RULE. Divide 1728, the solid inches in a foot +r the solidity to be cut off, by the area of the end tw inches, and the quotient will be the length in inches. 160 ORTON'S LIGHTNING CALCULATOR. 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 tim- ber measure. | | | | 9 | | 11 3 9 inches. I 144 | 36 | 16 | 9 | 5 | 4 | 2 | 2 | 1 | feet. 1 |2|3|4|5|6|7|8|9| side of the square. This table shows, that if the side of the square be 1 inch, the length must be 144 feet ; if 2 inches be the side of the square, the length must be 36 feet, to make a solid foot. PROBLEM V. To find the solidity of round or unsquared timber RULE I. Gird the timber round the middle with a string ; one-fourth part of this girt squared and multiplied by the length will give the solidity. NOTE. If the circumference be taken in inches, and the length in feet, divide the last product by 144. RULE II By the Table. Multiply the area cor- responding to the quarter-girt in inches, by the length of the piece of timber in feet, and the product will be the solidity. NOTE. If the quarter girt exceed the table, take half of it, and four times the content thus formed will be the answer. TIMBER MEASURE. 161 A TABLE FOR MEASURING TIMBER. Quarter Girt. Area. Quarter Girt. Area. Quarter Girt. Area. Inches. 6 6} 6| 6J " Feet. .250 .272 .294 .317 Inches. 12 12J 12} 12f Feet. 1.000 1.042 1.085 1.129 Indies. 18 18} 19 19} Feet. 2.250 2.376 2.506 2.640 7 7J 7J 7J .340 .364 .390 .417 13 13} 13} 13| 1.174 1.219 1.265 1.313 20 20} 21 21} 2.777 2.917 3.062 3.209 i-to r-ikM |r)< OO OO 00 00 .444 .472 .501 .531 14 14t 14} 14f 1.361 1.410 1.460 1.511 22 22} 23 23} 3.362 3.516 3.673 3.835 9 9J 91 9J .562 .594 .626 .659 15 15} 15} 15| 1.562 1.615 1.668 1.722 24 24} 25 25} 4.000 4.168 4'. 340 4.516 10 iot io| 10} .694 .730 .766 .803 16 16} 16} 16| 1.777 1.833 1.890 1.948 26 26} 27 27} 4.694 4.876 5.062 5.252 11 Hi 11} ll! .840 .878 .918 .959 17 m 17} 17| 2.006 2.066 2.126 2.187 28 28} 29 29} 30 5.444 5.640 5.840 6.044 6.250 RULE III By the Carpenters Rule. Measure Ihe circumference of the piece of timber in the middle wnd take a ?? trier of it in inches, call this the girt. 162 ORTON S LIGHTNING CALCULATOR. Then set 12 on i>, to the length in feet on c, and against the girt in inches on D, you will find the content in feet on c. EXAMPLE 1. If a piece of round timber be 18 feet long, and the quarter girt 24 inches, how many feet of timbei are contained therein ? 24 quarter girt. 24 96 48 576 square. 18 4608 576 By the Table. Against 24 stands 4.00 Length, 18 Product, Ans. 72 feet. 72.00 144)10368(72 feet 1008 288 288 By the Carpenters' Rule. 12 on D : 18 on c : 24 on D : 72 on c. in. CARPENTERS' AND JOINERS' WORK. The Carpenters' and Joiners' works, which are measurable, are flooring, partitioning, roofing wainscoting, etc. TIMBER MEASURE. 163 1. 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 posi- tion in which they are laid to form a floor, such as trimming joists, common joists, girders, binding joists, bridging joists and ceiling joists. Girders and joists of floors, designed to bear great weights, should be let into the walls at each end about two-thirds of the wall's thickness. 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. Deductions must be made for stair- cases, chimneys, etc. Example 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 ? By Decimals. 57.25 28.5 By Duodecimals. F. I. 57 : 3 28 : 6 28625 45800 11450 456 114 28 7 : 7 : 6 : : 100)1631.625 feet. Squares 16.31625 11 16:31 : 7 : 6 Ans. 16 squares and 31 feet, 164 ORTON'S LIGHTNING CALCULATOR. iv. OF BRICKLAYERS' WORK. The principal is tiling, slating, walling and chha- aey work. 1. Of Tiling or Slating. Tiling and slating are measured by the square of 100 feet, as flooring, partitioning and roofing were in the Carpenters' work ; so that there is not much difference between the roofing and tiling; yet the tiling will be the most ; for the bricklayers sometimes will require to have double measure for hips and valleys. When gutters are allowed double measure, the way is to measure the length along the ridge-tile, and add it to the content of the roof: this makes an allowance of one foot in breadth, the whole length of the hips or valleys. It is usual also to allow double measure at the eaves, so much as the projector is over the plate, which is commonly about 18 or 20 inches. Sky-lights and chimney shafts are generally de- ducted, if they be large, otherwise not. Example 1. There is a roof covered with tiles, whose depth on both sides (with the usual allow- ance at the eaves) is 37 feet 3 inches, and the length 45 feet; how many squares of tiling are contained therein ? BRICKLAYERS' WORK, 165 BT DUODECIMALS. FEET. INCHES. 37 3 45 185 148 11 3 BY DECIMALS. 