PART XII.
MATHEMATICS.
_Mathematics_ is the science which treats of all kinds of quantity whatever, that can be numbered, or measured.
_Arithmetic_ is that part which treats of numbering.
_Fractions_ treat of broken numbers, or parts of numbers.
_Algebra_ is the art of computing by symbols.
In this science, quantities of all kinds are represented by the letters of the alphabet.
_Geometry_ is the science relating to measurement. By the assistance of geometry, engineers, &c., conduct all their works, take the distances of places, and the measure of inaccessible objects, &c.
_Characters, marks, or signs_,
which are used in arithmetic, and algebra, to denote several of the operations, and propositions:
+ _signifies_ plus, or addition, - minus, or subtraction, × multiplication, ÷ division, : :: : proportion, = equality, √ square root, ∛ cube root, 4^2 _denotes that_ 4 is to be squared. 4^3 _denotes that_ 4 is to be cubed.
ARITHMETIC.
REDUCTION.
_Reduction_ is the method of converting numbers from one name, or denomination to another: or the method of finding the value of a quantity in terms of some other higher, or lower quantity.
_To reduce from a higher to a lower denomination._
_Rule._—Multiply the given number by as many of the lower denomination as make one of the greater;[40] adding to the product as many of the lower denomination as are expressed in the given sum.
_Example._—In £6 15_s._ 5_d._, how many pence?
£. _s._ _d._ 6 15 5 20 ---- 135 12 ---- 1625 _Answer._ ----
_To convert from a lower to a higher denomination._
_Rule._—Divide the given number by as many of the lower denomination as are required to make one of the greater.[41] Should there be any remainder, it will be of the same denomination as the dividend.
_Example._—Convert 1625 pence into pounds, shillings, and pence.
12) 1625 pence ---- 20) 135 5 -------- £6 15_s._ 5 _d._ _Answer._
THE RULE OF THREE, OR SIMPLE PROPORTION.
It is called the _Rule of Three_ because three numbers are given to find a fourth. It is also called _Simple Proportion_, because the 1st term bears the same proportion to the 2nd, as the 3rd does to the 4th. Of the three given numbers, two of them are always of the same kind, or name, and are to be the 1st, and 2nd terms of the question; the 3rd number is always of the same name, or kind as the 4th, or answer sought; and in stating the question it is always to be made the 3rd term. If the answer will be _greater_ than the 3rd term, place the least of the other two given quantities for the 1st term; but if the answer will be _less_ than the 3rd term, put the greater of the two numbers, or quantities, for the 1st term.
_Rule._—State the question according to the above directions, and multiply the 2nd and 3rd terms together, and divide this product by the 1st, for the 4th term, or answer sought.
If the 1st and 2nd terms are not of the same denomination, they must be reduced to it; and if the third term is a compound number, it must be reduced to its lowest denomination before the multiplication, or division of the term takes place.
_Note 1._—The operation may frequently be considerably abridged, by dividing the 1st and 2nd, or the 1st and 3rd terms, by any number which will exactly divide them, afterwards using the quotients, instead of the numbers themselves.*
_Example._—If 2 tons of iron for ordnance cost £40, how many tons may be bought for £360?
As £40 : £360 :: 2 tons : 18 tons. (Thus 360 × 2) ÷ 40 = 18. The answer.
* Or thus, 9 × 2 = 18. The answer.
_Note 2._—_A concise method of ascertaining the annual amount of a daily sum of money._
_Rule._—Bring the daily sum into pence, and then add together as many pounds, half pounds, groats, and pence, as there are pence in the daily sum, for the amount required. For leap year, add the rate for one day.
_Example._—Required the annual amount of 2_s._ 6_d._ per diem.
2_s._ 6_d._ = 30_d._ 30 pounds. 15 = 30 half pounds. 10_s._ = 30 groats. 2_s._ 6_d._ = 30 pence. ----------------------------- Annual amount (365 days) ... £45 12_s._ 6_d._
_Note 3._—To find the amount of any number of days’ pay, the daily rate (under twenty shillings) being given.
The price of any article being given, the value of any number may be ascertained in a similar manner.
_Rule 1._ When the rate (or price) is an even number, multiply the given number by half of the rate, doubling the first figure to the right hand for the shillings, the remainder of the product will be pounds.
_Example._ Required the amount of 243 days’ pay, at 4_s._ per diem.
4/2 = 2 243 2 ------- £48 12_s._ _Ans._
_Rule 2._ When the price is an odd number, find for the greatest number as before, to which add one-twentieth of the given number for the odd shilling.
_Example._ What is the price of 566 pairs of shoes, at 7_s._ per pair.
566 2/0 ) 56/6 3 ------- -------- 28 6 169 16_s._ ------- 28 6 -------- £198 2_s._ _Ans._
FRACTIONS.
_A fraction_ is a quantity which expresses a part, or parts of a unit, or integer. It is denoted by two numbers placed with a line between them.
_A Simple fraction_ consists of two numbers, called the numerator, and denominator; thus,
3 numerator, -- 5 denominator.
_The Denominator_ is placed below the numerator, and expresses the number of equal parts into which the integer is divided.
_The Numerator_ expresses the number of parts of the broken unit, or integer; or shows how many of the parts of the unit are expressed by the fraction.
_A Compound fraction_ is a fraction of a fraction, as ⅔ of ⅘.
_A Mixed number_ consists of a whole number with a fraction annexed to it, as 4⅖.
_An Improper fraction_ has the numerator greater than the denominator, as 6/5.
REDUCTION OF FRACTIONS.
is bringing them from one denomination to another.
_To reduce a fraction to its lowest terms._
_Rule._—Divide the numerator, and the denominator, by any number that exactly divides them, and the quotients by any other number, till they can be no longer divided by any whole number, when the fraction will be in its lowest terms.
_Example._—Reduce 4032/6048 to its lowest terms.
Thus, (4)4032/6048 = (12)/1008/1512 = (6)/84/126 = (7)/14/21 = 2/3. _Answer._
_To reduce an improper fraction to a whole, or mixed number._
_Rule._—Divide the numerator by the denominator, the quotient will be the whole number; and the remainder (if any) the numerator of the fraction, having the divisor for the denominator.
_Example._—Reduce 114/12 to a whole, or mixed number.
12 ) 114 -------- 9-6/12 _Answer._
_To reduce a mixed number to an improper fraction._
_Rule._—Multiply the whole number by the denominator, and add the numerator to the product, under which place the given denominator.
_Example._—Reduce 17⅝ to an improper fraction.
17⅝ 8 --- 141 --- 8 _Answer._
_To reduce a compound fraction to a simple fraction._
_Rule._—Multiply all the numerators together for the numerator, and all the denominators for the denominator.
_Example._—Reduce ⅜ of ⅙ of ½ of 9 to a simple fraction.
Numerators 3 × 1 × 1 × 9 27 9 ----- ----- = -- = -- _Answer._ Denominators 8 × 6 × 2 × 1 96 32
_To reduce fractions of different denominators to equivalent fractions, having a common denominator._
_Rule._—Multiply each numerator by all the denominators except its own for the new numerators, and multiply all the denominators together for a common denominator.[42]
_Example._—Reduce ⅜, ⅔, and ⅘ to fractions having a common denominator.
3 × 3 × 5 = 45 2 × 8 × 4 = 80 4 × 8 × 3 = 96 8 × 3 × 5 = 120 _Answer_, 45/120, 80/120, and 96/120
ADDITION OF FRACTIONS.
_Rule._—Bring compound fractions to simple fractions; reduce all the fractions to a common denominator, then add all the numerators together, and place their sum over the common denominator. When mixed numbers are given, find the sum of the fractions, to which add the whole numbers.
_Example._—Add together ⅚, ¾, and 6½.
5 × 4 × 2 = 40 40/48 + 36/48 + 24/48 + 6 = 8-4/48 3 × 6 × 2 = 36 or, by cancelling, and dividing,[43] 1 × 6 × 4 = 24 10/12 + 9/12 + 6/12 + 6 = 8-1/12. _Answer._ 6 × 4 × 2 = 48
SUBTRACTION OF FRACTIONS.
_Rule._—Prepare the quantities, as in addition of fractions. Place the less quantity under the greater. Then, if possible, subtract the lower numerator from the upper; under the remainder write the common denominator, and, if there be whole numbers, find their difference as in simple subtraction. But if the lower numerator exceed the upper, subtract it from the common denominator, and to the remainder add the upper numerator; write the common denominator under this sum, and carry 1 to the whole number in the lower line.
_Example._— From 54-5/6 or 54-25/30 Take 25-5/15 or 25-10/30 ------- 29-15/30 _Answer._
MULTIPLICATION OF FRACTIONS.
_Rule._—Reduce mixed numbers to equivalent fractions; then multiply all the numerators together for a numerator, and all the denominators together for a denominator, which will give the product required.
_Example._—Multiply ⅚, ⅜, and 2½ together. ⅚ × ⅜ × (2½ or) 5/2 = 75/96 _Answer._
DIVISION OF FRACTIONS.
_Rule._—Prepare the fractions, as for multiplication; then divide the numerator by the numerator, and the denominator by the denominator, if they will exactly divide; but if they will not do so, then invert the terms of the divisor, and multiply the dividend by it, as in multiplication.
_Example._—Divide 9/16 by 4½. 9/16 ÷ (4½ or) 9/2 = ⅛ _Answer._
RULE OF THREE IN FRACTIONS.
_Rule._—State the terms, as directed in “Simple proportion;” reduce them (if necessary) to improper, or simple fractions, and the _two first_ to the same denomination. Then multiply together the second and third terms, and the first with its parts inverted, as in division, for the answer.
_Example._—If 4⅕ cwt. of sugar cost £19⅞, how much may be bought for £59⅝?
As 19⅞ : 59⅝ :: 4⅕ Or, 159/8 : 477/8 :: 21/5 : 12⅗ _Answer._ 8/159 × 477/8 × 21/5 = 80136/6360 = 12⅗ cwt.
DECIMALS.
_A decimal fraction_ is that which has for its denominator an unit (1), with as many ciphers annexed as the numerator has places; and it is usually expressed by setting down the numerator only, with a point before it, on the left hand. Thus, 5/10 is ·5; 25/100 is ·25; 25/1000 is ·025; ciphers being _prefixed_, to make up as many places as are required by the ciphers in the denominator.
_A mixed number_ is made up of a whole number with some decimal fraction, the one being separated from the other by a point, thus 3·25 is the same as 3-25/100 or 325/100.
_Ciphers on the right hand of decimals_ make no alteration in their value; for ·5, ·50, ·500 are decimals having all the same value, each being = 5/10. But when they are placed on the left hand, they decrease the value in a tenfold proportion; thus, ·5 is 5/10; but ·05 is 5/100.
ADDITION OF DECIMALS.
_Rule._—Set the numbers under each other, according to the value of their places, in which state the decimal separating points will all stand exactly under each other. Then beginning at the right hand, add up all the columns of numbers as in integers, and point off as many places for decimals as are in the greatest number of decimal places in any of the lines that are added; or place the point directly below all the other points.
_Example._—Required the sum of 29·0146, 3146·5, 14·16, and 165.
29·0146 3146·5 14·16 165· --------- _Answer_ 3354·6746
SUBTRACTION OF DECIMALS.
_Rule._—Place the numbers under each other according to the value of their places. Then, beginning at the right hand, subtract as in whole numbers, and point off the decimals, as in addition.
_Example._—Subtract 4·90142 from 214·81.
214·81 4·90142 --------- _Answer_ 209·90858
MULTIPLICATION OF DECIMALS.
_Rule._—Place the factors, and multiply them together, the same as if they were whole numbers. Then point off in the product just as many places of decimals as there are decimals in both the factors. But, if there be not so many figures in the product, prefix ciphers to supply the deficiency.[44]
_Example._—Multiply 32·108 by 2·5.
32·108 2·5 ------ 160540 64216 ------- 80·2700 _Answer._
DIVISION OF DECIMALS.
_Rule._—Divide as in whole numbers, and point off in the quotient as many places for decimals as the decimal places in the dividend exceed those in the divisor. When the decimal places of the quotient are not so many as the above rule requires, the deficiency is to be supplied by prefixing ciphers. When there is a remainder after the division, or when the decimal places in the divisor are more than those in the dividend, then ciphers may be annexed to the dividend, and the quotient carried on as far as required.
_Example._—Divide 234·7052 by 64·25.
64·25 ) 234·7052 ( 3·65 _Answer._ 19275 ------ 41955 38550 ------ 34052 32125 ----- 1927 _Remainder._
REDUCTION OF DECIMALS.
_To reduce a vulgar fraction to its equivalent decimal._
_Rule._—Divide the numerator by the denominator, as in Division of Decimals, annexing ciphers to the numerator as far as necessary: and the quotient will be the decimal required.
_Example._—Reduce 7/24 to a decimal.
24 = 4 × 6. Then 4)7· ------- 6)1·75 ------- ·291666, &c.
_To find the value of a decimal, in terms of the inferior denominations._
_Rule._—Multiply the decimal by the number of parts in the next lower denomination, and cut off as many places to the right hand for a remainder, as there are places in the given decimal. Multiply that remainder by the parts in the next lower denomination, again cutting off for another remainder as before. Proceed in the same manner through all the parts of the integer; then the several denominations, separated on the left hand, will make up the answer.
_Example._—What is the value of ·775 pounds sterling.
·775 20 ------ Shillings 15·500 12 ----- Pence 6·000 _Answer_ 15_s._ 6_d._
_To convert integers, or decimals to equivalent decimals of higher denominations._
_Rule._—Divide by the number of parts in the next higher denomination, continuing the operation to as many higher denominations as may be necessary.
_When there are several numbers, all to be converted to the decimal of the highest_—
Set the given numbers directly under each other for dividends, proceeding from the lowest to the highest; opposite to each dividend, on the left hand, place such a number for a divisor as will bring it to the next higher name. Begin at the uppermost, and perform all the divisions, placing the quotient of each division, as decimal parts, on the right hand of the dividend next below it; so shall the last quotient be the decimal required.
_Example._—Convert 15_s._ 9¾_d._ to the decimal of a pound sterling.
4 | 3· 12 | 9·75 20 | 15·8125 £·790625 _Answer._
_Example._—Convert 1 dwt. to the decimal of a pound, Troy weight.
20 ) 1 ------- 12 ) ·05 oz. ----------- ·004166 lb., &c., _Answer._
RULE OF THREE IN DECIMALS.
_Rule._—Prepare the terms, by reducing the fractions to decimals; compound numbers to decimals of the higher denominations, or integers of the lower; also the first, and second terms to the same name. Then multiply, and divide, as in the Rule of Three, in whole numbers.
_Example._—If ⅜ of a yard of cloth cost £⅖, what will 5/16 of a yard cost?
yd. yd. £. _s. d._ ⅜ = ·375 As ·375 : ·3125 :: 4 : ·333 &c. or 6 8 4 ------ ⅖ = ·4 ·375)·12500 (·3333 &c. 1125 20 -------------- 5/16 = ·3215 1250_s._ 6·666 &c. 1125 12 -------------- _Answer_, 6_s._ 8_d._ 125_d._ 7·999 &c. nearly 8_d._
DUODECIMALS.
By Duodecimals, artificers, &c., compute the content of their works.
_Rule._—Set down the two dimensions to be multiplied together one under the other, so that feet may stand under feet, inches under inches, &c.
Multiply each term in the multiplicand, beginning at the lowest, by the feet in the multiplier, and set the result of each straight under its corresponding term, observing to carry 1 for every 12, from the inches to the feet. In like manner multiply all the multiplicand by the inches, and parts of the multiplier, and set the result of each term one place removed to the right hand of those in the multiplicand: omitting, however, what is below parts of inches, only carrying to these the proper number of units from the lowest denominations. Or, instead of multiplying by the inches, take such part of the multiplicand as those are of a foot.
Then add the two lines together for the content required.
_Example._—Multiply 14 feet 9 inches, by 4 feet 6 inches.
ft. in. 14 9 4 6 ------- 59 0 7 4½ ------- 66 4½ _Answer._ -------
TABLES OF WEIGHTS, AND MEASURES.
TROY WEIGHT.
24 grains 1 pennyweight. 480 20 1 ounce. 5760 240 12 1 pound.
AVOIRDUPOIS WEIGHT.
16 drams 1 ounce. 256 16 1 pound. 7168 448 28 1 quarter. 28672 1792 112 4 1 hundred weight. 573440 35840 2240 80 20 1 ton.
_Note._—1 lb. Avoirdupois weight equals 14 oz. 11 dwts. 15½ grs. Troy. 1 oz. ditto 18 dwts. 5½ do. 1 dr. ditto 27·34375 do.
APOTHECARIES’ WEIGHT.
20 grains 1 scruple. 60 3 1 dram. 480 24 8 1 ounce. 5760 288 96 12 1 pound.
WEIGHTS.
_To find the weight, for tonnage._
_Cattle_— Divide the number by 3, for weight in tons.
_Sheep_ Average 60 lb. each. Divide by 33, for weight in tons.
_Pigs_ Average 80 lb. Divide by 15, for tons.
_Beer, or Ale_— Barrel 3¼ cwt. Hogshead 5¼ cwt.
_Oats_ Sack—24 stone. Divide quarters by 5, for tons.
_Rum_— Divide gallons by 224, for tons.
_Wine_ Cask—12 cwt.
_Rule for ascertaining the weight of Hay._
Measure the length and breadth of the stack; then take its height from the ground to the eaves, and add to this last one-third of the height from the eaves to the top: Multiply the length by the breadth, and the product by the height, all expressed in feet; divide the amount by 27, to find the cubic yards, which multiply by the number of stones supposed to be in a cubic yard (viz., in a stack of new hay, six stones; if the stack has stood a considerable time, eight stones; and if old hay, nine stones), and you have the weight in stones. For example, suppose a stack to be 60 feet in length, 30 in breadth, 12 in height from the ground to the eaves, and 9 (the third of which is three) from the eaves to the top; then 60 × 30 × 15 = 27000; 27000 ÷ 27 = 1000; and 1000 × 9 = 9000 stones of old hay.
LONG MEASURE.
12 inches 1 foot. 36 3 1 yard. 198 16½ 5½ 1 pole, perch, or rod. 7920 660 220 40 1 furlong. 63360 5280 1760 320 8 1 mile.
LAND MEASURE (_Length_).
7·92 inches 1 link. 100 links, or 22 yards 1 chain. 80 chains 1 mile. 69·121 miles 1 geographical degree.
LAND MEASURE (_Surface, or Superficial_).
62·7264 square inches 1 square link. 625 square links 1 square pole, or perch. 10000 square links 1 square chain. 2500 square links 1 square rood, or pole. 10 square chains 1 square acre. 100000 square links 1 square acre.
NAUTICAL MEASURE.
1 nautical mile 6082·66 feet. 3 miles 1 league. 20 leagues 1 degree. 360 degrees the earth’s circumference.
SQUARE MEASURE.
144 s. inches 1 s. foot 1296 9 1 s. yard. 39204 272¼ 30¼ 1 s. pole. 1568160 10890 1210 40 1 rood. 6272640 43560 4840 160 4 1 acre.
CUBIC MEASURE (_Measure of solidity_).
1728 cubic inches 1 cubic foot. 27 cubic feet 1 cubic yard.
_Note._—A cubic foot is equal to 2200 cylindrical inches, or 3300 spherical inches, or 6600 conical inches.
_Timber._
40 feet of round, and 50 feet of hewn timber make 1 _Ton_; 16 cubic feet make 1 _Foot_ of wood; 8 feet of wood make 1 _Cord_.
_Water._
Maximum density 42 deg. Fahrenheit.
1 cubic foot of water 6¼ imperial gallons. 1 cylindric foot do. about 5 do. 1 cubic foot weighs 62·5 lb. avoirdupois. 1 cylindric do. do. 49·1 1 lineal do. (1 in. square) do. ·434 12·2 imperial gallons weigh 1 cwt. 224 do. do. 1 ton. 1·8 cubic feet do. 1 cwt. 35·84 do. do. 1 ton.
MEASURES OF CAPACITY.
69⅓ cubic in 2 pints 1 quart. 277¼ 8 4 1 gallon. 554½ 16 8 2 1 peck. 2218⅕ 64 32 8 4 1 bushel. 10¼ cubic ft. 512 256 64 32 8 1 quarter.
FRENCH MEASURES.
English cubic inches.
Millilitre ·06103 Centilitre ·61028 Decilitre 6·10279 Litre, or cubic decimetre 61·02791 Decalitre 610·27900 Hectolitre 6102·79000 Kylolitre 61027·90000 Myrialitre 610279·00000 1 litre is nearly 2⅛ wine pints. 1 kilolitre 1 tun 12¾ wine gallons. 1 stere, or cubic metre 35·3171
English feet.
Metre 3·281 ” French feet, 3·07844 Millimetre. ·03937 Centimetre ·39371 Decimetre 3·93708 Metre 39·37079 Decametre 393·70790 Hectometre 3937·07900 Kilometre 39370·79000 Myriametre 393707·90000 8 kilometres are nearly 5 miles. 1 inch is ·0254 metre. 100 feet are nearly 30·5 metres.
INVOLUTION.
_Involution_ is the raising of powers from any given number, as a root.
A _Power_ is a quantity produced by multiplying any given number, called the _Root_, a certain number of times continually by itself. Thus, 2 × 2 = 4, the 2nd power, or square of 2, expressed thus, 2^2.
_The index, or exponent of a power_ is the number denoting the height, or degree of that power. Thus, 2 is the index of the 2nd power.
Powers that are to be raised, are usually denoted by placing the index above the root, or first power.
Thus 2^2 = 4, the 2nd power of 2.
_Example._—What is the 2nd power of 45?
45 × 45 = 2025 _Answer._
EVOLUTION.
_Evolution_ is the reverse of Involution, being the extracting, or finding the roots of any given powers, or numbers.
_The Root_ of any number, or power, is such a number as being multiplied into itself a certain number of times, will produce that power.
