Every branch of the mathematics 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 mathematics 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, 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 90 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 of Lineal Measures

| Measure | Equivalent | |---------------|------------| | 12 Inches | 1 Foot | | 3 Feet | 1 Yard | | 6 Feet | 1 Fathom | | 5½ Yards | 1 Pole, Rod, or Perch | | 40 Poles | 1 Furlong | | 8 Furlongs | 1 Mile | | 3 Miles | 1 League | | 69½ Miles nearly | 1 Degree | | 360 degrees | The earth's circumference |

Note. An inch is supposed equal to three barleycorns in length.

4 Inches = 1 Hand, or handsbreadth. 5 Feet = 1 Geometrical Pace. 4 Poles or 66 Feet = 1 English chain. 100 links each 7½ inches = 1 English chain. 74 Feet = 1 Scots chain.

### Table of Square Measures

| Measure | Equivalent | |---------------|------------| | 144 Square Inches | 1 Foot square | | 9 Square Feet | 1 Yard | | 30¼ Square Yards | 1 Pole | | 40 Square Poles | 1 Rod | | 4 Rods or 160 Square Poles | 1 Acre |

10 Square Chains = 1 Acre. 640 Square Acres = 1 Square Mile.

Note. The Scots acre is to the English acre as 100000 to 78694.

### Table of Solid Measures

| Measure | Equivalent | |---------------|------------| | 1728 Cubic Inches | 1 Cubic Foot | | 27 Cubic Feet | 1 Cubic Yard |

Note. 282 Cubic inches make 1 Ale Gallon. 231 1 Wine Gallon. 2150.42 1 Wincheffler Buttel. 105 Cubic inches 1 Scots Pint. The Wheat Firlot contains 21¾ Scots Pints. The Barley Firlot 31 Scots Pints.

### SECTION I.

OF THE MENSURATION OF RIGHT LINES 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 given, might be considered as belonging to this part of mensuration. But as they 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 66 feet; but more expeditiously with measuring tapes of 50 or 100 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 3 inches of the truth in every 50 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 fig. 1. and fig. 2. If an angle of elevation is to be measured, as the angle contained by a horizontal line AC, 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 instrument 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 EG, cut off 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 Of Right eye is to be placed at A the centre of the instrument, Lines and Angles. and the measure of the angle is the arch EF.

But the most convenient instrument of any for observing angles, whether vertical or horizontal, is the Theodolite. This instrument is variously constructed, so as to admit of being sold at a higher or lower price, according to the degree of accuracy the purchaser may wish to attain in his observations with it. An instrument of this kind is represented in fig. 3. Its principal parts are, 1. A telescope and its level CC, D. 2. The vertical arc BB. 3. The horizontal limb and compass AA. The limb is generally about 7 inches in diameter. 4. The staff with its parallel plates E.

The telescope CC in the best instruments is generally of the achromatic kind, in order to obtain a larger field and greater magnifying power. In the focus of the eye glass are two very fine hairs or wires, at right angles to each other, whose intersection is in the plane of the vertical arc. The object glass may be moved to different distances from the eye glass by turning the milled nut a, and thus may be accommodated to the eye of the observer and distance of the object. The screws for moving and adjusting the cross hairs, are sunk a little within the eye tube. On the outside of the telescope are two metal rings which are ground perfectly true. These are to lie on the supports e, e, called Y's, which are fixed to the vertical arc. The vertical arc BB is firmly fixed to a long axis which is at right angles to the plane of the arc. This axis is sustained by, and moveable on, the two supports, which are fixed firmly to the horizontal plate. On the upper part of the vertical arc are the two Y's for holding the telescope; the inner sides of these are so framed as to be tangents to the cylindrical rings of the telescope, and therefore bear only on one part. The telescope is confined to the Y's by two loops which turn on a joint, and may therefore be readily opened and turned back when the two pins are taken out.

One side of the vertical arc is graduated to half degrees, which are subdivided to every minute of a degree by a nonius. It is numbered each way from the middle from 0 to 90°; towards the eye end for angles of altitude, and towards the object end for angles of depression. On the other side of the vertical arc are two ranges of divisions, one for taking the upright height of timber in 100th parts of the distance between the instrument and tree whose height is to be measured; and the other for reducing hypotenusal lines to such as are horizontal.

The vertical arc is cut with teeth or a rack, and may be moved regularly, and with ease, by turning the milled nut b.

The compass is fixed to the upper horizontal plate, its ring is divided into 360°, and the bottom of the box is divided into four parts or quadrants, each of which is subdivided into 15°. The magnetic needle is supported in the middle of the box upon a steel pin finely pointed, and there is a wire trigger for throwing the needle off the point when not in use.

The horizontal limb AA consists of two plates, one moveable on the other, the outermost edge of the upper plate is chamfered to serve as an index to the degrees on the lower. The upper plate, together with the compass, vertical arc, and telescope, are easily turned round by a pinion fixed to the screw c; d is a nut for fixing the index to any part of the limb, and thereby rendering it secure, while the instrument is moved from one station to another. The horizontal limb is divided into half degrees, and numbered from the right hand towards the left; the divisions are subdivided by the nonius scale to every minute of a degree.

On the upper plate, towards the nonius, are a few divisions similar to those on the vertical arc, giving the 100th parts, for measuring the diameter of trees, buildings, &c.

The whole instrument fits on the conical ferril of a strong brass-headed staff, with three substantial wooden legs. The top or head of the staff consists of two brass plates E, parallel to each other: four screws pass through the upper plate and rest on the lower plate; by the action of these the horizontal limb may be set truly level, and for this purpose a strong pin is fixed to the outside of the plate, and connected with a ball that fits into a socket in the lower plate; the axis of the pin and ball are so framed as to be perpendicular to the plate, and consequently to the horizontal limb.

There are three adjustments necessary before the instrument is applied to the mensuration of angles. In the first place, care must be taken that the line of collimation (that is, the line of vision passing through the cross hairs) be exactly in the centre of the cylindrical rings round the telescope; in the next place, that the level be parallel to this line; and, lastly, the horizontal limb must be so let, that when the vertical arc is at zero, and the upper part moved round, the bubble of the level will remain in the middle of the open space.

When these adjustments are made, and the instrument is to be applied to practice, the lower plate of the horizontal limb AA being supposed to remain unmoved and parallel to the horizon, the telescope is to be directed successively to the different objects, whose angular positions are to be determined, by means of the pinions at c and d; (the former of which turns the upper part of the instrument round in a horizontal plane, and the latter turns the arc BB in a vertical plane). Then, the angle which a line passing through the axis of the telescope and any object makes with the horizon, will be indicated by the arc of the vertical circle between 0° and the index engraved on the nonius scale H fixed to the upper plate of the horizontal limb of the instrument. Also, the horizontal angle contained by two vertical planes conceived to pass through any two objects and the centre of the instrument, will be shown by the arc of the lower plate of the horizontal limb over which the index engraved on the upper plate has passed by the direction of the telescope being changed from the one object to the other.