37.25 45 18625 14900 16 76.25 16 76 3 2. Of Walling. Bricklayers commonly measure their work by the rod of 16J feet, or 272J square feet. In some places it is a custom to allow 18 feet to the rod ; that is, 324 square feet. Sometimes the work is measured by the rod of 21 feet long and 3 feet high, that is, 63 square feet ; and then no regard is paid to the thickness of the wall in measuring! but the price is regulated according to the thick- ness. When you measure a piece of brick-work, the .first thing is to inquire by which of these ways it must be measured ; then, having multiplied the length and breadth in feet together, divide the pro- duct by the proper divisor, viz.: 272.25, 324 or fi3, according to the measure of the rod, and the quo- tient will be the answer in square rods cf that measure. But, commonly, brick walls that are measured by the rod are to be reduced to a standard thick- 166 ORTON'S LIGHTING CALCULATOR. ness of a brick and a -half, which may be done by the following RULE. Multiply the number of superficial feet that are contained in the wall by the number of half bricks which that wall is in thickness; one- third part of that product will be tJie content in feet. The dimensions of a building are 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 hight, gives the superficies. And to reduce it to the standard thickness, etc., proceed as above. All the vacuities, such as doors, windows, window backs, etc., must be deducted. To measure any arched way, arched window or door, etc., take the hight of the window or dooi from the crown or middle of the arch to the bot- tom 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 hight add twice the former, and multiply the sum by the width of the window, door, etc., and one-third of the pro- duct will be the area, sufficiently near for practice. Example 1. If a wall be 72 feet 6 inches long, and 19 feet 3 inches high, and 5J bricks thick, how many rods of brick work are contained therein, when reduced to the standard ? GLAZIERS WORK. Vii. GLAZIERS' WORK. Glaziers take their dimensions in feet, inches and eights or tenths, or else in feet and hundredth parts of a foot, and estimate their work by the square foot. Windows are sometimes measured by taking the dimensions of one pane, and multiplying its super- ficies by the number of panes. But, more gen- erally, they measure the length and breadth of the window over all the panes and their frames for the length and breadth of the glazing. Circular or oval windows, as fan lights, etc., are measured as if they were square, taking for their dimensions the greatest length and breadth, as a compensation for the waste of glass and labor in cutting it to the necessary forms. Example 1. If a pane of glass be 4 feet 8| inches long, and 1 foot 4J inches broad, how many feet of glass are in that pane ? BY DUODECIMALS. BY DECIMALS. FT. IN. p. 4.729 4 8 9 1.354 1 4 3 18916 4 8 9 23645 1 6 11 14187 1 2 2 3 4729 6 4 10 2 3 6.403066 Ans. 6 feet 4 inches. 168 ORTON S LIGHTNING CALCULATOR. VIII. PLUMBERS WORK. Plumbers' work is generally rated at so much per pound, or by the hundred weight of 112 pounds, and the price is regulated according to the value of lead at the time when the work is per- formed. Sheet lead, used in roofing, guttering, etc., weighs from 6 to 12 pounds per square foot, ac cording to the thickness, and leaden pipe varies in weight per yard, according to the diameter of its bore in inches. The following table shows the weight of a square foot of sheet lead, according to its thickness, reck- oned in parts of an inch, and the common weight of a yard of leaden pipe corresponding to the diameter of its bore in inches: Thickness of Lead. Pounds to a Square foot. Bore of Leaden Pipe. Pounds per yard. ft 5.899 1 10 i 6.554 i 12 i 7.373 ii 16 4 8.427 H 18 i 9.831 if 21 * 11.797 2 24 I MASON S WORK. 169 Example 1. A piece of sheet lead measures 10 feet 9 inches in length, and 6 feet 6 inches ia breadth ; what is its weight at 8^ pounds 10 a square foot? BY DUODECIMALS BY DECIMALS FEET. INCHES. 16 9 6 6 100 6 8 4 6 FEET. 16.75 6.5 8375 10050 108 10 6 108.875 feet. Then 1 foot : 8J pounds : : 108.875 feet I 898.21875 pounds=8 cwt. 2J pounds nearly. ix. MASON'S WORK Masons measure their work sometimes hy the foot solid, sometimes hy the foot superficial, and sometimes by the foot in length. In taking dimensions they girt all their moldings as joiners do. The solids consist of blocks of marble, stone pillars, columns, etc. The superficies are payo- inents, slabs, chimney-pieces, etc. 170 OBTON'S LIGHTNING CALCULATOR. V. PLASTERERS WORK. Plasterers' work is principally of two kinds ; namely, plastering upon laths, called ceiling, and plastering upon walls or partitions made of framed timber, called rendering. In plastering upon walls, no deductions are made except for doors and windows, because cornice, festoons, enriched moldings, etc., are put on after the room is plastered. In plastering timber partitions, in large ware- houses, etc., where several of the braces and larger timbers project from the plastering, a fifth part is commonly deducted. Plastering between their timbers is generally called rendering between quarters. Whitening and coloring are measured in tho Bame manner as plastering ; and in timbered par- titions, one-fourth, or one-fifth of the whole area is commonly added, for the trouble of coloring the sides of the quarters and braces. Plasterers' work is measured by the yard square, consisting of nine square feet. In arches, the girt round them, multiplied by the length, will give the superficies. Example 1. If a ceiling be 59 feet 6 inches iong, and 24 feet 6 inches broad ; how many yards does that ceiling contain ? CISTERNS. 