Thus, 2 is the square root, or 2nd root of 4, because, 2^2 = 2 × 2 = 4; and 3 is the cube root, or third root of 27. But there are many numbers of which a proposed root can never be exactly found; by means of decimals, however, the root may be very nearly ascertained.
_Any power of a given number, or root_, may be found exactly by multiplying the number continually into itself.
Those roots which only approximate are called _Surd-roots_; but those which can be found, quite exactly, are called _Rational-roots_. Thus, the square root of 3 is a surd root, but the square root of 4 is a rational root, being equal to 2; also the cube root of 8 is rational, being equal to 2, but the cube root of 9 is surd, or irrational. Roots are sometimes denoted by writing the character √ before the power with the index of the root against it. Thus, the 3rd, or cube root of 20 is expressed by ∛20. When the power is expressed by several numbers with the sign + or - between them, a line is drawn from the top of the sign over all the parts of it; thus the cube (or third) root of 45 - 12 is ∛(45 - 12) or thus ∛(45 - 12).
TO EXTRACT THE SQUARE ROOT.
_Rule._—Divide the given number into periods of two figures each, by setting a point over the place of units, and another over the place of hundreds, and so on over every second figure, both to the left hand in integers, and right hand in decimals. Find the greatest square in the first period on the left hand, and set its root on the right hand of the given number, after the manner of the quotient figure in division. Subtract the square thus found from the said period, and to the remainder annex the two figures of the next following period for a dividend. Double[45] the root above-mentioned for a divisor, and find how often it is contained in the said dividend, exclusive of its right-hand figure; and set that quotient figure both in the quotient, and divisor. Multiply the whole augmented divisor by this last quotient figure, and subtract the product from the said dividend, bringing down to it the next period of the given number, for a new dividend. Repeat the same process over again—viz., find another new divisor, by doubling all the figures now found in the root; from which, and the last dividend find the next figure of the root as before; and so on through all the periods to the last.
_To extract the square root of a fraction, or mixed number._
Reduce the fraction to a decimal, and extract its root.
Mixed numbers may be either reduced to improper fractions, and the root extracted; or the fraction may be reduced to a decimal, then joined to the integer, and the root of the whole extracted.
_Example._—To find the square root of 29506624.
29506624 ( 5432 The Root. 25 -------- 104 | 450 4 | 416 ------------- 1083 | 3466 3 | 3249 ------------- 10862 | 21724 2 | 21724
TO EXTRACT THE CUBE ROOT.
_Rule 1._—By trials, or by the table of roots (_vide page_ 280), take the nearest rational cube to the given number, whether it be greater, or less, and call it the assumed cube.
2.—Then (_by the Rule of Three_),
As the sum of the given number, and double the assumed cube, is to the sum of the assumed cube, and double the given number, so is the root of the assumed cube, to the root required, nearly.
3.—Or as the first sum,
is to the difference of the given, and assumed cube, so is the assumed root, to the difference of the roots, nearly.
4.—Again, by using, in like manner, the cube of the root last found as a new assumed cube, another root will be obtained still nearer. Repeat this operation as often as necessary, using always the cube of the last-found root, for the assumed root.
_Example._—To find the cube root of 21035·8.
By trials it will be found _first_, that the root lies between 20, and 30; and, _secondly_, between 27, and 28. Taking, therefore, 27, its cube is 19683, which will be the assumed cube. Then by No. 2 of the Rule
19683 21035·8 2 2 ----- -------- 39366 42071·6 21035·8 19683· ------ -------- As 60401·8 : 61754·6 :: 27 : 27·6047 the Root, nearly.
Again for a second operation, the cube of this root is 21035·318645155832, and the process by No. 3 of the Rule will be
21035·318645, &c. 2 ------------ 42070·637290 21035·8 21035·8 21035·318645, &c. ------------ ------------ As 63106·43729 : diff. ·481355 :: 27·6047 : : the diff. ·000210560 ---------- consequently the root required is 27·604910560
TABLE OF SQUARES, CUBES, AND ROOTS.
+----+-------+---------+------------+----------+ | No.| Sqr. | Cube. | Sqr. root. |Cube root.| +----+-------+---------+------------+----------+ | 1 | 1 | 1 | 1·0000000 | 1·000000 | | 2 | 4 | 8 | 1·4142136 | 1·259921 | | 3 | 9 | 27 | 1·7320508 | 1·442250 | | 4 | 16 | 64 | 2·0000000 | 1·587401 | | 5 | 25 | 125 | 2·2360680 | 1·709976 | | 6 | 36 | 216 | 2·4494897 | 1·817121 | | 7 | 49 | 343 | 2·6457513 | 1·912933 | | 8 | 64 | 512 | 2·8284271 | 2·000000 | | 9 | 81 | 729 | 3·0000000 | 2·080084 | | 10 | 100 | 1000 | 3·1622777 | 2·154435 | | 11 | 121 | 1331 | 3·3166248 | 2·223980 | | 12 | 144 | 1728 | 3·4641016 | 2·289428 | | 13 | 169 | 2197 | 3·6055513 | 2·351335 | | 14 | 196 | 2744 | 3·7416574 | 2·410142 | | 15 | 225 | 3375 | 3·8729833 | 2·466212 | | 16 | 256 | 4096 | 4·0000000 | 2·519842 | | 17 | 289 | 4913 | 4·1231056 | 2·571282 | | 18 | 324 | 5832 | 4·2426407 | 2·620741 | | 19 | 361 | 6859 | 4·3588989 | 2·668402 | | 20 | 400 | 8000 | 4·4721360 | 2·714418 | | 21 | 441 | 9261 | 4·5825757 | 2·758923 | | 22 | 484 | 10648 | 4·6904158 | 2·802039 | | 23 | 529 | 12167 | 4·7958315 | 2·843867 | | 24 | 576 | 13824 | 4·8989795 | 2·884499 | | 25 | 625 | 15625 | 5·0000000 | 2·924018 | | 26 | 676 | 17576 | 5·0990195 | 2·962496 | | 27 | 729 | 19683 | 5·1961524 | 3·000000 | | 28 | 784 | 21952 | 5·2915026 | 3·036589 | | 29 | 841 | 24389 | 5·3851648 | 3·072317 | | 30 | 900 | 27000 | 5·4772256 | 3·107232 | | 31 | 961 | 29791 | 5·5677644 | 3·141381 | | 32 | 1024 | 32768 | 5·6568542 | 3·174802 | | 33 | 1089 | 35937 | 5·7445626 | 3·207534 | | 34 | 1156 | 39304 | 5·8309519 | 3·239612 | | 35 | 1225 | 42875 | 5·9160798 | 3·271066 | | 36 | 1296 | 46656 | 6·0000000 | 3·301927 | | 37 | 1369 | 50653 | 6·0827625 | 3·332222 | | 38 | 1444 | 54872 | 6·1644140 | 3·361975 | | 39 | 1521 | 59319 | 6·2449980 | 3·391211 | | 40 | 1600 | 64000 | 6·3245553 | 3·419952 | | 41 | 1681 | 68921 | 6·4031242 | 3·448217 | | 42 | 1764 | 74088 | 6·4807407 | 3·476027 | | 43 | 1849 | 79507 | 6·5574385 | 3·503398 | | 44 | 1936 | 85184 | 6·6332496 | 3·530348 | | 45 | 2025 | 91125 | 6·7082039 | 3·556893 | | 46 | 2116 | 97336 | 6·7823300 | 3·583048 | | 47 | 2209 | 103823 | 6·8556546 | 3·608826 | | 48 | 2304 | 110592 | 6·9282032 | 3·634241 | | 49 | 2401 | 117649 | 7·0000000 | 3·659306 | | 50 | 2500 | 125000 | 7·0170678 | 3·684031 | | 51 | 2601 | 132651 | 7·1414284 | 3·708430 | | 52 | 2704 | 140608 | 7·2111026 | 3·732511 | | 53 | 2809 | 148877 | 7·2801099 | 3·756286 | | 54 | 2916 | 157464 | 7·3484692 | 3·779763 | | 55 | 3025 | 166375 | 7·4161985 | 3·802953 | | 56 | 3136 | 175616 | 7·4893148 | 3·825862 | | 57 | 3249 | 185193 | 7·5498344 | 3·848501 | | 58 | 3364 | 195112 | 7·6157731 | 3·870877 | | 59 | 3481 | 205379 | 7·6811457 | 3·892996 | | 60 | 3600 | 216000 | 7·7459667 | 3·914867 | | 61 | 3721 | 226981 | 7·8102497 | 3·936497 | | 62 | 3844 | 238328 | 7·8740079 | 3·957892 | | 63 | 3969 | 250047 | 7·9372539 | 3·979057 | | 64 | 4096 | 262144 | 8·0000000 | 4·000000 | | 65 | 4225 | 274625 | 8·0622577 | 4·020726 | | 66 | 4356 | 287496 | 8·1240384 | 4·041240 | | 67 | 4489 | 300763 | 8·1853528 | 4·061548 | | 68 | 4624 | 314432 | 8·2462113 | 4·081656 | | 69 | 4761 | 328509 | 8·3066239 | 4·101566 | | 70 | 4900 | 343000 | 8·3666003 | 4·121285 | | 71 | 5041 | 357911 | 8·4261498 | 4·140818 | | 72 | 5184 | 373248 | 8·4852814 | 4·160168 | | 73 | 5329 | 389017 | 8·5440037 | 4·179339 | | 74 | 5476 | 405224 | 8·6023253 | 4·198336 | | 75 | 5625 | 421875 | 8·6602540 | 4·217163 | | 76 | 5776 | 438976 | 8·7177979 | 4·235824 | | 77 | 5929 | 456533 | 8·7749644 | 4·254321 | | 78 | 6084 | 474552 | 8·8317609 | 4·272659 | | 79 | 6241 | 493039 | 8·8881944 | 4·290841 | | 80 | 6400 | 512000 | 8·9442719 | 4·308870 | | 81 | 6561 | 531441 | 9·0000000 | 4·326749 | | 82 | 6724 | 551368 | 9·0553851 | 4·344481 | | 83 | 6889 | 571787 | 9·1104336 | 4·362071 | | 84 | 7056 | 592704 | 9·1651514 | 4·379519 | | 85 | 7225 | 614125 | 9·2195445 | 4·396830 | | 86 | 7396 | 636056 | 9·2736185 | 4·414005 | | 87 | 7569 | 658503 | 9·3273791 | 4·431047 | | 88 | 7744 | 681472 | 9·3808315 | 4·447960 | | 89 | 7921 | 704969 | 9·4339811 | 4·464745 | | 90 | 8100 | 729000 | 9·4868330 | 4·481405 | | 91 | 8281 | 753571 | 9·5393920 | 4·497942 | | 92 | 8464 | 778688 | 9·5916630 | 4·514357 | | 93 | 8649 | 804357 | 9·6436508 | 4·530655 | | 94 | 8836 | 830584 | 9·6953597 | 4·546836 | | 95 | 9025 | 857375 | 9·7467943 | 4·562903 | | 96 | 9216 | 884736 | 9·7979590 | 4·578857 | | 97 | 9409 | 912673 | 9·8488578 | 4·594701 | | 98 | 9604 | 941192 | 9·8994949 | 4·610436 | | 99 | 9801 | 970299 | 9·9498744 | 4·626065 | |100 | 10000 | 1000000 | 10·0000000 | 4·641589 | +----+-------+---------+------------+----------+
PILING OF SHOT, AND SHELL.
Shot, and shells, are usually piled in horizontal courses, the base being either an equilateral triangle, a square, or a rectangle. The triangular, and square piles terminate each in a single ball, but the rectangular pile finishes in a row of balls.
_To find the number of balls in a complete pile._
_Rule._—Add the three parallel edges together; then one-third of the product of that sum, and of the number of balls in the triangular face, will be the number sought.
_Note 1._—_The parallel edges_ in a _rectangular pile_ are the two rows in length at the base, and the upper ridge. In the _square pile_ the same, except that the upper row is only a single ball. In the _triangular pile_, one side of the base, the single ball at top, and that at the back, are considered the parallel edges.
_Note 2._—_The number of balls in the triangular face_ is found by multiplying half the number in the breadth at the base, by the number in the breadth at the base _plus_ 1.
_Note 3._—In all piles the breadth of the bottom is equal to the number of courses. In the oblong pile, the top row is one more than the difference between the length, and breadth of the bottom.
_Example._—To find the shot in a triangular pile, the bottom row consisting of 12 shot.
[Illustration: (triangular pyramid)]
Parallel edges. { 12 { 1 12 ÷ 2 = 6 { 1 12 + 1 = 13 ------- --- Triangular face 78 3 ) 14 4⅔ --- ---- 4⅔ 312 --- 52 ---- _Answer_ 364
_Example._—To find the shot in a square pile, the bottom row consisting of 12 shot.
[Illustration: (square pyramid)]
12 12 ÷ 2 = 6 12 12 + 1 = 13 1 ----- ---- 78 3 ) 25 8⅓ ---- ----- 8⅓ 624 ---- 26 ----- _Answer_ 650
_Example._—To find the shot in an oblong pile, whose base consists of 18 shot in length, and 12 in breadth.
[Illustration: (rectangular pyramid)]
18 18 - 12 = 6 18 1 7 -- ---- 7 3 ) 43 ---- 14⅓ 12 ÷ 2 = 6 ---- 12 + 1 = 13 ---- 78 14⅓ ---- 312 78 26 ----- _Answer_ 1118 -----
_Triangular pile._
_Rule._—Multiply the base by the base _plus_ 1, this product by the base _plus_ 2, and divide by 6.
_Square pile._
_Rule._—Multiply the bottom row by the bottom row _plus_ 1, and this product by twice the bottom row _plus_ 1, and divide by 6.
_Rectangular, or oblong pile._
_Rule._—Multiply the breadth of the base by itself _plus_ 1; and this product by three times the length of the base _plus_ 1, _minus_ the breadth of the base, and divide by 6.
_In the following formulæ let the letter_ (L) _denote the number in the bottom row, or the length; and_ (B) _the breadth of the lowest course_.
_Triangular pile_ (L × (L + 1) × (L + 2)) / 6
_Square pile_ (L × (L + 1) × (2L + 1)) / 6
_Oblong pile_ (B × (B + 1) × (3L + 1 - B)) / 6
_The number of shot in any pile,_
(whose base does not exceed 21) may readily be ascertained by referring to _the following Table, page 284_.
_For the square pile._—Look for the number of shot in the base, in the first vertical column on the left hand, and also in the diagonal column; and at their angle of meeting will be found the content required.
Thus 20 base gives 2870.
_For the triangular pile._—Look for the number in the base row in the diagonal column, and opposite to it will be found the content.
Thus 18 base gives 1140.
_For the oblong pile._—Look for the number in the length of the base in the vertical column, and the breadth of the base in the diagonal column, and at their angle of meeting will be found the content required.
Thus 17 length, and 12 breadth, gives 1040.
_To find the number of balls in an incomplete pile._
Compute the number in the pile considered as complete; also the number in the upper pile, or part wanting; and the difference between the two piles thus found will be the number in the frustrum, or incomplete pile.
_Table for computing the Content of any Pile, whose base row does not exceed 21 balls._
+--+--+-------------------------------------------------------------- | 1| 2| 4 | +---+ | 2| 5| 3| 10 | +---+ | 3| 8| 14| 4| 20| | +---+ | 4|11| 20| 30| 5| 35 | +---+ | 5|14| 26| 40| 55| 6| 56 | +---+ | 6|17| 32| 50| 70| 91| 7| 84 | +---+ | 7|20| 38| 60| 85|112|140| 8|120 | +---+ | 8|23| 44| 70|100|133|168|204| 9|165 | +---+ | 9|26| 50| 80|115|154|196|240|285| 10| 220 | +----+ |10|29| 56| 90|130|175|224|276|330|385| 11| 286 | +----+ |11|32| 62|100|145|196|252|312|375|440| 506| 12| 364 | +----+ |12|35| 68|110|160|217|280|348|420|495| 572| 650| 13| 455 | +----+ |13|38| 74|120|175|238|308|384|465|550| 638| 728| 819| 14| 560 | +----+ |14|41| 80|130|190|259|336|420|510|605| 704| 806| 910|1015| 15| 680 | +----+ |15|44| 86|140|205|280|364|456|555|660| 770| 884|1001|1120|1240| 16| | + |16|47| 92|150|220|301|392|492|600|715| 836| 962|1092|1225|1360|1496| | |17|50| 98|160|235|322|420|528|645|770| 902|1040|1183|1330|1480|1632| | |18|53|104|170|250|343|448|564|690|825| 968|1118|1274|1435|1600|1768| | |19|56|110|180|265|364|476|600|735|880|1034|1196|1365|1540|1720|1904| | |20|59|116|190|280|385|504|636|780|935|1100|1274|1456|1645|1840|2040| | |21|62|122|200|295|406|532|672|825|990|1166|1352|1547|1750|1960|2176| ++--+--+---+---+---+---+---+---+---+--+----+----+----+----+----+----+
15... | 816 || +----+ || 16... | 17| 969 || | +----+ || 17... |1785| 18|1140 || | +----+ || 18... |1938|2109| 19|1330 || | +----+ || 19... |2091|2280|2470| 20|1540 || | +----+ || 20... |2244|2451|2660|2870| 21|1771 || | +----+ || 21... |2397|2622|2850|3080|3311| 22|2024|| -------+----+----+----+----+----+----+----++
CORDAGE.
_Ropes_, _cables_, and all other descriptions of cordage are distinguished by their circumference, thus a two-inch rope means a rope two inches in circumference.
_To find the weight of a rope._
_First method._—Multiply the length in fathoms by the square of the circumference, and divide the product by 480 for the weight in cwts.
_Example._—Required the weight of 110 fathoms of 3-inch rope.
3 × 3 × 110 = 990, which divided by 480, gives 2 cwt. 7 lb. Weight required.
_Second method._—Divide the square of the circumference by 4, the quotient will give the weight, in pounds, per fathom.
_Example._—What is the weight of a 3-inch rope per fathom?
3^2 ÷ 4 = 2¼ lb. Weight required.
_To find the strength of a rope, or the weight it will support._
_First method._—Square the circumference, and divide by 5, for the number of tons which it will bear suspended from it.[46]
_Example._—What weight will 3-inch rope of the best description support?
(3 × 3) / 5 = 9/5 = 1⅘ ton, or 4030 lb. Weight required.
_Second method._—Multiply the square of the circumference by 2, the product will give the _practical weight in cwts_. that may be lifted by it, or about half the breaking weight.
_Example._—What number of cwts. may be lifted by a 3-inch rope?
3^2 × 2 = 18 cwts. Weight required.
_The strain_, in pounds, _a rope will bear safely_ = girt^2 × 200} ” ” _a cable_ ” ” = girt^2 × 120} nearly.
CHAINS.
_To find the weight of chains._
The square of the diameter of the link, measured in eighths of inches, will give the weight of the chain, per fathom, in pounds.
_Example._—What is the weight per fathom of a ¾-inch chain?
¾-inch = 6/8; 6^2 = 36 lb. Weight per fathom.
_Or_, the weight per foot of the chain, multiplied by 24, will give the weight per fathom of the chain, _nearly_. A chain cable with a stay across the links will weigh about one-twelfth more than the foregoing examples.
_To find the weight that may be safely lifted by a chain._
Divide the square of the diameter of the links, taken in eighths of an inch by 8, and the quotient will give the number of tons that may be lifted by the chain.
_Example._—What number of tons will a chain made of ¾-inch iron carry with safety?
¾-inch = 6/8 6^2 = 36 36 / 8 = 4½ tons. Weight required.
_The safe strain_ is equal to about 8 tons, per square inch, of the iron of which the chain is made.
_The stay across the link of a chain_ increases its strength about one-sixth.
_When the chain is of great length_, a deduction, from the above rules, must be allowed for the weight of it.
IRON RODS.
_To find the weight of round iron rods._
Divide the square of the diameter, in quarter inches, by 2, and the quotient will give the weight in pounds, per yard.
_Example._—What is the weight of a yard of 1-inch round iron.
1 inch = 4 quarters 4^2 = 16 16 / 2 = 8 lb. Weight required.
_To find the weight of square rods._
The weight of round rods, of similar diameter, divided by ·7854 will give the weight of the square rods.
_To find the weight that may be sustained, or lifted by round iron rods._
Find the weight in pounds, per yard; two-thirds of which will give the safe load, in tons.
A round iron rod of average quality of iron, one inch in diameter, will be torn asunder by 16 tons; it will be perceptibly damaged by half this strain, or 8 tons; its safe load will be one-third, or 5·33 tons.
TIMBER.
_To find the area, or superficial content of a plank._
Multiply the length by the mean breadth.[47]
_Example._—Required the content of a board whose length is 11 feet 2 inches, and breadth 1 foot 10 inches
ft. in. ft. in. ft. in. 11 2 × 1 10 = 20 5. Content required.
_To find the solid content of squared, or four-sided timber._
Multiply the mean breadth by the mean thickness, and the product by the length, for the content, nearly.
_Note 1._—If the tree taper regularly from the one end to the other, either take the mean breadth, and thickness in the middle, or take the dimensions at the two ends, and half their sum will be the mean dimensions; which, multiplied as by the above rule, will give the content, nearly.
_Note 2._—If the piece do not taper regularly, take several different dimensions, add them all together, and divide their sum by the number of them, for the mean dimensions.
_Example._—Required the content of a piece of timber 16 feet long, and side of square 14 inches.
ft. in. ft. in. ft. ft. in. 1 2 × 1 2 × 16 = 21 9. Content required.
_To find the solidity of round, or unsquared timber._
1. Multiply the square of the quarter girt (or the square of ¼ of the mean circumference), by the length, for the content.
_Note._—When the tree is tapering, take the mean dimensions, either by girting it in the middle for the mean girt, or at the two ends, taking half the sum of the two; or by girting it in several places, then adding all the girts together, and dividing the sum by the number of them for the mean girt. But when the tree is very irregular, divide it into several lengths, and find the content of each part separately.