Having thus explained shortly the nature of the instruments by which accessible lines and angles are to be measured, and the manner of applying them, we shall now shew, by a few examples, how to find from these other lines which cannot be determined by a direct measurement.

Example 1. Having measured AE, a distance of Fig. 4. 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, tower, was measured by a quadrant, or theodolite placed at C, and found to be $47^\circ 30'$. 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 = $47^\circ 30'$. And since by the rules of Plane Trigonometry,

$$\text{rad} : \tan. BCD :: DC : DB;$$

By employing the logarithmic tables (see Logarithms), and proceeding as is taught in Plane Trigonometry, we shall find DB = 218.3 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.

Ex. 2. Suppose a 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 $35^\circ$ and $64^\circ$, 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 = $64^\circ$, and one of the inward angles A = $35^\circ$; hence the other inward angle ACB, which is their difference, is given, and $64^\circ - 35^\circ = 29^\circ$.

Now in the triangle CAB

$$\sin. ACB : \sin. A :: AB : BC,$$

and $\sin. ACB : \sin. B :: AB : AC.$

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

Again, in the right-angled triangle BCD

$$\text{rad} : \sin. B :: BC : CD,$$

Hence CD will be found = 936 feet.

Ex. 3. 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 means of a theodolite, the angles CBA and CAB were measured, and found to be $37^\circ 40'$ and $59^\circ 15'$ respectively. It is required thence to determine the breadth CD, and the distance AC between the objects A and C.

This example 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 ACD,

$$\text{rad} : \sin. A :: AC : CD.$$

Hence CD is found to be 211.6 yards.

Ex. 4. 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 $4^\circ 52'$; and at R, the bottom of the tower, its depression (namely, the angle NRS) was $4^\circ 2'$. Required AS the horizontal distance of the vessel; and also RA, 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 = $4^\circ 52'$ subtract the angle RSA = NRS = $4^\circ 2'$, and there remains the angle BSBR = $50'$. Also, from the angle HBA = $90^\circ$ subtract HBS = $4^\circ 52'$, and there remains SBR = $85^\circ 8'$.

In the triangle SBR,

$$\sin. BSR : \sin. SBR :: BR : SR;$$

Hence SR is found. Again, in the triangle SRA,

$$\text{rad} : \sin. RSA :: SR : AR,$$

and $$\text{rad} : \cos. RSA :: SR : AS.$$

From the first of these proportions we find AR = 260 feet; and from the second, AS = 3690 feet.

Ex. 5. To measure the height of an obelisk CD, Fig. 6, standing on the top of a declivity, two stations at A and B were taken, one at the distance of 40, and the other at the distance of 100 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 $41^\circ$ with the plane of the declivity; and at B, the more remote station, the like angle was found to be $23^\circ 45'$. Hence it is required to find the height of the obelisk.

From the angle CAD = $41^\circ$, subtract the angle B = $23^\circ 45'$, and there remains the angle BCA = $17^\circ 15'$.

In the triangle BCA,

$$\sin. BCA : \sin. B :: AB : AC.$$ Hence AC = 81.49 feet.

And in the triangle ACD,

$$AC + AD : AC - AD :: \tan. \frac{1}{2}(D+C) : \tan. \frac{1}{2}(D-C).$$

Hence $\frac{1}{2}(D-C) = 42^\circ 24'15''$, which, subtracted from $\frac{1}{2}(D+C)$, gives the angle ACD = $27^\circ 5'15''$.

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

At A $\left\{ \begin{array}{l} BAM = 40^\circ 16' \\ MAH = 57^\circ 40' \end{array} \right.$

At B $\left\{ \begin{array}{l} ABH = 42^\circ 22' \\ HBM = 71^\circ 7' \end{array} \right.$

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

In the triangle HAB we have the angle HBA = $42^\circ 22'$, the angle HAB ($= HAM + MAB$) = $97^\circ$.

Lines and Angles.

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 = 706.8 yards.

Again, in the triangle MAB we have the angle MAB = 40° 16', the angle ABM ( = ABH + HBM ) = 113° 26', 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,

\[ AM + AH : AM - AH : \tan \frac{1}{2}(AHM + AMH) : \tan \frac{1}{2}(AHM - AMH). \]

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 = 1174 yards, the answer to the question.

Ex. 7. There are three objects A, B, C, whose distances abeam are known to be as follows; namely, from A to B 1063, from A to C 202, and from B to C 131 fathoms. Now to determine the distance of D a fourth object, or station, from 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 about 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 ABC 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' 44", 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', Of Right the angle DCA (= DEA) = 75° 15' 25". Hence we get DC = 236.97 fathoms, which is the remaining distance sought.

Ex. 8. From a ship at sea a point of land was observed to bear E. by S. and after sailing N. E. 12 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 = 2 points, or 10° 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.

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, ABCD, 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 100, or 1000 equal parts.

To measure any altitude HK with this instrument. Fig. 13.

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 AI, the vertical line IK, and the line AK drawn from the eye to the top of the object.

If the line of the plummet passes 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 AKI, we have \( \frac{AB}{AE} : \frac{AI}{IK} \). 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 square opposite to the sights in F, then, in the similar triangles AJK, BCF, the angle AKI is equal to BFC thus we have BC : CF :: AI : 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 be a square, 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 ABCD, whose side AB is 10½ inches.

Here \(10\frac{1}{2} \times 10\frac{1}{2} = 10.5 \times 10.5 = 110.25\) 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 \(13.75 \times 9.5 = 130.625\) square chains is the area, which, when reduced to acres, &c., is 13 ac. o ro. 10 po.

Ex. 3. Required the area of a parallelogram KLMN, whose length KL is 37 feet, and perpendicular breadth NO is 5¼ or 5.25 feet.

In this example the area is \(37 \times 5.25 = 194.25\) 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 is therefore equal to \(\frac{\text{fin. } K}{\text{rad.}} \times KN\); therefore the area of the figure is \(\frac{\text{fin. } K}{\text{rad.}} \times KN \times KL\), 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 the table of logarithms given at the end of the article LOGARITHMS. The operation may stand thus,

\[ \begin{align*} \log_{10} \text{ rad.} & = 10.00000 \\ \log_{10} \text{ fin. } 58° & = 9.92842 \\ \log_{10} (36 \times 25.5) & = \log_{10} 36 + \log_{10} 25.5 = 2.96284 \\ \log_{10} \text{ of area} & = 2.89126 \\ \text{area} & = 778.5 \text{ square feet}. \end{align*} \]

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 hypotenuse.

