MENSURATION.

Every branch of the mathematies which has for its object the comparison of geometrical quantities, and the determination of their proportions to each other, may be comprehended under the general name Mensuration. So that, taking the term in its most extensive sense, whatever is delivered in this work under the titles Geometry, Trigonometry, Conic Sections, part of Algebra, and a very considerable portion of Fluxions, may be considered as constituting particular branches of this general theory.

The term mensuration, however, is also frequently used in a less extensive sense, and is applied to a system of rules and methods by which numerical measures of geometrical quantities are obtained. And it is to this limited view of the subject that we propose to confine our attention in the present treatise. In general, it will only be necessary to give the practical rules, as we have already explained their foundation when treating of Geometry, Conic Sections, and Fluxions ; but, in addition to the rules, in a few instances, we shall give their demonstrations.

In all practical applications of mathematies it is necessary to express magnitudes of every kind by numbers. For this purpose a line of some determinate length, as one inch, one foot, _&c., is assumed as the measuring unit of lines, and the number expressing how often this unit is contained in any line, is the numerical value or measure of that line.

A surface of some determinate figure and magnitude is assumed as the measuring unit of surfaces, and the number of units contained in any surface is the numerical measure of that surface, and is called its area. It is usual to assume, [7:14:562]as the measuring unit of surfaces, a square, whose side is the measuring unit of lines.

A solid of a determinate figure and magnitude is in like manner assumed as the measuring unit of solids, and the number of units contained in any solid is its solidity or content. The unit of solids is a cube, each of whose edges is the measuring unit of lines, and consequently each of its faces the measuring unit of surfaces.

A right angle is conceived to be divided into ninety equal angles; and one of these, called an angle of one degree, is assumed as the measuring unit of angles.

The measures generally employed in the application of mensuration to the common affairs of life, and their proportions to each other, are expressed in the following tables:
[table]

These measures were recognised and established by an act of Parliament in 1825; but they were the measures in common use before that time. The length of the standard yard is the distance between two points marked on a straight rod of brass commonly called Pirds Standard Yard of 17 6θ. The length of the yard is to that of a pendulum vibrating seconds of mean time in the latitude of London, in a vacuum at the level of the sea, as 36 to 39T393, the temperature of the brass rod being 62° of Fahrenheit’s thermometer.
[table]
[table]

Note. We have given these Scottish measures for the purpose of comparison with Imperial measure, by which the Scottish measures, being now set aside, will be gradually superseded.
[table]

Note 1. To convert Scotch acres into Imperial, to the number'of Scotch acres add its ½th, also its 5 ⅛th; and in addition to these, if accuracy be required, the - r ⅞^th part of the g ⅛th. Suppose, for example, 100,000 Scotch acres are to be converted into Imperial acres:
[table]

2. The same rule will apply to the conversion of rent or value, and produce of Scotch acres.
[table]

The Imperial gallon is the space occupied by 10 lbs. of water, weighing 70,000 grains.

1 Old English Ale Gallon = 282 Cubic Inches.

The Imperial ale gallon is now the same as the Imperial wine gallon.

The Winchester bushel measure is a cylinder-18½ inches wide, and eight inches deep, and contains 2150.4252 cubic inches.

The Imperial bushel contains 2218∙ 191 cubic inches.
[table]

The Standard Pint is the Stirling Jug, which contains 104∙2034 cubic inches, and holds 26,306*982 grains of distilled water, at 62° of Fahrenheit’s thermometer, the height of the barometer being 30 inches.

The Scotch Gallon = 3∙0065122 Imperial Gallons.
[table]

The Linlithgow Wheat Firlot is to the Imperial Standard Bushel as 0∙998256 to 1. This is not far from a ratio of equality, so that the Scotch Standard Wheat Boll, Firlot, Peck, and Lippie nearly correspond with the Imperial Coomb and its submultiples, the Bushel, Peck, and Pottle.

The Linlithgow Barley Firlot is to the Imperial Standard Bushel as 1∙4562794 to 1.

SECTION I. or THE MENSURATION OF RIGHT EINES AND

ANGLES.

The rules by which certain of the sides or angles of a triangle are to be found, when other sides and angles are [7:14:563]given, might be considered as belonging to this part of mensuration. But as these are fully investigated and explained in the article Plane Trigonometry, it is not necessary to deliver them also here. Referring therefore to that article, we shall employ the remainder of this section in the application of trigonometry to the mensuration of heights and distances.

Mensuration of Heights and Distances.

By the application of geometry the measurement of lines, which, on account of their position or other circumstances, are inaccessible, is reduced to the determination of angles, and of other lines which are accessible, and admit of being measured by methods sufficiently obvious.

A line considered as traced on the ground may be measured with rods or a Gunter’s chain of sixty-six feet; but more expeditiously with measuring tapes of fifty or one hundred feet. By these, if the ground be tolerably even, and the direction of the line be traced pretty correctly, a distance may, by using proper care, be measured within about three inches of the truth in every fifty feet, so that the error may not exceed the 200th part of the whole line.

Vertical angles may be measured with a quadrant furnished with a plummet and sights in the manner indicated by the annexed figures. Ifan angle of elevationis to be measured, as the angle contained by a horizontal line AC, (fig. 1,) and a line drawn from A to B the top of a tower, hill, or other eminence; or to a celestial body, as a star, &c.; the centre of the quadrant must be fixed at A, and the instrumenti moved about A, in the vertical plane, till to an eye placed at G, the object B be seen through the two sights D, d. Then will the arch EF, cut otf by the plumb-line AF, be the measure of the angle CAB.

An angle of depression CAB (fig. 2), is to be measured exactly in the same manner, except that here the eye is to be placed at A the centre of the instrument, and the measure of the angle is the arch EF.

But the most convenient instrument for measuring angles, whether vertical or horizontal, is the Theodolite. This instrument has undergone various alterations and improvements since it was described (we believe, for the first time) by Sisson. Its principal parts are,

1. The horizontal limb. This consists of two circular plates A A, the upper or vernier plate, and BB, the graduated limb. The former moves freely above the latter, without actual contact, and both have a horizontal motion about a vertical axis C, which consists of two parts, one external, fixed to the graduated limb B, and another internal, fixed to the vernier plate A. Their form is conical, and they are ground and accurately fitted into each other, so as to have a perfectly steady motion. The external centre also fits into a ball at D, and the parts are held together by a screw at the lower end of the internal axis.

The lower plate is broader than the upper, and its edge is chamfered off and covered with silver to receive the graduations.

2. The verniers. These arc short scales on the upper plate, and on opposite sides of it, or 180° asunder. They are chamfered, so as to form with the chamfered edge of the lower plate a continued conical surface, and are covered with silver, and graduated in such a manner as to subdivide the divisions of the lower plate into minutes. By means of microscopes which are generally fixed over the verniers, the half or even the fourth of a minute may be estimated.

3. The parallel plates F and G. These serve for levelling the instrument, and are held together by a ball and socket at D. Four milled-headed screws H, H, &c., enter and work in the lower plate, while their upper ends press against the lower side of the upper plate; and being set in pairs opposite to each other, they act in contrary directions, and by this means the instrument is set up level for observation.

Beneath the parallel plates is a female screw adapted to the staff-head, which is united by brass joints to three mahogany legs, so constructed, that when shut up they form one round staff, and are secured in that form by rings put on them. When opened out they make a very firm stand, and the parallel plates may be made horizontal, although the ground may be uneven.

The lower plate of the horizontal limb can be fixed in any position by tightening the clamping screw I, which causes the collar K to embrace the axis C, and prevents its moving; but the final adjustment is to be given by means of the slow-motion tangent screw L which is attached to the upper parallel plate. In like manner the upper or vernier plate can be fixed to the lower by a elamp Μ, which is also furnished with a slow motion, produced by a tangent screw.

4. Spirit levels N, N. There are two of these at right angles to each other, with proper adjusting screws, on the plane of the vernier plate. Their use is to determine when the horizontal limb is set truly level.

ð. The compass. This is on the middle of the vernier plate, but of course it cannot be shewn in a side view of the instrument. .

6. The vertical arc and telescope. The arc is placed on a horizontal axis, the ends of which are supported by two frames, (only one of which can be shewn in the figure.) At one end of the axis is an arm R, with a clamping screw S, by which the axis is held when the telescope is nearly in the proper position: and there is a tangent screw T, by which a slow angular motion is given to the vertical arc and telescope to direct the latter accurately to an object. One side of the vertical arc is inlaid with silver, and by means of its vernier V can be read off to single minutes. The other side shews the difference between the hypothenuse and the base of a right angled triangle, or the number of links to be deducted from each chain length to reduce hypothenusal to horizontal lines.

The level which is shewn under, and parallel to, the telescope is attached to it at one end by a joint, and at the other by a capstan headed screw for raising or lowering that end: there is another screw at the jointed end for lateral adjustment: by their united action the level is placed parallel to the optical axis or line of collimation of the telescope.

The telescope has two collars or rings of: bell metal, ground truly eylindrical, on which it rests on its supports X, X, called Y’s; and it is confined in its place by the clips Z, Z, which may be opened by removing the pins Y, Y, for the purpose of reversing the teleseope or allowing it a circular motion round the axis during the adjustment.

In the focus of the eye glass are placed three lines formed of spider’s web, one horizontal and two crossing it obliquely at equal angles so as to include a small angle between them. These are preferable to a single perpendicular wire, because a station staff at a distance can be made to bisect the angle formed by the oblique wires with more certainty than the staff can be bisected by a vertical wire.

Three of the four screws for adjusting the cross-wires arc shewn at a, a, a, two of which are placed at right angles to the other two, so that by easing one of each pair and tightening the other opposite to it, the intersection of the wires may be placed exactly in the line of collimation.
[7:14:564]

The object glass of the telescope is fixed in a tube which slides within the outer tube, and may be thrust out or drawn backward by turning the milled head b on the side of the telescope. By this motion the optical image of any object to which the telescope is directed may be brought to the intersection of the cross wires.

Before the theodolite can be applied to accurate practice its parts must be adapted to each other by means of the clamps and screws and levels. The operations by which this is performed are called adjustments.

The first adjustment is that of the line of collimation, which must coincide with the axis of the cylindrical rings on the telescope, and pass through the intersection of the cross wires. The second places the level attached to the telescope parallel to the rectified line of collimation. The third makes the axis of the horizontal limb truly vertical by means of the telescope level, which is most to be depended on; and then the levels on the vernier plate are adjusted by their screws so that their air bubbles may remain at rest in the middle of their tubes while it makes a complete revolution on its axis. When these adjustments are perfect the vernier of the vertical arc must be set so that its index may point to zero on the arc; or else its deviation from zero must be noted and applied as an index error.