_Example._—Required the content of a tree, whose mean girt is 3·15 feet, and length 14½ feet.
3·15 / 4 = ·7875 ·7875 × ·7875 = ·62015625. ·62015 × 14·5 = 8·9922 feet of solid timber. The content required.
2. Find the mean area of a round tree, and multiply it by the length for the content.
_To find the weight of a tree._
Find its content in feet, and multiply that by the specific gravity of the wood.
(_Vide_ GRAVITY, _and Table of Specific Gravities_. _Page_ 318.)
_Example._—Required the weight of an elm-tree; whose mean girt is 5 feet, and length 60 feet.
5/4 = 1·25 1·25 × 1·25 = 1·5625. 1·5625 × 60 = 93·75. Content in feet.
TONNAGE.
Table of Tonnage, and Weight of _one_ of the following Carriages, Carts, Waggons, Gyns, &c., used in land service.
+--------------------------------------+-------+---------+-------------+ | | Tonn- | Weight. | | | | age. | | | +--------------------------------------+-------+---------+-------------+ | |Tons. |Cwt. lb.| | | | ft.| qrs. | | | | | |} With | |Carri- {24 pounder | 6 0|34 0 22|} bullock | |ages. { 8 inch | 6 0|34 2 12|} pole and | | T{ {For Iron {18 pounder | 4 39|27 2 9|} chain, | | r{ {Ordnance.{12 ” 21 cwt. | 4 7|18 3 24|} weighing 2 | | a{ | | |} qrs. 19 lb.| | v{ {Howitzer {10 inch| 6 17|39 0 9| | | ” { e{ { 8 ” | 5 37|33 2 0| Do. do. do.| | ” { l{ {12 pounder | 5 33|22 0 10| | | ” { l{ { 9 ” | 5 1|20 2 14| | | ” { i{For Brass { 6 ” | 4 21|17 3 5| | | ” { n{Ordnance. { {32 pr.| 5 29|23 3 13| | | ” { g{ {Howitzer {24 ” | 5 6|21 0 17| | | ” { { { {12 ” | 4 21|18 3 14| | | ” { c{Ammunition waggon | 5 36|20 0 3| {For all | | ” { o{ | | { natures. | | ” { m{Forge | 5 38|19 1 0| | | ” { p{Store-waggon (without spare | 5 11|18 1 10| | | ” { l{ wheel) | | | | | ” { e{Small arm ammunition waggon | 4 36|14 2 16| | | ” { t{Rocket {12 pounder | 7 33|20 2 8| | | ” { e{ { 6 ” | 5 17|20 1 20| | | ” { .{Pontoon {Large | 3 30|42 2 13| | | ” { { {Small | |22 2 5| | | ” {Garrison, wood, common standing }| | | | | ” { for 32 pounder of 25 cwt. }| 1 8| 8 0 7| | |Capstan, crab | 0 31| 3 3 26| | | {Forge, cavalry | 4 32|11 2 3| | | {Hand | 1 10| 4 3 4| | | {Hospital, conveyance | 3 16|10 2 20| | |Carts {Sling | 3 38|16 1 17| | | {Store | 3 16| 9 1 0| | | {French | 1 32| 5 2 13| | |Drugs {Large | 2 7|17 1 24| | | {Small | 0 29| 5 0 4| | |Gyns, {Large | 1 23| 9 2 22| | |Triangle {Small | 1 2| 7 3 3| | | {For 32 pounder | | | | | { garrison carriage}| 0 26 | 6 0 12| Fir. | | {Madras {For traversing }| | | | |Platform. { { carriage with }| | | | | { { tail-piece }| 1 23 |14 2 0| Teak. | | {Mortar, Alderson’s pattern | 0 30 | 8 1 4| | |Portable forge, and pack saddle, | | | | in wooden case | 0 17 | 2 1 3| | | {Flanders | 5 0 |16 1 25| | |Waggons {Platform | 3 16 |21 3 18| | | {Sling | 8 11 |31 3 26| | |Waggons, hospital, Mr. Holmes’ {Large | 9 10 |21 0 0| | | pattern {Small.| 6 30 |18 0 4| | +--------------------------------------+-------+---------+-------------+
The calculation of tonnage for baggage, stores, &c., is by measurement: _a Ton_, consisting of 40 cubic feet; but metals, and very heavy articles are estimated by actual weight, without reference to bulk.
_To ascertain the tonnage of sailing vessels, the hold being clear._
_Rule._—Divide the length of the upper deck between the afterpart of the stem, and the forepart of the stern-post, into six equal parts.
_Depths._—At the foremost, the middle, and the aftermost of those points of division, measure in feet, and decimal parts of a foot, the depth from the under side of the upper deck to the ceiling at the limber strake. In the case of a break in the upper deck the depths are to be measured from a line stretched in a continuation of the deck.
_Breadths._—Divide each of those three depths into five equal parts, and measure the inside breadths at the following points—viz., at one-fifth, and at four-fifths from the upper deck of the foremost, and aftermost depths, and at two-fifths, and four-fifths from the upper deck of the midship depth.
_Length._—At half the midship depth measure the length of the vessel from the afterpart of the stem to the forepart of the stern-post; then to twice the midship depth add the foremost, and the aftermost depths for the sum of the depths; add together the upper, and lower breadths at the foremost division, three times the upper breadth, and the lower breadth at the midship division, and the upper, and twice the lower breadth at the after division, for the sum of the breadths: then multiply the sum of the depths by the sum of the breadths, and this product by the length, and divide the final product by 3500, which will give the number of tons for register.
If the vessel have a poop, or half deck, or a break in the upper deck, measure the inside mean length, breadth, and height of such part thereof as may be included within the bulkhead; multiply these three measurements together, and dividing the product by 92·4, the quotient will be the number of tons to be added to the result as above found.
In order to ascertain the tonnage of open vessels, the depths are to be measured from the upper edge of the upper strake.
_To ascertain the tonnage of steam vessels._
_Rule._—In addition to the foregoing rules, when applied for the purpose of ascertaining the tonnage of any ship or vessel propelled by steam, the tonnage due to the cubical content of the engine-room must be deducted from the total tonnage of the vessel, as determined by either of the rules aforesaid, and the remainder will be the true register tonnage of the said ship or vessel.
_To determine the tonnage due to the cubical content of the engine-room._
_Rule._—Measure the inside length of the engine-room in feet and decimal parts of a foot, from the foremost to the aftermost bulkhead, then multiply the said length by the depth of the ship or vessel at the midship division as aforesaid, and the product by the inside breadth of the same division at two-fifths of the depth from the deck, taken aforesaid, and divide the last product by 92·4, and the quotient will be the tonnage due to the cubical content of the engine-room.
_To ascertain the tonnage of vessels when laden._
_Rule._—Measure, _first_, the length on the upper deck between the afterpart of the stem, and the forepart of the stern-post; _secondly_, the inside breadth on the under side of the upper deck, at the middle point of the length; and, _thirdly_, the depth from the under side of the upper deck down the pump-well to the sink; multiply these three dimensions together, and divide the product by 130, and the quotient will be the amount of the register tonnage of such ships.
MECHANICS.
_Mechanics_ is the science of forces, and the effects they produce when applied to machines in the motion of bodies.
_Machine, or engine_, is any mechanical instrument contrived to move bodies.
_Equilibrium_ is an equality of action, or force, between two or more powers, or weights, acting against each other, by which they destroy each other’s effects, and remain at rest.
_The centre of motion_ is the fixed point about which a body moves.
_The axis of motion_ is the fixed line about which it moves.
_The centre of gravity_ is a certain point on which a body (being freely suspended) will rest, in any position.
_The whole momentum_ or quantity of force of a moving body, is the result of the quantity of matter multiplied by the velocity with which it is moved.
THE MECHANICAL POWERS.
_Power is compounded of the weight, or expansive force of a moving body multiplied into its velocity._
The power of a body, which weighs 40 lb., and moves with the velocity of 50 feet in a second, is the same as that of another body which weighs 80 lb., and moves with the velocity of 25 feet in a second: for the products of the respective weights, and velocities are the same.
40 × 50 = 2000; and 80 × 25 = 2000.
_Power cannot be increased by mechanical means._
Power is applied to mechanical purposes—
1. By the lever; 4. By the inclined plane; 2. By the wheel, and axle; 5. By the wedge; 3. By the pulley; 6. By the screw;
which are the simple elements of all machines.
The whole theory of these elements consists simply in causing the weight, which is to be raised, to pass through a greater or a less space than the power which raises it; for, as power is compounded of the weight, or mass of a moving body, multiplied into its velocity, a weight passing through a certain space may be made to raise, through a less space, a weight heavier than itself.
THE LEVER.
_The lever_ is the most simple of all machines, being only a straight bar of iron, wood, &c., supported on, and moveable round a prop, called the _fulcrum_.
_Case 1._—When the fulcrum of the lever is between the power, and the weight.
_Rule._—Divide the weight to be raised by the power to be applied; the quotient will give the difference of leverage necessary to support the weight in equilibrio. Hence, a small addition either of leverage, or weight, will cause the power to preponderate.
_Example 1._—A ball weighing 3 tons is to be raised by 4 men, who can exert a force of 12 cwt.; required the proportionate length of lever?
3 tons = 60 cwt.; and 60/12 = 5
In this example, the proportionate lengths of the lever to maintain the weight in equilibrio, are as 5 to 1. If, therefore, an additional pound be added to the power, the power side of the lever will preponderate, and the weight will be raised. But, although the ball is raised by a force of only one-fifth of its weight, no power is gained, for the weight passes through only one-fifth of the space. The products, therefore, arising from the multiplication of the respective weights, and velocities are the same.
_Example 2._—A weight of 1 ton is to be raised with a lever 8 feet in length, by a man who can exert, for a short time, a force of rather more than 4 cwt.; required at what part of the lever the fulcrum must be placed?
(20 cwt.) / (4 cwt.) = 5; that is, the weight is to the power as 5 to 1, therefore, 8 / (5 + 1) = 1 foot and a third, from the weight.
_Example 3._—A weight of 40 lb. is placed 1 foot from the fulcrum of a lever; required the power to raise the same, when the length of the lever on the other side of the fulcrum is 5 feet?
(40 × 1) / 5 = 8 lb. _Ans._
_Case 2._—_When the fulcrum is at one extremity of the lever, and the power at the other._
_Rule._—As the distance between the power, and the fulcrum is to the distance between the weight, and the fulcrum, so is the effect to the power.
_Example 1._—Required the power necessary to raise 120 lb., when the weight is placed 6 feet from the power, and 2 feet from the fulcrum?
As 8 : 2 :: 120 : 30 lb. _Ans._
_Example 2._—A beam, 20 feet in length, and supported at both ends, bears a weight of 2 tons at the distance of eight feet from one end; required the weight on each support?
(40 cwt. × 8 feet) / (20 feet) = 16 cwt. on the support that is furthest from the weight; and (40 × 12) / (20 feet) = 24 cwt. on the support nearest to the weight.
_Case 3._—_When the weight to be raised is at one end of the lever, the fulcrum at the other, and the power is applied between them._
_Rule._—As the distance between the power, and the fulcrum, is to the length of the lever, so is the weight, to the power.
_Example._—The length of the lever being 8 feet, and the weight at its extremity 60 lb., required the power to be applied 6 feet from the fulcrum to raise it?
As 6 : 8 :: 60 : 80 lb. _Ans._
_Velocity is gained at the expense of power by the lever, and wheel, and axle._
Note 1.—When two men are carrying a load on a pole between them, the strongest man should have the weight placed nearer to him than the other man.
Note 2.—_To carry guns, &c._—If the burden can be carried by four men; after having made it fast to the middle of a large lever, fix the extremities of this lever on two shorter levers, and place a man at each of the points, C, D, E, F. _Vide plate, Mechanics, Fig. 1._ In _Fig. 2_, the weight is equally divided between eight men, and in _Fig. 3_, between sixteen men.
THE WHEEL, AND AXLE.
The advantage gained is in proportion as the circumference of the wheel exceeds that of the axle; therefore, the larger the wheel, and the smaller the axle, the stronger is the power of this machine, but then the weight will rise proportionally slower. A winch may be used instead of a wheel, for in turning the winch the hand will describe a circle, and there is no difference in the result, whether an entire wheel be turned, or a single spoke which the winch as a lever represents.
_Rule._—As the radius of the wheel is to the radius of the axle, so is the effect, to the power.
_Example._—A weight of 50 lb. is exerted on the periphery of a wheel, whose radius is 10 feet; required the weight raised at the extremity of a cord wound round the axle, the radius being 20 inches.
(50 lb. × 10 feet × 12 inches) / (20 inches) = 300 lb. _Ans._
THE PULLEY.
_The pulley_ consists of a grooved wheel, called _a sheave_, moveable on an axis, or gudgeon, and enclosed in a frame, or case, called a _block_. By passing a cord over the pulley, a man will be enabled _to draw up_ a weight equal to that which his own body supplies in pulling downwards.
_By combining a number of pulleys_, as many assistants are obtained as there are wheels: thus, two pulleys will have double the power of one, because half the weight is sustained by the frame to which one end of the cord is attached; but then it requires _double the time_ to do the work. As the _friction of the pulley_ is very great, particular attention must be paid that all the turns or kinks of a rope be taken out, before it is made use of, and it should enter easily into the grooves of the sheaves.
_Rule._—Divide the weight to be raised by twice the number of pulleys in the lower block; the quotient will give the power necessary to raise the weight.
_Example._—What power is required to raise 600 lb., when the lower block contains six pulleys?
600 / (6 × 2) = 50 lb. _Ans._
TACKLES.
_Tackles_ are indispensable in the service of the artillery.
_The fall_ is the rope of which the tackle is composed; that end of it which is fixed to the block is called _the standing part_, or _end_; the other, which is pulled, or hauled on by the men, is called _the running part_, or _end_; and the parts which pass from one block to the other are called the _returns of the fall_.
_In all operations with tackles, the following directions should be attended to_:—
1st. Make fastenings stronger than appears _actually_ necessary.
2nd. Examine the straps, and hooks of the blocks carefully.
3rd. Consider whether the cordage is new, half-worn, or almost worn out.
4th. Attend to the seizings of the clinches, the sheet bends, the _proper_ stoppering of the fall, the belaying the fall with two half hitches.
5th. Be very distrustful of selvages applied on smooth worn rope.
6th. Do not allow ropes to be struck, or trampled on, when the weight is suspended.
7th. The men should _stand as safe_ as the proper performance of the various duties will permit.
8th. In pulling a rope, the men ought to place themselves in a right line, and haul together. The most advantageous position for pulling is down a slope, or in a descending position.
COMBINATION OF PULLEYS.
_A leading block_ is a fixed pulley, which alters the direction of the power, but does not increase it: Power = Weight. On account of friction the power must exceed the weight a little, in order to raise it.
_Vide plate, Mechanics, Fig. 1._
_A whip_ is one moveable pulley, which increases the power without altering the direction.
Power = ½ weight (or 2 to 1).—_Vide Fig. 2._
_A whip upon whip_ will afford the same purchase as a tackle having a single and double block, and with much less friction.
_A gun tackle_ consists of two single blocks with fall fixed to the one, then rove through the other, and then through the first. Power = ½ weight (or 2 to 1): or Power = ⅓ weight (or 3 to 1). _Vide Fig. 3, and 4._
Two double blocks are generally used for very heavy guns.
_A luff tackle, or half watch tackle_, consists of one double and one single block: the fall is fixed to the single, then rove through first sheave of the double, then through sheave of single, and lastly through second sheave of double block. Power = ⅓ weight (3 to 1): or Power = ¼ weight (4 to 1). _Vide Fig. 5, and 6._
_A runner tackle_ is the same as a luff tackle, applied to the end of a large rope, called a runner, which is rove through a single block attached to a fixed point, or to a body that is to be moved, or raised; the standing end of the runner being secured to another point.
Power is either 6 to 1, or 7 to 1, or 8 to 1.
_A gyn tackle_ consists of one triple and one double block: the fall is fixed to the double, then rove through first sheave of triple, then through first sheave of double, then through second sheave of triple, then through second sheave of double, and lastly through third sheave of triple block.
Power = ⅕ weight (5 to 1): or Power = ⅙ weight (6 to 1). _Vide Fig. 7._
If the moveable block of a tackle be strapped with a tail, it is called a _tail_, or _jigger block_: and the tackle a _tail_, or _jigger tackle_: a block with a hook strapped to it, and attached to a selvage, answers the same purpose.
_Two double blocks_, with fall fixed to one of them, and then rove through the sheaves of both blocks, will either give Power = ¼ weight (4 to 1): or Power = ⅕ weight (5 to 1). _Fig. 8._
_Two triple blocks_, with fall fixed to one of them, then rove through sheaves of both blocks, will either give power = ⅙ weight (6 to 1): or Power = ⅐ weight (7 to 1). _Fig. 9._
[Illustration: _System of Pullies._
Fig. 1 2 3 4 5 6 7 8 9]
[Illustration: _To carry Guns &c._
Fig. 1 2 3]
[Illustration: HEIGHTS AND DISTANCES.]
_In the system of pulleys_ (_vide plate, Mechanics_) the Power is shown at the hooks of the moveable blocks, which are to be applied to the bodies, or weights, requiring to be moved or raised. The strain is also shown at the fixed blocks.
_In Fig. 3_, there are _three parts of the rope engaged_ in supporting the weight—_viz._, _the parts marked 1, 1, 1_. Each of them, hence, sustains _one-third_ of it, and the fall of the rope to which the power is to be attached requires the Power = 1, if weight = 3. The same principle of calculation is applicable to all systems of pulleys having one fixed block, any number of moveable wheels, and a single rope over all the wheels. Hence, in such a system of pulleys, gravity being applied, there will be an equilibrium, when the weight is as many times the power as there are portions of the rope employed in sustaining the weight. _For example_, in a system consisting of six moveable sheaves, the same rope going over them all, there will be 12 portions of the rope engaged; and to produce an equilibrium the power must be equivalent to 1/12 the weight, no allowance being made for friction.
From the foregoing observations, and by referring to the plate, it will be seen that _each tackle has two applications_, differing in power one from the other; _for example_, if the double block of a luff tackle is fixed to a weight to be moved, and the single block to a picket, or other fastening, _Fig. 6_, then, if _one man_ haul on the fall, the power of _four men_ will be applied to the weight (4 to 1), and the power of _three men_ to the picket; but if the double block be fixed to the picket, _Fig. 5_, and the single, block to the weight, then the force of only _three men_ will be applied to the weight (3 to 1), and a power of _four men_ to the picket, or fastening.
When the moveable block of one tackle is fixed to the fall of another tackle, their respective powers are to be multiplied into each other for the power of the combination: thus, if one luff tackle is fixed to the fall of another luff tackle (the double blocks of both tackles being moveable), the power will be 4 × 4 = 16 (16 to 1): in this, the men haul through 16 feet to move the weight one foot; therefore if the combination be increased until the men haul through 100 feet to move the weight one foot, then the power would be 100 to 1.
The foregoing powers are, however, only true in theory, and are, therefore, called _theoretical powers_: for owing to the great friction of the pulleys, the stiffness of the ropes, &c., the actual _practical powers_ are far less; so much so, that with a combination giving a power of 48 to 1, a 24-pr. (2½ tons weight) suspended, can scarcely overhaul the fall, the friction being so very great.
THE INCLINED PLANE.
_The inclined plane_ forms simply a gradual and sloping instead of a sudden and perpendicular ascent, by which heavy bodies may be raised to certain heights. The power necessary for raising a weight depends on the difference between the length of the plane and the height to be ascended. If the height be one-third of the length, then one pound will lift three pounds. The force with which a rolling body descends on an inclined plane is to the force of its absolute gravity, as the height of the plane is to its length.
_Parbuckling a gun_ on skids unites the advantage of one moveable pulley with that of the inclined plane.
_Rule._—As the length of the plane is to its height, so is the weight to the power.
_Example._—Required the power necessary to raise 540 lb. up an inclined plane, five feet long, and two feet high.
As 5 : 2 :: 540 : 216 lb. _Ans._
THE WEDGE.
_The wedge_ may be considered as two equally inclined planes joined together at their bases. It has a great advantage over all the other powers, arising from the force of percussion, or blow, with which the back is struck; which is a force incomparably greater than any dead weight, or pressure, such as is employed in other machines. The largest masses of timber may by this means be riven, and vessels of war, weighing many thousand tons, are lifted from their supports by the power of a few men, exerted by blows of mallets on wedges inserted for that purpose.
_The power of the wedge_ increases in proportion as its angle is acute. In tools intended for cutting wood the angle is commonly about 30°; for iron from 50° to 60°; and for brass from 80° to 90°.
_Case 1._—_When two bodies are forced from one another, by means of a wedge, in a direction parallel to its back._
_Rule._—As the length of the wedge is to half its back, or head, so is the resistance, to the power.
_Example._—The breadth of the back, or head of the wedge, being three inches, and the length of either of its inclined sides 10 inches, required the power necessary to separate two substances, with a force of 150 lb.
As 10 : 1½ :: 150 : 22½ lb. _Ans._
_Case 2._—_When only one of the bodies is moveable._
_Rule._—As the length of the wedge, is to its back, or head, so is the resistance, to the power.
_Example._—The breadth, length, and force, the same as in the last example.
As 10 : 3 :: 150 : 45 lb. _Ans._
THE SCREW.
_The screw_ is a spiral thread or groove cut round a cylinder, and everywhere making the same angle with the length of it. The force of a power applied to turn a screw round is to the force with which it presses upward, or downward, setting aside the friction, as the distance between two threads is to the circumference where the power is applied; or the advantage gained is as much as the circumference of a circle described by the handle of the winch exceeds the interval, or distance, between the spirals of the screw. Hence the force of any machine turned by a screw can readily be computed; for instance, in a press driven by a screw, whose threads are each a quarter of an inch asunder, and with a handle, to turn the screw, four feet long; then, if the natural force of a man, by which he can lift, pull, or draw, be 150 lb., and it is required to determine with what force the screw will press when the man turns the handle with his whole force; the diameter of the handle (_power_) being 4 feet, or 48 inches, its circumference is 48 × 3·1416, or 150⅘ nearly; and the distance of the threads being one-fourth of an inch, therefore the power is to the pressure as 1 to (150⅘ × 4) = 603⅕, but the power is equal to 150 lb., therefore as 1 : 603⅕ :: 150 : 90480, and consequently the pressure is equal to a weight of 90480 lb. independent of friction.