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

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

From the square of the hypotenuse 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.

Example 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 hypotenuse BC?

Here \(33^2 + 56^2 = 33^2 + 1089 = 4225\), and \(\sqrt{4225} = 65\) feet, = the hypotenuse BC.

Ex. 2. Suppose the hypotenuse 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 \(\sqrt{3136} = 56\) 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, 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 \(40 \times 14.52\), is 580.8 square chains, the half of which, or \(290.4\) sq. ch. = 29 ac. o 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 (fig. 16) 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.

\[ \begin{align*} \log_{10} \text{ rad.} & = 10.00000 \\ \log_{10} (30 \times 40) & = \log_{10} 30 + \log_{10} 40 = 3.07918 \\ \log_{10} \text{ fin. } 28° 57′ & = 9.68489 \\ \log_{10} \text{ of twice area} & = 2.76407 \\ \text{twice area} & = 580.85 \\ \text{area} & = 290.42 \end{align*} \] RULE III.

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 remainders. 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 \( \triangle ABC \) then be any triangle; produce \( AB \), any one of its sides, and take \( BD \), and \( B'd \), each equal to \( BC \); join \( CD \) and \( C'd \), and through \( A \) draw a line parallel to \( BC \), meeting \( CD \) and \( C'd \) produced in \( E \) and \( e \); thus the angle \( AED \) will be equal to the angle \( BCD \) (Geometry, 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 \( BCd \), that is, to the angle \( B'dC \), or \( A'de \), therefore \( A = A'd \).

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 \( A'd \) or \( A'e \), describe another circle meeting \( AC \) in \( f \) and \( g \), and draw \( BH \) and \( B'h \) perpendicular to \( CD \) and \( C'd \). Then, because \( BD, BC, B'd \) are equal, the point \( C \) is in the circumference of a circle, of which \( D'd \) is the diameter, therefore \( CD \) and \( C'd \) are bisected at \( H \) and \( h \) (Sect. II. Theor. 6.) and the angle \( DCd \) is a right angle, (Sect. II. Theor. 17.), and hence the figure \( CHBk \) is a rectangle, so that \( Bh = CH = \frac{1}{2} CD \), and \( B'h = Ch = \frac{1}{2} C'd \).

Join \( BE \), and \( Be \), then the triangle \( BAC \) is equal to each of the triangles \( BEC, BeC \) (Sect. IV. Theor. 2. Cor. 2.) but the triangle \( BEC \) is equal to \( \frac{1}{4} EC \times BH \) (Sect. IV. Theor. 2.), that is to \( \frac{1}{4} EC \times C'd \); and in like manner the triangle \( BeC \) is equal to \( \frac{1}{4} eC \times B'h \), that is to \( \frac{1}{4} eC \times C'd \), therefore the triangle \( ABC \) is equal to \( \frac{1}{4} EC \times C'd \), and also to \( \frac{1}{4} eC \times C'd \).

Now since \( CD : C'd :: CE \times CD : Ce \times C'd \) (Sect. IV. and also \( CD : C'd :: Ce \times CD : Ce \times C'd \) Theor. 3.)

Therefore \( CE \times CD : Cl \times C'd :: Ce \times CD : Ce \times C'd \); that is, because \( CE \times CD = FC \times CG \), and \( Ce \times C'd = fC \times Cg \) (Sect. IV. Corollaries to Theor. 28 and 29.)

\[ FC \times CG : CE \times C'd : Ce \times CD : fC \times Cg ; \]

which last proportion (by taking one-fourth of each of its terms, and substituting the triangle \( ABC \) for its equivalent values \( \frac{1}{4} CE \times C'd \) and \( \frac{1}{4} Ce \times CD \)) gives us

\[ \frac{1}{4} FC \times \frac{1}{4} CG : \text{trian. } ABC :: \text{trian. } ABC :: \frac{1}{4} fC \times \frac{1}{4} Cg . \]

Now, if it be considered that the radius of the circle \( DGE \) is \( AB + BC \), and that the radius of the circle \( gde \) is \( AB - BC \), it will readily appear that, putting \( 2s \) for the perimeter of the triangle \( ABC \), we have

\[ FC = AB + BC + AC = 2s, \] \[ CG = AB + BC - AC = 2s - 2AC, \] \[ fC = AC + \left\{ AB - BC \right\} = 2s - 2BC, \] \[ gC = AC - \left\{ AB - BC \right\} = 2s - 2AB. \]

Put now \( a, b, c \) for the sides \( AC, BC, \) and \( AB \) respectively, then \( \frac{1}{4} FC = s - a, \frac{1}{4} GC = s - a, \frac{1}{4} fC = s - b, \frac{1}{4} gC = s - c \); thus the last proportion becomes

\[ s \times (s - a) : \text{trian. } ABC :: \text{trian. } ABC : (s - b) \times (s - c), \]

which conclusion, when expressed in words at length, is evidently the proportion 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 \( s \times (s - a) \) and \( (s - b) \times (s - c) \), is equal to

\[ \sqrt{\left\{ s \times (s - a) \times (s - b) \times (s - c) \right\}} \]

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.

And \( \frac{108}{2} = 54 = \) half that sum.

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

\[ 54 \times 30 \times 18 \times 6 = 174960. \]

And the square root of this product is

\[ \sqrt{174960} = 418.28 \text{ sq. ch. the area required.} \]

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, and half the product will be the area.

This rule is demonstrated in Geometry, Sect. IV. Theor. 7.

Example. Required the area of the trapezoid \( \triangle ABCD \) 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

\[ 19.75 \times 15.4 = 304.15; \]

and half this product is

\[ \frac{304.15}{2} = 152.075 \text{ sq. ch. = 15 ac. 33.2 po.} \]

the area required. **MENSURATION.**

**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 the diagonal from the other two opposite angles, the sum of these perpendiculars 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 perps. is \(16 + 18 = 34\), which multiplied by 42, and divided by 2 gives

\[ \frac{34 \times 42}{2} = 714 \text{ the area.} \]

**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 ABCDE, whose side AB, or BC, &c. is 25 feet, and perpendicular HK is 17.2 feet.

Here \(25 \times 5 = 125\) = the perimeter,

And \(125 = 17.2 = 2150\),

And its half \(1075 =\) 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 the table of logarithmic sines and tangents given in the article LOGARITHMS.

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

**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 3.1416, 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 1000000th part.

The proportion of the diameter of a circle to its circumference is investigated in GEOMETRY, Sect. VI., Prop. 6. Also in FLUXIONS, § 137 and § 140. 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,

\[ \frac{7}{22} : : 20 : \frac{20 \times 22}{7} = 62\frac{6}{7} \text{ the answer.} \]

Or by the third rule \(3.1416 \times 20 = 62.832\) the answer.