Problem I.

To determine the height of an object, the bottom of which is accessible.

Example. Having measured AE, a distance of 200 feet in a direct horizontal line from the bottom of a tower, the angle BCD, contained by the horizontal line CD: and a line drawn from C to the top of the tower, was measured by a quadrant, or theodolite placed at C, and found to be 47° 30'. The centre C of the instrument was five feet above the line AE at its extremity E. It is required hence to determine AB the height of the tower.

In the right-angled triangle CBD we have given the side CD = 200 feet, and the angle C=47° 30'. And since by the rules of Plane Trigonometry, rad.: tan. BCD :: DC: DB;

By employing logarithmic tables and proceeding as is taught in Plane Trigonometry, we shall find DB=2l8∙3 feet. To which add DA=EC=5 feet, the height of the instrument, and we have AB=223.3 feet, the height of the tower.

Problem II.

To determine the height and distance of an object when the bottom or point directly under it is inaccessible.

Ex. 1. Suppose a small cloud, or balloon C, is seen at the same time by two observers at A and B, and that these stations are in the same vertical plane with the object C, and on the same side of it. Also suppose that its angles of elevation, viz. the angles A and B are 35° and 64°, and that AB, the distance between the observers, is 880 feet. It is required hence to determine CD the height of the object, also AC, BC its distances from the two observers.

In the triangle CAB there are given the outward angle CBD=64°, and one of the inward angles A=35°; hence the other inward angle ACB, which is their difference, is given, and = 64° —35° =29°.

Now in the triangle CAB

Sin. ACB: sin. AAB: BC, and sin. ACB: sin. B :: AB: AC.

From these proportions, by actual calculation, BÇ will be found =1041 feet, and AC=1631 feet.

Again, in the right-angled triangle BCD rad.: sin. B :: BC: CD.

Hence CD will be found =936 feet.

Ex. 2. To measure the height of an obelisk CD, standing on the top of a declivity, two stations at A and B were taken, one at the distance of forty, and the other at the distance of one hundred feet from the centre of its base, which was in a straight line with the stations. At the nearer station A, a line drawn from it to the top of the obelisk was found to make an angle of 41° with the plane of the declivity; and at B, the more remote station, the like angle was found to be 23° 45'. Hence it is required to find the height of the obelisk.

From the angle CAD=41°, subtract the angle B=23° 45', and there remains the angle BCA=17° 15'.

In the triangle BCA,

Sin. BCA: sin. B :: AB: AC. Hence AC=81.49 feet. And in the triangle ACD,

AC + AD: AC—ÀD :: tan.⅛ (D + C): tan. ½ (D—C). Hence ½ (D—C)=40° 24'½, which, subtracted from ½ (D + C,) gives the angle ACD=27° 5'½.

Lastly, in the triangle ACD,

Sin. ACD: sin. A :: AD: DC.

Hence DC, the height required, will be found to be 57'62 feet.

Ex. 3. At B the top of a tower which stood on a hill near the sea shore, the angle of depression of a ship at anchor (viz. the angle HBS), was 4° 52'; and at R, the bottom of the tower, its depression (namely, the angle NRS) was 4° 2 ’. Required AS the horizontal distance of the vessel: and also R A, the height of the bottom of the tower above the level of the sea, supposing RB the height of the tower itself to be 54 feet.

From the angle BSA=HBS=4° 52', subtract the angle RSA=NRS=4° 2 , , and there remains the angle BRS= 50'. Also, from the angle HBA=90° subtract HBS=4 ’ ^ 52', and there remains SBR=85° 8'.

In the triangle SBR,

Sin. BSR: sin. SBR :: BR: SR;

Hence SR is found. Again, in the triangle SRA, rad.: sin. RS A :: SR: AR, and rad.: cos. RSA :: SR: AS.

From the first of these proportions we And AR=26θ feet; and from the second, AS=3690 feet.

Problem HI.

To find the distance of an object from either of two given stations: also from a line joining them, when each station and the object can be seen from the other station.

Ex. 1. Wanting to know the breadth CD of a river, and also the distance of an object A close by its side from another object C on its opposite side, a base AB of 400 yards was measured along the bank. Then, by meaus of a theodolite, the angles CBA and CAB were measured, and found to be 37° 40' and 59° 15' respectively. It is required thence to determine the breadth CD, and the distance AC between the objects A and C.

This problem differs from the last only by the given angles, and distances required, lying in a horizontal instead of a vertical plane.

In the triangle ABC we have the base AB, also the angles A and B, and consequently the angle C given. And by Plane Trigonometry,

Sin. ACB: sin. B :: AB: AC.

Hence AC is found to be 246.2 yards.

Also, in the right-angled triangle A CD, rad.: sin. A :: AC: CD.

Hence CD is found to be 211.6 yards.

Ex. 2. From a ship at sea a point of land was observed to bear E. by S.; and after sailing N. E. twelve miles, the same point was found to bear S. E. by E. How far was the last observation made from the point of land?

Let A be the first position of the ship, B the second, and C the point of land. In the triangle ABC we have given the angle A=5 points or 56° 15 , , the angle B=9 points, or [7:14:565]101° 15', and the angle C=2 points or 22° 30'. Also the side AB=12 miles. Hence (by Trigon.) the side BC is readily found to be 26.073 miles.

» Problem IV.

To determine the distance between two inaccessible objects, which can be seen from two given stations at a known distance from each other, supposing them all in the same plane.

Ex. Wanting to know the distance between two inaccessible objects II and Μ, a base AB of 670 yards was measured in the same plane with the objects, and the following angles were taken at its extremities.

. . f BAM=40° 16' n í ABH=42° 22' AtA (MAH=57 40 AtB [HBM=71 7

Hence it is required to determine HM, the distance between the objects.

In the triangle HAB we have the angle HBA=42° 22', the angle HAB (=HAM + MAB)=97° 56', and therefore the remaining angle AHB=39° 42'. We have also the side AB=670 yards. Hence, by this proportion,

Sin. AHB: sin. HBA :: AB: AH, we find AH=7θ6∙8 yards.

Again, in the triangle MAB we have the angle MAB =40° 16', the angle ABM (=ABn + HBM)=113° 29', and therefore the angle AMB=26° 15'. Hence from the proportion,

Sin. AMB: sin. ABM :: AB: AM, we get AM=1389 ’ 4.

In the triangle HAM, besides the angle HAM=57° 40' we have now the sides AH=706∙8, and AM= 1389’4 yards, to find the remaining side HM. Therefore, proceeding according to the rules of trigonometry, we state this proportion,

AM+AH: AM—AH tan. ⅛ (AHM +AMH):

tan. ½ (AHM—AMH).

Hence we find half the difference of the angles AHM and AMH to be 30° 36', which taken from 61° 10', half the sum, leaves 30° 34' for AMH the least of the two angles. Lastly, from the proportion

Sin. HMA: sin. HAM :: HA: HM, we get HM=1l74 yards, the answer to the question.

Problem V.

The mutual distances of three remote objects being given, and the angles which these distances subtend at a station, in their plane, to find the position of the station.

Example. There are three objects, A, B, C, whose distances asunder are known to be as follows: namely, from A to B 106½, from A to C 202, and from B to C 131 fathoms. Now to determine the distance of D a fourth object, or station, frem each of the other three, the angle ADB was measured with a theodolite, or other suitable instrument, and found to be 13° 30', and the angle CDB was found 29° 50'. Hence it is required to determine the distances DA, DB, and DC, supposing DB the least of the three.

Let a circle be described through the points A, D, and C; and let DB be produced to meet the circle again in E, and draw AE, CE.

In the triangle AEC there are given the side AC=202 fathoms, the angle ACE (=ADE. Geom. Sect. II. Theor.

15.) =13° 30', and the angle CAE (=CDE) 29° 50'. Hence (by Trigon.) we shall have AE=68∙716 fathoms.

In the triangle ABC, all its sides are given, and hence the angle BAC will be found=35° 35' 54"; to this, add the angle CAE, and the sum is the angle EAB=65° 25' 54".

In the triangle ABE, we have given AB=106∙5, AE =68∙716, the angle BAE=65° 25' 54"; hence we shall have the angle ABE=38° 43' 41", and the angle AEB= 75° 51' 25".

In the triangle ADE we have the side AE=68 - 716, the angle ADE=13°"30', and the angle AED=75° 51' 25". Hence we have AD=285.43 fathoms, which is one of the distances required.

In the triangle ABD we have AB=106 - 5, the angle ADB, = 13° 30', the angle DAB (=ABE-^ADB) 25° 13' 45". Hence BD, another of the distances sought, will be found = 194.45 fathoms.

Lastly, In the triangle ADC, there is given AC=202, the angle ADC (=ADB + BDC) =43° 20', the angle DCA(=DEA)=75° 15' 25". Hence we get DC=256∙97 fathoms, which is the remaining distance sought.

There are various other instruments and methods by which the heights or distances of objects may be found. One of the most simple instruments, both in respect of its construction and application, is a square, AB CD, made of some solid material, and furnished with two sights on AB, one of its edges, and a plummet fastened to A, one of its angles, and having the two sides BC, CD, which contain the opposite angle divided into 10, or 100, or 1000 equal parts.

To measure any altitude HK with this instrument. Let it be held in such a position that

K, the top of the object may be seen through the sights on its edge AB, while its plane is perpendicular to the horizon; then the plummet will cut off from the square a triangle similar to that formed by the horizontal line Aĭ, the vertical line IK, and the line AK drawn from the eye to the top o. the object.

If the line of the plummet pass through D the opposite angle of the square, then the height KI will be equal to AI, the distance of the eye from the vertical line to be measured. If it meet AD, the side of the square next the eye, in some point E between A and D, then the triangles ABE, AIK, being similar, and the angle ABE equal to the angle AKΓ, we have AE: ABAI: IK. Let us now suppose AD=AB to be divided into 1000 equal parts; then the length of AE will be expressed by a certain number of these parts; thus the proportion of AE to AB, and consequently that of AI to IK will be given; therefore if AI be determined by actual measurement, we may from the above proportion immediately find IK,

If again the line of the plummet meet DC the side of the [7:14:566]square opposite to the sights, in F, then, in the similar triangles AIK, BCF, the angle AKI is equal to BFC; thus we have BC: CF :: Al: IK. Hence IK is determined as before, and in each case by adding HI the height of the eye, we shall have HK the whole height required.

SECTION II. MENSURATION OF PLANE FIGURES.

Problem I.

To find the area of a parallelogram, whether it beasquare, a rectangle, a rhombus, or a rhomboid.