COMPOUND MACHINES.
Though each of the mechanical powers is capable of overcoming the greatest possible resistance in theory, yet in practice, if used singly for producing very great effects, they would frequently be so unwieldy and unmanageable as to render it impossible to apply them. For this reason it is generally found more advantageous to combine them together, by which means the power is more easily applied, and many other advantages are obtained. In all the mechanical powers, and their combinations, and in all machines, simple as well as compound, _what is gained in power is lost in time or velocity_; and _vice versâ_, or in other words, the product of the power, and the space through which it moves, is equal to the product of the weight, and the space through which it moves in the same plane. Suppose that a man, by means of a fixed pulley, raises a beam to the top of a house in two minutes, it is clear that he will be able to raise six beams in twelve minutes; but by means of a tackle with three lower pulleys, he will raise the six beams at once with the same ease as he before raised one, but then he will be six times as long about it, that is, twelve minutes; thus the work is performed in the same time whether the mechanical power is used, or not. But the convenience gained by the power is very great; for if the six beams are joined in one, they may be raised by the tackle, though it would be impossible to move them by the unassisted strength of one man. No real gain of force is obtained by mechanical contrivances; on the contrary, from friction and other causes, force is always lost; but by machines a more convenient direction can be given to the moving power, and so modify its energy as to obtain effects which it could not otherwise produce.
FRICTION.
_Friction_ arises from the irregularities of the surfaces which move upon one another. The surfaces of bodies of the same nature are moved with more facility over each other than those of a dissimilar nature. In proportion as the surfaces which are to be moved upon one another are rough, a greater force is requisite to produce motion. The same surfaces when under a greater pressure, are subject to still further friction. A double pressure doubles the amount of friction, a treble pressure trebles, and so on in nearly the same proportion. When surfaces are moving along each other in the direction of their grains, the friction is greater than when the direction of the grains is at right angles. Friction is little influenced by the velocity with which bodies move upon one another. Friction may be diminished in various ways, as will appear by the result of the following experiment with a block of square stone weighing 1080 lb.:—
1. In order to drag this stone along the floor of a quarry lb. roughly chiselled, it required a force equal to 758
2. Over a floor of planks, ditto 652
3. Placed on a platform of wood, and dragged over a floor of planks 606
4. After soaping the two surfaces of wood, which slide over each other 182
5. Placed on rollers of three inches diameter, and moved along the floor of the quarry 34
6. To drag it on these rollers over a wooden floor 28
7. Mounted on a wooden platform, and the same rollers placed between the platform, and a plank floor 22
One of the most remarkable instances of the application of rollers is the transport of the rock which now serves as the pedestal of the equestrian statue of Peter the Great at St. Petersburg. This rock is a single block of granite weighing 1217 tons. A railway was formed, consisting of two lines of timber, furnished with hard metal grooves; similar, and corresponding metal grooves were fixed to the under side of the sledge, or frame, on which the stone was laid, and between these grooves were placed spheres of hard brass, about six inches in diameter. On these spheres the frame with its enormous load was easily moved by sixty men working at capstans with triple purchase blocks.
UNGUENTS.
Mr. G. Rennie found, from a mean of experiments, with different unguents, on axles in motion, and under different pressures, that, with the unguent tallow under a pressure of from 1 to 5 cwt., the friction did not exceed 1/39th of the whole pressure; when soft soap was applied, it became 1/34th; and with the softer unguents applied, such as oil, hogs’ lard, &c., the ratio of the friction to the pressure increased; but with the harder unguents, as soft soap, tallow, and anti-attrition composition, the friction considerably diminished; consequently, to render an unguent of proper efficiency, the nature of the unguent must be measured by the pressure, or weight, tending to force the surfaces together.
TRANSVERSE STRENGTH OF MATERIALS.
When a beam, of any material, is loaded, the surface in contact with the load is _compressed_, and the opposite surface _extended_; and there is a line between these, which is neither compressed, nor extended, called _the neutral line_.
_If the depth of a beam be doubled_, the breadth, and length between supports remaining the same, _its strength will be increased four times_.
_If its breadth be doubled_, the other dimensions being as above, _its strength will be doubled_.
By increasing the distance between the supports of any beam, its strength is decreased in the same ratio; twice the distance between the supports will weaken the beam one-half; half the distance between the supports will enable it to bear twice the load.[48]
The same beam will bear twice the load, if, instead of being concentrated in the middle, it be equally distributed over the whole length of the beam.
If the load on a beam be placed near to one of the supports, instead of in the middle, its effect will decrease in the ratio of its proximity to the support.
Let S s represent the beam, W the load or weight in the middle, w the weight near s; then the load which the beam will carry at the point where w is placed will be found by the following proportion:—
As S w × w s : S W × W s :: W : w.
A beam, fixed at one end, and loaded at the other, will bear half the weight of one of the same length supported at each end.
If the end of a beam, instead of being only supported, be _fixed_, its strength will be in the proportion of 3 to 2.
From the foregoing results it will be seen that the strength of a rectangular beam varies, as the breadth multiplied by the depth squared, divided by the length, (b × d^2)/l and if the breaking weight of any material, 1 inch square, and 1 foot long, be found, it will represent a _constant multiplier_ for the above equation.
Thus the breaking weight of a beam of Riga fir, 1 inch square, and 1 foot long (_vide following_ TABLE), is ·164 of a ton; and to find the breaking weight of a beam of any other dimensions, the rule is simply
W = (b d^2)/l × ·164.
_Example._—What will be the breaking weight of a beam of Riga fir, 8 inches broad, 12 inches deep, and 20 feet long?
(8 × 12^2)/20 = 57·6 57·6 × ·164 = 9·44 tons, breaking weight.
_Table of constants_, for beams of different materials, being the breaking weights of such beams, 1 inch square, and 1 foot long.
Riga fir ·164 of a ton. English oak ·248 of a ton. Red pine ·199 ” Canadian do. ·261 ” Pitch pine ·242 ” Dantzic do. ·219 ” Beech ·231 ” Teak ·366 ” Elm ·150 ” Cast iron, mean. 1·000 ” Ash ·301 ” Wrought do. 1·083 ”
From the foregoing rules Length = (b d^2)/W × constant.
Breadth = (l W)/d^2 × constant.
Depth = √((l W)/b) × constant.
_The practical weight_ that a beam will carry _with safety_, permanently, should only be taken at one-fourth of the above computations.
ADHESION OF NAILS, AND SCREWS.
_The percussive force_ required to drive the common sixpenny nail (73 to the pound) to the depth of an inch and a half into deal, with a weight of six pounds and a quarter, is four blows, or strokes, falling freely the space of one foot; and _the steady pressure_ to produce the same effect is four hundred pounds. A _sixpenny nail_ driven into dry elm to the depth of one inch across the grain requires a force of 327 pounds to extract it; and the same nail, driven into the same wood endways, or longitudinally, can be extracted with a force of 257 pounds.
To extract a sixpenny nail from a depth of one inch out of dry oak requires 507 pounds, and out of dry beech 667 pounds. A sixpenny nail driven two inches into dry oak would require a steady force of more than half a ton to extract it.
_A common screw_ of one-fifth of an inch diameter has an adhesive force of about three times that of a sixpenny nail.
TRIGONOMETRY.
_Plane trigonometry_ treats of the relations, and calculations of the sides, and angles of plane triangles.
_The measure of an angle_ is an arc of any circle contained between the two lines which form that angle, the angular point being the centre; and it is estimated by the number of degrees contained in that arc. Hence a right angle being measured by a quadrant, or quarter of a circle, is an angle of 90 degrees. The sum of the three angles of every triangle is equal to 180 degrees, or two right angles; therefore, in a right-angled triangle, taking one of the acute angles from 90 degrees, leaves the other acute angle; and the sum of the two angles in any triangle, taken from 180 degrees, leaves the third angle; or one angle being taken from 180 degrees leaves the sum of the other two angles.
_Definitions._
_The sine of an arc_ is the line drawn from one extremity of the arc perpendicular to the diameter of the circle which passes through the other extremity.
_The supplement of an arc_ is the difference, in degrees, between the arc, and a semicircle, or 180 degrees.
_The complement of an arc_ is the difference, in degrees, between the arc, and a quadrant, or 90 degrees.
_The tangent of an arc_ is a line touching the circle in one extremity of that arc, continued from thence to meet a line drawn from the centre through the other extremity; which last line is called _the secant_ of the same arc.
The _cosine_, _cotangent_, and _cosecant_ of an arc are the sine, tangent, and secant of the complement of that arc, _the co_ being only a contraction of the word complement.
The sine, tangent, or secant of an angle is the sine, tangent, or secant of the arc by which the angle is measured, or of the degrees, &c., in the same arc, or angle. _Vide also Definitions_, PRACTICAL GEOMETRY.
_There are two Methods of resolving triangles_, or the cases of trigonometry—viz., _Construction_, and _Computation_.
_1st method._—The triangle is constructed by making the sides from a scale of equal parts, and laying down the angles from the protractor. Then, by measuring the unknown parts by the same scale, the solution will be obtained.
_2nd method._—Having stated the terms of the proposition, resolve it like any other proportion, in which a fourth term is to be found from three given terms, by multiplying the second and third terms together, and dividing the product by the first.
_Note._—Every triangle has six parts—viz., three sides, and three angles; and, in every case in trigonometry, there must be given three of these parts to find the other three. Also of the three parts that are given, one of them at least must be a side; because, with the same angles, the sides may be greater, or less, in any proportion.
_Computation._
_Case 1._—_When a side and its opposite angle are two of the given parts._
The sides of any triangle having the same proportion to each other, as the sines of their opposite angles; then—
As any one side, is to the sine of its opposite angle; so is any other side, to the sine of its opposite angle.
_To find an angle_, begin the proportion with a side, opposite to a given angle; and, _to find a side_, begin with an angle opposite to a given side.
_Case 2._—_When the three sides of a triangle are given, to find the angles._
Let fall a perpendicular from the greatest angle, on the opposite side, or base, dividing it into two segments; and the whole triangle into two right-angled triangles: then the proportion will be—
As the base or sum of the segments, is to the sum of the other two sides; so is the difference of those sides, to the difference of the segment of the bases: then add half the difference of the segments to the half sum, or the half base, for the greater segment; and subtract the same for the less segment. Hence, in each of the two right-angled triangles, there will be known two sides, and the right angle opposite to one of them, consequently the other angle will be found by the method in _Case 1_.
USEFUL THEOREMS, AND COROLLARIES.
1. When one line meets another, the angles, which it makes on the same side of the other, are together equal to two right angles.
2. All the angles, which can be made at any point (by any number of lines), on the same side of a right line, are, when taken all together, equal to two right angles: and, as all the angles that can be made, on the other side of the line, are also equal to two right angles; therefore all the angles that can be made quite round a point, by any number of lines, are equal to four right angles. Hence also the whole circumference of a circle, being the sum of all the angles that can be made about the centre, is the measure of four right angles.
3. When two lines intersect each other, the opposite angles are equal.
4. When one side of a triangle is produced, or extended, the outward angle is equal to the sum of the two inward opposite angles.
5. In any triangle, the sum of all the three angles is equal to two right angles (180°). Hence, if one angle of a triangle be a right angle, the sum of the other two angles will be equal to a right angle (90°).
6. In any quadrilateral, the sum of all the four inward angles is equal to four right angles.
7. In any right-angled triangle, the square of the hypothenuse (or side opposite to the right angle) is equal to the sum of the squares of the other two sides. Therefore, _to find the hypothenuse_, add together the squares of the other two sides, and extract the square root of that sum: and _to find one of the other sides_, subtract from the square of the hypothenuse the square of the other given side, and extract the square root of the remainder for the side required.
Or hypothenuse = √(base^2 + perpendicular^2)
Base = √((hypoth. + perpend.) × (hypoth. - perpend.))
Perpendicular = √((hypoth. + base) × (hypoth. - base.))
TRIGONOMETRY, WITHOUT LOGARITHMS.[49]
“In all the more elaborate, and refined operations of trigonometry, it is not only desirable, but necessary to employ some of the larger logarithmic tables, both to save time, and to ensure the requisite accuracy in the results. But in the more ordinary operations, as in those of common surveying, ascertaining inaccessible heights, and distances, reconnoitring, &c., where it is not very usual to measure a distance nearer than within about its thousandth part, or to ascertain an angle nearer than within two or three minutes, it is quite a useless labour to aim at greater accuracy in a numerical result. Why compute the length of a line to the fourth, or fifth place of decimals, when it must depend upon another line, whose accuracy cannot be ensured beyond the unit’s place? Or, why compute an angle to seconds, when the instrument employed does not ensure the angles in the data beyond the nearest minute? In the following Table are brought together the _natural sines, and cosines_, to every degree in the quadrant, and this table will be found sufficiently extensive, and correct for the various practical purposes above alluded to. The requisite proportions must, it is true, be worked by multiplication, and division, instead of by logarithms. Yet this by no means involves such a disadvantage as might seem, at first sight. For when the measured lines are expressed by three, or at most, four figures, the multiplications, and divisions are performed nearly as quick, and in some cases quicker, than by logarithms. Then as to accuracy, even in cases where the computer will have to take proportional parts for the minutes of a degree, the result may usually, if not always, be relied upon to within about a minute.”
TRIGONOMETRIC RATIOS.
_Natural sines, and cosines to every degree in the quadrant, radius being 1·000000._
+----+--------+--------+----++----+--------+--------+----+ |Deg.| Sines. |Cosines.| ||Deg.| Sines. |Cosines.| | +----+--------+--------+----++----+--------+--------+----+ | 0 | ·00000 |1·00000 | 90 || | | | | | 1 | ·01745 | ·99985 | 89 || 26 | ·43837 | ·89879 | 64 | | 2 | ·03490 | ·99939 | 88 || 27 | ·45399 | ·89101 | 63 | | 3 | ·05234 | ·99863 | 87 || 28 | ·46947 | ·88295 | 62 | | 4 | ·06976 | ·99756 | 86 || 29 | ·48481 | ·87462 | 61 | | 5 | ·08716 | ·99619 | 85 || 30 | ·50000 | ·86603 | 60 | +----+----------------------------+--------+--------+----+ | 6 | ·10453 | ·99452 | 84 || 31 | ·51504 | ·85717 | 59 | | 7 | ·12187 | ·99255 | 83 || 32 | ·52992 | ·84805 | 58 | | 8 | ·13917 | ·99027 | 82 || 33 | ·54464 | ·83867 | 57 | | 9 | ·15643 | ·98769 | 81 || 34 | ·55919 | ·82904 | 56 | | 10 | ·17365 | ·98481 | 80 || 35 | ·57358 | ·81915 | 55 | +----+--------+--------+----++----+--------+--------+----+ | 11 | ·19081 | ·98163 | 79 || 36 | ·58778 | ·80902 | 54 | | 12 | ·20791 | ·97815 | 78 || 37 | ·60181 | ·79863 | 53 | | 13 | ·22495 | ·97437 | 77 || 38 | ·61566 | ·78801 | 52 | | 14 | ·24192 | ·97030 | 76 || 39 | ·62932 | ·77715 | 51 | | 15 | ·25882 | ·96593 | 75 || 40 | ·64279 | ·76604 | 50 | +----+--------+--------+----++----+--------+--------+----+ | 16 | ·27564 | ·96126 | 74 || 41 | ·65606 | ·75471 | 49 | | 17 | ·29237 | ·95630 | 73 || 42 | ·66913 | ·74314 | 48 | | 18 | ·30902 | ·95106 | 72 || 43 | ·68200 | ·73135 | 47 | | 19 | ·32557 | ·94552 | 71 || 44 | ·69466 | ·71934 | 46 | | 20 | ·34202 | ·93969 | 70 || 45 | ·70711 | ·70711 | 45 | +----+--------+--------+----++----+--------+--------+----+ | 21 | ·35837 | ·93358 | 69 || | | | | | 22 | ·37461 | ·92718 | 68 || | | | | | 23 | ·39073 | ·92050 | 67 || | | | | | 24 | ·40674 | ·91355 | 66 || | | | | | 25 | ·42262 | ·90631 | 65 || | | | | +----+--------+--------+----++----+--------+--------+----+ | |Cosines.| Sines. |Deg.|| |Cosines.| Sines. |Deg.| +----+--------+--------+----++----+--------+--------+----+
“The preceding table is so arranged that for angles not exceeding 45 degrees, the sine, and cosine for any number of degrees will be found opposite to the proposed number in the left hand column, and in the column under the appropriate word. When the number of degrees in the arc, or angle, exceeds 45 degrees, that number must be found in the right hand column, and opposite to it in the column indicated by the appropriate word at the bottom of the table. Thus, the sine, and cosine of 36 degrees are ·58778 and ·80902 respectively, the radius of the table being unity, or 1. The taking of proportional parts for minutes can only be done correctly in those parts of the table where the differences between the successive sines, &c., run pretty uniformly. Suppose we want the natural sine of 20° 16′. The sine of 21 degrees is ·35837, that of 20 degrees is ·34202; their difference is ·1635. This divided by 60 gives 27·25 for the proportional part due to 1 minute, and that again multiplied by 16 gives 436 for the proportional part for 16 minutes. Hence the sum of ·34202 and 436, or ·34638, is very nearly the sine of 20° 16′. But the operation may often be contracted by recollecting that 10 minutes are ⅙, 15 minutes are ¼, 40 minutes are ⅔ of a degree, and so on. Observe, also, that for cosines the results of the operations for proportional parts are to be _deducted_ from the value of the required trigonometrical quantity in the preceding degree.”
APPLICATION OF TRIGONOMETRY, WITHOUT LOGARITHMS, to the determination of Heights, and Distances.
_Example 1._—Having measured a distance of 200 feet in a direct horizontal line from the bottom of a steeple, the angle of elevation of its top, taken at that distance, was found to be 47° 30′, from hence it is required to find the height of the steeple?
By deducting 47° 30′ from 90°, the angle opposite the given side will be found (42° 30′).
Then by _Case 1_. TRIGONOMETRY:—
As sine ∠ 42° 30′ : 200 :: sine ∠ 47° 30′: Or ·67556 : 200 :: ·73723 : 208·2, &c., height required.
_By construction_—
The triangle is constructed by making the side from a scale of equal parts, and laying down the angles from the protractor. Then by measuring the unknown parts by the same scale, the solution will be obtained.
_Example 2._—Being on the side of a river, and requiring the distance to a house on the other side, 200 yards were measured in a straight line by the side of the river, and at each end of this base line the angles with the house were 68° 2′, and 73° 15′—required the distance from each end of the base line to the house?
The sum of the given angles (68° 2′ + 73° 15′) subtracted from 180° will give the third angle (38° 43′).
Then by _Case 1_. TRIGONOMETRY:—
As sine ∠ 38° 43′ : 200 :: sine ∠ 68° 2′ ·62544 : 200 :: ·92739 : 296·5 first distance required. As sine ∠ 38° 43′ : 200 :: sine ∠ 73° 15′ ·62544 : 200 :: ·95753 : 306·1 second distance required.
Similarly to the preceding examples, HEIGHTS, AND DISTANCES may be rapidly (and for military purposes, sufficiently accurately) computed in the field, by means of the foregoing trigonometrical table, if proper attention is paid to the principles by which the unknown angles of triangles may be ascertained: a base line, and requisite angle, or angles, having been given.
It will, however, be necessary to use advantageously the methods in Cases 1, 2 (_vide_ Trigonometry), and also the properties in the subsequent theorems, and corollaries.[50]
TABLE,
_Showing the reduction in feet, and decimals upon 100 feet, for the following angles of elevation, and depression._
+--------+----------+--------+----------+--------+----------+ | Angle. |Reduction.| Angle. |Reduction.| Angle. |Reduction.| +--------+----------+--------+----------+--------+----------+ | ° ′ | | ° ′ | | ° ′ | | | 3 0 | ·14 | 9 0 | 1·22 | 15 0 | 3·40 | | | | 9 30 | 1·38 | 15 30 | 3·64 | | 4 0 | ·25 | 10 0 | 1·52 | 16 0 | 3·88 | | | | 10 30 | 1·68 | 16 30 | 4·12 | | 5 0 | ·38 | 11 0 | 1·84 | 17 0 | 4·37 | | | | 11 30 | 2·01 | 17 30 | 4·63 | | 6 0 | ·55 | 12 0 | 2·19 | 18 0 | 4·90 | | 6 30 | ·65 | 12 30 | 2·37 | 18 30 | 5·17 | | 7 0 | ·76 | 13 0 | 2·56 | 19 0 | 5·44 | | 7 30 | ·86 | 13 30 | 2·77 | 19 30 | 5·74 | | 8 0 | .98 | 14 0 | 2·97 | 20 0 | 6·08 | | 8 30 | 1·10 | 14 30 | 3·18 | 20 30 | 6·33 | +--------+----------+--------+----------+--------+----------+
The reduction for 100 feet (from the above table) multiplied by the number of times 100 feet measured, will give the quantity to be subtracted from the measured length of an inclination, to reduce it to a horizontal position.
[Illustration: _Surveying, and Reconnoitring._
_Fig. 1._ _Fig. 2._ _Fig. 3._ _Fig. 4._ _Fig. 5._ _Fig. 6._
_J. W. Lowry, sc._]
TABLE,
_showing the rate of inclination of inclined planes, for the following angles of elevation_.