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

By the second rule

\[ \frac{355}{113} : : 10 : \frac{113 \times 10}{355} = 3.1831 \text{ the answer.} \]

**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 is 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°, and that the length of an arch is proportional to the number of degrees it contains.

*Examples.* 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 hypotenuse AC = 9, and the side AD = 3; hence, by trigonometry, the angle ACE will be found to contain $19^\circ 28' \frac{7}{9}$ = 19.471 degrees. The double of this, or 38.942, is the number of degrees in the whole arch AEB. Then by the rule

$$\frac{180}{38.942} : 9 \times 3.1416 : 9 \times 3.1416 \times 38.942 = 6.11701$$

is the answer.

Rule II.

From 8 times the chord of half the arch subtract the chord of the whole arch, and $\frac{1}{2}$ of the remainder will be the length of the arch nearly.

This rule may be demonstrated briefly thus. Let $a$ denote an arch of a circle; then from the series expressing the sine of an arch in terms of the arch, (see Fluxions, § 70. Ex. 3, also Trigonometry) we have, putting rad. = 1,

$$\sin \frac{1}{2} a = \frac{1}{2} a - \frac{a^3}{48} + \frac{a^5}{3840} - \&c.$$

Therefore, if the arch $a$ be small, so that $a^3$ is a very small quantity, then

$$\sin \frac{1}{2} a = \frac{1}{2} a - \frac{a^3}{48} \text{ nearly.}$$

In like manner we have

$$\sin \frac{1}{2} a = \frac{1}{2} a - \frac{a^3}{384} \text{ nearly.}$$

By means of the two last equations eliminate the quantity $a^3$, and the resulting equation is

$$16 \sin \frac{1}{2} a - 2 \sin \frac{1}{2} a = 3 a.$$

But $16 \sin \frac{1}{2} a = 8 \text{ chord } \frac{1}{2} a$, and $2 \sin \frac{1}{2} a = \text{ chord } a$. Therefore 8 chord $\frac{1}{2} a$ — chord $a = 3 a$.

Here we have supposed the radius of the circle to be unity; but the same must evidently be true, whatever be the radius of the circle.

Example. Suppose as before, that the chord AB is 6, and the radius AC is 9. Then CD = $\sqrt{(CA^2 - AD^2)} = \sqrt{72} = 8.4852814$, and DE = $9 - 8.4852814 = 0.5147186$, and hence AE = $\sqrt{(AD^2 + DE^2)} = 3.043836$.

Then by the rule

$$\frac{3.043836 \times 8 - 6}{3} = 6.116896$$

is the length of the arch, nearly the same as before.

Problem X.

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 $1 \times 3.1416 \div 4 = .7854$. But circles are to one another as the squares of their diameters, (Prop. 4.) therefore, putting $d$ for the diameter of any circle, $1 : d^2 :: .7844 : .7854 d^2$ = the area of the circle whose diameter is $d$.

Example. What is the area of a circle whose diameter is 7?

By the second rule $7 \times 7 \times .7854 = 38.4846$ the area.

By the first rule $7 \times 3.1416 =$ the circumference.

Then $\frac{7 \times 3.1416 \times 7}{4} = 7 \times 7 \times .7854$ the area, the same as before.

Problem XI.

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,

$$\text{circum.} :: \text{arch of sect.} :: \text{rad.} \times \frac{1}{2} \text{circum.} :: \text{area of sect.}$$

Hence area of sect. = rad. $\times \frac{1}{2}$ 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.

1. By the first rule.

First $3.1416 \times 3 = 9.4248$ the circum.

And $360 : 18 :: 9.4248 : 47124$ the arch of sect.

Then $.47124 \times 3 \div 4 = .35343$ the area.

2. By the second rule.

First $.7854 \times 3^2 = 7.0686$ the area of the circle.

Then $360 : 18 :: 7.0686 : .35343$ the area.

PROBLEM XII. To find the area of a segment of a circle.

RULE I. 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.

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' = 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.

Now .7854 × 400 = 314.16 the area of the whole circle. Therefore 360° : 73° 74' :: 314.16 : 64.3504 = area of the sector CAEB.

Again \( \sqrt{(CA^2 - AD^2)} = \sqrt{(100 - 36)} = \sqrt{64} = 8 = DC \).

Therefore \( AD \times DC = 6 \times 8 = 48 = \) area of the triangle.

Hence sector ACBA — triangle ACB = 16.3504 the area of seg. AEBDA.

PROBLEM XIII. 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 \( \frac{1}{3} \) of the product for the area.

This rule is demonstrated in Prop. 12, Part. I. CONIC SECTIONS.

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 \( 10 \times 4 \times \frac{1}{3} = \frac{80}{3} = 26\frac{2}{3} \) the area.

PROBLEM XIV. 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 of an ellipse is equal to the area of a circle whose diameter is a mean proportional between the axes of the ellipse, (CONIC SECTIONS, Part II. Of Plane Prop. 22.) 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, ACBD, be Fig. 25 and 25. What is the area?

\( 35 \times 25 \times .7854 = 687.225 \) the area.

Note. As to hyperbolic areas, the mathematical reader will find formulas for their exact mensuration in FLUXIONS, § 152. Ex. 4. and 5.

PROBLEM XV. To find nearly the area of a figure bounded by any curve line Aa a'a" &c. P, and a 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 b a, 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''' b''', the second, fourth, sixth, &c.

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

And put D for Bb, or bb', &c. the common distance between the perpendiculars.

Then the area of the figure will be nearly equal to

\[ \frac{1}{3} D \times (F + 4 \times 4E + 2R); \]

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 Aaa' and the three straight lines AB, BB', b'a' will be made up of the trapezoid AB b'a', and the space bounded by the arch A a' and its chord A a'. Now if the arch 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 \( \frac{1}{3} \) of its circumfribring parallelogram. (CONIC SECTIONS, Part I. Prop. xii.)

Therefore the space A a' b' BA will be nearly equal to the sum of the trapezoid AB b'a' and \( \frac{2}{3} \) of the parallelogram AD, which sum is evidently equal to \( \frac{1}{3} \) of the trapezoid AB b'a' together with \( \frac{2}{3} \) of the trapezoid CB b'D.