171 PROBLEM L To find the solid content of a Dome, having the highi and the dimensions of its base given. RULE. Multiply the area of the base by the Tiight, and f of the product will be the solidity. Example 1. What is the solidity of a dome, in the form of a hemisphere, the diameter of the cir- cular base being 60 feet ? 60'X -7854=1:2827.44 area of the base. Then f (2827.44X30>=:56548.8 cubic feet, Ans. [PROBLEM II. To find the superficies of a dome, having the highi and dimensions of its base given. RULE. Multiply the area of the base by 2, and ihe product will be the superficial content required ; or, multiply the square of the diameter of the base by 1.5708. FOR AN ELLIPTICAL DOME. Multiply the two diameters of the base together, and that product by 1.5708, the last product will be the area, sufficiently correct for practical purposes. xi. CISTERNS. . Cisterns are large reservoirs constructed to hold water, and to be permanent, should be made either of brick or masonry. 172 ORTON'S LIGHTNING CALCULATOR. It frequently occurs that they are to be so con- structed as to hold given quantities of water, and it then becomes a useful aad practical problem tc calculate their exact dimensions. How do you find the number of hogsheads which a cistern of given dimensions will contain? 1st. Find the solid content of the cistern in cubic inches. 2d. Divide the content so found by 14553, and the quotient will be the number of hogsheads. If the hight of a cistern be given, how do you find the diameter, so that the cistern shall con- tain a given number of hogsheads ? 1st. Reduce the hight of the cistern to inches, arid the content to cubic inches. 2d. Multiply the hight by the decimal .7854. 2. Divide the content by the last result, and extract the square root of the quotient, which wiii be the diameter of the cistern in inches. EXAMPLE. If the diameter of a cistern be given, how do you find the hight, so that the cistern shall contain a given number of hogsheads ? 1st, Reduce the content to cubic inches. 2d. Reduce the diameter to inches, and then mul- tiply its square by the decimal .7854. BINS FOB GRAIN. 17 3d. Divide the content by the last result, and Sne quotient will be the hight in inches. XII. BINS FOR GRAIN. Having any number of bushels, how then will /ou find the corresponding number of cubic feet ? Increase "the number of bushels one -fourth itself, and the result will be the number of cubic feet. How will you find the number of bushels which a bin of a given size will hold ? Find the content of the bin in cubic feet ; then diminish the content by one-fifth, and the resul* will be the content in bushels. How will you find the dimensions of a bin which tshall contain a given number of bushels ? Increase the number of bushels one -fourth itself, and the result will show the number of cubic feet which the bin will contain. Then, when two dimensions of the bin are known, divide fhe last result by their product, and the quotient will be the )ther dimension. A Log Table. Showing the number of feet of /oards any log will make whose diameter is from 16 to 36 inches at the smallest end, and from 10 fco 15 feet in length. 1 74 ORTON'S LIGHTNING CALCULATOR Diametel 10 Feet Diametel ^ Diametel 12 Feet Diami'toi 13 Feet Diameter 14 Feet Diaiiirtei 15 Feet 3 13 p 3 a 3 M 0* p 5T p 5* 3 3 1 3 CT? o n 3 < 3 orq B en 3 a M of 1 er i er Cf tr 15 90 15 99 15 108 15 117 15 126 15 135 16 100 16 110 16 120 16 130 16 140 16 150 17 125 17 137 17 150 17 160 17 175 17 187 18 155 18 170 18 186 18 201 18 216 18 232 19 165 19 176 19 198 19 214 19 230 19 247 20 172 20 189 20 206 20 263 20 246 20 258 21 184 21 202 21 220 21 238 21 256 21 276 22 194 22 212 22 232 22 263 22 294 22 291 23 219 23 240 23 278 23 315 23 332 23 333 24 250 24 276 24 300 24 325 24 350 24 375 25 280 25 308 25 336 25 364 25 392 25 420 26 299 26 323 26 346 26 375 26 404 26 448 27 327 27 3u7 27 392 27 425 27 457 27 490 28 360 28 396 28 432 28 462 28 504 28 540 [ 29 376 29 414 29 451 29 .488 29 526 29 504 30 412 30 452 30 494 30 535 30 576 30 618 31 "428 31 471 31 513 31 558 31 602 31 642 32 451 32 496 32 541 32 587 32 631 32 676 33 490 33 539 33 588 33 637 33 686 33 735 34 532 34 585 |34 638 34 691 34 T44 34 798 35 582 35 640 35 698 35 752 35 805 45 863 36 593 36 657 36 717 36 821 36 836 36 889 TABLE FOR BANKING AND EQUATION. Showing the number of days from any date in one month to the same date in any other month. Example. How many days from the 2d of Feb- luary to the 2d of August? Look for February at the left hand, and August at the top in the angle is 181. In leap year, add one day if February be included. WEIGHTS AND MEASURES. 