+--------+--------+--------+--------+--------+--------+ | Angle. | One in | Angle. | One in | Angle. | One in | +--------+--------+--------+--------+--------+--------+ | ° ′ | | ° ′ | | ° ′ | | | 0 15 | 228 | 3 30 | 17 | 7 0 | 8 | | 0 30 | 114 | 3 45 | 16 | 7 30 | 7¼ | | 0 45 | 76 | 4 0 | 15 | 8 0 | 7 | | 1 0 | 56 | 4 15 | 14 | 9 0 | 6½ | | 1 15 | 46 | 4 30 | 13 | 10 0 | 6 | | 1 30 | 38 | 4 45 | 12 | 11 0 | 5¾ | | 1 45 | 32 | 5 0 | 11½ | 12 0 | 5½ | | 2 0 | 28 | 5 15 | 11 | 13 0 | 5 | | 2 15 | 26 | 5 30 | 10½ | 14 0 | 4½ | | 2 30 | 23 | 5 45 | 10 | 15 0 | 4 | | 2 45 | 21 | 6 0 | 9½ | 16 0 | 3¾ | | 3 0 | 19 | 6 30 | 9 | 17 0 | 3½ | | 3 15 | 18 | 6 45 | 8½ | 18 0 | 3¼ | +--------+--------+--------+--------+--------+--------+
SURVEYING, AND RECONNOITRING.
HEIGHTS, AND DISTANCES.
The accurate determination of heights, and distances of objects being required in various military operations, especially for the position of batteries, the following methods for their attainment will be found useful when the requisite instruments are at hand; by frequent practice, the eye should, however, be enabled to determine, _nearly_, either the height of, or distance from any object.
HEIGHTS.
1.—BY MEANS OF A “POCKET SEXTANT,”
_to ascertain the height of an object_.
When the sextant is used for taking the height of objects, it is to be held vertically, and the quicksilvered part of the horizon glass will be on the left hand of the observer, or on the left part of the transparent glass. Altitudes are measured in the same manner as horizontal angles, for if we conceive the horizontal triangle A B C (_vide Plate 2, Fig. 2_), to be raised on its base A C with the angle C next to the observer, then the perpendicular A B becomes the height of the object B; and supposing the object to stand on a horizontal plane, then the ground and the object form the right angle at A; therefore, _if the object is accessible_, the sextant need only be set at any of the angles mentioned for distances (_vide Art._ DISTANCES), and walking backward on the line A C until the top of the object is brought down to the height of the observer’s eye from the ground, then the distance from where the observer stands to the object will be in the same proportion to its height as the base was to the distance. Then add the height of the eye from the ground, and the height of the object will be ascertained. If the object is not accessible, the angle must be taken, and calculated by trigonometry.
2.—BY MEANS OF A PORTABLE BAROMETER, AND THERMOMETER,
_to ascertain the height of an object_.
Observe the altitude (B) of the mercurial column in inches, tenths, and hundredths, at the bottom of the hill, or other object, the height of which is required.
Observe, also, the altitude (b) of the mercurial column at the top of the object. Observe the temperature on Fahrenheit’s thermometer at the times of the two barometrical observations, and take the mean between them. Then 55000 × (B - b)/(B × b) = the height of the hill in feet, for the temperature of 55 degrees on Fahrenheit. Add 1/440 of this result for every degree which the mean temperature exceeds 55 degrees, and subtract as much for every degree below 55 degrees. This will be a good approximation when the height of the hill is below 2000 feet.
3.—BY MEANS OF THE RECONNOITRING PROTRACTOR,[51]
_to measure the height of an inaccessible object_.
[_Plate_, SURVEYING, AND RECONNOITRING, _Fig. 1_.]
Place yourself at a convenient distance from the object whose height is required, taking care to have a good base line to the second station. Hold the protractor vertically, with a steady hand, the tube side uppermost, and bring the top of the object in a line with the centre of the tube. Allow the arm (or index) to vibrate freely, and, when steady, note the angular height of the object (shown by the edge of the index on the marginal scale of degrees). By the aid of points taken through the tube, or by pickets, then pace, or measure a base in a direct line from the object; and, when arrived at the second station, again note the angular height of the object.
_Construction_—
Set off the angles, and draw the respective lines, which, by their intersection, will determine the height of the perpendicular, to which the height of the protractor above the ground must be added for the altitude of the object. By using the scale of the measured base line, the height required will be ascertained, or it may be calculated by “TRIGONOMETRY, WITHOUT LOGARITHMS.”—_Page 303._
_To measure the height of an accessible object._
[_Plate_, SURVEYING, AND RECONNOITRING, _Fig. 2_.]
At an appropriate distance from the object, take its angular height and measure the distance to its base.
_Construction_—
Draw a line representing this distance, at one end of which draw another line at the angle found, and at the other erect a perpendicular; the intersection of these lines will determine the altitude of the object.
_To measure the vertical height of a hill, or mountain._
[_Fig. 3, Plate_, SURVEYING, AND RECONNOITRING.]
From a station a short distance from the hill, take, and note down its angular height; then select a rear position for a base line, using the tube of the protractor to insure a straight direction; proceed to the requisite distance on the base, and again note the altitude of the hill.
_Construction_—
The intersection of lines drawn from each end of the base line, at the angles found, will determine the altitude; the perpendicular height of which, added to that of the protractor above the ground, will give the altitude required.
_To measure the altitude of a tower, &c., on a height._
[_Fig. 4, Plate_, SURVEYING, AND RECONNOITRING.]
From the first station, near the base, take the altitude of the hill, and also that of the tower above it, and note down these angles; proceed to another station in a straight line with the former one, measuring its length, and again observe the angular height of the hill, and also that of the top of the tower.
Similarly to the previously described mode, ascertain, first, the height of the hill; second, the height of the hill, and tower; deduct the first calculation from the second, which will leave the height of the tower.
In all the foregoing cases the heights may be correctly ascertained by trigonometrical calculations (_vide_ TRIGONOMETRY, WITHOUT LOGARITHMS, _page 303_).
4.—BY THE SHADOW OF THE OBJECT,
_to ascertain the height_.
Set up vertically a staff of known length, and measure the length of its shadow upon a horizontal, or other plane; measure also the length of the shadow of the object of which the altitude is required. Then, by the property of similar triangles,
As the length of the shadow of the staff is to the altitude of the staff, so is the length of the shadow of the object to the altitude of the object.
5.—WHEN THERE IS NO SHADOW,
_to ascertain the height_.
Place a staff (equal in length to the height of the observer’s eye) vertically at such a distance from the foot of the required altitude, that the observer, having laid himself upon his back, with his feet against the bottom of the stick, may see the top of the staff, and object in the same line. Then, by similar triangles, the height may be readily ascertained.
6.—BY MEANS OF THE TANGENT SCALE OF A GUN,
_to ascertain the height of an object, the distance being known_.
Lay the gun for the top of the object the height of which is required, then raise the tangent scale until the top of it, and the notch on the muzzle are in line with the bottom of the object: then, by similar triangles,
As the length of the gun is to the length of the raised part of the tangent scale, so is the distance from the gun to the object, to the height required.
[Illustration: _Plate 2._
HEIGHTS. _Fig. 1._ _Fig. 2._
DISTANCES. _Fig. 3._ _Fig. 4._ _Fig. 5._
PRACTICAL GEOMETRY. _Fig. ½._ _Fig. 21._ _Fig. 22._]
7.—BY MEANS OF TWO PICKETS,
_to ascertain the height of an object_.
[_Vide 2nd Plate_, HEIGHTS, AND DISTANCES, _Fig. 1_.]
Let two pickets C D (4 feet), E F (6 feet), be placed with their bases in the line C A passing through A the height required, and move them nearer to, or farther from each other, until the summit B of the object is seen in the same line as D, and F, the tops of the rods. Then, by the principles of similar triangles,
As D H (= C E) : F H :: D G (= C A) : B G. To which add A G = C D for the whole height A B.
Thus, supposing C E to be 6 feet, F H 2 feet, and C A 150 feet, the proportion will be,
As 6 : 2 :: 150 : 50 feet. Then 50 + C D will be the altitude required.
DISTANCES.
1.—BY MEANS OF THE SEXTANT,[52]
_to find the distance from an object, whose height is known_.
Let A B represent the height of the object; C your station; and C B the distance to be found.
[Illustration: (right angle triangle A B C)]
Take the angle B C A with the sextant,[52] and note it in minutes; then A B, in feet × 573 ÷ B C A, in minutes = A C in fathoms. Or A B in feet × 573 ÷ B C A, in minutes × 2 = A C in yards.
573 is a constant multiple.
This method requires no table of sines, &c., the number of minutes in the angle being used instead of the sine.
2.—BY MEANS OF A POCKET SEXTANT,
_to measure inaccessible distances_.
When used for taking the distance of objects, the sextant is to be held horizontally, and the quicksilvered part of the glass will be uppermost, or above the transparent part.
To ascertain the distance A B (_vide Plate 2, Fig. 2_), obtain, by observation, the direction A C perpendicular to A B, which is thus performed:—Set the instrument at 90°, and place yourself at the point A, with your right towards the point B; then look through the sextant, and direct a picket to be placed in the line A C at 100 yards, or feet, from you, so that the point B will appear right above it. Then set the sextant at 45°, and walk along the line towards C until you bring the points A, and B to coincide; the base and perpendicular will then be of equal length, and A C being known, or measured, the distance A B will also be ascertained. But if you cannot walk far enough to find angle C 45°, find it equal to 63° 26′, and then A C = ½ A B; at 71° 34′ = ⅓ A B; at 75° 58′ = ¼ A B; at 78° 41′ = ⅕ A B; at 80° 32′ = ⅙ A B; at 82° 52′ = ⅛ A B; and at 84° 17′ the distance will be ⅒ A B.
Should the object be far distant, it will be necessary to take a long base, and the side A B must be calculated, therefore, by trigonometry.
3.—BY MEANS OF THE PRISMATIC COMPASS,
_to measure inaccessible distances_.
Having fixed the instrument to the stand, place it over the station-point, spreading the legs so as to give sufficient firmness, and observing that the card is level enough to allow it to play freely; raise the prism by means of the slide, until the divisions of the compass-card are distinctly seen; then look through the slit, and turn the box round until the thread bisects the object whose distance is required; allow the card to settle, and the division on it, which coincides with the thread of the vane, will be the azimuth, or bearing of the object, reckoned from the north, or south point of the needle, when the card is divided into twice 180 degrees. The angular distance between any two objects will, of course, be the difference of their bearings; thus, suppose one to bear 15° N.E., and the other 165° S.E., the angular distance between them will be 150°.
In military sketching, the compass is often supported merely by the hands, using the little spring to check the vibrations of the card. In windy weather, the mean of these vibrations must be taken for the bearing sought.
The directions for surveying, &c., &c., by means of “The Reconnoitring Protractor,” apply similarly to the “Prismatic Compass.”
4.—BY MEANS OF “THE RECONNOITRING PROTRACTOR,”
_to ascertain the distance from inacessible objects_.
[_Plate_, SURVEYING, AND RECONNOITRING, _Fig. 6_.]
Select a good position for a base line; fix the protractor on the tripod at the first station, placing the instrument in a direct line between the first station and the point selected for the second station. Direct the index consecutively at the objects, the relative distances of which are to be ascertained, and note correctly their respective angles. When the object is above the horizontal line, the sliding-sight must be sufficiently raised to take its bearing; and, should the object be below the level of the protractor, its angle may be taken by observation through the upper holes of the near sight; or the feet of the tripod may be adjusted, by raising, or sinking them in the ground, so that the index may be correctly directed to the object. Then proceed to the second station, measuring, or carefully pacing the base line, at the end of which fix the protractor in a straight line between the two stations; direct the index at the objects previously noted at the first station, taking their respective angles as before.
_Construction_—
Draw the base of the length required, according to the scale; from each end of which set off the angles found, and draw the lines required; the intersection of these will determine the position of the several objects, and their relative distances may be ascertained by measurement on the scale of the base line; or they may be calculated trigonometrically.
5.—BY MEANS OF TWO PICKETS,
_to ascertain the distance from an object_.
Take two pickets of unequal lengths, drive the shortest into the ground, say close to the edge of a river; measure some paces back from it, and drive in the other, till you find, by looking over the tops of both, that your sight cuts the opposite bank. Pull up the first picket, measure the same distance from the second in any direction the most horizontal, and drive it as deep in the ground as before. Then, if you look over them again, and observe where the line of sight falls, or terminates, you will have the distance required. This method is only applicable to short distances.
6.—_To ascertain the distance of the object A from B._
[_Vide Plate 2, Fig. 3._]
Place a picket at B, and another at C at a few yards’ distance, making A B C a right angle, or B C perpendicular to A B.[53] Divide B C into 4, 5, or any number of equal parts, make another similar angle at C in a direction from the object, and walk along the line C D until you bring yourself in a line with the object A, and any of the divisions (say O) of the line B C. Then (having measured C D)
as C O : C D :: B O : B A. Or, as 10 : 53 :: 30 : 159 yards.
7.—_To find the distance between two objects_, C, _and_ D.
[_Vide Plate 2, Fig. 4._]
From any point A, taken in the line C D, erect the perpendicular A E, in which set off from A to E 40 yards, set off from E to G, in the prolongation of A E, 10 yards, at G raise the perpendicular G F, and produce it towards I, plant pickets at E, and G, then move with another picket on G F, till F is in a line with E, and D; and on the prolongation of the perpendicular F G place another picket at I in the line with E, and C: measure F I (54 yards), then—
as G E : A E :: F I : C D; Or, as 10 : 40 :: 54 : 216 yards.
8.—_To find the inaccessible length_, A, B, _of the front of a fortification_.
[_Plate 2, Fig. 5._]
Plant a picket at C, from whence both points may be seen; find the lengths C A, C B (by the method in No. 5); make C E one-fourth, or any part of C B, and make C D bear the same proportion to C A: measure D E; then
as C D : D E :: C A : A B.
Nearly in the same manner the distance from B to A may be ascertained, when the point B is accessible; for having measured the line C B, and made the angle C E D equal to C B A, the proportion will be as C E : D E :: C B : B A.
9.—BY MEANS OF THE TANGENT SCALE OF A GUN,
_to ascertain the distance, the height of the object at the required distance being known_.
Lay the gun by the line of metal for the top of the object; then raise the tangent scale till the top of it and the notch on the muzzle are in line with the foot of the object, and note what length of scale is required.
Then,—by similar triangles—
As the length of the raised part of the tangent scale is to the length of the gun; so is the height of the distant object to the distance required.
Thus, supposing the height of the object to be 9 feet, the length of that part of the tangent scale which is raised, 3 inches, and of the gun 6 feet, the proportion will be—
As 3 : 72 :: 108 : 2592 inches, or 216 feet.
10.—BY MEANS OF THE PEAK OF A CAP,
_to measure the breadth of a river_.
Place yourself at the edge of one bank, and lower the peak of your cap till you find the edge of it cut the other bank, then steady your head by placing your hand under your chin, and turn round gently to some level spot of ground on your side of the river, and observe where your eyes, and the edge of the peak again meet the ground; measure the distance, which will be _nearly_ the breadth of the river.
11.—BY THE REPORT OF FIRE-ARMS, TO ASCERTAIN THE DISTANCE OF ANY OBJECT, _vide_ SOUND, _page 316_.
_To estimate distances, in the field._
Good eyesight recognises masses of troops at 1700 yards; beyond this distance the glitter of arms may be observed. At 1300 yards infantry may be distinguished from cavalry, and the movement of troops may be seen; the horses of cavalry are not, however quite distinct, but that the men are on horseback is clear. A single individual detached from the rest of the corps may be seen at 1000 yards, but his head does not appear as a round ball until he has approached up to 700 yards; at which distance white cross-belts, and white trousers may be seen. At 500 yards the face may be observed as a light coloured spot; the head, body, arms, and their movements, as well as the uniform, and the firelocks (when bright barrels) can be made out. At between 200 and 250 yards all parts of the body are clearly visible, the details of the uniform are tolerably clear, and the officers may be distinguished from the men.
_Vide_ “UNITED SERVICE MAGAZINE.”—No. CCCXXXI.
BY MEANS OF THE RECONNOITRING PROTRACTOR,
_to traverse roads_.
[_Plate_, SURVEYING, AND RECONNOITRING, _Fig. 5_.]
Fix the protractor on the tripod at the first station, placing it so that the side tube may be in a direct line with the intended second station. From each end of the tube observe the objects in sight (or place pickets) in order to secure a straight line in pacing, or measuring, from the first to the second station. Mark the distance between the stations, and place the protractor, by means of the tube, in a direct line with the first station. Then select the third station, and direct the arm or index correctly to it (using the upper holes of the near sight for a declivity, or raising the sliding-sight for an ascent); note the angle thus found, and notice the objects in front, and rear (if any, if not, place pickets) for points to enable you to pace towards, and work with accuracy at the third station. Select station 4, place the tube in line with the third, and second stations; note the bearing of No. 4, and pace the distance to it. Proceed thus from station to station, entering the angles, and distances in your note-book, as well as the offsets (which must also be carefully measured) from the lines taken, until the survey is completed.
_Construction_—
The day’s work will be easily plotted on paper, by setting off the angles found, and drawing lines for the measured distances, according to scale.
SOUND.
The movement communicated to the particles of air by the vibrations of a sonorous body is the cause of the sensation of sound; and it is because the particles are driven from the point of vibration in every direction, as from a centre, that the sound is perceived at once, everywhere within the surface of a sphere of a certain extent.
The velocity of sound; or the space through which it is propagated in a given time, has been differently estimated by authors who have written on this subject. Roberval states it to be at the rate of 560 feet in a second; Gassendus at 1473; Mersenne at 1474; Duhamel at 1338; Newton at 960; Derham, in whose measure Flamsteed and Halley acquiesce, at 1142. By accounts in the Memoirs of the Royal Academy of Sciences, at Paris, 1738, where cannon were fired at various distances, under many varieties of weather, wind, and other circumstances, and where the measures of the different places had been settled with the utmost exactness, it was found that sound was propagated, on a medium, at the rate of 1038 French feet in a second of time, which is equivalent to 1107 _English feet_, the French foot being in proportion to the English as 15 to 16.
From various experiments made with great care by Dr. O. Gregory, it has been found that sound flies through the air uniformly at the rate of about 1100 feet per second, when the air is quiescent, and at a medium temperature. At the temperature of freezing, or a little below, the velocity is about 1120. The approximate velocities under different temperatures may be found by adding to 1100 _half a foot_ for every degree on Fahrenheit’s thermometer above the freezing point. The mean velocity may be taken at 370 yards per second, or a mile in 4-7/9 second. Hence, multiplying any time employed by sound in moving by 370, will give the corresponding space in yards, or dividing any space in yards by 370 will give the time which sound will occupy in passing uniformly over that space. If the wind blow briskly, as at the rate of 20 to 60 feet per second, in the direction in which the sound moves, the velocity of the sound will be proportionally augmented; if the direction of the wind is opposed to that of the sound, the difference of their velocities must be employed. The velocity of sound is not affected by its intensity, the smallest sound moving as rapidly as the loudest.
_To ascertain the distance of any object by the report of fire-arms._
(_Vide 11. Page 314._)
Multiply the number of seconds which elapse between the time of seeing the flash, and hearing the report by 1100, and the product will be the distance in feet, with sufficient accuracy for ordinary purposes. If greater accuracy be required, this rule must be modified, on account of the velocity, and direction of the wind, and state of the thermometer.
_Sound will be louder_ in proportion to the condensation of the air. Water is one of the greatest conductors of sound; it can be heard on water nearly twice as far as upon land.
GRAVITY.
Gravity is downward pressure, or weight, being the natural tendency of all bodies towards the centre of the earth. (_Vide Gravity_, MOTION, FORCES. _Page 320._)
_Absolute gravity_ denotes the whole force with which a body tends downwards, as when the body is in empty space.
_Specific gravity_ denotes the relative or comparative gravity of any body, in respect to that of another body of equal bulk, or magnitude.
_Centre of gravity_ is that point in a body, or system of bodies, on which, if rested, or suspended, the whole would remain in a state of equilibrium about that point.
_The centre of gravity_ of a circle, regular polygon, prism, cylinder, or sphere, is in its centre.
_The centre of gravity_ of a triangle is found by bisecting any two of its sides, and drawing lines from the points of bisection to the opposite angles; the intersection of these lines will be the centre of gravity.
_Force of gravity, or gravitation_, is an accelerated velocity, which bodies acquire in falling freely from a state of rest.
1. The space through which a body will fall in feet, in any given time equals the product of the square of the time multiplied by 16·0833.
_Example._—Required the space a falling body will pass through in five seconds?
16·0833 × 25 = 412·0825 feet.
2. The velocity in feet, which a body in descending freely will acquire in a given time, equals the product of the time in seconds multiplied by 32·1666.
_Example._—What is the velocity acquired at the end of seven seconds?
32·1666 × 7 = 225·1662 feet.
3. The velocity in feet per second that a body will acquire, in falling through a given space, equals the square root of the product of the time multiplied by 64·3333.
_Example._—The space through which a body has fallen is 201 feet; required its velocity at the end of the fall?
√(64·3333 × 201) = √(12931) = 1137 feet.
SPECIFIC GRAVITIES OF SEVERAL SOLID, AND FLUID BODIES.
Air,[54] in a mean state 1·232 Brass, cast 8000 Brick 2000 Coal[54] 1250 Copper 9000 Cork 240 Clay 2160 Earth, common 1984 Flint 2570 Gold, standard 18888 Gun metal 8784 Gunpowder—solid 1745 ” loose 868 Granite 3000 Iron, cast 7425 Lead 11325 Pitch 1150 Sand[54] 1520 Silver, standard 10535 Steel 7850 Stone, common 2520 Tin 7320 Water, rain 1000 [54] sea 1030 Wood—alder 800 ash, the trunk 845 beech 852 elm, and larch 540 fir, Riga, & maple 750 pine, pitch & red 660 oak 950 walnut 671
These numbers represent the weight of a cubic foot (or 1728 cubic inches) of each of the bodies in ounces (avoirdupois).