Of Plane CB b' D. Now the area of the trapezoid ABB'b' is

\[ \frac{AB + a'b'}{2} \times BB' \] (Geometry, Sect. IV. Theor. 7.)

\[ = \frac{AB + a'b'}{2} \times 2BB; \] and in like manner the area of the trapezoid CB b' D is

\[ \frac{CB + DB}{2} \times BB' = ab \times 2BB; \]

therefore the area of the figure Aa a'b'B is nearly

\[ \frac{1}{3} \times \frac{AB + a'b'}{2} \times 2BB + \frac{1}{3} \times ab \times 2BB \]

\[ = \frac{1}{3}(AB + 4ab + a'b') \times BB. \]

In the very same way it may be shewn that the area of the figure a'a''a''b''b'' is nearly

\[ \frac{1}{3}(a'b' + 4a''b'' + a'''b''') \times BB, \]

and that the area of the figure a''a''PQ b'' is nearly

\[ \frac{1}{3}(a'''b''' + 4a''b'' + PQ) \times BB. \]

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

\[ \frac{1}{3}BB \left\{ \frac{AB + PQ}{2} + 4(a'b' + a''b'' + a'''b''') + 2(a'b' + a''b'') \right\} \]

as was to be demonstrated.

Example 1. 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 perpendiculums mn, pq, rs. Now because CA = 1, therefore

\[ CD = \frac{1}{4}, \quad CR = \frac{1}{8}, \quad CP = \frac{1}{4}, \quad CM = \frac{1}{8}. \]

Hence \( DE = \sqrt{(EC^2 - CD^2)} = \sqrt{\left(1 - \frac{1}{4}\right)} = \frac{1}{2} \sqrt{3}; \) and in like manner

\[ rS = \frac{1}{8} \sqrt{55}, \quad pQ = \frac{1}{4} \sqrt{15}, \quad mN = \frac{1}{8} \sqrt{63}. \]

Therefore

\[ F + L + \frac{1}{4} \sqrt{3} = 1.8665 \] \[ 4E = \frac{1}{2} \sqrt{55} + \frac{1}{4} \sqrt{63} = 7.6767 \] \[ 2R = \frac{1}{4} \sqrt{15} = 1.9365 \]

The sum

\[ 11.4792 \]

Multiply by \( \frac{1}{4} \)

\[ 2.873 \]

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 semiorbinate MD = 12, and the transverse CF = 15.

Let FM be divided into five equal parts by the semiorbinate HI, mn, pq, rs. Thus CH = 17, CM = 19, CP = 21, CR = 23, CM = 25. Now, since from the nature of the curve, \( \sqrt{(CM^2 - CF^2)} : MD :: \sqrt{(CH^2 - CF^2)} : HI \) (Conic Sections, Part III. Prop. 19).

And Geometry, Sect. IV. Theor. 12.), that is, in numbers, 20 : 12 :: 8 : HI, therefore HI = \( \frac{8}{3} \). In like manner we find \( mn = \frac{8}{3} \sqrt{34}, pq = \frac{1}{3} \sqrt{6}, \) and \( rs = \frac{1}{3} \sqrt{19}. \)

Therefore

\[ F + 4(HI + MD) = 16.8 \] \[ 4E = \frac{1}{2} mn + 4pq = 68.8399 \] \[ 2R = 2pq = 17.6363 \]

The figure HIDM = 103.2762 \( \times \frac{1}{3} = 68.8508 \)

to which adding FH, 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 Crofs.

A statute acre of land being 160 square poles, the chain is made 4 poles or 66 feet in length, that 10 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 crofs 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 crofs 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 telecopic or other sights, which are used to determine how much one place is higher or lower than another. An offset-staff for measuring the offsets 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 stakes 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 \( A m B C D q \) represent a field to be measured. Let it be resolved into the triangles \( A m B, A B D, B C D, A q D \). Let all the sides of the large triangles \( A B D, B C D \), and the perpendiculars of the small ones \( A m B, A q D \) 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 curvilinear spaces may be measured by the rule given in Prob. XV. 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 \( A B C D E \), and its distances \( F A, F B, F C, \ldots \) 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 \( A, B, C, D, \ldots \) successively, and the lines \( F a, F b, F c, \ldots \) 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 a figure \( a b c d e \) 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 Base 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 \( A B \). 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 \( B A \), and let a straight line be drawn from \( b \) in the direction \( B C \). 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 \( A B C \) 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 each line 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 all together 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 considered as a prism of an indefinitely great number of sides.

Ex. 1. What is the surface of a cube, the length of Fig. 32., its side \( A B \) being 20 feet?

Here \( 4 \times 20 = 80 \) the perim. of end.

And \( 80 \times 20 = 1600 \) the four sides.

And \( 2 \times 20 \times 20 = 800 \) the top and bottom.

The sum \( 2400 = \) the area or surface.

Ex. 2. What is the convex surface of a cylinder whose length \( A B \) is 20 feet, and the circumference of its base 3 feet?

Here \( 3 \times 20 = 60 \) 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 cylinder of an indefinitely great number of sides.

Ex. 1. What is the upright surface of a triangular pyramid, \( A B C D \), the slant height, \( A E \), being 20 feet, and each side of the base 3 feet?

Here \( \frac{3 \times 3 \times 20}{2} = 90 \) feet, the surface.

Ex. 2. Required the convex surface of a cone, the slant height \( A B \) being 50 feet, and the diameter of its base 8 feet.

Here \( 8.5 \times 3.1416 = \) circum. of base.

And \( \frac{8.5 \times 3.1416 \times 50}{2} = 667.59 \), the answer. **MENSURATION**

**Problem III.** 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, whose slant height AE 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 \( \frac{3}{4} \times 4 = 13 \frac{1}{2} \) the per. of gr. end. And \( \frac{2}{5} \times 4 = 8 \frac{3}{5} \) the per. of lefs end. And their sum is 22 feet.

Therefore \( \frac{22 \times 10}{2} = 110 \) 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 II. Sect VIII. and Theor 2. Sect IX. Geometry.

**Ex. 1.** What is the solid content of a cube AG, the length of whose side is 24 inches?

Here \( 24 \times 24 = 576 \) sq. inches, the area of the end. And \( 576 \times 24 = 13824 \) cub. inches is the solidity.

**Ex. 2.** Required the content of a triangular prism, whose 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 3. of Prob. 3. Sect. II., to be

\[ \sqrt{(3 \times 3 \times 2 \times 1)} = 6 \text{ sq. feet}. \]

Therefore \( 6 \times 20 = 120 \) 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 Prob. 10. of Sect. II. the area of its base is

\[ .7854 \times 18.5^2 = 268.803 \text{ sq. inches}; \]

Therefore \( 268.803 \times 8 = 2150.424 \) 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 Theor. 16. Sect. VIII. and Theor. 3. Sect. IX. Geometry.

**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.

First, the area of the base, as found by Rule 3. of Prob. 3. Sect. II., is

\[ \sqrt{(4.5 \times 1.5 \times 1.5 \times 1.5)} = 3.89711. \]

Therefore \( \frac{3.89711 \times 30}{3} = 38.9711 \) cub. feet is the solid content.