175 From To 4 ft o> PH SH' c3 & *C ft "5 >-a bb 53 < +3 PH "d O > o K 8 A Jan 365 334 306 275 245 214 184 153 122 92 61 31 31 365 337 306 276 245 215 184 153 123 92 62 59 28 365 334 304 273 243 212 181 151 120 90 90 59 31 365 335 304 274 243 212 182 151 121 120 89 61 30 365 334 304 273 242 212 181 151 151 120 92 61 31 365 335 304 273 243 212 182 181 150 122 91 61 30 365 334 303 273 242 212 212 181 153 122 92 61 31 365 334 304 273 243 243 212 184 153 123 92 62 31 365 335 304 274 273 242 214 183 153 122 92 61 30 365 334 304 304 273 245 214 184 153 123 92 61 31 365 335 331 303 275 244 214 183 153 122 91 61 30 365 Feb March April May June July... Sept Oct Nov Dec TABLE SHOWING DIFFERENCE OF TIME AT 1? O'CLOCK (NOON) New York......... 12.00 N. Buffalo 11.40 A. M. Cincinnati 11.18 Chicago 11.07 St. Louis 10.55 San Francisco... 8.45 New Orleans 10.56 Washington 11.48 Charleston 11.36 Havana 11.25 AT NEW YORK. Boston 12.12 p. M. Quebec 12.12 Portland 12.15 London 4.55 Paris 5.05 Rome 5.45 Constantinople 6.41 Vienna 600 St. Petersburg.. 6.57 Pekin, night... 12.40A.M. TROT WEIGHT. By this weight gold, silver, platina and precious etones, except diamonds, are estimated. *20 Mites 1 Grain. 20 Grains.... 1 Penny wt. Any quantity of gold is supposed to be divided 20 Punnywts 1 Ounce. 12 Ounces 1 Pound. 176 ORTON'S LIGHTNING CALCULATOR. into 24 parts, called carats. If pure, it is said to be 24 carats fine; if there be 22 parts of pure gold and 2 parts of alloy, it is said to be 22 carats fine The standard of American coin is nine-tenths pure gold, and is worth $20.67. What is called the new standard, used for watch cases, etc., is 18 carats fine. The term carat is also applied to a weight of 3J grains troy, used in weighing diamonds ; it is divided into 4 parts, called grains ; 4 grains troy are thus equal to 5 grains diamond weight. APOTHECARIES' WEIGHT USED IN MEDICAL PRESCRIPTIONS. The pound and ounce of this weight are the same as the pound and ounce troy, but differently divided. 90 Grains Troy...l Scruple. 8 Scruples 1 Drachm. 8 Drachms...! Ounce Troy. 12 Ounces....! Pound Troy. Druggists buy their goods by avoirdupois weight. AVOIRDUPOIS WEIGHT. By this weight all goods are sold except those named under troy weight. Grains ................................ 1 Dram. 16' Drama ................................. 1 Ounce. 16 Ounces ................................. 1 Pound. 28 Pounds ................................... 1 Quarter. 4 Quarters or 100 pounds ............ 1 Hundred Weight. 20 Hundredweight ...................... 1 Ton. The grain avoirdupois, though never used, is the same as the grain in troy weight. 7,000 grains make the avoirdupois pound, and 5,760 grains tha WEIGHTS AND MEASURES. 177 troy pound. Therefore, the troy pound is less than the avoirdupois pound in the proportion of 14 to 17, nearly; but the troy ounce is greater than the avoirdupois ounce in the proportion of 79 to 72, nearly. In times past it was the custom to allow 112 pounds for a hundred weight, but usage, as well as the laws of a majority of the States, at the present time call 100 pounds a hundred weight. APOTHECARIES' FLUID MEASURE. 60 Minims 1 Fluid Drachm. 8 Fluid Drachms 1 Ounce (Troy). 16 Ounces (Troy) 1 Pint. 8 Pints 1 Gallon. MEASURE OP CAPACITY FOR ALL LIQUIDS. 6 Ounces Avoirdupois of water make 1 Gill. 4 Gills IPint = 34f Cubic Inches (nearly). 2 Pints 1 Quart = 69J do 4 Quarts 1 Gallon =277J do 31J Gallons 1 Barrel, 42 Gallons 1 Tierce. 63 Gallons, or 2 bbls 1 Hogshead. 2 Hogsheads 1 Pipe or Butt, 2 Pipes 1 Tun. The gallon must contain exactly 10 pounds avoir- dupois, of pure water, at a temperature of 62 , the barometer being at 30 inches. It is the standard unit of measure of capacity for liquids and dry goods of every description, and is J larger than the old wine measure, -g T 2 larger than the old 178 ORTON 3 LIGHTNING CALCULATOE. dry measure, and ^ less than the old ale measure The wine gallon must contain 231 cubic inches. MEASURE OF CAPACITY FOR ALL DRY GOODS. 4 Gills 1 pint = 34f cubic inchs( nearly) 2 Pints 1 quart = 69J cubic inches. 4 Quarts 1 gallon = 27 7 J cubic inches. 2 Gallons 1 peck = 554J cubic inches. 4 Pecks, or 8 gals. 1 bushel =2150 J cubic inches. 8 Bushels 1 quarter = 10 J cubic feet (nearly). When selling the following articles a barrel weighs as here stated : For rice, 600 Ibs.; flour, 196 Ibs.; powder, 25 Ibs.; corn, as bought and sold in Kentucky, Ten- nessee, etc., 5 bushels of shelled corn as bought and sold at New Orleans, a flour-barrel full of ears: potatoes, as sold in New York, a barrel contains 2| bushels; pork, a barrel is 200 Ibs., distinguished in quality by "clear," "mess," "prime;" a barrel of beef is the same weight. The legal bushel of America is the old Win- chester measure of 2,150.42 cubic inches. The imperial bushel of England is 2,218.142 cubic inches, so that 32 English bushels are about equa- to 33 of ours. Although we are all the time talking about the price of grain, etc., by the bushel, we sell by weight, as follows : Wheat, beans, potatoes, and clover-seed, 60 Ibs WEIGHTS AND MEASURES. 