Rule I.

Multiply the length by the perpendicular breadth, and the product will be the area.

This rule is demonstrated in Geometry, sect. iv. theor. 5.

Ex. 1. Required the area of a square AB CD, whose side AB is 10½ inches.

Here 10⅛ χ 10½ or 10∙5 × 10∙5 = 110 ’ 25 square inches is the area required.

Ex. 2. Required the area of a rectangle EFGH, whose length EF is 13*75 chains, and breadth FG is 9 ’ 5 chains.

Here 13*75 ×9∙5 = 130∙625 square chains is the area, which, when reduced to acres, &c. is 13 ac. 0 ro. 10 po.

Ex.⅛. Required the area of a parallelogram KLMN, whose length KL is 37 feet, and perpendicular breadth NO is 5½ or 5∙25 feet.

In this example the area is'37 × 5∙25=194∙25 square feet, or 21∙583 square yards.

Rule II.

As radius,

To the sine of any angle of the parallelogram, So is the product of the sides including the angle, To the area of the parallelogram.

To see the reason of this rule it is only necessary to observe, that in the parallelogram KLMN, the perpendicular breadth NO is a fourth proportional to radius, sine of the angle K, and the oblique line KN, (Trigonometry,) and . , , sin. K ____

is therefore equal to -— × KN; therefore the area rad.

of the figure ⅛-× KN × KL, which expression is the same as the result obtained by the above rule.

Ex. Suppose the sides KL and KN are 36 feet, and 25*5 feet, and the angle K is 58°, required the area.

Here it will be convenient to employ a table of logarithms. The operation may stand thus, log' rad. I0∙0000Q log. sin. 58° 9∙92842

log. (36 × 25∙5)=log. 36+log. 25∙5. 2∙96284

• log. of area 2∙89126

area=778∙5 square feet.

Problem II.

Having given any two sides of a right angled triangle, to find the remaining side.

Rule.

1. When the sides about the right angle are given, to find the hypothenuse.

Add together the squares of the sides about the right angle, and the square root of the sum will be the hypothenuse.

2. When the hypothenuse and one of the sides about the right angle is given, to find the other side.

From the square of the hypothenuse subtract the square of the given side, and the square root of the remainder will be the other side.

This rule is deduced from theor. 13. sect. iv. Geometry. Ex. 1. In a right-angled triangle ABC, the sides AB and AC, about the right angle, are 33 feet and 56 feet; what is the length of the hypothenuse BC?

Here 33 2 +56 2 =3136 +1089=4225, and √(4225)=65 feet, =the hypothenuse BC.

Ex. 2. Suppose the hypothenuse BC to be 65 feet, and AB one of the sides about the right angle to be 33 feet; what is the length of AC the other side?

Here 65 2 -33 2 =4225-1089=3136; and √ ’ (3316)=56 feet=the side AC.

Problem III.

To find the area of a triangle.

Rule I.

Multiply any one of its sides by the perpendicular let fall upon it from the opposite angle, and half the product will be the area.

The truth of this rule is proved in Geometry, Part I. sect. iv. theor. 6.

Example. What is the area of a triangle ABC, whose base AC is 40, and perpendicular BD is 14∙52 chains?

The product of the base by the perpendicular, or 40 × 14∙52, is 580∙8 square chains, the half of which, or 290∙4 sq. ch.=29 ac. 0 r. 6∙4 po. is the area of the triangle.

Rule II.

As radius,

To the sine of any angle of a triangle,

So is the product of the sides including the angle, To twice the area of the triangle.

This rule follows immediately from the second rule of Prob. I., by considering that the triangle KNL is half the parallelogram KNML.

Example. What is the area of a triangle ABC, whose two sides AB and AC are 30 and 40, and the included angle A is 28° 57'?

Operation by Logarithms. log. rad. I0∙00000 log. (30 × 40)=log. 30+log. 40. 3∙07918

log. sin. 28° 57' 9-68489 log. of twice area 2∙76407

twice area=580 ’ 85

area 290*42

Rule HI.

When the three sides are given, add together the three sides, and take half the sum. Next, subtract each side severally from the said half sum, thus obtaining three remain [7:14:567]ders. Lastly, multiply the said half sum, and those three remainders all together, and extract the square root of the last product for the area of the triangle.

This practical rule is deduced from the following geometrical theorem. The area of a triangle is a mean proportional between two rectangles, one of which is contained by half the perimeter of the triangle, and the excess of half the perimeter above any one of its sides; and the other is contained by the excesses of half the perimeter above each of the other two sides. As this theorem is not only remarkable, but also of great utility in mensuration, we shall here give its demonstration.

Let ABC then be any triangle; produce AB, any one of its sides, and take BD and B√, each equal to BC; join CD and Cd, and through A draw a line parallel to BC, meeting CD and Cd produced in E and e ; thus the angle AED will be equal to the angle BCD, (Geometry, Part I. sect. i. theor. 21), that is, to the angle BDC or ADC, (sect. i. theor. 11); and hence AE=AD, (sect. i. theor. 12); and in like manner, because the angle Aed is equal to the angle BC<Z, that is, to the angle B<ZC, or Ade, therefore AE= Ad.

On A as a centre, at the distance AD or AE, describe a circle meeting AC in F and G; and on the same centre, with the distance Ad or AE, describe another circle meeting AC in f and g, and draw BH and B7z perpendicular to CD and Cd. Then, because BD, BC, Bt7 are equal, the point C is in the circumference of a circle, of which De? is the diameter, therefore CD and Cd are bisected at H and h, (sect. ii. theor. 6), and the angle DC<Z is a right angle, (sect. ii. theor. 17), and hence the figure CHBΛ is a rectangle, so that B7i=CH=½ CD, and BH=C7i=½ Cr∕.

Join BE, and Be, then the triangle BAC is equal to each of the triangles BEC, BeC, (sect. iv. theor. 2, eor. 2); but the triangle BEC is equal to ½ EC × BH, (sect. iv. theor. 2), that is, to ½ EC × Cd; and in like manner the triangle BeC is equal to ½ e C × B7«, that is, to ½ e C × CD, therefore the triangle ABC is equal to ½ EC × Cd, and also to ⅛ eC × CD.

Now since CD: Cd :: CE× CD: CE× Cd J sect. iv. ∣ and also CD: Cd :: Ce X CD: Ce × Cc? ( theor. 3. ∫ Therefore CE × CD: CE × Cd :: Ce × CD: Ce × C¢7; that is, because CE × CD=FC × CG, and Ce × C<7= fC χCg, (sect. iv. corollaries to theor. 28 and 29).

FC×CG: CE×Cc?:: Ce×CD: fC×Cg; which last proportion (by taking one fourth of each of its terms, and substituting the triangle ABC for its equivalent values ½ CE × Cd and ½ Ce × CD) gives us

½ FC ×½ CG: trian. ABC :: trian, ABC: ½∕C ×½ Cg.

Now, if it be considered that the radius of the circle gde is AB—BC, it will readily appear that, putting 2 s for the perimeter of the triangle ABC, we have

FC(=AB + BC + AC) = 2 ä

CG(=AB+ BC — AC) = 2 í—2 AC,

∕C(=AC+ { AB—BC ∣)= 2 s— 2 BC, g C (= AC— { AB—BC } )= 2 . s -2 AB.

Put now a, b, c for the sides AC, BC, and AB respectively, then ½ FC=v, ½ GC=.s ’ — a, ½yC=v— b, ½ Cgz∑s — c; thus the last proportion becomes s×(s—α): triam. ABC :: trian. ABC: (s∙—6)×(.s ’ c ), which conclusion, when expressed in words at length, is evidently the proposition to be demonstrated.

And as a mean proportional between two quantities is found by taking the square root of the product, it follows that the area of the triangle ABC, which is a mean between s X (5—<z) and (v—6) × (.s , —c), is equal to J { 5 × («— α) × (s—b) × (s— c) I which formula, when expressed in words at length, gives the preceding rule.

Example. Required the area of a triangle whose three sides are 24, 36, and 48 chains respectively.

Here 24 + 36 + 48=108=the sum of the three sides, 108

And = 54=half that sum; Δ

Also 54—24=30, the first remainder; 54—36=18, the second remainder; and 54—48=6, the third remainder.

The product of the half sum and remainders is

54 × 30 × 18 × 6= 174960.

And the square root of this product is

4∕(174960)=418 - 28 sq. eh. the area required.

Problem IV.

To find the area of a trapezoid.

Rule.

Add together the two parallel sides, then multiply their sum by the perpendicular breadth, or distance between them, and half the product will be the area.

T⅛is rule is demonstrated in Geometry, Part I. sect∙iv. theor. 7.

Example. Required the area of the trapezoid AB CD, whose parallel sides AB and DC are 7’5 and 12∙25 chains and perpendicular breadth DE is 15∙4 chains.

The sum of the parallel sides is 7 ’ 5 + 12∙25=19 ’ 75; which multiplied by the breadth, is

19*75 × 15-4=304-15; and half this product is

^-^=152∙075 sq. ch.= 15 ac. 33∙2 po.

the area required.

Problem V.

To find the area of any trapezium.

Rule.

Divide the trapezium into two triangles by a diagonal, then find the areas of these triangles, and add them together.

Note. If two perpendiculars be let fall on either diagonal from the other two opposite angles, the sum of these per [7:14:568]pendiculars being multiplied by the diagonal, half the product will be the area of the trapezium. The reason of this rule is sufficiently obvious.

Example. In the trapezium ABCD the diagonal AC is 42, and the two perpendiculars BE, DF are 16 and 18: What is its area?

Here the sum of the perp, is 16+18=34, which multiplied by 42, and divided by 2, gives 34 × 42 n — =714, the area. δ Problem VI.

To find the area of an irregular polygon. Rule.

Draw diagonals dividing the proposed polygon into trapeziums and triangles; then find the areas of all these separately, and add them together for the content of the whole polygon. The reason of this rule, and the manner of applying it, are sufficiently obvious.

Problem VII.

To find the area of a regular polygon. Rule.

Multiply the perimeter of the polygon, or sum of its sides, by the perpendicular drawn from its centre on one of its sides, and take half the product for the area.

This rule is only in effect resolving the polygon into as many triangles as it has sides, by drawing lines from its centre to all its angles, then taking the sum of their areas for the area of the figure.

Example. Required the area of a regular pentagon ABCDĔ, whose side AB or BC, &c., is 25 feet, and perpendicular HK is 17*2 feet.

Here 25 ×5 = 125=the perimeter, And 125= 17’2=2150,

) f And its half 1O75=the area required.