**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.

Here \( .7854 \times \frac{3^2}{2} = 1.76715 \) is the area of the base in square feet.

And \( \frac{1.76715 \times 15}{3} = 8.8357 \) cub. feet is the solid content.

**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.** Let PABCD be any pyramid, and AG a frustum of it contained between ABCD its base, and EFGH a plane parallel to the base. Put \( a \) for the side of a square equal to AC 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 \( a^2 \) is the area of the base of the frustum; \( b^2 \) is the area of its top; \( \frac{1}{3} (a^2 + b^2 + c^2) \) is the solid content of the whole pyramid; (\( \text{Geom. Sect. VIII. Theor. 16.} \))

\( \frac{1}{3} b^2 c \) is the content of its upper part; and therefore

\[ \frac{1}{3} \left[ a^2 (h+c) - b^2 c \right] \]

is the solid content of the frustum itself. Now the base and top of the frustum being similar figures, (\( \text{Sect. VIII.} \))

Of solids. Theor. 13.) their areas are to one another as the squares of AB and EF their homologous sides, (Sect. IV. Theor. 27.) But AB : EF :: BP : PF (Sect. VII. Theor. 7. and Sect. IV. Theor. 20.) : PM : PL, (Sect. VII. Theor. 14.); therefore the area of the base of the frustum is to the area of its top as PM² : PL², that is, \(a^2 : b^2 :: (h+c)^2 : c^2\), and consequently \(a : b :: h+c : c\); hence \(ac = b(h+c)\), and \(c = \frac{b}{a-b}\).

Let these values of \(c\) and \(h+c\) be now substituted in the preceding expression for the content of the frustum, and it will become, by proper reduction,

\[ \frac{2}{3} \left( \frac{h^3 - l^3}{a-b} \right). \]

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,

\[ \frac{2}{3} h \left( a^2 + ab + b^2 \right), \]

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

\[ \sqrt{(41.569 \times 23.383)} = 31.177. \]

Hence the mean area is

\[ \frac{1}{3} (23.383 + 41.569 + 31.177) = 32.043. \]

And the solid content of the frustum is

\[ 32.043 \times 9 = 288.387 \text{ cubic feet}. \]

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.) \(5^2 \times \pi \times 78.54\), and the area of the lesser end is \(3^2 \times \pi \times 78.54\), and the mean proportional between them is

\[ \sqrt{(5^2 \times 3^2 \times 78.54^2)} = 5 \times 3 \times 78.54; \text{ therefore the mean area is} \]

\[ \frac{78.54}{3} \times (5^2 + 3^2 + 5 \times 3) = 12.8282. \]

And the content of the frustum

\[ 12.8282 \times 9 = 115.4538 \text{ cub. feet}. \]

PROBLEM VII.

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

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, §. 165. It may however be deduced from principles more elementary, by reasoning as follows. Let PCQ be a semicircle, and ABCDE several successive sides of a regular polygon inscribed in it. Conceive the semicircle to revolve about the diameter PQ as an axis, then the arch ABCDE 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 in L, and draw AF, LM, BG, CH, &c. perpendicular to PQ. For the sake of brevity, let circ. AF 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. Sect. VI. Prop. 4.), and because \(\frac{1}{2}(AF + BG) = LM\), therefore \(\frac{1}{2}(\text{circ. AF} + \text{circ. BG}) = \text{circ. LM}\). Now the area of the surface generated by the chord AB is \(\frac{1}{2}(\text{circ. AF} + \text{circ. BG})(\times AB)\) (Prob. 3.) therefore the same area is also equal to \((\text{circ. LM}) \times AB\). Draw AO parallel to FG, and draw LN to the centre of the circle. Then the triangles AOB, OLM are manifestly similar; therefore \(AB : AO :: NL : LM :: \text{circ. NL} : \text{circ. LM}\); and hence \(AO \times \text{circ. NL} = AB \times \text{circ. LM}\). But this last quantity has been proved equal to the surface generated by AB, therefore the same surface is equal to \(AO \times \text{circ. NL}\), or to \(FG \times \text{circ. NL}\), 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 \(GH \times \text{circ. LN}, HI \times \text{circ. LN}, IK \times \text{circ. LN}\). Therefore the whole surface generated by the chords AB, BC, CD, DE, &c. is equal to \((FG + GH + HI + IK) \times \text{circ. LN} = FK \times \text{circ. LN}\). 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 \(17 \times 3.1416 = 53.4072\) inches = the circum.

Then \(53.4072 \times 17 = 907.9224\) sq. inches = 6.305 square feet the answer.

Ex. 2. What is the convex surface of a segment 8 inches in height cut off from the same globe?

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 \( \frac{2}{3} \) 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, Sect. IX. Theor. 6. And the second is deduced from the first, thus: put \( d \) for the diameter of the sphere, then \( d^2 \times .7854 \) is the area of a great circle of the sphere, and by the first rule \( \frac{2}{3}d^3 \times .7854 = d^3 \times .5236 \) is its content.

Example. What is the content of a sphere whose diameter is 6 feet?

Answer \( 6^3 \times .5236 = 113.0976 \) cub. feet.

Problem IX. To find the solid content of a spherical segment.

Rule. From 3 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, § 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, Sect. 8. and 9.

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 DHHG; 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 (GEOMETRY, 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 2CE, or 2AF, the diameter of the circle, = \( d \), and AD, the height of the spherical segment, = \( h \). Then AC = \( \frac{1}{2}d \) and \( 2CA = 2AD = 2CD = d - 2h \). Let \( n \) denote the number .7854. Then the area of the base common to the conic frustum AH, and cylinder AG, is \( nd^2 \) (Sect. II. Prob. 10.), and the area of the top of the frustum is \( n(d - 2h)^2 \), and the mean proportional between these areas is \( n(d - 2h)d \). Therefore the solid content of the frustum is (by Prob. 3. of this sect.)

\[ \frac{1}{3} \left[ nd^2 + n(d - 2h)^2 + nd(d - 2h) \right] \times h. \]

Now the solid content of the cylinder is \( nd^2h \): (Prob. 1.) Therefore the solid content of the spherical segment, (which is equal to the difference between the cylinder AG and conic frustum AH) is equal to

\[ n^2d^2h - \left( nd^2h - 2nd^2h + \frac{4}{3}n^2h^3 \right), \]

that is, to \( 2nd^2h - \frac{4}{3}n^2h^3 \), or to

\[ \frac{2}{3} \left( 3d - 2h \right) h^3, \]

which expression, if it be considered that \( \frac{2}{3}n \) or \( \frac{2}{3} \times .7854 \) 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 \( 3 \times 21 - 2 \times 4.5 = 54 \).