179 to the bushel ; corn, rye, flax-seed, and onions, 56 Ibs.; corn on the cob, 70 Ibs.; buckwheat, 52 Ibs.; barley 48 Ibs.; hemp-seed, 44 Ibs.j timothy-seed, 45 Ibs.; castor beans, 46 Ibs.; oats, 35 Ibs.; bran, 20 Ibs.; blue-grass seed, 14 Ibs.; salt the real weight of coarse salt is 85 Ibs.; dried apples, 24 Ibs.; dried peaches, 33 Ibs., according to some rules, but others are 22 Ibs. for a bushel, while tn Indiana, dried apples and peaches are sold by the heaping bushel ; so are potatoes, turnips, onions, apples, etc., and in some sections oats are heaped. A bushel of corn in the ear is three heaped half bushels, or four even full. In Tennessee a hundred ears of corn is some- times counted as a bushel. A hoop 18J inches diameter, 8 inches deep, holds a Winchester bushel. A box, 12 inches eauare, 7 and 7^ deep, will hold half a busl el A heaping bushel is 2,815 cubic inches. CLOTH MEASURE. 2J- Inches 1 nail. 4 Nails 1 quarter of a yard, 4 Quarters 1 yard. FOREIGN CLOTH MEASURE. 2| Quarters 1 Ell Hamburgh. 3 Quarters 1 Ell Flemish. 5 Quarters 1 Ell English 6 Quarters 1 Ell French, 12 180 ORTON'S LIGHTNING CALCULATOR. MEASURE OF LENGTH. 12 Inches .................... 1 foot. 3 Feet ...................... 1 yard. 5J Yards ................... 1 rod, pole, or perch, 40 Poles ..................... 1 furlong. 8 Furlongs, or 1,760 yds, 1 mile. 69- 1 - Miles ) 1 degree of a great cir< '* j -- es '* ele of the earth. By scientific persons and revenue officers, the Inch is divided into tenths, liundredths^ etc. Among mechanics, the inch is divided into eighths. The division of the inch into 12 parts, called lines, is not now in use. A standard English mile, which is the measure that we use, is 5,280 feet in length, 1,760 yards, or 320 rods. A strip, one rod wide and one mile long, is two acres. By this it is easy to calculate the quantity of land taken up by roads, and also how much is wasted by fences. GUNTER'S CHAIN. USED FOR LAND MEASURE 7 /^Inches .............................. 1 Link. 100 Links, or 66 feet, or 4 poles ........ 1 Chain. 10 Chains long by 1 broad, or 10 ) ., A *? - J > 1 Acre. square chains ................... j 80 Chains ................ ................... 1 Mile. WEIGHTS AND MEASURES. 181 SURFACE MEASURE. 144 Sq. inches 1 sq. foot I 40 Sq. perches 1 rood 9 Sq. feet 1 sq. yard | 4 Roods 1 acre 80^ Sq. yards 1 sq. rdorprch j 640 Acres 1 sq. mile Measure 209 feet on each side, and you have a square acre, within an inch. The following gives the comparative size, in square yards, of acres in different countries : English acre, 4,840 square yards ; Scotch, 6,150; Irish, 7,840; Hamburgh, 11,545; Amsterdam, 9,722; Dantzic, 6,650 ; France (hectare), 11,960 ; Prussia (morgen), 3,053. This difference should he borne in mind in read- ing of the products per acre in different countries. Our land measure is that of England. GOVERNMENT LAND MEASURE. A Township 36 sections, each a mile square. A section 640 acres. A quarter section, half a mile square 160 acres. An eighth section, half a mile long, north and south, and a quarter of a mile wide 80 acres. A sixteenth section, a quarter of a mile square 40 acres. 182 ORTON'S LIGHTNING CALCULATOR. The sections are all numbered 1 to 36, com- mencing at the north-east corner, thus : 6 5 4 3 2 NWiX E SWJSE 7 8 9 10 11 12 18 17 16* 15 14 13 19 20 21 22 23 24 30 29 28 27 26 25 31 32 33 34 35 36 The sections are all divided in quarters, which are named by the cardinal points, as in section 1. The quarters are divided in the same way. The description of a forty-acre lot would read: The south half of the west half of the south-west quarter of section 1 in township 24, north of range 7 west, or as the case might be ; and sometimes will fall short, and sometimes overrun the number of acres it is supposed to contain. SQUARE MEASURE FOR CARPENTERS, MASONS, ETC 144 Sq Inches 1 Sq Foot. 9 Sq Ft, or 1,296 Sq In. 1 Sq Yard, 100 Sq Feet 1 Sq of Flooring, Roofing, eta 30J Sq Yards 1 Sq Rod. 86 Sq Yards 1 Rood of Building. School section. WEIGHTS AND MEASURES. GEOGRAPHICAL OR NAUTICAL MEASURE. 6 Feet .................. 1 Fathom. 110 Fathoms or 660 ft. 1 Furlong. 6075| Feet ............... 1 Nautical Mile. 3 Nautical Miles ...... 1 League. -n f The earth's circumference .Degrees .......... -j o^ OKKI -i ( =24,855 \ miles, nearly. The nautical mile is 795| feet longer than the common mile. MEASURE OF SOLIDITY. 1728 Cubic Inches ............. 1 Cubic Foot, 27 Cubic Feet ................. 1 Cubic Yard. 16 Cubic Feet ................. 1 Cord Foot, or a ft of wood. 8 Cord ft or 128 Cubic ft.. 1 Cord. 40 ft of round or 50 ft of hewn timber.. 