Note. If only the side of the polygon be given, Its perpendicular may be found by the following proportion:

As radius,

To the tan. of half the angle of the polygon,

So is half the side of the polygon, To the perpendicular.

And here, as well as in all other trigonometrical calculations, we may employ a table of logarithmic sines and tangents.

The angle of the polygon, that is, the angle contained by any two of its adjacent sides, will be found from this theorem. The sum of all its interior angles is equal to twice as many right angles, wanting four, as it has sides, which is demonstrated in Geometry, Part I. sect. i. theor. 25.

Problem VIII.

To find the diameter and circumference of a circle, the one from the other.

Rule I.

As 7 is to 22, so is the diameter to the circumference, nearly.

As 22 is to 7, so is the circumference to the diameter, nearly.

Rule II.

As 113 is to 355, so is the diameter to the circumference, nearly.

As 355 is to 113, so is the circumference to the diameter, nearly.

Rule III.

As 1 is to 3T416, so is the diameter to the circumference, nearly.

As 3∙1416 is to 1, so is the circumference to the diameter, nearly.

Note. The result obtained by the first rule, which is the least accurate of the three, will not differ from the true answer by so much as its 2400th part. But that obtained by the second rule, which is the most accurate, will not differ by so much as its 10,000,000th part.

The proportion of the diameter of a circle to its circumference, is investigated in Algebra, art. 273, and in Geometry, Part I. sect. vi. prop. 6. Also in Fluxions, sect. 38 and sect. 139∙ The manner of finding the first and second rules, and others of the same kind, is explained in Algebra, sect. xxi. But it is impossible to express exactly, by finite numbers, the proportion of the diameter of the circle to its circumference.

Example 1. To find the circumference of a circle whose diameter is 20.

By the first rule, 20 × 22

7: 22:: 20: —γ—=62∣. the answer.

Or by the third rule, 3∙1416 × 20=62∙832 the answer.

Ex. 2. The circumference of a circle is 10 feet, what is its diameter?

By the second rule,

355: 113 :: 10: —--—=3∙1831 the answer. «jÖÕ Problem IX.

To find the length of any arch of a circle.

Rule I.

As 180 is to the number of degrees in the arch, so ⅛ 3∙1416 times the radius to its length.

To see the reason of this rule it is only necessary to consider, that 3-1416 times the radius is (by last rule) equal to half the circumference, or to an arch of 180 c, , and that the length of an arch is proportional to the number of degrees it contains.

Example. Required the length of the arch AEB, whose chord AB is 6, the radius AC or CB being 9∙ Draw CD perpendicular to the chord, then CD will bisect the chord in D, and the arch in E. Now in the right-angled triangle ACD, there is given the hypothenuse AC=9, and the side AD=3; hence, by trigonometry, the angle ACE will be found to contain 19° 28'∕ 0 =19 ’ 471 degrees.

The double of this, or 38*942, is the number of degreu in the whole arch AEB. Then, by the rule, 180: 38∙942 :: 9×3-1416: — 3 ' 14 *θ× 38 ' 94 g =6∙11701 loD

the answer. i [7:14:569]Rule II.

As six times the distance of any chord from the centre together with nine times the radius of the circle, is to that distance together with fourteen times the radius; so is the chord to the arc nearly.

The investigation of this rule is as follows:

Let a denote any arc of a circle of which the radius=1. Putting A, B, C for indeterminate quantities, assume

1 + A cos. a _ a

B + C cos. a ~ sin. a '

Then, observing that sin. a cos. α=½ sin. 2 a, we have sin. α+½ A sin. 2 α=a B + α C, cos. α. (1.)

Now, by formulae investigated in Algebra, art. 286, α ' 5 β5

Sm.α= π --- + -—_&c.

r ,2

Cos. o=l--'+ —- r -⅛ c ∙

Therefore we have for sin. a, ½ sin. 2 a, a cos. a; these approximate values, which include the fifth power of the arc,

Q∙ α 3 α 5

Sin. α=α-y + η,θ nearl T> i ∙ o 4α 3 16α 5

½ sin. 2 a~a θ- + — nearly,

3 α 3 5 α 5 , a cos. «=α θ- + j2θ nearly.

Hence substituting these values in equation (1.) and dividing by a we obtain

(1 + A)-(1 + 4 A)≤- + (1 +16 A) ≤-=B + C-3 C≤- + 5 c 120^

Making now the coefficients of like terms on each side of this equation equal, we have

1+A=B + C, 1+4A=3C, 1+16A=5 C;

,. v λ 1 n 9 „6

and hence, A=—, B=—, C=—.

These values being substituted for A, B, C in the assumed equation, the result after reduction is

14+cos. a_ a 9+6cos.α sin.α* ’ Hence the rule.

Ex. 1. Required the length of ½th of the circumference of a circle whose radius is 10?

In this case the chord AB is equal to the radius AC= 10; and the perpendicular CD=√(CA i —AB 2 )= √(100-25)=√75^= 8.660254: hence 6 CD + 9 AC= 141-961524, and CD +14 AC=148∙660254. Wenowstate this proportion,

141-961524: 148∙660254 :: 10: 10∙47187=arc AEB.

This is a near approximation to the length of the arc: its true value is 10∙471976, &c.

Ex. 2. It is known that an arc of 57° 17 , 44"∙8 is very nearly equal to the radius. Now, let it be proposed to find its approximate length by the formula, the radius being supposed equal to 1 mile.

In this case by the Trigonometrical Tables the chord =•958851: its distance from the centre=∙8775826. Therefore the proportion is

14-2654956: 14∙8775826=∙958851: ∙9999922.

The error of the approximation is∙0000078 of a mile, which is equal to about half an inch.

Problem X.

To find a straight line nearly equal to a given arc of a circle.

TK 1 J 4 + cos. a a

The formula-—- c = - may be put under this

9+θcos. a sin. α ; r form 15 — (1 — cos, a) _ a

10+5cos.α—(1—cos∙a) - ' sin. a λ 3—⅜ ver. sin. a a

Or, ———— = .

2 +cos. a —I ver. sin. a sin. a

Hence we obtain the following simple geometrical construction for finding a straight line nearly equal to a given arc of a circle:

In DL produced take LH equal to the radius, and HK from H towards the centre equal to one fifth ; part of the versed sine DE: From K draw KF, KG through A and B the extremities of the arc AB to meet the tangent at the middle point of the arc in F and G: The intercepted portion FG of the tangent is equal to the arc AB nearly.

Problem XI.

To find the area of a circle.

Rule I.

Multiply half the circumference by half the diameter, and the product will be the area.

Rule II.

Multiply the square of the diameter by *7854, and the product will be the area.

The first of these rules has been demonstrated in Geometry, sect. vi. prop. 3. And the second rule is deduced from the first, as follows. It appears from prop. 6. sect. vi. Geometry, that the diameter of a circle being unity, its circumference is 3∙1416 nearly; therefore, by the first rule, its area is 1 × 3 , 1416÷4=∙7854. But circles are to one another as the squares of their diameters, (prop. 4.) therefore, putting d for the diameter of any circle, 1: d i :: •7854: ∙7854 <7 2 =the area of the circle whose diameter is d. Example. WTiat is the area of a circle whose diameter is 7∙

By the second rule 7×7 ×∙7854=38∙4846 the area.

By the first rule 7 ×3∙1416=the circumference.

7 y∙ 3∙1416 χ7

τhen - x --- 4 - * =7 ×7×∙7854=38∙4846 the area, the same as before.

Problem XII.

To find the area of any sector of a circle.

Rule I.

Multiply the radius by half the arch of the sector, and the product will be the area, as in the whole circle.

Rule II.

As 360 is to the degrees in the arc of the sector, so is the area of the whole circle to the area of the sector.

The first of these rules follows easily from the rule for the whole area, by considering that the whole circumference is to the arch of the sector, as the whole area to the area of the sector, that is, [7:14:570]circum.: arch of sect.:: rad. × ½ circum.: area of sect. Hence area of sect∙=rad. ×½ arch of sect.

The second rule is too obvious to need any formal proof.

Example. To find the area of a circular sector ACB whose arch AEB contains 18 degrees, the diameter being 3 feet. (See fig. to prob.ix.)

1. By the first rule.

First 3∙1416 × 3=9 ’ 4248 the circum.

And 360: 18 :: 9 ’ 4248: ∙47l24 the arch of sect.

Then ∙47l24 × 3-÷-4=∙35343 the area.

2. By the second rule.

First ∙7854 ×3 2 =7∙0686 the area of the circle.

Then 360: 18 :: 7∙0686: ∙35343 the area.

Problem XIII.

To find the area of a segment of a circle.

Rule.

Find the area of the sector having the same arch with the segment by the last problem. Find also the area contained by the chord of the segment and the two radii of the sector. Then take the sum of these two for the answer when the segment is greater than a semicircle, or take their difference when it is less than a semicircle. As is evident by inspection of the figure of a segment. (Fig. to prob, ix.)

Example. To find the area of the segment AEBDA, its chord AB being 12, and the radius AC or BC 10.

First, as AC: AD :: rad.: sin. 36° 52'½=36∙87 degrees, the degrees in the angle ACE or arch AE. And their double or 73*74=the degrees in the whole arch AEB.

Now7854 × 400=314*16 the area of the whole circle. Therefore 360°: 73*74 :: 314∙16: 64∙3504=area of the sector CAEB.

Again √ (CA 2 -AD 2 ) =√ (100—36) = √ 64 = 8 = DC.

Therefore AD × DC=6 × 8=48=area of the triangle. Hence sector ACBA—triangle ACB=16∙3504 the area of seg. AEBDA.

Problem XIV.

To find the area of any segment of a parabola, that is the space included by any arch of a parabola, and the straight line joining its extremities.

Rule.

Multiply the base of the segment by its height, and take ½ of the product for the area.

This rule is demonstrated in Conic Sections, part iv. sect. iii. prop. 2.

Example. The base AB of a parabolic segment ACB is 10, and its altitude CD, (that is, the greatest line that can be drawn in the segment perpendicular to the base AB) is 4: What is its area?

Here 10 X 4 × ½=^=26½ the area, o Problem XV.

To find the area of an ellipse.

Rule.

Multiply the product of the two axes by the number •7854 for the area of the ellipse.

For the area ofan ellipse is equal to the area of a circlewhose diameter is a mean proportional between the axes of the ellipse, (as is easily deduced from Conic Sections, cor. i. prop. 3. sect, iii part iv.) that is, to the area of a circle, the square of whose diameter is equal to the product of the axes. But by prob. x. the area of a circle is equal to the square of the diameter multiplied by *7854; therefore the area of an ellipse is equal to the product of the axes multiplied by the same number *7854.