Then \( 54 \times 4.5 \times 4.5 \times .5236 = 572.5566 \) 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 ABCD, 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 FHGE be the common section of this plane and the generating... Let \( P \) denote the parameter of the axis. Then since

\[ EG^2 = P \times AE, \]

and

\[ EH^2 = P \times EB, \]

\[ EG^2 + EH^2 = P \times AB = CB^2. \]

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, § 163.

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

Here \( 10 \times 7.854 = 78.54 \) the area of the base.

And \( \frac{1}{3} \times 7.854 \times 12 = 47.124 \) cub. feet is the solidity.

---

**Problem X.**

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 \( \frac{1}{3} A(A + c) \), and the content of the part cut off is \( \frac{1}{3} ac \); therefore the content of the frustum is

\[ \frac{1}{3} \left( A(A + c) - ac \right) = \frac{1}{3} A(A + c) - \frac{1}{3} ac. \]

But from the nature of the parabola, \( c : h + c :: a : A \); hence \( Ac = ab + ac \) and \( c = \frac{ah}{A-a} \).

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

\[ \frac{1}{3}(A(A + a)) = \frac{1}{3}h(A + a), \]

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 \( 58^2 \times 7.854 \), and \( 30^2 \times 7.854 \); therefore their sum is \( (58^2 + 30^2) \times 7.854 = 4264 \times 7.854 \). And the content of the figure is \( \frac{1}{3} \times 4264 \times 7.854 \times 18 = 30140.5104 \), the answer.

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**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 \( \frac{9}{15} \) of the product for the content of the solid.

For the investigation of this rule we must refer the reader to Fluxions, § 163. Ex. 2.

**Example.** The length of the parabolic spindle AEBc A is 60, and the middle diameter Ee 34; what is the solidity?

Here \( 34^2 \times 7.854 \) is area of the middle section.

Therefore \( 34^2 \times 7.854 \times 60 \times \frac{9}{15} = 29533.5168 \) is the solidity required.

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**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 \( \frac{1}{15} \) of .7854, will be the content.

For, referring the reader to Fluxions, § 163. Ex. 2, as before, and substituting \( h \) for \( AC = \frac{1}{2} b \), but, in other respects, retaining the figure and notation, we have this general expression for the segment APP,

\[ \frac{\pi x^3}{a^2} \left( \frac{4h^3}{3} - hx + \frac{x^3}{5} \right), \]

which, when \( x = h \) gives \( \frac{8\pi h^5}{15a^2} \) for the value of the semi-spindle. From this quantity let the former be subtracted, and there will remain

\[ \frac{8\pi h^5}{15a^2} - \frac{\pi x^3}{a^2} \left( \frac{4h^3}{3} - hx + \frac{x^3}{5} \right) \]

for the content of the frustum. In this expression let \( x \) be put instead of \( h - x \) or CD, and, denoting CE the radius of the greater end of the spindle by \( d \), let \( \frac{h^3}{d^3} \) be substituted instead of its value \( a \). Then we shall have the content of the frustum otherwise expressed by

\[ \frac{\pi d^3}{h^4} \left\{ h^4 - \frac{2h^3x^4}{3} + \frac{x^4}{5} \right\} \]

which value, by putting \( h \sqrt{\frac{d^3}{d}} \) in its two last terms instead

Demonstration. Let ABE be a quadrant of an ellipse, 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. 11.

\[ \frac{CE^2}{AF^2} : \frac{DB^2}{CA^2} = CD^2, \]

and by sim. tr. \( \frac{AF^2}{DH^2} : \frac{CA^2}{CD^2}. \)

Therefore (GEOMETRY, Sect. III. Theor. 8),

\[ AF^2 : DB^2 + DH^2 :: CA^2 : CA^2. \]

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 shown 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 \( 4n \times DG^2 \times CD \), and (Prob. 5.) the content of the cone is \( \frac{4}{3}n \times DH^2 \times CD \), and therefore the content of the frustum of the spheroid is

\[ 4n \times CD (DG^2 - \frac{1}{3}DH^2). \]

But it was shewn that \( DH^2 = DG^2 - DB^2 \); therefore the content of the frustum is also equal to

\[ \frac{4}{3}n \times CD (2CE^2 + DB^2), \]

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 \( 15^2 \times 7854 \), and the area of the less \( 9^2 \times 7854 \), therefore the content is

\[ 7854 (9^2 + 2 \times 15^2) \times \frac{4}{3} = 1390.158 \text{ cubic inches.} \]

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 \( \frac{3}{4} \) of the height, so is half the cylinder of the same base and altitude to the solidity of the hyperboloid.

Demonstration. Let BA be a hyperbola, A its transverse axis, C its centre, CF, C'f its asymptotes, FA a tangent at its vertex. Draw FE parallel to CA, and draw any straight line parallel to Ff, meeting the asymptotes in H and h, the curve in B and b, the axis in D, and the line FE in G. Then, because \( AF^2 = BH \times hB \) (CONIC SECTIONS, Part III. Prop. 11.) and...

and HB × hB = DH² - DB² (Geometry, Sect. IV. Theor. 12.), therefore AF² = DH² - DB² and DB² = HD² - DG². Hence it conceives 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 h, perpendicular to the axis, will be equal to the difference of the sections of the other two by the same plane. Therefore the hyperboloid B A b is equal to the difference between the conic frustum F H h f and the cylinder FG g f. Let A a the transverse axis be denoted by p, F f 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,

\[ \frac{CA}{CD} : \frac{Ff}{Hh} :: \frac{Ff}{Hh} : \frac{Ff}{Hh} \]

therefore \( \frac{Ff}{Hh} = \frac{CD}{CA} \times \frac{(p + h)q^2}{\frac{1}{2}p} = q^2 + \frac{2hq^2}{p} \)

Now \( Ff^2 = q^2 \), and \( Hh^2 = (Bb^2 + Ff^2) = b^2 + q^2 \), therefore putting n for .7854, we have (by Prob. 6.) the content of the conic frustum FH h f equal to

\[ \frac{n}{3} \left( q^2 + b^2 + q^2 + \frac{2hq^2}{p} \right) = \frac{n}{3} \left( 3q^2 + b^2 + \frac{2hq^2}{p} \right) \]

from this subtract \( nq^2 \), the expression for the content of the cylinder FG g f, and there will remain

\[ \frac{n}{3} \left( b^2 + \frac{2hq^2}{p} \right) \]

for the content of the hyperboloid. But from the nature of the hyperbola

\[ Aa^2 : Ff^2 :: AD \times Da : BD^2, \]

that is \( p : q :: (p + h) : \frac{1}{2}b^2; \)

therefore \( \frac{2hq^2}{p} = \frac{p}{2} \frac{b^2}{p + h}; \) and hence the content of the hyperboloid is also equal to

\[ \frac{n}{3} \left( b^2 + \frac{p}{2} \frac{b^2}{p + h} \right) = \frac{n}{2} \left( \frac{p + \frac{1}{2}h}{p + h} \right). \]

Now if it be considered that the quantity \( n \frac{h}{b} b^2 \) 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 \( 24^2 \times .7854 \times 10 \), therefore, we have the proportion,

\[ \frac{40}{3} : \frac{110}{3} :: \frac{24^2 \times .7854 \times 10}{2} : \]

\[ \frac{24^2 \times .7854 \times 10 \times 110}{40 \times 3 \times 2} = 2073.456, \] the content of the solid as required.