42 Cubic Feet ............... 1 Ton of Shipping. ft \ n T r... / J ANGULAR MEASURE, OR DIVISIONS OF THE CIRCLE. 60 Seconds 1 Minute. 60 Minutes 1 Degree. 360 Degrees. > Degrees 1 Sign. 90 Degrees 1 Quadrant, 1 Circumference. MEASURE OF TIME. 60 Seconds 1 Minute 60 Minutes 1 Hour. 24 Hours 1 Day. 7 Days 1 Week. 28 Days , 1 Lunar Month., 28 29, 30 or 31 Days 1 Cal. Month. 12 Cal. Months 1 Year. 865 Days 1 Com. Year. SG6 Days 1 Leap Year. S65|- Days 1 Julian Year. 365 D., 5 H., 48 M., 49 s 1 Solar Year. 365 D., 6 H., 9 M., 12 s 1 Siderial Year Q I 184 ORTON'S LIGHTNING CALCULATOR. ROPES AND CABLES. 6 Feet 1 Fathom 120 Feet 1 Cable Length. Miscellaneous Important Facts about Weights and Measures. BOARD MEASURE. Boards are sold by superficial measure, at so much per foot of one inch or less in thickness, adding one-fourth to the price for each quarter* inch thickness over an inch. GRAIN MEASURE IN BULK. Multiply the width and length of the pile to- gether, and that product by the hight, and divide by 2,150, and you have the contents in bushels. If you wish the contents of a pile of ears of corn, or roots, in heaped bushels, ascertain the cubic inches and divide by 2,818. A TON WEIGHT. In this country a ton is 2,000 pounds. In most places a ton of hay, etc., is 2,240 pounds, and in Borne places that foolish fashion still prevails of weighing all bulky articles sold by the tun, by the " long weight," or tare of 12 Ibs. per cwt. A tun of round timber :"s 40 feet; of square timber, 54 cubic feet. WEIQHTS AND MEASURES. 185 A quarter of corn or other grain sold by the bushel Is eight imperial bushels, or quarter of a tun. A ton of liquid measure is 252 gallons. BUTTER Is so\d by avoirdupois weight, which compares with troy weight as 144 to 175 ; the troy pound being that much the lightest. But 175 troy ounces equal 192 of avoirdupois. A firkin of butter is- 56 Ibs.; a tub of butter is 84 Ibs. THE KILOGRAMME OP FRANCE Is 1000 grammes, and equal to 2 Ibs. 2 oz. 4 grs. avoirdupois. A BALE OF COTTON, In Egypt, is 90 Ibs.; in America a commercial bale is 400 Ibs.; though put up to vary from 280 to 720, in different localities. A bale or bag of Sea Island cotton is 300 Ibs. WOOL. In England, wool is sold by the sack or boll, of 22 stones, which, at 14 Ibs. the stone, is 308 Ibs. A pack of wool is 17 stone, 2 Ibs., which is rated as a pack load for a horse. It is 240 Ibs. A tod of wool is 2 stones of 14 Ibs. A wey of wool is 6J tods. Two weys, a sack. A clove of wool is half & stone. 186 ORTON'S LIGHTNING CATCULATOB. THE STONE WEIGHT So often spoken of in English measures, is 14 Ibs., when weighing wool, feathers, hay, etc.; but a stone of beef, fish, butter, cheese, etc., is only 8 pounds. HAY. In England, a truss, when new, is 60 Ibs., or 56 Ibs. of old hay. A truss of straw, 40 Ibs. A load of hay is 36 trusses. In this country, a load is just what it may hap- pen to weigh ; and a tun of hay is either 2,000 Ibs. or 2,240 Ibs., according to the custom of the locality. A bale of hay is generally considered about 300 Ibs., but there is no regularity in the weight. A cube of a solid mow, 10 feet square, will weigh a tun. A LAST Is an English measure of various articles. A last of soap, ashes, herrings, and some snail lar things, is 2 barrels. A last of corn is 10 quarters. A last of gunpowder, 24 barrels. A last of flax or feathers, 1,700 Ibs. A last of wool, 12 sacks. A SCOTCH PINT Contains 105 cubic inches, and is equal to foul English pints. 21 J Scotch pints make a farlot of wheat. Tf EIGHTS AND MEASURES. 187 COAL. A chaldron is 58f cubic feet, or by measure, 36 heaped bushels. A heaped bushel of anthracite coal weighs 80 Ibs., making 2,880 Ibs. to a chaldron. WOOD. A cord of wood is 128 solid feet, in this country and England. In France it is 576-feet. We cord wood 4 feet long, in piles 4 feet by 8. In New Orleans wood is retailed by the pound, and to a limited extent here. It is also sold by the barrel. A load of wood in New York is 42| cubic feet, or one-third of a cord. Wood is sold in England by the stack, skid, quintal, billet, and bundle. A stack is 108 solid feet, and unusually piled 12 feet long, 3 feet high, and 3 feet wide. A quintal of wood is 100 Ibs. A skid is a round bundle of sticks, 4 feet long A one-notch skid girts 16 inches. A two-notch skid, 23 inches. A three-notch skid, 28 inches. A four-notch skid, 33 inches. A five-notch skid, 38 inches. A billet of wood is a bundle of sticks, 3 feet long, and girts 7, 10, or 14 inches, and these bun- dles sell by the score or hundred. A score is 20, and comes from the count by tally, or marks. 188 ORTON'S LIGHTNING CALCULATOR. Faggots of wood are bundles of brush 3 feet long, two round. A load of faggots is 50 bundles, All wood should be sold by the pound. CAPACITY OF CISTERNS OR WELLS. Tabular view of the number of gallons contained in the clear between the brickwork for each ten inches of depth : DT4.ME' 2 f e f 9 9. 