Example. If the axes of an ellipse, AB CD, be 35 and 25. WTiat is the area? 35 × 25 X ∙7854=687∙225 the area.

Note. As to hyperbolic areas, the reader will find formulas for their exact mensuration in Fluxions, § 156. Ex. 4 and 5.

Problem XVI.

To find nearly the area of a figure bounded by any curve line A a a' a", &c. P, the straight line BQ, and AB, PQ two other straight lines drawn from the extremities of the curve perpendicular to BQ.

Rule.

Let BQ, the base of the figure, be divided into any even number of equal parts by the perpendiculars ba, b' a', b" a", &c. which meet the curve in the points a, a', a", &c.

Let F and L

denote the first and last perpendiculars AB and PQ.

Let E denote the sum of all the remaining even perpendiculars, viz. a b, a" b", a v, δ lv , the second, fourth, sixth, &c.

Let R denote the sum of the remaining perpendiculars, viz. a' b', a"' b'", &c.

And put D for B b, or bb , , &c. the common distance between the perpendiculars.

Then the area of the figure will be nearly equal to ⅛D×(F + L + 4E+2R) 5 and the approximation will be so much the more accurate according as the number of perpendiculars is the greater.

Demonstration. Join the tops of the first and third perpendiculars by the line A a' meeting the second perpendicular in E, and draw CD through a so as to form the parallelogram A a' DC; then the space bounded by the curve line A a a' and the three straight lines AB, B b', b' a' will be made up of the trapezoid AB b’ a', and the space bounded by the arch A a a’ and its chord A a'. Now if the arch A a a' be small, this last space will be nearly two thirds of the parallelogram AD, for it will be nearly equal to the area contained by the straight line A a’, and an arch of a parabola passing through the points A, a, a', and having a b for a diameter, which area is ½ of its circumscribing parallelogram. (Conic Sections, part iv. sect. iii. prop. 2). Therefore the space Aaa' b’ BA will be nearly equal to the sum of the trapezoid AB b' a' and ½ of the parallelogram AD, which sum is evidently equal to ½ of the trapezoid AB b' a’, together with ½ of the trapezoid CB b’ D. Now the area

AB ∣ α ʹ D f of the trapezoid AB b' a' is - × Bδ ’ (Geometry,

, λ , . AB + a'b' τ , sect. iv. theor. ∕.) = ×2B5; and in like man-

Δ

ner the area of the trapezoid CB 6'D is × B6'=

2

ab × 2Bδ; therefore the area of the figure A a a' b' B is 1 1 AB + √Z>' n nearly ⅜× -ÿ × 2BZ» + f × ab × 2B5=⅜(AB+4αZ> A-α' δ ’) BA [7:14:571]In the very same way it may be shewn that the area of the figure a' a" a'" b"' b' is nearly

⅜(α' b' +4 a" b"+a" , b'") ×BZ>, and that the area of the figure α' ’ ' α iv PQ6" ’ is nearly ½(α"' 0"' + 4α i v δ iv +PQ)×B6.

Therefore, the area of the whole figure bounded by the curve line AP, and the straight lines AB, BQ, QP, is .nearly equal to the sum of these three expressions, namely, to

í AB+PQ )

½Bδ× -J +4(α bA-a'' b" + a^ 5 iv ) J- ( +2(α' b'+a'" b'") ) as was to be demonstrated.

Example I. Let it be required to find the area of the quadrant ABC, whereof the radius AC=1.

Let AC be bisected by the perpendicular DE, and let CD be divided into four equal parts by the perpendiculars mn, pq, rs. Now because CA=1, therefore CD=½, Cr =f, Cy>=½, Cm=½. Hence DE= √(EC 2 —CD 2 )= √(1—½)=½ x /3; and in like manner rs= ⅜√55, pq=⅛* t ∕ 15,wm=½ x ∕ 63. Therefore,

F+L=l+⅜√iΓ= l∙8660 4 E=½√55+½√63 = 7∙6767 2 R=⅜√15 = 1∙9365

The sum * 11∙4792

Multiply by ⅜ D= f ⅛

The product is * .4783

Subtract the triangle CDE .2165

There remains the sector CBE= .2618

The triple of which is the quadrant ABC= .7854

Ex. 2. To find the area of the hyperbola FDM, of which the absciss FM=10, the semiordinate MD=12, and semitransverse C F = 15.

Let FM be divided into five equal parts by the semiordinates HI, mn, pq, rs. Thus CH = 17, Cm =19, Cp = 21, Cr=23, CM=25. Now, since from the nature of the curve, √(CM 2 -CF 2 ): MD :: √(CH 2 -CF 2 ): HI (Conic Sections, part iii. prop. 22. and GEOM.part i. sect, iv∙theor. 12,) that is, in numbers, 20: 12 :: 8: HI, therefore HI=⅞ 4 . In like manner we find τnτ⅛=∣> x ∕34, pq∙=A∕ 6,andrs=⅛ 2 x ∕19∙ Therefore

F+L(=HI + MD) = 16∙8 , 4E(=4≡+4rs) = 68∙8399

2R(=2 w ) = 17∙6363

The figure HIDM = 103-2762 × 2 =68∙8508. to which adding FIH, considered as a portion of a parabola, we have 75∙245 for the area of the hyperbola.

OF LAND-SURVEYING.

The instruments most commonly employed in land-surveying, are the Chain, the Plane Table, and Cross.

A statute acre of land being 160 square poles, the chain is made four poles, or sixty-six feet in length, that ten square chains, or 100,000 square links) may be equal to an acre. Hence each link is 7 ’ 92 inches in length.

The plane table is used for drawing a plan of a field, and taking such angles as are necessary to calculate its area. It is of a rectangular form, and is surrounded by a moveable frame, by means of which a sheet of paper may be fixed to its surface. It is furnished with an index by which a line may be drawn on the paper in the direction of any object in the field, and with scales of equal parts by which such lines may be made proportional to the distances of the objects from the plane table when measured by the chain, and its frame is divided into degrees for observing angles.

The cross consists of two pair of sights set at right angles to each other upon a staff having a pike at the bottom to stick into the ground. Its use is to determine the points where a perpendicular drawn from any object to a line will meet that line; and this is effected by finding by trials a point in the line, such that the cross being fixed over it so that one pair of the sights may be in the direction of the line, the object from which the perpendicular is to be drawn may be seen through the other pair; then the point thus found will be the bottom of the perpendicular, as is evident.

A theodolite may also be applied with great advantage to land-surveying, more especially when the ground to be measured is of great extent.

In addition to these, there are other instruments employed in surveying, as the perambulator, which is used for measuring roads and other great distances. Levels, with telescopic or other sights, which are used to determine how much one place is higher or lower than another. An ofsett-staff for measuring the ofsets and other short distances. Ten small arrows, or rods of iron or wood, which are used to mark the end of every chain length. Pickets or staves with flags to be set up as marks or objects of direction; and, lastly, scales, compasses, &c. for protracting and measuring the plan upon paper.

The observations and measurements ^are to be regularly entered as they are taken, in a book which is called the field book, and which serves as a register of all that is done or occurs in the course of the survey.

To Measure a Field by the Chain.

Let ArøBCraDç represent a field to be measured. Let it be resolved into the triangles AwB, ABD, BCD, AgD. Let all the sides of the large triangles ABD, BCD, and the perpendiculars of the small ones AmB, AyD from their vertices m, q, be measured by the chain, and the areas calculated by the rules delivered in this section, and their amount is the area of the whole. But if, on account of the curvature of its sides the field cannot be wholly resolved into triangles, then, either a straight line may be drawn over the curve side, so that the parts cut off from the field, and those added to it may be nearly equal; or, without going beyond the bounds of the field, the curvilineal spaces may be measured by the rule given in prob. xvi. of this section.

To Measure a Field with the Plane Table.

Let the plane table be fixed at F, about the middle of the field ABCDE, and its distances FA, FB, FC, &c. from the several corners of the field measured by the chain. Let the index be directed from any point assumed on the paper to the points A, B, C, D, &c., successively, and the lines Fa, Fb, Fc, &c., drawn in these directions.

Let the angles contained by these lines be observed, and the lines themselves made proportional to the distances measured. Then their extremities being joined, there will be formed [7:14:572]a figure ab c d e similar to that of the field; and the area , of the field may be found by calculating the areas of the several triangles of which it consists.

To Plan a Field from a given Ease Line.

Let two stations A, B be taken within the field, but not in the same straight line with any of its corners; and let their distance be measured. Then the plane table being fixed at A, and the point a assumed on its surface directly above A, let its index be directed to B, and the straight line a b drawn along the side of it to represent AB. Also, let the index be directed from a to an object at the corner C, and an indefinite straight line drawn in that direction, and so of every other corner successively. Next, let the plane table be set at B, so that b may be directly over B, and b a in the same direction with BA, and let a straight-line be drawn from b in the direction BC. The intersection of this line with the former, it is evident will determine the point C, and the triangle a b c on the paper will be similar to ABC in the field. In this manner all the other points are to be determined, and these being joined there will be an exact representation of the field.

If the angles at both stations were observed, as the distance between them is given, the area of the field might be calculated from these data, but the operation is too tedious for practice. It is usual therefore to measure such lines in the figure that has been constructed as will render the calculation easy.

SECTION III. MENSURATION OF SOLIDS.

Problem I.

To find the surface of a right prism, or cylinder.

Rule.

Multiply the perimeter of the end 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.

The truth of this rule will be evident, if it be considered that the sides of a right prism are rectangles, whose common length is the same as the length of the solid, and their breadths taken altogether make up the perimeter of the ends of the prism. And as a cylinder may be considered as the limit of all the prisms which can be inscribed in or circumscribed about its base, so the surface of the cylinder will be the limit of the surfaces of these prisms, and the expression for that limit is evidently the product of the circular base by its height. Or a cylinder may be con-: sidered as a prism of an indefinitely great number of sides.

Ex. 1. What is the surface of a cube, the length of its side AB being 20 feet?

Here 4 × 20= 80 theperim. of end.

And 80 × 20= 1600 the four sides.

And 2 × 20 X 20 = ⅜00 the~top and bottom.

The sum 2400=theareaorsurfacc.

Ex. 2 What is the convex surface of a cylinder whose length AB is 20 feet, and the circumference of its base 3 feet? Here 3 × 20=60 feet, the answer.

Problem II.

To find the surface of a right pyramid or cone. Rule.

Multiply the perimeter of the base by the slant height or length of the side, and half the product will evidently be the surface of the sides, or the sum of the areas of all the triangles which form it. To which add the area of the end or base, if required.