OF GAUGING.

Gauging treats of the measuring of casks, and other things falling under the cognizance of the officers of Gauging, the excise, and it has received its name from a gauge 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 .00094, or divided by .1077 for ale gallons, or multiplied by .00117 or divided by 882 for wine gallons.

The truth of this rule may be proved thus. Put B Fig. 44. for FG, the bung diameter, 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 \( (2B^2 + H^2)L \times \frac{3}{8} \times .7854 \), which being divided by 282 (the cubic inches in an ale gallon) gives \( (2B^2 + H^2)L \times .00928371 \), or \( (2B^2 + H^2) \times \frac{1}{1077.157} \times L \), for the content in ale gallons. And being divided by 231, (the cubic inches in a wine gallon) gives \( (2B^2 + H^2)L \times .00113333L \), or \( (2B^2 + H^2) \times \frac{1}{882.355} \times L \), for the content in wine gallons.

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

Here \( (2 \times 32^2 + 24^2) \times 40 \times .00094 = 97.44 \) ale gallons, is the content required.

And \( (2 \times 32^2 + 24^2) \times 40 \times .00117 = 118.95 \) wine gallons is the same content.

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 \( \frac{3}{5} \) or \( \frac{4}{5} \) of

Of Gauging 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 numbers as in the rule to last problem, will give the content.

For by Problem 12, the content in inches is

\[ \frac{8B^2 + 4BH + 3H^2}{15} \times 7854 L; \]

and this formula may be otherwise expressed thus,

\[ \left(2B^2 + H^2 - \frac{1}{3}(B-H)^2\right) \times \frac{7854}{3} \times L, \]

and hence is derived the rule, the multipliers or divisors being evidently the same as in last problem.

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

Answer. \((2 \times 3^2 + 2^2 - \frac{1}{3} \times 8^2) \times 40 \times .0009 = 96.49\) the content in ale gallons.

And \(1033.6 \times .0011 = 117.79\) the content in wine gallons.

PROBLEM III.

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 .0014, or divided by 718, the result will be the content in ale gallons; or if it be multiplied by .0017, or divided by 588, the result will be the content in wine gallons.

For by Problem 10, the content in inches is \(\frac{1}{2}(B^2 + H^2) \times 7854 L\); and this expression being divided by 282 gives \((B^2 + H^2) \times .00139255 L\) or \((B^2 + H^2) \times \frac{1}{718.125} \times L\) for the content in ale gallons; and divided by 231 gives \((B^2 + H^2) \times .0017 L\) or \((B^2 + H^2) \times \frac{1}{588.233}\) for the content in wine gallons.

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

Answer. \((3^2 + 2^2) \times 40 \times .0014 = 89.1\) the content in ale gallons.

And \(6400 \times .0017 = 108.8\) the content in wine 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 .000233 or divide it by 4308 for the content in ale gallons; or multiply the result by .00283, or divide it by 3529, for the content in wine gallons.

For by Problem 6, the content in inches is \(\frac{1}{3}(B^2 + BH + H^2) \times 7854 L\) which expression is equivalent to

\[ \left\{\frac{1}{3}(B+H)^2 + (B-H)^2\right\} \times \frac{7854}{12} L. \]

Now \(\frac{7854}{12}\) divided by 282 gives .000233 or \(\frac{1}{4308.628}\) the multiplier for ale gallons, and divided by 231 gives .00028333 or \(\frac{1}{3529.42}\) the multiplier for wine gallons.

Ex. Supposing the dimensions of a cask as before, what is its content?

Answer. \((3 \times 5^2 + 8^2) \times 40 \times .000233 = 87.93\), the content in ale gallons.

And \(37883 \times .00028333 = 107.35\), is the content in wine gallons.

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

RULE.

Add into one sum 39 times the square of the bung diameter, 25 times the square of the head diameter, and 26 times the product of the diameters; multiply the sum by the length, and the product by .000343; then the last product divided by 9 will give the wine gallons, and divided by 11 will give the ale gallons.

In investigating this rule the ingenious 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 \(A\), \(ABE\) will be a tangent to a parabola \(BF\), and therefore \(FI = \frac{1}{2} EI\). But \(BI = \frac{1}{2} AK\), and hence, by sim. triangles \(EI = \frac{1}{2} EK\); consequently \(FI = \frac{1}{2} EI = \frac{1}{2} EK = \frac{1}{2} FK = \frac{1}{2} (B-H)\); so that the common denominator \(BL = FG = 2FI = B - \frac{1}{2}(B-H) = \frac{1}{2}(B-H)\), which call \(c\). Now by the rules for parabolic spindles and conic frustrums we obtain (putting \(n\) for .7854)

\[ \frac{8B^2 + 4BH + 3C^2}{15} \times \frac{Ln}{3} = \frac{3.78B^2 + 44BH + 3H^2}{25 \times 45} \times Ln \text{ for the parabolic or middle part; and } \frac{C^2 + CH + H^2}{3} \times Ln \text{ for the two ends.} \]

Of Gauging gonds, and the sum of these two gives after proper reduction \((39B^2 + 26BH + 25H^2) \times \frac{L}{90}\), nearly, for the content in inches. And the quantity \(\frac{n}{90} = \frac{7854}{90}\).

Being divided by 231 gives \(0.0034\) the multiplier for wine gallons; and since 231 is to 282 as 9 to 11 nearly,

\[ \frac{0.0034}{11} \]

will be the multiplier for ale gallons as in the rule.

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

Here \((39^2 \times 3^2 + 26 \times 3^2 \times 24 + 25 \times 2^2) \times 40 \times 0.0034 = 1010.5\); which being divided by 9 and by 11 we obtain 112.5 wine gallons or 95.9 ale gallons for the content required.

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MENSTRUAL, or MENSTRUOUS, in Physiology, is applied to the blood which flows from women in their ordinary monthly purgations. See MIDWIFERY and MEDICINE Index.

MENSTRUUM, in Chemistry, any body which in a fluid or tublized state is capable of interfering its small parts betwixt the small parts of other bodies, so as to divide them subtly, and form a new uniform compound of the two.