9 ?. 71 < PER. et eqi i GALLONS. ial 19 DIAME 8 f e 9 8} ! 9 } 11 12 13 14 15 < 20 < 25 ' TER. et eqi i i i i < GALLONS. ial 313 30 353 44 396 60 461 78 489 97 692 122 705 148 827 176 959 207 1101 240 1958 .. 275 < . .. 3059 TO MEASURE CORN IN THE CRIB. Corn is generally put up in cribs made of rails, but the rule will apply to a crib of any size or kind. Two cubic feet of good, sound, dry corn in th* ear, will make a bushel of shelled corn. To get then, the quantity of shelled corn in a crib of core in the ear, measure the length, breadth, and high* of the crib, inside the rail; multiply the length by WEIGHTS AND MEASURES. 189 the breadth, and the product by the night ; then divide the result by two, and you have the number of bushels of shelled corn in the crib. In measuring the hight, of course, the hight of the corn is intended. And there will be found to be a difference in measuring corn in this mode, between fall and spring, because it shrinks very much in the winter and spring, and settles down. KULES FOR DETERMINING THE WEIGHT OF LIVE CATTLE. Measure in inches the girth round the breast, just behind the shoulder-blade, and the length of the back from the tail to the forepart of the shoul- der-blade. Multiply the girth by the length, and divide by 144. If the girth is less than three feet, multiply the quotient by 11 ; if between three feet and five feet, multiply by 16 ; if between five feet and seven feet, multiply by 23 ; if between seven and nine feet, multiply by 31. If the animal is lean, deduct ^ from the result. Take the girth and length in feet, multiply the square of the girth by the length, and multiply the product by 3.36. The result will be the an- swer in pounds. The live weight, multiplied by .605 gives a near approximation to the net weight 190 ORTON'S LIGHTNING CALCULATOR. ASTRONOMICAL CALCULATIONS. A scientific method of telling immediately what day of the wecJc any date transpired or will transpirej from the commencement of the Christian Era* for the term of three thousand years. MONTHLY TABLE. The ratio to add for each month will be found in the following table: Ratio of June is Ratio of September is 1 Ratio of December is.......l Ratio of April is 2 Ratio of July is 2 Ratio of January is 3 Ratio of October is 3 Ratio of May is 4 Ratio of August is 5 Ratio of March is 6 Ratio of February is 6 Ratio of November is 6 NOTE. On Leap Year the ratio of January is 2, and the ratio of February is 5. The ratio of the other ten months do not change on Leap Years. CENTENNIAL TABLE. The ratio to add for. each century will be found in the following table: Q 200, 900, 1800, 2200, 2600, 3000, ratio is ! 300, 1000, ratio is 6 | 400, 1100, 1900, 2300, 2700, ratio is 5 * 600 1200, 1600, 2000, 2400, 2800, ratio is 4 3 600 1300, ratio is 3 000, 700, 1400, 1700, 2100, 2500, 2900, ratio is 2 100, 800, 1500, ratio is 1 ASTRONOMICAL CALCULATIONS. 191 NOTE. The figure opposite each century is its ratio; thus the ratio for 200, 900, etc., is 0. To find the ratio of any century, first find the century in the above table, then run the eye along the line until you arrive at the end; the small figure at the end is its ratio. METHOD OP OPERATION. RULE.* To the given year add its fourth part } rejecting the fractions ; to this sum add the day of the month; then add the ratio of the month and the ratio of the century. Divide this sum by 7 ; the remainder is the day of the week, counting Sunday as the first, Monday as the second, Tuesday as the third, Wednesday as the fourth, Thursday as the fifth, Friday as the sixth, Saturday as the seventh; the remainder for Saturday will be or zero. EXAMPLE 1. Required the day of the week for the 4th of July, 1810. To the given year, which is 10 Add its fourth part, rejecting fractions 2 Now add the day of the month, which is 4 Now add the ratio of July, which is... 2 Now add the ratio of 1800, which is.. ( Divide the .whole sum by 7. 7 | J.8 4 ~2 We have 4 for a remainder, which signifies the fourth day of the week, or Wednesday. *When dividing the year by 4, always leave off the centuries. Va divide by 4 to find the number of Leap Years. 192 ORTON'S LIGHTNING CALCULATOR. NOTE. In finding the day of the week for the present century, no attention need be paid vo the centennial ratio^ as it is 0. V EXAMPLE 2. Required the day of the week for the 2d of June, 1805. To the given year, which is 5 Add its fourth part, rejecting fractions 1 Now add the day of the month, which is 2 Now add the ratio of June, which is Divide the whole sum by 7. 7 | 81 i "We have 1 for a remainder, which signifies the first day of the week, or Sunday. The Declaration of American Independence was signed July 4, 1776. Required the day of the week. To the given year, which is 76 Add its fourth part, rejecting fractions 19 Now add the day of the month, which is 4 Now add the ratio of July, which is 2 Now add the ratio of 1700, which is 2 Divide the whole sum by 7. 