Note. Here a cone is -considered as a pyramid of an indefinitely great number of sides.

Ex. 1. What is the upright surface of a triangular pyramid, A BCD, the slant height AE, being 20 feet, and each side of the base 3 feet? ττ 3 × 3 × 20 ,

Here =90 feet, the surface.

zι

Ex. 2. Required the convex surface of a cone, the slant height AB being 50 feet, and the diameter of its base 8⅜ feet.

Here 8∙5 ×3∙1416=circum. of base. . a j 8∙5×3∙1416×50 _ 1

And =6667*59, the answer.

Problem HI.

To find the surface of the frustum of a right pyramid or cone, being the lower part, when the top is cut off by a plane parallel to the base.

Rule.

Add together the perimeters of the two ends, and multiply their sum by the slant height, and take half the product for the answer.

The truth of this rule will be evident if it be considered that the sides of the frustum are trapezoids, whose parallel sides bound its top and base, and whose common breadth is its slant height.

Example. How many square feet are in the surface of a frustum AG of a square pyramid, (see fig. to prob, vi.) whose slant height RE is 10 feet; also each side of the greater end AC is 3 feet 4 inches, and each side of the lesser end EG 2 feet 2 inches?

Here 3½×4=13½ the per. of gr. end.

And 2½ × 4=8∣ the per. of less end.

And their sum is 22 feet.

22 X 10

Therefore —-—=110 feet, is the answer.

Problem IV.

To find the solid content of any prism or cylinder. Rule.

Find the area of the base or end of the figure, and multiply it by the height or length,, and the product will be the area.

This rule follows immediately from theor. 11, and theor. 2, part ii. sect. iii. Geometry. Ex. 1. What is the solid content of a cube AG, the length of whose sides is 24 inches?

Here 24 × 24=576 sq. inches, the area of the end.

And 576 × 24=13824 cub. inches is the solidity.

Ex. 2. Required the content of a triangular prism, whose [7:14:573]length AD is 20 feet, and the sides of its triangular base ABC, are 3, 4, and 5 feet.

First, the area of the triangular base is found by Rule iii. of Prob. 3, sect. ii. to be

√(6 × 3 × 2 × 1 )=6 sq. feet.

Therefore 6 × 20=120 cub. feet, the solidity.

Ex. 3. The Winchester bushel is a cylinder 18½ inches in diameter, and eight inches deep. How many cubic inches does it contain?

By prop. 10 of sect. ii. the area of its base is

•7854 × 18∙5 2 =268∙803 sq. inches;

Therefore 268∙803 × 8=2150∙424 is the solid content.

Problem V.

To find the solid content of any pyramid or cone. Rule.

Find the area of the base, and multiply that area by the height, and one-third of the product will be the content of the solid.

This rule is demonstrated in Geometry, part ii. sect. ii. prop, xvii. cor. 1, and sect. iii. theor. 3.

Ex. 1. What is the content of a triangular pyramid ABCD, whose perpendicular height AF is 30 feet, and each side of its base BCD is three feet? (See fig. in prob, ii. sect, iii.)

First, the area of the base, as found by rule iii. of prob. Q QPPt. 11- iς

√(4∙5 × 1∙5 × 1∙5 × 1∙5)=3∙89711.

Therefore ———=38 ’ 9711 cub. feet is the solid con-

o tent.

Ex. 2. What is the solid content of a cone, the radius BC of its base being nine inches, and its height AC 15 feet? (See fig. in prob. ii. sect, iii∙)

3 2

Here ∙7854 × — =1-76715 is the area of the base in square feet.

And ———=8∙8357 cub. feet is the solid content.

o Problem VI.

To find the solidity of the frustum of a cone or pyramid. Rule.

Add into one sum the areas of the two ends, and the mean proportional between them, that is, the square root of their product, and one-third of that sum will be a mean area, which being multiplied by the perpendicular height or length of the frustum will give the content.

Demonstration. LetPABCD be any pyramid, and AG a frustum of it contained between A BCD its base, and EFGH, a plane parallel to the base. Put a for the side of a square equal to ÁC the base of the frustum; b for the side of a square equal to EG its top; h for LM the height of the frustum, and c for PL the height of the part of the pyramid above the frustum. Then α 2 is the area of the base of the frustum; b 2 is the area of its top; ½ α β (Ä+<?) is the solid content of the whole pyramid, (Geom. part ii. sect. ii. theor. 17), ⅜ δ 2 c is the content of its upper part; and therefore

⅜∣α 2 (δ+c)-δ 2 cj-

is the solid content of the frustum itself. Now the base and top of the frustum being similar figures, (Geom. part ii. sect. ii. thcor. 13), their areas are to one another as the squares of AB and EF, their homologous sides, (part i. sect. iv. theor. 27). But AB: EF:: BP: PF (sect. iv. theor. 20.) :: PM: PL; therefore the area of the base of the frustum is to the area of its top as PM 2 : PL 2 , that is, 2 : δ 2 :: (A+c) 2 .∙ c 2 , and consequently a : b :: Jι4 r c : c; hence a c—b hA∙b c, and c—— ∖ , and A + c=-^-A. Let a — o a — b these values of c and 7⅛+cbe now substituted in the preceding expression for the content of the frustum, and it will become by proper reduction,

3 ' l a "

Let the numerator of the fractional part of this formula be actually divided by its denominator, and we shall obtain for the area of the frustum this more simple expression, ⅜ h (α 2 + α bA-b 2 ), which formula, when expressed in words, is the rule. And as a cone may be considered as the limit of all the pyramids that can be inscribed in it, when the number of sides is conceived indefinitely increased, it is evident that the rule will apply alike to the cone and pyramid.

Ex. 1. Required the solidity of the frustum of a hexagonal pyramid, the side of whose greater end is four feet, and that of its lesser end is three feet, and its height nine feet.

First, by prob. 7, sect. ii. the area of the base of the frustum is found to be 41∙569, and the area of its lesser end 23∙383 square feet. And the mean proportional between these is

√ (41∙569×23∙383)=31∙177.

Hence the mean area is

⅜ (23∙383+41∙569+31∙177)=32∙043∙ And the solid content of the frustum is

32∙043×9=288∙387 cubic feet.

Ex. 2. What is the solidity of the frustum of a cone, the diameter of the greater end being five feet, that of the lesser end three feet, and the altitude nine feet?

Here the area of the greater end is (by prob. 10, sect, ii.) 5 2 × ∙7854, and the area of the lesser end is 3 2 × ∙7854, and the mean proportional between them is √(5 2 × 3 2 × +854 2 ) =5 ×3× - 7854; therefore the mean area is

× (5 2 +3 2 +5 ×3)=12∙8282. ó

And the content of the frustum

12∙8282 ×9=H5∙4538 cub. feet.

Problem VII

To find the surface of a sphere, or of any segment or zone of it.

Rule.

Multiply the circumference of the sphere by the height of the part required, and the product will be the curve surface, whether it be a segment, a zone, or the whole sphere.

Note. The height of the whole sphere is its diameter.

The truth of this rule has been already shown in the article Fluxions, sect.

165. It may, however, be deduced from principles more elementary, by reasoning as follows: Let PCQ be a semicircle, and AB, BC,CD, &c., several successive sides of a regular polygon inscribed in it. Conceive[7:14:574] the semicircle to revolve about the diameter PQ as an axis, then the arch ABODE will generate a portion of the surface of a sphere, and the chords AB, BC, CD, &c., will generate The surfaces of frustums of cones; and it is easy to see that the number of chords may be so great that the surface which they generate shall differ from the surface generated by the arch ACE by a quantity which is less than any assigned quantity. Bisect AB inL, and draw AF, LM, BG, CH, &c., perpendicular to PQ. For the sake of brevity, let cire. A F denote the circumference of a circle whose radius is AF. Then because AF, BG, LM, are to each other respectively as circ. AF, circ. BG, circ. LM, (Geom. part i. sect. vi. prop. 4. cor. i.), and because ½ (AF + BG)=LM, therefore ½ (circ. NFA-circ. BG)=circ. LM. Now the area of the surface generated by the chord AB is ½ (cire. AF A-circ. BG) ×AB, (prob. 3), therefore the same area is also equal to (circ. LM) ×AB. Draw AO parallel to FG, and draw LN to the centre of the circle. Then the triangles AOB, LMN are manifestly similar; therefore AB: AO :: NL: LM :: cire. NL: cire. LM; and hence AO ×czrc. NL=AB×circ. LM. But this last quantity has been proved equal to the surface generated by AB, therefore the same surface is equal to AO ×cfre. NL, or to FG ×circ. NL, that is, to the rectangle contained by FG and the circumference of a circle inscribed in the polygon. In the same way it may be shown that the surfaces generated by BC, CD, DE, are respectively equal to Gü×circ. LN, HI ×circ. LN, IK×czrc. LN. Therefore the whole surface generated by the chords AB, BC, CD, DE, &c., is equal to(FG+GH + HI+IK) × circ. LN=FK × circ. LN. Conceive now the number of chords between A and E to be indefinitely increased; then, observing that the limit of the surface generated by the chords is the surface generated by the arch ABCDE, and that the limit of NL is NP, the radius of the generating circle, it follows that the spherical surface or zone generated by the arch ACE is equal to the product of the circumference of the sphere by FK the height of the zone.

Ex 1. What is the superficies of a globe whose diameter is 17 inches?

First 17 × 3∙1416=53∙4072 inches=the circum.

Then 53∙4072 X 17=907 ’ 9224 sq. inches=6∙305 sq. feet, the answer.

Ex. 2. What is the convex surface of a segment eight inches in height cut off from the same globe?

Here 53∙4072 × 8=427 ’ 2576 sq. inches=2 ’ 967 sq. feet, the answer.

Problem VIII.

To find the solidity of a sphere.

Rule I.

Multiply the area of a great circle of the sphere by its diameter, and take f of the product for the content.

Rule II.

Multiply the cube of the diameter by the decimal -5236 for the content.

The first of these rules is demonstrated in Geometry, part i. sect. iii. theor. 6. And the second is deduced from the first, thus: put d for the diameter of the sphere, then d 2 × •7854 is the area of a great circle of the sphere, and by the first rule fd×d 2 × ∙7854=cf 3 ×∙5236 is its content.

Example. What is the content of a sphere whose diameter is six feet?

Answer 6 3 × *5236= 113O976 cub. feet.

Problem IX.

To find the solid content of a spherical segment.

Rule.

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 the product by the decimal •5236 for the content.