_To find the magnitude of any body from its weight._
As the tabular specific gravity of the body is to its weight in avoirdupois ounces; so is one cubic foot (or 1728 cubic inches) to its content in feet, or inches, respectively.
_To find the weight of a body, from its magnitude._
As one cubic foot (1728 cubic inches) is to the content of the body; so is its tabular specific gravity to the weight of the body.
_To find the specific gravity of a body._
1.—_When the body is heavier than water._
Weigh it both in water, and out of water, and take the difference:
Then,—As the weight lost in water is to the whole or absolute weight; so is the specific gravity of water to the specific gravity of the body.
2.—_When the body is lighter than water_, so that it will not sink, annex to it another body heavier than water, so that the mass compounded of the two may sink together. Weigh the denser body, and the compound mass separately, both in water, and out of it; then find how much each loses in water, by subtracting its weight in water from its weight in air; and subtract the less of these remainders from the greater.
Then,—As the last remainder is to the weight of the light body in air; so is the specific gravity of water to the specific gravity of the body.
3.—_For a fluid of any sort._
Take a piece of a body of known specific gravity, weigh it both in, and out of the fluid, finding the loss of weight by taking the difference of the two:
Then,—As the whole or absolute weight is to the loss of weight; so is the specific gravity of the solid to the specific gravity of the fluid.
_To find the quantities of two ingredients in a given compound._
Take the three differences of every pair of the three specific gravities, namely, the specific gravities of the compound, and each ingredient, and multiply each specific gravity by the difference of the other two:
Then,—As the greatest product is to the whole weight of the compound; so is each of the other two products to the weights of the two ingredients.
_To find the diameter of any small sphere, or globule, whose specific gravity is given_ (_or can be found in the Table_) _and weight known._
Divide its weight in grains by the number expressing its specific gravity; extract the cube root of this quotient, and multiply it by 1·9612 for the diameter.
WEIGHT OF A CUBIC FOOT OF THE FOLLOWING MATERIALS,
_in pounds_.
Ash 49 Beech 43 Birch 49 Box 60 Cork 15 Elm 36 Fir 30 Mahogany, Spanish 50 Pine, red 41 Teak 41 Walnut 41 Coke 46 Clay 125 Earth, loose 95 Gravel 120 Granite 166 Brick, common 98 Chalk 145 Coal, Newcastle 78 Antimony 418 Brass, cast 525 Copper 538 Gold, pure 1203 Iron, cast, variable 444 Lead 717 Silver, standard 644 Tin 455
By means of the foregoing table, the weight of any quantity of the materials specified (in cubic feet) may readily be found.
MOTION, FORCES, &c.
_Body_ is the mass or quantity of matter in any material substance, and it is always proportional to its weight, or gravity, whatever its figure may be.
_Density_ is the proportional weight, or quantity of matter in any body.
_Velocity, or celerity_, is an affection of motion by which a body passes over a certain space in a certain time.
_Momentum, or quantity of motion_, is the power, or force, in moving bodies.
_Force_ is a power exerted on a body to move it, or to stop it. If the force act constantly, it is a _permanent force_, like pressure, or the force or gravity; but if it act instantaneously, or for an imperceptibly short time, it is called _impulse_, or _percussion_, like the smart blow of a hammer.
_A motive, or moving force_, is the power of an agent to produce motion.
_Accelerative, or retardative force_, is that which affects the velocity only, or it is that by which the velocity is accelerated, or retarded.
The change, or alteration of motion by any external force, is always proportional to that force, and in the direction of the right line in which it acts.
_If a body be projected_ in free space, either parallel to the horizon, or in an oblique direction, by the force of gunpowder, or any other impulse: it will, by this motion, in conjunction with the action of gravity, describe the curve line of a parabola.
_A parabola_ is the section formed by cutting a cone, with a plane, parallel to the side of the cone.
_Gravity_ (_vide page 316_) is a force of such a nature that all bodies, whether light or heavy, fall perpendicularly through equal spaces in the same time, abstracting the resistance of the air; as lead, and a feather, which, in an exhausted receiver, fall from the top to the bottom in the same time. The velocities acquired by descending, are in the exact proportion of the times of descent, and the spaces descended are proportional to the squares of the times, and, therefore, to the squares of the velocities. Hence, then, it follows that the weights, or gravities of bodies near the surface of the earth are proportional to the quantities of matter contained in them; and that the spaces, times, and velocities generated by gravity, have the relations contained in the three general proportions before laid down.
A body in the latitude of London falls nearly 16-1/12 feet in the first second of time, and consequently, at the end of that time, it has acquired a velocity double, or of 32⅙ feet.
The times being as the velocities, and the spaces as the squares of either; therefore,
if the times be as the Nos. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10; the velocities will also be as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10; and the spaces as their squares 1, 4, 9, 16, 25, 36, 49, 64, 81, 100; and the spaces for each time, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.
Namely, as the series of the odd numbers, which are the differences of the squares denoting the whole spaces. So that if the first series of natural numbers be seconds of time,
namely: the times in seconds 1 2 3 4 &c. the velocities in feet will be 32⅙ 64⅓ 96½ 128⅔, &c. the spaces in the whole times 16-1/12 64⅓ 144¾ 257⅓, &c. and the space for each second 16-1/12 48¼ 80-5/12 112-7/12, &c.
of which spaces the common difference is 32⅙ feet, the natural and obvious measure of the force of gravity.
Thus, a body falling from a state of rest acquires a velocity to pass through 9 spaces in the fifth second of time; 7 in the fourth; 5 in the third; 3 in the second; and 1 in the first. Thus it is 9 + 7 + 5 + 3 + 1 = 25, which shows that the whole spaces passed through in 5 seconds equal the square of 5.
_The momentum_, or force, of a body falling through the atmosphere is the mass or weight, multiplied by the square root of the height it has fallen through, multiplied by 8·021.
Suppose a weight of 10 tons to be raised 9 feet, and to drop thence suddenly on a bridge; the momentum is 10 × (3 × 8·021) = 240·63 tons. That is, a weight of 10 tons, so falling, would exert as great a strain to break down the bridge, as the pressure of 240·63 tons of dead weight.
Thus, a one-ounce ball falling from a height of 400 feet, would strike the earth with a momentum of
oz. feet. oz. lb. 1 × (20 × 8·021) = 160·42 = 10·026.
By experiments to ascertain the effect of Carnot’s vertical fire, it was found that 4-oz. balls only penetrated ½0 of an inch into deal board, and from 2 to 3 inches into meadow ground.
_Amplitude_ signifies the range of a projectile, or the right line upon the ground, subtending the curvilinear path in which it moves.
_The time of flight_ of different shot, and shells is equal to the time a heavy body takes to descend freely from the highest point described by the curve of the projectile.
_To find the time of descent_:
Divide the given height, or altitude, by 16-1/12, and the square root of the quotient will be the time required. Thus, if the altitude is 1200 feet, and the time of descent is required,
1200 ÷ 16-1/12 = 74·61, the square root of which is 8·637, the time required.
When a body is projected vertically _downwards_ with a given velocity, the space described is equal to the time multiplied by the velocity, together with the product of 16-1/12 by the square of the time; but, if the body is projected _upwards_, the latter product must be subtracted from the former.
PRACTICAL GEOMETRY.
DEFINITIONS.[55]
_A line is perpendicular to another_ when it inclines not more on the one side than on the other, the angles on both sides being equal.
_Parallel lines_ are those which have no inclination to each other, being everywhere equi-distant, however far produced, or extended.
_An angle_ is the inclination, or opening of two lines, which meet in a point called the _vertex, or angular point_: and the two lines are called the _legs, or sides_ of the angle.
_The measure of an angle_ is estimated by the number of degrees contained in the arc between its two legs.
_A rectilinear angle_ has its legs or sides, _right_, or straight lines.
_A curvilinear angle_ has its legs _curves_.
_A right angle_ is formed by one line perpendicular to another; the measure of which is an arc of 90°.
_An acute angle_ is less than a right angle, or than 90°.
_An obtuse angle_ is greater than a right angle.
_An oblique angle_ may be either acute, or obtuse.
_The circumference, or periphery of a circle_ is the curved line which bounds it, being everywhere equally distant from the _centre_. The circumference is supposed to be divided into 360 degrees (marked thus °); each degree into 60 minutes, each minute (′) into 60 seconds (″).
_An arc_ is any part of the circumference of a circle.
_A chord, or subtense_, is a right line joining the extremities of an arc.
_The radius of a circle_ is a right line drawn from the centre to the circumference.
_The diameter of a circle_ is a right line drawn through the centre, and terminated by the circumference.
_A semicircle_ (180°) is that part of a circle which is contained between the diameter, and half the circumference.
_A quadrant_ is the fourth part of a circle, being contained between two radii, and an arc of 90°.
_A segment_ is that part of a circle which is cut off by a chord.
_A sector_ is that part of a circle contained between two radii, and an arc.
_A secant_ is a line which cuts a circle, lying partly within, and partly without it.
_A tangent_ is a line which touches a circle, or curve, without cutting it.
_The point of contact_ is where a tangent touches an arc.
_Triangles_ are figures having three sides, and three angles.
_An equilateral triangle_ has its three sides equal.
_An isosceles triangle_ has only two equal sides.
_A scalene triangle_ has all its sides unequal.
_A rectangular, or right-angled triangle_ has one of its angles a right one, or 90°; and the square of the side opposite the right angle is equal to the sum of the squares of the sides containing that angle; hence a triangle, having its sides proportional to the numbers 3, 4, 5, will be right-angled.
_The hypothenuse_ is the side opposite the right angle in a rectangular triangle.
_An obtuse-angled triangle_ has one of its angles obtuse.
_An acute-angled triangle_ has all its angles acute.
_The three angles of any triangle_, taken together, are equal to two right angles, or 180°.
_The difference of the squares of two sides of a triangle_ is equal to the product of their sum and difference.
_The sides of a triangle are proportional_ to the sines of their opposite angles.
_Quadrangles, or quadrilaterals_, are plane figures bounded by four right lines.
_A square_ is a quadrilateral having all its sides equal, and all its angles right angles. The _diagonal of a square_ is equal to the square root of twice the square of its sides: and _the side of the square_ is equal to the square root of half the square of its diagonal.
_The diagonal_ is a right line drawn across a quadrilateral figure, from one angle to another. The sum of the squares of the two diagonals of every parallelogram is equal to the sum of the squares of the four sides.
_A parallelogram_ is a quadrilateral, whose opposite sides are parallel.
_A rectangle_ is a parallelogram having four right angles.
_A rhomboid_ is an oblique-angled parallelogram.
_A rhombus, or lozenge_, is a quadrilateral, whose sides are all equal but its angles oblique.
_A trapezium_ is a quadrilateral, which has none of its sides parallel to each other.
_A trapezoid_ is a quadrilateral, which has only two of its sides parallel.
_Polygons_ are plane figures bounded by more than four sides.
_A regular polygon_ has all its sides, and angles equal.
_The perimeter_ of a figure is the sum of all its sides.
_To bisect_—is to divide into two equal parts.
_To trisect_—is to divide into three equal parts.
_To inscribe_—is to draw one figure within another, so that all the angles of the inner figure touch either the angles, sides, or planes of the external figure.
_To circumscribe_—is to draw a figure round another, so that either the angles, sides, or planes of the circumscribing figure touch all the angles of the figure within it.
LINES, ANGLES, AND FIGURES.
_To divide a given right line into two equal parts._
From the extremities of the line as centres, and with any opening in the compasses, greater than half the given line, as a radius, describe arcs intersecting each other above, and below the given line. A line being drawn through these intersections will divide the given line into two equal parts.
_An arc of a circle_ is bisected in the same manner.
_To bisect an angle._
From the angular point, measure equal distances on the two lines (forming the angle), and from these points, with the same distance as radius, describe arcs intersecting each other. A line drawn from their intersections to the angular point will bisect the angle.
_To erect a perpendicular._
From the point A set off any length 4 times to C; from A as a centre with 3 of those parts describe an arc at B, and from C with 5 of them cut the arc at B. Draw A B, which will be the perpendicular required. Any equimultiples of these numbers, 3, 4, 5, may be used for erecting a perpendicular. _Plate 2_, HEIGHTS AND DISTANCES, and PRACTICAL GEOMETRY, _Fig. ½_.
_To erect a perpendicular._
Set off on each side of the point A, any two equal distances, A D, A E. From D and E as centres, and with any radius greater than half D E, describe two arcs intersecting each other in F. Through A, and F draw the line A F, and it will be the perpendicular required.
_Fig. 1.—Plate_, PRACTICAL GEOMETRY.
_To let fall a perpendicular._
From D as a centre, and with any radius, describe an arc intersecting the given line. From the points of intersection C, and E, with any radius greater than half, describe two arcs, cutting each other at F. Through D, and F draw a line, and D F will be the perpendicular required. _Fig. 2._
_To draw a line parallel to a given line._
From any point D in the given line with the radius D C, describe the arc C E, and from C with the same radius describe the arc D F. Take E C, and set it off from D to F. Through C, and F draw C F for the parallel required. _Fig. 3._
_To divide an angle into two equal parts._
From B as a centre with any radius describe an arc A C. From A, and C with any radius describe arcs intersecting each other in D. Then draw B D, and it will bisect the angle. _Fig. 4._
[Illustration: PRACTICAL GEOMETRY.
_Fig. 1-9._]
[Illustration:_Fig. 10-14._]
_To divide a right angle into three equal parts._
From B as a centre with any radius describe the arc A C. From A with the radius A B cut the arc A C in D, and with the same radius from C cut it in E. Then through the intersections D, and E draw the lines B D, B E, and they will trisect, or divide the angle into three equal parts. _Fig. 5._
_To find the centre of a circle._
Draw any chord A B, and bisect it by the perpendicular C D. Divide C D into two equal parts, and the point of bisection O will be the centre required. _Fig. 6._
_To describe an equilateral triangle._
From the points A, B, as centres, and with A B as radius, describe arcs intersecting each other in C. Draw C A, C B, and the figure A B C will be the triangle required. _Fig. 7._
_To describe a square._
From the point B, draw B C perpendicular, and equal to A B. On A, and C, with the radius A B, describe arcs cutting each other in D. Draw the lines D A, D C, and the figure A B C D will be the square required. _Fig. 8._
_To inscribe a square in a circle._
Draw the diameters A B, C D perpendicular to each other. Then draw the lines A D, A C, B D, B C; and A B C D will be the square required. _Fig. 9._
_To inscribe an octagon in a circle._
Bisect any two arcs A C, B C of the square A B C D in G, and E. Through the points G, and E, and the centre O draw lines, which produce to F, and H. Join A F, F D, D H, &c. and they will form the octagon required. _Fig. 9._
_On a line to describe all the several polygons, from the hexagon to the dodecagon._
Bisect A B by the perpendicular C D. From A as a centre, and with A B as a radius, describe the arc B E, which divide into six equal parts; and from E as a centre describe the arcs 5 F, 4 G, 3 H, &c. Then from the intersection E as a centre, and with E A as a radius, describe the circle A I D B, which will contain A B six times. From F in like manner as a centre, and with F A as radius, describe the circle A K L B, which will contain A B seven times; and so on for the other polygons. _Fig. 10._
_To inscribe in a circle an equilateral triangle._
From any point D in the circumference as a centre, and with the radius D O of the given circle, describe an arc A O B cutting the circumference in A, and B. Through D, and O draw D C. Then, join A B, A C, B C; and the figure A B C will be the triangle required. _Fig. 11._
_To inscribe a hexagon in a circle._
Bisect the arcs A C, B C in E, and F, and join A D, D B, B F, &c., which will form the hexagon. Or carry the radius six times round the circumference, and the hexagon will be obtained. _Fig. 11._
_To inscribe a dodecagon in a circle._
Bisect the arc A D of the hexagon in G, and A G being carried twelve times round the circumference, will form the dodecagon. _Fig. 11._
_To inscribe a pentagon, hexagon, or decagon, in a circle._
Draw the diameter A B, and make the radius D C perpendicular to A B. Bisect D B in E. From E as a centre, and with E C as radius, describe an arc cutting A D in F. Join C F, which will be the side of the pentagon, C D that of the hexagon, and D F that of the decagon. _Fig. 12._
_To find the angles at the centre, and circumference of a regular polygon._
Divide 360 by the number of the sides of the given polygon, and the quotient will be the angle at the centre; and this angle being subtracted from 180, the difference will be the angle, at the circumference, required.
_Table, showing the angles at the centre, and circumference._
Names. No. of Angles Angles at sides. at centre. circumference. Trigon 3 120° 60° Tetragon 4 90° 90° Pentagon 5 72° 108° Hexagon 6 60° 120° Heptagon 7 51° 25-5/7′ 128° 34-2/7′ Octagon 8 45° 135° Nonagon 9 40° 140° Decagon 10 36° 144°
_To inscribe any regular polygon in a circle._
From the centre C draw the radii C A, C B, making an angle equal to that at the centre of the proposed polygon, as contained in the preceding table. Then the distance A B will be one side of the polygon, which, being carried round the circumference the proper number of times, will complete the polygon required. _Fig. 13._
[Illustration: _Fig. 15-20._]
_To circumscribe a circle about a triangle._
Bisect any two of the given sides, A B, B C by the perpendiculars E F, D F. From the intersection F as a centre, and with the distance of any of the angles, as a radius, describe the circle required. _Fig. 14._
_To circumscribe a circle about a square._
Draw the two diagonals A C, B D intersecting each other in O. From O as a centre, and with O A, or O B, as a radius, describe the required circle. _Fig. 15._
_To circumscribe a square about a circle._
Draw the two diameters A B, C D perpendicular to each other, through the points A, C, B, D, draw the tangents E F, E G, G H, F H, and E G H F will be the square required. _Fig. 16._
_To reduce a map, or plan, from one scale to another._
Divide the given figure A C by cross lines, forming as many squares as may be thought necessary. Draw a line E F, on which set off as many parts from the scale M, as A B contains parts of the scale N. Draw E H, and F G perpendicular to E F, and each equal to the proportional parts contained in A D, or B C. Join H G, and divide the figure E G into the same number of squares as the original A C. Describe in every square what is contained in the corresponding square of the given figure; and E F G H will be the reduced plan required. The same operation will serve either to reduce, or enlarge any map, plan, drawing, or painting. _Fig. 17._
MENSURATION OF PLANES, AND SOLIDS.
_Mensuration is of three kinds_, viz., lineal, superficial, and solid.
_Lineal measure_ has reference to length only.
_Superficial measure_ (_or the surface_) includes length, and breadth.
_Solid measure_ (_or the content_) comprehends length, breadth, and thickness.
MENSURATION OF PLANES.
_The area_ of any plane figure is the superficial measure contained within its extremes, or bounds. This area is estimated by the number of small squares that may be contained in it, the side of these measuring squares being an inch, a foot, or any other fixed quantity, and hence the area is said to be so many square inches, square feet, &c. _Vide Table, Square measure._ _Page 275._
_To find the area of a parallelogram, whether a square, rectangle, &c._
Multiply the length by the breadth, or perpendicular height, for the area required.
_Example._—Required the area of a rectangle, whose length is 9 feet, and breadth 4 feet.
9 × 4 = 36 feet. The required area, or surface.
_To find the area of a triangle, its base, and perpendicular height being given._
Multiply the base by the perpendicular height, and half the product will be the area.
_Example._—Required the number of square yards contained in a triangle, whose base is 20 yards, and perpendicular height 14 yards.
(20 × 14) / 2 = 140 square yards. Area required.
_To find the area of a triangle, whose three sides are given._
From half the sum of the three sides, subtract each side severally; multiply the half sum, and the three remainders together, and the square root of the product will be the area required.
_Example._—Required the area of a triangle, whose sides are 50, 40, and 30 feet.
(50 + 40 + 30) / 2 = 60, half the sum of the three sides. 60 - 30 = 30 First difference. 60 - 40 = 20 Second difference. 60 - 50 = 10 Third difference. 30 × 20 × 10 × 60 = 360000. Square root of 360000 = 600. Area required.
_Two sides of a right-angled triangle being given, to find the third side._
1. When the two sides forming the right angle are given, to find the hypothenuse, or side opposite the right angle.
Take the square root of the sum of the two sides squared for the side required.
_Example._—Required the length of the interior slope of a rampart, whose perpendicular height is 17 feet, and the base of the slope 20 feet.
17 × 17 = 289 20 × 20 = 400 ----- The square root of 689 = 26·24. The length required.
2. When the hypothenuse, and one of the perpendicular sides are given.
From the square of the hypothenuse, subtract the square of the given side, and the square root of the remainder will be the side required.
_Example._—The hypothenuse being 5 yards, and the base 4 yards, required the other side.
5 × 5 = 25 4 × 4 = 16 ------------ The square root of 9 = 3 yards. The side required.
_To find the area of a trapezium, A B C D._
Draw the diagonal A C, upon which let fall from its opposite angles B, and D, the perpendiculars B F, D E. Find by measurement the diagonal A C, and the perpendiculars B F, D E, then multiply the sum of the perpendiculars by the diagonal, and half the product will be the area of the trapezium. _Fig. 18._
_Example._—Required the area of the trapezium, whose diagonal A C is 100 feet, and perpendiculars B F 30 feet, and D E 40 feet.
((30 + 40) × 100) / 2 = 3500 square feet. Area required.
Or, divide the trapezium into two triangles by a diagonal, then find the areas of these triangles, and add them together.
_To find the area of a trapezoid, A B C D._
Multiply the sum of the parallel sides A B, D C by the perpendicular distance E C, and half the product will be the area. _Fig. 19._
_Example._—Required the area of the trapezoid A B C D, of which the parallel sides A B, D C are 120 feet, and 90 feet, and the perpendicular distance E C 40 feet.
((120 + 90) × 40) / 2 = 4200 square feet. Area required.
_To find the area of an irregular figure, or polygon._
Draw diagonals dividing the figure into trapeziums, and triangles; then, having found the area of each, add them together, and the sum will be the area required.