7 | 1035 ~14 TVe have 5 for a remainder, which signifies the fifth day of the week, or Thursday. The Pilgrim Fathers landed on Plymouth Rock Dec. 20, 1620. Required the day of the week. ASTRONOMICAL CALCULATIONS. 193 . To the given year, which is 20 Add its fourth part, rejecting fractions 5 Now add the day of the month, which is 20 Now add the ratio of December, which is 1 Now add the ratio of 1600, which is 4 Divide the whole sum by 7. 7 |_50 1 * 7 We have 1 for a remainder, which signifies tha first day of the week, or Sunday. On what day will happen the 8th of January, 1815? An*. Sunday. On what day will happen the 4th of May, 1810? On what day will happen the 3d of December, 1423? An*. Friday. On what day of the week were you born? The earth revolves round the sun once in 365 days, 5 hours, 48 minutes, 48 seconds; this period is, therefore, a Solar year. In order to keep pace with the solar year, in our reckoning, we make every fourth to contain 366 days, and call it Leap Year. Still greater accuracy requires, however, that the leap day be dispensed with three times in every 400 years. Hence, every year (except the centennial years) that is divisible by 4 is a Leap Year, and every centennial year that is divisible by 400 is also a Leap Year. The next centennial year that will be a Leap Year h 2000 194 ORTON'S LIGHTNING CALCULATOR. For the practical convenience of those who have occasion to refer to mensuration, we have arranged the following useful .able of multiples. It covers the whole ground of practical geometry, and should be studied carefully by those who wish to be skilled in this beautiful branch of mathematics: TABLE OF MULTIPLES.' Diameter of a circle X 3-1416 Circumference. Radius of a circle X 6.283185 Circumference. Square of the radius of a circle X 3.1416 Area. Square of the diameter of a circle X 0.7854 Area. Bquare of the circumference of a circle X 0.07958 = Area. Half the circumference of a circle X D 7 nalf its diameter Area, Circumference of a circle X 0.159155 Radius. Bquare root of the area of a circle X 0.56419 Radius. Circumference of a circle X 0.31831 Diameter. Bquare root of the area of a circle X 1.12838 Diameter. Diameter of a circle X .86 Side of inscribed equilateral triangle. Diameter of a circle X 0.7071 Side of an inscribed square. Circumference of a circle X - 225 Sid of *& inscribed square. Circumference of a circle X 0.282 Side of an equal square. Diameter of a circle X 0.8862 Side of an equal square. Base of a triangle X by % * ne altitude Area. Multiplying both diameters and .7854 together Area of an ellipse. Surface of a sphere X bv % of its diameter Solidity. Circumference of a sphere X D 7 ^ s diameter Surface. Square of the diameter of a sphere X 3.1416 -= Surface. Square of the circumference of a sphere X 0.3183 = Surface. Cube of the diameter of a sphere X 0.5236 Solidity. Cube of the radius of a sphere X 4.1888 Solidity. Cube of the circumference of a sphere X 0.016887 Solidity. Square root of the surface of a sphere X 0.56419 Diameter. Square root of the surface of a sphere X 1.772454 Circumference Cube root of the solidity of a sphere X 1-2407 Diameter. Cube root of the solidity of a sphere X 3.8978 Circumference, Radius of a sphere X 1.1547 -= Side of inscribed cube. Bquare root of (% of the square of) the diameter of a sphere - Side of inscribed cube, irea of its tase X by % of its altitude Solidity of a cone or pyr- amid, whether round, square, or triangular. 4rea of one of its sides X 6 = Surface of a cube. 41Utu.de of Tiapezoid X V* the sum of its parallel aides Area. i THIS BOOK IS DUE ON THE LAST DATE STAMPED BELOW AN INITIAL PINE 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.OO ON THE SEVENTH DAY OVERDUE. YA 02396 M306O26 Q 07 THE UNIVERSITY OF CALIFORNIA LIBRARY CONTENTS. 1. THREE NEW PROCESSES OF ADDITION. 2. Two PROCESSES OF PROVING SUBTRACTION. 3. LIGHTKINC METHOD OF MULTIPLICATION. 4. RAP.D PROCESS OF SQUARING AND CUBING. 5. THREE NEW FORMS o* CONTRACTIONS IN DIVISION. 6. MENTAL OPERATIONS -x FRACTJ. 7. THE CANCELING SYSTEM THOROUGHLY EXPLAINED. 8. LIGHTNING METHOD OF COMPUTING INTEREST. 9. PARTIAL PAYMENTS ON NOTES AND^BONDS. Vwo PROCESSES OF AVERAGING ACCOUNTS. 11. EXTRACTIONS OF THE ROOTS MENTALLY. 12. To MEASURE CORN IN THE CRIB. 13. "^\pir PROCESS OF MEASURING WOOD, BARK, OR COAL. 14. ..-.PII* PROCESS OF MEASURING ALL KINDS OF TIMBER. 15. HOW TO TELL THE No. OP INCH BOARDS IN A LOG. 16. WEIGHTS AND MEASURES. 17. ASTRONOMICAL CALCULATE MULTIPLICATION - : D VALUABLE INFORMATI-X TH BOOH-K peculiarities which tl.s student will here fin'd in , Interest, Mensuration, and in Averagin -.', not all new. Indeed there can be nothini ;-le ; but, as far as the author's knowledge ex i ware that these A! r bet . any Arithmetical work. The >n se uthat the poople could not corn preciate Mathematical beauty otherwise, presents this brie' . with the full assurance that w' pay vine attentior to the subject will l>-. ; abundantly i>