This rule has been investigated in Fluxions, sect. 163. But it may be proved in a more elementary manner by means of the following axiom. If two solids be contained between two parallel planes, and if the sections of these solids by a third plane parallel to the other two, at any altitude, be always equal to one another, then the solids themselves are equal. Or more generally thus. If two solids between two parallel planes be such, that any sections of them by a third plane parallel to the other two have always to each other the same given ratio, then the solids themselves are to one another in that ratio. We have given this proposition in the form of an axiom, for the sake of brevity, but its truth may be strictly demonstrated, as has been done when treating of pyramids and the sphere, in Geometry, part ii. sect. ii. and iii.

Let us now suppose ABE to be a quadrant, C the centre of the circle, AFEC a square described about the quadrant, and CF the diagonal of the square. Suppose the figures to revolve about AC as an axis, then the quadrant will generate a hemisphere, the triangle ACF will generate a cone, and the square AE a cylinder. Let these three solids be cut by a plane perpendicular to the axis, and meeting the plane of the square, in the line DHBG, and join CB. Then because CDB is a right-angled triangle, a circle described with CB as a radius is equal to two circles described with CD and DB as radii. (Geom. part i. sect. vi. prop. 4, cor. 2). But CB=DG, and since CA=AF, therefore CD=DH; therefore the circle described with the radius DG, is equal to the sum of the circles described with the radii DH, DB; that is, the section of the cylinder at any altitude, is equal to the corresponding sections of the sphere and cone taken together. Consequently, by the foregoing axiom, the cylinder is equal to the hemisphere and cone taken together, and also the segment of the cylinder between the planes AF, DG is equal to the sum of the segments of the hemisphere and cone contained between the same planes. Put 2 CE, or 2 AF, the diameter of the circle, —d, and AD, the height of the spherical segment, =h. Then AC=½ <7 and 2 CA—2 AD=2 CD=<7-2 A. Let n denote the number ∙7∙854. Then the area of the base common to the conic frustum AH, and cylinder AG, is η cP, (sect. ii. prob. 10), and the area of the top of the frustum is n (d —2Λ) β , and the mean proportional between these areas is n (d —2 h)d. Therefore the solid content of the frustum is (by prob. 6 of this sect.)

⅜ nd z A-n{d — 2h) s -γnd(d — 2ä) ; ×A.

~η d a h — 2 n d A 2 +£ n h 3 .

Now the solid content of the cylinder is n d* 2 h: (prob. 1). Therefore the solid content of the spherical segment, (which is equal to the difference between the cylinder AG and the conic frustum AH), is equal to n d q h —(n <7 2 h —2 n d h 2 + I n h z ), that is, to 2 n d h q — f n h z , or to

Q A⅛ ∑Ti(3 íZ—2 A) A 2 , ó

2 7z 2 *7854 which expression, if it be considered that —or

3 3 [7:14:575]is equal to ∙5236, is evidently the same as that given by the rule.

Example. In a sphere whose diameter is 21, what is the solidity of a segment whose height is 4∙5 inches?

First 3 × 21— 2×4∙5=54.

Then 54×4∙5 ×4∙5 × ∙5236=572∙5566 inches, the solidity required.

Problem X.

To find the solid content of a paraboloid, or solid, produced by the rotation of a parabola about its axis.

Rule.

Multiply the area of the base by the height, and take half the product for the content.

To demonstrate this rule, let AGC and BHD be two equal semi-parabolas lying in contrary directions, and having their vertices at the extremity of the line AB. Let AD, BC be ordinates to the curves. Complete the rectangle AB CD, and conceive it to revolve about AB as an axis; then the rectangle will generate a cylinder, the radius of whose base will be AD, and the two semi-parabolas will generate two equal paraboloids having the same base and altitude as the cylinder. Let a plane be drawn perpendicular to the axis, and let FHGÉ be the common section of this plane and the generating figure. Let P denote the parameter of the axis. Then since

EG 2 =P×AE, and EH 2 =P×EB, EG 2 + EH 2 =P XAB=CB 2 Hence it appears, as in the demonstration of the preceding rule, that of the solids described by ADCB, ACB, ADB between the same parallel planes, the section of the cylinder at any altitude is equal to the corresponding sections of the paraboloids taken together. Consequently (by the axiom) the cylinder is equal to both the paraboloids taken together; hence each is half a cylinder of the same base and altitude agreeing with the rule.

The same thing is also proved in Fluxions, sect. 163.

Example. If the diameter of the base of a paraboloid be 10 and its height 12 feet, what is its content?

Here 10 ×∙7854=7 ’ 854 the area of the base.

And ½×7 ’ 854× 12=47*124 cub. feet is the solidity.

Problem XI.

To find the solid content of a frustum of a paraboloid. Rule.

Add together the areas of the circular ends, then multiply that sum by the height of the frustum, and take half the product for its solid content.

To prove this rule put A and a for the greater and lesser ends of the frustum, and h for its height; also let c denote the height of the portion cut off from the complete paraboloid, so as to form the frustum. Then, by the last problem, the content of the complete paraboloid is ½ A (½+c), and the content of the part cut off is ½ a c, therefore the content of the frustum is

½-£ A (/*+<?)— a c) = ½ { A A + c(A—α) j.

But from the nature of the parabola, c : h+c :: a : A; hence * 1! 4 aĭl

A c=a k-↑-a c and c= i -.

A— a

Let this value of c be substituted instead of it in the above expression for the content of the frustum, and it becomes

½(A h+a A)=½Λ(A+α), and hence is derived the rule.

Example. Required the solidity of the frustum of a paraboloid, the diameter of the greater end being 58, and that of the lesser end 30, and the height 18.

First, (by prob. 10, sect, ii.) we find the areas of the ends to be 58 2 ×∙7854, and 30 2 × , 7854; therefore their sum is (58 2 +30 2 ) X ∙7854=4264 × ∙7854; and the content of the figure is ½ × 4264 × ∙7854×18=03140∙5104, the answer.

Problem XI.

To find the solid content of a parabolic spindle or solid generated by the rotation of AEB an arc of a parabola, about AB an ordinate to the axis.

Rule.

Multiply the area of the middle section by the length, and take r ⅛ of the product for the content of the solid.

For the investigation of this rule we must refer the reader to Fluxions, § 163. Ex. 4.

Example. The length of the parabolic spindle AEBeA is 60, and the middle diameter E e 34; what is the solidity?

Here 34 2 × ∙7854 is area of the middle section.

Therefore 34 2 × ∙7854 × 60 χ τ ⅛=29053∙5168. is the solidity required.

Problem XII.

To find the solid content of the frustum of a parabolic spindle, one of the ends of the frustum passing through the centre of the spindle.

Rule.

Add into one sum eight times the square of the diameter of the greater end, and three times the square of the lesser end, and four times the product of the diameters; multiply the sum by the length, and the product multiplied by •05236, or τ 1 3 ∙ of ∙7854, will be the content.

For, referring the reader to Fluxions, § 163. Ex. 4. as before, and substituting lι for AC=½ b, but, in other respects, retaining the figure and notation, we have this general expression for the segment AP p, πx)Pι-x 2 ∖ which, when x~h gives - π ^∖, for the value of the semi- 15 d 2 spindle. From this quantity let the former be subtracted, and there will remain

8πA 5 π ae 3 /4 ∕z 2 x 2 ∖ i6^~-ςrC3 — + M

for the content of the frustum. In this expression let z be put instead of h — χ or CD, and, denoting CE the radius of

7 i 2

the greater end of the spindle by d, let — be substituted instead of its value a. Then we shall have the content of the frustum otherwise expressed by

τrrf 2 z f 7 , 2Λ 2 z* , zl· )

- τ -∣ a∙—r + τ∫

which value, by putting K∕A~~^) in its two last terms in. stead of z i is changed to [7:14:576]15 and hence is derived the preceding rule.

Example. Suppose the diameter of the greater end to be 8, and the diameter of the lesser end 6, and the length 10, required the content?

First8×8 +3 × 6 2 +4 × 8 × 6=812.

Then, 812 × 10 × ∙05236=425∙1632, the content.

Problem XIII.

To find the solid content of a spheroid, or solid generated by the rotation of an ellipse about either axis.

Rule.

Multiply continually together the fixed axis, and the square of the revolving axis, and the number ∙5236 or ½ of 3-1416, and the last product will be the solidity.

For, let the semiellipse ADB, and semicircle AEB, revolve about the same fixed axis AB, and thus generate a spheroid and sphere. Let CD the revolving semiaxis of the ellipse meet the circle in E, and draw QP any ordinate to the fixed axis meeting the circle in R. Then, from the nature of the ellipse PQ 2 : PR 2 :: CD 2 : CE 2 or CA 2 (Conic Sections, part ii. prop. 13, cor. 3.) Hence it follows, (Geom. part i. sect. vi. prop. 4.), that every section of the spheroid is to the corresponding section of the sphere in the same given ratio, namely, that of the square of the revolving axis to the square of the fixed axis; therefore (Axiom in the dem. of prob. 9∙) the whole spheroid is to the whole sphere in the same ratio. That is, (because the content of die sphere is AB 3 ×∙5236) AB 2 : (2CD 2 ) :: AB 3 ×-5236': (the cont. of spheroid.) Hence the content of the spheroid is AB×(2CD) 2 ×∙5236.

Ex. 1. What is the solid content of an oblong spheroid, or solid generated by the rotation of an ellipse about its greater axis, the axes being 50, and 30?

Here 50 × 30 2 × ∙5236=23562, the content.

Ex. What is the solid content of an oblate spheroid, or solid generated by the rotation of an ellipse about its lesser axis, the two axes being as before?

Here 30 × 50 2 × ∙5236=39270 the ansiver.

Problem XIV.

To find the solid content of the frustum of a spheroid, its ends being perpendicular to the fixed axis, and one of them passing through the centre.

Rule.

To the area of the less end add twice that of the greater, multiply the sum by the altitude of the frustum, and ½ of the product will be the content.

Note. This rule will also apply to the sphere.

Demonstration. Let ABE be a quadrant of an ellipses C its centre, CAFE its circumscribed rectangle, and CF its diagonal. Draw any straight line DG parallel to CE, meeting AC, CF, ABE and EF in D, H, B, and G. Then by Conic Sections, part ii. prop. 13.

CE 2 or AF 2 : DB 2 :: CA 2 : CA 2 -CD 2 , and by sim. tr. AF 2 : DH 2 :: CA 2 : CD 2 .

Therefore (Geometry, part i. sect. iii. theor. 10.),

AF 2 : DB 2 + DH 2 :: CA 2 ∖CA β ;

Hence DB 2 + DH 2 =AF 2 =DG 2 .