_To find the area of a figure, having a part bounded by a curve._
Draw a right line joining the extremities of the curve, then find the area of the trapezium. On the right line let fall as many perpendiculars as the several windings of the curve may require. Find their lengths, and divide their sum by the number of perpendiculars, and the quotient will be the mean breadth; which being multiplied by the length of the right line, will give the area of the curved part. This area being added to that of the trapezium will give the area of the required figure.
_To measure long irregular figures._
Measure the breadth at both ends, and at several places _at equal distances_. Add together all these intermediate breadths, and half the two extremes, which sum multiply by the length, and divide by the number of parts for the area. If the perpendiculars, or breadths, _be not at equal distances_, compute all the parts separately, as so many trapezoids, and add them all together for the whole area.
_Example._—The breadths of an irregular figure at five equi-distant places being 8, 2, 7, 9, 4, and the whole length 40, required the area.
8 + 4 = 12 12 ÷ 2 = 6 6 + 2 + 7 + 9 = 24 (24 × 40) / 4 = 240. Area required.
_To find the number of square acres in any of the preceding figures._[56]
Divide the superficial content in feet by 43560, and the quotient will be the number required.
_To bring square chains to acres._
Of square chains strike off two decimal places to the right, and the rest of the figures will be acres.
_To bring square links to acres._
Of square links cut off five of the figures on the right hand, for decimals, and the rest will be acres; then multiply these decimals by 4, for roods, cutting off five figures as before; and the decimals of these again by 40, for perches, when five figures are again to be struck off.
_To find the area of a regular polygon._
Multiply the _perimeter_ (or sum of the sides) of the polygon by the perpendicular drawn from its centre on one of its sides, and take half the product for the area.
Or, multiply the area of one of the triangles by the number of sides of the polygon, and the product will be the area of it.
_Example._—Required, the area of a regular hexagon, whose side is 40 feet, and the perpendicular 34·64 feet.
40 × 6 = 240 the perimeter. (240 × 34·64) / 2 = 4156·8 square feet. Area required.
_To find the diameter, and circumference of any circle, the one from the other._
Use either of the following proportions:
as 7 is to 22 } { so is the diameter or as 1 is to 3·1416 } { to the circumference.
as 22 is to 7 } { so is the circumference or, as 3·1416 is to 1 } { to the diameter.
or, instead of dividing the diameter by 3·1416, multiply it by ·3183, for the circumference.
_Example 1._—Required, the circumference of a circle, whose diameter is 20 feet.
As 7 : 22 :: 20 : 62·857 feet. Circumference required.
_Example 2._—Required, the diameter of a circle, whose circumference is 36 inches.
As 22 : 7 :: 36 : 11·45 inches. Diameter required.
_To find the diameter of a circle, the area being given._
Divide the area by ·7854, and the square root of the quotient will be the diameter required.
_Example._—Required, the diameter of a circle, whose area is 176·715 square feet.
176·715 ÷ ·7854 = 225. Square root of 225 = 15 feet. Diameter required.
_To find the area of a circle._
1. Multiply half the circumference by half the diameter, or multiply the whole circumference by the whole diameter, and take ¼ of the product.
2. Or, square the diameter, and multiply that square by ·7854 for the area.
3. Or, square the circumference, and multiply that square by 0·7958.
_Example 1._—Required the area of a circle, whose circumference is 55·548 inches, and its diameter 18 inches.
55·548 / 2 = 27·774 half circumference. 18 / 2 = 9 half diameter. 27·774 × 9 = 249·966, square inches. Area required.
_Example 2._—Required the area of a circle whose diameter is 12 feet.
12 × 12 = 144, square of the diameter. ·7854 × 144 = 113·0976 square feet. Area required.
_Example 3._—Required the area of a circle, whose circumference is 22 feet.
22 × 22 = 484. 484 × ·07958 = 38·51672 square feet. Area required.
_To find the area of a circular ring_,
or space included between the circumferences of two circles, the one within the other.
1. Subtract the square of the less diameter from the square of the greater, and multiply their difference by ·7854.
2. Or, find the area of each circle separately, and subtract one from the other, for the area required.
3. Or, multiply the sum of the diameters by the difference of the same, and that product by ·7854 for the area.
_Example._—Required the area of a ring, the diameters of whose bounding circles are 10, and 20.
_By Rule 3._
20 + 10 = 30, sum of diameters. 20 - 10 = 10, difference of diameters. 30 × 10 × ·7854 = 235·62. The area.
_To find the length of any arc of a circle._
1. As 360° is to the number of degrees in the arc, so is the circumference to the length of the arc.
2. Or, multiply the degrees in the given arc by the radius of the circle, and the product by ·01745 for the length of the arc.
_Example._—_Rule 2._—Required the length of an arc of 30°, the radius being 9 feet.
30 × 9 × ·01745 = 4·7115. Length of arc.
_To find the area of the sector of a circle._
Multiply the radius by the arc, and half the product will be the area.
_Example._—Required the area of the sector, whose radius is 30 inches, and the length of the arc 36·6 inches.
(36·6 × 30) / 2 = 549 square inches. Area required.
_To find the area of the segment of a circle._
Find the area of the sector, by the preceding rule. Then find the area of the triangle formed by the chord of the segments, and the radii of the sector. Then, if the segment be less than a semicircle, subtract the area of the triangle from it; or, if the segment be greater than a semicircle, add the area of the triangle to it; for the area of the segment.
_Example._—Required the area of a segment less than a semicircle, the radius being 20 inches, the chord 22·42 inches, the length of the arc 24·43 inches, and the perpendicular 16·56 inches.
(24·43 × 20) / 2 = 244·3 square inches. Area of the sector. (22·42 × 16·56) / 2 = 185·6376 square inches. Area of the triangle. 244·3 - 185·6376 = 58·6624 square inches. Area required.
_To find the area of a semicircle._
1. Multiply ¼ of the circumference by the radius, and the product will be the area.
2. Or, multiply the square of the diameter by ·7854, and half the product will be the area.
_Example._—_Rule 2._—Required the area of a semicircle, the diameter being 50 inches.
(50 × 50 × ·7854) / 2 = 981·75 square inches. Area required.
_To find the area of an ellipsis, or oval._
Multiply the longest diameter, or axis, by the shortest, then multiply the product by ·7854 for the area.
_Example._—Required the area of the ellipse, whose diameters are 25 inches, and 18 inches.
25 × 18 × ·7854 = 353·43 square inches. Area required.
_To find the area of a parabola, or its segment._
Multiply the base by the perpendicular height, and take two-thirds of the product for the area.
_Example._—Required the area of a parabola, whose base is 20 feet, and height 12 feet.
20 × 12 = 240 ⅔ of 240 = 160 square feet. Area required.
MENSURATION OF SOLIDS.
_A solid_ is a body containing length, breadth, and thickness.
_Solids are measured_ by cubes, whose sides are each an inch, a foot, a yard, &c., and the solidity, capacity, or content of any figure is computed by the number of such cubes as are contained in it.—_Vide Cubic measure, page 276._
_A cube_ is a solid contained by six equal square sides.
_A pyramid_ is a solid whose sides are all triangles meeting together in a point, the base being any plane figure whatever. It is called a triangular pyramid when its base is a triangle; a square pyramid when its base is a square, &c.
_The segment of a pyramid, cone, or any other solid_ is a part of D E F G cut off from the top by a plane D E F, parallel to the base A B C.—_Vide Fig. 21, Plate 2_, HEIGHTS, DISTANCES, and PRACTICAL GEOMETRY.
_A frustrum, or trunk_, is a part A B C D E F, that remains at the bottom after the segment is cut off.
_A cone_ is a round pyramid, of which the base is a circle.
_The axis of a solid_ is a line from the vertex (or point) to the centre of the base, or through the centres of the two ends. When the axis is perpendicular to the base, it is a right prism, pyramid, or cone; otherwise it is oblique.
_A sphere_ is a solid contained under one convex surface, and is described by the revolution of a semicircle about its diameter, which remains fixed.
_The centre of the sphere_ is such a point within the solid as is everywhere equally distant from the convex surface, or circumference of it.
_The diameter_ (_or axis_) _of a sphere_ is a straight line, which passes through the centre, and is terminated by the convex surface.
_A segment of a sphere_ is a part cut off by a plane, the section of which is always a circle, called the _base of the segment_.
_A sector of a sphere_ is that which is composed of a segment (less than an hemisphere) and of a cone.
_A prism_ is a solid, the sides of which are parallelograms, having its ends equal, and similar plane figures.
_Prisms are named_ according to the number of angles in the base.
_A cylinder_ is a solid, the two ends of which are circular; and it is described, or formed, by the revolution of a right-angled parallelogram about one of its sides, which remains fixed.
_To find the superficies of a prism, or cylinder._
Multiply the perimeter of one end of the prism by the length, or height of the solid, and the product will be the surface of all its sides. To which add also the area of the two ends of the prism, when required.
Or, compute the areas of all the sides, and ends separately, and add them all together.
_Example._—Required the surface of a cube, whose sides are each 5 inches.
5 + 5 + 5 + 5 = 20, perimeter of one end. 20 × 5 = 100, surface of sides. 5 × 5 = 25, area of one end. 100 + 25 + 25 = 150 square inches. Surface of cube.
_To find the surface of a pyramid, or cone._
Multiply the perimeter of the base by the slant height, or length of the side, and half the product will be the surface of the sides; to which add the area of the base when required.
_Example._—Required the upright surface of a triangular pyramid, the slant height being 20 feet, and each side of the base 3 feet.
3 + 3 + 3 = 9, perimeter of base. (9 × 20) / 2 = 90 feet. Surface required.
_To find the surface of the frustrum of a pyramid, or cone._
Add together the perimeters of the two ends, multiply their sum by the slant height, and take half the product.
_Example._—How many square feet are in the surface of the frustrum of a square pyramid, whose slant height is 10 feet, each side of the base 3 feet, and each side of the less end 2 feet.
3 + 3 + 3 + 3 = 12, perimeter of base. 2 + 2 + 2 + 2 = 8, perimeter of less end. ((12 + 8) × 10) / 2 = 100 feet. Surface required.
_To find the solid content of a prism or cylinder._
Find the area of the base, or end, and multiply it by the length of the prism, or cylinder. Fora cube, multiply its side twice by itself; and for a parallelopipedon, multiply the length, breadth, and depth together for the content.
_Example._—Required the solid content of a cube, whose side is 24 inches.
24 × 24 × 24 = 13824 square inches. Content required.
_To find the content of the solid part of a hollow cylinder._
From the content of the whole cylinder considered as a solid, subtract the content of the hollow part, also considered as a solid, and the difference will be the solidity required.
_Example._—Required the content of the solid part of the hollow cylinder whose exterior diameter is 12 inches, the interior diameter 8 inches, and height 20 inches.
12 × 12 × ·7854 = 113·0976, area of base of cylinder. 113·0976 × 20 = 2261·952, solidity of whole cylinder. 8 × 8 × ·7854 = 50·2656, area of base of hollow cylinder. 50·2656 × 20 = 1005·312, content of hollow part. 2261·952 - 1005·312 = 1256·64 cubic inches. Solidity required.
_To find the solidity of the frustrum of a cylinder._
Multiply the area of the base by half the greatest, and the least lengths, and the product will be the solidity.
_Example._—Required the solidity of a frustrum, whose diameter is 24 inches, the greatest length 36 inches, and the least length 20 inches.
24 × 24 = 576. Square of the diameter. 576 × ·7854 = 452·3904. Area of the base. 452·3904 × (36 + 20) / 2 = 12666·9312 Cubic inches. Solidity required.
_To find the content of a pyramid, or cone._
Find the area of the base, and multiply that area by the perpendicular height, and take ⅓ of the product.
_Example._—Required the solidity of a square pyramid, each side of its base being 30, and its perpendicular height 25.
30 × 30 = 900, area of base. (900 × 25) / 3 = 7500, solidity required.
_To find the solidity of the frustrum of a cone, or pyramid._
Add into one sum the areas of the two ends, and the mean proportional between them: take ⅓ of that sum for the mean area, which multiply by the perpendicular height, or length of the frustrum.
_Note._—_To find a mean proportional._
As one of the sides of the base is to the homologous, or corresponding side of the other end, so is the area of the base to the mean proportional required.
_Example._—Required the number of solid feet in a piece of timber, whose bases are squares, each side of the greater end being 15 inches, and each side of the less end 6 inches; also the length of the perpendicular altitude 24 feet.
15 × 15 = 225, area of the base. 6 × 6 = 36, area of the top. As 15: 6 :: 225: 90, mean proportional. 24 feet = 288 inches. ((225 + 36 + 90) × 288) / 3 = 33696 cubic inches = 19½ cubic feet.
_To find the surface of a sphere, or any segment._
Multiply the circumference of the sphere by its diameter, which will give the whole surface.
Or, square the diameter, and multiply by 3·1416. Or, square the circumference, and multiply by ·3183; or divide by 3·1416.
_Note._—_For the surface of the segment, or frustrum_, multiply the whole circumference of the sphere by the height of the part required.
_Example._—Required the superficies of a globe whose diameter is 24 inches.
24 × 24 × 3·1416 = 1809·5616 square inches.
_To find the solidity of a sphere, or globe._
1. Multiply the surface by the diameter, and take ⅙ of the product.
Or, multiply the square of the diameter by the circumference, and take ⅙ of the product.
2. Cube the diameter, and multiply by ·5236.
3. Cube the circumference, and multiply by ·01688.
_Example._—Required the content of a sphere, whose axis is 12.
12 × 12 × 12 × ·5236 = 904·7808. Content required.
_To find the solidity of an hemisphere._
Find the solidity of the sphere, and half the content will be that of the hemisphere.
_Note 1._—Any sphere, or globe _twice_ the diameter of another contains _four times_ the superficies, or area of the other, and _eight times_ the solid content. Hence the superficies of spheres are as the squares, and the solidity as the cubes of their diameters.
_Note 2._—The cube of the diameter of a sphere in inches, multiplied by ·00188, will give the number of _imperial gallons it will contain_.
_To find the solid content of a spherical segment._
1. From three times the diameter of the sphere, take double the height of the segment; then multiply the remainder by the square of the height, and this product by ·5236.
2. Or, to three times the square of the radius of the segment’s base add the square of its height; then multiply the sum by the height, and the product by ·5236.
_Example._—Required the content of a spherical segment 2 feet in height, cut from a sphere of 8 feet diameter.
(3 × 8) - (2 × 2) = 20 20 × 2^2 × ·5236 = 41·888 cubic feet. Content required.
_To find the diameter of a sphere, its solidity being given._
Divide the solidity by ·5236, and take the cube root of the quotient.
_Example._—The solidity of a sphere being 113·0976 solid inches, what will be its diameter?
113·0976 / ·5236 = 216, the cube root of which is 6 inches, the diameter required.
_To find the weight of an iron shot, its diameter being given._
Take ⅛ of the cube of the diameter, and ⅛ of that eighth, and the sum of these two quotients will be the weight in pounds.
Or, as 64 is to 9 lb. so is the diameter cubed to its weight.
_Example._—Required the weight of an iron shot whose diameter is 3·5 inches?
3·5 cubed = 42·875, cube of diameter. 42·875 / 8 = 5·359 5·359 / 8 = ·669 5·359 + ·669 = 6·028 pounds. Weight required.
_To find the weight of a leaden ball, its diameter being given._
Take ⅓ of the cube of the diameter, and from it subtract ⅓ of this third, and the remainder will be the weight, nearly.
Or, take 3/14 of the cube of the diameter.
_Example._—What is the weight of a leaden ball whose diameter is 3·3 inches?
3·3 cubed = 35·937, cube of diameter. 35·937 / 3 = 11·979 11·979 / 3 = 3·993 11·979 - 3·993 = 7·986 pounds. Weight required.
_To find the diameter of an iron shot, its weight being given._
Multiply the cube root of the shot’s weight by 1·923 for the diameter.
Pr. Cube root. Diameter. 42 3·4760, &c. { Multiplied } 6·684, &c. 32 3·1748 { by 1·923, } 6·103 ” 24 2·8844 { diameter } 5·545 ” 18 2·6207 { of a } 5·038 ” 12 2·2894 { 1 lb. } 4·401 ” 9 2·0800 { shot. } 3·999 ” 6 1·8171 { } 3·494 ” 3 1·4422 { } 2·772 ”
_To find the diameter of a leaden ball, its weight being given._
To 4 times the weight add half the weight, and 3/100 of half the weight; and the cube root of this sum will be the diameter in inches, nearly.
_Example._—What is the diameter of a leaden ball, whose weight is 8 pounds?
8 x 4 = 32 8 / 2 = 4 3 / 100 of 4 = ·12. 32 + 4 + ·12 = 36·12, of which the cube root is 3·3 inches, nearly. Diameter required.
_To find the weight of an iron shell, its interior and exterior diameter being given._
Take 9/64 of the difference of the cubes of the external and internal diameters, for the weight of the shell in pounds.
_Example._—What is the weight of a shell whose exterior diameter is 12·85 inches, and interior diameter 8·75 inches?
12·85 cubed = 2121·8241, 8·75 cubed = 669·9218. 2121·8241 - 669·9218 = 1451·9022. 9/64 of 1451·9022 = 204·1737 pounds. Weight required.
_To find the quantity of powder a shell will contain._
Divide the cube of the interior diameter in inches by 57·3, and the quotient will be the weight in pounds.
Or, multiply the cube of the diameter by 11, and divide by 21 for the quantity in half ounces.
_Example._—How much powder will fill a shell, whose internal diameter is 7 inches?
7 cubed = 343. 343 / 57·3 = 6 pounds nearly. Powder required.
_To find the side of a cubical box to contain a given quantity of powder._[57]
Multiply the weight in pounds by 30, and the cube root of the product will be the side of the box in inches.
_Example._—Required the side of a cubical box to hold 50 pounds of powder?
50 × 30 = 1500, the cube root of which is 11·44, which will be the side of the box in inches.
_To find the quantity of powder to fill the chamber of a mortar, or howitzer._
Multiply the content of the chamber in inches by 55, and divide the product by 1728, and the quotient will be the quantity of powder in pounds.
_Note._—The chamber of a mortar, or howitzer, is formed of a hollow frustrum of a right cone, and of a hollow hemisphere.
_Example._—Required the quantity of powder to fill the chamber of a 13-inch mortar in which the diameter A B is 9·5 inches, the diameter C E 6·5 inches, and the length D G 21·5 inches. _Vide Fig. 22. Plate 2._ HEIGHTS AND DISTANCES, and PRACTICAL GEOMETRY.
The content of the chamber must be found by finding the content of the hollow frustrum of the cone, and that of the hemisphere (_vide preceding rules_): which in this example will be 999·9741875.
Then (999·9741875 × 55) / 1728 = 31 pounds, nearly.
_To find the quantity of powder to fill a rectangular box._
Divide the content (viz., length × breadth × depth) of the box in inches by 30 for the pounds of powder.
_Example._—How much powder will fill a box, the length being 15 inches, the breadth 12, and the depth 10 inches.
15 × 12 × 10 / 30 = 1800 / 30 = 60 pounds. Number required.
_To find the quantity of powder to fill a cylinder._
Multiply the square of the diameter by the length, then divide by 38·2 for the pounds of powder.
_Example._—How much powder will the cylinder contain, whose diameter is 10 inches, and length 20 inches?
(10 × 10 × 20) / 38·2 = 52⅓ pounds, nearly.
_To find the size of a shell, to contain a given weight of powder._
Multiply the pounds of powder by 57·3, and the cube root of the product will be the diameter in inches.
_Example._—Required the diameter of a shell to contain 6 lb. of powder?
6 × 57·3 = 343·8, the cube root of which is 7, the diameter required, in inches.
_To find what length of a cylinder (or bore of a gun) will be filled by a given weight of powder._
Multiply the weight in pounds by 38·2, and divide the product by the square of the diameter in inches, for the length.
_Example._—What length of a cylinder 8 inches in diameter will be filled with 20 lb. of powder?
(20 × 38·2) / (8 × 8) = 11-15/16 inches.
_To find the content, and weight of a piece of ordnance._
Divide the length of the gun into as many sections as may be found necessary. Find the content of each (_by preceding rules_) and from their sum subtract the content of a cylinder, whose length is equal to that of the bore, and its diameter equal to that of the calibre of the piece; multiply the difference (if it be a brass gun) by 5·0833, (if an iron gun) by 4·2968, and the product will be the weight in ounces.
_Note._—A cubic inch of gun metal weighs 5·0833 ounces. Ditto of _cast_ iron 4·2968 ounces.
_To find the content of a cask._
Multiply half the sum of the areas of the two interior circles, viz. at the head, and bung, by the interior length, for the content.
Or, to the area of the head add twice the area at the bung, multiply that sum by the length, and take one-third of the product.
_Example._—Required the content of a cask, its greatest interior diameter being 24 inches, its least interior diameter 20 inches, and the interior length 30 inches.
24 × 24 × ·7854 = 452·3904, area of large circle. 20 × 20 × ·7854 = 314·1600, area of small circle. (452·3904 + 314·1600) / 2 = 383·2752, half sum.
Then 383·2752 × 30 = 11498·256, the content; which being divided by 277¼ (the number of cubic inches in a gallon) will give the number of gallons contained in the cask.
Thus 11498·256 / 277·25 = 41·4725, &c. Number of gallons required.
_Note._-The content of any vessel in cubic feet, multiplied by 6·232 (or if in inches by ·003607) will give the number of _imperial gallons it will contain_.
EPITOME OF MENSURATION.
OF THE CIRCLE, CYLINDER, SPHERE, ETC.
1. The circle contains a greater area than any other plane figure, bounded by an equal perimeter, or outline.
2. The areas of circles are to each other as the squares of their diameters; any circle twice the diameter of another contains four times the area of the other.
3. The diameter of a circle being 1, its circumference equals 3·1416.
4. The diameter of a circle is equal to ·31831 of its circumference.
5. The square of the diameter of a circle being 1, its area equals ·7854.
6. The square root of the area of a circle, multiplied by 1·12837, equals its diameter.
7. The diameter of a circle, multiplied by ·8862, or the circumference multiplied by ·2821, equals the side of a square of equal area.
8. The sum of the squares of half the chord, and versed sine, divided by the versed sine, the quotient equals the diameter of the corresponding circle.
9. The chord of the whole arc of a circle taken from eight times the chord of half the arc, one-third of the remainder equals the length of the arc.
10. Or, the number of degrees contained in the arc of a circle, multiplied by the diameter of the circle, and by ·008727, the product equals the length of the arc in equal terms of unity.
11. The length of the arc of the sector of a circle multiplied by its radius, half the product is the area.
12. The area of the segment of a circle equals the area of the sector, minus the area of a triangle whose vertex is the centre; and base equals the chord of the segment.
13. The sum of the diameters of two concentric circles multiplied by their difference, and by ·7854, equals the area of the ring, or space contained between them.
14. The sum of the thickness, and internal diameter of a cylindric ring multiplied by the square of its thickness, and by 2·4674, equals its solidity.
15. The circumference of a cylinder multiplied by its length, or height, equals its convex surface.
16. The area of the end of a cylinder multiplied by its length, equals its solid content.
17. The area of the internal diameter of a cylinder multiplied by its depth, equals its cubical capacity.
18. The square of the diameter of a cylinder multiplied by its length, and divided by any other required length, the square root of the quotient equals the diameter of the other cylinder of equal solidity, or capacity.
19. The square of the diameter of a sphere multiplied by 3·1416 equals its convex surface.
20. The cube of the diameter of a sphere multiplied by ·5236, equals its solid content.
21. The height of any spherical segment, or zone, multiplied by the diameter of the sphere, of which it is a part, and by 3·1416, equals the area, or convex surface of the segment;
22. Or, the height of the segment multiplied by the circumference of the sphere of which it is a part, equals the area.
23. The solidity of any spherical segment is equal to three times the square of the radius of its base, plus the square of its height, and multiplied by its height, and by ·5236.
24. The solidity of a spherical zone equals the sum of the squares of the radii of its two ends, and one-third the square of its height, multiplied by the height, and by 1·5708.
25. The solidity of the middle zone of a sphere equals the sum of the square of either end, and two-thirds the square of the height, multiplied by the height, and by ·7854.
26. The capacity of a cylinder 1 foot in diameter, and 1 foot in length, equals 4·895 imperial gallons.
27. The capacity of a cylinder 1 inch in diameter, and 1 foot in length, equals ·034 of an imperial gallon.
28. The capacity of a cylinder 1 inch in diameter, and 1 inch in length, equals ·002832 of an imperial gallon.
29. The capacity of a sphere 1 foot in diameter, equals 3·263 imperial gallons.
30. The capacity of a sphere 1 inch in diameter, equals ·001888 of an imperial gallon.
31. Hence the capacity of any other cylinder in imperial gallons is obtained by multiplying the square of its diameter by its length; or the capacity of any other sphere by the cube of its diameter, and by the number of imperial gallons contained as above in the unity of its measurement.
OF THE SQUARE, RECTANGLE, CUBE, ETC.
1. The side of a square equals the square root of its area.
2. The area of a square equals the square of one of its sides.
3. The diagonal of a square equals the square root of twice the square of its side.
4. The side of a square is equal to the square root of half the square of its diagonal.
5. The side of a square, equal to the diagonal of a given square, contains double the area of the given square.
6. The area of a rectangle equals its length multiplied by its breadth.
7. The length of a rectangle equals the area divided by the breadth; or the breadth equals the area divided by the length.
8. The side, or end of a rectangle, equals the square root of the sum of the diagonal, and opposite side to that required, multiplied by their difference.
9. The diagonal in a rectangle equals the square root of the sum of the squares of the base, and perpendicular.
10. The solidity of a cube equals the area of one of its sides multiplied by the length of one of its edges.
11. The edge of a cube equals the cube root of its solidity.
12. The capacity of a 12-inch cube equals 6·232 gallons.
_Surfaces, and solidities of the regular bodies, when the linear edge is 1._
+-------------+--------------+------------+-----------+ |No. of Sides.| Names. | Surfaces. | Solids. | +-------------+--------------+------------+-----------+ | 4 | Tetrahedron | 1·7320508 | 0·1178513 | | 6 | Hexahedron | 6· | 1· | | 8 | Octahedron | 3·4641016 | 0·4714045 | | 12 | Dodecahedron | 20·6457788 | 7·6631189 | | 20 | Icosahedron | 8·6602540 | 2·1816950 | +-------------+--------------+------------+-----------+
The tabular surface multiplied by the square of the linear edge, the product equals the surface required:
Or, the tabular solidity, multiplied by the cube of the linear edge, the product is the solidity required.
OF TRIANGLES, POLYGONS, ETC.
1. The complement of an angle is its defect from a right angle.
2. The supplement of an angle is its defect from two right angles.
3. The sine, tangent, and secant of an angle, are the cosine, cotangent and cosecant of the complement of that angle.
4. The hypothenuse of a right-angled triangle being made radii, its sides become the sines of the opposite angles, or the cosines of the adjacent angles.
5. The three angles of every triangle are equal to two right angles; hence the oblique angles of a right-angled triangle are each other’s complements.
6. The sum of the squares of the two given sides of a right-angled triangle is equal to the square of the hypothenuse.
7. The difference between the square of the hypothenuse, and given side of a right-angled triangle is equal to the square of the required side.
8. The area of a triangle equals half the product of the base multiplied by the perpendicular height;
9. Or, the area of a triangle equals half the product of the two sides, and the natural sine of the contained angle.
10. The side of any regular polygon multiplied by its apothem, or perpendicular, and by the number of its sides, half the product is the area.
_Table of the areas of regular polygons whose sides are unity._
+----------+------+-----------+-----------+------------+------------+ | Name of |No. of| Apothem, | Area, when| Interior | Central | | polygon. |sides.| or perpen-|side is one| angle. | angle. | | | | dicular. | or unity. | | | +----------+------+-----------+-----------+------------+------------+ | | | | | ° ′ | ° ′ | | Triangle | 3 | 0·2886751 | 0·4330127 | 60 0 | 120 0 | | Square | 4 | 0·5 | 1· | 90 0 | 90 0 | | Pentagon | 5 | 0·6881910 | 1·7204774 | 108 0 | 72 0 | | Hexagon | 6 | 0·8660254 | 2·5980762 | 120 0 | 60 0 | | Heptagon | 7 | 1·0382607 | 3·6339124 | 128 34-2/7 | 51 25-5/7 | | Octagon | 8 | 1·2071068 | 4·8284271 | 135 0 | 45 0 | | Nonagon | 9 | 1·3737387 | 6·1818242 | 140 0 | 40 0 | | Decagon | 10 | 1·5388418 | 7·6942088 | 144 0 | 36 0 | | Undecagon| 11 | 1·7028436 | 9·3656399 | 147 16-4/11| 32 43-7/11| | Dodecagon| 12 | 1·8660254 |11·1961524 | 150 0 | 30 0 | +----------+------+-----------+-----------+------------+------------+
The tabular area of the corresponding polygon multiplied by the square of the side of the given polygon, equals the area of the given polygon.
OF ELLIPSES, CONES, FRUSTRUMS, ETC.
1. The square root of half the sum of the squares of the two diameters of an ellipse multiplied by 3·1416 equals its circumference.
2. The product of the two axes of an ellipse multiplied by ·7854 equals its area.
3. The curve surface of a cone is equal to half the product of the circumference of its base multiplied by its slant side, to which, if the area of the base be added, the sum is the whole surface.
4. The solidity of a cone equals one-third of the product of its base multiplied by its altitude, or height.
5. The squares of the diameters of the two ends of the frustrum of a cone added to the product of the two diameters, and that sum multiplied by its height, and by ·2618, equals its solidity.
THE END.
LONDON:
PRINTED BY W. CLOWES AND SONS, STAMFORD STREET AND CHARING CROSS.
FOOTNOTES:
[1] _Vide page_ vi.
[2] The Articles omitted consist chiefly of directions, &c., or are not generally required.
[3] NOTE. _In the “Exercise and Movements.”_
Commander’s Words are printed in SMALL CAPITALS. Executive Small print. Directions, &c. _Italics_.
[4] _Note._—Vide “Motion,” “Forces,” &c., Velocity, Gravity, and Amplitude.
[5] Vide “Tables,” “Excentric Shot, Experiments.”
[6] _When a shot is jammed in a gun, and cannot be rammed home to the cartridge_, destroy the charge by pouring water down the vent, and muzzle until the ingredients are dissolved, and cleared out of the bore; then introduce a small quantity of powder through the vent, and blow out the shot.
[7] The recoil of guns on sleighs varies from four to five feet when on rough ground or in deep snow; to twenty or thirty yards when on glare ice. In the latter case it is of course necessary to send the ammunition sleighs further to the rear; but the recoil may be considerably lessened by placing a small chain round each of the runners.
Ice of eight inches thick will bear with safety a weight of 1115 lb. (or nearly half a ton) on the square foot.
[8] Old pattern.
[9] Further information relative to mixing the composition, and filling combustibles, &c., &c., may be obtained from the “Aide Mémoire,” under the head, “Pyrotechny, Military.”
[10] This will be discontinued when Shrapnell Diaphragm shells are generally introduced into the Service.
[11] _Vide_ Practice Tables for Ranges, Elevations, &c.
[12] The composition for French cannon tubes is two parts of fulminate of mercury and two of mealed powder, mixed together: then formed into a paste with distilled water, slightly impregnated with gum arabic.
[13] Extracted from “Instructions and Regulations for Field Battery Exercise and Movements” for the Royal Regiment of Artillery: the Sections, &c., being similarly numbered.
Commander’s Words are printed in SMALL CAPITALS. Executive Common type. Directions, &c. _Italics._
[14] The Sections, of which merely the heads are given, consist chiefly of details too long for the limited size of the Manual, and they are therefore necessarily omitted.
[15] When Guns are in action, and “CEASE FIRING” is given, all Guns then loaded are to be fired off, and on no account is a Gun to be limbered up, or to move whilst loaded.
[16] The Commanding officer’s Word of command is always to be repeated by the officers.
[17] From “Field Battery Exercise.”
[18] From “Field Battery Exercise,” &c.
[19] From “Field Battery Exercise,” &c.
[20] In the transport of horses to Turkey (July, 1854,) in the Himalaya and Simla steamers, the distance between the upright posts was 2 feet 1 inch in the clear per horse, and the length 9 feet.
[21] “For the guidance of the Farriers of the Royal Artillery. Suggested by Charles Percival, Veterinary Surgeon; and approved of by the Right Honourable the Master-General, and Honourable Board of Ordnance.”
[22] In administering draughts to horses, the greatest possible care and attention are required; should the horse cough, or make an attempt to do so, his head must be instantly lowered, otherwise a portion of the drink will be apt to find its way into the trachea or windpipe, which will produce most distressing symptoms, and often be followed by death. In lowering the head, a can or vessel of any kind should be held under the mouth to catch the drink as it escapes.
[23] From “INSTRUCTIONS FOR THE SERVICE OF HEAVY ORDNANCE.”—_Article 15._
[24] _Words of command_—SMALL CAPITALS.
[25] From “Instructions, and Regulations for the Service, and Management of Heavy Ordnance, for the Royal Regiment of Artillery.” Fourth edition. The Parts, and Articles are numbered in conformity thereto.
[26] _Words of command_—SMALL CAPITALS.
[27] Vide PART 12, “ARTILLERIST’S MANUAL,” etc., The Mechanical powers. The Lever.
[28] By the ballistic experiment, conducted in May, 1837, it was found that, with a heavy 6-pounder gun, a charge of 1½ lb. gave a velocity of 1740 feet, and a charge of 2 lb. a velocity of 1892 feet per second. The shot employed were of a high gauge, windage only ·078 inch, and the powder was of the strongest quality; the weight of the pendulum fired into was 58 cwt. 3 qrs. 16 lb. A light 6-pounder, two feet shorter than the heavy 6-pounder, with similar charges, gave velocities of about 190 feet less.
[29] Extracted from PART 2, and APPENDIX of General Sir Howard Douglas’ highly valued work, entitled “A TREATISE ON NAVAL GUNNERY.” 3rd edition.
[30] _On wads for Heavy Ordnance._
The presence of a compressible body, between the powder and the ball, is necessary for the preservation of the gun. The results of the experiments at Fere, in 1844; at Ruelle in 1844, and 1846; and at Gavres in 1848; with cast iron 30, and 24-pounders, proved that all the pieces, fired without a thin piece of cork interposed between the powder and the ball, burst before 500 discharges were made; whilst those, with which this precaution was taken, sustained 1800 and 2000 discharges without any damage, except an enlargement of the vent. _Vide_ United Service Magazine, September, 1855.
[31] _Vide_ “TREATISE ON NAVAL GUNNERY.” 3rd Edition. By General Sir H. Douglas.
[32] _In Extreme training of a gun to the Right_: Nos. 3, 5, 7, 11, 13, are placed outside; Nos. 8, 6, inside the tackle. No. 13 keeps the end of the fall coiled up.
_In Extreme training to the Left_: Nos 4, 6, 8, 2, are placed outside; Nos. 13, 7, 5, inside the tackle. No. 2 keeps the end of the fall coiled up.
[33] _In running out the right guns_, Nos. 3, 5, 5, 7, man the left tackle; Nos. 4, 6, 6, 2, man the right tackle.
_In running out the left guns_, Nos. 3, 5, 7, 5, man the left tackle; Nos. 4, 6, 6, 2, man the right tackle.
[34] _Note._—When the direction of the gun is to be altered, the word “Traverse” is to be given, if the gun is in, and “Point,” when the gun is out.
[35] _Vide_ Sir Howard Douglas’s highly-valued publication, entitled “A TREATISE ON NAVAL GUNNERY.” Fourth edition.
[36] _Vide_—“United Service Magazine,” No. CCCVIII.
[37] _Vide_ FIELD FORTIFICATION, pages 246, 247.
[38] _Vide_ Preface.
[39] For a square, the length of the perpendicular is ⅛th the exterior side; for a pentagon ⅐th; for the hexagon, and other polygons, ⅙th.
[40] _Vide Tables of Weights, and Measures._
[41] _Vide Tables_ of Weights, and Measures.
[42] In reducing fractions to a common denominator, and in multiplication of fractions, the work may be considerably diminished by cancelling any figures, which are in all the multiples; or by dividing a figure in each of them by any figure which can divide all without any remainder.
[43] See Note, page 268.
[44] _To multiply decimals by 1, with any number of ciphers, as 10, 100_, &c.—This is done by only removing the decimal point so many places farther to the right hand, as there are ciphers in the multiplier, and subjoining ciphers, if need be.
[45] The best way of doubling the root, to form the new divisor, is by adding the last figure always to the last divisor, as appears in the following example.
After the figures belonging to the given number are all exhausted, the operation may be continued into decimals, by adding any number of periods of ciphers, two in each period.
[46] _This rule is only applicable to the very best-made new cordage. The circumference squared should be divided by 6 instead of 5 for the description of rope generally employed._
[47] When the board is tapering, add the breadths at the two ends together, and take half the sum for the mean breadth. _Or else_, take the mean breadth in the middle.
[48] _To strengthen a beam, &c. which is required to support a great weight over a cavity, or ditch._—Place a prop, or short skid, under the centre of the beam, and pass a strong rope, or chain, over the beam lengthways, and under the skid, hauling it very tight, and making fast.
[49] In Lieut.-Colonel B. Jackson’s scientific “Treatise on Military Surveying, &c., &c., &c.,” _Portable trigonometry without logarithms_, is thus introduced—
“The following useful application of Trigonometry, by means of the natural sines, tangents, &c., is taken from an early number of that valuable periodical, ‘The Mechanics’ Magazine,’ and will be found particularly suited to the purposes of the military surveyor.”
[50] For further information on Surveying, and Reconnoitring, reference should be made to the highly-valued publication, entitled “A TREATISE ON MILITARY SURVEYING, INCLUDING SKETCHING IN THE FIELD, PLAN DRAWING, LEVELLING, MILITARY RECONNOISSANCE, &c.,” by Lieut.-Colonel Basil Jackson, containing a full account of every surveying instrument, and the right adaptation of them.
[51] 1. The Reconnoitring protractor is not intended to supply the place of the Theodolite, or other expensive instruments, when very great accuracy is required in surveying, or in trigonometrical observations; but, in the hands of officers accustomed to the use of it, bearings may be rapidly taken, heights and distances ascertained, roads traversed, &c., &c., with sufficient accuracy for a military survey, or reconnoissance.
The protractor has a tripod, on which it is to be steadily fixed for taking angles, &c.; but the instrument can nevertheless be used without the tripod; and mounted officers may, after a little practice, make a reconnoissance with the protractor alone, especially if they are able to measure, or calculate the distance of base lines, by the length of the paces of their horses.
2. A survey, &c., may be very rapidly taken in the field, by laying drawing-paper on the face of the protractor, under the marginal scale, fixing it firmly by means of drawing-pins in the sides, and using, at the first station, the edge of the index as a ruler to set off on the paper, at once, by observation through the sights, the angles of the objects whose distance is required; drawing a base line parallel to the tube side of the instrument, and also lines at the angles found. At the second station, the paper must be moved a few inches, for a base line to be drawn; at the termination of which (the second station) the index is to be directed to the objects, as before, and lines are to be produced until they intersect those drawn at the first station: thus the position of the objects will be obtained; and, by using the scale on the index for the length drawn for the measured base line, as well as for the lines directed to the objects, their respective distances will be ascertained.
3. The reconnoitring protractor, and all other instruments for surveying, &c., &c. can be readily obtained from Messrs. Elliott, 56, Strand, London.
[52] Or Reconnoitring protractor.
[53] To erect a perpendicular, _vide_ “Practical Geometry.”
[54]
3 inch cube full of air floats 1 lb. in water. 3 inch cube of water weighs 1 lb. in air. 1 cubic foot of water weighs 64 lb. in air. 1 ditto coal ditto 80 - 64 = 16 in water. 1 ditto sand ditto 95 - 64 = 31 in water.
A suit of clothes and a pair of boots, which weigh 7 lb. in air, when well saturated with water, only weigh in water 1 lb.
[55] _Vide also Definitions_—TRIGONOMETRY, page 301.
[56] _Gunter’s chain_ is in length 4 poles = 22 yards = 66 feet, and is divided into 100 links. Each link is therefore 22/100 of a yard, or 66/100 of a foot, or 7·92 inches. _Land is estimated_ in acres, roods, and perches. _An acre_ contains 10 square chains, or as much as 10 chains in length and 1 chain in breadth; or in yards it is 220 × 22 = 4840; or in poles it is 40 × 4 = 160 square poles; or in links it is 1000 X 100 = 100,000 square links. An acre is divided into 4 parts called roods, and a rood into 40 parts called perches, which are square poles, or the square of a pole of 5½ yards long, or the square of a quarter of a chain, or of 25 links, which is 625 links. Thus the divisions of land measure are—
625 square links = 1 pole, or perch. 40 perches = 1 rood. 4 roods = 1 acre.
The length of lines, measured with a chain, should be set down in links as integers, instead of in chains, and decimals. Therefore, after the content is found, it will be in square links.
[57] 57·3 is the number of pounds of powder contained in a cubic foot, when shaken; and 55 pounds when not shaken. According to the first case, one pound of powder will occupy 30 cubic inches; and according to the second case one pound will occupy 31·4182 cubic inches.
TRANSCRIBER’S NOTE
Footnote [37] is referenced six times from page 237; footnote [52] is referenced twice from page 311; footnote [54] is referenced four times from pages 317 and page 318.
Obvious typographical errors and punctuation errors have been corrected after careful comparison with other occurrences within the text and consultation of external sources.
Some hyphens in words have been silently removed, some added, when a predominant preference was found in the original book.
Some { bracketing in some tables has been adjusted or removed for readability.
Except for those changes noted below, all misspellings in the text, and inconsistent or archaic usage, have been retained.
Pg viii: page number ‘8’ replaced by ‘48’. Pg xvi: Added new section ‘CONGREVE ROCKETS.’ to the ToC. Pg xviii: ‘Embrasures’ replaced by ‘Embrazures’. Pg xxi: ‘312, 313, 314’ replaced by ‘312–4’. Pg 61: ‘thirty ronnds of’ replaced by ‘thirty rounds of’. Pg 69: in the table header ‘Fore ... Diameter’ replaced by ‘Fore ... Hind’. Pg 72: in second column of the table ‘3¾’ replaced by ‘2¾’. Pg 80: ‘4⅗ inch Mortar’ replaced by ‘4⅖ inch Mortar’. Pg 85: in the table ‘5½ in c’ replaced by ‘5½ inch’. Pg 91: in LEVERS section ‘lb. oz.’ replaced by ‘ft. in.’ Pg 171: in the table ‘1’ replaced by ‘10’. Pg 171: in the table ‘1209’ replaced by ‘1200’. Pg 176: ‘to facilite the’ replaced by ‘to facilitate the’. Pg 187: ‘assist 2 at’ replaced by ‘assists 2 at’. Pg 195: ‘at an elevavation’ replaced by ‘at an elevation’. Pg 220: the Remarks column has been moved under the table to conserve table space. Pg 222: ‘a longe range’ replaced by ‘a long range’. Pg 226: ‘they ricoched and’ replaced by ‘they ricocheted and’. Pg 226: the italic markup on the small table has been removed. Pg 227: ‘the same is in’ replaced by ‘the same as in’. Pg 234: ‘left betweeen the’ replaced by ‘left between the’. Pg 235: ‘placed at tho top’ replaced by ‘placed at the top’. Pg 238: ‘embrasures should be’ replaced by ‘embrazures should be’. Pg 291: ‘(5 × 1)’ replaced by ‘(5 + 1)’. Pg 318: the footnote in the original book ‘See note, p. 317’ was redundant and has been removed. Pg 320: ‘is is a’ replaced by ‘it is a’.
Footnote [7]: ‘to five eet’ replaced by ‘to five feet’.