Conceive now the figure to revolve about AC as an axis, so that the elliptic quadrant may generate the half of a spheroid, the rectangle AE a cylinder, and the triangle ACF a cone; then it is evident (as was shewn in the case of the sphere in prob. 9∙) that every section of the first of these solids by a plane perpendicular to the axis is equal to the difference of the sections of the other two, and consequently that the frustum of the spheroid between CE and DG is equal to the difference between the cylinder having DG or CE for the radius of its base, and CD for its altitude, and the cone having DH for the radius of its base, and DC for its altitude.

Put n for the number 3∙1416, then (prob. 4.) the content of the cylinder is 4 n × DG 2 × CD, and (prob. 5.) the con-tentjofthe cone is f n × DH 2 × CD, and therefore the content of the frustum of the spheroid is

4n×CD (DG 2 —½ DH 2 ).

But it was shewn that DH 2 =DG 2 —DB 2 ; therefore the content of the frustum is also CE 2 equal to f n×CD (2CE 2 + DB 2 ), and hence is derived the rule.

Ex. Suppose the greater end of the frustum to be 15, the less end 9, and the length 10 inches, required the content?

The area of the greater end is 15 2 ×∙7854, and the area of the less 9 2 × ∙7854, therefore the content is •7854 (9 2 + 2× 15 2 ) ×⅞ fi =1390 ’ 158 cubic inches.

Problem XV.

To find the solid content of a hyperboloid, or’solid generated by the rotation of a hyperbola about its transverse axis.

Rule.

As the sum of the transverse axis and the height of the solid is to the sum of the said transverse axis and ∙∣ of the height so is half the cylinder of the same base and altitude to the solidity of the hyperboloid.

Demonstration. Let B Ab be a hyperbola, Aa its transverse axis, C its centre, CF, Cfits asymptotes, FA∕ a tangent at its vertex. Draw FE parallel to CA, and draw any straight line parallel to F∕i meeting the asymptotes in II and Ä, the curve in B and b, the axis in D, and the line FE in G. Then, because AF 2 =BH×⅛B (Conic Sections, partiii. prop. 16, cor. 4), and HB × AB=DH 2 —

DB 2 (Geom. part i. sect. iv. theor. 12), therefore AF β = DH 2 —DB 2 , and DB 2 =HD 2 —DG 2 . Hence it appears, that if the figure be conceived to revolve about CA as an axis, so that the hyperbolic arc AB may generate a hyperboloid, the triangle DCH a cone, and the rectangle DAFG a cylinder, any section of the first of these solids by a plane H A, perpendicular to the axis, will be equal to the difference of the sections of the other two by the same plane. Therefore the hyperboloid BA b is equal to the difference between the conic frustum FH∕yfand the cylinder FGg/î Let A a, the transverse axis, be denoted by p, Ff≈ its conjugate axis by q, AD the height of the solid by h, B b its base by b. Then, because by similar triangles, &c.

CΛ: CD:: F/: H h :: F ∕ 2 : F∕×H h, therefore F∕ × H × F∕ 2 =^⅛^- = 7 a +

CA ⅛p 2 hq 2 P [7:14:577] Now F∕ 2 =<7 2 , and Hh 2 (=Bδ 2 +F∕ 2 )=5 2 +<∕ 2 , therefore putting n for ∙7854, we have (by prob. 6.) the content of the conic frustum FII∕∕∕ equal to from this subtract n h q 2 i the expression for the content of the cylinder FG⅛∕⅛ and there will remain ; i*(i ∙ + ≡⅛Λ

3∖ p J for the content of the hyperboloid. But from the nature the hyperbola

A α 2 : F∕ 2 :: AD×D a l’BD 2 , that is, j» 2 : q 2 :: (/?+AE) h : ½ô 2 ; , . ς ahq 2 pb 2 therefore =<77—rτ∖ ’ and * hence the content of the P 2 (p+li) hyperbola is also equal to nil ∕ι 3 , pb 2 ∖ nhb 2 p + lh

3∖ + 2(; t > + ∕ 1 √=-×^+A∙

Now if it be considered that the quantity n h b 2 is the expression for the content of a cylinder whose base is b and height h, it will appear evident, that this last formula is the same as would result from the foregoing rule.

Ex. Suppose the height of the hyperboloid to be 10, the radius of its base 12, and its transverse axis 30. What is its content?

1. Because a cylinder of the same base and altitude is 24 2 × ’ 7854 χ 10, therefore, we have the proportion,

110 24 2 ×∙7854 × 10 .

’ 3 2 :

24 2 × k 7854× 10 × 110 n 1

40><3×2 =2073∙456, the content of the solid as required.

OF GUAGING.

Gūaging treats of the measuring of casks, and other things falling under the cognizance of the officers of the excise, and it has received its name from a guage or rod used by the practitioners of the art.

From the way in which casks are constructed, they are evidently solids of no determinate geometrical figure. It is, however, usual to consider them as having one or other of the four following forms:

1. The middle frustum of a spheroid.

2. The middle frustum of a parabolic spindle.

3. The two equal frustums of a paraboloid.

4. The two equal frustums of a cone.

We have already given rules by which the content of each of these solids may be found in cubic feet, inches, &c. But as it is usual to express the contents of casks in gallons, we shall give the rules again in a form suited to that mode of estimating capacity. Observing that in each case the lineal dimensions of the cask are supposed to be taken in inches.

Problem I.

To find the content of a cask of the first, or spheroidal variety.

Rule.

To the square of the head diameter add double the square of the bung diameter, and multiply the sum by the length of the cask. Then let the product be multiplied by •0009442, or divided by 1059*1 for imperial gallons.

The truth of this rule may be proved thus. Put B for FG, the bung diameter, (see fig. next page) H for AH the head diameter, and L for AD, the length of the cask, then (by prob. 14.) the content of the cask is (2B 2 +∏ 2 )L ×-⅛-> which being divided by 277*274 (the cubic inches in an imperial gallon) gives (2B 2 + H 2 ) L× •0009442, or (2B 2 +H 2 )L× —- ι-- for the content in imperial gallons.

Ex. Suppose the bung and head diameters to be 32 and 24, and the length 40 inches. Required the contènt?

Here (2 × 32 2 +24 2 ) × 40 × ∙0009442=99 1 103 imperial gallons.

Problem II.

To find the content of a cask of the second, or parabolic spindle form.

Rule.

To the square of the head diameter add double that of the bung diameter, and from the sum take f, or 1 ⅛∙ of the square of the difference of the said diameters. Then multiply the remainder by the length, and the product multiplied, or divided by the same number as in the rule to last problem, will give the content.

For by problem 12, the content in inches is 8_B.-_+4BH + 3m x . 78ä4L; lö

and this formula may be otherwise expressed thus,

12 B≈ + H≈-f (B-H)∙ ) xψxL, and hence is derived the rule, the multiplier or divisor being evidently the same as in last problem.

Ex. The dimensions of a cask being the same as in last problem; required the contents?

Ans. (2 × 32 2 + 24 2 - 2 × 8 2 )×40 × ∙0009442=98∙136, the content in imperial gallons.

Problem HI.

To find the content of a cask of the third or paraboloidal variety.

Rule.

To the square of the bung diameter add the square of the head diameter, and multiply the sum by the length; then, if the product be multiplied by ∙001416, or divided by706∙2, the result will be the content in imperial gallons.

For by problem 10, the content in inches is ½(B 2 +H 2 ) ×∙7854 L; and this expression being divided by 277 ’ 274 gives (B 2 + ∏ 1 )χ∙001416 L, or (B 2 + ∏ 2 )×-L ×L ι uo — for the content in imperial gallons.

Ex. Suppose the dimensions of a cask, as before; required the content?

Ans. (32 2 + 24 2 )×40 ×∙001416=90’62, the content in imperial gallons.

Problem IV.

To find the content of a cask of the fourth or conical variety.

Rule.

To three times the square of the sum of the diameters add the square of the difference of the diameters; multiply the sum by the length; and multiply the result by .00236 or divide it by 4237'3, for the content in imperial gallons.
[7:14:578]

For by problem 6, the content in inches is ⅜ (B 2 +BH + ∏ 2 ) × , 7854 L, which expression is equivalent to

{3 (B + H)≡ + (B-H)* } X'^- 4 L∙

Now divided by 277*274 gives ∙000236≈-* the multiplier for imperial gallons.

Ex. Supposing the dimensions of a cask as before; what is its contents?

Ans. (3×56 2 + 8 2 )×40×∙000236≈=89∙415 the content in imperial gallons.

As these four forms of casks are merely hypothetical, it may be expected that some uncertainty will attend the application of the rules to actual measurement. The following rule, however, given by Dr. Hutton in his Treatise on Mensuration, will apply equally to any cask whatever; and as its truth had been proved by several casks which have been actually filled with a true gallon measure, after their contents were computed by it, we presume it is more to be depended upon than the others.

Rule.

Add into one sum thirty-nine times the square of the bung diameter, twenty-five times the square of the head diameter, and twenty-six times the product of the diameters; multiply the sum by the length, and the product by •00003147; the last result gives the content in Imperial gallons.

In investigating this rule the author assumed as a hypothesis, that one-third of a cask at each end is nearly the frustum of a cone, and that the middle part may be taken as the middle frustum of a parabolic spindle. This being supposed, let AB and CD be the two right-lined parts and BC the parabolic part; produce AB and DC to meet in E, and draw lines as in the figure. Let L, B, and H denote the same as before. Then, since AB has the same direction as EB at B, ABE will be a tangent to a parabola BF, and therefore FI=½ El.

But BI=½ AK, and hence, by sim. triangles, EI=½ EK; consequently FI=½ EI=⅜ EK=½ FK= τ ⅛ (B—H); so that the common diameter BL=FG—2 FI=B j(B H)=

∣(4 B + H), which call C. Now by the rules of parabolic spindles and conic frustums we obtain (putting n for ∙7854), 8 B 2 + 4BC + 3C 2 Ln_ 328 B 2 + 44 BH + 3 H 2 15 × 3 “25×45

C 2 i CH I H g

× L n for the parabolic or middle part; and —— -—

• * ∙

2L» 160 B 2 X 280 BH + 310H 2 τ fi .

× — = 25 χ 45 ĭ × L η for the two ends, and the sum of these two gives, after proper reduction, (39B 2 +26Bn + 25H 2 ) × nearly, for the content y U

in inches. And the quantity or ^ 8 θ 4 being divided by 277∙274 gives ∙00003147, the multiplier for converting’the inches into Imperial gallons.

Ex. Suppose a cask to have the same dimensions as in the four former rules; required the content.

Here (39 × 32 2 + 26 × 32 × 24 + 25 × 24 2 ) × 40 × •00003147 gives 93∙5 Imperial gallons for the content required.